- cat nip
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/create-a-text-field.md b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/create-a-text-field.md
index e1a515b7168..1f4cdc30a8c 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/create-a-text-field.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/create-a-text-field.md
@@ -9,7 +9,7 @@ dashedName: create-a-text-field
# --description--
-现在让我们来创建一个 `form` 表单。
+现在让我们来创建一个 Web 表单。
`input` 输入框可以让你轻松获得用户的输入。
@@ -17,15 +17,15 @@ dashedName: create-a-text-field
``
-**注意：**`input` 输入框是没有结束标签的。
+注意 `input` 输入框是没有结束标签的。
# --instructions--
-在列表下面创建一个 `type` 为 `text` 的 `input` 输入框。
+在列表下面创建一个类型为 `text` 的 `input` 输入框。
# --hints--
-网页中应存在一个 `type` 为 `text` 的 `input` 输入框。
+网页中应存在一个类型为 `text` 的 `input` 输入框。
```js
assert($('input[type=text]').length > 0);
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/create-an-ordered-list.md b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/create-an-ordered-list.md
index 6e4de46e846..bd4be8849cc 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/create-an-ordered-list.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/create-an-ordered-list.md
@@ -11,22 +11,22 @@ dashedName: create-an-ordered-list
HTML 中有用于创建有序列表的特定元素。
-有序列表以 `` 开始，中间包含一个或多个 `- ` 元素，最后以 `
` 结束。
+有序列表以 `` 开始，中间包含一个或多个 `- ` 元素。 最后以 `
` 结束。
例如:
```html
- - 加菲猫
- - 哆啦A梦
+ - Garfield
+ - Sylvester
```
-将会创建一个包含加菲猫和哆啦 A 梦的有序列表。
+将创建一个包含 `Garfield` 和 `Sylvester` 的编号列表。
# --instructions--
-请创建一个有序列表，内容是猫咪最讨厌的三样东西（`Top 3 things cats hate:`），内容可以任意指定。
+请创建一个有序列表，内容是猫咪最讨厌的三样东西。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/declare-the-doctype-of-an-html-document.md b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/declare-the-doctype-of-an-html-document.md
index da11e50f3a9..26c56c4168e 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/declare-the-doctype-of-an-html-document.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/declare-the-doctype-of-an-html-document.md
@@ -9,15 +9,15 @@ dashedName: declare-the-doctype-of-an-html-document
# --description--
-到目前为止，我们已经学习了一些 HTML 标签。还有一些标签是用来组成网页整体结构的，每个 HTML 文档中都有这些标签的身影。
+之前的挑战涵盖了一些 HTML 元素和它们的用法。 另外有些元素为页面提供了整体结构，每个 HTML 文档中都应该有这些元素。
-在文档的顶部，我们需要告诉浏览器网页所使用的 HTML 的版本。HTML 是一个在不停发展的语言，大部分浏览器都支持 HTML 的最新标准，也就是 HTML5。但是一些陈旧的网页可能使用的是老版本的 HTML。
+在文档的顶部，我们需要告诉浏览器网页所使用的 HTML 的版本。 HTML 是一个在不停发展的语言，大部分浏览器都支持 HTML 的最新标准，也就是 HTML5。 大部分主流浏览器都支持最新的 HTML5 规范。 但是一些陈旧的网页可能使用的是老版本的 HTML。
-你可以通过 `` 来告诉浏览器页面上使用的 HTML 版本，"`...`" 部分就是版本号。`` 对应的就是 HTML5。
+你可以通过 `` 来告诉浏览器页面上使用的 HTML 版本，"`...`" 部分就是版本号。 `` 对应的就是 HTML5。
-`!` 和大写的 `DOCTYPE` 是很重要的，尤其是对那些老的浏览器。但 `html` 无论大写小写都可以。
+`!` 和大写的 `DOCTYPE` 是很重要的，尤其是对那些老的浏览器。 但 `html` 无论大写小写都可以。
-所有的 HTML 代码都必须位于 `html` 标签中。其中 `` 位于 `` 之后，`` 位于网页的结尾。
+所有的 HTML 代码都必须位于 `html` 标签中。 其中 `` 位于 `` 之后，`` 位于网页的结尾。
以下是网页结构的一个例子：
@@ -30,7 +30,7 @@ dashedName: declare-the-doctype-of-an-html-document
# --instructions--
-请在代码编辑器的顶部添加一个 `DOCTYPE（文档类型）` 为 HTML5 的声明，然后添加一个 `html` 元素，再添加一个 `h1` 元素，你可以随意填写 `h1` 的内容文本。
+请在代码编辑器的顶部添加一个 `DOCTYPE` 为 HTML5 的声明。 在它下面添加 `html` 开始和结束标签，其中包裹一个 `h1` 元素。 标题可以包括任何文本。
# --hints--
@@ -57,6 +57,7 @@ assert(code.match(/\s*?\s*?.*?\s*?<\/h1>\s*?<\/html>/gi));
## --seed-contents--
```html
+
```
# --solutions--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/define-the-head-and-body-of-an-html-document.md b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/define-the-head-and-body-of-an-html-document.md
index 0f17fbbf0bc..c474470fc3c 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/define-the-head-and-body-of-an-html-document.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/define-the-head-and-body-of-an-html-document.md
@@ -9,7 +9,7 @@ dashedName: define-the-head-and-body-of-an-html-document
# --description--
-`html` 的结构主要分为两大部分：`head` 和 `body`。网页的描述应放入 `head` 标签，网页的内容则应放入 `body` 标签。
+`html` 的结构主要分为两大部分：`head` 和 `body`。 网页的描述应放入 `head` 标签， 网页的内容（向用户展示的）则应放入 `body` 标签。
比如 `link`、`meta`、`title` 和 `style` 都应放入 `head` 标签。
@@ -29,48 +29,83 @@ dashedName: define-the-head-and-body-of-an-html-document
# --instructions--
-请给网页添加 `head` 和 `body`，`head` 元素应包含 `title`；`body` 元素应该包含 `h1` 和 `p`。
+标记文本的结构主要分为两大部分：`head` 和 `body`。 `head` 元素应只包含 `title`，`body` 元素应该包含 `h1` 和 `p`。
# --hints--
网页应只有一个 `head` 元素。
```js
-assert($('head').length == 1);
+const headElems = code.replace(/\n/g, '').match(/\.*?\<\/head\s*>/g);
+assert(headElems && headElems.length === 1);
```
网页应只有一个 `body` 元素。
```js
-assert($('body').length == 1);
+const bodyElems = code.replace(/\n/g, '').match(/.*?<\/body\s*>/g);
+assert(bodyElems && bodyElems.length === 1);
```
`head` 应为 `html` 的子元素。
```js
-assert($('html').children('head').length == 1);
+const htmlChildren = code
+ .replace(/\n/g, '')
+ .match(/(?.*)<\/html\s*>/);
+let foundHead;
+if (htmlChildren) {
+ const { children } = htmlChildren.groups;
+
+ foundHead = children.match(/.*<\/head\s*>/);
+}
+assert(foundHead);
```
`body` 应为 `html` 的子元素。
```js
-assert($('html').children('body').length == 1);
+const htmlChildren = code
+ .replace(/\n/g, '')
+ .match(/(?.*?)<\/html\s*>/);
+let foundBody;
+if (htmlChildren) {
+ const { children } = htmlChildren.groups;
+ foundBody = children.match(/.*<\/body\s*>/);
+}
+assert(foundBody);
```
`title` 应为 `head` 的子元素。
```js
-assert(code.match(/\s*?\s*?.*?\s*?<\/title>\s*?<\/head>/gi));
+const headChildren = code
+ .replace(/\n/g, '')
+ .match(/<head\s*>(?<children>.*?)<\/head\s*>/);
+let foundTitle;
+if (headChildren) {
+ const { children } = headChildren.groups;
+ foundTitle = children.match(/<title\s*>.*?<\/title\s*>/);
+}
+assert(foundTitle);
```
`h1` 和 `p` 都应为 `body` 的子元素。
```js
-assert(
- code.match(
- /<body>\s*?(((<h1>\s*?.*?\s*?<\/h1>\s*?)(<p>(.*\s*)*?<\/p>\s*?))|((<p>\s*?.*?\s*?<\/p>\s*?)(<h1>(.*\s*)*?<\/h1>\s*?)))<\/body>/gi
- )
-);
+const bodyChildren = code
+ .replace(/\n/g, '')
+ .match(/<body\s*>(?<children>.*?)<\/body\s*>/);
+let foundElems;
+if (bodyChildren) {
+ const { children } = bodyChildren.groups;
+ const h1s = children.match(/<h1\s*>.*<\/h1\s*>/g);
+ const ps = children.match(/<p\s*>.*<\/p\s*>/g);
+ const numH1s = h1s ? h1s.length : 0;
+ const numPs = ps ? ps.length : 0;
+ foundElems = numH1s === 1 && numPs === 1;
+}
+assert(foundElems);
```
# --seed--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/fill-in-the-blank-with-placeholder-text.md b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/fill-in-the-blank-with-placeholder-text.md
index 21942435a14..a36fc867dfe 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/fill-in-the-blank-with-placeholder-text.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/fill-in-the-blank-with-placeholder-text.md
@@ -9,17 +9,19 @@ dashedName: fill-in-the-blank-with-placeholder-text
# --description--
-Web 开发者通常用 [lorem ipsum text](https://baike.baidu.com/item/Lorem%20ipsum/3684081) 来做占位符，占位符就是一些用于占位的文字，没有实际意义。
+Web 开发者通常用 <dfn>lorem ipsum text</dfn> 来做占位符。 lorem ipsum text 是从古罗马西塞罗的一段著名经文中随机抽取的。
-之所以叫 `lorem ipsum text`，是因为 `lorem ipsum` 是古罗马西塞罗谚语的前两个单词。
+Lorem ipsum text 自 16 世纪以来就在排版中被用作占位符，这一习惯也在 Web 开发中得以延续。
+
+五个世纪已经很久远了。 因为我们正在构建一个 CatPhotoApp，所以我们使用 “kitty ipsum” 文本。
# --instructions--
-请把 `p` 元素的内容文本替换为：`Kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff.`
+`p` 元素的内容文本应包含 `Kitty ipsum`。
# --hints--
-`p` 元素的内容文本应包含 `Kitty ipsum`。
+你的 `p` 元素应包含 “kitty ipsum” 文本的前面几个词。
```js
assert.isTrue(/Kitty(\s)+ipsum/gi.test($('p').text()));
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/headline-with-the-h2-element.md b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/headline-with-the-h2-element.md
index 0059fb31abb..59edafa06eb 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/headline-with-the-h2-element.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/headline-with-the-h2-element.md
@@ -13,11 +13,11 @@ dashedName: headline-with-the-h2-element
这节课中，我们将会为页面引入作为第二级标题的 `h2` 元素。
-这些元素用来告诉浏览器网站的结构是什么样子。`h1` 元素通常被用作主标题，`h2` 元素通常被用作副标题，还有 `h3`、`h4`、`h5`、`h6` 元素，它们分别用作不同级别的标题。
+这些元素用来告诉浏览器网站的结构是什么样子。 `h1` 元素通常被用作主标题，`h2` 元素通常被用作副标题， 还有 `h3`、`h4`、`h5`、`h6` 元素，它们分别用作不同级别的标题。
# --instructions--
-在内容为 "Hello World" 的 `h1` 元素下面创建一个 `h2` 元素，`h2` 元素的内容为 `CatPhotoApp`。
+在内容为 "Hello World" 的 `h1` 元素下面创建一个 `h2` 元素，其内容为 “CatPhotoApp”。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/inform-with-the-paragraph-element.md b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/inform-with-the-paragraph-element.md
index 55f5b92d6c4..f55220985d5 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/inform-with-the-paragraph-element.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/inform-with-the-paragraph-element.md
@@ -9,7 +9,7 @@ dashedName: inform-with-the-paragraph-element
# --description--
-`p` 是 `paragraph` 的缩写，通常用来创建一个段落，就和你写作文一样。
+`p` 元素是网站上段落文本使用的元素。 `p` 是“paragraph（段落）”的缩写。
你可以像这样创建一个段落：
@@ -17,7 +17,9 @@ dashedName: inform-with-the-paragraph-element
# --instructions--
-请在 `h2` 元素下方添加一个 `p` 元素，元素内容是 `Hello Paragraph`。
+在 `h2` 元素下方添加一个 `p` 元素，其内容是 `Hello Paragraph`。
+
+**注意：**按照惯例，所有 HTML 标签都应该是小写字母，例如应使用 `<p></p>`，而不会使用`<P></P>`。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/introduction-to-html5-elements.md b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/introduction-to-html5-elements.md
index b56557bd27c..da27348729d 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/introduction-to-html5-elements.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/introduction-to-html5-elements.md
@@ -9,11 +9,9 @@ dashedName: introduction-to-html5-elements
# --description--
-HTML5 引入了很多更具描述性的 HTML 元素，例如：`header`、`footer`、`nav`、`video`、`article`、`section` 等等。
+HTML5 引入了很多更具描述性的 HTML 元素， 包括 `main`、`header`、`footer`、`nav`、`video`、`article`、`section` 等等。
-这些元素让 HTML 更易读，同时有助于搜索引擎优化和无障碍访问。
-
-`main` 元素让搜索引擎和开发者能很快地找到网页的主要内容。
+这些元素让 HTML 更易读，同时有助于搜索引擎优化和无障碍访问。 `main` 元素让搜索引擎和开发者能很快地找到网页的主要内容。
举个例子，下面的 `main` 元素嵌套了两个子元素：
@@ -28,7 +26,7 @@ HTML5 引入了很多更具描述性的 HTML 元素，例如：`header`、`foote
# --instructions--
-请在现有的段落之后创建一个新的段落，段落内容为：`Purr jump eat the grass rip the couch scratched sunbathe, shed everywhere rip the couch sleep in the sink fluffy fur catnip scratched.`
+请在现有的 `p` 之后创建一个新的 `p` 元素，内容为：`Purr jump eat the grass rip the couch scratched sunbathe, shed everywhere rip the couch sleep in the sink fluffy fur catnip scratched.`
然后，请添加一个 `main` 元素，作为现有的两个 `p` 元素的父级元素。
@@ -49,7 +47,7 @@ assert(
);
```
-新建的段落应包含关键词 `Purr jump eat`。
+新建的`p` 元素应包含 `kitty ipsum text` 的前面几个词。
```js
assert.isTrue(/Purr\s+jump\s+eat/gi.test($('p').text()));
@@ -61,7 +59,7 @@ assert.isTrue(/Purr\s+jump\s+eat/gi.test($('p').text()));
assert($('main').length === 1);
```
-`main` 元素应有两个 `p` 元素作为它的子元素。
+`main` 元素应有两个段落元素作为它的子元素。
```js
assert($('main').children('p').length === 2);
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/link-to-external-pages-with-anchor-elements.md b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/link-to-external-pages-with-anchor-elements.md
index 054a6f152b7..b16eb2b4643 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/link-to-external-pages-with-anchor-elements.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/link-to-external-pages-with-anchor-elements.md
@@ -15,7 +15,7 @@ dashedName: link-to-external-pages-with-anchor-elements
`<a href="https://freecodecamp.org">this links to freecodecamp.org</a>`
-在浏览器中，以上的标签会将文字 **"链接到 freeCodeCamp"** 展示成一个可点击的超链接。 点击该文本就会跳转到 **`https://freecodecamp.org`**。
+浏览器将显示文本 `this links to freecodecamp.org`，这是一个可点击的链接。 你可以通过这个链接访问 `https://www.freecodecamp.org`。
# --instructions--
@@ -23,7 +23,7 @@ dashedName: link-to-external-pages-with-anchor-elements
# --hints--
-`a` 元素的锚文本应为 “cat photos”。
+`a` 元素应有锚文本 `cat photos`。
```js
assert(/cat photos/gi.test($('a').text()));
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/link-to-internal-sections-of-a-page-with-anchor-elements.md b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/link-to-internal-sections-of-a-page-with-anchor-elements.md
index 19f7d75b38f..133a98d88b9 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/link-to-internal-sections-of-a-page-with-anchor-elements.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/link-to-internal-sections-of-a-page-with-anchor-elements.md
@@ -9,9 +9,11 @@ dashedName: link-to-internal-sections-of-a-page-with-anchor-elements
# --description--
-`a` 元素还可以用来实现页面内不同区域的跳转，只需要把 `a` 元素的 `href` 值设置为井号 `#` 加欲跳转区域所对应的 `id` 属性值即可。`id` 是描述网页元素的一个属性，它的值在整个页面中唯一。
+`a`（*anchor*）元素也可以用于创建内部链接，跳转到网页内的各个不同部分。
-下面是用来创建内部 `a` 的例子：
+要创建内部链接，你需要将链接的 `href` 属性值设置为一个哈希符号 `#` 加上你想内部链接到的元素的 `id`，通常是在网页下方的元素。 然后你需要将相同的 `id` 属性添加到你链接到的元素中。 `id` 是描述网页元素的一个属性，它的值在整个页面中唯一。
+
+当用户点击了 `Contacts` 链接，页面就会跳转到网页的 **Contacts** 区域。
```html
<a href="#contacts-header">Contacts</a>
@@ -19,37 +21,37 @@ dashedName: link-to-internal-sections-of-a-page-with-anchor-elements
<h2 id="contacts-header">Contacts</h2>
```
-当用户点击了 `Contacts` 链接，页面就会跳转到网页的 **Contacts** 区域。
+当用户点击 `Contacts` 链接，可以访问网页中带有 **Contacts** 标题元素的部分。
# --instructions--
-请修改 `href` 属性值为 `#footer` 来将外部链接更改为内部链接，同时修改文本 `cat photos` 为 `Jump to Bottom`。
-
-移除 `target="\_blank"` 属性，这个属性的存在会让链接在新标签页中打开。
+通过修改 `href` 属性值为 `"#footer"`，同时修改文本 `cat photos` 为 `Jump to Bottom`，来更改外部链接为内部链接。
然后添加一个 `<footer>` 元素，并将它的 `id` 属性值设置为 `footer`。
+然后给页面底部的 `<footer>` 元素添加一个 `id` 属性，值为 `footer`。
+
# --hints--
-页面中应只存在一个 `a` 元素。
+页面中应只存在一个 `footer` 元素。
```js
assert($('a').length == 1);
```
-页面中应只存在一个 `footer` 元素。
+`a` 的 `href` 属性值应为 `#footer`。
```js
assert($('footer').length == 1);
```
-`a` 的 `href` 属性值应为 `#footer`。
+`a` 不应有 `target` 属性。
```js
assert($('a').eq(0).attr('href') == '#footer');
```
-`a` 不应有 `target` 属性。
+`a` 的内容文本应为 `Jump to Bottom`。
```js
assert(
@@ -58,7 +60,7 @@ assert(
);
```
-`a` 的内容文本应为 `Jump to Bottom`。
+`footer` 元素的 `id` 属性值应为 `footer`。
```js
assert(
@@ -69,7 +71,7 @@ assert(
);
```
-`footer` 元素的 `id` 属性值应为 `footer`。
+`footer` 标签应该有一个 `id` 属性，值为 “footer”。
```js
assert($('footer').eq(0).attr('id') == 'footer');
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/make-dead-links-using-the-hash-symbol.md b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/make-dead-links-using-the-hash-symbol.md
index 139346f5784..440a7caabf1 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/make-dead-links-using-the-hash-symbol.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/make-dead-links-using-the-hash-symbol.md
@@ -9,13 +9,13 @@ dashedName: make-dead-links-using-the-hash-symbol
# --description--
-有时你想为网站添加一个 `a` 元素，但还不确定要将它链接到哪里。这时，你可以使用链接占位符。
+有时你想为网站添加一个 `a` 元素，但还不确定要将它链接到哪里。
-链接占位符还有一个应用场景，就是实现在点击 `a` 元素时执行 `JavaScript` 代码。我们会在后面的课程中学到。
+当你使用 `JavaScript` 更改链接的指向时，这也很方便，我们将在后面的课程中介绍。
# --instructions--
-目前，`href` 的属性值是 "`https://freecatphotoapp.com`"。请将 `href` 属性的值替换为`#`，以此来创建链接占位符。
+目前，`href` 的属性值是 "`https://freecatphotoapp.com`"。 请将 `href` 属性的值替换为`#`，以此来创建链接占位符。
例如: `href="#"`
@@ -49,9 +49,9 @@ assert($('a').attr('href') === '#');
<h2>CatPhotoApp</h2>
<main>
<p>Click here to view more <a href="#" target="_blank">cat photos</a>.</p>
-
+
<img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back.">
-
+
<p>Kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff.</p>
<p>Purr jump eat the grass rip the couch scratched sunbathe, shed everywhere rip the couch sleep in the sink fluffy fur catnip scratched.</p>
</main>
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/nest-an-anchor-element-within-a-paragraph.md b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/nest-an-anchor-element-within-a-paragraph.md
index c7fea7d3644..ad4acbb7cb4 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/nest-an-anchor-element-within-a-paragraph.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/nest-an-anchor-element-within-a-paragraph.md
@@ -17,17 +17,17 @@ dashedName: nest-an-anchor-element-within-a-paragraph
</p>
```
-让我们来拆解一下这个例子： 通常，文本是被包裹在 `p` 元素内：
-`<p> Here's a ... for you to follow. </p>` 接下来是 `<a>` 元素（它需要结束标签 `</a>`）：
-`<a> ... </a>` `target` 是 `a` 元素的属性，它用来指定链接的打开方式。属性值 `"_blank"` 表示链接会在新标签页打开。`href` 是 `a` 的另一个属性，它用来指定链接的 URL：
-`<a href="https://freecodecamp.org"> ... </a>` `a` 元素内的内容文本 **"link to freecodecamp.org"**，会显示为一个可以点击的链接：
-`<a href=" ... ">link to freecodecamp.org</a>` 此示例的最终输出结果是这样：
+我们来分解一下这个示例：普通文本被包裹在 `p` 元素中，例如
+`<p> Here's a ... for you to follow. </p>` 然后是 *anchor* 元素 `<a>`（要求有一个结束标签 `</a>`）：
+`<a> ... </a>` `target` 是一个锚点标签属性，用来指定在哪里打开链接，`_blank` 值指定在一个新的窗口打开链接，`href` 是一个锚点标签属性，包含链接的 URL 地址
+`<a href="http://freecodecamp.org"> ... </a>`。`a` 元素中的文本，**“链接到 freecodecamp.org”**，叫作 `anchor text`，将显示一个可点击的链接：
+`<a href=" ... ">link to freecodecamp.org</a>`。 这个示例最后输出的结果像这样：
-Here's a [link to freecodecamp.org](http://freecodecamp.one) for you to follow.
+你可以点击这个 [freecodecamp.org 链接](http://freecodecamp.one)。
# --instructions--
-创建一个新的段落（`p`）标签来包裹 `main` 元素里的 `a` 节点。新段落标签的内容为："View more cat photos"，其中 "cat photos" 是一个链接，其余是纯文本。
+创建一个新的段落（`p`）标签来包裹 `main` 元素里的 `a` 节点。 新段落应有文本 `View more cat photos`，其中 `cat photos` 是一个链接，其余是纯文本。
# --hints--
@@ -40,7 +40,7 @@ assert(
);
```
-`a` 元素的内容文本应为 "cat photos"。
+`a` 元素应有锚文本 `cat photos`。
```js
assert(
@@ -50,7 +50,7 @@ assert(
);
```
-你应该在 `a` 标签的外部创建一个新的 `p` 标签。页面中应至少包含 3 个 `p` 标签。
+你应该在 `a` 标签的外部创建一个新的 `p` 标签。 页面中应至少包含 3 个 `p` 标签。
```js
assert($('p') && $('p').length > 2);
@@ -65,7 +65,7 @@ assert(
);
```
-`p` 元素应该包含文本 "View more "（请注意，more 之后有一个空格）。
+`p` 元素应有文本 `View more`（后面有一个空格）。
```js
assert(
@@ -80,7 +80,7 @@ assert(
);
```
-`a` 元素中不应包含文本 "View more"。
+`a` 元素 <em>不</em> 应有文本 `View more`。
```js
assert(
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/nest-many-elements-within-a-single-div-element.md b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/nest-many-elements-within-a-single-div-element.md
index 63a5a2038b0..68993bddc3f 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/nest-many-elements-within-a-single-div-element.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/nest-many-elements-within-a-single-div-element.md
@@ -9,7 +9,7 @@ dashedName: nest-many-elements-within-a-single-div-element
# --description--
-`div` 元素也叫 Content Division Element（内容划分元素）元素，是一个包裹其他元素的通用容器。
+`div` 元素也叫内容划分元素，是一个包裹其他元素的通用容器。
它也是 HTML 中出现频率最高的元素。
@@ -17,9 +17,9 @@ dashedName: nest-many-elements-within-a-single-div-element
# --instructions--
-请把无序列表、有序列表和段落都嵌套进同一个 `div` 元素。
+将你的列表“猫喜欢的三件事”和“猫最讨厌的三件事”放入同一个 `div` 元素中。
-提示：你可以在第一个 `<p>` 之前插入 `div` 开始标记，在 `</ol>` 之后插入 `div` 结束标签。这样，所有的列表都会位于 `div` 之内。
+提示：你可以在第一个 `<p>` 之前插入 `div` 开始标记，在 `</ol>` 之后插入 `div` 结束标签。 这样，所有的列表都会位于 `div` 之内。
# --hints--
@@ -92,7 +92,7 @@ assert(
<h2>CatPhotoApp</h2>
<main>
<p>Click here to view more <a href="#">cat photos</a>.</p>
-
+
<a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
<div>
<p>Things cats love:</p>
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/say-hello-to-html-elements.md b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/say-hello-to-html-elements.md
index d33a9e93a24..84a66e63198 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/say-hello-to-html-elements.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/say-hello-to-html-elements.md
@@ -9,29 +9,33 @@ dashedName: say-hello-to-html-elements
# --description--
-欢迎参加 freeCodeCamp 的编程挑战赛，这些挑战将会帮助你逐步掌握 Web 开发。
+欢迎访问 freeCodeCamp 的 HTML 编程挑战。 这些挑战将会帮助你逐步掌握 Web 开发。
-HTML 是英文 Hyper Text Markup Language（超文本标记语言）的缩写。首先，我们来用 HTML 制作一个简单的网页，你可以直接在网页内置的代码编辑器中编辑代码。
+首先，我们来用 HTML 制作一个简单的网页。 你可以直接在网页内置的代码编辑器中编辑代码。
-看到代码编辑器中的 `<h1>Hello</h1>` 了吗? 那就是一个 HTML 元素。
+你看到编辑器中的 `<h1>Hello</h1>` 了吗？ 那是一个 HTML 元素。
-大部分 HTML 元素都有一个`开始标签`和一个`结束标签`。
+大部分 HTML 元素都有一个开始标签和一个结束标签。
-开始标签像这样：`<h1>`
+开始标签像这样：
-结束标签像这样：`</h1>`
+`<h1>`
-开始标签和结束标签的唯一区别就是结束标签多了一个 `/`。
+结束标签像这样：
-每个挑战都有测试，任何时候点击**运行测试**按钮就可以运行测试。如果代码通过测试，将会弹出一个窗口，你就可以进入下一个挑战。反之，测试区会显示你没有通过测试的原因。
+`</h1>`
+
+开始标签和结束标签的唯一区别就是结束标签多了一个斜杠。
+
+每个挑战都有测试，任何时候你点击“运行测试”按钮，就可以运行测试。 如果代码通过测试，将会弹出一个窗口，你就可以进入下一个挑战。
# --instructions--
-请把 `h1` 元素的内容改为 `Hello World`。
+要通过这个挑战的测试，需要修改 `h1` 元素的文本为 `Hello World`。
# --hints--
-`h1` 元素的内容文本应为 `Hello World`。
+`h1` 元素应有文本 `Hello World`。
```js
assert.isTrue(/hello(\s)+world/gi.test($('h1').text()));
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/turn-an-image-into-a-link.md b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/turn-an-image-into-a-link.md
index de67103f937..8384bb88d79 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/turn-an-image-into-a-link.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/turn-an-image-into-a-link.md
@@ -11,21 +11,21 @@ dashedName: turn-an-image-into-a-link
你可以通过把元素嵌套进 `a` 里使其变成一个链接。
-如果我们要把图片嵌套进 `a` 元素，可以这样写：
+如果我们要把图片嵌套进 `a` 元素， 可以这样写：
-`<a href="#"><img src="https://bit.ly/fcc-running-cats" alt="三只萌萌的小猫"></a>`
+`<a href="#"><img src="https://bit.ly/fcc-running-cats" alt="Three kittens running towards the camera."></a>`
-如果把 `a` 的 `href` 属性值设置为 `#`，就可以创建固定链接。
+如果把 `a` 的 `href` 属性值设置为 `#`，创建的是一个死链接（不跳转到其他画面）。
# --instructions--
-请把现存的图片嵌套进 `a` 中。
+请把现存的图片嵌套进 `a`（ *锚点*）元素中。
-如果鼠标悬停在图片上时，鼠标的光标如果从箭头指针变成手形指针，那么此时图片就是一个链接了。
+完成后，请你把鼠标光标悬停在你的图像上， 鼠标光标将变成点击光标。 于是图片就变成了链接。
# --hints--
-应将图片嵌套进 `a` 元素中。
+应将 `img` 嵌套进 `a` 元素中。
```js
assert($('a').children('img').length > 0);
@@ -69,9 +69,9 @@ assert(
<h2>CatPhotoApp</h2>
<main>
<p>Click here to view more <a href="#">cat photos</a>.</p>
-
+
<a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
-
+
<p>Kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff.</p>
<p>Purr jump eat the grass rip the couch scratched sunbathe, shed everywhere rip the couch sleep in the sink fluffy fur catnip scratched.</p>
</main>
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/use-html5-to-require-a-field.md b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/use-html5-to-require-a-field.md
index 89828582f87..7feb9a76c35 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/use-html5-to-require-a-field.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/use-html5-to-require-a-field.md
@@ -17,6 +17,8 @@ dashedName: use-html5-to-require-a-field
请给 `input` 元素加上 `required` 属性，这样用户就必须先在输入框里填入内容，然后才可以提交表单。
+然后尝试在不输入任何文本的情况下提交表单， 看看 HTML5 表单是如何通知你这个字段是必填的。
+
# --hints--
`input` 元素应有 `required` 属性。
@@ -61,9 +63,9 @@ assert($('input').prop('required'));
<h2>CatPhotoApp</h2>
<main>
<p>Click here to view more <a href="#">cat photos</a>.</p>
-
+
<a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
-
+
<p>Things cats love:</p>
<ul>
<li>cat nip</li>
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/use-the-value-attribute-with-radio-buttons-and-checkboxes.md b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/use-the-value-attribute-with-radio-buttons-and-checkboxes.md
index 9b2de3a6ab3..2518f784900 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/use-the-value-attribute-with-radio-buttons-and-checkboxes.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/basic-html-and-html5/use-the-value-attribute-with-radio-buttons-and-checkboxes.md
@@ -8,26 +8,26 @@ dashedName: use-the-value-attribute-with-radio-buttons-and-checkboxes
# --description--
-提交表单时，所选项的值会发送给服务端。`radio` 和 `checkbox` 的 `value` 属性值决定了发送到服务端的实际内容。
+提交表单时，所选项的值会发送给服务端。 `radio` 和 `checkbox` 的 `value` 属性值决定了发送到服务端的实际内容。
例如：
```html
-<label for="indoor">
- <input id="indoor" value="indoor" type="radio" name="indoor-outdoor">Indoor
+<label for="indoor">
+ <input id="indoor" value="indoor" type="radio" name="indoor-outdoor">Indoor
</label>
-<label for="outdoor">
- <input id="outdoor" value="outdoor" type="radio" name="indoor-outdoor">Outdoor
+<label for="outdoor">
+ <input id="outdoor" value="outdoor" type="radio" name="indoor-outdoor">Outdoor
</label>
```
-这里有两个 `radio` 单选框。当用户提交表单时，如果 `indoor` 选项被选中，表单数据会包含：`indoor-outdoor=indoor`。也就是所选项的 `name` 和 `value` 属性值。
+这里有两个 `radio` 单选框。 当用户提交表单时，如果 `indoor` 选项被选中，表单数据会包含：`indoor-outdoor=indoor`。 也就是所选项的 `name` 和 `value` 属性值。
-如果没有指明 `value` 属性值，则会使用默认值做为表单数据提交，也就是 `on`。在这种情况下，如果用户选中 "indoor" 选项然后提交表单，表单数据则会包含 `indoor-outdoor=on`。这样的表单数据看起来不够直观，因此最好将 `value` 属性值设置为一些有意义的内容。
+如果没有指明 `value` 属性值，则会使用默认值做为表单数据提交，也就是 `on`。 在这种情况下，如果用户选中 "indoor" 选项然后提交表单，表单数据则会包含 `indoor-outdoor=on`。 这样的表单数据看起来不够直观，因此最好将 `value` 属性值设置为一些有意义的内容。
# --instructions--
-给每一个 `radio` 和 `checkbox` 输入框添加 `value` 属性，属性值为 `input` 对应 `label` 中文本的小写形式。如 Outdoor 的 `value` 应设置为 `"outdoor"`，即 `value="outdoor"`。
+给每一个 `radio` 和 `checkbox` 输入框添加 `value` 属性。 使用输入标签的文本，小写形式，作为属性的值。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/add-flex-superpowers-to-the-tweet-embed.md b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/add-flex-superpowers-to-the-tweet-embed.md
index 639ac5f606c..054466ef9e3 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/add-flex-superpowers-to-the-tweet-embed.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/add-flex-superpowers-to-the-tweet-embed.md
@@ -9,16 +9,22 @@ dashedName: add-flex-superpowers-to-the-tweet-embed
# --description--
-我们以右边的嵌入推文为例，一些元素换一个布局方式或许更好看。上一个挑战演示了 `display: flex`，现在你需要把它添加到推文内嵌的多个组件中，调整它们的位置。
+我们以右边的嵌入推文为例， 一些元素换一个布局方式或许更好看。 上一个挑战演示了 `display: flex`， 现在你需要把它添加到推文内嵌的多个组件中，调整它们的位置。
# --instructions--
-请为下列项目添加 CSS 属性 `display: flex`。注意，以下 CSS 选择器已为你写好：
+请为下列项目添加 CSS 属性 `display: flex`。 注意，以下 CSS 选择器已为你写好：
`header`、header 中的 `.profile-name`、header 中的 `.follow-btn`、header 中的 `h3` 和 `h4`、`footer` 以及 footer 中的 `.stats`。
# --hints--
+`.follow-btn` 应在页面上呈现。 请关闭广告拦截器等任何扩展。
+
+```js
+assert($('.follow-btn').length > 0 && $('.follow-btn').css('display') !== 'none');
+```
+
`header` 的 `display` 属性值应为 `flex`。
```js
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/align-elements-using-the-align-items-property.md b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/align-elements-using-the-align-items-property.md
index 5baeb3ac25c..7774d3ef964 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/align-elements-using-the-align-items-property.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/align-elements-using-the-align-items-property.md
@@ -9,22 +9,21 @@ dashedName: align-elements-using-the-align-items-property
# --description--
-`align-items` 属性与 `justify-content` 类似。回忆一下，`justify-content` 属性使 flex 子元素沿主轴排列。行的主轴是水平线，列的主轴是垂直线。
+`align-items` 属性与 `justify-content` 类似。 回忆一下，`justify-content` 属性使 flex 子元素沿主轴排列。 行的主轴是水平线，列的主轴是垂直线。
-Flex 容器中，与主轴垂直的叫做 **cross axis（交叉轴）**。行的交叉轴是垂直的，列的交叉轴是水平的。
+Flex 容器中，与主轴垂直的叫做 **cross axis（交叉轴）**。 行的交叉轴是垂直的，列的交叉轴是水平的。
-CSS 中的 `align-items` 属性用来定义 flex 子元素沿交叉轴的对齐方式。对行来说，定义的是元素的上下对齐方式；对列来说，是定义元素的左右对齐方式。
+CSS 中的 `align-items` 属性用来定义 flex 子元素沿交叉轴的对齐方式。 对行来说，定义的是元素的上下对齐方式； 对列来说，是定义元素的左右对齐方式。
`align-items` 的可选值包括：
-<ul><li><code>flex-start</code>：从 flex 容器的起始位置开始对齐项目。对行来说，把项目移至容器顶部；对列来说，是把项目移至容器左边。</li><li><code>flex-end</code>：从 flex 容器的终止位置开始对齐项目。对行来说，把项目移至容器底部；对列来说，把项目移至容器右边。</li><li><code>center</code>：把项目居中放置。对行来说，垂直居中（项目距离顶部和底部的距离相等）；对列来说，水平居中（项目距离左边和右边的距离相等）。</li><li><code>stretch</code>：拉伸项目，填满 flex 容器。例如，排成行的项目从容器顶部拉伸到底部。如未设置<code>align-items</code>的值，那么默认值是<code>stretch</code>。</li><li><code>baseline</code>：沿基线对齐。基线是文本相关的概念，可以认为它是字母排列的下端基准线。</li></ul>
+<ul><li><code>flex-start</code>：从 flex 容器的起始位置开始对齐项目。 对行来说，把项目移至容器顶部； 对列来说，是把项目移至容器左边。</li><li><code>flex-end</code>：从 flex 容器的终止位置开始对齐项目。 对行来说，把项目移至容器底部； 对列来说，把项目移至容器右边。</li><li><code>center</code>：把项目居中放置。 对行来说，垂直居中（项目距离顶部和底部的距离相等）； 对列来说，水平居中（项目距离左边和右边的距离相等）。</li><li><code>stretch</code>：拉伸项目，填满 flex 容器。 例如，排成行的项目从容器顶部拉伸到底部。 如未设置<code>align-items</code>的值，那么这就是默认值。</li><li><code>baseline</code>：沿基线对齐。 基线是文本相关的概念，可以认为它是字母排列的下端基准线。</li></ul>
# --instructions--
-这个例子可以帮助你理解这个属性。请在 `#box-container` 里添加 CSS 属性 `align-items` 并将值设为 `center`。
+这个例子可以帮助你理解这个属性。 请在 `#box-container` 里添加 CSS 属性 `align-items` 并将值设为 `center`。
-**提示：**
-请在编辑器里试试 `align-items` 的其他值，看看它们之间的区别。但要通过挑战，你必须把属性值设为 `center`。
+**提示：** 请在编辑器里试试 `align-items` 的其他值，看看它们之间的区别。 但要通过挑战，你必须把属性值设为 `center`。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/align-elements-using-the-justify-content-property.md b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/align-elements-using-the-justify-content-property.md
index 5085116f252..1ef791a288c 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/align-elements-using-the-justify-content-property.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/align-elements-using-the-justify-content-property.md
@@ -9,22 +9,22 @@ dashedName: align-elements-using-the-justify-content-property
# --description--
-flex 子元素有时不能充满整个 flex 容器，所以我们经常需要告诉 CSS 以什么方式排列 flex 子元素，以及调整它们的间距。幸运的是，我们可以通过 `justify-content` 属性的不同值来实现。在介绍 `justify-content` 的可选值之前，我们要先理解一些重要术语。
+flex 子元素有时不能充满整个 flex 容器， 所以我们经常需要告诉 CSS 以什么方式排列 flex 子元素，以及调整它们的间距。 幸运的是，我们可以通过 `justify-content` 属性的不同值来实现。 在介绍属性的可选值之前，我们要先理解一些重要术语。
[这张图片的元素横着排列，很好地描述了下面的概念。](https://www.w3.org/TR/css-flexbox-1/images/flex-direction-terms.svg)
-回忆一下，如果把 flex 容器设为一个行，它的子元素会从左到右逐个排列；如果把 flex 容器设为一个列，它的子元素会从上到下逐个排列。子元素排列的方向被称为 **main axis（主轴）**。对于行，主轴水平贯穿每一个项目；对于列，主轴垂直贯穿每一个项目。
+回忆一下，如果把 flex 容器设为一个行，它的子元素会从左到右逐个排列； 如果把 flex 容器设为一个列，它的子元素会从上到下逐个排列。 子元素排列的方向被称为 **main axis（主轴）**。 对于行，主轴水平贯穿每一个项目； 对于列，主轴垂直贯穿每一个项目。
-关于 flex 子元素在主轴的排列方式，很常用的是 `justify-content: center;`：即 flex 子元素在 flex 容器中居中排列。其他可选值还有：
+对于如何沿主轴线排放 flex 项目，有几种选择。 关于 flex 子元素在主轴的排列方式，很常用的是 `justify-content: center;`：即 flex 子元素在 flex 容器中居中排列。 其他可选值还有：
-<ul><li><code>flex-start</code>：从 flex 容器的起始位置开始排列项目。对行来说是把项目移至左边，对于列是把项目移至顶部。如未设置<code>justify-content</code>的值，那么默认值是<code>flex-start</code>。</li><li><code>flex-end</code>：从 flex 容器的终止位置开始排列项目。对行来说是把项目移至右边，对于列是把项目移至底部。</li><li><code>space-between</code>：项目间保留一定间距地沿主轴居中排列。第一个和最后一个项目被放置在容器边沿。例如，在行中第一个项目会紧贴着容器左边，最后一个项目会紧贴着容器右边，然后其他项目均匀排布。</li><li><code>space-around</code>：与<code>space-between</code>相似，但头尾两个项目不会紧贴容器边缘，所有项目之间的空间均匀排布。</li></ul>
+<ul><li><code>flex-start</code>：从 flex 容器的起始位置开始排列项目。 对行来说是把项目移至左边， 对于列是把项目移至顶部。 如未设置 <code>justify-content</code> 的值，那么这就是默认值。</li><li><code>flex-end</code>：从 flex 容器的终止位置开始排列项目。 对行来说是把项目移至右边， 对于列是把项目移至底部。</li><li><code>space-between</code>：项目间保留一定间距地沿主轴居中排列。 第一个和最后一个项目被放置在容器边沿。 例如，在行中第一个项目会紧贴着容器左边，最后一个项目会紧贴着容器右边，然后其他项目均匀排布。</li><li><code>space-around</code>：与<code>space-between</code>相似，但头尾两个项目不会紧贴容器边缘，所有项目之间的空间均匀排布。</li><li><code>space-evenly</code>：头尾两个项目不会紧贴容器边缘，所有项目之间的空间均匀排布。</li></ul>
# --instructions--
-这个例子可以帮助你理解这个属性。请为 `#box-container` 元素添加 CSS 属性 `justify-content`，并将其属性值设置为 `center`。
+这个例子可以帮助你理解这个属性。 请为 `#box-container` 元素添加 CSS 属性 `justify-content`，并将其属性值设置为 `center`。
**提示：**
-在编辑器里试试 `justify-content` 的其他可用值，看看它们之间的区别。但要通过挑战，你必须把值设为 `center`。
+在编辑器里试试 `justify-content` 的其他可用值，看看它们之间的区别。 但要通过挑战，你必须把值设为 `center`。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/apply-the-flex-direction-property-to-create-a-column-in-the-tweet-embed.md b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/apply-the-flex-direction-property-to-create-a-column-in-the-tweet-embed.md
index c8119b38d41..9dae457efbb 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/apply-the-flex-direction-property-to-create-a-column-in-the-tweet-embed.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/apply-the-flex-direction-property-to-create-a-column-in-the-tweet-embed.md
@@ -9,15 +9,21 @@ dashedName: apply-the-flex-direction-property-to-create-a-column-in-the-tweet-em
# --description--
-在之前的挑战中，我们把嵌入推文的 `header` 和 `footer` 的 `flex-direction` 属性值设为 row（行）。相似地，把 `.profile-name` 选择器中的元素竖着排列会好看一点。
+在之前的挑战中，我们把嵌入推文的 `header` 和 `footer` 的 `flex-direction` 属性值设为 row（行）。 相似地，把 `.profile-name` 选择器中的元素竖着排列会好看一点。
# --instructions--
-请给 header 的 `.profile-name` 元素添加 CSS 属性 `flex-direction`，并将其值设为 column。
+请给标题的 `.profile-name` 元素添加 CSS 属性 `flex-direction`，并将其值设为 `column`。
# --hints--
-`.profile-name` 所选择的元素应有 `flex-direction` 属性，且其属性值应为 column。
+`.follow-btn` 应在页面上呈现。 请关闭广告拦截器等任何扩展。
+
+```js
+assert($('.follow-btn').length > 0 && $('.follow-btn').css('display') !== 'none');
+```
+
+`.profile-name` 元素应有一个值为 `column` 的`flex-direction` 属性。
```js
assert($('.profile-name').css('flex-direction') == 'column');
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/apply-the-flex-direction-property-to-create-rows-in-the-tweet-embed.md b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/apply-the-flex-direction-property-to-create-rows-in-the-tweet-embed.md
index f27b5b4ac0e..21ddb0e79a5 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/apply-the-flex-direction-property-to-create-rows-in-the-tweet-embed.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/apply-the-flex-direction-property-to-create-rows-in-the-tweet-embed.md
@@ -9,21 +9,27 @@ dashedName: apply-the-flex-direction-property-to-create-rows-in-the-tweet-embed
# --description--
-嵌入推文示例中的 `header` 和 `footer` 有自己的子元素，使用 `flex-direction` 属性可以把这些子元素排成行。这个属性告诉 CSS 需要将这些子元素水平排列。
+嵌入推文示例中的 `header` 和 `footer` 有自己的子元素，使用 `flex-direction` 属性可以把这些子元素排成行。 这个属性告诉 CSS 需要将这些子元素水平排列。
# --instructions--
-为 `header` 和 `footer` 添加 CSS 属性 `flex-direction` 并把值设为 row。
+为 `header` 和 `footer` 添加 CSS 属性 `flex-direction`，并把值设为 `row`。
# --hints--
-`header` 应有 `flex-direction` 属性，其值应为 row。
+`.follow-btn` 应在页面上呈现。 请关闭广告拦截器等任何扩展。
+
+```js
+assert($('.follow-btn').length > 0 && $('.follow-btn').css('display') !== 'none');
+```
+
+`header` 应有 `flex-direction` 属性，其值应为 `row`。
```js
assert(code.match(/header\s*?{[^}]*?flex-direction:\s*?row;/g));
```
-`footer` 应有 `flex-direction` 属性，其值应为 row。
+`footer` 应有 `flex-direction` 属性，其值应为 `row`。
```js
assert(code.match(/footer\s*?{[^}]*?flex-direction:\s*?row;/g));
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-display-flex-to-position-two-boxes.md b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-display-flex-to-position-two-boxes.md
index 923e9bafd9e..8b656add18f 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-display-flex-to-position-two-boxes.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-display-flex-to-position-two-boxes.md
@@ -9,7 +9,7 @@ dashedName: use-display-flex-to-position-two-boxes
# --description--
-这节我们会使用不同的挑战方式来学习如何使用 CSS 更灵活地布局元素。首先我们会通过一个挑战来解释原理，然后通过操作一个简单的推文组件来应用弹性盒子（flexbox）。
+这节我们会使用不同的挑战方式来学习如何使用 CSS 更灵活地布局元素。 首先我们会通过一个挑战来解释原理，然后通过操作一个简单的推文组件来应用弹性盒子（flexbox）。
只要在一个元素的 CSS 中添加 `display: flex;`，就可以使用其它 flex 属性来构建响应式页面了。
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-align-items-property-in-the-tweet-embed.md b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-align-items-property-in-the-tweet-embed.md
index 31be7cf9606..9d82414f859 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-align-items-property-in-the-tweet-embed.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-align-items-property-in-the-tweet-embed.md
@@ -9,15 +9,21 @@ dashedName: use-the-align-items-property-in-the-tweet-embed
# --description--
-在上一个挑战中，我们介绍了 `align-items` 属性并给出了示例。我们可以对推文的几个嵌入元素使用这个属性，以调整其中 flex 元素的对齐方式。
+在上一个挑战中，我们介绍了 `align-items` 属性并给出了示例。 我们可以对推文的几个嵌入元素使用这个属性，以调整其中 flex 元素的对齐方式。
# --instructions--
-为 header 中的 `.follow-btn` 元素添加 CSS 属性 `align-items`，并将其属性值设为 `center`。
+为 header 中的 `.follow-btn` 元素添加 CSS 属性 `align-items`， 并将其属性值设为 `center`。
# --hints--
-`.follow-btn` 选取的元素应有 `align-items` 属性，且其属性值应为 `center`。
+`.follow-btn` 应在页面上呈现。 请关闭广告拦截器等任何扩展。
+
+```js
+assert($('.follow-btn').length > 0 && $('.follow-btn').css('display') !== 'none');
+```
+
+`.follow-btn` 元素应有值为 `center` 的 `align-items` 属性。
```js
assert($('.follow-btn').css('align-items') == 'center');
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-align-self-property.md b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-align-self-property.md
index 2eddf90a252..79931d31f0f 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-align-self-property.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-align-self-property.md
@@ -9,13 +9,13 @@ dashedName: use-the-align-self-property
# --description--
-flex 子项目的最后一个属性是 `align-self`。这个属性允许你调整单个子元素自己的对齐方式，而不会影响到全部子元素。因为 `float`、`clear` 和 `vertical-align` 等调整对齐方式的属性都不能应用于 flex 子元素，所以这个属性显得十分有用。
+flex 子项目的最后一个属性是 `align-self`。 这个属性允许你调整单个子元素自己的对齐方式，而不会影响到全部子元素。 因为 `float`、`clear` 和 `vertical-align` 等调整对齐方式的属性都不能应用于 flex 子元素，所以这个属性显得十分有用。
`align-self` 可设置的值与 `align-items` 的一样，并且它会覆盖 `align-items` 所设置的值。
# --instructions--
-请为 `#box-1` 和 `#box-2` 添加 CSS 属性 `align-self`。将 `#box-1` 的 `align-self` 属性值设为 center，将 `#box-2` 的设为 `flex-end`。
+请为 `#box-1` 和 `#box-2` 添加 CSS 属性 `align-self`。 将 `#box-1` 的 `align-self` 属性值设为 center，将 `#box-2` 的设为 `flex-end`。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-basis-property-to-set-the-initial-size-of-an-item.md b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-basis-property-to-set-the-initial-size-of-an-item.md
index f699c355995..6d725d0c4fa 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-basis-property-to-set-the-initial-size-of-an-item.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-basis-property-to-set-the-initial-size-of-an-item.md
@@ -11,11 +11,11 @@ dashedName: use-the-flex-basis-property-to-set-the-initial-size-of-an-item
`flex-basis` 属性定义了在使用 CSS 的 `flex-shrink` 或 `flex-grow` 属性对元素进行调整前，元素的初始大小。
-`flex-basis` 属性的单位与其他表示尺寸的属性的单位一致（`px`、`em`、`%` 等）。如果值为 `auto`，则项目的尺寸随内容调整。
+`flex-basis` 属性的单位与其他表示尺寸的属性的单位一致（`px`、`em`、`%` 等）。 如果值为 `auto`，则项目的尺寸随内容调整。
# --instructions--
-使用 `flex-basis` 为盒子设置初始值。请给 `#box-1` 和 `#box-2` 添加 CSS 属性 `flex-basis`。设置 `#box-1` 的尺寸初始值为 `10em`，`#box-2` 的尺寸初始值为 `20em`。
+使用 `flex-basis` 为盒子设置初始值。 请给 `#box-1` 和 `#box-2` 添加 CSS 属性 `flex-basis`。 设置 `#box-1` 的尺寸初始值为 `10em`，`#box-2` 的尺寸初始值为 `20em`。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-direction-property-to-make-a-column.md b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-direction-property-to-make-a-column.md
index ac1261ff2e8..aca2cebcd85 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-direction-property-to-make-a-column.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-direction-property-to-make-a-column.md
@@ -9,7 +9,7 @@ dashedName: use-the-flex-direction-property-to-make-a-column
# --description--
-在之前两个挑战中，我们使用了 `flex-direction` 属性创建行（row）。这个属性还能创建一个列，让子元素竖直排列在 flex 容器中。
+在之前两个挑战中，我们使用了 `flex-direction` 属性，值为 `row`。 这个属性还能创建一个列，让子元素竖直排列在 flex 容器中。
# --instructions--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-direction-property-to-make-a-row.md b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-direction-property-to-make-a-row.md
index 6457987d42f..2c46d081d34 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-direction-property-to-make-a-row.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-direction-property-to-make-a-row.md
@@ -9,11 +9,11 @@ dashedName: use-the-flex-direction-property-to-make-a-row
# --description--
-给元素添加 `display: flex` 属性可以让它变成 flex 容器。只要给父元素添加 `flex-direction` 属性，并把属性值设置为 row 或 column，即可横向排列或纵向排列它的所有子元素。设为 row 可以让子元素横向排列，设为 column 可以让子元素纵向排列。
+给元素添加 `display: flex` 属性可以让它变成 flex 容器， 然后可以让元素的项目排列成行或列。 只要给父元素添加 `flex-direction` 属性，并把属性值设置为 row 或 column，即可横向排列或纵向排列它的所有子元素。 创建一行将使子项水平对齐，创建一列将使子项垂直对齐。
`flex-direction` 的其他可选值还有 `row-reverse` 和 `column-reverse`。
-**注意：**`flex-direction` 的默认值为 row。
+**注意：**`flex-direction` 的默认值为 `row`。
# --instructions--
@@ -21,7 +21,7 @@ dashedName: use-the-flex-direction-property-to-make-a-row
# --hints--
-`#box-container` 应有 `flex-direction` 属性，其属性值应为 row-reverse。
+`#box-container` 元素应有 `flex-direction` 属性，其属性值应为 `row-reverse`。
```js
assert($('#box-container').css('flex-direction') == 'row-reverse');
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-grow-property-to-expand-items.md b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-grow-property-to-expand-items.md
index 0f696a7c147..147f028c98f 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-grow-property-to-expand-items.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-grow-property-to-expand-items.md
@@ -3,19 +3,19 @@ id: 587d78ae367417b2b2512afc
title: 使用 flex-grow 属性定义 flex 子元素的增长系数
challengeType: 0
videoUrl: 'https://scrimba.com/p/pVaDAv/c2p78cg'
-forumTopicId: 1301111
+forumTopicId: 301111
dashedName: use-the-flex-grow-property-to-expand-items
---
# --description--
-与 `flex-shrink` 相对的是 `flex-grow`。你应该还记得，`flex-shrink` 会在容器太小时对子元素作出调整。相应地，`flex-grow` 会在容器太大时对子元素作出调整。
+与 `flex-shrink` 相对的是 `flex-grow`。 你应该还记得，`flex-shrink` 会在容器太小时对子元素作出调整。 相应地，`flex-grow` 会在容器太大时对子元素作出调整。
-例子与上一个挑战相似，如果一个项目的 `flex-grow` 属性值为 `1`，另一个项目的 `flex-grow` 属性值为 `3`，那么后者会较前者扩大 3 倍。
+例子与上一个挑战相似，如果一个项目的 `flex-grow` 属性值为 `1`，另一个项目的 `flex-grow` 属性值为 `3`，那么值为 `3` 的一个会较另一个扩大 3 倍。
# --instructions--
-请为 `#box-1` 和 `#box-2` 添加 CSS 属性 `flex-grow`；将 `#box-1` 的 `flex-grow` 属性值设为 `1`，`#box-2` 的属性值设为 `2`。
+请为 `#box-1` 和 `#box-2` 添加 CSS 属性 `flex-grow`。 将 `#box-1` 的属性值设为 `1`，`#box-2` 的属性值设为 `2`。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-shorthand-property.md b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-shorthand-property.md
index b14a5490613..29d97dbbeb8 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-shorthand-property.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-shorthand-property.md
@@ -9,7 +9,7 @@ dashedName: use-the-flex-shorthand-property
# --description--
-上面几个 flex 属性有一个简写方式。`flex-grow`、`flex-shrink` 和 `flex-basis` 属性可以在 `flex` 中一并设置。
+上面几个 flex 属性有一个简写方式。 `flex-grow`、`flex-shrink` 和 `flex-basis` 属性可以在 `flex` 中一并设置。
例如，`flex: 1 0 10px;` 会把项目属性设为 `flex-grow: 1;`、`flex-shrink: 0;` 以及 `flex-basis: 10px;`。
@@ -17,9 +17,9 @@ dashedName: use-the-flex-shorthand-property
# --instructions--
-请给 `#box-1` 和 `#box-2` 添加 `flex` 属性。设置 `#box-1` 的 `flex-grow` 属性值为 `2`、`flex-shrink` 属性值为 `2`、`flex-basis` 属性值为 `150px`；设置 `#box-2` 的 `flex-grow` 属性值为 `1`、`flex-shrink` 属性值为 `1`、`flex-basis` 属性值为 `150px`。
+请给 `#box-1` 和 `#box-2` 添加 `flex` 属性。 设置 `#box-1` 的 `flex-grow` 属性值为 `2`、`flex-shrink` 属性值为 `2`、`flex-basis` 属性值为 `150px`。 设置 `#box-2` 的 `flex-grow` 属性值为 `1`、`flex-shrink` 属性值为 `1`、`flex-basis` 属性值为 `150px`。
-通过上面的设置，在容器大于 300px 时，`#box-1` 扩大的空间会是 `#box-2` 扩大空间的两倍；在容器小于 300px 时，`#box-1` 缩小的空间会是 `#box-2` 缩小空间的两倍。300px 是两个盒子的 `flex-basis` 属性值之和。
+通过上面的设置，在容器大于 300px 时，`#box-1` 扩大的空间会是 `#box-2` 扩大空间的两倍；在容器小于 300px 时，前者缩小的空间会是 `#box-2` 缩小空间的两倍。 300px 是两个盒子的 `flex-basis` 属性值之和。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-shrink-property-to-shrink-items.md b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-shrink-property-to-shrink-items.md
index dbe6872b0ce..6345a9670c3 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-shrink-property-to-shrink-items.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-shrink-property-to-shrink-items.md
@@ -9,15 +9,15 @@ dashedName: use-the-flex-shrink-property-to-shrink-items
# --description--
-目前为止，挑战里提到的属性都是应用于 flex 容器（flex 子元素的父元素）的。除此之外，flex 子元素也有很多实用属性。
+目前为止，挑战里提到的属性都是应用于 flex 容器（flex 子元素的父元素）的。 除此之外，flex 子元素也有很多实用属性。
-首先介绍的是 `flex-shrink` 属性。使用之后，如果 flex 容器太小，则子元素会自动缩小。当容器的宽度小于里面所有子元素的宽度之和时，所有子元素都会自动压缩。
+首先介绍的是 `flex-shrink` 属性。 使用之后，如果 flex 容器太小，则子元素会自动缩小。 当容器的宽度小于里面所有子元素的宽度之和时，所有子元素都会自动压缩。
-子元素的 `flex-shrink` 接受数值作为属性值。数值越大，则该元素与其他元素相比会被压缩得更厉害。比如，一个项目的 `flex-shrink` 属性值为 `1`，另一个项目的 `flex-shrink` 属性值为 `3`，那么后者相比前者会受到 `3` 倍压缩。
+子元素的 `flex-shrink` 接受数值作为属性值。 数值越大，则该元素与其他元素相比会被压缩得更厉害。 比如，一个项目的 `flex-shrink` 属性值为 `1`，另一个项目的 `flex-shrink` 属性值为 `3`，那么后者相比前者会受到 `3` 倍压缩。
# --instructions--
-请为 `#box-1` 和 `#box-2` 添加 CSS 属性 `flex-shrink`，将 `#box-1` 的 `flex-shrink` 属性值设为 `1`，将 `#box-2` 的属性值设为 `2`。
+请为 `#box-1` 和 `#box-2` 添加 CSS 属性 `flex-shrink`。 将 `#box-1` 的属性值设为 `1`，将 `#box-2` 的属性值设为 `2`。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-wrap-property-to-wrap-a-row-or-column.md b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-wrap-property-to-wrap-a-row-or-column.md
index 0055463d1de..38464a43b15 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-wrap-property-to-wrap-a-row-or-column.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-flex-wrap-property-to-wrap-a-row-or-column.md
@@ -9,9 +9,9 @@ dashedName: use-the-flex-wrap-property-to-wrap-a-row-or-column
# --description--
-CSS flexbox 有一个把 flex 子元素拆分为多行（或多列）的特性。默认情况下，flex 容器会调整项目大小，把它们都塞到一起。对于行来说，所有项目都会在一条直线上。
+CSS flexbox 有一个把 flex 子元素拆分为多行（或多列）的特性。 默认情况下，flex 容器会调整项目大小，把它们都塞到一起。 对于行来说，所有项目都会在一条直线上。
-不过，使用 `flex-wrap` 属性可以使项目换行展示。这意味着多出来的子元素会被移到新的行或列。换行发生的断点由子元素和容器的大小决定。
+不过，使用 `flex-wrap` 属性可以使项目换行展示。 这意味着多出来的子元素会被移到新的行或列。 换行发生的断点由子元素和容器的大小决定。
换行方向的可选值有这些：
@@ -19,7 +19,7 @@ CSS flexbox 有一个把 flex 子元素拆分为多行（或多列）的特性
# --instructions--
-现在的布局中，一行里面的元素太多了。请为 `#box-container` 元素添加 CSS 属性 `flex-wrap`，把将其属性值设为 `wrap`。
+现在的布局中，一行里面的元素太多了。 请为 `#box-container` 元素添加 CSS 属性 `flex-wrap`，把将其属性值设为 `wrap`。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-justify-content-property-in-the-tweet-embed.md b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-justify-content-property-in-the-tweet-embed.md
index 122647a19e5..48a8f52c253 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-justify-content-property-in-the-tweet-embed.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-justify-content-property-in-the-tweet-embed.md
@@ -9,7 +9,7 @@ dashedName: use-the-justify-content-property-in-the-tweet-embed
# --description--
-上一项挑战展示了 `justify-content` 属性的作用。如果我们想对齐推文内的子元素，可以把 `justify-content` 应用在 `.profile-name` 上。
+上一项挑战展示了 `justify-content` 属性的作用。 如果我们想对齐推文内的子元素，可以把给 `.profile-name` 元素应用这个属性。
# --instructions--
@@ -17,7 +17,13 @@ dashedName: use-the-justify-content-property-in-the-tweet-embed
# --hints--
-`.profile-name` 元素的 `justify-content` 可选用以下属性值：`center`、`flex-start`、`flex-end`、`space-between`、`space-around`。
+`.follow-btn` 应在页面上呈现。 请关闭广告拦截器等任何扩展。
+
+```js
+assert($('.follow-btn').length > 0 && $('.follow-btn').css('display') !== 'none');
+```
+
+`.profile-name` 元素的 `justify-content` 属性可选用以下属性值：`center`、`flex-start`、`flex-end`、`space-between`、`space-around` 或 `space-evenly`。
```js
assert(
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-order-property-to-rearrange-items.md b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-order-property-to-rearrange-items.md
index 2ffa498981c..eca0e55513f 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-order-property-to-rearrange-items.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-flexbox/use-the-order-property-to-rearrange-items.md
@@ -9,11 +9,11 @@ dashedName: use-the-order-property-to-rearrange-items
# --description--
-`order` 属性告诉 CSS flex 容器里子元素的顺序。默认情况下，项目排列顺序与源 HTML 文件中顺序相同。`order` 属性接受数字作为参数，可以使用负数。
+`order` 属性告诉 CSS flex 容器里子元素的顺序。 默认情况下，项目排列顺序与源 HTML 文件中顺序相同。 这个属性接受数字作为参数，可以使用负数。
# --instructions--
-请给 `#box-1` 和 `#box-2` 添加 CSS 属性 `order`，并将 `#box-1` 的`order` 属性值设为 `2`，`#box-2` 的属性值设为 `1`。
+请给 `#box-1` 和 `#box-2` 添加 CSS 属性 `order`， 并将 `#box-1` 的属性值设为 `2`，`#box-2` 的属性值设为 `1`。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/add-columns-with-grid-template-columns.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/add-columns-with-grid-template-columns.md
index 6f316855934..9d1fbd6dfd9 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/add-columns-with-grid-template-columns.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/add-columns-with-grid-template-columns.md
@@ -9,7 +9,7 @@ dashedName: add-columns-with-grid-template-columns
# --description--
-简单地添加一个网格元素并不会有任何明显的效果。你还需要明确网格的结构。在一个网格容器中使用 `grid-template-columns` 属性可以添加一些列，示例如下：
+简单地添加一个网格元素并不会有任何明显的效果。 你还需要明确网格的结构。 在一个网格容器中使用 `grid-template-columns` 属性可以添加一些列，示例如下：
```css
.container {
@@ -18,7 +18,7 @@ dashedName: add-columns-with-grid-template-columns
}
```
-上面的代码会在网格容器中添加两列，宽度均为 50px。`grid-template-columns` 属性值的个数表示网格的列数，每个值表示相应的列宽度。
+上面的代码会在网格容器中添加两列，宽度均为 50px。 `grid-template-columns` 属性值的个数表示网格的列数，每个值表示相应的列宽度。
# --instructions--
@@ -55,7 +55,7 @@ assert(
display: grid;
/* Only change code below this line */
-
+
/* Only change code above this line */
}
</style>
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/add-gaps-faster-with-grid-gap.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/add-gaps-faster-with-grid-gap.md
index 486ed0901dd..7e23a33b23f 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/add-gaps-faster-with-grid-gap.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/add-gaps-faster-with-grid-gap.md
@@ -3,13 +3,13 @@ id: 5a9036ee38fddaf9a66b5d37
title: 使用 grid-gap 为网格添加间距
challengeType: 0
videoUrl: 'https://scrimba.com/p/pByETK/ca2qVtv'
-forumTopicId: 1301118
+forumTopicId: 301118
dashedName: add-gaps-faster-with-grid-gap
---
# --description--
-`grid-gap` 属性是前两个挑战中出现的 `grid-row-gap` 和 `grid-column-gap` 的简写属性，它更方便使用。如果 `grid-gap` 只有一个值，那么这个值表示行与行之间、列与列之间的间距均为这个值。如果 `grid-gap` 有两个值，那么第一个值表示行间距，第二个值表示列间距。
+`grid-gap` 属性是前两个挑战中出现的 `grid-row-gap` 和 `grid-column-gap` 的简写属性，它更方便使用。 如果 `grid-gap` 只有一个值，那么这个值表示行与行之间、列与列之间的间距均为这个值。 如果有两个值，那么第一个值表示行间距，第二个值表示列间距。
# --instructions--
@@ -49,7 +49,7 @@ assert(
grid-template-rows: 1fr 1fr 1fr;
/* Only change code below this line */
-
+
/* Only change code above this line */
}
</style>
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/add-rows-with-grid-template-rows.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/add-rows-with-grid-template-rows.md
index 287140aba82..65215880e24 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/add-rows-with-grid-template-rows.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/add-rows-with-grid-template-rows.md
@@ -9,7 +9,7 @@ dashedName: add-rows-with-grid-template-rows
# --description--
-在上个挑战中，你创建的网格会自动设置行数。你可以用 `grid-template-rows` 设置网格的行，就像用 `grid-template-columns` 设置网格的列那样。
+在上个挑战中，你创建的网格会自动设置行数。 你可以用 `grid-template-rows` 设置网格的行，就像用 `grid-template-columns` 设置网格的列那样。
# --instructions--
@@ -47,7 +47,7 @@ assert(
grid-template-columns: 100px 100px 100px;
/* Only change code below this line */
-
+
/* Only change code above this line */
}
</style>
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/align-all-items-horizontally-using-justify-items.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/align-all-items-horizontally-using-justify-items.md
index 19e6455582b..3ed1ba09b33 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/align-all-items-horizontally-using-justify-items.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/align-all-items-horizontally-using-justify-items.md
@@ -9,11 +9,11 @@ dashedName: align-all-items-horizontally-using-justify-items
# --description--
-有时你想让 CSS 网格中的网格项共享对齐方式。你可以像之前学习的那样分别设置它们的对齐方式，也可以对网格容器使用 `justify-items` 使它们一次性沿水平轴对齐。这个属性能接受我们在之前两个挑战中学到的所有值作为属性值，但与之前不同的是，它会将网格中**所有**的网格项按所设置的方式对齐。
+有时你想让 CSS 网格中的网格项共享对齐方式。 你可以像之前学习的那样分别设置它们的对齐方式，也可以对网格容器使用 `justify-items` 使它们一次性沿水平轴对齐。 这个属性能接受我们在之前两个挑战中学到的所有值作为属性值，但与之前不同的是，它会将网格中**所有**的网格项按所设置的方式对齐。
# --instructions--
-请使用 `justify-items` 属性设置所有网格项水平居中。
+请使用这个属性设置所有网格项水平居中。
# --hints--
@@ -50,7 +50,7 @@ assert(
grid-gap: 10px;
/* Only change code below this line */
-
+
/* Only change code above this line */
}
</style>
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/align-all-items-vertically-using-align-items.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/align-all-items-vertically-using-align-items.md
index 60ec19e6275..a8fb7ecbfa6 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/align-all-items-vertically-using-align-items.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/align-all-items-vertically-using-align-items.md
@@ -13,7 +13,7 @@ dashedName: align-all-items-vertically-using-align-items
# --instructions--
-请使用 `align-items` 属性将所有网格项移动到单元格的末尾。
+请使用这个属性将所有网格项移动到单元格的末尾。
# --hints--
@@ -48,7 +48,7 @@ assert(
grid-gap: 10px;
/* Only change code below this line */
-
+
/* Only change code above this line */
}
</style>
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/align-an-item-horizontally-using-justify-self.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/align-an-item-horizontally-using-justify-self.md
index cc2a6bc4d45..d5a4c454b6f 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/align-an-item-horizontally-using-justify-self.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/align-an-item-horizontally-using-justify-self.md
@@ -9,7 +9,7 @@ dashedName: align-an-item-horizontally-using-justify-self
# --description--
-在 CSS 网格中，每个网格项的内容分别位于被称为<dfn>单元格（cell）</dfn>的框内。你可以使用网格项的 `justify-self` 属性，设置其内容的位置在单元格内沿水平轴的对齐方式。默认情况下，这个属性的值是 `stretch`，这将使内容占满整个单元格的宽度。该 CSS 网格属性也可以使用其他的值：
+在 CSS 网格中，每个网格项的内容分别位于被称为<dfn>单元格（cell）</dfn>的框内。 你可以使用网格项的 `justify-self` 属性，设置其内容的位置在单元格内沿水平轴的对齐方式。 默认情况下，这个属性的值是 `stretch`，这将使内容占满整个单元格的宽度。 该 CSS 网格属性也可以使用其他的值：
`start`：使内容在单元格左侧对齐，
@@ -43,7 +43,7 @@ assert(
background: LightSalmon;
/* Only change code below this line */
-
+
/* Only change code above this line */
}
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/align-an-item-vertically-using-align-self.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/align-an-item-vertically-using-align-self.md
index 72960cbf41a..69f8c977407 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/align-an-item-vertically-using-align-self.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/align-an-item-vertically-using-align-self.md
@@ -9,7 +9,7 @@ dashedName: align-an-item-vertically-using-align-self
# --description--
-正如能设置网格项沿水平方向的对齐方式一样，我们也可以设置网格项沿竖直方向的对齐方式。为此，我们可以对网格项使用 `align-self` 属性来实现。在上一个挑战中适用于 `justify-self` 属性的属性值同样也可用于 `align-self`。
+正如能设置网格项沿水平方向的对齐方式一样，我们也可以设置网格项沿竖直方向的对齐方式。 为此，我们可以对网格项使用 `align-self` 属性来实现。 在上一个挑战中适用于 `justify-self` 属性的属性值同样也可用于这个属性。
# --instructions--
@@ -36,7 +36,7 @@ assert(code.match(/.item3\s*?{[\s\S]*align-self\s*?:\s*?end\s*?;[\s\S]*}/gi));
background: PaleTurquoise;
/* Only change code below this line */
-
+
/* Only change code above this line */
}
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-a-column-gap-using-grid-column-gap.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-a-column-gap-using-grid-column-gap.md
index b277213115e..4dbf64ad47b 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-a-column-gap-using-grid-column-gap.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-a-column-gap-using-grid-column-gap.md
@@ -9,7 +9,7 @@ dashedName: create-a-column-gap-using-grid-column-gap
# --description--
-目前为止，在你所创建的网格中，每列都相互紧挨着。如果需要在列与列之间添加一些间距，我们可以使用 `grid-column-gap`：
+目前为止，在你所创建的网格中，每列都相互紧挨着。 有时候你想要列之间有一个间距。 如果需要在列与列之间添加一些间距，我们可以使用 `grid-column-gap`：
```css
grid-column-gap: 10px;
@@ -55,7 +55,7 @@ assert(
grid-template-rows: 1fr 1fr 1fr;
/* Only change code below this line */
-
+
/* Only change code above this line */
}
</style>
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-a-row-gap-using-grid-row-gap.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-a-row-gap-using-grid-row-gap.md
index 921da61ee24..e324d66c7af 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-a-row-gap-using-grid-row-gap.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-a-row-gap-using-grid-row-gap.md
@@ -47,7 +47,7 @@ assert(
grid-template-rows: 1fr 1fr 1fr;
/* Only change code below this line */
-
+
/* Only change code above this line */
}
</style>
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-flexible-layouts-using-auto-fill.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-flexible-layouts-using-auto-fill.md
index 162e6a5867a..10fe096ba37 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-flexible-layouts-using-auto-fill.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-flexible-layouts-using-auto-fill.md
@@ -9,19 +9,17 @@ dashedName: create-flexible-layouts-using-auto-fill
# --description--
-`repeat` 方法带有一个名为<dfn>自动填充（auto-fill）</dfn>的功能。它的功能是根据容器的大小，尽可能多地放入指定大小的行或列。你可以通过结合 `auto-fill` 和 `minmax` 来更灵活地布局。
-
-在最右侧的预览区中，`grid-template-columns` 被设置为：
+repeat 方法带有一个名为<dfn>自动填充（auto-fill）</dfn>的功能。 它的功能是根据容器的大小，尽可能多地放入指定大小的行或列。 你可以通过结合 `auto-fill` 和 `minmax` 来更灵活地布局。
```css
repeat(auto-fill, minmax(60px, 1fr));
```
-上面的代码效果是这样：首先，列的宽度会随容器大小改变。其次，只要容器宽度不足以插入一个宽为 60px 的列，当前行的所有列就都会一直拉伸。请自己调整宽度，动手试一下就不难理解了。**注意：**如果容器宽度不足以将所有网格项放在同一行，余下的网格项将会移至新的一行。
+上面的代码效果是这样：首先，列的宽度会随容器大小改变。其次，只要容器宽度不足以插入一个宽为 60px 的列，当前行的所有列就都会一直拉伸。请自己调整宽度，动手试一下就不难理解了。 **注意：**如果容器宽度不足以将所有网格项放在同一行，余下的网格项将会移至新的一行。
# --instructions--
-在第一个网格中，请使用 `auto-fill` 和 `repeat` 来填充网格。其中列宽的最小值为 `60px`，最大值为 `1fr`。你可以调整最右侧的预览区大小，查看自动填充的效果。
+在第一个网格中，请使用 `auto-fill` 和 `repeat` 来填充网格。 其中列宽的最小值为 `60px`，最大值为 `1fr`。 你可以调整最右侧的预览区大小，查看自动填充的效果。
# --hints--
@@ -98,7 +96,7 @@ assert(
.item3{background:PaleTurquoise;}
.item4{background:LightPink;}
.item5{background:PaleGreen;}
-
+
.container {
font-size: 40px;
min-height: 100px;
@@ -106,14 +104,14 @@ assert(
background: LightGray;
display: grid;
/* Only change code below this line */
-
+
grid-template-columns: repeat(auto-fill, minmax(60px, 1fr));
-
+
/* Only change code above this line */
grid-template-rows: 1fr 1fr 1fr;
grid-gap: 10px;
}
-
+
.container2 {
font-size: 40px;
min-height: 100px;
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-flexible-layouts-using-auto-fit.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-flexible-layouts-using-auto-fit.md
index 95022829a04..2228489dfeb 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-flexible-layouts-using-auto-fit.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-flexible-layouts-using-auto-fit.md
@@ -9,13 +9,13 @@ dashedName: create-flexible-layouts-using-auto-fit
# --description--
-`auto-fit` 效果几乎和 `auto-fill` 一样。不同点仅在于，当容器的大小大于各网格项之和时，`auto-fill` 会持续地在一端放入空行或空列，这样就会使所有网格项挤到另一边；而 `auto-fit` 则不会在一端放入空行或空列，而是会将所有网格项拉伸至合适的大小。
+`auto-fit` 效果几乎和 `auto-fill` 一样。 不同点仅在于，当容器的大小大于各网格项之和时，`auto-fill` 会持续地在一端放入空行或空列，这样就会使所有网格项挤到另一边；而 `auto-fit` 则不会在一端放入空行或空列，而是会将所有网格项拉伸至合适的大小。
**注意：**如果容器宽度不足以将所有网格项放在同一行，余下的网格项将会移至新的一行。
# --instructions--
-在第二个网格中，请用 `auto-fit` 和 `repeat` 来填充网格，其中列宽的最小值为 `60px`，最大值为`1fr`。你可以调整最右侧的预览区来查看效果。
+在第二个网格中，请用 `auto-fit` 和 `repeat` 来填充网格，其中列宽的最小值为 `60px`，最大值为`1fr`。 你可以调整最右侧的预览区来查看效果。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-grids-within-grids.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-grids-within-grids.md
index 134d244f2f8..2f0bda4c818 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-grids-within-grids.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-grids-within-grids.md
@@ -2,14 +2,13 @@
id: 5a94fe8569fb03452672e464
title: 在网格中创建网格
challengeType: 0
-videoUrl: 'https://scrimba.com/p/pByETK/c6N78Ap'
forumTopicId: 301128
dashedName: create-grids-within-grids
---
# --description--
-将元素转换为网格只会影响其子元素（即直接后代元素，英文为 `direct descendants`。意思是一个元素的所有后代元素中，父级元素为该元素的所有元素）。因此，如果我们把某个子元素设置为网格，就会得到一个嵌套的网格。
+将元素转换为网格只会影响其子元素（即直接后代元素，英文为 direct descendants。意思是一个元素的所有后代元素中，父级元素为该元素的所有元素）。 因此，如果我们把某个子元素设置为网格，就会得到一个嵌套的网格。
例如，如果我们设置 class 为 `item3` 的元素的 `display` 和 `grid-template-columns` 属性，就会得到一个嵌套的网格。
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-your-first-css-grid.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-your-first-css-grid.md
index d33b7f84359..9378e6c1ea1 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-your-first-css-grid.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/create-your-first-css-grid.md
@@ -9,17 +9,17 @@ dashedName: create-your-first-css-grid
# --description--
-通过将属性 `display` 的值设为 `grid`，HTML 元素就可以变为网格容器。通过前面的操作，你可以对该容器使用与 CSS 网格（CSS Grid）相关的属性。
+通过将属性 `display` 的值设为 `grid`，HTML 元素就可以变为网格容器。 通过前面的操作，你可以对该容器使用与 CSS 网格（CSS Grid）相关的属性。
**注意：**在 CSS 网格中，父元素称为<dfn>容器（container）</dfn>，它的子元素称为<dfn>项（items）</dfn>。
# --instructions--
-请将 class 为 `container` 的 `div` 的 `display` 属性值设置为 `grid`。
+请将 class 为 `container` 的 div 的 display 属性值设置为 `grid`。
# --hints--
-class 为 `container` 的 `div` 元素应具有 `display` 属性且属性值应为 `grid`。
+`container` class 应具有 `display` 属性，属性值应为 `grid`。
```js
assert(code.match(/.container\s*?{[\s\S]*display\s*?:\s*?grid\s*?;[\s\S]*}/gi));
@@ -43,7 +43,7 @@ assert(code.match(/.container\s*?{[\s\S]*display\s*?:\s*?grid\s*?;[\s\S]*}/gi));
background: LightGray;
/* Only change code below this line */
-
+
/* Only change code above this line */
}
</style>
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/divide-the-grid-into-an-area-template.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/divide-the-grid-into-an-area-template.md
index 5c74cce3cc5..9666f8b6d3d 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/divide-the-grid-into-an-area-template.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/divide-the-grid-into-an-area-template.md
@@ -9,7 +9,7 @@ dashedName: divide-the-grid-into-an-area-template
# --description--
-你可以将网格中的一些单元格组合成一个<dfn>区域（area）</dfn>，并为该区域指定一个自定义名称。可以通过给容器加上 `grid-template-areas` 来实现：
+你可以将网格中的一些单元格组合成一个<dfn>区域（area）</dfn>，并为该区域指定一个自定义名称。 可以通过给容器加上 `grid-template-areas` 来实现：
```css
grid-template-areas:
@@ -18,7 +18,7 @@ grid-template-areas:
"footer footer footer";
```
-上面的代码将顶部三个单元格合并成一个名为 `header` 的区域，将底部三个单元格合并为一个名为 `footer` 的区域，并在中间行创建了两个区域：`advert` 和 `content`。**注意：**在代码中，每个单词代表一个网格单元格，每对引号代表一行。除了自定义标签，你还能使用句点（`.`）来表示一个空单元格。
+上面的代码将顶部三个单元格合并成一个名为 `header` 的区域，将底部三个单元格合并为一个名为 `footer` 的区域，并在中间行创建了两个区域：`advert` 和 `content`。 **注意：**在代码中，每个单词代表一个网格单元格，每对引号代表一行。 除了自定义标签，你还能使用句点（`.`）来表示一个空单元格。
# --instructions--
@@ -26,13 +26,15 @@ grid-template-areas:
# --hints--
-class 为 `container` 的元素应具有 `grid-template-areas` 属性，在其属性值中，应使用 `.` 代替 `.advert`。
+class 为 `container` 的元素应具有 `grid-template-areas` 属性，在其属性值中，应使用 `.` 代替 `advert`。
```js
assert(
- code.match(
- /.container\s*?{[\s\S]*grid-template-areas\s*?:\s*?"\s*?header\s*?header\s*?header\s*?"\s*?"\s*?.\s*?content\s*?content\s*?"\s*?"\s*?footer\s*?footer\s*?footer\s*?"\s*?;[\s\S]*}/gi
- )
+ __helpers
+ .removeCssComments(code)
+ .match(
+ /.container\s*?{[\s\S]*grid-template-areas\s*?:\s*?"\s*?header\s*?header\s*?header\s*?"\s*?"\s*?.\s*?content\s*?content\s*?"\s*?"\s*?footer\s*?footer\s*?footer\s*?"\s*?;[\s\S]*}/gi
+ )
);
```
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/limit-item-size-using-the-minmax-function.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/limit-item-size-using-the-minmax-function.md
index 2198ddb6fee..ff3e5578010 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/limit-item-size-using-the-minmax-function.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/limit-item-size-using-the-minmax-function.md
@@ -9,7 +9,7 @@ dashedName: limit-item-size-using-the-minmax-function
# --description--
-此外，内置函数 `minmax` 也可用于设置 `grid-template-columns` 和 `grid-template-rows` 的值。它的作用是在网格容器改变大小时限制网格项的大小。为此，你需要指定网格项允许的尺寸范围。例如：
+此外，内置函数 `minmax` 也可用于设置 `grid-template-columns` 和 `grid-template-rows` 的值。 它的作用是在网格容器改变大小时限制网格项的大小。 为此，你需要指定网格项允许的尺寸范围。 例如：
```css
grid-template-columns: 100px minmax(50px, 200px);
@@ -19,7 +19,7 @@ grid-template-columns: 100px minmax(50px, 200px);
# --instructions--
-请用 `minmax` 函数替换 `repeat` 函数中的 `1fr`，限定其最小值为 `90px`，最大值为`1fr`。你可以调整最右侧的预览面板查看效果。
+请用 `minmax` 函数替换 `repeat` 函数中的 `1fr`，限定其最小值为 `90px`，最大值为`1fr`。 你可以调整最右侧的预览面板查看效果。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/place-items-in-grid-areas-using-the-grid-area-property.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/place-items-in-grid-areas-using-the-grid-area-property.md
index 03e0438bb63..473c4e294ad 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/place-items-in-grid-areas-using-the-grid-area-property.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/place-items-in-grid-areas-using-the-grid-area-property.md
@@ -9,7 +9,7 @@ dashedName: place-items-in-grid-areas-using-the-grid-area-property
# --description--
-像上一个挑战那样，在为网格添加区域模板后，你可以通过引用你所定义的区域的名称，将元素放入相应的区域。为此，你需要对网格项使用 `grid-area`：
+像上一个挑战那样，在为网格添加区域模板后，你可以通过引用你所定义的区域的名称，将元素放入相应的区域。 为此，你需要对网格项使用 `grid-area`：
```css
.item1 {
@@ -17,7 +17,7 @@ dashedName: place-items-in-grid-areas-using-the-grid-area-property
}
```
-这样，class 为 `item1` 的网格项就被放到了 `header` 区域里。在这个示例中，网格项将占用第一行整行，因为这一整行都被命名为 `header` 区域。
+这样，class 为 `item1` 的网格项就被放到了 `header` 区域里。 在这个示例中，网格项将占用第一行整行，因为这一整行都被命名为标题区域。
# --instructions--
@@ -28,7 +28,11 @@ dashedName: place-items-in-grid-areas-using-the-grid-area-property
class 为 `item5` 的元素应具有 `grid-area` 属性且属性值应为 `footer`。
```js
-assert(code.match(/.item5\s*?{[\s\S]*grid-area\s*?:\s*?footer\s*?;[\s\S]*}/gi));
+assert(
+ __helpers
+ .removeCssComments(code)
+ .match(/.item5\s*?{[\s\S]*grid-area\s*?:\s*?footer\s*?;[\s\S]*}/gi)
+);
```
# --seed--
@@ -46,7 +50,7 @@ assert(code.match(/.item5\s*?{[\s\S]*grid-area\s*?:\s*?footer\s*?;[\s\S]*}/gi));
background: PaleGreen;
/* Only change code below this line */
-
+
/* Only change code above this line */
}
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/reduce-repetition-using-the-repeat-function.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/reduce-repetition-using-the-repeat-function.md
index 2fa5e61b16b..c9768187492 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/reduce-repetition-using-the-repeat-function.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/reduce-repetition-using-the-repeat-function.md
@@ -11,7 +11,7 @@ dashedName: reduce-repetition-using-the-repeat-function
使用 `grid-template-columns` 或 `grid-template-rows` 定义网格结构时，你需要为添加的每一行或每一列都输入一个值。
-如果一个网格共有 100 行且每行高度相同，那我们就需要输入 100 个值，这显然不太实际。为此，更好的方式是使用 `repeat` 方法指定行或列的重复次数，后面加上逗号以及需要重复的值。
+如果一个网格共有 100 行且每行高度相同， 那我们就需要输入 100 个值，这显然不太实际。 为此，更好的方式是使用 `repeat` 方法指定行或列的重复次数，后面加上逗号以及需要重复的值。
以下为添加 100 行网格的例子，每行高度均为 50px：
@@ -19,7 +19,7 @@ dashedName: reduce-repetition-using-the-repeat-function
grid-template-rows: repeat(100, 50px);
```
-你还可以用 repeat 方法重复多个值，并在定义网格结构时与其他值一起使用。比如：
+你还可以用 repeat 方法重复多个值，并在定义网格结构时与其他值一起使用。 比如：
```css
grid-template-columns: repeat(2, 1fr 50px) 20px;
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-css-grid-units-to-change-the-size-of-columns-and-rows.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-css-grid-units-to-change-the-size-of-columns-and-rows.md
index 51457886fb8..f7f9d3b637d 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-css-grid-units-to-change-the-size-of-columns-and-rows.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-css-grid-units-to-change-the-size-of-columns-and-rows.md
@@ -9,7 +9,7 @@ dashedName: use-css-grid-units-to-change-the-size-of-columns-and-rows
# --description--
-在 CSS 网格中，可以使用绝对单位（如 `px`）或相对单位（如 `em`）来定义行或列的大小。下面的单位也可以使用：
+在 CSS 网格中，可以使用绝对单位（如 `px`）或相对单位（如 `em`）来定义行或列的大小。 下面的单位也可以使用：
`fr`：设置列或行占剩余空间的比例，
@@ -23,7 +23,7 @@ dashedName: use-css-grid-units-to-change-the-size-of-columns-and-rows
grid-template-columns: auto 50px 10% 2fr 1fr;
```
-这段代码添加了五个列。第一列的宽与它的内容宽度相等；第二列宽 50px；第三列宽是它容器的 10%；最后两列，将剩余的宽度平均分成三份，第四列占两份，第五列占一份。
+这段代码添加了五个列。 第一列的宽与它的内容宽度相等；第二列宽 50px；第三列宽是它容器的 10%；最后两列，将剩余的宽度平均分成三份，第四列占两份，第五列占一份。
# --instructions--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-grid-area-without-creating-an-areas-template.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-grid-area-without-creating-an-areas-template.md
index 46ae3544d68..74e8557439b 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-grid-area-without-creating-an-areas-template.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-grid-area-without-creating-an-areas-template.md
@@ -9,27 +9,27 @@ dashedName: use-grid-area-without-creating-an-areas-template
# --description--
-我们在上一次挑战中学到的 `grid-area` 属性还有另一种使用方式。如果网格中没有定义区域模板，你也可以像这样为它添加一个模板：
+我们在上一次挑战中学到的 `grid-area` 属性还有另一种使用方式。 如果网格中没有定义区域模板，你也可以像这样为它添加一个模板：
```css
item1 { grid-area: 1/1/2/4; }
```
-这里使用了你之前学习的网格线编号来定义网格项的区域。上例中数字代表这些值：
+这里使用了你之前学习的网格线编号来定义网格项的区域。 上例中数字代表这些值：
```css
-grid-area: 起始的水平网格线 / 起始的垂直网格线 / 结束的水平网格线 / 结束的垂直网格线;
+grid-area: horizontal line to start at / vertical line to start at / horizontal line to end at / vertical line to end at;
```
因此，示例中的网格项将占用第 1 条水平网格线（起始）和第 2 条水平网格线（终止）之间的行，及第 1 条垂直网格线（起始）和第 4 条垂直网格线（终止）之间的列。
# --instructions--
-请用 `grid-area` 属性将 `class` 为 `item5` 的元素放置在第 3 条和第 4 条水平网格线，以及第 1 条和第 4 条垂直网格线之间的区域内。
+请用 `grid-area` 属性将 class 为 `item5` 的元素放置在第 3 条和第 4 条水平网格线，以及第 1 条和第 4 条垂直网格线之间的区域内。
# --hints--
-`class` 为 `item5` 的元素应具有 `grid-area` 属性且属性值应为 `3/1/4/4`。
+class 为 `item5` 的元素应具有 `grid-area` 属性，且位于水平第三和第四条线、垂直第一和第四条线之间的区域。
```js
assert(
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-grid-column-to-control-spacing.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-grid-column-to-control-spacing.md
index 3e23df08556..2c5e79996f6 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-grid-column-to-control-spacing.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-grid-column-to-control-spacing.md
@@ -9,13 +9,13 @@ dashedName: use-grid-column-to-control-spacing
# --description--
-到目前为止，所有的讨论都是围绕网格容器的。`grid-column` 属性是我们要讨论的，第一个用于网格项本身的属性。
+到目前为止，所有的讨论都是围绕网格容器的。 `grid-column` 属性是我们要讨论的，第一个用于网格项本身的属性。
-网格中，假想的水平线和垂直线被称为<dfn>线（lines）</dfn>。这些线在网格的左上角从 1 开始编号，垂直线向右、水平线向下累加计数。
+网格中，假想的水平线和垂直线被称为<dfn>线（lines）</dfn>。 这些线在网格的左上角从 1 开始编号，垂直线向右、水平线向下累加计数。
这是一个 3x3 网格的线条：
-<div style='position:relative;margin:auto;background:Gainsboro;display:block;margin-top:100px;margin-bottom:50px;width:200px;height:200px;'><p style='left:25%;top:-30%;font-size:130%;position:absolute;color:RoyalBlue;'>column lines</p><p style='left:0%;top:-15%;font-size:130%;position:absolute;color:RoyalBlue;'>1</p><p style='left:30%;top:-15%;font-size:130%;position:absolute;color:RoyalBlue;'>2</p><p style='left:63%;top:-15%;font-size:130%;position:absolute;color:RoyalBlue;'>3</p><p style='left:95%;top:-15%;font-size:130%;position:absolute;color:RoyalBlue;'>4</p><p style='left:-40%;top:45%;font-size:130%;transform:rotateZ(-90deg);position:absolute;'>row lines</p><p style='left:-10%;top:-10%;font-size:130%;position:absolute;'>1</p><p style='left:-10%;top:21%;font-size:130%;position:absolute;'>2</p><p style='left:-10%;top:53%;font-size:130%;position:absolute;'>3</p><p style='left:-10%;top:85%;font-size:130%;position:absolute;'>4</p><div style='left:0%;top:0%;width:5%;height:100%;background:RoyalBlue;position:absolute;'></div><div style='left:31%;top:0%;width:5%;height:100%;background:RoyalBlue;position:absolute;'></div><div style='left:63%;top:0%;width:5%;height:100%;background:RoyalBlue;position:absolute;'></div><div style='left:95%;top:0%;width:5%;height:100%;background:RoyalBlue;position:absolute;'></div><div style='left:0%;top:0%;width:100%;height:5%;background:black;position:absolute;'></div><div style='left:0%;top:31%;width:100%;height:5%;background:black;position:absolute;'></div><div style='left:0%;top:63%;width:100%;height:5%;background:black;position:absolute;'></div><div style='left:0%;top:95%;width:100%;height:5%;background:black;position:absolute;'></div></div>
+<div style='position:relative;margin:auto;background:Gainsboro;display:block;margin-top:100px;margin-bottom:50px;width:200px;height:200px;'><p style='left:25%;top:-30%;font-size:130%;position:absolute;color:RoyalBlue;'>列线</p><p style='left:0%;top:-15%;font-size:130%;position:absolute;color:RoyalBlue;'>1</p><p style='left:30%;top:-15%;font-size:130%;position:absolute;color:RoyalBlue;'>2</p><p style='left:63%;top:-15%;font-size:130%;position:absolute;color:RoyalBlue;'>3</p><p style='left:95%;top:-15%;font-size:130%;position:absolute;color:RoyalBlue;'>4</p><p style='left:-40%;top:45%;font-size:130%;transform:rotateZ(-90deg);position:absolute;'>行线</p><p style='left:-10%;top:-10%;font-size:130%;position:absolute;'>1</p><p style='left:-10%;top:21%;font-size:130%;position:absolute;'>2</p><p style='left:-10%;top:53%;font-size:130%;position:absolute;'>3</p><p style='left:-10%;top:85%;font-size:130%;position:absolute;'>4</p><div style='left:0%;top:0%;width:5%;height:100%;background:RoyalBlue;position:absolute;'></div><div style='left:31%;top:0%;width:5%;height:100%;background:RoyalBlue;position:absolute;'></div><div style='left:63%;top:0%;width:5%;height:100%;background:RoyalBlue;position:absolute;'></div><div style='left:95%;top:0%;width:5%;height:100%;background:RoyalBlue;position:absolute;'></div><div style='left:0%;top:0%;width:100%;height:5%;background:black;position:absolute;'></div><div style='left:0%;top:31%;width:100%;height:5%;background:black;position:absolute;'></div><div style='left:0%;top:63%;width:100%;height:5%;background:black;position:absolute;'></div><div style='left:0%;top:95%;width:100%;height:5%;background:black;position:absolute;'></div></div>
你可以用 `grid-column` 属性定义网格项开始和结束的位置，进而控制每个网格项占用的列数。
@@ -37,9 +37,8 @@ class 为 `item5` 的元素应具有 `grid-column` 属性。
```js
assert(
- $('style')
- .text()
- .replace(/\s/g, '')
+ __helpers
+ .removeWhiteSpace($('style').text())
.match(/\.item5{.*grid-column:.*}/g)
);
```
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-grid-row-to-control-spacing.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-grid-row-to-control-spacing.md
index a4fa692943e..cfeb52aa6d9 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-grid-row-to-control-spacing.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-grid-row-to-control-spacing.md
@@ -9,7 +9,7 @@ dashedName: use-grid-row-to-control-spacing
# --description--
-和设置一个网格项占用多列类似，你也可以设置它占用多行。你可以使用 `grid-row` 属性来定义一个网格项开始和结束的水平线。
+和设置一个网格项占用多列类似，你也可以设置它占用多行。 你可以使用 `grid-row` 属性来定义一个网格项开始和结束的水平线。
# --instructions--
@@ -21,10 +21,7 @@ class 为 `item5` 的元素应具有 `grid-row` 属性。
```js
assert(
- $('style')
- .text()
- .replace(/\s/g, '')
- .match(/\.item5{.*grid-row:.*}/g)
+ __helpers.removeWhiteSpace($('style').text()).match(/\.item5{.*grid-row:.*}/g)
);
```
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-media-queries-to-create-responsive-layouts.md b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-media-queries-to-create-responsive-layouts.md
index 6eee6947190..0d247e8420b 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-media-queries-to-create-responsive-layouts.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/css-grid/use-media-queries-to-create-responsive-layouts.md
@@ -11,7 +11,7 @@ dashedName: use-media-queries-to-create-responsive-layouts
将 CSS 网格与使用媒体查询结合使用，如使用媒体查询重新排列网格区域、更改网格尺寸以及重新排列网格项位置，我们可以让制作出的网站更具响应性。
-在右侧的预览区中，当网页可视区域的宽不小于 300px 时，列数从 1 变为 2。并且，广告（advertisement）区域会完全占据左列。
+在右侧的预览区中，当网页可视区域的宽不小于 300px 时，列数从 1 变为 2。 并且，广告（advertisement）区域会完全占据左列。
# --instructions--
@@ -23,9 +23,11 @@ dashedName: use-media-queries-to-create-responsive-layouts
```js
assert(
- code.match(
- /@media\s*?\(\s*?min-width\s*?:\s*?400px\s*?\)[\s\S]*.container\s*?{[\s\S]*grid-template-areas\s*?:\s*?"\s*?header\s*?header\s*?"\s*?"\s*?advert\s*?content\s*?"\s*?"\s*?footer\s*?footer\s*?"\s*?;[\s\S]*}/gi
- )
+ __helpers
+ .removeCssComments(code)
+ .match(
+ /@media\s*?\(\s*?min-width\s*?:\s*?400px\s*?\)[\s\S]*.container\s*?{[\s\S]*grid-template-areas\s*?:\s*?"\s*?header\s*?header\s*?"\s*?"\s*?advert\s*?content\s*?"\s*?"\s*?footer\s*?footer\s*?"\s*?;[\s\S]*}/gi
+ )
);
```
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-principles/create-a-media-query.md b/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-principles/create-a-media-query.md
index 1e923543233..963fc4fff1d 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-principles/create-a-media-query.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-principles/create-a-media-query.md
@@ -2,16 +2,16 @@
id: 587d78b0367417b2b2512b08
title: 创建一个媒体查询
challengeType: 0
-videoUrl: 'https://scrimba.com/c/cPp7VfD'
-forumTopicId: 1
+videoUrl: 'https://scrimba.com/p/pzrPu4/cqwKrtm'
+forumTopicId: 301139
dashedName: create-a-media-query
---
# --description--
-媒体查询是 CSS3 中引入的一项新技术，它可以根据不同的视口大小调整内容的布局。视口是指浏览器中，用户可见的网页内容。视口会随访问网站的设备不同而改变。
+媒体查询是 CSS3 中引入的一项新技术，它可以根据不同的视口大小调整内容的布局。 视口是指浏览器中，用户可见的网页内容。 视口会随访问网站的设备不同而改变。
-媒体查询由媒体类型组成，如果媒体类型与展示网页的设备类型匹配，则应用对应的样式。你可以在媒体查询中使用各种选择器和样式。
+媒体查询由媒体类型组成，如果媒体类型与展示网页的设备类型匹配，则应用对应的样式。 你可以在媒体查询中使用各种选择器和样式。
下面是一个媒体查询的例子，当设备宽度小于或等于 100px 时返回内容：
@@ -25,7 +25,7 @@ dashedName: create-a-media-query
# --instructions--
-请添加一条媒体查询规则，当设备的高度小于等于 800px 时，`p` 标签的 `font-size` 为 `10px`。
+请添加一条媒体查询规则，当设备的高度小于等于 `800px` 时，`p` 标签的 `font-size` 为 `10px`。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-principles/make-an-image-responsive.md b/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-principles/make-an-image-responsive.md
index 1ee9bf9fcd3..d150c5d7df5 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-principles/make-an-image-responsive.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-principles/make-an-image-responsive.md
@@ -2,14 +2,14 @@
id: 587d78b1367417b2b2512b09
title: 使图片自适应设备尺寸
challengeType: 0
-videoUrl: 'https://scrimba.com/c/cPp7VfD'
-forumTopicId: 1
+videoUrl: 'https://scrimba.com/p/pzrPu4/cz763UD'
+forumTopicId: 301140
dashedName: make-an-image-responsive
---
# --description--
-用 CSS 来让图片自适应其实很简单。你只需要给图片添加这些属性:
+用 CSS 来让图片自适应其实很简单。 你只需要给图片添加这些属性:
```css
img {
@@ -22,7 +22,7 @@ img {
# --instructions--
-给 `responsive-img` 添加样式规则，使其成为响应式的图片。它不应该超出父容器（在本例中，即预览窗口）的范围，并保持宽高比不变。添加代码后，拖动浏览器窗口，看看图片发生什么变化。
+给 `responsive-img` 添加样式规则，使其成为响应式的图片。 它不应该超出父容器（在本例中，即预览窗口）的范围，并保持宽高比不变。 添加代码后，拖动浏览器窗口，看看图片发生什么变化。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-principles/make-typography-responsive.md b/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-principles/make-typography-responsive.md
index 2f2bf3f702b..49a8ae704c8 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-principles/make-typography-responsive.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-principles/make-typography-responsive.md
@@ -2,26 +2,26 @@
id: 587d78b1367417b2b2512b0c
title: 使排版根据设备尺寸自如响应
challengeType: 0
-videoUrl: 'https://scrimba.com/c/cPp7VfD'
-forumTopicId: 1
+videoUrl: 'https://scrimba.com/p/pzrPu4/crzN7T8'
+forumTopicId: 301141
dashedName: make-typography-responsive
---
# --description--
-除了使用 `em` 或 `px` 设置文本大小，你还可以用视窗单位来做响应式排版。视窗单位和百分比都是相对单位，但它们是基于不同的参照物。视窗单位是相对于设备的视窗尺寸（宽度或高度），百分比是相对于父级元素的大小。
+除了使用 `em` 或 `px` 设置文本大小，你还可以用视窗单位来做响应式排版。 视窗单位和百分比都是相对单位，但它们是基于不同的参照物。 视窗单位是相对于设备的视窗尺寸（宽度或高度），百分比是相对于父级元素的大小。
四个不同的视窗单位分别是：
<ul><li><code>vw</code>：如 <code>10vw</code> 的意思是视窗宽度的 10%。</li><li><code>vh：</code> 如 <code>3vh</code> 的意思是视窗高度的 3%。</li><li><code>vmin：</code> 如 <code>70vmin</code> 的意思是视窗的高度和宽度中较小一个的 70%。</li><li><code>vmax：</code> 如 <code>100vmax</code> 的意思是视窗的高度和宽度中较大一个的 100%。</li></ul>
-下面这个例子是设置 body 标签的宽度为视窗宽度的 30%。
+下面这个例子是设置 `body` 标签的宽度为视窗宽度的 30%。
`body { width: 30vw; }`
# --instructions--
-请将 `h2` 标签的 `width` 设置为视窗宽度的 80%，`p` 标签的 `width` 为视窗高度和宽度中较小值的 75%。
+请将 `h2` 标签的 `width` 设置为视窗宽度的 80%，将段落的 `width` 设置为视窗高度和宽度中较小值的 75%。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-principles/use-a-retina-image-for-higher-resolution-displays.md b/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-principles/use-a-retina-image-for-higher-resolution-displays.md
index 16fb26f7872..5d76079b6d1 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-principles/use-a-retina-image-for-higher-resolution-displays.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-principles/use-a-retina-image-for-higher-resolution-displays.md
@@ -2,16 +2,16 @@
id: 587d78b1367417b2b2512b0a
title: 针对高分辨率屏幕应使用视网膜图片
challengeType: 0
-videoUrl: 'https://scrimba.com/c/cPp7VfD'
-forumTopicId: 1
+videoUrl: 'https://scrimba.com/p/pzrPu4/cVZ4Rfp'
+forumTopicId: 301142
dashedName: use-a-retina-image-for-higher-resolution-displays
---
# --description--
-随着联网设备的增加，设备间的区别不仅发生在尺寸和规格上，还发生在用于显示的设备上。像素密度就是区分不同显示设备的一个指标，它一般会以 PPI（Pixel Per Inch，即每英寸像素）或 DPI（每英寸点数）为计量单位。最著名的显示器就是 Apple MacBook Pro 笔记本电脑上的“视网膜显示屏”（现亦用于 iMac）。由于“视网膜显示屏”和“非视网膜显示屏”显示器之间像素密度的不同，某些未考虑高分辨率显示器的图像在高分辨率显示器上渲染时，可能因出现“像素化”而不够清晰。
+随着联网设备的增加，设备间的区别不仅发生在尺寸和规格上，还发生在用于显示的设备上。 像素密度就是区分不同显示设备的一个指标，它一般会以 PPI（Pixel Per Inch，即每英寸像素）或 DPI（每英寸点数）为计量单位。 最著名的显示器就是 Apple MacBook Pro 笔记本电脑上的“视网膜显示屏”（现亦用于 iMac）。 由于“视网膜显示屏”和“非视网膜显示屏”显示器之间像素密度的不同，某些未考虑高分辨率显示器的图像在高分辨率显示器上渲染时，可能因出现“像素化”而不够清晰。
-要让图像在高分辨率显示器（例如 MacBook Pro 的“视网膜显示屏”）上正常显示，最简单的方法是将其 width 和 height 值设置为原始文件的一半，如下所示：
+让图像正确出现在高分辨率显示器（例如 MacBook Pros “Revistina display”）上的最简单方式， 是定义它们的 `width` 和 `height` 值为原始值的一半。 下面是一个仅使用原始高度和宽度一半的图像示例：
```html
<style>
@@ -22,20 +22,20 @@ dashedName: use-a-retina-image-for-higher-resolution-displays
# --instructions--
-请将 `img` 标签的 `width` 和 `height` 设置为它们原始宽高的一半。在这个例子中，原始 `height` 和 `width` 的值都为 200px。
+设置 `img` 标签的 `width` 和 `height` 为它们的原始值的一半。 在这个例子中，原始 `height` 和原始 `width` 的值都为 `200px`。
# --hints--
`img` 标签的 `width` 属性值应为 100px。
```js
-assert($('img').css('width') == '100px');
+assert(document.querySelector('img').width === 100);
```
`img` 标签的 `height` 属性值应为 100px。
```js
-assert($('img').css('height') == '100px');
+assert(document.querySelector('img').height === 100);
```
# --seed--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-personal-portfolio-webpage.md b/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-personal-portfolio-webpage.md
index 4c5b113e8ab..81edb8c0c21 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-personal-portfolio-webpage.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-personal-portfolio-webpage.md
@@ -10,25 +10,25 @@ dashedName: build-a-personal-portfolio-webpage
**目标：**在 [CodePen.io](https://codepen.io) 上创建一个与这个功能类似的 app：<https://codepen.io/freeCodeCamp/full/zNBOYG>。
-在满足以下[需求](https://en.wikipedia.org/wiki/User_story)并能通过所有测试的前提下，你可以根据自己的喜好来美化你的 app。
+在满足以下[需求](https://en.wikipedia.org/wiki/User_story)并能通过所有测试的前提下， 你可以根据自己的喜好来美化你的 app。
-你可以使用 HTML、JavaScript 以及 CSS 来完成项目。由于目前你只学到了 CSS 课程，所以我们建议你只使用 CSS 来完成这个项目，同时巩固一下你之前所学的内容。你也可以使用 Bootstrap 或者 SASS。我们不推荐你在这个项目中使用其他技术（比如 jQuery、React、Angular 或 Vue）。否则，如果在编码中出现问题，你需要自行解决。在后续的其他项目中，你将有机会使用像是 React 等其他技术栈。如果你在使用上述推荐的技术栈过程中遇到问题，请提交给我们来处理。祝你编码愉快！
+你可以使用 HTML、JavaScript 以及 CSS 来完成项目。 由于目前你只学到了 CSS 课程，所以我们建议你只使用 CSS 来完成这个项目，同时巩固一下你之前所学的内容。 你也可以使用 Bootstrap 或者 SASS。 我们不推荐你在这个项目中使用其他技术（比如 jQuery、React、Angular 或 Vue）。 在后续的其他项目中，你将有机会使用像是 React 等其他技术栈。 我们会接受并尝试修复你在使用推荐技术栈创建项目时报告的问题。 祝你编码愉快！
-**需求 1：**此 app 中应存在一个 `id` 为 `welcome-section` 的欢迎区。
+**需求 1：**此 app 中应存在一个 id 为 `welcome-section` 的欢迎区。
**需求 2：**欢迎区内应存在一个包含标题文本的 `h1` 元素。
-**需求 3：**此 app 中应存在一个 `id` 为 `projects` 的项目展示区。
+**需求 3：**此 app 中应存在一个 id 为 `projects` 的项目展示区。
-**需求 4：**项目展示区应至少包含一个 `class` 为 `project-tile` 的元素来展示项目。
+**需求 4：**项目展示区应至少包含一个 class 为 `project-tile` 的元素来展示项目。
**需求 5：**项目展示区应至少包含一个链接到项目的超链接元素。
-**需求 6：**此 app 中应存在一个 `id` 为 `navbar` 的导航栏。
+**需求 6：**此 app 中应存在一个 id 为 `navbar` 的导航栏。
**需求 7：**导航栏中应包含一个可以滚动到本页面不同区域的链接。
-**需求 8：**此 app 中应包含一个 `id` 为 `profile-link` 的链接。点击这个链接时，它应在浏览器的新标签页内打开我的 GitHub 或者 FCC 作品集页面。
+**需求 8：**此 app 中应包含一个 id 为 `profile-link` 的链接。 点击这个链接时，它应在浏览器的新标签页内打开我的 GitHub 或者 FCC 作品集页面。
**需求 9：**在此 app 中，应至少使用一次媒体查询。
@@ -36,7 +36,7 @@ dashedName: build-a-personal-portfolio-webpage
**需求 11：**导航栏应始终保持在视口顶部。
-你可以通过 fork [这个项目](http://codepen.io/freeCodeCamp/pen/MJjpwO)来构建你的项目，也可以使用此 CDN 链接在任何你喜欢的环境中运行测试：`https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`。
+你可以通过 fork [这个 CodePen 项目](https://codepen.io/freeCodeCamp/pen/MJjpwO) 来构建你的项目， 也可以使用此 CDN 链接在任何你喜欢的环境中运行测试：`https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`。
完成项目并通过所有测试后，请输入你的项目在 CodePen 上的链接并提交。
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-product-landing-page.md b/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-product-landing-page.md
index 96d323e6e9a..7c843140f79 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-product-landing-page.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-product-landing-page.md
@@ -10,31 +10,31 @@ dashedName: build-a-product-landing-page
**目标：**在 [CodePen.io](https://codepen.io) 上创建一个与这个功能类似的 app：<https://codepen.io/freeCodeCamp/full/RKRbwL>。
-在满足以下[需求](https://en.wikipedia.org/wiki/User_story)并能通过所有测试的前提下，你可以根据自己的喜好来美化你的 app。
+在满足以下[需求](https://en.wikipedia.org/wiki/User_story)并能通过所有测试的前提下， 你可以根据自己的喜好来美化你的 app。
-你可以使用 HTML、JavaScript 以及 CSS 来完成项目。由于目前你只学到了 CSS 课程，所以我们建议你只使用 CSS 来完成这个项目，同时巩固一下你之前所学的内容。你也可以使用 Bootstrap 或者 SASS。我们不推荐你在这个项目中使用其他技术（比如 jQuery、React、Angular 或 Vue）。否则，如果在编码中出现问题，你需要自行解决。在后续的其他项目中，你将有机会使用像是 React 等其他技术栈。如果你在使用上述推荐的技术栈过程中遇到问题，请提交给我们来处理。祝你编码愉快！
+你可以使用 HTML、JavaScript 以及 CSS 来完成项目。 由于目前你只学到了 CSS 课程，所以我们建议你只使用 CSS 来完成这个项目，同时巩固一下你之前所学的内容。 你也可以使用 Bootstrap 或者 SASS。 我们不推荐你在这个项目中使用其他技术（比如 jQuery、React、Angular 或 Vue）。 在后续的其他项目中，你将有机会使用像是 React 等其他技术栈。 我们会接受并尝试修复你在使用推荐技术栈创建项目时报告的问题。 祝你编码愉快！
-**需求 1：**产品登陆页应存在 `id` 为 `"header"` 的 `header` 元素。
+**需求 1：**产品登陆页应存在 `id="header"` 的 `header` 元素。
-**需求 2：**在 `header` 元素内应存在 `id` 为 `"header-img"` 的图像，这里通常用来放置公司的 logo。
+**需求 2：**在 `header` 元素内应存在 `id="header-img"` 的图像， 这里通常用来放置公司的 logo。
-**需求 3：**在 `#header` 元素内，应存在一个 `id` 为 `"nav-bar"` 的 `nav` 元素。
+**需求 3：**在 `#header` 元素内，应存在一个 `id="nav-bar"` 的 `nav` 元素。
-**需求 4：**`nav` 元素中应至少包含三个 `class` 为 `nav-link` 且可点击的元素。
+**需求 4：**`nav` 元素中应至少包含三个 class 为 `nav-link` 且可点击的元素。
**需求 5：**当点击 `nav` 内的 `.nav-link` 按钮时，应滚动到产品主页上相应的部分。
-**需求 6：**页面上应存在 `id` 为 `"video"` 的嵌入式视频播放区域。
+**需求 6：**页面上应存在 `id="video"` 的嵌入式视频播放区域。
-**需求 7：**产品登陆页应存在一个 `id` 为 `"form"` 的 `form` 元素。
+**需求 7：**产品登陆页应存在一个 `id="form"` 的 `form` 元素。
-**需求 8：**在表单元素 `form` 中，应存在一个 `id` 为 `"email"` 的 `input` 输入框供用户填写邮箱。
+**需求 8：**在表单中，应存在一个 `id="email"` 的 `input` 输入框供用户填写邮箱。
-**需求 9：**`#email` 输入框内应存在描述该区域用途的占位符文字（`placeholder`）。
+**需求 9：**`#email` 输入框内应存在描述该区域用途的占位符文字。
**需求 10：**`#email` 输入框应使用 HTML5 验证来确认输入的内容是否为邮箱。
-**需求 11：**在表单中，应存在一个 `id` 为 `"submit"` 的 `input` 提交按钮。
+**需求 11：**在表单中，应存在一个 `id="submit"` 的 `input` 提交按钮。
**需求 12：**当点击 `#submit` 元素时，应将邮箱信息提交到一个静态页面（请使用这个模拟的 URL：<https://www.freecodecamp.com/email-submit>）。
@@ -44,7 +44,7 @@ dashedName: build-a-product-landing-page
**需求 15：**在此 app 中，应至少使用一次 CSS 的 flexbox 布局。
-你可以通过 fork [这个项目](http://codepen.io/freeCodeCamp/pen/MJjpwO)来构建你的项目，也可以使用此 CDN 链接在任何你喜欢的环境中运行测试：`https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`。
+你可以通过 fork [这个 CodePen 项目](https://codepen.io/freeCodeCamp/pen/MJjpwO) 来构建你的项目， 也可以使用此 CDN 链接在任何你喜欢的环境中运行测试：`https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`。
完成项目并通过所有测试后，请输入你的项目在 CodePen 上的链接并提交。
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-survey-form.md b/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-survey-form.md
index 87e5d43dab6..b6be04cc3df 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-survey-form.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-survey-form.md
@@ -10,43 +10,43 @@ dashedName: build-a-survey-form
**目标：**在 [CodePen.io](https://codepen.io) 上创建一个与这个功能类似的 app：<https://codepen.io/freeCodeCamp/full/VPaoNP>。
-在满足以下[需求](https://en.wikipedia.org/wiki/User_story)并能通过所有测试的前提下，你可以根据自己的喜好来美化你的 app。
+在满足以下 [需求](https://en.wikipedia.org/wiki/User_story) 并能通过所有测试的前提下， 你可以根据自己的喜好来美化你的 app。
-你可以使用 HTML、JavaScript 以及 CSS 来完成项目。由于目前你只学到了 CSS 课程，所以我们建议你只使用 CSS 来完成这个项目，同时巩固一下你之前所学的内容。你也可以使用 Bootstrap 或者 SASS。我们不推荐你在这个项目中使用其他技术（比如 jQuery、React、Angular 或 Vue）。否则，如果在编码中出现问题，你需要自行解决。在后续的其他项目中，你将有机会使用像是 React 等其他技术栈。如果你在使用上述推荐的技术栈过程中遇到问题，请提交给我们来处理。祝你编码愉快！
+你可以使用 HTML、JavaScript 以及 CSS 来完成项目。 由于目前你只学到了 CSS 课程，所以我们建议你只使用 CSS 来完成这个项目，同时巩固一下你之前所学的内容。 你也可以使用 Bootstrap 或者 SASS。 我们不推荐你在这个项目中使用其他技术（比如 jQuery、React、Angular 或 Vue）。 在后续的其他项目中，你将有机会使用像是 React 等其他技术栈。 我们会接受并尝试修复你在使用推荐技术栈创建项目时报告的问题。 祝你编码愉快！
-**需求 1：**此 app 中应存在一个 `id` 为 `"title"` 的 `h1` 元素作为标题。
+**需求 1：**此 app 中应存在一个 `id="title"` 的大小为 H1 的标题。
-**需求 2：**此 app 中应存在一个 `id` 为 `"description"` 的 `p` 元素，其中应包含一段简短的介绍。
+**需求 2：**此 app 中应存在一段 `id="description"` 的大小为 P 的短小的描述。
-**需求 3：**此 app 中应存在一个 `id` 为 `"survey-form"` 的 `form` 元素。
+**需求 3：**此 app 中应存在一个 `id="survey-form"` 的 `form` 元素。
-**需求 4：**在表单元素内，应存在 `id` 为 `"name"` 的输入框（必填项），以便用户输入姓名。
+**需求 4：**在表单元素内，应存在 `id="name"` 的输入框（必填项），以便用户输入姓名。
-**需求 5：**在表单元素内，应存在 `id` 为 `"email"` 的输入框（必填项），以便用户输入邮箱。
+**需求 5：**在表单元素内，应存在 `id="email"` 的输入框（必填项），以便用户输入邮箱。
**需求 6：**如果用户输入了格式不正确的邮箱，则应出现来自 HTML5 表单数据校验的错误信息。
-**需求 7：**在表单内，用户应可以在 `id` 为 `"number"` 的输入框中输入数字。
+**需求 7：**在表单内，用户应可以在 `id="number"` 的输入框中输入数字。
**需求 8：**如果用户在数字输入框内输入非数字内容，则应出现来自 HTML5 表单数据校验的错误信息。
**需求 9：**如果用户输入的数字超出了使用 `min` 和 `max` 属性定义的范围，则应出现来自 HTML5 表单数据校验的错误信息。
-**需求 10：**表单中的姓名、邮箱和数字输入框需有对应的包含描述输入框用途的 `label`。这些 `label` 的 `id` 应分别为 `"name-label"`、`"email-label"`、`"number-label"`。
+**需求 10：**表单中的姓名、邮箱和数字输入框需有对应的包含描述输入框用途的标签。这些标签的 id 应分别为 `id="name-label"`、`id="email-label"` 和 `id="number-label"`。
-**需求 11：**表单中的姓名、邮箱和数字输入框需有对应的描述文字作为占位符（`placeholder`）。
+**需求 11：**表单中的姓名、邮箱和数字输入框需有对应的描述文字作为占位符。
-**需求 12：**在表单元素内，应存在一个 `id` 为 `"dropdown"` 的下拉列表，用户可以从中选取一个选项。
+**需求 12：**在表单元素内，应存在一个 `id="dropdown"` 的下拉列表，用户可以从中选取一个选项。
-**需求 13：**在表单元素内，应至少存在一组单选按钮，且应使用 `name` 属性进行分组。用户可以从中选取一个选项。
+**需求 13：**在表单元素内，应至少存在一组单选按钮，用户可以从中选取一个选项。 每组应使用 `name` 属性进行分组。
-**需求 14：**在表单元素内，应存在几个复选框，且每个复选框都应有 `value` 属性。用户可以选中这些复选框。
+**需求 14：**在表单元素内，应存在几个复选框，且每个复选框都应有 `value` 属性。
**需求 15：**在表单元素的最后，应存在一个 `textarea` 元素，以便用户输入额外的批注。
-**需求 16：**在表单元素内，应存在一个 `id` 为 `"submit"` 的按钮，以便用户提交表单。
+**需求 16：**在表单元素内，应存在一个 `id="submit"` 的按钮，以便用户提交表单。
-你可以通过 fork [这个项目](http://codepen.io/freeCodeCamp/pen/MJjpwO)来构建你的项目，也可以使用此 CDN 链接在任何你喜欢的环境中运行测试：`https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`。
+你可以 fork [这个 CodePen 项目](https://codepen.io/freeCodeCamp/pen/MJjpwO)来创建你的项目。 也可以使用此 CDN 链接在任何你喜欢的环境中运行测试：`https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`。
完成项目并通过所有测试后，请输入你的项目在 CodePen 上的链接并提交。
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-technical-documentation-page.md b/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-technical-documentation-page.md
index 2a45cce102f..eb347199c8e 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-technical-documentation-page.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-technical-documentation-page.md
@@ -10,17 +10,17 @@ dashedName: build-a-technical-documentation-page
**目标：**在 [CodePen.io](https://codepen.io) 上创建一个与这个功能类似的 app：<https://codepen.io/freeCodeCamp/full/NdrKKL>。
-在满足以下[需求](https://en.wikipedia.org/wiki/User_story)并能通过所有测试的前提下，你可以根据自己的喜好来美化你的 app。
+在满足以下 [需求](https://en.wikipedia.org/wiki/User_story) 并能通过所有测试的前提下， 你可以根据自己的喜好来美化你的 app。
-你可以使用 HTML、JavaScript 以及 CSS 来完成项目。由于目前你只学到了 CSS 课程，所以我们建议你只使用 CSS 来完成这个项目，同时巩固一下你之前所学的内容。你也可以使用 Bootstrap 或者 SASS。我们不推荐你在这个项目中使用其他技术（比如 jQuery、React、Angular 或 Vue）。否则，如果在编码中出现问题，你需要自行解决。在后续的其他项目中，你将有机会使用像是 React 等其他技术栈。如果你在使用上述推荐的技术栈过程中遇到问题，请提交给我们来处理。祝你编码愉快！
+你可以使用 HTML、JavaScript 以及 CSS 来完成项目。 由于目前你只学到了 CSS 课程，所以我们建议你只使用 CSS 来完成这个项目，同时巩固一下你之前所学的内容。 你也可以使用 Bootstrap 或者 SASS。 我们不推荐你在这个项目中使用其他技术（比如 jQuery、React、Angular 或 Vue）。 在后续的其他项目中，你将有机会使用像是 React 等其他技术栈。 我们会接受并尝试修复你在使用推荐技术栈创建项目时报告的问题。 祝你编码愉快！
-**需求 1：**此 app 中应存在一个 `id` 为 `"main-doc"` 的 `main` 元素，它包含页面的主要内容（技术文档）。
+**需求 1：**此 app 中应存在一个 `id="main-doc"` 的 `main` 元素，它包含页面的主要内容（技术文档）。
-**需求 2：**在 `#main-doc` 元素内，应存在至少 5 个 `section` 元素，每个元素的 class 都应为 `main-section`。
+**需求 2：**在 `#main-doc` 元素内，应有一些 `section` 元素，每个元素的 class 都应为 `main-section`。 应存在至少 5 个这样的元素。
**需求 3：**每个 `.main-section` 内的第一个元素应为 `header` 元素，其中包含描述该部分主题的内容文本。
-**需求 4：**类名为 `main-section` 的每个 `section` 元素都应有一个与包含在其中的每个`header`的文本相对应的 id，用下划线替换文本中的所有空格（例如，包含标题 "Javascript and Java" 的`section`应该有对应`id="Javascript_and_Java"`）。
+**需求 4：**类名为 `main-section` 的每个 `section` 元素都应有一个与包含在其中的每个 `header` 的文本相对应的 id， 所有空格都应该被替换为下划线（例如，包含标题 “JavaScript and Java” 的 `section` 应有一个相应的 `id="JavaScript_and_Java"`）。
**需求 5：**所有 `.main-section` 元素内总计应有至少 10 个 `p` 元素。
@@ -32,19 +32,19 @@ dashedName: build-a-technical-documentation-page
**需求 9：**navbar 元素内应有一个 `header` 元素，其中包含描述技术文档主题的内容文本。
-**需求 10：**此外，navbar 元素应包含 `class` 为 `nav-link` 的 `a` 元素，每个 class 为 `main-section` 的元素都需要有对应的 `a` 元素。
+**需求 10：**此外，navbar 元素应包含 class 为 `nav-link` 的 `a` 元素， 每个 class 为 `main-section` 的元素都需要有一个。
**需求 11：**navbar 中的 `header` 元素应置于 navbar 中所有 `a` 元素之前。
-**需求 12：**所有 class 为 `nav-link` 的 `a` 元素都需要包含与其 `header` 相应的内容文本。例如，对于一个文本为 "Hello world" 的 `section` 或 `header`，你的navbar 中也应存在一个内容文本为 "Hello world" 的 `a` 元素。
+**需求 12：**所有 class 为 `nav-link` 的元素都需要包含与 `section` 中 `header` 相应的内容文本。例如，对于一个文本为 "Hello world" 的一节或标题，你的 navbar 中也应存在一个内容文本为 "Hello world" 的元素。
-**需求 13：**当点击 navbar 中一个 `a` 元素时，页面应滚动到 `main-doc` 中的相应部分。例如，点击文本为 "Hello world" 的 `nav-link` 元素的时候，页面应滚动到包含相同内容的 header 和 id 所处的 `section` 元素。
+**需求 13：**当点击 navbar 中一个元素时，页面应滚动到 `main-doc` 中的相应部分。例如，点击文本为 "Hello world" 的 `nav-link` 元素的时候，页面应滚动到包含相同内容的 `header` 和 id 所处的 `section` 元素。
-**需求 14：**在常规尺寸的设备上（如笔记本电脑和台式机），`id` 为 `"navbar"` 的元素应显示在屏幕左侧，且始终对用户可见。
+**需求 14：**在常规尺寸的设备上（如笔记本电脑和台式机），`id="navbar"` 的元素应显示在屏幕左侧，且始终对用户可见。
**需求 15：**在此 app 中，应至少使用一次媒体查询。
-你可以通过 fork [这个项目](http://codepen.io/freeCodeCamp/pen/MJjpwO)来构建你的项目，也可以使用此 CDN 链接在任何你喜欢的环境中运行测试：`https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`。
+你可以 fork [这个 CodePen 项目](https://codepen.io/freeCodeCamp/pen/MJjpwO) 来构建你的项目。 也可以使用此 CDN 链接在任何你喜欢的环境中运行测试：`https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`。
完成项目并通过所有测试后，请输入你的项目在 CodePen 上的链接并提交。
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-tribute-page.md b/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-tribute-page.md
index 6240a323374..631fdbd6ac0 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-tribute-page.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/responsive-web-design-projects/build-a-tribute-page.md
@@ -10,29 +10,29 @@ dashedName: build-a-tribute-page
**目标：**在 [CodePen.io](https://codepen.io) 上创建一个与这个功能类似的 app：<https://codepen.io/freeCodeCamp/full/zNqgVx>。
-在满足以下[需求](https://en.wikipedia.org/wiki/User_story)并能通过所有测试的前提下，你可以根据自己的喜好来美化你的 app。
+在满足以下 [需求](https://en.wikipedia.org/wiki/User_story) 并能通过所有测试的前提下， 你可以根据自己的喜好来美化你的 app。
-你可以使用 HTML、JavaScript 以及 CSS 来完成项目。由于目前你只学到了 CSS 课程，所以我们建议你只使用 CSS 来完成这个项目，同时巩固一下你之前所学的内容。你也可以使用 Bootstrap 或者 SASS。我们不推荐你在这个项目中使用其他技术（比如 jQuery、React、Angular 或 Vue）。否则，如果在编码中出现问题，你需要自行解决。在后续的其他项目中，你将有机会使用像是 React 等其他技术栈。如果你在使用上述推荐的技术栈过程中遇到问题，请提交给我们来处理。祝你编码愉快！
+你可以使用 HTML、JavaScript 以及 CSS 来完成项目。 由于目前你只学到了 CSS 课程，所以我们建议你只使用 CSS 来完成这个项目，同时巩固一下你之前所学的内容。 你也可以使用 Bootstrap 或者 SASS。 我们不推荐你在这个项目中使用其他技术（比如 jQuery、React、Angular 或 Vue）。 在后续的其他项目中，你将有机会使用像是 React 等其他技术栈。 我们会接受并尝试修复你在使用推荐技术栈创建项目时报告的问题。 祝你编码愉快！
-**需求 1：**此 app 中应存在一个 `id` 为 `"main"` 的元素，页面上的所有元素都应置于这个元素中。
+**需求 1：**此 app 中应存在一个 `id="main"` 的元素，页面上的所有元素都应置于这个元素中。
-**需求 2：**此 app 中应存在一个 `id` 为 `"title"` 的元素，其中包含描述致敬页主题的字符串文本，如 "Dr. Norman Borlaug"。
+**需求 2：**此 app 中应存在一个 `id="title"` 的元素，其中包含描述致敬页主题的字符串文本，如 "Dr. Norman Borlaug"。
-**需求 3：**此 app 中应存在一个 `id` 为 `"img-div"` 的 `div` 元素。
+**需求 3：**此 app 中应存在一个 `id="img-div"` 的 `div` 元素。
-**需求 4：**在 `img-div` 元素内，应存在一个 `id` 为 `"image"` 的 `img` 元素。
+**需求 4：**在 `img-div` 元素内，应存在一个 `id="image"` 的 `img` 元素。
-**需求 5：**在 `img-div` 元素内，应存在一个相应的 `id` 为 `"img-caption"` 的元素，其中包含对 `img-div` 中图像的描述文本。
+**需求 5：**在 `img-div` 元素内，应存在一个相应的 `id="img-caption"` 的元素，其中包含对 `img-div` 中图像的描述文本。
-**需求 6：**此 app 中应存在一个 `id` 为 `"tribute-info"` 的元素，其中应包含描述致敬页主题的内容文本。
+**需求 6：**此 app 中应存在一个 `id="tribute-info"` 的元素，其中应包含描述致敬页主题的内容文本。
-**需求 7：**此 app 中应存在一个 `id` 为 `"tribute-link"` 的 `a` 元素，它应链接到一个包含有关致敬页主题额外信息的外部网页。提示：你必须为 `a` 元素提供 `target` 属性，并将其属性值设置为 `_blank`（即 `target="_blank"`），这样才可以在新选项卡中打开链接。
+**需求 7：**此 app 中应存在一个 `id="tribute-link"` 的 `a` 元素，它应链接到一个包含有关致敬页主题额外信息的外部网页。 提示：你必须为 a 元素提供 `target` 属性，并将其属性值设置为 `_blank`（即 `target="_blank"`），这样才可以在新选项卡中打开链接。
**需求 8：**`img` 元素应相对于其父元素的宽度自动调整大小，但不超过图片的原始大小。
**需求 9：**`img` 应在其父元素内居中。
-你可以通过 fork [这个项目](http://codepen.io/freeCodeCamp/pen/MJjpwO)来构建你的项目，也可以使用此 CDN 链接在任何你喜欢的环境中运行测试：`https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`。
+你可以 fork [这个 CodePen 项目](https://codepen.io/freeCodeCamp/pen/MJjpwO) 来构建你的项目。 也可以使用此 CDN 链接在任何你喜欢的环境中运行测试：`https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`。
完成项目并通过所有测试后，请输入你的项目在 CodePen 上的链接并提交。
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/boo-who.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/boo-who.md
index 1c6a09d83d2..4a3296deb59 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/boo-who.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/boo-who.md
@@ -1,6 +1,6 @@
---
id: a77dbc43c33f39daa4429b4f
-title: 基本类型布尔值的检查
+title: Boo who
challengeType: 5
forumTopicId: 16000
dashedName: boo-who
@@ -8,67 +8,67 @@ dashedName: boo-who
# --description--
-检查一个值是否是[基本类型](https://developer.mozilla.org/zh-CN/docs/Glossary/Primitive)中的布尔值（boolean）类型。函数应返回 true 或者 false。
+Check if a value is classified as a boolean primitive. Return true or false.
-基本类型中的布尔值为 true 或者 false。
+Boolean primitives are true and false.
# --hints--
-`booWho(true)` 应返回 true。
+`booWho(true)` should return true.
```js
assert.strictEqual(booWho(true), true);
```
-`booWho(false)` 应返回 true。
+`booWho(false)` should return true.
```js
assert.strictEqual(booWho(false), true);
```
-`booWho([1, 2, 3])` 应返回 false。
+`booWho([1, 2, 3])` should return false.
```js
assert.strictEqual(booWho([1, 2, 3]), false);
```
-`booWho([].slice)` 应返回 false。
+`booWho([].slice)` should return false.
```js
assert.strictEqual(booWho([].slice), false);
```
-`booWho({ "a": 1 })` 应返回 false。
+`booWho({ "a": 1 })` should return false.
```js
assert.strictEqual(booWho({ a: 1 }), false);
```
-`booWho(1)` 应返回 false。
+`booWho(1)` should return false.
```js
assert.strictEqual(booWho(1), false);
```
-`booWho(NaN)` 应返回 false。
+`booWho(NaN)` should return false.
```js
assert.strictEqual(booWho(NaN), false);
```
-`booWho("a")` 应返回 false。
+`booWho("a")` should return false.
```js
assert.strictEqual(booWho('a'), false);
```
-`booWho("true")` 应返回 false。
+`booWho("true")` should return false.
```js
assert.strictEqual(booWho('true'), false);
```
-`booWho("false")` 应返回 false。
+`booWho("false")` should return false.
```js
assert.strictEqual(booWho('false'), false);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/chunky-monkey.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/chunky-monkey.md
index 3abf1a66289..9abbde07ab9 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/chunky-monkey.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/chunky-monkey.md
@@ -1,6 +1,6 @@
---
id: a9bd25c716030ec90084d8a1
-title: 分割数组
+title: Chunky Monkey
challengeType: 5
forumTopicId: 16005
dashedName: chunky-monkey
@@ -8,11 +8,11 @@ dashedName: chunky-monkey
# --description--
-请编写一个函数，该函数将一个数组（第一个参数）拆分成若干长度为 `size`（第二个参数）的子数组，并将它们作为二维数组返回。
+Write a function that splits an array (first argument) into groups the length of `size` (second argument) and returns them as a two-dimensional array.
# --hints--
-`chunkArrayInGroups(["a", "b", "c", "d"], 2)` 应返回 `[["a", "b"], ["c", "d"]]`。
+`chunkArrayInGroups(["a", "b", "c", "d"], 2)` should return `[["a", "b"], ["c", "d"]]`.
```js
assert.deepEqual(chunkArrayInGroups(['a', 'b', 'c', 'd'], 2), [
@@ -21,7 +21,7 @@ assert.deepEqual(chunkArrayInGroups(['a', 'b', 'c', 'd'], 2), [
]);
```
-`chunkArrayInGroups([0, 1, 2, 3, 4, 5], 3)` 应返回 `[[0, 1, 2], [3, 4, 5]]`。
+`chunkArrayInGroups([0, 1, 2, 3, 4, 5], 3)` should return `[[0, 1, 2], [3, 4, 5]]`.
```js
assert.deepEqual(chunkArrayInGroups([0, 1, 2, 3, 4, 5], 3), [
@@ -30,7 +30,7 @@ assert.deepEqual(chunkArrayInGroups([0, 1, 2, 3, 4, 5], 3), [
]);
```
-`chunkArrayInGroups([0, 1, 2, 3, 4, 5], 2)` 应返回 `[[0, 1], [2, 3], [4, 5]]`。
+`chunkArrayInGroups([0, 1, 2, 3, 4, 5], 2)` should return `[[0, 1], [2, 3], [4, 5]]`.
```js
assert.deepEqual(chunkArrayInGroups([0, 1, 2, 3, 4, 5], 2), [
@@ -40,7 +40,7 @@ assert.deepEqual(chunkArrayInGroups([0, 1, 2, 3, 4, 5], 2), [
]);
```
-`chunkArrayInGroups([0, 1, 2, 3, 4, 5], 4)` 应返回 `[[0, 1, 2, 3], [4, 5]]`。
+`chunkArrayInGroups([0, 1, 2, 3, 4, 5], 4)` should return `[[0, 1, 2, 3], [4, 5]]`.
```js
assert.deepEqual(chunkArrayInGroups([0, 1, 2, 3, 4, 5], 4), [
@@ -49,7 +49,7 @@ assert.deepEqual(chunkArrayInGroups([0, 1, 2, 3, 4, 5], 4), [
]);
```
-`chunkArrayInGroups([0, 1, 2, 3, 4, 5, 6], 3)` 应返回 `[[0, 1, 2], [3, 4, 5], [6]]`。
+`chunkArrayInGroups([0, 1, 2, 3, 4, 5, 6], 3)` should return `[[0, 1, 2], [3, 4, 5], [6]]`.
```js
assert.deepEqual(chunkArrayInGroups([0, 1, 2, 3, 4, 5, 6], 3), [
@@ -59,7 +59,7 @@ assert.deepEqual(chunkArrayInGroups([0, 1, 2, 3, 4, 5, 6], 3), [
]);
```
-`chunkArrayInGroups([0, 1, 2, 3, 4, 5, 6, 7, 8], 4)` 应返回 `[[0, 1, 2, 3], [4, 5, 6, 7], [8]]`。
+`chunkArrayInGroups([0, 1, 2, 3, 4, 5, 6, 7, 8], 4)` should return `[[0, 1, 2, 3], [4, 5, 6, 7], [8]]`.
```js
assert.deepEqual(chunkArrayInGroups([0, 1, 2, 3, 4, 5, 6, 7, 8], 4), [
@@ -69,7 +69,7 @@ assert.deepEqual(chunkArrayInGroups([0, 1, 2, 3, 4, 5, 6, 7, 8], 4), [
]);
```
-`chunkArrayInGroups([0, 1, 2, 3, 4, 5, 6, 7, 8], 2)` 应返回 `[[0, 1], [2, 3], [4, 5], [6, 7], [8]]`。
+`chunkArrayInGroups([0, 1, 2, 3, 4, 5, 6, 7, 8], 2)` should return `[[0, 1], [2, 3], [4, 5], [6, 7], [8]]`.
```js
assert.deepEqual(chunkArrayInGroups([0, 1, 2, 3, 4, 5, 6, 7, 8], 2), [
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/confirm-the-ending.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/confirm-the-ending.md
index 2261da50e70..2ee2ca37128 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/confirm-the-ending.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/confirm-the-ending.md
@@ -1,6 +1,6 @@
---
id: acda2fb1324d9b0fa741e6b5
-title: 检查字符串结尾
+title: Confirm the Ending
challengeType: 5
forumTopicId: 16006
dashedName: confirm-the-ending
@@ -8,31 +8,31 @@ dashedName: confirm-the-ending
# --description--
-检查字符串（第一个参数 `str`）是否以给定的目标字符串（第二个参数 `target`）结束。
+Check if a string (first argument, `str`) ends with the given target string (second argument, `target`).
-这个挑战*可以*用 ES2015 引入的 `.endsWith()` 方法来解决。但在这个挑战中，请使用 JavaScript 的字符串子串方法或正则表达式来通过挑战，而不要使用 `.endsWith()` 方法。
+This challenge *can* be solved with the `.endsWith()` method, which was introduced in ES2015. But for the purpose of this challenge, we would like you to use one of the JavaScript substring methods instead.
# --hints--
-`confirmEnding("Bastian", "n")` 应返回 true。
+`confirmEnding("Bastian", "n")` should return true.
```js
assert(confirmEnding('Bastian', 'n') === true);
```
-`confirmEnding("Congratulation", "on")` 应返回 true。
+`confirmEnding("Congratulation", "on")` should return true.
```js
assert(confirmEnding('Congratulation', 'on') === true);
```
-`confirmEnding("Connor", "n")` 应返回 false。
+`confirmEnding("Connor", "n")` should return false.
```js
assert(confirmEnding('Connor', 'n') === false);
```
-`confirmEnding("Walking on water and developing software from a specification are easy if both are frozen", "specification")` 应返回 false。
+`confirmEnding("Walking on water and developing software from a specification are easy if both are frozen", "specification")` should return false.
```js
assert(
@@ -43,31 +43,31 @@ assert(
);
```
-`confirmEnding("He has to give me a new name", "name")` 应返回 true。
+`confirmEnding("He has to give me a new name", "name")` should return true.
```js
assert(confirmEnding('He has to give me a new name', 'name') === true);
```
-`confirmEnding("Open sesame", "same")` 应返回 true。
+`confirmEnding("Open sesame", "same")` should return true.
```js
assert(confirmEnding('Open sesame', 'same') === true);
```
-`confirmEnding("Open sesame", "sage")` 应返回 false。
+`confirmEnding("Open sesame", "sage")` should return false.
```js
assert(confirmEnding('Open sesame', 'sage') === false);
```
-`confirmEnding("Open sesame", "game")` 应返回 false。
+`confirmEnding("Open sesame", "game")` should return false.
```js
assert(confirmEnding('Open sesame', 'game') === false);
```
-`confirmEnding("If you want to save our world, you must hurry. We dont know how much longer we can withstand the nothing", "mountain")` 应返回 false。
+`confirmEnding("If you want to save our world, you must hurry. We dont know how much longer we can withstand the nothing", "mountain")` should return false.
```js
assert(
@@ -78,13 +78,13 @@ assert(
);
```
-`confirmEnding("Abstraction", "action")` 应该返回 true。
+`confirmEnding("Abstraction", "action")` should return true.
```js
assert(confirmEnding('Abstraction', 'action') === true);
```
-不应使用内置方法 `.endsWith()` 来完成挑战。
+Your code should not use the built-in method `.endsWith()` to solve the challenge.
```js
assert(!/\.endsWith\(.*?\)\s*?;?/.test(code) && !/\['endsWith'\]/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/convert-celsius-to-fahrenheit.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/convert-celsius-to-fahrenheit.md
index c9ec18284eb..616f44a9879 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/convert-celsius-to-fahrenheit.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/convert-celsius-to-fahrenheit.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244b3
-title: 将摄氏度转换为华氏度
+title: Convert Celsius to Fahrenheit
challengeType: 1
forumTopicId: 16806
dashedName: convert-celsius-to-fahrenheit
@@ -8,43 +8,43 @@ dashedName: convert-celsius-to-fahrenheit
# --description--
-将摄氏度转换为华氏度的计算方式为：摄氏度乘以 `9/5` 然后加上 `32`。
+The algorithm to convert from Celsius to Fahrenheit is the temperature in Celsius times `9/5`, plus `32`.
-输入参数 `celsius` 代表一个摄氏度的温度。请根据上述转换公式，将已定义好的 `fahrenheit` 变量赋值为相应的华氏度的温度值。
+You are given a variable `celsius` representing a temperature in Celsius. Use the variable `fahrenheit` already defined and assign it the Fahrenheit temperature equivalent to the given Celsius temperature. Use the algorithm mentioned above to help convert the Celsius temperature to Fahrenheit.
# --hints--
-`convertToF(0)` 应返回一个数字。
+`convertToF(0)` should return a number
```js
assert(typeof convertToF(0) === 'number');
```
-`convertToF(-30)` 应返回 `-22`。
+`convertToF(-30)` should return a value of `-22`
```js
assert(convertToF(-30) === -22);
```
-`convertToF(-10)` 应返回 `14`。
+`convertToF(-10)` should return a value of `14`
```js
assert(convertToF(-10) === 14);
```
-`convertToF(0)` 应返回 `32`。
+`convertToF(0)` should return a value of `32`
```js
assert(convertToF(0) === 32);
```
-`convertToF(20)` 应返回 `68`。
+`convertToF(20)` should return a value of `68`
```js
assert(convertToF(20) === 68);
```
-`convertToF(30)` 应返回 `86`。
+`convertToF(30)` should return a value of `86`
```js
assert(convertToF(30) === 86);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/factorialize-a-number.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/factorialize-a-number.md
index a336cdf0028..06d29b3b133 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/factorialize-a-number.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/factorialize-a-number.md
@@ -1,6 +1,6 @@
---
id: a302f7aae1aa3152a5b413bc
-title: 计算整数的阶乘
+title: Factorialize a Number
challengeType: 5
forumTopicId: 16013
dashedName: factorialize-a-number
@@ -8,43 +8,43 @@ dashedName: factorialize-a-number
# --description--
-返回一个给定整数的阶乘计算结果。
+Return the factorial of the provided integer.
-对于整数 n，n 的阶乘就是所有小于等于 n 的正整数的乘积。
+If the integer is represented with the letter n, a factorial is the product of all positive integers less than or equal to n.
-`n` 的阶乘通常用符号 `n!` 来表示。
+Factorials are often represented with the shorthand notation `n!`
-例如：`5! = 1 * 2 * 3 * 4 * 5 = 120`
+For example: `5! = 1 * 2 * 3 * 4 * 5 = 120`
-在这个挑战中，只有非负整数会作为参数传入函数。
+Only integers greater than or equal to zero will be supplied to the function.
# --hints--
-`factorialize(5)` 应返回一个数字。
+`factorialize(5)` should return a number.
```js
assert(typeof factorialize(5) === 'number');
```
-`factorialize(5)` 应返回 120。
+`factorialize(5)` should return 120.
```js
assert(factorialize(5) === 120);
```
-`factorialize(10)` 应返回 3628800。
+`factorialize(10)` should return 3628800.
```js
assert(factorialize(10) === 3628800);
```
-`factorialize(20)` 应返回 2432902008176640000。
+`factorialize(20)` should return 2432902008176640000.
```js
assert(factorialize(20) === 2432902008176640000);
```
-`factorialize(0)` 应返回 1。
+`factorialize(0)` should return 1.
```js
assert(factorialize(0) === 1);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/falsy-bouncer.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/falsy-bouncer.md
index dbeb044ee3f..32ef7c39011 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/falsy-bouncer.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/falsy-bouncer.md
@@ -1,6 +1,6 @@
---
id: adf08ec01beb4f99fc7a68f2
-title: 过滤数组中的假值
+title: Falsy Bouncer
challengeType: 5
forumTopicId: 16014
dashedName: falsy-bouncer
@@ -8,33 +8,33 @@ dashedName: falsy-bouncer
# --description--
-从数组中移除所有假值（falsy values）。
+Remove all falsy values from an array.
-JavaScript 中的假值有 `false`、`null`、`0`、`""`、`undefined`、`NaN`。
+Falsy values in JavaScript are `false`, `null`, `0`, `""`, `undefined`, and `NaN`.
-提示：可以考虑将每个值都转换为布尔值（boolean）。
+Hint: Try converting each value to a Boolean.
# --hints--
-`bouncer([7, "ate", "", false, 9])` 应返回 `[7, "ate", 9]`。
+`bouncer([7, "ate", "", false, 9])` should return `[7, "ate", 9]`.
```js
assert.deepEqual(bouncer([7, 'ate', '', false, 9]), [7, 'ate', 9]);
```
-`bouncer(["a", "b", "c"])` 应返回 `["a", "b", "c"]`。
+`bouncer(["a", "b", "c"])` should return `["a", "b", "c"]`.
```js
assert.deepEqual(bouncer(['a', 'b', 'c']), ['a', 'b', 'c']);
```
-`bouncer([false, null, 0, NaN, undefined, ""])` 应返回 `[]`。
+`bouncer([false, null, 0, NaN, undefined, ""])` should return `[]`.
```js
assert.deepEqual(bouncer([false, null, 0, NaN, undefined, '']), []);
```
-`bouncer([1, null, NaN, 2, undefined])`应返回 `[1, 2]`。
+`bouncer([null, NaN, 1, 2, undefined])` should return `[1, 2]`.
```js
assert.deepEqual(bouncer([null, NaN, 1, 2, undefined]), [1, 2]);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/find-the-longest-word-in-a-string.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/find-the-longest-word-in-a-string.md
index e545f5c9950..1e587b787c3 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/find-the-longest-word-in-a-string.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/find-the-longest-word-in-a-string.md
@@ -1,6 +1,6 @@
---
id: a26cbbe9ad8655a977e1ceb5
-title: 找出字符串中的最长单词
+title: Find the Longest Word in a String
challengeType: 5
forumTopicId: 16015
dashedName: find-the-longest-word-in-a-string
@@ -8,13 +8,13 @@ dashedName: find-the-longest-word-in-a-string
# --description--
-返回给出的句子中，最长单词的长度。
+Return the length of the longest word in the provided sentence.
-函数的返回值应是一个数字。
+Your response should be a number.
# --hints--
-`findLongestWordLength("The quick brown fox jumped over the lazy dog")` 应返回一个数字。
+`findLongestWordLength("The quick brown fox jumped over the lazy dog")` should return a number.
```js
assert(
@@ -24,7 +24,7 @@ assert(
);
```
-`findLongestWordLength("The quick brown fox jumped over the lazy dog")` 应返回 6。
+`findLongestWordLength("The quick brown fox jumped over the lazy dog")` should return 6.
```js
assert(
@@ -32,19 +32,19 @@ assert(
);
```
-`findLongestWordLength("May the force be with you")` 应返回 5。
+`findLongestWordLength("May the force be with you")` should return 5.
```js
assert(findLongestWordLength('May the force be with you') === 5);
```
-`findLongestWordLength("Google do a barrel roll")` 应返回 6。
+`findLongestWordLength("Google do a barrel roll")` should return 6.
```js
assert(findLongestWordLength('Google do a barrel roll') === 6);
```
-`findLongestWordLength("What is the average airspeed velocity of an unladen swallow")` 应返回 8。
+`findLongestWordLength("What is the average airspeed velocity of an unladen swallow")` should return 8.
```js
assert(
@@ -54,7 +54,7 @@ assert(
);
```
-`findLongestWordLength("What if we try a super-long word such as otorhinolaryngology")` 应返回 19。
+`findLongestWordLength("What if we try a super-long word such as otorhinolaryngology")` should return 19.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/finders-keepers.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/finders-keepers.md
index 76e97c355b0..6760acd6603 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/finders-keepers.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/finders-keepers.md
@@ -1,6 +1,6 @@
---
id: a6e40f1041b06c996f7b2406
-title: 按参数过滤数组
+title: Finders Keepers
challengeType: 5
forumTopicId: 16016
dashedName: finders-keepers
@@ -8,11 +8,11 @@ dashedName: finders-keepers
# --description--
-请写一个函数来检查数组（第一个参数 `arr`）中的元素，并返回数组中第一个通过校验测试的元素。其中，“通过校验测试”指的是对于数组中的一个元素 `x`，若 `func(x)` 返回的结果为 `true`，则校验测试通过。如果没有元素通过测试，请返回 `undefined`。
+Create a function that looks through an array `arr` and returns the first element in it that passes a 'truth test'. This means that given an element `x`, the 'truth test' is passed if `func(x)` is `true`. If no element passes the test, return `undefined`.
# --hints--
-`findElement([1, 3, 5, 8, 9, 10], function(num) { return num % 2 === 0; })` 应返回 8。
+`findElement([1, 3, 5, 8, 9, 10], function(num) { return num % 2 === 0; })` should return 8.
```js
assert.strictEqual(
@@ -23,7 +23,7 @@ assert.strictEqual(
);
```
-`findElement([1, 3, 5, 9], function(num) { return num % 2 === 0; })` 应返回 undefined。
+`findElement([1, 3, 5, 9], function(num) { return num % 2 === 0; })` should return undefined.
```js
assert.strictEqual(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/mutations.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/mutations.md
index 33be0a8da78..a3b60172381 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/mutations.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/mutations.md
@@ -1,6 +1,6 @@
---
id: af2170cad53daa0770fabdea
-title: 比较字符串
+title: Mutations
challengeType: 5
forumTopicId: 16025
dashedName: mutations
@@ -8,89 +8,83 @@ dashedName: mutations
# --description--
-输入参数是一个数组，其中包含两个字符串元素。如果数组里的第一个字符串包含了第二个字符串中的所有字母，则返回 true。
+Return true if the string in the first element of the array contains all of the letters of the string in the second element of the array.
-例如，`["hello", "Hello"]` 应该返回 true。因为在忽略大小写的情况下，第一个字符串中包含了第二个字符串里出现的所有字母。
+For example, `["hello", "Hello"]`, should return true because all of the letters in the second string are present in the first, ignoring case.
-而 `["hello", "hey"]` 应该返回 false。因为第一个字符串 "hello" 没有包含字母 "y"。
+The arguments `["hello", "hey"]` should return false because the string "hello" does not contain a "y".
-最后，`["Alien", "line"]` 应该返回 true。因为 "line" 中的所有字母都出现在了 "Alien" 中。
+Lastly, `["Alien", "line"]`, should return true because all of the letters in "line" are present in "Alien".
# --hints--
-`mutation(["hello", "hey"])` 应返回 false。
+`mutation(["hello", "hey"])` should return false.
```js
assert(mutation(['hello', 'hey']) === false);
```
-`mutation(["hello", "Hello"])` 应返回 true。
+`mutation(["hello", "Hello"])` should return true.
```js
assert(mutation(['hello', 'Hello']) === true);
```
-`mutation(["zyxwvutsrqponmlkjihgfedcba", "qrstu"])` 应返回 true。
+`mutation(["zyxwvutsrqponmlkjihgfedcba", "qrstu"])` should return true.
```js
assert(mutation(['zyxwvutsrqponmlkjihgfedcba', 'qrstu']) === true);
```
-`mutation(["Mary", "Army"])` 应返回 true。
+`mutation(["Mary", "Army"])` should return true.
```js
assert(mutation(['Mary', 'Army']) === true);
```
-`mutation(["Mary", "Aarmy"])` 应返回 true。
+`mutation(["Mary", "Aarmy"])` should return true.
```js
assert(mutation(['Mary', 'Aarmy']) === true);
```
-`mutation(["Alien", "line"])` 应返回 true。
+`mutation(["Alien", "line"])` should return true.
```js
assert(mutation(['Alien', 'line']) === true);
```
-`mutation(["floor", "for"])` 应返回 true。
+`mutation(["floor", "for"])` should return true.
```js
assert(mutation(['floor', 'for']) === true);
```
-`mutation(["hello", "neo"])` 应返回 false。
+`mutation(["hello", "neo"])` should return false.
```js
assert(mutation(['hello', 'neo']) === false);
```
-`mutation(["voodoo", "no"])` 应返回 false。
+`mutation(["voodoo", "no"])` should return false.
```js
assert(mutation(['voodoo', 'no']) === false);
```
-`mutation(["voodoo", "no"])` 应返回 false。
-
-```js
-assert(mutation(['voodoo', 'no']) === false);
-```
-
-`mutation(["ate", "date"]` 应返回 false。
+`mutation(["ate", "date"]` should return false.
```js
assert(mutation(['ate', 'date']) === false);
```
-`mutation(["Tiger", "Zebra"])` 应返回 false。
+`mutation(["Tiger", "Zebra"])` should return false.
```js
assert(mutation(['Tiger', 'Zebra']) === false);
```
-`mutation(["Noel", "Ole"])` 应返回 true。
+`mutation(["Noel", "Ole"])` should return true.
```js
assert(mutation(['Noel', 'Ole']) === true);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/repeat-a-string-repeat-a-string.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/repeat-a-string-repeat-a-string.md
index f3fae9643d4..f00c1a25367 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/repeat-a-string-repeat-a-string.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/repeat-a-string-repeat-a-string.md
@@ -1,6 +1,6 @@
---
id: afcc8d540bea9ea2669306b6
-title: 重复输出字符串
+title: Repeat a String Repeat a String
challengeType: 5
forumTopicId: 16041
dashedName: repeat-a-string-repeat-a-string
@@ -8,55 +8,53 @@ dashedName: repeat-a-string-repeat-a-string
# --description--
-为`num` times（第二个参数）重复给定的字符串`str` （第一个参数）。如果`num`不是正数，则返回空字符串。如果卡住，请记得使用[Read-Search-Ask](https://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck-coding/19514) 。编写自己的代码。
-
-将一个给定的字符串 `str`（第一个参数）重复输出 `num`（第二个参数）次。如果 `num` 不是正数，返回空字符串。在这个挑战中，请不要使用 JavaScript 内置的 `.repeat()` 方法。
+Repeat a given string `str` (first argument) for `num` times (second argument). Return an empty string if `num` is not a positive number. For the purpose of this challenge, do *not* use the built-in `.repeat()` method.
# --hints--
-`repeatStringNumTimes("*", 3)` 应返回 `"***"`。
+`repeatStringNumTimes("*", 3)` should return `"***"`.
```js
assert(repeatStringNumTimes('*', 3) === '***');
```
-`repeatStringNumTimes("abc", 3)` 应返回 `"abcabcabc"`。
+`repeatStringNumTimes("abc", 3)` should return `"abcabcabc"`.
```js
assert(repeatStringNumTimes('abc', 3) === 'abcabcabc');
```
-`repeatStringNumTimes("abc", 4)` 应返回 `"abcabcabcabc"`。
+`repeatStringNumTimes("abc", 4)` should return `"abcabcabcabc"`.
```js
assert(repeatStringNumTimes('abc', 4) === 'abcabcabcabc');
```
-`repeatStringNumTimes("abc", 1)` 应返回 `"abc"`。
+`repeatStringNumTimes("abc", 1)` should return `"abc"`.
```js
assert(repeatStringNumTimes('abc', 1) === 'abc');
```
-`repeatStringNumTimes("*", 8)` 应返回 `"********"`。
+`repeatStringNumTimes("*", 8)` should return `"********"`.
```js
assert(repeatStringNumTimes('*', 8) === '********');
```
-`repeatStringNumTimes("abc", -2)` 应返回 `""`。
+`repeatStringNumTimes("abc", -2)` should return `""`.
```js
assert(repeatStringNumTimes('abc', -2) === '');
```
-不应使用内置的 `repeat()` 方法。
+The built-in `repeat()` method should not be used.
```js
assert(!/\.repeat/g.test(code));
```
-`repeatStringNumTimes("abc", 0)` 应返回 `""`。
+`repeatStringNumTimes("abc", 0)` should return `""`.
```js
assert(repeatStringNumTimes('abc', 0) === '');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/return-largest-numbers-in-arrays.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/return-largest-numbers-in-arrays.md
index ba1b9df8f39..2148b0cbd2b 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/return-largest-numbers-in-arrays.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/return-largest-numbers-in-arrays.md
@@ -1,6 +1,6 @@
---
id: a789b3483989747d63b0e427
-title: 找出多个数组中的最大数字
+title: Return Largest Numbers in Arrays
challengeType: 5
forumTopicId: 16042
dashedName: return-largest-numbers-in-arrays
@@ -8,13 +8,13 @@ dashedName: return-largest-numbers-in-arrays
# --description--
-请返回一个数组，该数组由参数中每个子数组中的最大数字组成。为简单起见，给出的数组总会包含 4 个子数组。
+Return an array consisting of the largest number from each provided sub-array. For simplicity, the provided array will contain exactly 4 sub-arrays.
-别忘了，你可以通过 for 循环遍历一个数组，并用 `arr[i]` 的写法来访问数组中的元素。
+Remember, you can iterate through an array with a simple for loop, and access each member with array syntax `arr[i]`.
# --hints--
-`largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]])` 应返回一个数组。
+`largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]])` should return an array.
```js
assert(
@@ -27,7 +27,7 @@ assert(
);
```
-`largestOfFour([[13, 27, 18, 26], [4, 5, 1, 3], [32, 35, 37, 39], [1000, 1001, 857, 1]])` 应返回 `[27, 5, 39, 1001]`。
+`largestOfFour([[13, 27, 18, 26], [4, 5, 1, 3], [32, 35, 37, 39], [1000, 1001, 857, 1]])` should return `[27, 5, 39, 1001]`.
```js
assert.deepEqual(
@@ -41,7 +41,7 @@ assert.deepEqual(
);
```
-`largestOfFour([[4, 9, 1, 3], [13, 35, 18, 26], [32, 35, 97, 39], [1000000, 1001, 857, 1]])` 应返回 `[9, 35, 97, 1000000]`。
+`largestOfFour([[4, 9, 1, 3], [13, 35, 18, 26], [32, 35, 97, 39], [1000000, 1001, 857, 1]])` should return `[9, 35, 97, 1000000]`.
```js
assert.deepEqual(
@@ -55,7 +55,7 @@ assert.deepEqual(
);
```
-`largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]])` 应返回 `[25, 48, 21, -3]`。
+`largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]])` should return `[25, 48, 21, -3]`.
```js
assert.deepEqual(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/reverse-a-string.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/reverse-a-string.md
index b5890c35e88..3e44678ceb6 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/reverse-a-string.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/reverse-a-string.md
@@ -1,6 +1,6 @@
---
id: a202eed8fc186c8434cb6d61
-title: 反转字符串
+title: Reverse a String
challengeType: 5
forumTopicId: 16043
dashedName: reverse-a-string
@@ -8,33 +8,33 @@ dashedName: reverse-a-string
# --description--
-请反转传入函数的字符串。
+Reverse the provided string.
-在反转字符串之前，你可能需要将其切分成包含字符的数组。
+You may need to turn the string into an array before you can reverse it.
-函数的返回结果应为字符串。
+Your result must be a string.
# --hints--
-`reverseString("hello")` 应返回一个字符串。
+`reverseString("hello")` should return a string.
```js
assert(typeof reverseString('hello') === 'string');
```
-`reverseString("hello")` 应返回 `"olleh"`。
+`reverseString("hello")` should become `"olleh"`.
```js
assert(reverseString('hello') === 'olleh');
```
-`reverseString("Howdy")` 应返回 `"ydwoH"`。
+`reverseString("Howdy")` should become `"ydwoH"`.
```js
assert(reverseString('Howdy') === 'ydwoH');
```
-`reverseString("Greetings from Earth")` 应返回 `"htraE morf sgniteerG"`。
+`reverseString("Greetings from Earth")` should return `"htraE morf sgniteerG"`.
```js
assert(reverseString('Greetings from Earth') === 'htraE morf sgniteerG');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/slice-and-splice.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/slice-and-splice.md
index 7198e04d561..aa06a9e5fa5 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/slice-and-splice.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/slice-and-splice.md
@@ -1,6 +1,6 @@
---
id: 579e2a2c335b9d72dd32e05c
-title: Slice 与 Splice
+title: Slice and Splice
challengeType: 5
forumTopicId: 301148
dashedName: slice-and-splice
@@ -8,29 +8,29 @@ dashedName: slice-and-splice
# --description--
-本挑战的输入参数为两个数组和一个索引值。
+You are given two arrays and an index.
-请使用数组的 `slice` 和 `splice` 方法，将第一个数组中的所有元素依次复制到第二个数组中。
+Copy each element of the first array into the second array, in order.
-请注意，你需要从第二个数组索引值为 `n`（函数的第三个参数）的地方开始插入。
+Begin inserting elements at index `n` of the second array.
-最后，请返回插入元素后的数组。作为输入参数的两个数组在函数执行前后应保持不变。
+Return the resulting array. The input arrays should remain the same after the function runs.
# --hints--
-`frankenSplice([1, 2, 3], [4, 5], 1)` 应返回 `[4, 1, 2, 3, 5]`。
+`frankenSplice([1, 2, 3], [4, 5], 1)` should return `[4, 1, 2, 3, 5]`.
```js
assert.deepEqual(frankenSplice([1, 2, 3], [4, 5], 1), [4, 1, 2, 3, 5]);
```
-`frankenSplice([1, 2], ["a", "b"], 1)` 应返回 `["a", 1, 2, "b"]`。
+`frankenSplice([1, 2], ["a", "b"], 1)` should return `["a", 1, 2, "b"]`.
```js
assert.deepEqual(frankenSplice(testArr1, testArr2, 1), ['a', 1, 2, 'b']);
```
-`frankenSplice(["claw", "tentacle"], ["head", "shoulders", "knees", "toes"], 2)` 应返回 `["head", "shoulders", "claw", "tentacle", "knees", "toes"]`。
+`frankenSplice(["claw", "tentacle"], ["head", "shoulders", "knees", "toes"], 2)` should return `["head", "shoulders", "claw", "tentacle", "knees", "toes"]`.
```js
assert.deepEqual(
@@ -43,20 +43,20 @@ assert.deepEqual(
);
```
-第一个数组中的所有元素都应按原始顺序添加到第二个数组中。
+All elements from the first array should be added to the second array in their original order.
```js
assert.deepEqual(frankenSplice([1, 2, 3, 4], [], 0), [1, 2, 3, 4]);
```
-函数运行后，第一个数组应保持不变。
+The first array should remain the same after the function runs.
```js
frankenSplice(testArr1, testArr2, 1);
assert.deepEqual(testArr1, [1, 2]);
```
-函数运行后，第二个数组应保持不变。
+The second array should remain the same after the function runs.
```js
frankenSplice(testArr1, testArr2, 1);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/title-case-a-sentence.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/title-case-a-sentence.md
index 7aa05bc55a3..4d155143d31 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/title-case-a-sentence.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/title-case-a-sentence.md
@@ -1,6 +1,6 @@
---
id: ab6137d4e35944e21037b769
-title: 句中单词首字母大写
+title: Title Case a Sentence
challengeType: 5
forumTopicId: 16088
dashedName: title-case-a-sentence
@@ -8,31 +8,31 @@ dashedName: title-case-a-sentence
# --description--
-请将传入的字符串中，每个单词的第一个字母变成大写并返回。注意除首字母外，其余的字符都应是小写的。
+Return the provided string with the first letter of each word capitalized. Make sure the rest of the word is in lower case.
-另外请注意，像是 “the”、“of” 之类的连接词的首字母也要大写。
+For the purpose of this exercise, you should also capitalize connecting words like "the" and "of".
# --hints--
-`titleCase("I'm a little tea pot")` 应返回一个字符串。
+`titleCase("I'm a little tea pot")` should return a string.
```js
assert(typeof titleCase("I'm a little tea pot") === 'string');
```
-`titleCase("I'm a little tea pot")` 应返回 `I'm A Little Tea Pot`。
+`titleCase("I'm a little tea pot")` should return `I'm A Little Tea Pot`.
```js
assert(titleCase("I'm a little tea pot") === "I'm A Little Tea Pot");
```
-`titleCase("sHoRt AnD sToUt")` 应返回 `Short And Stout`。
+`titleCase("sHoRt AnD sToUt")` should return `Short And Stout`.
```js
assert(titleCase('sHoRt AnD sToUt') === 'Short And Stout');
```
-`titleCase("HERE IS MY HANDLE HERE IS MY SPOUT")` 应返回 `Here Is My Handle Here Is My Spout`。
+`titleCase("HERE IS MY HANDLE HERE IS MY SPOUT")` should return `Here Is My Handle Here Is My Spout`.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/truncate-a-string.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/truncate-a-string.md
index 586412482d4..c32c069878b 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/truncate-a-string.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/truncate-a-string.md
@@ -1,6 +1,6 @@
---
id: ac6993d51946422351508a41
-title: 截断字符串
+title: Truncate a String
challengeType: 5
forumTopicId: 16089
dashedName: truncate-a-string
@@ -8,11 +8,11 @@ dashedName: truncate-a-string
# --description--
-如果传入的字符串（第一个参数）的长度大于传入的值（第二个参数），请在这个位置截断它并在后面加上 `...`，然后返回结果。
+Truncate a string (first argument) if it is longer than the given maximum string length (second argument). Return the truncated string with a `...` ending.
# --hints--
-`truncateString("A-tisket a-tasket A green and yellow basket", 8)` 应返回 "A-tisket..."。
+`truncateString("A-tisket a-tasket A green and yellow basket", 8)` should return "A-tisket...".
```js
assert(
@@ -21,7 +21,7 @@ assert(
);
```
-`truncateString("Peter Piper picked a peck of pickled peppers", 11)` 应返回 "Peter Piper..."。
+`truncateString("Peter Piper picked a peck of pickled peppers", 11)` should return "Peter Piper...".
```js
assert(
@@ -30,7 +30,7 @@ assert(
);
```
-`truncateString("A-tisket a-tasket A green and yellow basket", "A-tisket a-tasket A green and yellow basket".length)` 应返回 "A-tisket a-tasket A green and yellow basket"。
+`truncateString("A-tisket a-tasket A green and yellow basket", "A-tisket a-tasket A green and yellow basket".length)` should return "A-tisket a-tasket A green and yellow basket".
```js
assert(
@@ -41,7 +41,7 @@ assert(
);
```
-`truncateString("A-tisket a-tasket A green and yellow basket", "A-tisket a-tasket A green and yellow basket".length + 2)` 应返回 "A-tisket a-tasket A green and yellow basket"。
+`truncateString("A-tisket a-tasket A green and yellow basket", "A-tisket a-tasket A green and yellow basket".length + 2)` should return "A-tisket a-tasket A green and yellow basket".
```js
assert(
@@ -52,13 +52,13 @@ assert(
);
```
-`truncateString("A-", 1)` 应返回 "A..."。
+`truncateString("A-", 1)` should return "A...".
```js
assert(truncateString('A-', 1) === 'A...');
```
-`truncateString("Absolutely Longer", 2)` 应返回 "Ab..."。
+`truncateString("Absolutely Longer", 2)` should return "Ab...".
```js
assert(truncateString('Absolutely Longer', 2) === 'Ab...');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/where-do-i-belong.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/where-do-i-belong.md
index 8b1765cbe84..20f9427e64d 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/where-do-i-belong.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-algorithm-scripting/where-do-i-belong.md
@@ -1,6 +1,6 @@
---
id: a24c1a4622e3c05097f71d67
-title: 找出元素在排序后数组中的索引
+title: Where do I Belong
challengeType: 5
forumTopicId: 16094
dashedName: where-do-i-belong
@@ -8,105 +8,105 @@ dashedName: where-do-i-belong
# --description--
-第一个参数的数组在排序后，将一个值（第二个参数）插入该数组，并使数组保持有序。返回这个新插入元素的**最小**索引值（应为一个数字）。
+Return the lowest index at which a value (second argument) should be inserted into an array (first argument) once it has been sorted. The returned value should be a number.
-例如，`getIndexToIns([1,2,3,4], 1.5)` 应该返回 `1` 因为 `1.5` 大于 `1`（索引为 0）且小于 `2`（索引为 1）。
+For example, `getIndexToIns([1,2,3,4], 1.5)` should return `1` because it is greater than `1` (index 0), but less than `2` (index 1).
-同样地，`getIndexToIns([20,3,5], 19)` 应该返回 `2`。因为数组排序后会变成 `[3,5,20]`，而 `19` 小于 `20`（索引为 2）且大于 `5`（索引为 1）。
+Likewise, `getIndexToIns([20,3,5], 19)` should return `2` because once the array has been sorted it will look like `[3,5,20]` and `19` is less than `20` (index 2) and greater than `5` (index 1).
# --hints--
-`getIndexToIns([10, 20, 30, 40, 50], 35)` 应返回 `3`。
+`getIndexToIns([10, 20, 30, 40, 50], 35)` should return `3`.
```js
assert(getIndexToIns([10, 20, 30, 40, 50], 35) === 3);
```
-`getIndexToIns([10, 20, 30, 40, 50], 35)` 应返回一个数字。
+`getIndexToIns([10, 20, 30, 40, 50], 35)` should return a number.
```js
assert(typeof getIndexToIns([10, 20, 30, 40, 50], 35) === 'number');
```
-`getIndexToIns([10, 20, 30, 40, 50], 30)` 应返回 `2`。
+`getIndexToIns([10, 20, 30, 40, 50], 30)` should return `2`.
```js
assert(getIndexToIns([10, 20, 30, 40, 50], 30) === 2);
```
-`getIndexToIns([10, 20, 30, 40, 50], 30)` 应返回一个数字。
+`getIndexToIns([10, 20, 30, 40, 50], 30)` should return a number.
```js
assert(typeof getIndexToIns([10, 20, 30, 40, 50], 30) === 'number');
```
-`getIndexToIns([40, 60], 50)` 应返回 `1`。
+`getIndexToIns([40, 60], 50)` should return `1`.
```js
assert(getIndexToIns([40, 60], 50) === 1);
```
-`getIndexToIns([40, 60], 50)` 应返回一个数字。
+`getIndexToIns([40, 60], 50)` should return a number.
```js
assert(typeof getIndexToIns([40, 60], 50) === 'number');
```
-`getIndexToIns([3, 10, 5], 3)` 应返回 `0`。
+`getIndexToIns([3, 10, 5], 3)` should return `0`.
```js
assert(getIndexToIns([3, 10, 5], 3) === 0);
```
-`getIndexToIns([3, 10, 5], 3)` 应返回一个数字。
+`getIndexToIns([3, 10, 5], 3)` should return a number.
```js
assert(typeof getIndexToIns([3, 10, 5], 3) === 'number');
```
-`getIndexToIns([5, 3, 20, 3], 5)` 应返回 `2`。
+`getIndexToIns([5, 3, 20, 3], 5)` should return `2`.
```js
assert(getIndexToIns([5, 3, 20, 3], 5) === 2);
```
-`getIndexToIns([5, 3, 20, 3], 5)` 应返回一个数字。
+`getIndexToIns([5, 3, 20, 3], 5)` should return a number.
```js
assert(typeof getIndexToIns([5, 3, 20, 3], 5) === 'number');
```
-`getIndexToIns([2, 20, 10], 19)` 应返回 `2`。
+`getIndexToIns([2, 20, 10], 19)` should return `2`.
```js
assert(getIndexToIns([2, 20, 10], 19) === 2);
```
-`getIndexToIns([2, 20, 10], 19)` 应返回一个数字。
+`getIndexToIns([2, 20, 10], 19)` should return a number.
```js
assert(typeof getIndexToIns([2, 20, 10], 19) === 'number');
```
-`getIndexToIns([2, 5, 10], 15)` 应返回 `3`。
+`getIndexToIns([2, 5, 10], 15)` should return `3`.
```js
assert(getIndexToIns([2, 5, 10], 15) === 3);
```
-`getIndexToIns([2, 5, 10], 15)` 应返回一个数字。
+`getIndexToIns([2, 5, 10], 15)` should return a number.
```js
assert(typeof getIndexToIns([2, 5, 10], 15) === 'number');
```
-`getIndexToIns([], 1)`应该返回 `0`。
+`getIndexToIns([], 1)` should return `0`.
```js
assert(getIndexToIns([], 1) === 0);
```
-`getIndexToIns([], 1)` 应返回一个数字。
+`getIndexToIns([], 1)` should return a number.
```js
assert(typeof getIndexToIns([], 1) === 'number');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/access-an-arrays-contents-using-bracket-notation.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/access-an-arrays-contents-using-bracket-notation.md
index 1665e1958b2..482d2c52363 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/access-an-arrays-contents-using-bracket-notation.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/access-an-arrays-contents-using-bracket-notation.md
@@ -1,6 +1,6 @@
---
id: 5a661e0f1068aca922b3ef17
-title: 使用方括号访问数组的元素
+title: Access an Array's Contents Using Bracket Notation
challengeType: 1
forumTopicId: 301149
dashedName: access-an-arrays-contents-using-bracket-notation
@@ -8,55 +8,55 @@ dashedName: access-an-arrays-contents-using-bracket-notation
# --description--
-所有数据结构的基本特性是，它们不仅可以存储数据，还可以让我们按需访问存放在其中的数据。我们已经学习了如何创建数组，现在让我们来学习如何访问数组中的数据。
+The fundamental feature of any data structure is, of course, the ability to not only store data, but to be able to retrieve that data on command. So, now that we've learned how to create an array, let's begin to think about how we can access that array's information.
-我们先定义一个包含 3 个元素的数组：
+When we define a simple array as seen below, there are 3 items in it:
```js
let ourArray = ["a", "b", "c"];
```
-在数组中，内部的每个元素都有一个与之对应的<dfn>索引</dfn>（<dfn>index</dfn>）。索引既是该元素在数组中的位置，也是我们访问该元素的参考。需要注意的是，JavaScript 数组的索引是从 0 开始的（这种从 0 开始的规则叫做 <dfn>zero-indexed</dfn>），即数组的第一个元素是在数组中的***第 0 个***位置，而不是第 1 个位置。要从数组中获取一个元素，我们可以在数组字面量后面加一个用方括号（`[]`）括起来的索引。不过习惯上，我们会通过表示数组的变量名来访问，而不是直接通过字面量。这种从数组中读取元素的方式叫做<dfn>方括号表示法</dfn>（<dfn>bracket notation</dfn>）。如果我们要从数组 `ourArray` 中获取数据元素 `"a"` 并将其赋值给另一个变量，可以这样写：
+In an array, each array item has an <dfn>index</dfn>. This index doubles as the position of that item in the array, and how you reference it. However, it is important to note, that JavaScript arrays are <dfn>zero-indexed</dfn>, meaning that the first element of an array is actually at the ***zeroth*** position, not the first. In order to retrieve an element from an array we can enclose an index in brackets and append it to the end of an array, or more commonly, to a variable which references an array object. This is known as <dfn>bracket notation</dfn>. For example, if we want to retrieve the `"a"` from `ourArray` and assign it to a variable, we can do so with the following code:
```js
let ourVariable = ourArray[0];
-// ourVariable 的值为 "a"
+// ourVariable equals "a"
```
-除了使用索引来获取某个元素值以外，你还可以通过类似的写法来*设置*一个索引位置的元素值：
+In addition to accessing the value associated with an index, you can also *set* an index to a value using the same notation:
```js
ourArray[1] = "not b anymore";
-// ourArray 现在的值为 ["a", "not b anymore", "c"];
+// ourArray now equals ["a", "not b anymore", "c"];
```
-在上面的代码中，我们用方括号表示法把索引为 1 的元素从 `"b"` 改成了 `"not b anymore"`。
+Using bracket notation, we have now reset the item at index 1 from `"b"`, to `"not b anymore"`.
# --instructions--
-在本挑战中，请将 `myArray` 中的第二个元素（索引为 `1`）设置为除了 `"b"` 以外的任意值。
+In order to complete this challenge, set the 2nd position (index `1`) of `myArray` to anything you want, besides `"b"`.
# --hints--
-`myArray[0]` 应为 `"a"`。
+`myArray[0]` should be equal to `"a"`
```js
assert.strictEqual(myArray[0], 'a');
```
-`myArray[1]` 不应为 `"b"`。
+`myArray[1]` should not be equal to `"b"`
```js
assert.notStrictEqual(myArray[1], 'b');
```
-`myArray[2]` 应为 `"c"`。
+`myArray[2]` should be equal to `"c"`
```js
assert.strictEqual(myArray[2], 'c');
```
-`myArray[3]` 应为 `"d"`。
+`myArray[3]` should be equal to `"d"`
```js
assert.strictEqual(myArray[3], 'd');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/access-property-names-with-bracket-notation.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/access-property-names-with-bracket-notation.md
index d7d84ecd37e..32bf0bb356f 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/access-property-names-with-bracket-notation.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/access-property-names-with-bracket-notation.md
@@ -1,6 +1,6 @@
---
id: 587d7b7c367417b2b2512b1a
-title: 使用方括号访问属性名称
+title: Access Property Names with Bracket Notation
challengeType: 1
forumTopicId: 301150
dashedName: access-property-names-with-bracket-notation
@@ -8,28 +8,28 @@ dashedName: access-property-names-with-bracket-notation
# --description--
-在关于对象的第一个挑战中，我们提到可以在一对方括号中用一个变量作为属性名来访问属性的值。假设一个超市收银台程序中有一个 `foods` 对象，并且有一个函数会设置 `selectedFood`；如果我们需要查询 `foods` 对象中，某种食物是否存在，可以这样实现：
+In the first object challenge we mentioned the use of bracket notation as a way to access property values using the evaluation of a variable. For instance, imagine that our `foods` object is being used in a program for a supermarket cash register. We have some function that sets the `selectedFood` and we want to check our `foods` object for the presence of that food. This might look like:
```js
let selectedFood = getCurrentFood(scannedItem);
let inventory = foods[selectedFood];
```
-上述代码会先读取 `selectedFood` 变量的值，并返回 `foods` 对象中以该值命名的属性所对应的属性值。若没有以该值命名的属性，则会返回 `undefined`。有时候对象的属性名在运行之前是不确定的，或者我们需要动态地访问对象的属性值。在这些场景下，方括号表示法就变得十分有用。
+This code will evaluate the value stored in the `selectedFood` variable and return the value of that key in the `foods` object, or `undefined` if it is not present. Bracket notation is very useful because sometimes object properties are not known before runtime or we need to access them in a more dynamic way.
# --instructions--
-我们已经定义了 `checkInventory` 函数，它接受一个被扫描到的商品名作为输入参数。请让这个函数返回 `foods` 对象中，以 `scannedItem` 的值所命名的属性对应的属性值。在本挑战中，只有合理有效的属性名会作为参数传入 `checkInventory`，因此你不需要处理参数无效的情况。
+We've defined a function, `checkInventory`, which receives a scanned item as an argument. Return the current value of the `scannedItem` key in the `foods` object. You can assume that only valid keys will be provided as an argument to `checkInventory`.
# --hints--
-`checkInventory` 应是一个函数。
+`checkInventory` should be a function.
```js
assert.strictEqual(typeof checkInventory, 'function');
```
-`foods` 对象应只包含以下键值对：`apples: 25`、`oranges: 32`、`plums: 28`、`bananas: 13`、`grapes: 35`、`strawberries: 27`。
+The `foods` object should have only the following key-value pairs: `apples: 25`, `oranges: 32`, `plums: 28`, `bananas: 13`, `grapes: 35`, `strawberries: 27`.
```js
assert.deepEqual(foods, {
@@ -42,19 +42,19 @@ assert.deepEqual(foods, {
});
```
-`checkInventory("apples")` 应返回 `25`。
+`checkInventory("apples")` should return `25`.
```js
assert.strictEqual(checkInventory('apples'), 25);
```
-`checkInventory("bananas")` 应返回 `13`。
+`checkInventory("bananas")` should return `13`.
```js
assert.strictEqual(checkInventory('bananas'), 13);
```
-`checkInventory("strawberries")` 应返回 `27`。
+`checkInventory("strawberries")` should return `27`.
```js
assert.strictEqual(checkInventory('strawberries'), 27);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/add-items-to-an-array-with-push-and-unshift.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/add-items-to-an-array-with-push-and-unshift.md
index 9cbc81af84c..afd8dff1242 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/add-items-to-an-array-with-push-and-unshift.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/add-items-to-an-array-with-push-and-unshift.md
@@ -1,6 +1,6 @@
---
id: 587d78b2367417b2b2512b0e
-title: 使用 push() 和 unshift() 为数组添加元素
+title: Add Items to an Array with push() and unshift()
challengeType: 1
forumTopicId: 301151
dashedName: add-items-to-an-array-with-push-and-unshift
@@ -8,30 +8,28 @@ dashedName: add-items-to-an-array-with-push-and-unshift
# --description--
-数组的长度与数组能包含的数据类型一样，都是不固定的。数组可以包含任意数量的元素，可以不限次数地往数组中添加元素或者从中移除元素。总之，数组是<dfn>可变的</dfn>（<dfn>mutable</dfn>）。在本挑战中，我们要学习两种修改数组的方法：`Array.push()` 和 `Array.unshift()`。
+An array's length, like the data types it can contain, is not fixed. Arrays can be defined with a length of any number of elements, and elements can be added or removed over time; in other words, arrays are <dfn>mutable</dfn>. In this challenge, we will look at two methods with which we can programmatically modify an array: `Array.push()` and `Array.unshift()`.
-这两个方法都接收一个或多个元素作为参数，并会将参数中的元素添加到该数组中。`push()` 方法会将元素插入到数组的末尾，而 `unshift()` 方法会将元素插入到数组的开头。请看以下例子：
+Both methods take one or more elements as parameters and add those elements to the array the method is being called on; the `push()` method adds elements to the end of an array, and `unshift()` adds elements to the beginning. Consider the following:
```js
let twentyThree = 'XXIII';
let romanNumerals = ['XXI', 'XXII'];
romanNumerals.unshift('XIX', 'XX');
-// 数组现在为 ['XIX', 'XX', 'XXI', 'XXII']
+// now equals ['XIX', 'XX', 'XXI', 'XXII']
romanNumerals.push(twentyThree);
-// 数组现在为 ['XIX', 'XX', 'XXI', 'XXII', 'XXIII']
+// now equals ['XIX', 'XX', 'XXI', 'XXII', 'XXIII']Notice that we can also pass variables, which allows us even greater flexibility in dynamically modifying our array's data.
```
-**注意：**我们甚至可以传入变量，这在动态改变数组数据的场景下十分有用。
-
# --instructions--
-我们已经定义了一个 `mixedNumbers` 函数，它接收一个数组作为参数。请你修改这个函数，使用 `push()` 和 `unshift()` 来将 `'I', 2, 'three'` 插入到数组开头，将 `7, 'VIII', 9` 插入到数组的末尾，使得这个函数返回一个依次包含 1 至 9。
+We have defined a function, `mixedNumbers`, which we are passing an array as an argument. Modify the function by using `push()` and `unshift()` to add `'I', 2, 'three'` to the beginning of the array and `7, 'VIII', 9` to the end so that the returned array contains representations of the numbers 1-9 in order.
# --hints--
-`mixedNumbers(["IV", 5, "six"])` 应返回 `["I", 2, "three", "IV", 5, "six", 7, "VIII", 9]`。
+`mixedNumbers(["IV", 5, "six"])` should now return `["I", 2, "three", "IV", 5, "six", 7, "VIII", 9]`
```js
assert.deepEqual(mixedNumbers(['IV', 5, 'six']), [
@@ -47,13 +45,13 @@ assert.deepEqual(mixedNumbers(['IV', 5, 'six']), [
]);
```
-`mixedNumbers` 函数中应调用 `push()` 方法。
+The `mixedNumbers` function should utilize the `push()` method
```js
assert(mixedNumbers.toString().match(/\.push/));
```
-`mixedNumbers` 函数中应调用 `unshift()` 方法。
+The `mixedNumbers` function should utilize the `unshift()` method
```js
assert(mixedNumbers.toString().match(/\.unshift/));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/add-items-using-splice.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/add-items-using-splice.md
index 62b27e1e878..04a8878555b 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/add-items-using-splice.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/add-items-using-splice.md
@@ -1,6 +1,6 @@
---
id: 587d78b3367417b2b2512b11
-title: 使用 splice() 添加元素
+title: Add Items Using splice()
challengeType: 1
forumTopicId: 301152
dashedName: add-items-using-splice
@@ -8,7 +8,7 @@ dashedName: add-items-using-splice
# --description--
-还记得在上个挑战中我们提到 `splice()` 方法最多可以接收 3 个参数吗？除了移除元素，我们还可以利用它的第三个参数来向数组中*添加*元素。第三个参数可以是一个或多个元素，这些元素会被添加到数组中。这样，我们能够便捷地将数组中的一个或多个连续元素换成其他的元素。例如：
+Remember in the last challenge we mentioned that `splice()` can take up to three parameters? Well, you can use the third parameter, comprised of one or more element(s), to add to the array. This can be incredibly useful for quickly switching out an element, or a set of elements, for another.
```js
const numbers = [10, 11, 12, 12, 15];
@@ -16,20 +16,20 @@ const startIndex = 3;
const amountToDelete = 1;
numbers.splice(startIndex, amountToDelete, 13, 14);
-// 第二个 12 被移除，在第二个 12 的索引处添加 13、14。
+// the second entry of 12 is removed, and we add 13 and 14 at the same index
console.log(numbers);
-// 返回 [ 10, 11, 12, 13, 14, 15 ]
+// returns [ 10, 11, 12, 13, 14, 15 ]
```
-在上面的代码中，数组一开始包含了若干数字。接着，我们调用 `splice()` 方法，索引为 (3) 的地方开始删除元素，删除的元素数量是 (1)。然后，(13, 14) 是在删除位置插入的元素，可以在 `amountToDelete` 后面传入任意数量的元素（以逗号分隔），每个都会被插入到数组中。
+Here, we begin with an array of numbers. Then, we pass the following to `splice()`: The index at which to begin deleting elements (3), the number of elements to be deleted (1), and the remaining arguments (13, 14) will be inserted starting at that same index. Note that there can be any number of elements (separated by commas) following `amountToDelete`, each of which gets inserted.
# --instructions--
-我们已经定义了一个 `htmlColorNames` 函数，它以一个 HTML 颜色的数组作为输入参数。请修改这个函数，使用 `splice()` 来移除数组中的前两个元素，并在对应的位置上添加 `'DarkSalmon'` 和 `'BlanchedAlmond'`。
+We have defined a function, `htmlColorNames`, which takes an array of HTML colors as an argument. Modify the function using `splice()` to remove the first two elements of the array and add `'DarkSalmon'` and `'BlanchedAlmond'` in their respective places.
# --hints--
-`htmlColorNames` 应返回 `["DarkSalmon", "BlanchedAlmond", "LavenderBlush", "PaleTurqoise", "FireBrick"]`。
+`htmlColorNames` should return `["DarkSalmon", "BlanchedAlmond", "LavenderBlush", "PaleTurquoise", "FireBrick"]`
```js
assert.deepEqual(
@@ -50,19 +50,19 @@ assert.deepEqual(
);
```
-`htmlColorNames` 函数中应调用 `splice()` 方法。
+The `htmlColorNames` function should utilize the `splice()` method
```js
assert(/.splice/.test(code));
```
-不应使用 `shift()` 或 `unshift()`。
+You should not use `shift()` or `unshift()`.
```js
assert(!/shift|unshift/.test(code));
```
-不应使用数组的方括号表示法。
+You should not use array bracket notation.
```js
assert(!/\[\d\]\s*=/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/add-key-value-pairs-to-javascript-objects.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/add-key-value-pairs-to-javascript-objects.md
index adca71461fa..f5adc6e51b4 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/add-key-value-pairs-to-javascript-objects.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/add-key-value-pairs-to-javascript-objects.md
@@ -1,6 +1,6 @@
---
id: 587d7b7c367417b2b2512b18
-title: 将键值对添加到对象中
+title: Add Key-Value Pairs to JavaScript Objects
challengeType: 1
forumTopicId: 301153
dashedName: add-key-value-pairs-to-javascript-objects
@@ -8,7 +8,7 @@ dashedName: add-key-value-pairs-to-javascript-objects
# --description--
-对象（object）本质上是<dfn>键值对（key-value pair）</dfn>的集合。或者说，一系列被映射到唯一标识符的数据就是对象；习惯上，唯一标识符叫做<dfn>属性（property）</dfn>或者<dfn>键（key）</dfn>）；数据叫做<dfn>值（value）</dfn>。让我们来看一个简单的例子：
+At their most basic, objects are just collections of <dfn>key-value</dfn> pairs. In other words, they are pieces of data (<dfn>values</dfn>) mapped to unique identifiers called <dfn>properties</dfn> (<dfn>keys</dfn>). Take a look at an example:
```js
const tekkenCharacter = {
@@ -18,19 +18,19 @@ const tekkenCharacter = {
};
```
-上面的代码定义了一个叫做 `tekkenCharacter` 的“铁拳”游戏人物对象。它有三个属性，每个属性都对应一个特定的值。如果我们想为它再添加一个叫做 `origin` 的属性，可以这样写：
+The above code defines a Tekken video game character object called `tekkenCharacter`. It has three properties, each of which map to a specific value. If you want to add an additional property, such as "origin", it can be done by assigning `origin` to the object:
```js
tekkenCharacter.origin = 'South Korea';
```
-上面的代码中，我们使用了<dfn>点号表示法（dot notation）</dfn>。如果我们现在输出这个对象，便可以看到它具有 `origin` 属性。接下来，因为这个人物在游戏中有着与众不同的橘色头发，我们可以通过方括号表示法来为它添加这个属性，像这样：
+This uses dot notation. If you were to observe the `tekkenCharacter` object, it will now include the `origin` property. Hwoarang also had distinct orange hair. You can add this property with bracket notation by doing:
```js
tekkenCharacter['hair color'] = 'dyed orange';
```
-如果要设置的属性中存在空格，或者要设置的属性是一个变量，那我们必须使用<dfn>方括号表示法（bracket notation）</dfn>来为对象添加属性。在上面的代码中，我们把属性 `hair color` 放到引号里，以此来表示整个字符串都是需要设置的属性。如果我们不加上引号，那么中括号里的内容会被当作一个变量来解析，这个变量对应的值就会作为要设置的属性，请看这段代码：
+Bracket notation is required if your property has a space in it or if you want to use a variable to name the property. In the above case, the property is enclosed in quotes to denote it as a string and will be added exactly as shown. Without quotes, it will be evaluated as a variable and the name of the property will be whatever value the variable is. Here's an example with a variable:
```js
const eyes = 'eye color';
@@ -38,7 +38,7 @@ const eyes = 'eye color';
tekkenCharacter[eyes] = 'brown';
```
-执行以上所有示例代码后，对象会变成这样：
+After adding all the examples, the object will look like this:
```js
{
@@ -53,35 +53,35 @@ tekkenCharacter[eyes] = 'brown';
# --instructions--
-我们已经为你创建了 `foods` 对象。请使用上述任意语法，来为 `foods` 对象添加如下三个键值对：`bananas` 属性，值为 `13`；`grapes` 属性，值为 `35`；`strawberries` 属性，值为 `27`。
+A `foods` object has been created with three entries. Using the syntax of your choice, add three more entries to it: `bananas` with a value of `13`, `grapes` with a value of `35`, and `strawberries` with a value of `27`.
# --hints--
-`foods` 应为一个对象。
+`foods` should be an object.
```js
assert(typeof foods === 'object');
```
-`foods` 应有一个值为 `13` 的 `"bananas"` 属性。
+The `foods` object should have a key `"bananas"` with a value of `13`.
```js
assert(foods.bananas === 13);
```
-`foods` 应有一个值为 `35` 的 `"grapes"` 属性。
+The `foods` object should have a key `"grapes"` with a value of `35`.
```js
assert(foods.grapes === 35);
```
-`foods` 应有一个值为 `27` 的 `"strawberries"` 属性。
+The `foods` object should have a key `"strawberries"` with a value of `27`.
```js
assert(foods.strawberries === 27);
```
-应使用点号表示法或方括号表示法来设置对象的属性。
+The key-value pairs should be set using dot or bracket notation.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/check-for-the-presence-of-an-element-with-indexof.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/check-for-the-presence-of-an-element-with-indexof.md
index e64201560be..4bc028927f1 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/check-for-the-presence-of-an-element-with-indexof.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/check-for-the-presence-of-an-element-with-indexof.md
@@ -1,6 +1,6 @@
---
id: 587d7b7b367417b2b2512b14
-title: 使用 indexOf() 检查元素是否存在
+title: Check For The Presence of an Element With indexOf()
challengeType: 1
forumTopicId: 301154
dashedName: check-for-the-presence-of-an-element-with-indexof
@@ -8,25 +8,31 @@ dashedName: check-for-the-presence-of-an-element-with-indexof
# --description--
-由于数组随时都可以修改或发生“突变”（*mutated*），我们很难保证某个数据始终处于数组中的特定位置，甚至不能保证该元素是否还存在于该数组中。好消息是，JavaScript 为我们提供了内置方法 `indexOf()`。这个方法让我们可以方便地检查某个元素是否存在于数组中。`indexOf()` 方法接受一个元素作为输入参数，并返回该元素在数组中的位置（索引）；若该元素不存在于数组中则返回 `-1`。
+Since arrays can be changed, or *mutated*, at any time, there's no guarantee about where a particular piece of data will be on a given array, or if that element even still exists. Luckily, JavaScript provides us with another built-in method, `indexOf()`, that allows us to quickly and easily check for the presence of an element on an array. `indexOf()` takes an element as a parameter, and when called, it returns the position, or index, of that element, or `-1` if the element does not exist on the array.
-例如：
+For example:
```js
let fruits = ['apples', 'pears', 'oranges', 'peaches', 'pears'];
-fruits.indexOf('dates'); // 返回 -1
-fruits.indexOf('oranges'); // 返回 2
-fruits.indexOf('pears'); // 返回 1，因为第一个出现在数组中的 'pears' 元素索引为 1
+fruits.indexOf('dates'); // returns -1
+fruits.indexOf('oranges'); // returns 2
+fruits.indexOf('pears'); // returns 1, the first index at which the element exists
```
# --instructions--
-`indexOf()` 在快速检查一个数组中是否存在某个元素时非常有用。我们已经定义了一个 `quickCheck` 函数，它接受一个数组和一个元素作为输入参数。请修改这个函数，通过 `indexOf()` 方法，使得当参数数组中包含第二个参数的元素时返回 `true`，不包含时返回 `false`。
+`indexOf()` can be incredibly useful for quickly checking for the presence of an element on an array. We have defined a function, `quickCheck`, that takes an array and an element as arguments. Modify the function using `indexOf()` so that it returns `true` if the passed element exists on the array, and `false` if it does not.
# --hints--
-`quickCheck(["squash", "onions", "shallots"], "mushrooms")` 应返回 `false`。
+The `quickCheck` function should return a boolean (`true` or `false`), not a string (`"true"` or `"false"`)
+
+```js
+assert.isBoolean(quickCheck(['squash', 'onions', 'shallots'], 'mushrooms'));
+```
+
+`quickCheck(["squash", "onions", "shallots"], "mushrooms")` should return `false`
```js
assert.strictEqual(
@@ -35,7 +41,7 @@ assert.strictEqual(
);
```
-`quickCheck(["squash", "onions", "shallots"], "onions")` 应返回 `true`。
+`quickCheck(["onions", "squash", "shallots"], "onions")` should return `true`
```js
assert.strictEqual(
@@ -44,19 +50,19 @@ assert.strictEqual(
);
```
-`quickCheck([3, 5, 9, 125, 45, 2], 125)` 应返回 `true`。
+`quickCheck([3, 5, 9, 125, 45, 2], 125)` should return `true`
```js
assert.strictEqual(quickCheck([3, 5, 9, 125, 45, 2], 125), true);
```
-`quickCheck([true, false, false], undefined)` 应返回 `false`。
+`quickCheck([true, false, false], undefined)` should return `false`
```js
assert.strictEqual(quickCheck([true, false, false], undefined), false);
```
-`quickCheck` 函数中应使用 `indexOf()` 方法。
+The `quickCheck` function should utilize the `indexOf()` method
```js
assert.notStrictEqual(quickCheck.toString().search(/\.indexOf\(/), -1);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/check-if-an-object-has-a-property.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/check-if-an-object-has-a-property.md
index e4ad433f63a..e2ac0392eb2 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/check-if-an-object-has-a-property.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/check-if-an-object-has-a-property.md
@@ -1,6 +1,6 @@
---
id: 587d7b7d367417b2b2512b1c
-title: 检查对象是否具有某个属性
+title: Check if an Object has a Property
challengeType: 1
forumTopicId: 301155
dashedName: check-if-an-object-has-a-property
@@ -8,21 +8,21 @@ dashedName: check-if-an-object-has-a-property
# --description--
-我们已经学习了如果添加、修改和移除对象中的属性。但如果我们想知道一个对象中是否包含某个属性呢？JavaScript 为我们提供了两种不同的方式来实现这个功能：一个是通过 `hasOwnProperty()` 方法，另一个是使用 `in` 关键字。假如我们有一个 `users` 对象，为检查它是否含有 `Alan` 属性，可以这样写：
+Now we can add, modify, and remove keys from objects. But what if we just wanted to know if an object has a specific property? JavaScript provides us with two different ways to do this. One uses the `hasOwnProperty()` method and the other uses the `in` keyword. If we have an object `users` with a property of `Alan`, we could check for its presence in either of the following ways:
```js
users.hasOwnProperty('Alan');
'Alan' in users;
-// 都返回 true
+// both return true
```
# --instructions--
-我们已经定义了一个包含若干用户信息的 `users` 对象和一个 `isEveryoneHere` 函数，该函数接收 `users` 对象作为参数。请完成该函数使其在 `users` 对象中同时包含 `Alan`、`Jeff`、`Sarah`、`Ryan` 四个属性时才返回 `true`，否则返回 `false`。
+We've created an object, `users`, with some users in it and a function `isEveryoneHere`, which we pass the `users` object to as an argument. Finish writing this function so that it returns `true` only if the `users` object contains all four names, `Alan`, `Jeff`, `Sarah`, and `Ryan`, as keys, and `false` otherwise.
# --hints--
-`users` 对象应该只包含 `Alan`、`Jeff`、`Sarah`、`Ryan` 4 个属性。
+The `users` object should only contain the keys `Alan`, `Jeff`, `Sarah`, and `Ryan`
```js
assert(
@@ -34,13 +34,13 @@ assert(
);
```
-`isEveryoneHere` 函数在 `users` 对象包含 `Alan`、`Jeff`、`Sarah`、`Ryan` 4 个属性时应返回 `true`。
+The function `isEveryoneHere` should return `true` if `Alan`, `Jeff`, `Sarah`, and `Ryan` are properties on the `users` object
```js
assert(isEveryoneHere(users) === true);
```
-`isEveryoneHere` 函数在 `users` 对象不包含 `Alan`、`Jeff`、`Sarah`、`Ryan` 4 个属性时应返回 `false`。
+The function `isEveryoneHere` should return `false` if `Alan` is not a property on the `users` object
```js
assert(
@@ -51,7 +51,7 @@ assert(
);
```
-如果 `users` 对象中不包含属性 `Jeff`，则函数 `isEveryoneHere` 应返回 `false`。
+The function `isEveryoneHere` should return `false` if `Jeff` is not a property on the `users` object
```js
assert(
@@ -62,7 +62,7 @@ assert(
);
```
-如果 `users` 对象中不包含属性 `Sarah`，则函数 `isEveryoneHere` 应返回 `false`。
+The function `isEveryoneHere` should return `false` if `Sarah` is not a property on the `users` object
```js
assert(
@@ -73,7 +73,7 @@ assert(
);
```
-如果 `users` 对象中不包含属性 `Ryan`，则函数 `isEveryoneHere` 应返回 `false`。
+The function `isEveryoneHere` should return `false` if `Ryan` is not a property on the `users` object
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/combine-arrays-with-the-spread-operator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/combine-arrays-with-the-spread-operator.md
index 8c298c78c8a..0a27f7bb090 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/combine-arrays-with-the-spread-operator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/combine-arrays-with-the-spread-operator.md
@@ -1,6 +1,6 @@
---
id: 587d7b7b367417b2b2512b17
-title: 使用展开运算符合并数组
+title: Combine Arrays with the Spread Operator
challengeType: 1
forumTopicId: 301156
dashedName: combine-arrays-with-the-spread-operator
@@ -8,30 +8,30 @@ dashedName: combine-arrays-with-the-spread-operator
# --description--
-<dfn>展开语法</dfn>的另一个重要用途是合并数组，或者将某个数组的所有元素插入到另一个数组的任意位置。我们也可以使用 ES5 的语法连接两个数组，但只能让它们首尾相接。而展开语法可以让这样的操作变得极其简单：
+Another huge advantage of the <dfn>spread</dfn> operator, is the ability to combine arrays, or to insert all the elements of one array into another, at any index. With more traditional syntaxes, we can concatenate arrays, but this only allows us to combine arrays at the end of one, and at the start of another. Spread syntax makes the following operation extremely simple:
```js
let thisArray = ['sage', 'rosemary', 'parsley', 'thyme'];
let thatArray = ['basil', 'cilantro', ...thisArray, 'coriander'];
-// thatArray 现在是 ['basil', 'cilantro', 'sage', 'rosemary', 'parsley', 'thyme', 'coriander']
+// thatArray now equals ['basil', 'cilantro', 'sage', 'rosemary', 'parsley', 'thyme', 'coriander']
```
-使用展开语法，我们像这样就可以实现一个用传统方法会写得很复杂且冗长的操作。
+Using spread syntax, we have just achieved an operation that would have been more complex and more verbose had we used traditional methods.
# --instructions--
-我们已经定义了一个返回 `sentence` 变量的 `spreadOut` 函数。请修改这个函数，利用<dfn>展开语法</dfn>使该函数返回数组 `['learning', 'to', 'code', 'is', 'fun']`。
+We have defined a function `spreadOut` that returns the variable `sentence`. Modify the function using the <dfn>spread</dfn> operator so that it returns the array `['learning', 'to', 'code', 'is', 'fun']`.
# --hints--
-`spreadOut` 应返回 `["learning", "to", "code", "is", "fun"]`。
+`spreadOut` should return `["learning", "to", "code", "is", "fun"]`
```js
assert.deepEqual(spreadOut(), ['learning', 'to', 'code', 'is', 'fun']);
```
-`spreadOut` 函数里应用到展开语法。
+The `spreadOut` function should utilize spread syntax
```js
assert.notStrictEqual(spreadOut.toString().search(/[...]/), -1);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/copy-an-array-with-the-spread-operator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/copy-an-array-with-the-spread-operator.md
index fd96b8083b2..ca3aa490b5a 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/copy-an-array-with-the-spread-operator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/copy-an-array-with-the-spread-operator.md
@@ -1,6 +1,6 @@
---
id: 587d7b7b367417b2b2512b13
-title: 使用展开运算符复制数组
+title: Copy an Array with the Spread Operator
challengeType: 1
forumTopicId: 301157
dashedName: copy-an-array-with-the-spread-operator
@@ -8,24 +8,24 @@ dashedName: copy-an-array-with-the-spread-operator
# --description--
-`slice()` 可以让我们从一个数组中选择一些元素来复制到新数组中，而 ES6 中又引入了一个简洁且可读性强的语法：<dfn>展开运算符（spread operator）</dfn>，它能让我们方便地复制数组中的*所有*元素。展开语法写出来是这样：`...`
+While `slice()` allows us to be selective about what elements of an array to copy, among several other useful tasks, ES6's new <dfn>spread operator</dfn> allows us to easily copy *all* of an array's elements, in order, with a simple and highly readable syntax. The spread syntax simply looks like this: `...`
-我们可以用展开运算符来复制数组：
+In practice, we can use the spread operator to copy an array like so:
```js
let thisArray = [true, true, undefined, false, null];
let thatArray = [...thisArray];
-// thatArray 的值现在也是 [true, true, undefined, false, null]
-// thisArray 保持不变。现在 thatArray 所包含的值与 thisArray 完全相同
+// thatArray equals [true, true, undefined, false, null]
+// thisArray remains unchanged and thatArray contains the same elements as thisArray
```
# --instructions--
-我们已经定义了一个 `copyMachine` 函数，它接受 `arr`（一个数组）和 `num`（一个数字）作为输入参数。该函数需要返回一个由 `num` 个 `arr` 组成的新的二维数组。同时，我们写好了大致的流程，只是细节实现还没有写完。请修改这个函数，使用展开语法，使该函数能正常工作（提示：我们已经学到过的一个方法很适合用在这里）！
+We have defined a function, `copyMachine` which takes `arr` (an array) and `num` (a number) as arguments. The function is supposed to return a new array made up of `num` copies of `arr`. We have done most of the work for you, but it doesn't work quite right yet. Modify the function using spread syntax so that it works correctly (hint: another method we have already covered might come in handy here!).
# --hints--
-`copyMachine([true, false, true], 2)` 应返回 `[[true, false, true], [true, false, true]]`。
+`copyMachine([true, false, true], 2)` should return `[[true, false, true], [true, false, true]]`
```js
assert.deepEqual(copyMachine([true, false, true], 2), [
@@ -34,7 +34,7 @@ assert.deepEqual(copyMachine([true, false, true], 2), [
]);
```
-`copyMachine([1, 2, 3], 5)` 应返回 `[[1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3]]`。
+`copyMachine([1, 2, 3], 5)` should return `[[1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3]]`
```js
assert.deepEqual(copyMachine([1, 2, 3], 5), [
@@ -46,13 +46,13 @@ assert.deepEqual(copyMachine([1, 2, 3], 5), [
]);
```
-`copyMachine([true, true, null], 1)` 应返回 `[[true, true, null]]`。
+`copyMachine([true, true, null], 1)` should return `[[true, true, null]]`
```js
assert.deepEqual(copyMachine([true, true, null], 1), [[true, true, null]]);
```
-`copyMachine(["it works"], 3)` 应返回 `[["it works"], ["it works"], ["it works"]]`。
+`copyMachine(["it works"], 3)` should return `[["it works"], ["it works"], ["it works"]]`
```js
assert.deepEqual(copyMachine(['it works'], 3), [
@@ -62,10 +62,10 @@ assert.deepEqual(copyMachine(['it works'], 3), [
]);
```
-`copyMachine` 函数中应对 `arr` 使用`展开运算符`。
+The `copyMachine` function should utilize the `spread operator` with array `arr`
```js
-assert(removeJSComments(code).match(/\.\.\.arr/));
+assert(__helpers.removeJSComments(code).match(/\.\.\.arr/));
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/copy-array-items-using-slice.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/copy-array-items-using-slice.md
index 489ea95be57..54ba5bade9c 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/copy-array-items-using-slice.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/copy-array-items-using-slice.md
@@ -1,6 +1,6 @@
---
id: 587d7b7a367417b2b2512b12
-title: 使用 slice() 复制数组元素
+title: Copy Array Items Using slice()
challengeType: 1
forumTopicId: 301158
dashedName: copy-array-items-using-slice
@@ -8,25 +8,25 @@ dashedName: copy-array-items-using-slice
# --description--
-接下来我们要介绍 `slice()` 方法。`slice()` 不会修改数组，而是会复制，或者说*提取（extract）*给定数量的元素到一个新数组。同时，调用方法的数组保持不变。`slice()` 只接收 2 个输入参数：第一个是开始提取元素的位置（索引），第二个是提取元素的结束位置（索引）。`slice()` 提取的元素中不包括第二个参数所对应的元素。请看以下例子：
+The next method we will cover is `slice()`. Rather than modifying an array, `slice()` copies or *extracts* a given number of elements to a new array, leaving the array it is called upon untouched. `slice()` takes only 2 parameters — the first is the index at which to begin extraction, and the second is the index at which to stop extraction (extraction will occur up to, but not including the element at this index). Consider this:
```js
let weatherConditions = ['rain', 'snow', 'sleet', 'hail', 'clear'];
let todaysWeather = weatherConditions.slice(1, 3);
-// todaysWeather 等于 ['snow', 'sleet'];
-// weatherConditions 仍然等于 ['rain', 'snow', 'sleet', 'hail', 'clear']
+// todaysWeather equals ['snow', 'sleet'];
+// weatherConditions still equals ['rain', 'snow', 'sleet', 'hail', 'clear']
```
-在上面的代码中，我们从一个数组中提取了一些元素，并用这些元素创建了一个新数组。
+In effect, we have created a new array by extracting elements from an existing array.
# --instructions--
-我们已经定义了一个 `forecast` 函数，它接受一个数组作为参数。请修改这个函数，利用 `slice()` 从输入的数组中提取信息，最终返回一个包含元素 `'warm'` 和 `'sunny'` 的新数组。
+We have defined a function, `forecast`, that takes an array as an argument. Modify the function using `slice()` to extract information from the argument array and return a new array that contains the elements `'warm'` and `'sunny'`.
# --hints--
-`forecast` 应返回 `["warm", "sunny"]`。
+`forecast` should return `["warm", "sunny"]`
```js
assert.deepEqual(
@@ -35,7 +35,7 @@ assert.deepEqual(
);
```
-`forecast` 函数中应使用 `slice()` 方法。
+The `forecast` function should utilize the `slice()` method
```js
assert(/\.slice\(/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/create-complex-multi-dimensional-arrays.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/create-complex-multi-dimensional-arrays.md
index b505c28c404..d15e24bcd4c 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/create-complex-multi-dimensional-arrays.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/create-complex-multi-dimensional-arrays.md
@@ -1,6 +1,6 @@
---
id: 587d7b7b367417b2b2512b16
-title: 创建复杂的多维数组
+title: Create complex multi-dimensional arrays
challengeType: 1
forumTopicId: 301159
dashedName: create-complex-multi-dimensional-arrays
@@ -8,37 +8,37 @@ dashedName: create-complex-multi-dimensional-arrays
# --description--
-很好！你现在已经学到很多关于数组的知识了，但这些只是个开始。我们将在接下来的中挑战中学到更多与数组相关的知识。在继续学习<dfn>对象</dfn>（<dfn>Objects</dfn>）之前，让我们再花一点时间了解下更复杂的数组嵌套。
+Awesome! You have just learned a ton about arrays! This has been a fairly high level overview, and there is plenty more to learn about working with arrays, much of which you will see in later sections. But before moving on to looking at <dfn>Objects</dfn>, lets take one more look, and see how arrays can become a bit more complex than what we have seen in previous challenges.
-数组的一个强大的特性是，它可以包含其他数组，甚至完全由其他数组组成。在上一个挑战中，我们已经接触到了包含数组的数组，但它还算是比较简单的。数组中的数组还可以再包含其他数组，即可以嵌套任意多层数组。习惯上，我们称这种数据结构为<dfn>多维（multi-dimensional）数组</dfn>或嵌套（nested）数组。请看如下的示例：
+One of the most powerful features when thinking of arrays as data structures, is that arrays can contain, or even be completely made up of other arrays. We have seen arrays that contain arrays in previous challenges, but fairly simple ones. However, arrays can contain an infinite depth of arrays that can contain other arrays, each with their own arbitrary levels of depth, and so on. In this way, an array can very quickly become very complex data structure, known as a <dfn>multi-dimensional</dfn>, or nested array. Consider the following example:
```js
-let nestedArray = [ // 顶层，或第 1 层，即最外层数组
- ['deep'], // 数组中的数组，第 2 层
+let nestedArray = [ // top, or first level - the outer most array
+ ['deep'], // an array within an array, 2 levels of depth
[
- ['deeper'], ['deeper'] // 第 3 层，元素为嵌套的两个数组
+ ['deeper'], ['deeper'] // 2 arrays nested 3 levels deep
],
[
[
- ['deepest'], ['deepest'] // 第 4 层，元素为嵌套的两个数组
+ ['deepest'], ['deepest'] // 2 arrays nested 4 levels deep
],
[
[
- ['deepest-est?'] // 第 5 层，元素为嵌套的一个数组
+ ['deepest-est?'] // an array nested 5 levels deep
]
]
]
];
```
-虽然这个例子看起来错综复杂，不过，尤其是在处理大量数据的时候，这种数据结构还是经常会用到的。尽管结构复杂，不过我们仍可以通过方括号表示法来访问嵌套得最深的数组：
+While this example may seem convoluted, this level of complexity is not unheard of, or even unusual, when dealing with large amounts of data. However, we can still very easily access the deepest levels of an array this complex with bracket notation:
```js
console.log(nestedArray[2][1][0][0][0]);
// logs: deepest-est?
```
-既然我们知道数据的位置，当然，我们也可以修改它：
+And now that we know where that piece of data is, we can reset it if we need to:
```js
nestedArray[2][1][0][0][0] = 'deeper still';
@@ -49,11 +49,11 @@ console.log(nestedArray[2][1][0][0][0]);
# --instructions--
-我们已经定义了一个叫做 `myNestedArray` 的数组变量。请修改 `myNestedArray`，使用<dfn>字符串（string）</dfn>、<dfn>数字（number）</dfn>或<dfn>布尔值（boolean）</dfn>作为数组的元素，并让 `myNestedArray` 刚好有 5 层（注意，最外层的数组是第 1 层）。同时，请在第 3 层的数组中包含字符串 `'deep'`；在第 4 层的数组中包含字符串 `'deeper'`，在第 5 层的数组中包含字符串 `'deepest'`。
+We have defined a variable, `myNestedArray`, set equal to an array. Modify `myNestedArray`, using any combination of <dfn>strings</dfn>, <dfn>numbers</dfn>, and <dfn>booleans</dfn> for data elements, so that it has exactly five levels of depth (remember, the outer-most array is level 1). Somewhere on the third level, include the string `'deep'`, on the fourth level, include the string `'deeper'`, and on the fifth level, include the string `'deepest'`.
# --hints--
-`myNestedArray` 中的数据元素应只包含字符串、数字或者布尔值。
+`myNestedArray` should contain only numbers, booleans, and strings as data elements
```js
assert.strictEqual(
@@ -77,7 +77,7 @@ assert.strictEqual(
);
```
-`myNestedArray` 应刚好包含 5 层嵌套数组。
+`myNestedArray` should have exactly 5 levels of depth
```js
assert.strictEqual(
@@ -100,7 +100,7 @@ assert.strictEqual(
);
```
-`myNestedArray` 中应只有一个字符串 `"deep"`，并且应出现在第 3 层数组中。
+`myNestedArray` should contain exactly one occurrence of the string `"deep"` on an array nested 3 levels deep
```js
assert(
@@ -129,7 +129,7 @@ assert(
);
```
-`myNestedArray` 中应只有一个字符串 `"deeper"`，并且应出现在第 4 层数组中。
+`myNestedArray` should contain exactly one occurrence of the string `"deeper"` on an array nested 4 levels deep
```js
assert(
@@ -158,7 +158,7 @@ assert(
);
```
-`myNestedArray` 中应只有一个字符串 `"deepest"`，并且应出现在第 5 层数组中。
+`myNestedArray` should contain exactly one occurrence of the string `"deepest"` on an array nested 5 levels deep
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/generate-an-array-of-all-object-keys-with-object.keys.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/generate-an-array-of-all-object-keys-with-object.keys.md
index 6d75c47078c..df6c8f99963 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/generate-an-array-of-all-object-keys-with-object.keys.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/generate-an-array-of-all-object-keys-with-object.keys.md
@@ -1,6 +1,6 @@
---
id: 587d7b7d367417b2b2512b1e
-title: 使用 Object.keys() 生成由对象的所有属性组成的数组
+title: Generate an Array of All Object Keys with Object.keys()
challengeType: 1
forumTopicId: 301160
dashedName: generate-an-array-of-all-object-keys-with-object-keys
@@ -8,15 +8,15 @@ dashedName: generate-an-array-of-all-object-keys-with-object-keys
# --description--
-我们可以给 `Object.keys()` 方法传入一个对象作为参数，这会返回一个由对象中所有属性（字符串）组成的数组。需要注意的是，数组中元素的顺序是不确定的。
+We can also generate an array which contains all the keys stored in an object using the `Object.keys()` method and passing in an object as the argument. This will return an array with strings representing each property in the object. Again, there will be no specific order to the entries in the array.
# --instructions--
-请完成 `getArrayOfUsers` 函数的实现，使其返回一个由输入对象中的所有属性所组成的数组。
+Finish writing the `getArrayOfUsers` function so that it returns an array containing all the properties in the object it receives as an argument.
# --hints--
-`users` 对象应该只包含 `Alan`、`Jeff`、`Sarah`、`Ryan` 这 4 个属性。
+The `users` object should only contain the keys `Alan`, `Jeff`, `Sarah`, and `Ryan`
```js
assert(
@@ -28,7 +28,7 @@ assert(
);
```
-`getArrayOfUsers` 函数应返回一个包含 `users` 对象中所有属性的数组。
+The `getArrayOfUsers` function should return an array which contains all the keys in the `users` object
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/iterate-through-all-an-arrays-items-using-for-loops.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/iterate-through-all-an-arrays-items-using-for-loops.md
index b101ad1943a..6674711f213 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/iterate-through-all-an-arrays-items-using-for-loops.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/iterate-through-all-an-arrays-items-using-for-loops.md
@@ -1,6 +1,6 @@
---
id: 587d7b7b367417b2b2512b15
-title: 使用 for 循环遍历数组中的全部元素
+title: Iterate Through All an Array's Items Using For Loops
challengeType: 1
forumTopicId: 301161
dashedName: iterate-through-all-an-arrays-items-using-for-loops
@@ -8,9 +8,9 @@ dashedName: iterate-through-all-an-arrays-items-using-for-loops
# --description--
-使用数组时，我们经常需要遍历数组的所有元素来找出我们需要的一个或多个元素，抑或是对数组执行一些特定的操作。JavaScript 为我们提供了几个内置的方法，它们以不同的方式遍历数组，以便我们可以用于不同的场景（如 `every()`、`forEach()`、`map()` 等等）。然而，最简单的 `for` 循环不仅能实现上述这些方法的功能，而且相比之下也会更加灵活。
+Sometimes when working with arrays, it is very handy to be able to iterate through each item to find one or more elements that we might need, or to manipulate an array based on which data items meet a certain set of criteria. JavaScript offers several built in methods that each iterate over arrays in slightly different ways to achieve different results (such as `every()`, `forEach()`, `map()`, etc.), however the technique which is most flexible and offers us the greatest amount of control is a simple `for` loop.
-请看以下的例子：
+Consider the following:
```js
function greaterThanTen(arr) {
@@ -24,18 +24,18 @@ function greaterThanTen(arr) {
}
greaterThanTen([2, 12, 8, 14, 80, 0, 1]);
-// 返回 [12, 14, 80]
+// returns [12, 14, 80]
```
-在这个函数中，我们用一个 `for` 循环来遍历数组，逐一对其中的元素进行判断。通过上面的代码，我们可以找出数组中大于 `10` 的所有元素，并返回一个包含这些元素的新数组。
+Using a `for` loop, this function iterates through and accesses each element of the array, and subjects it to a simple test that we have created. In this way, we have easily and programmatically determined which data items are greater than `10`, and returned a new array containing those items.
# --instructions--
-我们已经定义了 `filteredArray` 函数，它接受一个嵌套的数组 `arr` 和一个 `elem` 作为参数，并要返回一个新数组。`arr` 数组中嵌套的数组里可能包含 `elem` 元素，也可能不包含。请修改该函数，用一个 `for` 循环来做筛选，使函数返回一个由 `arr` 中不包含 `elem` 的数组所组成的新数组。
+We have defined a function, `filteredArray`, which takes `arr`, a nested array, and `elem` as arguments, and returns a new array. `elem` represents an element that may or may not be present on one or more of the arrays nested within `arr`. Modify the function, using a `for` loop, to return a filtered version of the passed array such that any array nested within `arr` containing `elem` has been removed.
# --hints--
-`filteredArray([[10, 8, 3], [14, 6, 23], [3, 18, 6]], 18)` 应返回 `[ [10, 8, 3], [14, 6, 23] ]`。
+`filteredArray([[10, 8, 3], [14, 6, 23], [3, 18, 6]], 18)` should return `[ [10, 8, 3], [14, 6, 23] ]`
```js
assert.deepEqual(
@@ -54,7 +54,7 @@ assert.deepEqual(
);
```
-`filteredArray([ ["trumpets", 2], ["flutes", 4], ["saxophones", 2] ], 2)` 应返回 `[ ["flutes", 4] ]`。
+`filteredArray([ ["trumpets", 2], ["flutes", 4], ["saxophones", 2] ], 2)` should return `[ ["flutes", 4] ]`
```js
assert.deepEqual(
@@ -70,7 +70,7 @@ assert.deepEqual(
);
```
-`filteredArray([ ["amy", "beth", "sam"], ["dave", "sean", "peter"] ], "peter")` 应返回 `[ ["amy", "beth", "sam"] ]`。
+`filteredArray([ ["amy", "beth", "sam"], ["dave", "sean", "peter"] ], "peter")` should return `[ ["amy", "beth", "sam"] ]`
```js
assert.deepEqual(
@@ -85,7 +85,7 @@ assert.deepEqual(
);
```
-`filteredArray([[3, 2, 3], [1, 6, 3], [3, 13, 26], [19, 3, 9]], 3)` 应返回 `[]`。
+`filteredArray([[3, 2, 3], [1, 6, 3], [3, 13, 26], [19, 3, 9]], 3)` should return `[ ]`
```js
assert.deepEqual(
@@ -102,7 +102,7 @@ assert.deepEqual(
);
```
-`filteredArray` 函数中应使用 `for` 循环。
+The `filteredArray` function should utilize a `for` loop
```js
assert.notStrictEqual(filteredArray.toString().search(/for/), -1);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/iterate-through-the-keys-of-an-object-with-a-for...in-statement.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/iterate-through-the-keys-of-an-object-with-a-for...in-statement.md
index 42616dc8dc2..390132b5041 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/iterate-through-the-keys-of-an-object-with-a-for...in-statement.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/iterate-through-the-keys-of-an-object-with-a-for...in-statement.md
@@ -1,6 +1,6 @@
---
id: 587d7b7d367417b2b2512b1d
-title: 使用 for...in 语句遍历对象
+title: Iterate Through the Keys of an Object with a for...in Statement
challengeType: 1
forumTopicId: 301162
dashedName: iterate-through-the-keys-of-an-object-with-a-for---in-statement
@@ -8,25 +8,25 @@ dashedName: iterate-through-the-keys-of-an-object-with-a-for---in-statement
# --description--
-如果我们想要遍历对象中的所有属性，只需要使用 JavaScript 中的 <dfn>for...in</dfn> 语句即可。以遍历 `users` 对象的属性为例：
+Sometimes you may need to iterate through all the keys within an object. This requires a specific syntax in JavaScript called a <dfn>for...in</dfn> statement. For our `users` object, this could look like:
```js
for (let user in users) {
console.log(user);
}
-// 输出：
+// logs:
Alan
Jeff
Sarah
Ryan
```
-在上面的代码中，我们定义了一个 `user` 变量。可以观察到，这个变量在遍历对象的 `for...in` 语句执行过程中会一直被重置并赋予新值，结果就是不同的用户名打印到了 console 中。**注意：**对象中的键是无序的，这与数组不同。因此，一个对象中某个属性的位置，或者说它出现的相对顺序，在引用或访问该属性时是不确定的。
+In this statement, we defined a variable `user`, and as you can see, this variable was reset during each iteration to each of the object's keys as the statement looped through the object, resulting in each user's name being printed to the console. **NOTE:** Objects do not maintain an ordering to stored keys like arrays do; thus a key's position on an object, or the relative order in which it appears, is irrelevant when referencing or accessing that key.
# --instructions--
-我们已经定义了一个 `countOnline` 函数，请在其中使用 <dfn>for...in</dfn> 语句来遍历 `users` 对象中的用户，并返回 `online` 属性为 `true` 的用户数量。以下是一个传入 `countOnline` 函数的对象示例，注意每个用户都有 `online` 属性，其属性值为 `true` 或 `false`：
+We've defined a function `countOnline` which accepts one argument (a users object). Use a <dfn>for...in</dfn> statement within this function to loop through the users object passed into the function and return the number of users whose `online` property is set to `true`. An example of a users object which could be passed to `countOnline` is shown below. Each user will have an `online` property with either a `true` or `false` value.
```js
{
@@ -44,29 +44,29 @@ Ryan
# --hints--
-函数 `countOnline` 中应使用 `for in` 语句遍历传入的对象。
+The function `countOnline` should use a `for in` statement to iterate through the object keys of the object passed to it.
```js
assert(
code.match(
- /for\s*\(\s*(var|let)\s+[a-zA-Z_$]\w*\s+in\s+[a-zA-Z_$]\w*\s*\)\s*{/
+ /for\s*\(\s*(var|let|const)\s+[a-zA-Z_$]\w*\s+in\s+[a-zA-Z_$]\w*\s*\)/
)
);
```
-当传入 `{ Alan: { online: false }, Jeff: { online: true }, Sarah: { online: false } }` 时，函数 `countOnline` 应该返回 `1`。
+The function `countOnline` should return `1` when the object `{ Alan: { online: false }, Jeff: { online: true }, Sarah: { online: false } }` is passed to it
```js
assert(countOnline(usersObj1) === 1);
```
-当传入 `{ Alan: { online: true }, Jeff: { online: false }, Sarah: { online: true } }` 时，函数 `countOnline` 应该返回 `2`。
+The function `countOnline` should return `2` when the object `{ Alan: { online: true }, Jeff: { online: false }, Sarah: { online: true } }` is passed to it
```js
assert(countOnline(usersObj2) === 2);
```
-当传入 `{ Alan: { online: false }, Jeff: { online: false }, Sarah: { online: false } }` 时，函数 `countOnline` 应该返回 `0`。
+The function `countOnline` should return `0` when the object `{ Alan: { online: false }, Jeff: { online: false }, Sarah: { online: false } }` is passed to it
```js
assert(countOnline(usersObj3) === 0);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/modify-an-array-stored-in-an-object.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/modify-an-array-stored-in-an-object.md
index f406d97893b..c3e256070eb 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/modify-an-array-stored-in-an-object.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/modify-an-array-stored-in-an-object.md
@@ -1,6 +1,6 @@
---
id: 587d7b7d367417b2b2512b1f
-title: 修改存储在对象中的数组
+title: Modify an Array Stored in an Object
challengeType: 1
forumTopicId: 301163
dashedName: modify-an-array-stored-in-an-object
@@ -8,21 +8,21 @@ dashedName: modify-an-array-stored-in-an-object
# --description--
-我们已经学习了 JavaScript 对象的这些基本操作：添加、修改、移除键值对、检查某个属性是否存在、遍历对象的所有属性。在继续学习 JavaScript 的过程中，我们会了解对象的更多用法。另外，在之后的数据结构课程中，我们还会学习 ES6 的 <dfn>Map</dfn> 和 <dfn>Set</dfn>。这两种数据结构与我们现在学到的对象十分类似，但它们在对象的基础上提供了一些额外的功能。目前，我们已经学习了数组和对象的基础知识，让我们试着来用所学的知识解决一些更复杂的问题。
+Now you've seen all the basic operations for JavaScript objects. You can add, modify, and remove key-value pairs, check if keys exist, and iterate over all the keys in an object. As you continue learning JavaScript you will see even more versatile applications of objects. Additionally, the Data Structures lessons located in the Coding Interview Prep section of the curriculum also cover the ES6 <dfn>Map</dfn> and <dfn>Set</dfn> objects, both of which are similar to ordinary objects but provide some additional features. Now that you've learned the basics of arrays and objects, you're fully prepared to begin tackling more complex problems using JavaScript!
# --instructions--
-请看一下代码编辑器中我们为你写好的对象。`user` 对象包含 3 个属性；`data` 对象包含 5 个属性，其中包含一个叫做 `friends` 的数组。这就是对象作为数据结构所展现出的灵活性。我们已经写好了 `addFriend` 函数的一部分，请你完成这个函数，使其接受一个 `user` 对象，将 `friend` 参数中的名字添加到 `user.data.friends` 数组中并返回该数组。
+Take a look at the object we've provided in the code editor. The `user` object contains three keys. The `data` key contains five keys, one of which contains an array of `friends`. From this, you can see how flexible objects are as data structures. We've started writing a function `addFriend`. Finish writing it so that it takes a `user` object and adds the name of the `friend` argument to the array stored in `user.data.friends` and returns that array.
# --hints--
-`user` 对象应该包含 `name`、`age` 和 `data` 三个属性。
+The `user` object should have `name`, `age`, and `data` keys.
```js
assert('name' in user && 'age' in user && 'data' in user);
```
-`addFriend` 函数应该接受一个 `user` 对象和一个 `friend` 字符串作为输入参数，并将这个字符串插入到 `user` 对象的 `friends` 数组中。
+The `addFriend` function should accept a `user` object and a `friend` string as arguments and add the friend to the array of `friends` in the `user` object.
```js
assert(
@@ -35,7 +35,7 @@ assert(
);
```
-`addFriend(user, "Pete")` 应该返回 `["Sam", "Kira", "Tomo", "Pete"]`。
+`addFriend(user, "Pete")` should return `["Sam", "Kira", "Tomo", "Pete"]`.
```js
assert.deepEqual(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/modify-an-object-nested-within-an-object.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/modify-an-object-nested-within-an-object.md
index 17724b01437..731f95ff6ae 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/modify-an-object-nested-within-an-object.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/modify-an-object-nested-within-an-object.md
@@ -1,6 +1,6 @@
---
id: 587d7b7c367417b2b2512b19
-title: 修改嵌套在对象中的对象
+title: Modify an Object Nested Within an Object
challengeType: 1
forumTopicId: 301164
dashedName: modify-an-object-nested-within-an-object
@@ -8,7 +8,7 @@ dashedName: modify-an-object-nested-within-an-object
# --description--
-现在我们来看一个稍复杂的对象。在对象中，我们也可以嵌套任意层数的对象，对象的属性值可以是 JavaScript 支持的任意类型，包括数组和其他对象。请看以下例子：
+Now let's take a look at a slightly more complex object. Object properties can be nested to an arbitrary depth, and their values can be any type of data supported by JavaScript, including arrays and even other objects. Consider the following:
```js
let nestedObject = {
@@ -19,13 +19,14 @@ let nestedObject = {
online: 80,
onlineStatus: {
active: 67,
- away: 13
+ away: 13,
+ busy: 8
}
}
};
```
-`nestedObject` 有 3 个属性：`id`（属性值为数字）、`date`（属性值为字符串）、`data`（属性值为嵌套的对象）。虽然对象中的数据可能很复杂，我们仍能使用上一个挑战中讲到的写法来访问我们需要的信息。如果我们想把嵌套在 `onlineStatus` 中 `busy` 的属性值改为 `10`，可以用点号表示法来这样实现：
+`nestedObject` has three properties: `id` (value is a number), `date` (value is a string), and `data` (value is an object with its nested structure). While structures can quickly become complex, we can still use the same notations to access the information we need. To assign the value `10` to the `busy` property of the nested `onlineStatus` object, we use dot notation to reference the property:
```js
nestedObject.data.onlineStatus.busy = 10;
@@ -33,11 +34,11 @@ nestedObject.data.onlineStatus.busy = 10;
# --instructions--
-我们已经定义了一个 `userActivity` 对象，它包含了另一个对象。请将 `online` 的属性值改为 `45`。
+Here we've defined an object `userActivity`, which includes another object nested within it. Set the value of the `online` key to `45`.
# --hints--
-`userActivity` 应包含 `id`、`date` 和 `data` 属性。
+`userActivity` should have `id`, `date` and `data` properties.
```js
assert(
@@ -45,19 +46,19 @@ assert(
);
```
-`userActivity` 应包含 `data` 属性，其属性值应为包含 `totalUsers` 和 `online` 属性的对象。
+`userActivity` should have a `data` key set to an object with keys `totalUsers` and `online`.
```js
assert('totalUsers' in userActivity.data && 'online' in userActivity.data);
```
-`userActivity` 的 `data` 属性值中的 `online` 属性值应被改为 `45`。
+The `online` property nested in the `data` key of `userActivity` should be set to `45`
```js
assert(userActivity.data.online === 45);
```
-应使用点号表示法或方括号表示法来修改 `online` 属性值。
+The `online` property should be set using dot or bracket notation.
```js
assert.strictEqual(code.search(/online: 45/), -1);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/remove-items-from-an-array-with-pop-and-shift.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/remove-items-from-an-array-with-pop-and-shift.md
index 9c6c6541fb8..9cb6b79038f 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/remove-items-from-an-array-with-pop-and-shift.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/remove-items-from-an-array-with-pop-and-shift.md
@@ -1,6 +1,6 @@
---
id: 587d78b2367417b2b2512b0f
-title: 使用 pop() 和 shift() 从数组中删除元素
+title: Remove Items from an Array with pop() and shift()
challengeType: 1
forumTopicId: 301165
dashedName: remove-items-from-an-array-with-pop-and-shift
@@ -8,35 +8,35 @@ dashedName: remove-items-from-an-array-with-pop-and-shift
# --description--
-`push()` 和 `unshift()` 都有一个与它们作用相反的函数：`pop()` 和 `shift()`。与插入元素相反，`pop()` 会从数组的末尾*移除*一个元素，而 `shift()` 会从数组的开头移除一个元素。`pop()` 和 `shift()` 与 `push()` 和 `unshift()` 的关键区别在于，用于删除元素的方法不接收参数，而且每次只能删除数组中的一个元素。
+Both `push()` and `unshift()` have corresponding methods that are nearly functional opposites: `pop()` and `shift()`. As you may have guessed by now, instead of adding, `pop()` *removes* an element from the end of an array, while `shift()` removes an element from the beginning. The key difference between `pop()` and `shift()` and their cousins `push()` and `unshift()`, is that neither method takes parameters, and each only allows an array to be modified by a single element at a time.
-让我们来看以下的例子：
+Let's take a look:
```js
let greetings = ['whats up?', 'hello', 'see ya!'];
greetings.pop();
-// 数组现在是 ['whats up?', 'hello']
+// now equals ['whats up?', 'hello']
greetings.shift();
-// 数组现在是 ['hello']
+// now equals ['hello']
```
-这些用于删除数组元素的方法会返回被删除的元素：
+We can also return the value of the removed element with either method like this:
```js
let popped = greetings.pop();
-// 返回 'hello'，即 popped 的值为 'hello'
-// greetings 数组现在为 []
+// returns 'hello'
+// greetings now equals []
```
# --instructions--
-我们已经定义了一个 `popShift` 函数，它接收一个数组作为输入参数并返回一个新的数组。请修改这个函数，使用 `pop()` 和 `shift()` 来移除输入的数组中的第一个元素和最后一个元素，并将这两个被移除的元素分别赋值给对应的变量，使得最终返回的数组里包含这两个值。
+We have defined a function, `popShift`, which takes an array as an argument and returns a new array. Modify the function, using `pop()` and `shift()`, to remove the first and last elements of the argument array, and assign the removed elements to their corresponding variables, so that the returned array contains their values.
# --hints--
-`popShift(["challenge", "is", "not", "complete"])` 应返回 `["challenge", "complete"]`。
+`popShift(["challenge", "is", "not", "complete"])` should return `["challenge", "complete"]`
```js
assert.deepEqual(popShift(['challenge', 'is', 'not', 'complete']), [
@@ -45,13 +45,13 @@ assert.deepEqual(popShift(['challenge', 'is', 'not', 'complete']), [
]);
```
-`popShift` 函数中应使用 `pop()` 方法。
+The `popShift` function should utilize the `pop()` method
```js
assert.notStrictEqual(popShift.toString().search(/\.pop\(/), -1);
```
-`popShift` 函数中应使用 `shift()` 方法。
+The `popShift` function should utilize the `shift()` method
```js
assert.notStrictEqual(popShift.toString().search(/\.shift\(/), -1);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/remove-items-using-splice.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/remove-items-using-splice.md
index f248c7ec060..ce579da8e77 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/remove-items-using-splice.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/remove-items-using-splice.md
@@ -1,6 +1,6 @@
---
id: 587d78b2367417b2b2512b10
-title: 使用 splice() 删除元素
+title: Remove Items Using splice()
challengeType: 1
forumTopicId: 301166
dashedName: remove-items-using-splice
@@ -8,40 +8,42 @@ dashedName: remove-items-using-splice
# --description--
-在之前的挑战中，我们已经学习了如何用 `shift()` 和 `pop()` 从数组的开头或末尾移除元素。但如果我们想删除数组中间的一个元素，或者想一次删除多个元素，该如何操作呢？这时候我们就需要使用 `splice()` 方法了，`splice()` 可以让我们从数组中的任意位置**连续删除任意数量的元素**。
+Ok, so we've learned how to remove elements from the beginning and end of arrays using `shift()` and `pop()`, but what if we want to remove an element from somewhere in the middle? Or remove more than one element at once? Well, that's where `splice()` comes in. `splice()` allows us to do just that: **remove any number of consecutive elements** from anywhere in an array.
-`splice()` 最多可以接受 3 个参数，但现在我们先关注前两个。`splice()` 接收的前两个参数以调用 `splice()` 数组中的元素索引作为参考。别忘了，数组的索引是*从 0 开始的*，所以我们要用 `0` 来表示数组中的第一个元素。`splice()` 的第一个参数代表从数组中的哪个索引开始移除元素，而第二个参数表示要从数组中的这位位置删除多少个元素。例如：
+`splice()` can take up to 3 parameters, but for now, we'll focus on just the first 2. The first two parameters of `splice()` are integers which represent indexes, or positions, of the array that `splice()` is being called upon. And remember, arrays are *zero-indexed*, so to indicate the first element of an array, we would use `0`. `splice()`'s first parameter represents the index on the array from which to begin removing elements, while the second parameter indicates the number of elements to delete. For example:
```js
let array = ['today', 'was', 'not', 'so', 'great'];
array.splice(2, 2);
-// 从索引为 2 的位置（即第三个元素）开始移除 2 个元素
-// array 的值现在是 ['today', 'was', 'great']
+// remove 2 elements beginning with the 3rd element
+// array now equals ['today', 'was', 'great']
```
-`splice()` 不仅会修改调用该方法的数组，还会返回一个包含被移除元素的数组：
+`splice()` not only modifies the array it's being called on, but it also returns a new array containing the value of the removed elements:
```js
let array = ['I', 'am', 'feeling', 'really', 'happy'];
let newArray = array.splice(3, 2);
-// newArray 的值是 ['really', 'happy']
+// newArray equals ['really', 'happy']
```
# --instructions--
-我们已经定义了数组 `arr`。请使用 `splice()` 从 `arr` 里移除元素，使剩余的元素之和为 `10`。
+We've initialized an array `arr`. Use `splice()` to remove elements from `arr`, so that it only contains elements that sum to the value of `10`.
# --hints--
-不应修改这一行 `const arr = [2, 4, 5, 1, 7, 5, 2, 1];`。
+You should not change the original line of `const arr = [2, 4, 5, 1, 7, 5, 2, 1];`.
```js
-assert(code.replace(/\s/g, '').match(/constarr=\[2,4,5,1,7,5,2,1\];?/));
+assert(
+ __helpers.removeWhiteSpace(code).match(/constarr=\[2,4,5,1,7,5,2,1\];?/)
+);
```
-`ahr` 的剩余元素之和应为 `10`。
+`arr` should only contain elements that sum to `10`.
```js
assert.strictEqual(
@@ -50,16 +52,18 @@ assert.strictEqual(
);
```
-应对 `arr` 调用 `splice()` 方法。
+Your code should utilize the `splice()` method on `arr`.
```js
-assert(code.replace(/\s/g, '').match(/arr\.splice\(/));
+assert(__helpers.removeWhiteSpace(code).match(/arr\.splice\(/));
```
-splice 应只删除 `arr` 里面的元素，不能给 `arr` 添加元素。
+The splice should only remove elements from `arr` and not add any additional elements to `arr`.
```js
-assert(!code.replace(/\s/g, '').match(/arr\.splice\(\d+,\d+,\d+.*\)/g));
+assert(
+ !__helpers.removeWhiteSpace(code).match(/arr\.splice\(\d+,\d+,\d+.*\)/g)
+);
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/use-an-array-to-store-a-collection-of-data.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/use-an-array-to-store-a-collection-of-data.md
index 7840b67d8d8..61f8b4f1c33 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/use-an-array-to-store-a-collection-of-data.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/use-an-array-to-store-a-collection-of-data.md
@@ -1,6 +1,6 @@
---
id: 587d7b7e367417b2b2512b20
-title: 使用数组存储不同类型的数据
+title: Use an Array to Store a Collection of Data
challengeType: 1
forumTopicId: 301167
dashedName: use-an-array-to-store-a-collection-of-data
@@ -8,15 +8,15 @@ dashedName: use-an-array-to-store-a-collection-of-data
# --description--
-以下是最简单的<dfn>数组（Array）</dfn>示例：这是一个<dfn>一维数组（one-dimensional array）</dfn>，它只有一层，或者说它里面没有包含其它数组。可以观察到，这个数组中只包含了<dfn>布尔值（booleans）</dfn>、<dfn>字符串（strings）</dfn>、<dfn>数字（numbers）</dfn>以及 JavaScript 中的其他数据类型：
+The below is an example of the simplest implementation of an array data structure. This is known as a <dfn>one-dimensional array</dfn>, meaning it only has one level, or that it does not have any other arrays nested within it. Notice it contains <dfn>booleans</dfn>, <dfn>strings</dfn>, and <dfn>numbers</dfn>, among other valid JavaScript data types:
```js
let simpleArray = ['one', 2, 'three', true, false, undefined, null];
console.log(simpleArray.length);
-// 输出 7
+// logs 7
```
-所有数组都有一个表示长度的属性，我们可以通过 `Array.length` 来访问它。接下来是一个稍复杂的数组示例：这是一个<dfn>多维数组（multi-dimensional Array）</dfn>，或者说是一个包含了其他数组的数组。可以观察到，在这个数组内部还包含了 JavaScript 的<dfn>对象（objects）</dfn>，我们会在后面的挑战中详细讨论该数据结构。现在，你只需要知道数组能够存储复杂的对象数据。
+All arrays have a length property, which as shown above, can be very easily accessed with the syntax `Array.length`. A more complex implementation of an array can be seen below. This is known as a <dfn>multi-dimensional array</dfn>, or an array that contains other arrays. Notice that this array also contains JavaScript <dfn>objects</dfn>, which we will examine very closely in our next section, but for now, all you need to know is that arrays are also capable of storing complex objects.
```js
let complexArray = [
@@ -45,35 +45,35 @@ let complexArray = [
# --instructions--
-我们已经定义了一个名为 `yourArray` 的变量。请修改代码，将一个含有至少 5 个元素的数组赋值给 `yourArray` 变量。你的数组中应包含至少一个 <dfn>string</dfn> 类型的数据、一个 <dfn>number</dfn> 类型的数据和一个 <dfn>boolean</dfn> 类型的数据。
+We have defined a variable called `yourArray`. Complete the statement by assigning an array of at least 5 elements in length to the `yourArray` variable. Your array should contain at least one <dfn>string</dfn>, one <dfn>number</dfn>, and one <dfn>boolean</dfn>.
# --hints--
-yourArray 应为数组。
+`yourArray` should be an array.
```js
assert.strictEqual(Array.isArray(yourArray), true);
```
-`yourArray` 应包含至少 5 个元素。
+`yourArray` should be at least 5 elements long.
```js
assert.isAtLeast(yourArray.length, 5);
```
-`yourArray` 应包含至少一个 `boolean`。
+`yourArray` should contain at least one `boolean`.
```js
assert(yourArray.filter((el) => typeof el === 'boolean').length >= 1);
```
-`yourArray` 应包含至少一个 `number`。
+`yourArray` should contain at least one `number`.
```js
assert(yourArray.filter((el) => typeof el === 'number').length >= 1);
```
-`yourArray` 应包含至少一个 `string`。
+`yourArray` should contain at least one `string`.
```js
assert(yourArray.filter((el) => typeof el === 'string').length >= 1);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/use-the-delete-keyword-to-remove-object-properties.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/use-the-delete-keyword-to-remove-object-properties.md
index 7724f7b7fbf..575e41f90aa 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/use-the-delete-keyword-to-remove-object-properties.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-data-structures/use-the-delete-keyword-to-remove-object-properties.md
@@ -1,6 +1,6 @@
---
id: 587d7b7c367417b2b2512b1b
-title: 使用 delete 关键字删除对象属性
+title: Use the delete Keyword to Remove Object Properties
challengeType: 1
forumTopicId: 301168
dashedName: use-the-delete-keyword-to-remove-object-properties
@@ -8,11 +8,11 @@ dashedName: use-the-delete-keyword-to-remove-object-properties
# --description--
-现在我们已经学习了什么是对象以及对象的基本特性和用途。总之，对象是以键值对的形式，灵活、直观地存储结构化数据的一种方式，***而且***，通过对象的属性查找属性值是速度很快的操作。在本章余下的挑战中，我们来了解一下对象的几种常用操作，这样你能更好地在代码中使用这个十分有用的数据结构：对象。
+Now you know what objects are and their basic features and advantages. In short, they are key-value stores which provide a flexible, intuitive way to structure data, ***and***, they provide very fast lookup time. Throughout the rest of these challenges, we will describe several common operations you can perform on objects so you can become comfortable applying these useful data structures in your programs.
-在之前的挑战中，我们已经试过添加和修改对象中的键值对。现在我们来看看如何从一个对象中*移除*一个键值对。
+In earlier challenges, we have both added to and modified an object's key-value pairs. Here we will see how we can *remove* a key-value pair from an object.
-我们再来回顾一下上一个挑战中的 `foods` 对象。如果我们想移除 `apples` 属性，可以像这样使用 `delete` 关键字：
+Let's revisit our `foods` object example one last time. If we wanted to remove the `apples` key, we can remove it by using the `delete` keyword like this:
```js
delete foods.apples;
@@ -20,11 +20,11 @@ delete foods.apples;
# --instructions--
-请使用 `delete` 关键字来移除 `foods` 中的 `oranges`、`plums` 和 `strawberries` 属性。
+Use the delete keyword to remove the `oranges`, `plums`, and `strawberries` keys from the `foods` object.
# --hints--
-`foods` 对象应只包含 3 个属性：`apples`、`grapes` 和 `bananas`。
+The `foods` object should only have three keys: `apples`, `grapes`, and `bananas`.
```js
assert(
@@ -35,7 +35,7 @@ assert(
);
```
-应使用 `delete` 关键字来移除 `oranges`、`plums` 和 `strawberries` 属性。
+The `oranges`, `plums`, and `strawberries` keys should be removed using `delete`.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/access-array-data-with-indexes.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/access-array-data-with-indexes.md
index b5045febb53..0a1b7081087 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/access-array-data-with-indexes.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/access-array-data-with-indexes.md
@@ -1,6 +1,6 @@
---
id: 56bbb991ad1ed5201cd392ca
-title: 通过索引访问数组中的数据
+title: Access Array Data with Indexes
challengeType: 1
videoUrl: 'https://scrimba.com/c/cBZQbTz'
forumTopicId: 16158
@@ -9,28 +9,30 @@ dashedName: access-array-data-with-indexes
# --description--
-我们可以使用索引（`indexes`）来访问数组中的数据。
+We can access the data inside arrays using <dfn>indexes</dfn>.
-数组索引与字符串索引都需要通过方括号表示法使用。区别仅仅在于，字符串通过索引会得到一个字符，而数组通过索引得到的是一个元素。数组索引与字符串索引一样是从 0 开始的，所以数组中第一个元素的索引是 `0`。
+Array indexes are written in the same bracket notation that strings use, except that instead of specifying a character, they are specifying an entry in the array. Like strings, arrays use <dfn>zero-based</dfn> indexing, so the first element in an array has an index of `0`.
-**示例**
+<br>
+
+**Example**
```js
var array = [50,60,70];
-array[0]; // 返回 50
-var data = array[1]; // 值为 60
+array[0]; // equals 50
+var data = array[1]; // equals 60
```
-**注意**
-数组名与方括号之间不应有空格，不推荐像是 `array [0]` 的写法。尽管 JavaScript 能够正确处理，但可能会让其他人感到困惑。
+**Note**
+There shouldn't be any spaces between the array name and the square brackets, like `array [0]`. Although JavaScript is able to process this correctly, this may confuse other programmers reading your code.
# --instructions--
-创建一个名为 `myData` 的变量，并把 `myArray` 的第一个值赋给它。
+Create a variable called `myData` and set it to equal the first value of `myArray` using bracket notation.
# --hints--
-变量 `myData` 的值应为 `myArray` 中的第一个元素值。
+The variable `myData` should equal the first value of `myArray`.
```js
assert(
@@ -48,7 +50,7 @@ assert(
);
```
-应使用方括号表示法访问变量 `myArray` 中的元素。
+The data in variable `myArray` should be accessed using bracket notation.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/access-multi-dimensional-arrays-with-indexes.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/access-multi-dimensional-arrays-with-indexes.md
index 01bb076ab8c..d7ca5ed6411 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/access-multi-dimensional-arrays-with-indexes.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/access-multi-dimensional-arrays-with-indexes.md
@@ -1,6 +1,6 @@
---
id: 56592a60ddddeae28f7aa8e1
-title: 使用索引访问多维数组
+title: Access Multi-Dimensional Arrays With Indexes
challengeType: 1
videoUrl: 'https://scrimba.com/c/ckND4Cq'
forumTopicId: 16159
@@ -9,9 +9,9 @@ dashedName: access-multi-dimensional-arrays-with-indexes
# --description--
-我们可以把<dfn>多维</dfn>数组看作成是*数组中的数组*。使用方括号表示法访问数组时，第一个方括号访问的是数组的第一层，第二个方括号访问的是数组的第二层，以此类推。
+One way to think of a <dfn>multi-dimensional</dfn> array, is as an *array of arrays*. When you use brackets to access your array, the first set of brackets refers to the entries in the outer-most (the first level) array, and each additional pair of brackets refers to the next level of entries inside.
-**示例**
+**Example**
```js
var arr = [
@@ -20,33 +20,30 @@ var arr = [
[7,8,9],
[[10,11,12], 13, 14]
];
-arr[3]; // 返回 [[10,11,12], 13, 14]
-arr[3][0]; // 返回 [10,11,12]
-arr[3][0][1]; // 返回 11
+arr[3]; // equals [[10,11,12], 13, 14]
+arr[3][0]; // equals [10,11,12]
+arr[3][0][1]; // equals 11
```
-**注意**
-数组名与方括号之间不应有空格，不推荐像是 `array [0][0]` 或 `array [0] [0]` 的写法。尽管 JavaScript 能够正确处理，但可能会让其他人感到困惑。
+**Note**
+There shouldn't be any spaces between the array name and the square brackets, like `array [0][0]` and even this `array [0] [0]` is not allowed. Although JavaScript is able to process this correctly, this may confuse other programmers reading your code.
# --instructions--
-使用方括号表示法从 `myArray` 中读取元素，使得 `myData` 的值为 `8`。
+Using bracket notation select an element from `myArray` such that `myData` is equal to `8`.
# --hints--
-`myData` 的值应为 `8`。
+`myData` should be equal to `8`.
```js
assert(myData === 8);
```
-应使用方括号表示法从 `myArray` 中取值。
+You should be using bracket notation to read the correct value from `myArray`.
```js
-assert(
- /myArray\[2\]\[1\]/g.test(code) &&
- !/myData\s*=\s*(?:.*[-+*/%]|\d)/g.test(code)
-);
+assert(/myData=myArray\[2\]\[1\]/.test(__helpers.removeWhiteSpace(code)));
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-nested-arrays.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-nested-arrays.md
index 3121dec2bbd..59d912c1cf2 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-nested-arrays.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-nested-arrays.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244cd
-title: 访问嵌套数组
+title: Accessing Nested Arrays
challengeType: 1
videoUrl: 'https://scrimba.com/c/cLeGDtZ'
forumTopicId: 16160
@@ -9,9 +9,9 @@ dashedName: accessing-nested-arrays
# --description--
-在之前的挑战中，我们学习了在对象中嵌套对象和数组。与访问嵌套的对象一样，我们可以用方括号表示法来访问嵌套数组。
+As we have seen in earlier examples, objects can contain both nested objects and nested arrays. Similar to accessing nested objects, Array bracket notation can be chained to access nested arrays.
-下面是访问嵌套数组的例子：
+Here is an example of how to access a nested array:
```js
var ourPets = [
@@ -38,17 +38,17 @@ ourPets[1].names[0]; // "Spot"
# --instructions--
-请使用点号表示法和方括号表示法来检索变量 `myPlants` 中的第二棵树，即返回 `type` 为 `trees` 的数组中的第二个元素。
+Retrieve the second tree from the variable `myPlants` using object dot and array bracket notation.
# --hints--
-`secondTree` 的值应为 "pine"。
+`secondTree` should equal "pine".
```js
assert(secondTree === 'pine');
```
-应使用点号表示法和方括号表示法来检索变量 `myPlants`。
+Your code should use dot and bracket notation to access `myPlants`.
```js
assert(/=\s*myPlants\[1\].list\[1\]/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-nested-objects.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-nested-objects.md
index 3d1a48c39b1..d8a7470d1f4 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-nested-objects.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-nested-objects.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244cc
-title: 访问嵌套对象
+title: Accessing Nested Objects
challengeType: 1
videoUrl: 'https://scrimba.com/c/cRnRnfa'
forumTopicId: 16161
@@ -9,9 +9,9 @@ dashedName: accessing-nested-objects
# --description--
-我们可以通过连续使用点号表示法和方括号表示法来访问对象的嵌套属性。
+The sub-properties of objects can be accessed by chaining together the dot or bracket notation.
-这是一个访问嵌套对象的示例：
+Here is a nested object:
```js
var ourStorage = {
@@ -32,17 +32,17 @@ ourStorage.desk.drawer; // "stapler"
# --instructions--
-请把 `myStorage` 对象中 `glove box` 的属性值赋给变量 `gloveBoxContents`。请优先使用点号表示法来访问属性；对于点号表示法不能处理的情况，再考虑使用方括号表示法。
+Access the `myStorage` object and assign the contents of the `glove box` property to the `gloveBoxContents` variable. Use dot notation for all properties where possible, otherwise use bracket notation.
# --hints--
-`gloveBoxContents` 的值应为 "maps"。
+`gloveBoxContents` should equal "maps".
```js
assert(gloveBoxContents === 'maps');
```
-应使用点号表示法和方括号表示法来访问 `myStorage`。
+Your code should use dot and bracket notation to access `myStorage`.
```js
assert(/=\s*myStorage\.car\.inside\[\s*("|')glove box\1\s*\]/g.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-object-properties-with-bracket-notation.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-object-properties-with-bracket-notation.md
index 61f4ed13766..a92f2eddcd0 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-object-properties-with-bracket-notation.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-object-properties-with-bracket-notation.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244c8
-title: 通过方括号表示法访问对象属性
+title: Accessing Object Properties with Bracket Notation
challengeType: 1
videoUrl: 'https://scrimba.com/c/cBvmEHP'
forumTopicId: 16163
@@ -9,11 +9,11 @@ dashedName: accessing-object-properties-with-bracket-notation
# --description--
-访问对象的第二种方式是方括号表示法（`[]`）。如果你想访问的属性名中包含空格，就必须使用方括号表示法来获取它的属性值。
+The second way to access the properties of an object is bracket notation (`[]`). If the property of the object you are trying to access has a space in its name, you will need to use bracket notation.
-当然，如果属性名不包含空格，我们也可以使用方括号表示法。
+However, you can still use bracket notation on object properties without spaces.
-这是一个使用方括号表示法（`[]`）读取对象属性的例子：
+Here is a sample of using bracket notation to read an object's property:
```js
var myObj = {
@@ -26,39 +26,39 @@ myObj['More Space']; // Spock
myObj["NoSpace"]; // USS Enterprise
```
-注意，属性名中如果包含空格，就必须把属性名称用单引号或双引号包裹起来。
+Note that property names with spaces in them must be in quotes (single or double).
# --instructions--
-请使用方括号表示法读取对象 `testObj` 中 `an entree` 和 `the drink` 的属性值，并分别赋值给 `entreeValue` 和 `drinkValue`。
+Read the values of the properties `"an entree"` and `"the drink"` of `testObj` using bracket notation and assign them to `entreeValue` and `drinkValue` respectively.
# --hints--
-`entreeValue` 应为一个字符串。
+`entreeValue` should be a string
```js
assert(typeof entreeValue === 'string');
```
-`entreeValue` 的值应为 `"hamburger"`。
+The value of `entreeValue` should be `"hamburger"`
```js
assert(entreeValue === 'hamburger');
```
-`drinkValue` 应为一个字符串。
+`drinkValue` should be a string
```js
assert(typeof drinkValue === 'string');
```
-`drinkValue` 的值应为 `"water"`。
+The value of `drinkValue` should be `"water"`
```js
assert(drinkValue === 'water');
```
-你应使用两次方括号表示法。
+You should use bracket notation twice
```js
assert(code.match(/testObj\s*?\[('|")[^'"]+\1\]/g).length > 1);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-object-properties-with-dot-notation.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-object-properties-with-dot-notation.md
index 3c8eb1bfc5d..af90aafcbc0 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-object-properties-with-dot-notation.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-object-properties-with-dot-notation.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244c7
-title: 通过点号表示法访问对象属性
+title: Accessing Object Properties with Dot Notation
challengeType: 1
videoUrl: 'https://scrimba.com/c/cGryJs8'
forumTopicId: 16164
@@ -9,11 +9,11 @@ dashedName: accessing-object-properties-with-dot-notation
# --description--
-访问对象属性有两种方式：点号表示法（`.`）和方括号表示法（`[]`）。
+There are two ways to access the properties of an object: dot notation (`.`) and bracket notation (`[]`), similar to an array.
-如果我们已经提前知道要访问的属性名，使用点号表示法是最方便的。
+Dot notation is what you use when you know the name of the property you're trying to access ahead of time.
-以下是使用点号表示法读取对象属性的例子：
+Here is a sample of using dot notation (`.`) to read an object's property:
```js
var myObj = {
@@ -26,35 +26,35 @@ var prop2val = myObj.prop2; // val2
# --instructions--
-请使用点号表示法读取对象 `testObj`，把 `hat` 的属性值赋给变量 `hatValue`，把 `shirt` 的属性值赋给 `shirtValue`。
+Read in the property values of `testObj` using dot notation. Set the variable `hatValue` equal to the object's property `hat` and set the variable `shirtValue` equal to the object's property `shirt`.
# --hints--
-`hatValue` 应为一个字符串。
+`hatValue` should be a string
```js
assert(typeof hatValue === 'string');
```
-`hatValue` 的值应为 `"ballcap"`。
+The value of `hatValue` should be `"ballcap"`
```js
assert(hatValue === 'ballcap');
```
-`shirtValue` 应为一个字符串。
+`shirtValue` should be a string
```js
assert(typeof shirtValue === 'string');
```
-`shirtValue` 的值应为 `"jersey"`。
+The value of `shirtValue` should be `"jersey"`
```js
assert(shirtValue === 'jersey');
```
-你应使用两次点号表示法。
+You should use dot notation twice
```js
assert(code.match(/testObj\.\w+/g).length > 1);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-object-properties-with-variables.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-object-properties-with-variables.md
index e888e6820a8..05713220d67 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-object-properties-with-variables.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/accessing-object-properties-with-variables.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244c9
-title: 通过变量访问对象属性
+title: Accessing Object Properties with Variables
challengeType: 1
videoUrl: 'https://scrimba.com/c/cnQyKur'
forumTopicId: 16165
@@ -9,9 +9,9 @@ dashedName: accessing-object-properties-with-variables
# --description--
-方括号表示法的另一个用途就是访问一个存储在变量中的值所对应的属性值。当你需要遍历对象的所有属性，或者根据一个变量的值查找对应的属性值时，这种写法尤其适用。
+Another use of bracket notation on objects is to access a property which is stored as the value of a variable. This can be very useful for iterating through an object's properties or when accessing a lookup table.
-以下是一个使用变量来访问属性的例子：
+Here is an example of using a variable to access a property:
```js
var dogs = {
@@ -22,7 +22,7 @@ var myBreed = dogs[myDog];
console.log(myBreed); // "Doberman"
```
-使用此写法的另一种场景是程序执行期间所动态获取的属性名称，如下所示：
+Another way you can use this concept is when the property's name is collected dynamically during the program execution, as follows:
```js
var someObj = {
@@ -32,49 +32,49 @@ function propPrefix(str) {
var s = "prop";
return s + str;
}
-var someProp = propPrefix("Name"); // someProp 的值现在是 'propName'
+var someProp = propPrefix("Name"); // someProp now holds the value 'propName'
console.log(someObj[someProp]); // "John"
```
-注意，当我们通过变量名访问属性时，不需要给变量名包裹引号。因为我们实际需要获取的是变量的值，而不是变量的名称。
+Note that we do *not* use quotes around the variable name when using it to access the property because we are using the *value* of the variable, not the *name*.
# --instructions--
-首先，请将 `playerNumber` 的值设置为 `16`。然后通过 `playerNumber` 变量，使用方括号表示法获取 `testObj` 中 `16` 的属性值，然后把这个属性值赋给变量 `player`。
+Set the `playerNumber` variable to `16`. Then, use the variable to look up the player's name and assign it to `player`.
# --hints--
-`playerNumber` 应为一个数字。
+`playerNumber` should be a number
```js
assert(typeof playerNumber === 'number');
```
-`player` 应为一个字符串。
+The variable `player` should be a string
```js
assert(typeof player === 'string');
```
-`player` 的值应为 "Montana"。
+The value of `player` should be "Montana"
```js
assert(player === 'Montana');
```
-应使用方括号表示法访问 `testObj`。
+You should use bracket notation to access `testObj`
```js
assert(/testObj\s*?\[.*?\]/.test(code));
```
-不应直接将 `Montana` 赋值给 `player`。
+You should not assign the value `Montana` to the variable `player` directly.
```js
assert(!code.match(/player\s*=\s*"|\'\s*Montana\s*"|\'\s*;/gi));
```
-应在中括号里使用 `playerNumber` 变量。
+You should be using the variable `playerNumber` in your bracket notation
```js
assert(/testObj\s*?\[\s*playerNumber\s*\]/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/add-new-properties-to-a-javascript-object.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/add-new-properties-to-a-javascript-object.md
index eb971fae59e..975b0206be1 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/add-new-properties-to-a-javascript-object.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/add-new-properties-to-a-javascript-object.md
@@ -1,6 +1,6 @@
---
id: 56bbb991ad1ed5201cd392d2
-title: 给对象添加新属性
+title: Add New Properties to a JavaScript Object
challengeType: 1
videoUrl: 'https://scrimba.com/c/cQe38UD'
forumTopicId: 301169
@@ -9,19 +9,19 @@ dashedName: add-new-properties-to-a-javascript-object
# --description--
-你也可以像更改属性一样给对象添加属性。
+You can add new properties to existing JavaScript objects the same way you would modify them.
-我们可以像这样给 `ourDog` 添加 `"bark"` 属性：
+Here's how we would add a `"bark"` property to `ourDog`:
`ourDog.bark = "bow-wow";`
-或者
+or
`ourDog["bark"] = "bow-wow";`
-现在当我们访问 `ourDog.bark` 时会得到它 bark 的属性值 "bow-wow".
+Now when we evaluate `ourDog.bark`, we'll get his bark, "bow-wow".
-代码示例:
+Example:
```js
var ourDog = {
@@ -36,17 +36,17 @@ ourDog.bark = "bow-wow";
# --instructions--
-请给 `myDog` 添加一个 `"bark"` 属性，将它的值设置为狗的叫声，如 "woof"。你可以使用点号表示法或方括号表示法来完成此挑战。
+Add a `"bark"` property to `myDog` and set it to a dog sound, such as "woof". You may use either dot or bracket notation.
# --hints--
-应给 `myDog` 添加 `"bark"` 属性。
+You should add the property `"bark"` to `myDog`.
```js
assert(myDog.bark !== undefined);
```
-不应在初始化 myDog 的时候添加 `"bark"` 属性。
+You should not add `"bark"` to the setup section.
```js
assert(!/bark[^\n]:/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/add-two-numbers-with-javascript.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/add-two-numbers-with-javascript.md
index fb78ac82a83..255cd3a3ed3 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/add-two-numbers-with-javascript.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/add-two-numbers-with-javascript.md
@@ -1,6 +1,6 @@
---
id: cf1111c1c11feddfaeb3bdef
-title: 加法运算
+title: Add Two Numbers with JavaScript
challengeType: 1
videoUrl: 'https://scrimba.com/c/cM2KBAG'
forumTopicId: 16650
@@ -9,31 +9,31 @@ dashedName: add-two-numbers-with-javascript
# --description--
-`Number` 是 JavaScript 中的一种数据类型，用来表示数值。
+`Number` is a data type in JavaScript which represents numeric data.
-现在我们来尝试在 JavaScript 中做加法运算。
+Now let's try to add two numbers using JavaScript.
-JavaScript 中，我们通过符号 `+` 来进行加法运算。
+JavaScript uses the `+` symbol as an addition operator when placed between two numbers.
-**示例：**
+**Example:**
```js
-myVar = 5 + 10; // 赋值 15
+myVar = 5 + 10; // assigned 15
```
# --instructions--
-请改变数字 `0` 让变量 sum 的值为 `20`。
+Change the `0` so that sum will equal `20`.
# --hints--
-`sum` 的值应为 `20`。
+`sum` should equal `20`.
```js
assert(sum === 20);
```
-应使用 `+` 运算符。
+You should use the `+` operator.
```js
assert(/\+/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/adding-a-default-option-in-switch-statements.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/adding-a-default-option-in-switch-statements.md
index d766f8d5ee4..51b62017b09 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/adding-a-default-option-in-switch-statements.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/adding-a-default-option-in-switch-statements.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244de
-title: 在 Switch 语句中添加默认选项
+title: Adding a Default Option in Switch Statements
challengeType: 1
videoUrl: 'https://scrimba.com/c/c3JvVfg'
forumTopicId: 16653
@@ -9,9 +9,9 @@ dashedName: adding-a-default-option-in-switch-statements
# --description--
-在 `switch` 语句中你可能无法用 case 来指明所有情况，这时我们可以添加 default 语句来解决这个问题。在无法找到匹配的 case 时，default 语句就会执行。这非常类似于 if/else 语句中的 else 语句。
+In a `switch` statement you may not be able to specify all possible values as `case` statements. Instead, you can add the `default` statement which will be executed if no matching `case` statements are found. Think of it like the final `else` statement in an `if/else` chain.
-注意，`default` 语句应该在所有的 `case` 之后。
+A `default` statement should be the last case.
```js
switch (num) {
@@ -30,7 +30,7 @@ switch (num) {
# --instructions--
-请实现根据下面的条件来设置 `answer` 的 `switch` 语句：
+Write a switch statement to set `answer` for the following conditions:
`"a"` - "apple"
`"b"` - "bird"
`"c"` - "cat"
@@ -38,49 +38,49 @@ switch (num) {
# --hints--
-`switchOfStuff("a")` 的值应为 "apple"。
+`switchOfStuff("a")` should have a value of "apple"
```js
assert(switchOfStuff('a') === 'apple');
```
-`switchOfStuff("b")` 的值应为 "bird"。
+`switchOfStuff("b")` should have a value of "bird"
```js
assert(switchOfStuff('b') === 'bird');
```
-`switchOfStuff("c")` 的值应为 "cat"。
+`switchOfStuff("c")` should have a value of "cat"
```js
assert(switchOfStuff('c') === 'cat');
```
-`switchOfStuff("d")` 的值应为 "stuff"。
+`switchOfStuff("d")` should have a value of "stuff"
```js
assert(switchOfStuff('d') === 'stuff');
```
-`switchOfStuff("4")` 的值应为 "stuff"。
+`switchOfStuff(4)` should have a value of "stuff"
```js
assert(switchOfStuff(4) === 'stuff');
```
-不应使用 `if` 或 `else` 语句。
+You should not use any `if` or `else` statements
```js
assert(!/else/g.test(code) || !/if/g.test(code));
```
-应有一个 `default` 语句。
+You should use a `default` statement
```js
assert(switchOfStuff('string-to-trigger-default-case') === 'stuff');
```
-应至少有 3 个 `break` 语句。
+You should have at least 3 `break` statements
```js
assert(code.match(/break/g).length > 2);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/appending-variables-to-strings.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/appending-variables-to-strings.md
index 0993d538f9f..be826ab034d 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/appending-variables-to-strings.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/appending-variables-to-strings.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244ed
-title: 将变量追加到字符串
+title: Appending Variables to Strings
challengeType: 1
videoUrl: 'https://scrimba.com/c/cbQmZfa'
forumTopicId: 16656
@@ -9,30 +9,30 @@ dashedName: appending-variables-to-strings
# --description--
-我们不仅可以通过字符串的<dfn>字面量</dfn>创建多行字符串，还可以使用加法赋值运算符（`+=`）来将变量追加到字符串。
+Just as we can build a string over multiple lines out of string <dfn>literals</dfn>, we can also append variables to a string using the plus equals (`+=`) operator.
-代码示例：
+Example:
```js
var anAdjective = "awesome!";
var ourStr = "freeCodeCamp is ";
ourStr += anAdjective;
-// ourStr 的值现在为 "freeCodeCamp is awesome!"
+// ourStr is now "freeCodeCamp is awesome!"
```
# --instructions--
-请将变量 `someAdjective` 的设置为包含至少三个字符的字符串，并使用 `+=` 运算符把它追加到变量 `myStr` 上。
+Set `someAdjective` to a string of at least 3 characters and append it to `myStr` using the `+=` operator.
# --hints--
-`someAdjective` 应为至少包含三个字符的字符串。
+`someAdjective` should be set to a string at least 3 characters long.
```js
assert(typeof someAdjective !== 'undefined' && someAdjective.length > 2);
```
-应使用 `+=` 操作符把 `someAdjective` 追加到 `myStr` 的末尾。
+You should append `someAdjective` to `myStr` using the `+=` operator.
```js
assert(code.match(/myStr\s*\+=\s*someAdjective\s*/).length > 0);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/assigning-the-value-of-one-variable-to-another.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/assigning-the-value-of-one-variable-to-another.md
index 6f8e8278c29..2938f087ee5 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/assigning-the-value-of-one-variable-to-another.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/assigning-the-value-of-one-variable-to-another.md
@@ -1,6 +1,6 @@
---
id: 5ee127a03c3b35dd45426493
-title: 将变量的值赋给另一个变量
+title: Assigning the Value of One Variable to Another
challengeType: 1
videoUrl: ''
forumTopicId: 418265
@@ -9,7 +9,7 @@ dashedName: assigning-the-value-of-one-variable-to-another
# --description--
-使用<dfn>赋值</dfn>运算符将值赋给变量后，我们可以继续使用<dfn>赋值</dfn>运算符将这个变量的值赋给其它变量。
+After a value is assigned to a variable using the <dfn>assignment</dfn> operator, you can assign the value of that variable to another variable using the <dfn>assignment</dfn> operator.
```js
var myVar;
@@ -18,27 +18,27 @@ var myNum;
myNum = myVar;
```
-在上面的代码中，我们声明了没有初始值的 `myVar` 变量，然后给它赋值 `5`。接下来，我们又声明了没有初始值的变量 `myNum`。之后，`myVar` 的值 `5` 被赋给了变量 `myNum`。现在，`myNum` 的值也是 `5`。
+The above declares a `myVar` variable with no value, then assigns it the value `5`. Next, a variable named `myNum` is declared with no value. Then, the contents of `myVar` (which is `5`) is assigned to the variable `myNum`. Now, `myNum` also has the value of `5`.
# --instructions--
-请把 `a` 的值赋给 `b`。
+Assign the contents of `a` to variable `b`.
# --hints--
-不应修改注释上方的代码。
+You should not change code above the specified comment.
```js
assert(/var a;/.test(code) && /a = 7;/.test(code) && /var b;/.test(code));
```
-`b` 的值应为 7。
+`b` should have a value of 7.
```js
assert(typeof b === 'number' && b === 7);
```
-`a` 的值应通过 `=` 运算符赋给 `b`。
+`a` should be assigned to `b` with `=`.
```js
assert(/b\s*=\s*a\s*/g.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/assignment-with-a-returned-value.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/assignment-with-a-returned-value.md
index 3ae6f575fda..15a61d6e5d3 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/assignment-with-a-returned-value.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/assignment-with-a-returned-value.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244c3
-title: 用返回值来赋值
+title: Assignment with a Returned Value
challengeType: 1
videoUrl: 'https://scrimba.com/c/ce2pEtB'
forumTopicId: 16658
@@ -9,30 +9,30 @@ dashedName: assignment-with-a-returned-value
# --description--
-在[使用赋值运算符存储值](/learn/javascript-algorithms-and-data-structures/basic-javascript/storing-values-with-the-assignment-operator)这一挑战中我们曾提到，赋值发生之前会先完成等号右边的操作。这意味着我们可把一个函数的返回值赋给一个变量。
+If you'll recall from our discussion of [Storing Values with the Assignment Operator](/learn/javascript-algorithms-and-data-structures/basic-javascript/storing-values-with-the-assignment-operator), everything to the right of the equal sign is resolved before the value is assigned. This means we can take the return value of a function and assign it to a variable.
-假设我们已经定义了函数 `sum`，它的作用就是将两个数字相加，那么：
+Assume we have pre-defined a function `sum` which adds two numbers together, then:
`ourSum = sum(5, 12);`
-会调用 `sum` 函数，函数返回数值 `17`，然后把它赋值给 `ourSum` 变量。
+will call `sum` function, which returns a value of `17` and assigns it to `ourSum` variable.
# --instructions--
-请调用 `processArg` 函数并传入 `7` 作为参数，然后把返回值赋值给变量 `processed`。
+Call the `processArg` function with an argument of `7` and assign its return value to the variable `processed`.
# --hints--
-`processed` 的值应为 `2`。
+`processed` should have a value of `2`
```js
assert(processed === 2);
```
-应将 `processArg` 的返回值赋给 `processed`。
+You should assign `processArg` to `processed`
```js
-assert(/processed\s*=\s*processArg\(\s*7\s*\)\s*;/.test(code));
+assert(/processed\s*=\s*processArg\(\s*7\s*\)/.test(code));
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/build-javascript-objects.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/build-javascript-objects.md
index 9c881020d6b..9289d9ff584 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/build-javascript-objects.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/build-javascript-objects.md
@@ -1,6 +1,6 @@
---
id: 56bbb991ad1ed5201cd392d0
-title: 创建 JavaScript 对象
+title: Build JavaScript Objects
challengeType: 1
videoUrl: 'https://scrimba.com/c/cWGkbtd'
forumTopicId: 16769
@@ -9,13 +9,13 @@ dashedName: build-javascript-objects
# --description--
-你可能听说过对象（`object`）这个概念。
+You may have heard the term `object` before.
-对象和数组很相似。区别在于，数组是通过索引来访问和修改数据，而对象是通过`属性`来访问和修改数据。
+Objects are similar to `arrays`, except that instead of using indexes to access and modify their data, you access the data in objects through what are called `properties`.
-对象可以让存储的数据结构化，它可以让我们更好地描述真实世界的物体，比如一只猫。
+Objects are useful for storing data in a structured way, and can represent real world objects, like a cat.
-这是一个对象的示例：
+Here's a sample cat object:
```js
var cat = {
@@ -26,7 +26,7 @@ var cat = {
};
```
-在这个示例中，属性都是以字符串的形式储存，例如 `"name"`、`"legs"`、`"tails"`。然而，我们也可以使用数字作为属性，甚至可以省略字符串属性的引号，如下所示：
+In this example, all the properties are stored as strings, such as - `"name"`, `"legs"`, and `"tails"`. However, you can also use numbers as properties. You can even omit the quotes for single-word string properties, as follows:
```js
var anotherObject = {
@@ -36,17 +36,17 @@ var anotherObject = {
};
```
-但是，如果你的对象包含任何非字符串属性，JavaScript 也会将它们自动转换为字符串。
+However, if your object has any non-string properties, JavaScript will automatically typecast them as strings.
# --instructions--
-请创建一个名为 `myDog` 的对象，它应包含这些属性：`"name"`、`"legs"`、`"tails"`、`"friends"`。
+Make an object that represents a dog called `myDog` which contains the properties `"name"` (a string), `"legs"`, `"tails"` and `"friends"`.
-请确保 `"name"` 属性值是字符串、`"legs"` 和 `"tails"` 属性值是数字、`"friends"` 属性值是数组。只要满足这些条件，你可以随意设定这些属性值。
+You can set these object properties to whatever values you want, as long as `"name"` is a string, `"legs"` and `"tails"` are numbers, and `"friends"` is an array.
# --hints--
-`myDog` 应包含 `name` 属性，且属性值是一个字符串。
+`myDog` should contain the property `name` and it should be a `string`.
```js
assert(
@@ -64,7 +64,7 @@ assert(
);
```
-`myDog` 应包含 `legs` 属性，且属性值是一个数字。
+`myDog` should contain the property `legs` and it should be a `number`.
```js
assert(
@@ -82,7 +82,7 @@ assert(
);
```
-`myDog` 应包含 `tails` 属性，且属性值是一个数字。
+`myDog` should contain the property `tails` and it should be a `number`.
```js
assert(
@@ -100,7 +100,7 @@ assert(
);
```
-`myDog` 应包含 `friends` 属性，且属性值是一个数组。
+`myDog` should contain the property `friends` and it should be an `array`.
```js
assert(
@@ -118,7 +118,7 @@ assert(
);
```
-`myDog` 应只包含给定的属性。
+`myDog` should only contain all the given properties.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/chaining-if-else-statements.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/chaining-if-else-statements.md
index 8ceeb4b5e36..f68a2f94dbb 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/chaining-if-else-statements.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/chaining-if-else-statements.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244dc
-title: 多个 if else 语句
+title: Chaining If Else Statements
challengeType: 1
videoUrl: 'https://scrimba.com/c/caeJgsw'
forumTopicId: 16772
@@ -9,7 +9,7 @@ dashedName: chaining-if-else-statements
# --description--
-将 `if/else` 语句串联在一起可以实现复杂的逻辑，这是多个 `if/else if` 语句串联在一起的<dfn>伪代码</dfn>：
+`if/else` statements can be chained together for complex logic. Here is <dfn>pseudocode</dfn> of multiple chained `if` / `else if` statements:
```js
if (condition1) {
@@ -26,89 +26,89 @@ if (condition1) {
# --instructions--
-请将 `if`/`else if` 语句串联起来实现下面的逻辑：
+Write chained `if`/`else if` statements to fulfill the following conditions:
-`num < 5` - 返回 "Tiny"
-`num < 10` - 返回 "Small"
-`num < 15` - 返回 "Medium"
-`num < 20` - 返回 "Large"
-`num >= 20` - 返回 "Huge"
+`num < 5` - return "Tiny"
+`num < 10` - return "Small"
+`num < 15` - return "Medium"
+`num < 20` - return "Large"
+`num >= 20` - return "Huge"
# --hints--
-应至少有 4 个 `else` 语句。
+You should have at least four `else` statements
```js
assert(code.match(/else/g).length > 3);
```
-应至少有 4 个 `if` 语句。
+You should have at least four `if` statements
```js
assert(code.match(/if/g).length > 3);
```
-应至少有 1 个 `return` 语句。
+You should have at least one `return` statement
```js
assert(code.match(/return/g).length >= 1);
```
-`testSize(0)` 应返回 "Tiny"。
+`testSize(0)` should return "Tiny"
```js
assert(testSize(0) === 'Tiny');
```
-`testSize(4)` 应返回 "Tiny"。
+`testSize(4)` should return "Tiny"
```js
assert(testSize(4) === 'Tiny');
```
-`testSize(5)` 应返回 "Small"。
+`testSize(5)` should return "Small"
```js
assert(testSize(5) === 'Small');
```
-`testSize(8)` 应返回 "Small"。
+`testSize(8)` should return "Small"
```js
assert(testSize(8) === 'Small');
```
-`testSize(10)` 应返回 "Medium"。
+`testSize(10)` should return "Medium"
```js
assert(testSize(10) === 'Medium');
```
-`testSize(14)` 应返回 "Medium"。
+`testSize(14)` should return "Medium"
```js
assert(testSize(14) === 'Medium');
```
-`testSize(15)` 应返回 "Large"。
+`testSize(15)` should return "Large"
```js
assert(testSize(15) === 'Large');
```
-`testSize(17)` 应返回 "Large"。
+`testSize(17)` should return "Large"
```js
assert(testSize(17) === 'Large');
```
-`testSize(20)` 应返回 "Huge"。
+`testSize(20)` should return "Huge"
```js
assert(testSize(20) === 'Huge');
```
-`testSize(25)` 应返回 "Huge"。
+`testSize(25)` should return "Huge"
```js
assert(testSize(25) === 'Huge');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comment-your-javascript-code.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comment-your-javascript-code.md
index 98b759a99a0..e828a1ba0e6 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comment-your-javascript-code.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comment-your-javascript-code.md
@@ -1,6 +1,6 @@
---
id: bd7123c9c441eddfaeb4bdef
-title: 给代码添加注释
+title: Comment Your JavaScript Code
challengeType: 1
videoUrl: 'https://scrimba.com/c/c7ynnTp'
forumTopicId: 16783
@@ -9,39 +9,39 @@ dashedName: comment-your-javascript-code
# --description--
-被注释的代码块在 JavaScript 之中是不会执行的。在代码中写注释是一个非常好的方式让你自己和其他人理解代码。
+Comments are lines of code that JavaScript will intentionally ignore. Comments are a great way to leave notes to yourself and to other people who will later need to figure out what that code does.
-JavaScript 中的注释方式有以下两种：
+There are two ways to write comments in JavaScript:
-使用`//`注释掉当前行的代码
+Using `//` will tell JavaScript to ignore the remainder of the text on the current line:
```js
// This is an in-line comment.
```
-你也可以使用多行注释来注释你的代码，以<code>/<em></em></code>*开始，用\`\`*`/`来结束，就像下面这样：
+You can make a multi-line comment beginning with `/*` and ending with `*/`:
```js
/* This is a
multi-line comment */
```
-**最佳实践**
-写代码的时候，要定期添加注释对部分代码块进行解释。适当的注释能让别人和你自己更容易看懂代码。
+**Best Practice**
+As you write code, you should regularly add comments to clarify the function of parts of your code. Good commenting can help communicate the intent of your code—both for others *and* for your future self.
# --instructions--
-尝试创建这两种类型的注释。
+Try creating one of each type of comment.
# --hints--
-创建一个`//`样式的注释, 被注释的文本至少要包含 5 个字符。
+You should create a `//` style comment that contains at least five letters.
```js
assert(code.match(/(\/\/)...../g));
```
-创建一个`/* */`样式的注释, 被注释的文本至少要包含 5 个字符。
+You should create a `/* */` style comment that contains at least five letters.
```js
assert(code.match(/(\/\*)([^\/]{5,})(?=\*\/)/gm));
@@ -50,7 +50,6 @@ assert(code.match(/(\/\*)([^\/]{5,})(?=\*\/)/gm));
# --seed--
## --seed-contents--
-
```js
```
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-equality-operator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-equality-operator.md
index 84d7f42e256..8fd2086992d 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-equality-operator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-equality-operator.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244d0
-title: 相等运算符
+title: Comparison with the Equality Operator
challengeType: 1
videoUrl: 'https://scrimba.com/c/cKyVMAL'
forumTopicId: 16784
@@ -9,9 +9,9 @@ dashedName: comparison-with-the-equality-operator
# --description--
-在 JavaScript 中，有很多<dfn>相互比较的操作</dfn>。所有这些操作符都返回一个`true`或`false`值。
+There are many <dfn>comparison operators</dfn> in JavaScript. All of these operators return a boolean `true` or `false` value.
-最基本的运算符是相等运算符：`==`。相等运算符比较两个值，如果它们是同等，返回`true`，如果它们不等，返回`false`。值得注意的是相等运算符不同于赋值运算符（`=`），赋值运算符是把等号右边的值赋给左边的变量。
+The most basic operator is the equality operator `==`. The equality operator compares two values and returns `true` if they're equivalent or `false` if they are not. Note that equality is different from assignment (`=`), which assigns the value on the right of the operator to a variable on the left.
```js
function equalityTest(myVal) {
@@ -22,7 +22,7 @@ function equalityTest(myVal) {
}
```
-如果`myVal`等于`10`，相等运算符会返回`true`，因此大括号里面的代码会被执行，函数将返回`"Equal"`。否则，函数返回`"Not Equal"`。 在 JavaScript 中，为了让两个不同的`数据类型`（例如`数字`和`字符串`）的值可以作比较，它必须把一种类型转换为另一种类型。然而一旦这样做，它可以像下面这样来比较：
+If `myVal` is equal to `10`, the equality operator returns `true`, so the code in the curly braces will execute, and the function will return `"Equal"`. Otherwise, the function will return `"Not Equal"`. In order for JavaScript to compare two different <dfn>data types</dfn> (for example, `numbers` and `strings`), it must convert one type to another. This is known as "Type Coercion". Once it does, however, it can compare terms as follows:
```js
1 == 1 // true
@@ -33,29 +33,29 @@ function equalityTest(myVal) {
# --instructions--
-把`相等运算符`添加到指定的行，这样当`val`的值为`12`的时候，函数会返回"Equal"。
+Add the equality operator to the indicated line so that the function will return "Equal" when `val` is equivalent to `12`.
# --hints--
-`testEqual(10)`应该返回 "Not Equal"。
+`testEqual(10)` should return "Not Equal"
```js
assert(testEqual(10) === 'Not Equal');
```
-`testEqual(12)`应该返回 "Equal"。
+`testEqual(12)` should return "Equal"
```js
assert(testEqual(12) === 'Equal');
```
-`testEqual("12")`应该返回 "Equal"。
+`testEqual("12")` should return "Equal"
```js
assert(testEqual('12') === 'Equal');
```
-你应该使用`==`运算符。
+You should use the `==` operator
```js
assert(code.match(/==/g) && !code.match(/===/g));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-greater-than-operator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-greater-than-operator.md
index 371271f1a40..840dc21d816 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-greater-than-operator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-greater-than-operator.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244d4
-title: 大于运算符
+title: Comparison with the Greater Than Operator
challengeType: 1
videoUrl: 'https://scrimba.com/c/cp6GbH4'
forumTopicId: 16786
@@ -9,11 +9,11 @@ dashedName: comparison-with-the-greater-than-operator
# --description--
-使用大于运算符（`>`）来比较两个数字。如果大于运算符左边的数字大于右边的数字，将会返回`true`。否则，它返回`false`。
+The greater than operator (`>`) compares the values of two numbers. If the number to the left is greater than the number to the right, it returns `true`. Otherwise, it returns `false`.
-与相等运算符一样，大于运算符在比较的时候，会转换值的数据类型。
+Like the equality operator, greater than operator will convert data types of values while comparing.
-**例如**
+**Examples**
```js
5 > 3 // true
@@ -24,53 +24,53 @@ dashedName: comparison-with-the-greater-than-operator
# --instructions--
-添加`大于`运算符到指定的行，使得返回的语句是有意义的。
+Add the greater than operator to the indicated lines so that the return statements make sense.
# --hints--
-`testGreaterThan(0)`应该返回 "10 or Under"。
+`testGreaterThan(0)` should return "10 or Under"
```js
assert(testGreaterThan(0) === '10 or Under');
```
-`testGreaterThan(10)`应该返回 "10 or Under"。
+`testGreaterThan(10)` should return "10 or Under"
```js
assert(testGreaterThan(10) === '10 or Under');
```
-`testGreaterThan(11)`应该返回 "Over 10"。
+`testGreaterThan(11)` should return "Over 10"
```js
assert(testGreaterThan(11) === 'Over 10');
```
-`testGreaterThan(99)`应该返回 "Over 10"。
+`testGreaterThan(99)` should return "Over 10"
```js
assert(testGreaterThan(99) === 'Over 10');
```
-`testGreaterThan(100)`应该返回 "Over 10"。
+`testGreaterThan(100)` should return "Over 10"
```js
assert(testGreaterThan(100) === 'Over 10');
```
-`testGreaterThan(101)`应该返回 "Over 100"。
+`testGreaterThan(101)` should return "Over 100"
```js
assert(testGreaterThan(101) === 'Over 100');
```
-`testGreaterThan(150)`应该返回 "Over 100"。
+`testGreaterThan(150)` should return "Over 100"
```js
assert(testGreaterThan(150) === 'Over 100');
```
-你应该使用`>`运算符至少两次。
+You should use the `>` operator at least twice
```js
assert(code.match(/val\s*>\s*('|")*\d+('|")*/g).length > 1);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-greater-than-or-equal-to-operator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-greater-than-or-equal-to-operator.md
index 0850dd9d52d..0cb1d1d46a7 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-greater-than-or-equal-to-operator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-greater-than-or-equal-to-operator.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244d5
-title: 大于或等于运算符
+title: Comparison with the Greater Than Or Equal To Operator
challengeType: 1
videoUrl: 'https://scrimba.com/c/c6KBqtV'
forumTopicId: 16785
@@ -9,11 +9,11 @@ dashedName: comparison-with-the-greater-than-or-equal-to-operator
# --description--
-使用`大于等于`运算符（`>=`）来比较两个数字的大小。如果大于等于运算符左边的数字比右边的数字大或者相等，它会返回`true`。否则，它会返回`false`。
+The greater than or equal to operator (`>=`) compares the values of two numbers. If the number to the left is greater than or equal to the number to the right, it returns `true`. Otherwise, it returns `false`.
-与相等运算符相似，`大于等于`运算符在比较的时候会转换值的数据类型。
+Like the equality operator, `greater than or equal to` operator will convert data types while comparing.
-**例如**
+**Examples**
```js
6 >= 6 // true
@@ -24,53 +24,53 @@ dashedName: comparison-with-the-greater-than-or-equal-to-operator
# --instructions--
-添加`大于等于`运算符到指定行，使得函数的返回语句有意义。
+Add the greater than or equal to operator to the indicated lines so that the return statements make sense.
# --hints--
-`testGreaterOrEqual(0)`应该返回 "Less than 10"。
+`testGreaterOrEqual(0)` should return "Less than 10"
```js
assert(testGreaterOrEqual(0) === 'Less than 10');
```
-`testGreaterOrEqual(9)`应该返回 "Less than 10"。
+`testGreaterOrEqual(9)` should return "Less than 10"
```js
assert(testGreaterOrEqual(9) === 'Less than 10');
```
-`testGreaterOrEqual(10)`应该返回 "10 or Over"。
+`testGreaterOrEqual(10)` should return "10 or Over"
```js
assert(testGreaterOrEqual(10) === '10 or Over');
```
-`testGreaterOrEqual(11)`应该返回 "10 or Over"。
+`testGreaterOrEqual(11)` should return "10 or Over"
```js
assert(testGreaterOrEqual(11) === '10 or Over');
```
-`testGreaterOrEqual(19)`应该返回 "10 or Over"。
+`testGreaterOrEqual(19)` should return "10 or Over"
```js
assert(testGreaterOrEqual(19) === '10 or Over');
```
-`testGreaterOrEqual(100)`应该返回 "20 or Over"。
+`testGreaterOrEqual(100)` should return "20 or Over"
```js
assert(testGreaterOrEqual(100) === '20 or Over');
```
-`testGreaterOrEqual(21)`应该返回 "20 or Over"。
+`testGreaterOrEqual(21)` should return "20 or Over"
```js
assert(testGreaterOrEqual(21) === '20 or Over');
```
-你应该使用`>=`运算符至少两次。
+You should use the `>=` operator at least twice
```js
assert(code.match(/val\s*>=\s*('|")*\d+('|")*/g).length > 1);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-inequality-operator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-inequality-operator.md
index 72d077088b4..24d6c5fbec0 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-inequality-operator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-inequality-operator.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244d2
-title: 不等运算符
+title: Comparison with the Inequality Operator
challengeType: 1
videoUrl: 'https://scrimba.com/c/cdBm9Sr'
forumTopicId: 16787
@@ -9,9 +9,9 @@ dashedName: comparison-with-the-inequality-operator
# --description--
-不相等运算符（`!=`）与相等运算符是相反的。这意味着不相等运算符中，如果“不为真”并且返回`false`的地方，在相等运算符中会返回`true`，*反之亦然*。与相等运算符类似，不相等运算符在比较的时候也会转换值的数据类型。
+The inequality operator (`!=`) is the opposite of the equality operator. It means "Not Equal" and returns `false` where equality would return `true` and *vice versa*. Like the equality operator, the inequality operator will convert data types of values while comparing.
-**例如**
+**Examples**
```js
1 != 2 // true
@@ -23,41 +23,41 @@ dashedName: comparison-with-the-inequality-operator
# --instructions--
-在`if`语句中，添加不相等运算符`!=`，这样函数在当`val`不等于 `99`的时候，会返回 "Not Equal"。
+Add the inequality operator `!=` in the `if` statement so that the function will return "Not Equal" when `val` is not equivalent to `99`
# --hints--
-`testNotEqual(99)`应该返回 "Equal"。
+`testNotEqual(99)` should return "Equal"
```js
assert(testNotEqual(99) === 'Equal');
```
-`testNotEqual("99")`应该返回 "Equal"。
+`testNotEqual("99")` should return "Equal"
```js
assert(testNotEqual('99') === 'Equal');
```
-`testNotEqual(12)`应该返回 "Not Equal"。
+`testNotEqual(12)` should return "Not Equal"
```js
assert(testNotEqual(12) === 'Not Equal');
```
-`testNotEqual("12")`应该返回 "Not Equal"。
+`testNotEqual("12")` should return "Not Equal"
```js
assert(testNotEqual('12') === 'Not Equal');
```
-`testNotEqual("bob")`应该返回 "Not Equal"。
+`testNotEqual("bob")` should return "Not Equal"
```js
assert(testNotEqual('bob') === 'Not Equal');
```
-你应该使用`!=`运算符。
+You should use the `!=` operator
```js
assert(code.match(/(?!!==)!=/));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-less-than-operator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-less-than-operator.md
index 7fff8b708a6..5fd6bc8d55f 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-less-than-operator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-less-than-operator.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244d6
-title: 小于运算符
+title: Comparison with the Less Than Operator
challengeType: 1
videoUrl: 'https://scrimba.com/c/cNVRWtB'
forumTopicId: 16789
@@ -9,9 +9,9 @@ dashedName: comparison-with-the-less-than-operator
# --description--
-使用<dfn>小于</dfn>运算符（`<`）比较两个数字的大小。如果小于运算符左边的数字比右边的数字小，它会返回`true`。否则会返回`false`。与相等运算符类似，<dfn>小于</dfn> 运算符在做比较的时候会转换值的数据类型。
+The <dfn>less than</dfn> operator (`<`) compares the values of two numbers. If the number to the left is less than the number to the right, it returns `true`. Otherwise, it returns `false`. Like the equality operator, <dfn>less than</dfn> operator converts data types while comparing.
-**例如**
+**Examples**
```js
2 < 5 // true
@@ -23,47 +23,47 @@ dashedName: comparison-with-the-less-than-operator
# --instructions--
-添加`小于`运算符到指定行，使得函数的返回语句有意义。
+Add the less than operator to the indicated lines so that the return statements make sense.
# --hints--
-`testLessThan(0)`应该返回 "Under 25"。
+`testLessThan(0)` should return "Under 25"
```js
assert(testLessThan(0) === 'Under 25');
```
-`testLessThan(24)`应该返回 "Under 25"。
+`testLessThan(24)` should return "Under 25"
```js
assert(testLessThan(24) === 'Under 25');
```
-`testLessThan(25)`应该返回 "Under 55"。
+`testLessThan(25)` should return "Under 55"
```js
assert(testLessThan(25) === 'Under 55');
```
-`testLessThan(54)`应该返回 "Under 55"。
+`testLessThan(54)` should return "Under 55"
```js
assert(testLessThan(54) === 'Under 55');
```
-`testLessThan(55)`应该返回 "55 or Over"。
+`testLessThan(55)` should return "55 or Over"
```js
assert(testLessThan(55) === '55 or Over');
```
-`testLessThan(99)`应该返回 "55 or Over"。
+`testLessThan(99)` should return "55 or Over"
```js
assert(testLessThan(99) === '55 or Over');
```
-你应该使用`<`运算符至少两次。
+You should use the `<` operator at least twice
```js
assert(code.match(/val\s*<\s*('|")*\d+('|")*/g).length > 1);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-less-than-or-equal-to-operator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-less-than-or-equal-to-operator.md
index e05f7ce8501..5174b635b4f 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-less-than-or-equal-to-operator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-less-than-or-equal-to-operator.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244d7
-title: 小于或等于运算符
+title: Comparison with the Less Than Or Equal To Operator
challengeType: 1
videoUrl: 'https://scrimba.com/c/cNVR7Am'
forumTopicId: 16788
@@ -9,9 +9,9 @@ dashedName: comparison-with-the-less-than-or-equal-to-operator
# --description--
-使用`小于等于`运算符（`<=`）比较两个数字的大小。如果在小于等于运算符左边的数字小于或者等于右边的数字，它会返回`true`。如果在小于等于运算符左边的数字大于右边的数字，它会返回`false`。与相等运算符类似，`小于等于`运算符会转换数据类型。
+The less than or equal to operator (`<=`) compares the values of two numbers. If the number to the left is less than or equal to the number to the right, it returns `true`. If the number on the left is greater than the number on the right, it returns `false`. Like the equality operator, `less than or equal to` converts data types.
-**例如**
+**Examples**
```js
4 <= 5 // true
@@ -23,53 +23,53 @@ dashedName: comparison-with-the-less-than-or-equal-to-operator
# --instructions--
-添加`小于等于`运算符到指定行，使得函数的返回语句有意义。
+Add the less than or equal to operator to the indicated lines so that the return statements make sense.
# --hints--
-`testLessOrEqual(0)`应该返回 "Smaller Than or Equal to 12"。
+`testLessOrEqual(0)` should return "Smaller Than or Equal to 12"
```js
assert(testLessOrEqual(0) === 'Smaller Than or Equal to 12');
```
-`testLessOrEqual(11)`应该返回 "Smaller Than or Equal to 12"。
+`testLessOrEqual(11)` should return "Smaller Than or Equal to 12"
```js
assert(testLessOrEqual(11) === 'Smaller Than or Equal to 12');
```
-`testLessOrEqual(12)`应该返回 "Smaller Than or Equal to 12"。
+`testLessOrEqual(12)` should return "Smaller Than or Equal to 12"
```js
assert(testLessOrEqual(12) === 'Smaller Than or Equal to 12');
```
-`testLessOrEqual(23)`应该返回 "Smaller Than or Equal to 24"。
+`testLessOrEqual(23)` should return "Smaller Than or Equal to 24"
```js
assert(testLessOrEqual(23) === 'Smaller Than or Equal to 24');
```
-`testLessOrEqual(24)`应该返回 "Smaller Than or Equal to 24"。
+`testLessOrEqual(24)` should return "Smaller Than or Equal to 24"
```js
assert(testLessOrEqual(24) === 'Smaller Than or Equal to 24');
```
-`testLessOrEqual(25)`应该返回 "More Than 24"。
+`testLessOrEqual(25)` should return "More Than 24"
```js
assert(testLessOrEqual(25) === 'More Than 24');
```
-`testLessOrEqual(55)`应该返回 "More Than 24"。
+`testLessOrEqual(55)` should return "More Than 24"
```js
assert(testLessOrEqual(55) === 'More Than 24');
```
-你应该使用`<=`运算符至少两。
+You should use the `<=` operator at least twice
```js
assert(code.match(/val\s*<=\s*('|")*\d+('|")*/g).length > 1);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-strict-equality-operator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-strict-equality-operator.md
index e71fd6ab530..b8d73b9aa42 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-strict-equality-operator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-strict-equality-operator.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244d1
-title: 严格相等运算符
+title: Comparison with the Strict Equality Operator
challengeType: 1
videoUrl: 'https://scrimba.com/c/cy87atr'
forumTopicId: 16790
@@ -9,44 +9,44 @@ dashedName: comparison-with-the-strict-equality-operator
# --description--
-严格相等运算符（`===`）是相对相等操作符（`==`）的另一种比较操作符。与相等操作符不同的是，它会同时比较元素的值和`数据类型`。
+Strict equality (`===`) is the counterpart to the equality operator (`==`). However, unlike the equality operator, which attempts to convert both values being compared to a common type, the strict equality operator does not perform a type conversion.
-如果比较的值类型不同，那么在严格相等运算符比较下它们是不相等的，会返回 false 。
+If the values being compared have different types, they are considered unequal, and the strict equality operator will return false.
-**示例**
+**Examples**
```js
3 === 3 // true
3 === '3' // false
```
-`3`是一个`数字`类型的，而`'3'`是一个`字符串`类型的，所以 3 不全等于 '3'。
+In the second example, `3` is a `Number` type and `'3'` is a `String` type.
# --instructions--
-在`if`语句值使用严格相等运算符，这样当`val`严格等于7的时候，函数会返回"Equal"。
+Use the strict equality operator in the `if` statement so the function will return "Equal" when `val` is strictly equal to `7`
# --hints--
-`testStrict(10)`应该返回 "Not Equal"。
+`testStrict(10)` should return "Not Equal"
```js
assert(testStrict(10) === 'Not Equal');
```
-`testStrict(7)`应该返回 "Equal"。
+`testStrict(7)` should return "Equal"
```js
assert(testStrict(7) === 'Equal');
```
-`testStrict("7")`应该返回 "Not Equal"。
+`testStrict("7")` should return "Not Equal"
```js
assert(testStrict('7') === 'Not Equal');
```
-你应该使用`===`运算符。
+You should use the `===` operator
```js
assert(code.match(/(val\s*===\s*\d+)|(\d+\s*===\s*val)/g).length > 0);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-strict-inequality-operator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-strict-inequality-operator.md
index efd8c067a8d..9d4b49785c4 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-strict-inequality-operator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-strict-inequality-operator.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244d3
-title: 严格不等运算符
+title: Comparison with the Strict Inequality Operator
challengeType: 1
videoUrl: 'https://scrimba.com/c/cKekkUy'
forumTopicId: 16791
@@ -9,9 +9,9 @@ dashedName: comparison-with-the-strict-inequality-operator
# --description--
-严格不相等运算符（`!==`）与全等运算符是相反的。这意味着严格不相等并返回`false`的地方，用严格相等运算符会返回`true`，*反之亦然*。严格不相等运算符不会转换值的数据类型。
+The strict inequality operator (`!==`) is the logical opposite of the strict equality operator. It means "Strictly Not Equal" and returns `false` where strict equality would return `true` and *vice versa*. Strict inequality will not convert data types.
-**示例**
+**Examples**
```js
3 !== 3 // false
@@ -21,35 +21,35 @@ dashedName: comparison-with-the-strict-inequality-operator
# --instructions--
-在`if`语句中，添加严格不相等运算符`!==`，这样如果`val`与`17`严格不相等的时候，函数会返回 "Not Equal"。
+Add the strict inequality operator to the `if` statement so the function will return "Not Equal" when `val` is not strictly equal to `17`
# --hints--
-`testStrictNotEqual(17)`应该返回 "Equal"。
+`testStrictNotEqual(17)` should return "Equal"
```js
assert(testStrictNotEqual(17) === 'Equal');
```
-`testStrictNotEqual("17")`应该返回 "Not Equal"。
+`testStrictNotEqual("17")` should return "Not Equal"
```js
assert(testStrictNotEqual('17') === 'Not Equal');
```
-`testStrictNotEqual(12)`应该返回 "Not Equal"。
+`testStrictNotEqual(12)` should return "Not Equal"
```js
assert(testStrictNotEqual(12) === 'Not Equal');
```
-`testStrictNotEqual("bob")`应该返回 "Not Equal"。
+`testStrictNotEqual("bob")` should return "Not Equal"
```js
assert(testStrictNotEqual('bob') === 'Not Equal');
```
-应该使用 `!==` 运算符。
+You should use the `!==` operator
```js
assert(code.match(/(val\s*!==\s*\d+)|(\d+\s*!==\s*val)/g).length > 0);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparisons-with-the-logical-and-operator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparisons-with-the-logical-and-operator.md
index 2cd3ea70874..bf9c2a6c1ae 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparisons-with-the-logical-and-operator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparisons-with-the-logical-and-operator.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244d8
-title: 逻辑与运算符
+title: Comparisons with the Logical And Operator
challengeType: 1
videoUrl: 'https://scrimba.com/c/cvbRVtr'
forumTopicId: 16799
@@ -9,9 +9,9 @@ dashedName: comparisons-with-the-logical-and-operator
# --description--
-有时你需要在一次判断中做多个操作。当且仅当运算符的左边和右边都是`true`，<dfn>逻辑与</dfn> 运算符（`&&`）才会返回`true`。
+Sometimes you will need to test more than one thing at a time. The <dfn>logical and</dfn> operator (`&&`) returns `true` if and only if the <dfn>operands</dfn> to the left and right of it are true.
-同样的效果可以通过 if 语句的嵌套来实现：
+The same effect could be achieved by nesting an if statement inside another if:
```js
if (num > 5) {
@@ -22,7 +22,7 @@ if (num > 5) {
return "No";
```
-只有当`num`的值在 6 和 9 之间（包括 6 和 9）才会返回 "Yes"。相同的逻辑可被写为：
+will only return "Yes" if `num` is greater than `5` and less than `10`. The same logic can be written as:
```js
if (num > 5 && num < 10) {
@@ -33,65 +33,65 @@ return "No";
# --instructions--
-请使用逻辑与运算符把两个 if 语句合并为一个 if 语句，如果`val`小于或等于`50`并且大于或等于`25`，返回`"Yes"`。否则，将返回`"No"`。
+Replace the two if statements with one statement, using the && operator, which will return `"Yes"` if `val` is less than or equal to `50` and greater than or equal to `25`. Otherwise, will return `"No"`.
# --hints--
-你应该使用`&&`运算符一次。
+You should use the `&&` operator once
```js
assert(code.match(/&&/g).length === 1);
```
-你应该只有一个`if`表达式。
+You should only have one `if` statement
```js
assert(code.match(/if/g).length === 1);
```
-`testLogicalAnd(0)`应该返回 "No"。
+`testLogicalAnd(0)` should return "No"
```js
assert(testLogicalAnd(0) === 'No');
```
-`testLogicalAnd(24)`应该返回 "No"。
+`testLogicalAnd(24)` should return "No"
```js
assert(testLogicalAnd(24) === 'No');
```
-`testLogicalAnd(25)`应该返回 "Yes"。
+`testLogicalAnd(25)` should return "Yes"
```js
assert(testLogicalAnd(25) === 'Yes');
```
-`testLogicalAnd(30)`应该返回 "Yes"。
+`testLogicalAnd(30)` should return "Yes"
```js
assert(testLogicalAnd(30) === 'Yes');
```
-`testLogicalAnd(50)`应该返回 "Yes"。
+`testLogicalAnd(50)` should return "Yes"
```js
assert(testLogicalAnd(50) === 'Yes');
```
-`testLogicalAnd(51)`应该返回 "No"。
+`testLogicalAnd(51)` should return "No"
```js
assert(testLogicalAnd(51) === 'No');
```
-`testLogicalAnd(75)`应该返回 "No"。
+`testLogicalAnd(75)` should return "No"
```js
assert(testLogicalAnd(75) === 'No');
```
-`testLogicalAnd(80)`应该返回 "No"。
+`testLogicalAnd(80)` should return "No"
```js
assert(testLogicalAnd(80) === 'No');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparisons-with-the-logical-or-operator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparisons-with-the-logical-or-operator.md
index 2aabcbf5803..cd5d09d36f9 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparisons-with-the-logical-or-operator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/comparisons-with-the-logical-or-operator.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244d9
-title: 逻辑或运算符
+title: Comparisons with the Logical Or Operator
challengeType: 1
videoUrl: 'https://scrimba.com/c/cEPrGTN'
forumTopicId: 16800
@@ -9,11 +9,11 @@ dashedName: comparisons-with-the-logical-or-operator
# --description--
-只要<dfn>逻辑或</dfn>运算符`||`两边任何一个为`true`，那么它就返回`true`；否则返回`false`。
+The <dfn>logical or</dfn> operator (`||`) returns `true` if either of the <dfn>operands</dfn> is `true`. Otherwise, it returns `false`.
-<dfn>逻辑或</dfn>运算符由两个管道符号（|）组成。这个按键位于退格键和回车键之间。
+The <dfn>logical or</dfn> operator is composed of two pipe symbols: (`||`). This can typically be found between your Backspace and Enter keys.
-下面这样的语句你应该很熟悉：
+The pattern below should look familiar from prior waypoints:
```js
if (num > 10) {
@@ -25,7 +25,7 @@ if (num < 5) {
return "Yes";
```
-只有当`num`大于等于 5 或小于等于 10 时，函数返回"Yes"。相同的逻辑可以简写成：
+will return "Yes" only if `num` is between `5` and `10` (5 and 10 included). The same logic can be written as:
```js
if (num > 10 || num < 5) {
@@ -36,65 +36,65 @@ return "Yes";
# --instructions--
-请使用逻辑或运算符把两个 if 语句合并为一个 if 语句，如果`val`不在 10 和 20 之间(包括 10 和 20)，返回`"Outside"`。反之，返回`"Inside"`。
+Combine the two `if` statements into one statement which returns `"Outside"` if `val` is not between `10` and `20`, inclusive. Otherwise, return `"Inside"`.
# --hints--
-你应该使用一次`||`操作符。
+You should use the `||` operator once
```js
assert(code.match(/\|\|/g).length === 1);
```
-你应该只有一个`if`表达式。
+You should only have one `if` statement
```js
assert(code.match(/if/g).length === 1);
```
-`testLogicalOr(0)`应该返回 "Outside"。
+`testLogicalOr(0)` should return "Outside"
```js
assert(testLogicalOr(0) === 'Outside');
```
-`testLogicalOr(9)`应该返回 "Outside"。
+`testLogicalOr(9)` should return "Outside"
```js
assert(testLogicalOr(9) === 'Outside');
```
-`testLogicalOr(10)`应该返回 "Inside"。
+`testLogicalOr(10)` should return "Inside"
```js
assert(testLogicalOr(10) === 'Inside');
```
-`testLogicalOr(15)`应该返回 "Inside"。
+`testLogicalOr(15)` should return "Inside"
```js
assert(testLogicalOr(15) === 'Inside');
```
-`testLogicalOr(19)`应该返回 "Inside"。
+`testLogicalOr(19)` should return "Inside"
```js
assert(testLogicalOr(19) === 'Inside');
```
-`testLogicalOr(20)`应该返回 "Inside"。
+`testLogicalOr(20)` should return "Inside"
```js
assert(testLogicalOr(20) === 'Inside');
```
-`testLogicalOr(21)`应该返回 "Outside"。
+`testLogicalOr(21)` should return "Outside"
```js
assert(testLogicalOr(21) === 'Outside');
```
-`testLogicalOr(25)`应该返回 "Outside"。
+`testLogicalOr(25)` should return "Outside"
```js
assert(testLogicalOr(25) === 'Outside');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/compound-assignment-with-augmented-addition.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/compound-assignment-with-augmented-addition.md
index 92756ad3687..0a2ab049922 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/compound-assignment-with-augmented-addition.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/compound-assignment-with-augmented-addition.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244af
-title: 复合赋值之 +=
+title: Compound Assignment With Augmented Addition
challengeType: 1
videoUrl: 'https://scrimba.com/c/cDR6LCb'
forumTopicId: 16661
@@ -9,13 +9,13 @@ dashedName: compound-assignment-with-augmented-addition
# --description--
-在编程当中，通常通过赋值来修改变量的内容。记住，赋值时 Javascript 会先计算`=`右边的内容，所以我们可以写这样的语句：
+In programming, it is common to use assignments to modify the contents of a variable. Remember that everything to the right of the equals sign is evaluated first, so we can say:
`myVar = myVar + 5;`
-以上是最常见的运算赋值语句，即先运算、再赋值。还有一类操作符是一步到位既做运算也赋值的。
+to add `5` to `myVar`. Since this is such a common pattern, there are operators which do both a mathematical operation and assignment in one step.
-其中一种就是`+=`运算符。
+One such operator is the `+=` operator.
```js
var myVar = 1;
@@ -25,35 +25,35 @@ console.log(myVar); // Returns 6
# --instructions--
-使用`+=`操作符实现同样的效果。
+Convert the assignments for `a`, `b`, and `c` to use the `+=` operator.
# --hints--
-`a`应该等于`15`。
+`a` should equal `15`.
```js
assert(a === 15);
```
-`b`应该等于`26`。
+`b` should equal `26`.
```js
assert(b === 26);
```
-`c`应该等于`19`。
+`c` should equal `19`.
```js
assert(c === 19);
```
-你应该对每个变量使用`+=`操作符。
+You should use the `+=` operator for each variable.
```js
assert(code.match(/\+=/g).length === 3);
```
-不要修改注释上面的代码。
+You should not modify the code above the specified comment.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/compound-assignment-with-augmented-division.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/compound-assignment-with-augmented-division.md
index 17a37bee900..92b7d49cb42 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/compound-assignment-with-augmented-division.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/compound-assignment-with-augmented-division.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244b2
-title: 复合赋值之 /=
+title: Compound Assignment With Augmented Division
challengeType: 1
videoUrl: 'https://scrimba.com/c/c2QvKT2'
forumTopicId: 16659
@@ -9,45 +9,45 @@ dashedName: compound-assignment-with-augmented-division
# --description--
-`/=`操作符是让变量与另一个数相除并赋值。
+The `/=` operator divides a variable by another number.
`myVar = myVar / 5;`
-变量`myVar`等于自身除以`5`的值。等价于:
+Will divide `myVar` by `5`. This can be rewritten as:
`myVar /= 5;`
# --instructions--
-使用`/=`操作符实现同样的效果。
+Convert the assignments for `a`, `b`, and `c` to use the `/=` operator.
# --hints--
-`a`应该等于`4`。
+`a` should equal `4`.
```js
assert(a === 4);
```
-`b`应该等于`27`。
+`b` should equal `27`.
```js
assert(b === 27);
```
-`c`应该等于`3`。
+`c` should equal `3`.
```js
assert(c === 3);
```
-应该对每个变量使用`/=`操作符。
+You should use the `/=` operator for each variable.
```js
assert(code.match(/\/=/g).length === 3);
```
-不要修改注释上面的代码。
+You should not modify the code above the specified comment.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/compound-assignment-with-augmented-multiplication.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/compound-assignment-with-augmented-multiplication.md
index 9c09c9e26f2..c05680ba89d 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/compound-assignment-with-augmented-multiplication.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/compound-assignment-with-augmented-multiplication.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244b1
-title: 复合赋值之 *=
+title: Compound Assignment With Augmented Multiplication
challengeType: 1
videoUrl: 'https://scrimba.com/c/c83vrfa'
forumTopicId: 16662
@@ -9,45 +9,45 @@ dashedName: compound-assignment-with-augmented-multiplication
# --description--
-`*=`操作符是让变量与一个数相乘并赋值。
+The `*=` operator multiplies a variable by a number.
`myVar = myVar * 5;`
-变量`myVar`等于自身与数值`5`相乘的值。也可以写作这样的形式:
+will multiply `myVar` by `5`. This can be rewritten as:
`myVar *= 5;`
# --instructions--
-使用`*=`操作符实现同样的效果。
+Convert the assignments for `a`, `b`, and `c` to use the `*=` operator.
# --hints--
-`a`应该等于`25`。
+`a` should equal `25`.
```js
assert(a === 25);
```
-`b`应该等于`36`。
+`b` should equal `36`.
```js
assert(b === 36);
```
-`c`应该等于`46`。
+`c` should equal `46`.
```js
assert(c === 46);
```
-应该对每个变量使用`*=`操作符。
+You should use the `*=` operator for each variable.
```js
assert(code.match(/\*=/g).length === 3);
```
-不要修改注释上面的代码。
+You should not modify the code above the specified comment.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/compound-assignment-with-augmented-subtraction.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/compound-assignment-with-augmented-subtraction.md
index c976c37bfd5..add12120cee 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/compound-assignment-with-augmented-subtraction.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/compound-assignment-with-augmented-subtraction.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244b0
-title: 复合赋值之 -=
+title: Compound Assignment With Augmented Subtraction
challengeType: 1
videoUrl: 'https://scrimba.com/c/c2Qv7AV'
forumTopicId: 16660
@@ -9,45 +9,45 @@ dashedName: compound-assignment-with-augmented-subtraction
# --description--
-与`+=`操作符类似，`-=`操作符用来对一个变量进行减法赋值操作。
+Like the `+=` operator, `-=` subtracts a number from a variable.
`myVar = myVar - 5;`
-变量`myVar`等于自身减去`5`的值。也可以写成这种形式：
+will subtract `5` from `myVar`. This can be rewritten as:
`myVar -= 5;`
# --instructions--
-使用`-=`操作符实现同样的效果。
+Convert the assignments for `a`, `b`, and `c` to use the `-=` operator.
# --hints--
-`a`应该等于`5`。
+`a` should equal `5`.
```js
assert(a === 5);
```
-`b`应该等于`-6`。
+`b` should equal `-6`.
```js
assert(b === -6);
```
-`c`应该等于`2`。
+`c` should equal `2`.
```js
assert(c === 2);
```
-应该对每个变量使用`-=`操作符。
+You should use the `-=` operator for each variable.
```js
assert(code.match(/-=/g).length === 3);
```
-不要修改注释上面的代码。
+You should not modify the code above the specified comment.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/concatenating-strings-with-plus-operator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/concatenating-strings-with-plus-operator.md
index 9da4270047b..ec94ec73ed2 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/concatenating-strings-with-plus-operator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/concatenating-strings-with-plus-operator.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244b7
-title: 用加号运算符连接字符串
+title: Concatenating Strings with Plus Operator
challengeType: 1
videoUrl: 'https://scrimba.com/c/cNpM8AN'
forumTopicId: 16802
@@ -9,42 +9,49 @@ dashedName: concatenating-strings-with-plus-operator
# --description--
-在 JavaScript 中，当对一个`String`类型的值使用`+`操作符的时候，它被称作 <dfn>拼接操作符</dfn>。你可以通过<dfn>拼接</dfn>其他字符串来创建一个新的字符串。
+In JavaScript, when the `+` operator is used with a `String` value, it is called the <dfn>concatenation</dfn> operator. You can build a new string out of other strings by <dfn>concatenating</dfn> them together.
-**示例**
+**Example**
```js
'My name is Alan,' + ' I concatenate.'
```
-**提示**
-注意空格。拼接操作不会在两个字符串之间添加空格，所以想加上空格的话，你需要自己在字符串里面添加。
+**Note**
+Watch out for spaces. Concatenation does not add spaces between concatenated strings, so you'll need to add them yourself.
+
+Example:
+
+```js
+var ourStr = "I come first. " + "I come second.";
+// ourStr is "I come first. I come second."
+```
# --instructions--
-使用`+`操作符，把字符串`"This is the start. "`和`"This is the end."`连接起来并赋值给变量`myStr`。
+Build `myStr` from the strings `"This is the start. "` and `"This is the end."` using the `+` operator.
# --hints--
-`myStr`的值应该是`This is the start. This is the end.`。
+`myStr` should have a value of `This is the start. This is the end.`
```js
assert(myStr === 'This is the start. This is the end.');
```
-使用`+`操作符构建`myStr`。
+You should use the `+` operator to build `myStr`.
```js
-assert(code.match(/(["']).*(["'])\s*\+\s*(["']).*(["'])/g).length > 1);
+assert(code.match(/(["']).*\1\s*\+\s*(["']).*\2/g));
```
-`myStr`应该被`var`关键字声明。
+`myStr` should be created using the `var` keyword.
```js
assert(/var\s+myStr/.test(code));
```
-确保有给`myStr`赋值。
+You should assign the result to the `myStr` variable.
```js
assert(/myStr\s*=/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/concatenating-strings-with-the-plus-equals-operator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/concatenating-strings-with-the-plus-equals-operator.md
index eaad41dcc5e..e1bf7235458 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/concatenating-strings-with-the-plus-equals-operator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/concatenating-strings-with-the-plus-equals-operator.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244b8
-title: 用 += 运算符连接字符串
+title: Concatenating Strings with the Plus Equals Operator
challengeType: 1
videoUrl: 'https://scrimba.com/c/cbQmmC4'
forumTopicId: 16803
@@ -9,31 +9,35 @@ dashedName: concatenating-strings-with-the-plus-equals-operator
# --description--
-我们还可以使用`+=`运算符来<dfn>concatenate</dfn>（拼接）字符串到现有字符串的结尾。对于那些被分割成几段的长的字符串来说，这一操作是非常有用的。
+We can also use the `+=` operator to <dfn>concatenate</dfn> a string onto the end of an existing string variable. This can be very helpful to break a long string over several lines.
-**提示**
-注意空格。拼接操作不会在两个字符串之间添加空格，所以如果想要加上空格的话，你需要自己在字符串里面添加。
+**Note**
+Watch out for spaces. Concatenation does not add spaces between concatenated strings, so you'll need to add them yourself.
+
+Example:
+
+```js
+var ourStr = "I come first. ";
+ourStr += "I come second.";
+// ourStr is now "I come first. I come second."
+```
# --instructions--
-通过使用`+=`操作符来连接这两个字符串：
-`"This is the first sentence. "`和`"This is the second sentence."`并赋给变量`myStr`。
+Build `myStr` over several lines by concatenating these two strings: `"This is the first sentence. "` and `"This is the second sentence."` using the `+=` operator. Use the `+=` operator similar to how it is shown in the editor. Start by assigning the first string to `myStr`, then add on the second string.
# --hints--
-`myStr`的值应该是`This is the first sentence. This is the second sentence.`。
+`myStr` should have a value of `This is the first sentence. This is the second sentence.`
```js
assert(myStr === 'This is the first sentence. This is the second sentence.');
```
-使用`+=`操作符创建`myStr`变量。
+You should use the `+=` operator to build `myStr`.
```js
-assert(
- code.match(/\w\s*\+=\s*["']/g).length > 1 &&
- code.match(/\w\s*\=\s*["']/g).length > 1
-);
+assert(code.match(/myStr\s*\+=\s*(["']).*\1/g));
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/constructing-strings-with-variables.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/constructing-strings-with-variables.md
index b21f215d53c..183aa48865e 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/constructing-strings-with-variables.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/constructing-strings-with-variables.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244b9
-title: 用变量构造字符串
+title: Constructing Strings with Variables
challengeType: 1
videoUrl: 'https://scrimba.com/c/cqk8rf4'
forumTopicId: 16805
@@ -9,21 +9,29 @@ dashedName: constructing-strings-with-variables
# --description--
-有时候你需要创建一个类似[Mad Libs](https://en.wikipedia.org/wiki/Mad_Libs)(填词游戏）风格的字符串。通过使用连接运算符`+`，你可以插入一个或多个变量来组成一个字符串。
+Sometimes you will need to build a string, [Mad Libs](https://en.wikipedia.org/wiki/Mad_Libs) style. By using the concatenation operator (`+`), you can insert one or more variables into a string you're building.
+
+Example:
+
+```js
+var ourName = "freeCodeCamp";
+var ourStr = "Hello, our name is " + ourName + ", how are you?";
+// ourStr is now "Hello, our name is freeCodeCamp, how are you?"
+```
# --instructions--
-把你的名字赋值给变量`myName`，然后把变量`myName`插入到字符串`"My name is "`和`" and I am well!"`之间，并把连接后的结果赋值给变量`myStr`。
+Set `myName` to a string equal to your name and build `myStr` with `myName` between the strings `"My name is "` and `" and I am well!"`
# --hints--
-`myName`至少要包含三个字符。
+`myName` should be set to a string at least 3 characters long.
```js
assert(typeof myName !== 'undefined' && myName.length > 2);
```
-使用两个`+`操作符创建包含`myName`的`myStr`变量。
+You should use two `+` operators to build `myStr` with `myName` inside it.
```js
assert(code.match(/["']\s*\+\s*myName\s*\+\s*["']/g).length > 0);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/count-backwards-with-a-for-loop.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/count-backwards-with-a-for-loop.md
index 551a829be07..9edc5f1cb99 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/count-backwards-with-a-for-loop.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/count-backwards-with-a-for-loop.md
@@ -1,6 +1,6 @@
---
id: 56105e7b514f539506016a5e
-title: 使用 For 循环反向遍历数组
+title: Count Backwards With a For Loop
challengeType: 1
videoUrl: 'https://scrimba.com/c/c2R6BHa'
forumTopicId: 16808
@@ -9,40 +9,40 @@ dashedName: count-backwards-with-a-for-loop
# --description--
-for循环也可以逆向迭代，只要我们定义好合适的条件。
+A for loop can also count backwards, so long as we can define the right conditions.
-为了让每次倒数递减 2，我们需要改变我们的`初始化`，`条件判断`和`计数器`。
+In order to decrement by two each iteration, we'll need to change our `initialization`, `condition`, and `final-expression`.
-我们让`i = 10`，并且当`i > 0`的时候才继续循环。我们使用`i -= 2`来让`i`每次循环递减 2。
+We'll start at `i = 10` and loop while `i > 0`. We'll decrement `i` by 2 each loop with `i -= 2`.
```js
var ourArray = [];
-for (var i=10; i > 0; i-=2) {
+for (var i = 10; i > 0; i -= 2) {
ourArray.push(i);
}
```
-循环结束后，`ourArray`的值为`[10,8,6,4,2]`。 让我们改变`初始化`和`计数器`，这样我们就可以按照奇数从后往前两两倒着数。
+`ourArray` will now contain `[10,8,6,4,2]`. Let's change our `initialization` and `final-expression` so we can count backward by twos by odd numbers.
# --instructions--
-使用一个`for`循环，把 9 到 1 的奇数添加进`myArray`。
+Push the odd numbers from 9 through 1 to `myArray` using a `for` loop.
# --hints--
-你应该使用`for`循环。
+You should be using a `for` loop for this.
```js
-assert(code.match(/for\s*\(/g).length > 1);
+assert(/for\s*\([^)]+?\)/.test(code));
```
-你应该使用数组方法`push`。
+You should be using the array method `push`.
```js
assert(code.match(/myArray.push/));
```
-`myArray`应该等于`[9,7,5,3,1]`。
+`myArray` should equal `[9,7,5,3,1]`.
```js
assert.deepEqual(myArray, [9, 7, 5, 3, 1]);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/counting-cards.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/counting-cards.md
index ba553d29a07..4d98fdecfba 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/counting-cards.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/counting-cards.md
@@ -1,6 +1,6 @@
---
id: 565bbe00e9cc8ac0725390f4
-title: 21点游戏
+title: Counting Cards
challengeType: 1
videoUrl: 'https://scrimba.com/c/c6KE7ty'
forumTopicId: 16809
@@ -9,22 +9,25 @@ dashedName: counting-cards
# --description--
-在赌场 21 点游戏中，玩家可以通过计算牌桌上已经发放的卡牌的高低值来让自己在游戏中保持优势，这就叫[21 点算法](https://www.douban.com/note/273781969/)。
+In the casino game Blackjack, a player can gain an advantage over the house by keeping track of the relative number of high and low cards remaining in the deck. This is called [Card Counting](https://en.wikipedia.org/wiki/Card_counting).
-根据下面的表格，每张卡牌都分配了一个值。如果卡牌的值大于 0，那么玩家应该追加赌注。反之，追加少许赌注甚至不追加赌注。
+Having more high cards remaining in the deck favors the player. Each card is assigned a value according to the table below. When the count is positive, the player should bet high. When the count is zero or negative, the player should bet low.
<table class='table table-striped'><thead><tr><th>Count Change</th><th>Cards</th></tr></thead><tbody><tr><td>+1</td><td>2, 3, 4, 5, 6</td></tr><tr><td>0</td><td>7, 8, 9</td></tr><tr><td>-1</td><td>10, 'J', 'Q', 'K', 'A'</td></tr></tbody></table>
-你需要写一个函数实现 21 点算法，它根据参数<code>card</code>的值来递增或递减变量<code>count</code>，函数返回一个由当前<code>count</code>和<code>Bet</code>（<code>count>0</code>）或<code>Hold</code>（<code>count<=0</code>）拼接的字符串。注意<code>count</code>和<code>"Bet"</code>或<code>Hold</code>应该用空格分开。</count>
-**例如：**
-<code>-3 Hold<br>5 Bet</code>
+You will write a card counting function. It will receive a `card` parameter, which can be a number or a string, and increment or decrement the global `count` variable according to the card's value (see table). The function will then return a string with the current count and the string `Bet` if the count is positive, or `Hold` if the count is zero or negative. The current count and the player's decision (`Bet` or `Hold`) should be separated by a single space.
-**提示**
-既然 card 的值为 7、8、9 时，count 值不变，那我们就可以忽略这种情况。
+**Example Output**
+`-3 Hold`
+`5 Bet`
+
+**Hint**
+Do NOT reset `count` to 0 when value is 7, 8, or 9. Do NOT return an array.
+Do NOT include quotes (single or double) in the output.
# --hints--
-Cards Sequence 2, 3, 4, 5, 6 应该返回`5 Bet`。
+Cards Sequence 2, 3, 4, 5, 6 should return `5 Bet`
```js
assert(
@@ -43,7 +46,7 @@ assert(
);
```
-Cards Sequence 7, 8, 9 应该返回 `0 Hold`。
+Cards Sequence 7, 8, 9 should return `0 Hold`
```js
assert(
@@ -60,7 +63,7 @@ assert(
);
```
-Cards Sequence 10, J, Q, K, A 应该返回 `-5 Hold`。
+Cards Sequence 10, J, Q, K, A should return `-5 Hold`
```js
assert(
@@ -79,7 +82,7 @@ assert(
);
```
-Cards Sequence 3, 7, Q, 8, A 应该返回 `-1 Hold`。
+Cards Sequence 3, 7, Q, 8, A should return `-1 Hold`
```js
assert(
@@ -98,7 +101,7 @@ assert(
);
```
-Cards Sequence 2, J, 9, 2, 7 应该返回 `1 Bet`。
+Cards Sequence 2, J, 9, 2, 7 should return `1 Bet`
```js
assert(
@@ -117,7 +120,7 @@ assert(
);
```
-Cards Sequence 2, 2, 10 应该返回 `1 Bet`。
+Cards Sequence 2, 2, 10 should return `1 Bet`
```js
assert(
@@ -134,7 +137,7 @@ assert(
);
```
-Cards Sequence 3, 2, A, 10, K 应该返回 `-1 Hold`。
+Cards Sequence 3, 2, A, 10, K should return `-1 Hold`
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/create-decimal-numbers-with-javascript.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/create-decimal-numbers-with-javascript.md
index 3edc704f7e1..de35aa22d38 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/create-decimal-numbers-with-javascript.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/create-decimal-numbers-with-javascript.md
@@ -1,6 +1,6 @@
---
id: cf1391c1c11feddfaeb4bdef
-title: 创建一个小数
+title: Create Decimal Numbers with JavaScript
challengeType: 1
videoUrl: 'https://scrimba.com/c/ca8GEuW'
forumTopicId: 16826
@@ -9,24 +9,24 @@ dashedName: create-decimal-numbers-with-javascript
# --description--
-我们也可以把小数存储到变量中。小数也被称作<dfn>浮点数</dfn> 。
+We can store decimal numbers in variables too. Decimal numbers are sometimes referred to as <dfn>floating point</dfn> numbers or <dfn>floats</dfn>.
-**提示**
-不是所有的实数都可以用 <dfn>浮点数</dfn> 来表示。因为可能存在四舍五入的错误，[详情查看](https://en.wikipedia.org/wiki/Floating_point#Accuracy_problems)。
+**Note**
+Not all real numbers can accurately be represented in <dfn>floating point</dfn>. This can lead to rounding errors. [Details Here](https://en.wikipedia.org/wiki/Floating_point#Accuracy_problems).
# --instructions--
-创建一个变量`myDecimal`并给它赋值一个浮点数。(例如`5.21`)。
+Create a variable `myDecimal` and give it a decimal value with a fractional part (e.g. `5.7`).
# --hints--
-`myDecimal`应该是一个数字。
+`myDecimal` should be a number.
```js
assert(typeof myDecimal === 'number');
```
-`myDecimal`应该包含小数点。
+`myDecimal` should have a decimal point
```js
assert(myDecimal % 1 != 0);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/declare-javascript-variables.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/declare-javascript-variables.md
index 0537c988f25..3713daa19e6 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/declare-javascript-variables.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/declare-javascript-variables.md
@@ -1,6 +1,6 @@
---
id: bd7123c9c443eddfaeb5bdef
-title: 声明变量
+title: Declare JavaScript Variables
challengeType: 1
videoUrl: 'https://scrimba.com/c/cNanrHq'
forumTopicId: 17556
@@ -9,38 +9,35 @@ dashedName: declare-javascript-variables
# --description--
-在计算机科学中，<dfn>数据</dfn>就是一切，它对于计算机意义重大。JavaScript 提供七种不同的<dfn>数据类型</dfn>，它们是`undefined`（未定义）， `null`（空），`boolean`（布尔型），`string`（字符串），`symbol`(符号)，`number`（数字），和`object`（对象）。
+In computer science, <dfn>data</dfn> is anything that is meaningful to the computer. JavaScript provides eight different <dfn>data types</dfn> which are `undefined`, `null`, `boolean`, `string`, `symbol`, `bigint`, `number`, and `object`.
-例如，计算机区分数字和字符集合的字符串，例如数字`12`和字符串`"12"`，`"dog"`或`"123 cats"`。计算机可以对数字执行数学运算，但不能对字符串执行数学运算。
+For example, computers distinguish between numbers, such as the number `12`, and `strings`, such as `"12"`, `"dog"`, or `"123 cats"`, which are collections of characters. Computers can perform mathematical operations on a number, but not on a string.
-<dfn>变量</dfn>允许计算机以一种动态的形式来存储和操作数据，通过操作指向数据的指针而不是数据本身来避免了内存泄露，以上的七种数据类型都可以存储到一个变量中。
+<dfn>Variables</dfn> allow computers to store and manipulate data in a dynamic fashion. They do this by using a "label" to point to the data rather than using the data itself. Any of the eight data types may be stored in a variable.
-`变量`非常类似于你在数学中使用的 x，y 变量，都是以一个简单命名的名称来代替我们赋值给它的数据。计算机中的`变量`与数学中的变量不同的是，计算机可以在不同的时间存储不同类型的变量。
+`Variables` are similar to the x and y variables you use in mathematics, which means they're a simple name to represent the data we want to refer to. Computer `variables` differ from mathematical variables in that they can store different values at different times.
-通过在变量的前面使用关键字`var`，声明一个变量，例如：
+We tell JavaScript to create or <dfn>declare</dfn> a variable by putting the keyword `var` in front of it, like so:
```js
var ourName;
```
-上面代码的意思是创建一个名为`ourName`的`变量`，在 JavaScript 中我们以分号结束语句。 `变量`名称可以由数字、字母、美元符号`$` 或者 下划线`_`组成，但是不能包含空格或者以数字为开头。
+creates a `variable` called `ourName`. In JavaScript we end statements with semicolons. `Variable` names can be made up of numbers, letters, and `$` or `_`, but may not contain spaces or start with a number.
# --instructions--
-使用`var` 关键字来创建一个名为`myName`的变量。
+Use the `var` keyword to create a variable called `myName`.
-**提示**
-如果遇到困难了，请看下`ourName`的例子是怎么写的。
+**Hint**
+Look at the `ourName` example above if you get stuck.
# --hints--
-你需要使用`var`关键字定义一个变量`myName`，并使用分号结尾。
+You should declare `myName` with the `var` keyword, ending with a semicolon
```js
-assert(
- /var\s+myName\s*;/.test(code),
- '你需要使用<code>var</code>关键字定义一个变量<code>myName</code>。并使用分号结尾。'
-);
+assert(/var\s+myName\s*;/.test(code));
```
# --seed--
@@ -52,7 +49,6 @@ if(typeof myName !== "undefined"){(function(v){return v;})(myName);}
```
## --seed-contents--
-
```js
```
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/declare-string-variables.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/declare-string-variables.md
index 15f5a50da2f..440fa546fc0 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/declare-string-variables.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/declare-string-variables.md
@@ -1,6 +1,6 @@
---
id: bd7123c9c444eddfaeb5bdef
-title: 声明字符串变量
+title: Declare String Variables
challengeType: 1
videoUrl: 'https://scrimba.com/c/c2QvWU6'
forumTopicId: 17557
@@ -9,19 +9,19 @@ dashedName: declare-string-variables
# --description--
-之前我们写过这样的代码：
+Previously we have used the code
`var myName = "your name";`
-`"your name"`被称作<dfn>字符串变量</dfn>。字符串是用单引号或双引号包裹起来的一连串的零个或多个字符。
+`"your name"` is called a <dfn>string</dfn> <dfn>literal</dfn>. It is a string because it is a series of zero or more characters enclosed in single or double quotes.
# --instructions--
-创建两个新的`字符串变量`：`myFirstName`和`myLastName`，并用你的姓和名分别为它们赋值。
+Create two new `string` variables: `myFirstName` and `myLastName` and assign them the values of your first and last name, respectively.
# --hints--
-`myFirstName`应该是一个字符串,并且至少包含一个字符。
+`myFirstName` should be a string with at least one character in it.
```js
assert(
@@ -39,7 +39,7 @@ assert(
);
```
-`myLastName`应该是一个字符串,并且至少包含一个字符。
+`myLastName` should be a string with at least one character in it.
```js
assert(
@@ -66,7 +66,6 @@ if(typeof myFirstName !== "undefined" && typeof myLastName !== "undefined"){(fun
```
## --seed-contents--
-
```js
```
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/decrement-a-number-with-javascript.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/decrement-a-number-with-javascript.md
index 232e8fe8180..2f71a24523e 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/decrement-a-number-with-javascript.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/decrement-a-number-with-javascript.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244ad
-title: 数字递减
+title: Decrement a Number with JavaScript
challengeType: 1
videoUrl: 'https://scrimba.com/c/cM2KeS2'
forumTopicId: 17558
@@ -9,30 +9,30 @@ dashedName: decrement-a-number-with-javascript
# --description--
-使用自减符号`--`，你可以很方便地对一个变量执行<dfn>自减</dfn>或者`-1`运算。
+You can easily <dfn>decrement</dfn> or decrease a variable by one with the `--` operator.
`i--;`
-等效于
+is the equivalent of
`i = i - 1;`
-**提示**
-`i--;`这种写法，省去了书写`=`符号的必要。
+**Note**
+The entire line becomes `i--;`, eliminating the need for the equal sign.
# --instructions--
-重写代码，使用`--`符号对`myVar`执行自减操作。
+Change the code to use the `--` operator on `myVar`.
# --hints--
-`myVar`应该等于`10`。
+`myVar` should equal `10`.
```js
assert(myVar === 10);
```
-`myVar = myVar - 1;`语句应该被修改。
+`myVar = myVar - 1;` should be changed.
```js
assert(
@@ -40,13 +40,13 @@ assert(
);
```
-对`myVar`使用`--`运算符。
+You should use the `--` operator on `myVar`.
```js
assert(/[-]{2}\s*myVar|myVar\s*[-]{2}/.test(code));
```
-不要修改注释上面的代码。
+You should not change code above the specified comment.
```js
assert(/var myVar = 11;/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/delete-properties-from-a-javascript-object.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/delete-properties-from-a-javascript-object.md
index dd37ebf1293..21c8366962d 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/delete-properties-from-a-javascript-object.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/delete-properties-from-a-javascript-object.md
@@ -1,6 +1,6 @@
---
id: 56bbb991ad1ed5201cd392d3
-title: 删除对象的属性
+title: Delete Properties from a JavaScript Object
challengeType: 1
videoUrl: 'https://scrimba.com/c/cDqKdTv'
forumTopicId: 17560
@@ -9,26 +9,51 @@ dashedName: delete-properties-from-a-javascript-object
# --description--
-我们同样可以删除对象的属性，例如：
+We can also delete properties from objects like this:
`delete ourDog.bark;`
+Example:
+
+```js
+var ourDog = {
+ "name": "Camper",
+ "legs": 4,
+ "tails": 1,
+ "friends": ["everything!"],
+ "bark": "bow-wow"
+};
+
+delete ourDog.bark;
+```
+
+After the last line shown above, `ourDog` looks like:
+
+```js
+{
+ "name": "Camper",
+ "legs": 4,
+ "tails": 1,
+ "friends": ["everything!"]
+}
+```
+
# --instructions--
-删除`myDog`对象的`"tails"`属性。
+Delete the `"tails"` property from `myDog`. You may use either dot or bracket notation.
# --hints--
-从`myDog`中删除`"tails"`属性。
+You should delete the property `"tails"` from `myDog`.
```js
assert(typeof myDog === 'object' && myDog.tails === undefined);
```
-不要修改`myDog`的初始化。
+You should not modify the `myDog` setup.
```js
-assert(code.match(/"tails": 1/g).length > 1);
+assert(code.match(/"tails": 1/g).length > 0);
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/divide-one-decimal-by-another-with-javascript.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/divide-one-decimal-by-another-with-javascript.md
index 0e3b63affc1..cb96b3ff715 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/divide-one-decimal-by-another-with-javascript.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/divide-one-decimal-by-another-with-javascript.md
@@ -1,6 +1,6 @@
---
id: bd7993c9ca9feddfaeb7bdef
-title: 两个小数相除
+title: Divide One Decimal by Another with JavaScript
challengeType: 1
videoUrl: 'https://scrimba.com/c/cBZe9AW'
forumTopicId: 18255
@@ -9,27 +9,27 @@ dashedName: divide-one-decimal-by-another-with-javascript
# --description--
-现在让我们将一个小数除以另一个小数。
+Now let's divide one decimal by another.
# --instructions--
-改变数值`0.0`的值让变量`quotient`的值等于`2.2`.
+Change the `0.0` so that `quotient` will equal to `2.2`.
# --hints--
-`quotient`的值应该等于`2.2`。
+The variable `quotient` should equal `2.2`
```js
assert(quotient === 2.2);
```
-使用`/`运算符将 4.4 除以 2。
+You should use the `/` operator to divide 4.4 by 2
```js
assert(/4\.40*\s*\/\s*2\.*0*/.test(code));
```
-quotient 变量应该只被赋值一次。
+The quotient variable should only be assigned once
```js
assert(code.match(/quotient/g).length === 1);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/divide-one-number-by-another-with-javascript.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/divide-one-number-by-another-with-javascript.md
index 87a379a91ca..7cd4e25b806 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/divide-one-number-by-another-with-javascript.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/divide-one-number-by-another-with-javascript.md
@@ -1,6 +1,6 @@
---
id: cf1111c1c11feddfaeb6bdef
-title: 除法运算
+title: Divide One Number by Another with JavaScript
challengeType: 1
videoUrl: 'https://scrimba.com/c/cqkbdAr'
forumTopicId: 17566
@@ -9,11 +9,11 @@ dashedName: divide-one-number-by-another-with-javascript
# --description--
-我们可以在 JavaScript 中做除法运算。
+We can also divide one number by another.
-JavaScript 中使用`/`符号做除法运算。
+JavaScript uses the `/` symbol for division.
-**示例**
+**Example**
```js
myVar = 16 / 2; // assigned 8
@@ -21,17 +21,17 @@ myVar = 16 / 2; // assigned 8
# --instructions--
-改变数值`0`来让变量`quotient`的值等于`2`。
+Change the `0` so that the `quotient` is equal to `2`.
# --hints--
-要使`quotient`的值等于 2。
+The variable `quotient` should be equal to 2.
```js
assert(quotient === 2);
```
-使用`/`运算符。
+You should use the `/` operator.
```js
assert(/\d+\s*\/\s*\d+/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/escape-sequences-in-strings.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/escape-sequences-in-strings.md
index c8d54d4f8a2..700cf22c266 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/escape-sequences-in-strings.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/escape-sequences-in-strings.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244b6
-title: 字符串中的转义序列
+title: Escape Sequences in Strings
challengeType: 1
videoUrl: 'https://scrimba.com/c/cvmqRh6'
forumTopicId: 17567
@@ -9,33 +9,36 @@ dashedName: escape-sequences-in-strings
# --description--
-引号不是字符串中唯一可以被<dfn>转义</dfn>的字符。使用转义字符有两个原因：首先是可以让你使用无法输入的字符，例如退格。其次是可以让你在一个字符串中表示多个引号，而不会出错。我们在之前的挑战中学到了这个。
+Quotes are not the only characters that can be <dfn>escaped</dfn> inside a string. There are two reasons to use escaping characters:
-<table class='table table-striped'><thead><tr><th>代码</th><th>输出</th></tr></thead><tbody><tr><td><code>\'</code></td><td>单引号</td></tr><tr><td><code>\"</code></td><td>双引号</td></tr><tr><td><code>\\</code></td><td>反斜杠</td></tr><tr><td><code>\n</code></td><td>换行符</td></tr><tr><td><code>\r</code></td><td>回车符</td></tr><tr><td><code>\t</code></td><td>制表符</td></tr><tr><td><code>\b</code></td><td>退格</td></tr><tr><td><code>\f</code></td><td>换页符</td></tr></tbody></table>
+1. To allow you to use characters you may not otherwise be able to type out, such as a carriage return.
+2. To allow you to represent multiple quotes in a string without JavaScript misinterpreting what you mean.
-*请注意，必须对反斜杠本身进行转义才能显示为反斜杠。*
+We learned this in the previous challenge.
+
+<table class='table table-striped'><thead><tr><th>Code</th><th>Output</th></tr></thead><tbody><tr><td><code>\'</code></td><td>single quote</td></tr><tr><td><code>\"</code></td><td>double quote</td></tr><tr><td><code>\\</code></td><td>backslash</td></tr><tr><td><code>\n</code></td><td>newline</td></tr><tr><td><code>\r</code></td><td>carriage return</td></tr><tr><td><code>\t</code></td><td>tab</td></tr><tr><td><code>\b</code></td><td>word boundary</td></tr><tr><td><code>\f</code></td><td>form feed</td></tr></tbody></table>
+
+*Note that the backslash itself must be escaped in order to display as a backslash.*
# --instructions--
-使用转义字符将下面三行文本字符串赋给变量`myStr`。
+Assign the following three lines of text into the single variable `myStr` using escape sequences.
<blockquote>FirstLine<br> \SecondLine<br>ThirdLine</blockquote>
-你需要使用转义字符正确地插入特殊字符，确保间距与上面文本一致并且单词或转义字符之间没有空格。
+You will need to use escape sequences to insert special characters correctly. You will also need to follow the spacing as it looks above, with no spaces between escape sequences or words.
-像这样用转义字符写出来：
-
-"FirstLine```换行符``制表符``反斜杠```SecondLine`换行符`ThirdLine"
+**Note:** The indentation for `SecondLine` is achieved with the tab escape character, not spaces.
# --hints--
-`myStr`不能包含空格。
+`myStr` should not contain any spaces
```js
assert(!/ /.test(myStr));
```
-`myStr`应该包含字符串`FirstLine`, `SecondLine` and `ThirdLine` （记得区分大小写）。
+`myStr` should contain the strings `FirstLine`, `SecondLine` and `ThirdLine` (remember case sensitivity)
```js
assert(
@@ -43,31 +46,31 @@ assert(
);
```
-`FirstLine`后面应该是一个新行`\n`。
+`FirstLine` should be followed by the newline character `\n`
```js
assert(/FirstLine\n/.test(myStr));
```
-`myStr`应该包含制表符`\t`并且制表符要在换行符后面。
+`myStr` should contain a tab character `\t` which follows a newline character
```js
assert(/\n\t/.test(myStr));
```
-`SecondLine`前面应该是反斜杠`\\`。
+`SecondLine` should be preceded by the backslash character <code>\\</code>
```js
-assert(/\SecondLine/.test(myStr));
+assert(/\\SecondLine/.test(myStr));
```
-`SecondLine`和`ThirdLine`之间应该是换行符。
+There should be a newline character between `SecondLine` and `ThirdLine`
```js
assert(/SecondLine\nThirdLine/.test(myStr));
```
-`myStr` 应该只包含介绍里面展示的字符串。
+`myStr` should only contain characters shown in the instructions
```js
assert(myStr === 'FirstLine\n\t\\SecondLine\nThirdLine');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/escaping-literal-quotes-in-strings.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/escaping-literal-quotes-in-strings.md
index ca8bb072696..ce8571d5fd5 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/escaping-literal-quotes-in-strings.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/escaping-literal-quotes-in-strings.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244b5
-title: 转义字符串中的引号
+title: Escaping Literal Quotes in Strings
challengeType: 1
videoUrl: 'https://scrimba.com/c/c2QvgSr'
forumTopicId: 17568
@@ -9,34 +9,34 @@ dashedName: escaping-literal-quotes-in-strings
# --description--
-定义一个字符串必须要用单引号或双引号来包裹它。那么当你的字符串里面包含：`"`或者`'`时该怎么办呢?
+When you are defining a string you must start and end with a single or double quote. What happens when you need a literal quote: `"` or `'` inside of your string?
-在 JavaScript 中，你可以通过在引号前面使用<dfn>反斜杠</dfn>（`\`）来转义引号。
+In JavaScript, you can <dfn>escape</dfn> a quote from considering it as an end of string quote by placing a <dfn>backslash</dfn> (<code>\\</code>) in front of the quote.
`var sampleStr = "Alan said, \"Peter is learning JavaScript\".";`
-有了转义符号，JavaScript 就知道这个单引号或双引号并不是字符串的结尾，而是字符串内的字符。所以，上面的字符串打印到控制台的结果为：
+This signals to JavaScript that the following quote is not the end of the string, but should instead appear inside the string. So if you were to print this to the console, you would get:
`Alan said, "Peter is learning JavaScript".`
# --instructions--
-使用<dfn>反斜杠</dfn>将一个字符串赋值给变量`myStr`，打印到控制台，输出为：
+Use <dfn>backslashes</dfn> to assign a string to the `myStr` variable so that if you were to print it to the console, you would see:
`I am a "double quoted" string inside "double quotes".`
# --hints--
-你的代码中应该包含两个双引号 (`"`) 以及四个转义的双引 (`\"`)。
+You should use two double quotes (`"`) and four escaped double quotes (`\"`).
```js
assert(code.match(/\\"/g).length === 4 && code.match(/[^\\]"/g).length === 2);
```
-变量 myStr 应该包含字符串`I am a "double quoted" string inside "double quotes".`。
+Variable myStr should contain the string: `I am a "double quoted" string inside "double quotes".`
```js
-assert(myStr === 'I am a "double quoted" string inside "double quotes".');
+assert(/I am a "double quoted" string inside "double quotes(\."|"\.)$/.test(myStr));
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/find-the-length-of-a-string.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/find-the-length-of-a-string.md
index a7a8668aff0..03bc3f0071f 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/find-the-length-of-a-string.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/find-the-length-of-a-string.md
@@ -1,6 +1,6 @@
---
id: bd7123c9c448eddfaeb5bdef
-title: 查找字符串的长度
+title: Find the Length of a String
challengeType: 1
videoUrl: 'https://scrimba.com/c/cvmqEAd'
forumTopicId: 18182
@@ -9,19 +9,19 @@ dashedName: find-the-length-of-a-string
# --description--
-你可以通过在字符串变量或字符串后面写上`.length`来获得字符串变量值的长度。
+You can find the length of a `String` value by writing `.length` after the string variable or string literal.
`"Alan Peter".length; // 10`
-例如，我们创建了一个变量`var firstName = "Charles"`，我们就可以通过使用`firstName.length`来获得`"Charles"`字符串的长度。
+For example, if we created a variable `var firstName = "Charles"`, we could find out how long the string `"Charles"` is by using the `firstName.length` property.
# --instructions--
-使用`.length`属性来获得变量`lastName`的长度，并把它赋值给变量`lastNameLength`。
+Use the `.length` property to count the number of characters in the `lastName` variable and assign it to `lastNameLength`.
# --hints--
-不能改变 `// Setup` 部分声明的变量。
+You should not change the variable declarations in the `// Setup` section.
```js
assert(
@@ -30,13 +30,13 @@ assert(
);
```
-`lastNameLength`应该等于 8。
+`lastNameLength` should be equal to eight.
```js
assert(typeof lastNameLength !== 'undefined' && lastNameLength === 8);
```
-你应该使用 `.length` 获取 `lastName` 的长度，像这样 `lastName.length`。
+You should be getting the length of `lastName` by using `.length` like this: `lastName.length`.
```js
assert(code.match(/=\s*lastName\.length/g) && !code.match(/lastName\s*=\s*8/));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/finding-a-remainder-in-javascript.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/finding-a-remainder-in-javascript.md
index e934ea79347..c811162a834 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/finding-a-remainder-in-javascript.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/finding-a-remainder-in-javascript.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244ae
-title: 求余运算
+title: Finding a Remainder in JavaScript
challengeType: 1
videoUrl: 'https://scrimba.com/c/cWP24Ub'
forumTopicId: 18184
@@ -9,42 +9,42 @@ dashedName: finding-a-remainder-in-javascript
# --description--
-<dfn>remainder</dfn>求余运算符`%`返回两个数相除得到的余数
+The <dfn>remainder</dfn> operator `%` gives the remainder of the division of two numbers.
-**示例**
+**Example**
-<blockquote>5 % 2 = 1 因为<br>Math.floor(5 / 2) = 2 （商）<br>2 * 2 = 4<br>5 - 4 = 1 （余数）</blockquote>
+<blockquote>5 % 2 = 1 because<br>Math.floor(5 / 2) = 2 (Quotient)<br>2 * 2 = 4<br>5 - 4 = 1 (Remainder)</blockquote>
-**用法**
-在数学中，判断一个数是奇数还是偶数，只需要判断这个数除以 2 得到的余数是 0 还是 1。
+**Usage**
+In mathematics, a number can be checked to be even or odd by checking the remainder of the division of the number by `2`.
-<blockquote>17 % 2 = 1（17 是奇数）<br>48 % 2 = 0（48 是偶数）</blockquote>
+<blockquote>17 % 2 = 1 (17 is Odd)<br>48 % 2 = 0 (48 is Even)</blockquote>
-**提示**
-余数运算符(<dfn>remainder</dfn>)有时被错误地称为“模数”运算符。它与模数非常相似，但不能用于负数的运算。\*\*\*\*
+**Note**
+The <dfn>remainder</dfn> operator is sometimes incorrectly referred to as the "modulus" operator. It is very similar to modulus, but does not work properly with negative numbers.
# --instructions--
-使用`%`运算符，计算 11 除以 3 的余数，并把余数赋给变量 remainder。
+Set `remainder` equal to the remainder of `11` divided by `3` using the <dfn>remainder</dfn> (`%`) operator.
# --hints--
-变量`remainder`应该被初始化。
+The variable `remainder` should be initialized
```js
assert(/var\s+?remainder/.test(code));
```
-`remainder`的值应该等于`2`。
+The value of `remainder` should be `2`
```js
assert(remainder === 2);
```
-你应该使用`%`运算符。
+You should use the `%` operator
```js
-assert(/\s+?remainder\s*?=\s*?.*%.*;/.test(code));
+assert(/\s+?remainder\s*?=\s*?.*%.*;?/.test(code));
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/generate-random-fractions-with-javascript.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/generate-random-fractions-with-javascript.md
index 352fcd68834..7565d935adf 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/generate-random-fractions-with-javascript.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/generate-random-fractions-with-javascript.md
@@ -1,6 +1,6 @@
---
id: cf1111c1c11feddfaeb9bdef
-title: 使用 JavaScript 生成随机分数
+title: Generate Random Fractions with JavaScript
challengeType: 1
videoUrl: 'https://scrimba.com/c/cyWJJs3'
forumTopicId: 18185
@@ -9,32 +9,32 @@ dashedName: generate-random-fractions-with-javascript
# --description--
-随机数非常适合用来创建随机行为。
+Random numbers are useful for creating random behavior.
-`Math.random()`用来生成一个在`0`（包括 0）到`1`（不包括 1）之间的随机小数，因此`Math.random()`可能返回 0 但绝不会返回 1。
+JavaScript has a `Math.random()` function that generates a random decimal number between `0` (inclusive) and not quite up to `1` (exclusive). Thus `Math.random()` can return a `0` but never quite return a `1`
-**提示**
-[使用赋值运算符存储值](storing-values-with-the-assignment-operator)这一节讲过，所有函数调用将在`return`执行之前解析，因此我们可以返回`Math.random()`函数的值。
+**Note**
+Like [Storing Values with the Equal Operator](/learn/javascript-algorithms-and-data-structures/basic-javascript/storing-values-with-the-assignment-operator), all function calls will be resolved before the `return` executes, so we can `return` the value of the `Math.random()` function.
# --instructions--
-更改`randomFraction`使其返回一个随机数而不是`0`。
+Change `randomFraction` to return a random number instead of returning `0`.
# --hints--
-`randomFraction`应该返回一个随机数。
+`randomFraction` should return a random number.
```js
assert(typeof randomFraction() === 'number');
```
-`randomFraction`应该返回一个小数。
+The number returned by `randomFraction` should be a decimal.
```js
assert((randomFraction() + '').match(/\./g));
```
-需要使用`Math.random`生成随机的小数。
+You should be using `Math.random` to generate the random decimal number.
```js
assert(code.match(/Math\.random/g).length >= 0);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/generate-random-whole-numbers-with-javascript.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/generate-random-whole-numbers-with-javascript.md
index afeaa3a5e7f..42f91172df6 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/generate-random-whole-numbers-with-javascript.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/generate-random-whole-numbers-with-javascript.md
@@ -1,6 +1,6 @@
---
id: cf1111c1c12feddfaeb1bdef
-title: 使用 JavaScript 生成随机整数
+title: Generate Random Whole Numbers with JavaScript
challengeType: 1
videoUrl: 'https://scrimba.com/c/cRn6bfr'
forumTopicId: 18186
@@ -9,25 +9,25 @@ dashedName: generate-random-whole-numbers-with-javascript
# --description--
-生成随机小数很棒，但随机数更有用的地方在于生成随机整数。
+It's great that we can generate random decimal numbers, but it's even more useful if we use it to generate random whole numbers.
-<ol><li>用<code>Math.random()</code>生成一个随机小数。</li><li>把这个随机小数乘以<code>20</code>。</li><li>用<code>Math.floor()</code>向下取整 获得它最近的整数。</li></ol>
+<ol><li>Use <code>Math.random()</code> to generate a random decimal.</li><li>Multiply that random decimal by <code>20</code>.</li><li>Use another function, <code>Math.floor()</code> to round the number down to its nearest whole number.</li></ol>
-记住`Math.random()`永远不会返回`1`。同时因为我们是在用`Math.floor()`向下取整，所以最终我们获得的结果不可能有`20`。这确保了我们获得了一个在0到19之间的整数。
+Remember that `Math.random()` can never quite return a `1` and, because we're rounding down, it's impossible to actually get `20`. This technique will give us a whole number between `0` and `19`.
-把操作连缀起来，代码类似于下面：
+Putting everything together, this is what our code looks like:
`Math.floor(Math.random() * 20);`
-我们先调用`Math.random()`，把它的结果乘以20，然后把上一步的结果传给`Math.floor()`，最终通过向下取整获得最近的整数。
+We are calling `Math.random()`, multiplying the result by 20, then passing the value to `Math.floor()` function to round the value down to the nearest whole number.
# --instructions--
-生成一个`0`到`9`之间的随机整数。
+Use this technique to generate and return a random whole number between `0` and `9`.
# --hints--
-`myFunction`的结果应该是一个整数。
+The result of `randomWholeNum` should be a whole number.
```js
assert(
@@ -39,13 +39,13 @@ assert(
);
```
-需要使用`Math.random`生成随机数字。
+You should use `Math.random` to generate a random number.
```js
-assert(code.match(/Math.random/g).length > 1);
+assert(code.match(/Math.random/g).length >= 1);
```
-你应该将`Math.random`的结果乘以 10 来生成 0 到 9 之间的随机数。
+You should have multiplied the result of `Math.random` by 10 to make it a number that is between zero and nine.
```js
assert(
@@ -54,10 +54,10 @@ assert(
);
```
-你需要使用`Math.floor`移除数字中的小数部分。
+You should use `Math.floor` to remove the decimal part of the number.
```js
-assert(code.match(/Math.floor/g).length > 1);
+assert(code.match(/Math.floor/g).length >= 1);
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/generate-random-whole-numbers-within-a-range.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/generate-random-whole-numbers-within-a-range.md
index 3693cc25b4d..a78f88bc641 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/generate-random-whole-numbers-within-a-range.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/generate-random-whole-numbers-within-a-range.md
@@ -1,6 +1,6 @@
---
id: cf1111c1c12feddfaeb2bdef
-title: 生成某个范围内的随机整数
+title: Generate Random Whole Numbers within a Range
challengeType: 1
videoUrl: 'https://scrimba.com/c/cm83yu6'
forumTopicId: 18187
@@ -9,39 +9,39 @@ dashedName: generate-random-whole-numbers-within-a-range
# --description--
-我们之前生成的随机数是在0到某个数之间，现在我们要生成的随机数是在两个指定的数之间。
+Instead of generating a random whole number between zero and a given number like we did before, we can generate a random whole number that falls within a range of two specific numbers.
-我们需要定义一个最小值和一个最大值。
+To do this, we'll define a minimum number `min` and a maximum number `max`.
-下面是我们将要使用的方法，仔细看看并尝试理解这行代码到底在干嘛：
+Here's the formula we'll use. Take a moment to read it and try to understand what this code is doing:
`Math.floor(Math.random() * (max - min + 1)) + min`
# --instructions--
-创建一个叫`randomRange`的函数，参数为 myMin 和 myMax，返回一个在`myMin`（包括 myMin）和`myMax`（包括 myMax）之间的随机数。
+Create a function called `randomRange` that takes a range `myMin` and `myMax` and returns a random whole number that's greater than or equal to `myMin`, and is less than or equal to `myMax`, inclusive.
# --hints--
-`randomRange`返回的随机数应该大于或等于`myMin`。
+The lowest random number that can be generated by `randomRange` should be equal to your minimum number, `myMin`.
```js
assert(calcMin === 5);
```
-`randomRange`返回的随机数应该小于或等于`myMax`。
+The highest random number that can be generated by `randomRange` should be equal to your maximum number, `myMax`.
```js
assert(calcMax === 15);
```
-`randomRange`应该返回一个随机整数, 而不是小数。
+The random number generated by `randomRange` should be an integer, not a decimal.
```js
assert(randomRange(0, 1) % 1 === 0);
```
-`randomRange`应该使用`myMax`和`myMin`, 并且返回两者之间的随机数。
+`randomRange` should use both `myMax` and `myMin`, and return a random number in your range.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/global-scope-and-functions.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/global-scope-and-functions.md
index cd476659399..72306dcf1de 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/global-scope-and-functions.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/global-scope-and-functions.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244be
-title: 全局作用域和函数
+title: Global Scope and Functions
challengeType: 1
videoUrl: 'https://scrimba.com/c/cQM7mCN'
forumTopicId: 18193
@@ -9,37 +9,37 @@ dashedName: global-scope-and-functions
# --description--
-在 JavaScript 中，<dfn>作用域</dfn>涉及到变量的作用范围。在函数外定义的变量具有 <dfn>全局</dfn> 作用域。这意味着，具有全局作用域的变量可以在代码的任何地方被调用。
+In JavaScript, <dfn>scope</dfn> refers to the visibility of variables. Variables which are defined outside of a function block have <dfn>Global</dfn> scope. This means, they can be seen everywhere in your JavaScript code.
-这些没有使用`var`关键字定义的变量，会被自动创建在全局作用域中，形成全局变量。当在代码其他地方无意间定义了一个变量，刚好变量名与全局变量相同，这时会产生意想不到的后果。因此你应该总是使用var关键字来声明你的变量。
+Variables which are used without the `var` keyword are automatically created in the `global` scope. This can create unintended consequences elsewhere in your code or when running a function again. You should always declare your variables with `var`.
# --instructions--
-在函数外声明一个`全局`变量`myGlobal`，并给它一个初始值`10`
+Using `var`, declare a global variable named `myGlobal` outside of any function. Initialize it with a value of `10`.
-在函数`fun1`的内部，**不**使用`var`关键字来声明`oopsGlobal`，并赋值为`5`。
+Inside function `fun1`, assign `5` to `oopsGlobal` ***without*** using the `var` keyword.
# --hints--
-应定义`myGlobal`。
+`myGlobal` should be defined
```js
assert(typeof myGlobal != 'undefined');
```
-`myGlobal`的值应为`10`。
+`myGlobal` should have a value of `10`
```js
assert(myGlobal === 10);
```
-应使用`var`关键字定义`myGlobal`。
+`myGlobal` should be declared using the `var` keyword
```js
assert(/var\s+myGlobal/.test(code));
```
-`oopsGlobal`应为全局变量且值为`5`。
+`oopsGlobal` should be a global variable and have a value of `5`
```js
assert(typeof oopsGlobal != 'undefined' && oopsGlobal === 5);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/global-vs.-local-scope-in-functions.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/global-vs.-local-scope-in-functions.md
index b39d59a54b8..4aa4e70ab0c 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/global-vs.-local-scope-in-functions.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/global-vs.-local-scope-in-functions.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244c0
-title: 函数中的全局作用域和局部作用域
+title: Global vs. Local Scope in Functions
challengeType: 1
videoUrl: 'https://scrimba.com/c/c2QwKH2'
forumTopicId: 18194
@@ -9,9 +9,9 @@ dashedName: global-vs--local-scope-in-functions
# --description--
-一个程序中有可能具有相同名称的<dfn>局部</dfn>变量 和<dfn>全局</dfn>变量。在这种情况下，`局部`变量将会优先于`全局`变量。
+It is possible to have both <dfn>local</dfn> and <dfn>global</dfn> variables with the same name. When you do this, the `local` variable takes precedence over the `global` variable.
-下面为例：
+In this example:
```js
var someVar = "Hat";
@@ -21,27 +21,27 @@ function myFun() {
}
```
-函数`myFun`将会返回`"Head"`，因为`局部变量`优先级更高。
+The function `myFun` will return `"Head"` because the `local` version of the variable is present.
# --instructions--
-给`myOutfit`添加一个局部变量来覆盖`outerWear`的值为`"sweater"`。
+Add a local variable to `myOutfit` function to override the value of `outerWear` with `"sweater"`.
# --hints--
-不要修改全局变量`outerWear`的值。
+You should not change the value of the global `outerWear`.
```js
assert(outerWear === 'T-Shirt');
```
-`myOutfit`应该返回`"sweater"`。
+`myOutfit` should return `"sweater"`.
```js
assert(myOutfit() === 'sweater');
```
-不要修改`return`语句。
+You should not change the return statement.
```js
assert(/return outerWear/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/golf-code.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/golf-code.md
index 70752bfae64..c53081daa98 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/golf-code.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/golf-code.md
@@ -1,6 +1,6 @@
---
id: 5664820f61c48e80c9fa476c
-title: 高尔夫代码
+title: Golf Code
challengeType: 1
videoUrl: 'https://scrimba.com/c/c9ykNUR'
forumTopicId: 18195
@@ -9,77 +9,77 @@ dashedName: golf-code
# --description--
-在高尔夫[golf](https://en.wikipedia.org/wiki/Golf)游戏中，每个洞都有自己的标准杆数`par`，代表着距离。根据你把球打进洞所挥杆的次数`strokes`，可以计算出你的高尔夫水平。
+In the game of [golf](https://en.wikipedia.org/wiki/Golf) each hole has a `par` meaning the average number of `strokes` a golfer is expected to make in order to sink the ball in a hole to complete the play. Depending on how far above or below `par` your `strokes` are, there is a different nickname.
-函数将会传送 2 个参数，分别是标准杆数`par`和挥杆次数`strokes`，根据下面的表格返回正确的水平段位。
+Your function will be passed `par` and `strokes` arguments. Return the correct string according to this table which lists the strokes in order of priority; top (highest) to bottom (lowest):
<table class='table table-striped'><thead><tr><th>Strokes</th><th>Return</th></tr></thead><tbody><tr><td>1</td><td>"Hole-in-one!"</td></tr><tr><td><= par - 2</td><td>"Eagle"</td></tr><tr><td>par - 1</td><td>"Birdie"</td></tr><tr><td>par</td><td>"Par"</td></tr><tr><td>par + 1</td><td>"Bogey"</td></tr><tr><td>par + 2</td><td>"Double Bogey"</td></tr><tr><td>>= par + 3</td><td>"Go Home!"</td></tr></tbody></table>
-`par`和`strokes`必须是数字而且是正数。
+`par` and `strokes` will always be numeric and positive. We have added an array of all the names for your convenience.
# --hints--
-`golfScore(4, 1)`应该返回 "Hole-in-one!"。
+`golfScore(4, 1)` should return "Hole-in-one!"
```js
assert(golfScore(4, 1) === 'Hole-in-one!');
```
-`golfScore(4, 2)`应该返回 "Eagle"。
+`golfScore(4, 2)` should return "Eagle"
```js
assert(golfScore(4, 2) === 'Eagle');
```
-`golfScore(5, 2)`应该返回 "Eagle"。
+`golfScore(5, 2)` should return "Eagle"
```js
assert(golfScore(5, 2) === 'Eagle');
```
-`golfScore(4, 3)`应该返回 "Birdie"。
+`golfScore(4, 3)` should return "Birdie"
```js
assert(golfScore(4, 3) === 'Birdie');
```
-`golfScore(4, 4)`应该返回 "Par"。
+`golfScore(4, 4)` should return "Par"
```js
assert(golfScore(4, 4) === 'Par');
```
-`golfScore(1, 1)`应该返回 "Hole-in-one!"。
+`golfScore(1, 1)` should return "Hole-in-one!"
```js
assert(golfScore(1, 1) === 'Hole-in-one!');
```
-`golfScore(5, 5)`应该返回 "Par"。
+`golfScore(5, 5)` should return "Par"
```js
assert(golfScore(5, 5) === 'Par');
```
-`golfScore(4, 5)`应该返回 "Bogey"。
+`golfScore(4, 5)` should return "Bogey"
```js
assert(golfScore(4, 5) === 'Bogey');
```
-`golfScore(4, 6)`应该返回 "Double Bogey"。
+`golfScore(4, 6)` should return "Double Bogey"
```js
assert(golfScore(4, 6) === 'Double Bogey');
```
-`golfScore(4, 7)`应该返回 "Go Home!"。
+`golfScore(4, 7)` should return "Go Home!"
```js
assert(golfScore(4, 7) === 'Go Home!');
```
-`golfScore(5, 9)`应该返回 "Go Home!"。
+`golfScore(5, 9)` should return "Go Home!"
```js
assert(golfScore(5, 9) === 'Go Home!');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/increment-a-number-with-javascript.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/increment-a-number-with-javascript.md
index 87f5871f03d..f2d9b912958 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/increment-a-number-with-javascript.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/increment-a-number-with-javascript.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244ac
-title: 数字递增
+title: Increment a Number with JavaScript
challengeType: 1
videoUrl: 'https://scrimba.com/c/ca8GLT9'
forumTopicId: 18201
@@ -9,45 +9,44 @@ dashedName: increment-a-number-with-javascript
# --description--
-使用`++`，我们可以很容易地对变量进行自增或者`+1`运算。
+You can easily <dfn>increment</dfn> or add one to a variable with the `++` operator.
`i++;`
-等效于
+is the equivalent of
`i = i + 1;`
-**注意**
-`i++;`这种写法，省去了书写`=`符号的必要。
+**Note**
+The entire line becomes `i++;`, eliminating the need for the equal sign.
# --instructions--
-重写代码，使用`++`来对变量`myVar`进行自增操作。
-
-**提示**
-了解更多关于`++`运算符[Arithmetic operators - Increment (++)](https://developer.mozilla.org/zh-CN/docs/Web/JavaScript/Reference/Operators/Arithmetic_Operators#%E9%80%92%E5%A2%9E_()).
+Change the code to use the `++` operator on `myVar`.
# --hints--
-`myVar`应该等于`88`。
+`myVar` should equal `88`.
```js
assert(myVar === 88);
```
-`myVar = myVar + 1;`语句应该被修改。
+You should not use the assignment operator.
```js
-assert(/var\s*myVar\s*=\s*87;\s*\/*.*\s*myVar\+\+;/.test(code));
+assert(
+ /var\s*myVar\s*=\s*87;\s*\/*.*\s*([+]{2}\s*myVar|myVar\s*[+]{2});/.test(code)
+);
```
-使用`++`运算符。
+You should use the `++` operator.
```js
assert(/[+]{2}\s*myVar|myVar\s*[+]{2}/.test(code));
```
-不要修改注释上方的代码。
+You should not change code above the specified comment.
```js
assert(/var myVar = 87;/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/initializing-variables-with-the-assignment-operator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/initializing-variables-with-the-assignment-operator.md
index c9043e14b6e..7e933ed620b 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/initializing-variables-with-the-assignment-operator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/initializing-variables-with-the-assignment-operator.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244a9
-title: 使用赋值运算符初始化变量
+title: Initializing Variables with the Assignment Operator
challengeType: 1
videoUrl: 'https://scrimba.com/c/cWJ4Bfb'
forumTopicId: 301171
@@ -9,22 +9,22 @@ dashedName: initializing-variables-with-the-assignment-operator
# --description--
-通常在声明变量的时候会给变量<dfn>初始化</dfn>一个初始值。
+It is common to <dfn>initialize</dfn> a variable to an initial value in the same line as it is declared.
`var myVar = 0;`
-创建一个名为`myVar`的变量并指定一个初始值`0`。
+Creates a new variable called `myVar` and assigns it an initial value of `0`.
# --instructions--
-通过关键字`var`定义一个变量`a`并且给它一个初始值`9`。
+Define a variable `a` with `var` and initialize it to a value of `9`.
# --hints--
-你需要初始化`a`的值为`9`。
+You should initialize `a` to a value of `9`.
```js
-assert(/var\s+a\s*=\s*9\s*/.test(code));
+assert(/var\s+a\s*=\s*9(\s*;?\s*)$/.test(code));
```
# --seed--
@@ -36,7 +36,6 @@ if(typeof a !== 'undefined') {(function(a){return "a = " + a;})(a);} else { (fun
```
## --seed-contents--
-
```js
```
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/introducing-else-if-statements.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/introducing-else-if-statements.md
index 257ee0da789..9ce7a4d8c6b 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/introducing-else-if-statements.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/introducing-else-if-statements.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244db
-title: 介绍 else if 语句
+title: Introducing Else If Statements
challengeType: 1
videoUrl: 'https://scrimba.com/c/caeJ2hm'
forumTopicId: 18206
@@ -9,7 +9,7 @@ dashedName: introducing-else-if-statements
# --description--
-如果你有多个条件语句，你可以通过`else if`语句把`if`语句链起来。
+If you have multiple conditions that need to be addressed, you can chain `if` statements together with `else if` statements.
```js
if (num > 15) {
@@ -23,53 +23,57 @@ if (num > 15) {
# --instructions--
-使用`if else`实现同样的效果。
+Convert the logic to use `else if` statements.
# --hints--
-你应该至少有两个`else`表达式。
+You should have at least two `else` statements
```js
assert(code.match(/else/g).length > 1);
```
-你应该至少有两个`if`表达式。
+You should have at least two `if` statements
```js
assert(code.match(/if/g).length > 1);
```
-`testElseIf(0)`应该返回 "Smaller than 5"。
+You should have closing and opening curly braces for each `if else` code block.
+
+```js
+assert(
+ code.match(
+ /if\s*\((.+)\)\s*\{[\s\S]+\}\s*else\s+if\s*\((.+)\)\s*\{[\s\S]+\}\s*else\s*\{[\s\S]+\s*\}/
+ )
+);
+```
+
+`testElseIf(0)` should return "Smaller than 5"
```js
assert(testElseIf(0) === 'Smaller than 5');
```
-`testElseIf(5)`应该返回 "Between 5 and 10"。
+`testElseIf(5)` should return "Between 5 and 10"
```js
assert(testElseIf(5) === 'Between 5 and 10');
```
-`testElseIf(7)`应该返回 "Between 5 and 10"。
+`testElseIf(7)` should return "Between 5 and 10"
```js
assert(testElseIf(7) === 'Between 5 and 10');
```
-`testElseIf(10)`应该返回 "Between 5 and 10"。
+`testElseIf(10)` should return "Between 5 and 10"
```js
assert(testElseIf(10) === 'Between 5 and 10');
```
-`testElseIf(12)`应该返回 "Greater than 10"。
-
-```js
-assert(testElseIf(12) === 'Greater than 10');
-```
-
-`testElseIf(12)` 应该返回 "Greater than 10"。
+`testElseIf(12)` should return "Greater than 10"
```js
assert(testElseIf(12) === 'Greater than 10');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/introducing-else-statements.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/introducing-else-statements.md
index 6780f1b5f6d..24e5b172254 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/introducing-else-statements.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/introducing-else-statements.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244da
-title: 介绍 else 语句
+title: Introducing Else Statements
challengeType: 1
videoUrl: 'https://scrimba.com/c/cek4Efq'
forumTopicId: 18207
@@ -9,7 +9,7 @@ dashedName: introducing-else-statements
# --description--
-当`if`语句的条件为真，大括号里的代码执行，那如果条件为假呢？正常情况下什么也不会发生。使用else语句，可以执行当条件为假时相应的代码。
+When a condition for an `if` statement is true, the block of code following it is executed. What about when that condition is false? Normally nothing would happen. With an `else` statement, an alternate block of code can be executed.
```js
if (num > 10) {
@@ -21,47 +21,47 @@ if (num > 10) {
# --instructions--
-请把多个`if`语句合并为一个`if/else`语句。
+Combine the `if` statements into a single `if/else` statement.
# --hints--
-你应该只有一个`if`表达式。
+You should only have one `if` statement in the editor
```js
assert(code.match(/if/g).length === 1);
```
-你应该使用一个`else`表达式。
+You should use an `else` statement
```js
assert(/else/g.test(code));
```
-`testElse(4)`应该返回 "5 or Smaller"。
+`testElse(4)` should return "5 or Smaller"
```js
assert(testElse(4) === '5 or Smaller');
```
-`testElse(5)`应该返回 "5 or Smaller"。
+`testElse(5)` should return "5 or Smaller"
```js
assert(testElse(5) === '5 or Smaller');
```
-`testElse(6)`应该返回 "Bigger than 5"。
+`testElse(6)` should return "Bigger than 5"
```js
assert(testElse(6) === 'Bigger than 5');
```
-`testElse(10)`应该返回 "Bigger than 5"。
+`testElse(10)` should return "Bigger than 5".
```js
assert(testElse(10) === 'Bigger than 5');
```
-不要修改上面和下面的代码。
+You should not change the code above or below the specified comments.
```js
assert(/var result = "";/.test(code) && /return result;/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-odd-numbers-with-a-for-loop.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-odd-numbers-with-a-for-loop.md
index a11fad71f24..bbcc4049a87 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-odd-numbers-with-a-for-loop.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-odd-numbers-with-a-for-loop.md
@@ -1,6 +1,6 @@
---
id: 56104e9e514f539506016a5c
-title: 使用 For 循环遍历数组的奇数
+title: Iterate Odd Numbers With a For Loop
challengeType: 1
videoUrl: 'https://scrimba.com/c/cm8n7T9'
forumTopicId: 18212
@@ -9,11 +9,9 @@ dashedName: iterate-odd-numbers-with-a-for-loop
# --description--
-for循环可以按照我们指定的顺序来迭代，通过更改我们的`计数器`，我们可以按照偶数顺序来迭代。
+For loops don't have to iterate one at a time. By changing our `final-expression`, we can count by even numbers.
-初始化`i = 0`，当`i < 10`的时候继续循环。
-
-`i += 2`让`i`每次循环之后增加2。
+We'll start at `i = 0` and loop while `i < 10`. We'll increment `i` by 2 each loop with `i += 2`.
```js
var ourArray = [];
@@ -22,21 +20,21 @@ for (var i = 0; i < 10; i += 2) {
}
```
-循环结束后，`ourArray`的值为`[0,2,4,6,8]`。 改变`计数器`，这样我们可以用奇数来数。
+`ourArray` will now contain `[0,2,4,6,8]`. Let's change our `initialization` so we can count by odd numbers.
# --instructions--
-写一个`for`循环，把从 1 到 9 的奇数添加到`myArray`。
+Push the odd numbers from 1 through 9 to `myArray` using a `for` loop.
# --hints--
-你应该使用`for`循环。
+You should be using a `for` loop for this.
```js
-assert(code.match(/for\s*\(/g).length > 1);
+assert(/for\s*\([^)]+?\)/.test(code));
```
-`myArray`应该等于`[1,3,5,7,9]`。
+`myArray` should equal `[1,3,5,7,9]`.
```js
assert.deepEqual(myArray, [1, 3, 5, 7, 9]);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-through-an-array-with-a-for-loop.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-through-an-array-with-a-for-loop.md
index fc7ec11bcfe..9d1ae06aa21 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-through-an-array-with-a-for-loop.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-through-an-array-with-a-for-loop.md
@@ -1,6 +1,6 @@
---
id: 5675e877dbd60be8ad28edc6
-title: 使用 For 循环遍历数组
+title: Iterate Through an Array with a For Loop
challengeType: 1
videoUrl: 'https://scrimba.com/c/caeR3HB'
forumTopicId: 18216
@@ -9,7 +9,7 @@ dashedName: iterate-through-an-array-with-a-for-loop
# --description--
-迭代输出一个数组的每个元素是 JavaScript 中的常见需求，`for`循环可以做到这一点。下面的代码将输出数组 `arr`的每个元素到控制台：
+A common task in JavaScript is to iterate through the contents of an array. One way to do that is with a `for` loop. This code will output each element of the array `arr` to the console:
```js
var arr = [10, 9, 8, 7, 6];
@@ -18,36 +18,36 @@ for (var i = 0; i < arr.length; i++) {
}
```
-记住数组的索引从零开始的，这意味着数组的最后一个元素的下标是：数组的长度 -1。我们这个循环的 <dfn>条件</dfn>是`i < arr.length`，当`i`的值为 长度 -1 的时候循环就停止了。在这个例子中，最后一个循环是 i === 4，也就是说，当i的值等于arr.length时，结果输出 6。
+Remember that arrays have zero-based indexing, which means the last index of the array is `length - 1`. Our condition for this loop is `i < arr.length`, which stops the loop when `i` is equal to `length`. In this case the last iteration is `i === 4` i.e. when `i` becomes equal to `arr.length` and outputs `6` to the console.
# --instructions--
-声明并初始化一个变量`total`为`0`。使用`for`循环，使得`total`的值为`myArr`的数组中的每个元素的值的总和。
+Declare and initialize a variable `total` to `0`. Use a `for` loop to add the value of each element of the `myArr` array to `total`.
# --hints--
-`total`应该被声明, 并且初始化值为 0。
+`total` should be declared and initialized to 0.
```js
-assert(code.match(/var.*?total\s*=\s*0.*?;/));
+assert(code.match(/(var|let|const)\s*?total\s*=\s*0.*?;?/));
```
-`total`应该等于 20。
+`total` should equal 20.
```js
assert(total === 20);
```
-你应该使用`for`循环在`myArr`中遍历。
+You should use a `for` loop to iterate through `myArr`.
```js
-assert(code.match(/for\s*\(/g).length > 1 && code.match(/myArr\s*\[/));
+assert(/for\s*\(/g.test(code) && /myArr\s*\[/g.test(code));
```
-不能直接把`total`设置成 20。
+You should not attempt to directly assign the value 20 to `total`.
```js
-assert(!code.match(/total[\s\+\-]*=\s*(\d(?!\s*[;,])|[1-9])/g));
+assert(!__helpers.removeWhiteSpace(code).match(/total[=+-]0*[1-9]+/gm));
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-with-javascript-do...while-loops.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-with-javascript-do...while-loops.md
index 7def961d901..f4dfc98ba7a 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-with-javascript-do...while-loops.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-with-javascript-do...while-loops.md
@@ -1,6 +1,6 @@
---
id: 5a2efd662fb457916e1fe604
-title: do...while 循环
+title: Iterate with JavaScript Do...While Loops
challengeType: 1
videoUrl: 'https://scrimba.com/c/cDqWGcp'
forumTopicId: 301172
@@ -9,7 +9,7 @@ dashedName: iterate-with-javascript-do---while-loops
# --description--
-这一节我们将要学习的是`do...while`循环，它会先执行`do`里面的代码，如果`while`表达式为真则重复执行，反之则停止执行。我们来看一个例子。
+The next type of loop you will learn is called a `do...while` loop. It is called a `do...while` loop because it will first `do` one pass of the code inside the loop no matter what, and then continue to run the loop `while` the specified condition evaluates to `true`.
```js
var ourArray = [];
@@ -20,7 +20,7 @@ do {
} while (i < 5);
```
-这看起来和其他循环语句差不多，返回的结果是`[0, 1, 2, 3, 4]`，`do...while`与其他循环不同点在于，初始条件为假时的表现，让我们通过实际的例子来看看。 这是一个普通的 while 循环，只要`i < 5`，它就会在循环中运行代码。
+The example above behaves similar to other types of loops, and the resulting array will look like `[0, 1, 2, 3, 4]`. However, what makes the `do...while` different from other loops is how it behaves when the condition fails on the first check. Let's see this in action: Here is a regular `while` loop that will run the code in the loop as long as `i < 5`:
```js
var ourArray = [];
@@ -31,7 +31,7 @@ while (i < 5) {
}
```
-注意，我们首先将`i`的值初始化为 5。执行下一行时，注意到`i`不小于 5，循环内的代码将不会执行。所以`ourArray`最终没有添加任何内容，因此示例中的所有代码执行完时，`ourArray`仍然是`[]`。 现在，看一下`do...while`循环。
+In this example, we initialize the value of `ourArray` to an empty array and the value of `i` to 5. When we execute the `while` loop, the condition evaluates to `false` because `i` is not less than 5, so we do not execute the code inside the loop. The result is that `ourArray` will end up with no values added to it, and it will still look like `[]` when all of the code in the example above has completed running. Now, take a look at a `do...while` loop:
```js
var ourArray = [];
@@ -42,30 +42,30 @@ do {
} while (i < 5);
```
-在这里，和使用 while 循环时一样，我们将`i`的值初始化为 5。执行下一行时，没有检查`i`的值，直接执行花括号内的代码。数组会添加一个元素，并在进行条件检查之前递增`i`。然后，在条件检查时因为`i`等于 6 不符合条件`i < 5`，所以退出循环。最终`ourArray`的值是`[5]`。 本质上，`do...while`循环确保循环内的代码至少运行一次。 让我们通过`do...while`循环将值添加到数组中。
+In this case, we initialize the value of `i` to 5, just like we did with the `while` loop. When we get to the next line, there is no condition to evaluate, so we go to the code inside the curly braces and execute it. We will add a single element to the array and then increment `i` before we get to the condition check. When we finally evaluate the condition `i < 5` on the last line, we see that `i` is now 6, which fails the conditional check, so we exit the loop and are done. At the end of the above example, the value of `ourArray` is `[5]`. Essentially, a `do...while` loop ensures that the code inside the loop will run at least once. Let's try getting a `do...while` loop to work by pushing values to an array.
# --instructions--
-将代码中的`while`循环更改为`do...while`循环，实现数字 10 添加到`myArray`中，代码执行完时，`i`等于`11`。
+Change the `while` loop in the code to a `do...while` loop so the loop will push only the number `10` to `myArray`, and `i` will be equal to `11` when your code has finished running.
# --hints--
-你应该使用`do...while`循环。
+You should be using a `do...while` loop for this exercise.
```js
assert(code.match(/do/g));
```
-`myArray`应该等于`[10]`。
+`myArray` should equal `[10]`.
```js
assert.deepEqual(myArray, [10]);
```
-`i`应该等于`11`。
+`i` should equal `11`
```js
-assert.deepEqual(i, 11);
+assert.equal(i, 11);
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-with-javascript-for-loops.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-with-javascript-for-loops.md
index 32ef9e8dbc5..fdd4e3ee80f 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-with-javascript-for-loops.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-with-javascript-for-loops.md
@@ -1,6 +1,6 @@
---
id: cf1111c1c11feddfaeb5bdef
-title: for 循环
+title: Iterate with JavaScript For Loops
challengeType: 1
videoUrl: 'https://scrimba.com/c/c9yNVCe'
forumTopicId: 18219
@@ -9,21 +9,21 @@ dashedName: iterate-with-javascript-for-loops
# --description--
-你可以使用循环多次执行相同的代码。
+You can run the same code multiple times by using a loop.
-JavaScript 中最常见的循环就是 “`for循环`”。
+The most common type of JavaScript loop is called a `for` loop because it runs "for" a specific number of times.
-for循环中的三个表达式用分号隔开：
+For loops are declared with three optional expressions separated by semicolons:
-`for ([初始化]; [条件判断]; [计数器])`
+`for ([initialization]; [condition]; [final-expression])`
-`初始化`语句只会在执行循环开始之前执行一次。它通常用于定义和设置你的循环变量。
+The `initialization` statement is executed one time only before the loop starts. It is typically used to define and setup your loop variable.
-`条件判断`语句会在每一轮循环的开始执行，只要条件判断为`true`就会继续执行循环。当条件为`false`的时候，循环将停止执行。这意味着，如果条件在一开始就为`false`，这个循环将不会执行。
+The `condition` statement is evaluated at the beginning of every loop iteration and will continue as long as it evaluates to `true`. When `condition` is `false` at the start of the iteration, the loop will stop executing. This means if `condition` starts as `false`, your loop will never execute.
-`计数器`是在每一轮循环结束时执行，通常用于递增或递减。
+The `final-expression` is executed at the end of each loop iteration, prior to the next `condition` check and is usually used to increment or decrement your loop counter.
-在下面的例子中，先初始化`i = 0`，条件`i < 5`为真，进入第一次循环，执行大括号里的代码，第一次循环结束。递增`i`的值，条件判断，就这样依次执行下去，直到条件判断为假，整个循环结束。
+In the following example we initialize with `i = 0` and iterate while our condition `i < 5` is true. We'll increment `i` by `1` in each loop iteration with `i++` as our `final-expression`.
```js
var ourArray = [];
@@ -32,21 +32,21 @@ for (var i = 0; i < 5; i++) {
}
```
-最终`ourArray`的值为`[0,1,2,3,4]`.
+`ourArray` will now contain `[0,1,2,3,4]`.
# --instructions--
-使用`for`循环把从 1 到 5 添加进`myArray`中。
+Use a `for` loop to work to push the values 1 through 5 onto `myArray`.
# --hints--
-你应该使用`for`循环。
+You should be using a `for` loop for this.
```js
-assert(code.match(/for\s*\(/g).length > 1);
+assert(/for\s*\([^)]+?\)/.test(code));
```
-`myArray`应该等于`[1,2,3,4,5]`。
+`myArray` should equal `[1,2,3,4,5]`.
```js
assert.deepEqual(myArray, [1, 2, 3, 4, 5]);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-with-javascript-while-loops.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-with-javascript-while-loops.md
index dc94de17742..3b9dfd40484 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-with-javascript-while-loops.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/iterate-with-javascript-while-loops.md
@@ -1,6 +1,6 @@
---
id: cf1111c1c11feddfaeb1bdef
-title: while 循环
+title: Iterate with JavaScript While Loops
challengeType: 1
videoUrl: 'https://scrimba.com/c/c8QbnCM'
forumTopicId: 18220
@@ -9,9 +9,9 @@ dashedName: iterate-with-javascript-while-loops
# --description--
-你可以使用循环多次执行相同的代码。
+You can run the same code multiple times by using a loop.
-我们将学习的第一种类型的循环称为 "`while`" 循环，因为它规定，当 "while" 条件为真，循环才会执行，反之不执行。
+The first type of loop we will learn is called a `while` loop because it runs "while" a specified condition is true and stops once that condition is no longer true.
```js
var ourArray = [];
@@ -22,26 +22,26 @@ while(i < 5) {
}
```
-在上面的代码里，`while` 循环执行 5 次把 0 到 4 的数字添加到 `ourArray` 数组里。
+In the code example above, the `while` loop will execute 5 times and append the numbers 0 through 4 to `ourArray`.
-让我们通过 while 循环将值添加到数组中。
+Let's try getting a while loop to work by pushing values to an array.
# --instructions--
-通过一个`while`循环，把从 0 到 4 的值添加到`myArray`中。
+Add the numbers 5 through 0 (inclusive) in descending order to `myArray` using a `while` loop.
# --hints--
-你应该使用`while`循环。
+You should be using a `while` loop for this.
```js
assert(code.match(/while/g));
```
-`myArray`应该等于`[0,1,2,3,4]`。
+`myArray` should equal `[5,4,3,2,1,0]`.
```js
-assert.deepEqual(myArray, [0, 1, 2, 3, 4]);
+assert.deepEqual(myArray, [5, 4, 3, 2, 1, 0]);
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/local-scope-and-functions.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/local-scope-and-functions.md
index 0b92992bd41..14d1ca237c1 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/local-scope-and-functions.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/local-scope-and-functions.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244bf
-title: 局部作用域和函数
+title: Local Scope and Functions
challengeType: 1
videoUrl: 'https://scrimba.com/c/cd62NhM'
forumTopicId: 18227
@@ -9,9 +9,9 @@ dashedName: local-scope-and-functions
# --description--
-在一个函数内声明的变量，以及该函数的参数都是局部变量，意味着它们只在该函数内可见。
+Variables which are declared within a function, as well as the function parameters have <dfn>local</dfn> scope. That means, they are only visible within that function.
-这是在函数`myTest`内声明局部变量`loc`的例子：
+Here is a function `myTest` with a local variable called `loc`.
```js
function myTest() {
@@ -22,27 +22,33 @@ myTest(); // logs "foo"
console.log(loc); // loc is not defined
```
-在函数外，`loc`是未定义的。
+`loc` is not defined outside of the function.
# --instructions--
-在函数`myFunction`内部声明一个局部变量`myVar`，并删除外部的 console.log。
+The editor has two `console.log`s to help you see what is happening. Check the console as you code to see how it changes. Declare a local variable `myVar` inside `myLocalScope` and run the tests.
-**提示：**
-如果你遇到了问题，可以先尝试刷新页面。
+**Note:** The console will still have 'ReferenceError: myVar is not defined', but this will not cause the tests to fail.
# --hints--
-未找到全局的`myVar`变量。
+The code should not contain a global `myVar` variable.
```js
-assert(typeof myVar === 'undefined');
+function declared() {
+ myVar;
+}
+assert.throws(declared, ReferenceError);
```
-需要定义局部的`myVar`变量。
+You should add a local `myVar` variable.
```js
-assert(/var\s+myVar/.test(code));
+assert(
+ /functionmyLocalScope\(\)\{.+(var|let|const)myVar[\s\S]*}/.test(
+ __helpers.removeWhiteSpace(code)
+ )
+);
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/logical-order-in-if-else-statements.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/logical-order-in-if-else-statements.md
index 584ff5762ed..55c074ae990 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/logical-order-in-if-else-statements.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/logical-order-in-if-else-statements.md
@@ -1,6 +1,6 @@
---
id: 5690307fddb111c6084545d7
-title: if else 语句中的逻辑顺序
+title: Logical Order in If Else Statements
challengeType: 1
videoUrl: 'https://scrimba.com/c/cwNvMUV'
forumTopicId: 18228
@@ -9,13 +9,13 @@ dashedName: logical-order-in-if-else-statements
# --description--
-`if`、`else if`语句中代码的执行顺序是很重要的。
+Order is important in `if`, `else if` statements.
-在条件判断语句中，代码的执行顺序是从上到下，所以你需要考虑清楚先执行哪一句，后执行哪一句。
+The function is executed from top to bottom so you will want to be careful of what statement comes first.
-这有两个例子。
+Take these two functions as an example.
-第一个例子：
+Here's the first:
```js
function foo(x) {
@@ -29,7 +29,7 @@ function foo(x) {
}
```
-第二个例子更改了代码的执行顺序：
+And the second just switches the order of the statements:
```js
function bar(x) {
@@ -43,7 +43,7 @@ function bar(x) {
}
```
-这两个函数看起来几乎一模一样，我们传一个值进去看看它们有什么区别。
+While these two functions look nearly identical if we pass a number to both we get different outputs.
```js
foo(0) // "Less than one"
@@ -52,23 +52,23 @@ bar(0) // "Less than two"
# --instructions--
-更改函数的逻辑顺序以便通过所有的测试用例。
+Change the order of logic in the function so that it will return the correct statements in all cases.
# --hints--
-`orderMyLogic(4)`应该返回 "Less than 5"。
+`orderMyLogic(4)` should return "Less than 5"
```js
assert(orderMyLogic(4) === 'Less than 5');
```
-`orderMyLogic(6)`应该返回 "Less than 10"。
+`orderMyLogic(6)` should return "Less than 10"
```js
assert(orderMyLogic(6) === 'Less than 10');
```
-`orderMyLogic(11)`应该返回 "Greater than or equal to 10"。
+`orderMyLogic(11)` should return "Greater than or equal to 10"
```js
assert(orderMyLogic(11) === 'Greater than or equal to 10');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulate-arrays-with-pop.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulate-arrays-with-pop.md
index 4026f2ecbc7..ae2cc7fab30 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulate-arrays-with-pop.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulate-arrays-with-pop.md
@@ -1,6 +1,6 @@
---
id: 56bbb991ad1ed5201cd392cc
-title: 使用 pop() 操作数组
+title: Manipulate Arrays With pop()
challengeType: 1
videoUrl: 'https://scrimba.com/c/cRbVZAB'
forumTopicId: 18236
@@ -9,11 +9,11 @@ dashedName: manipulate-arrays-with-pop
# --description--
-改变数组中数据的另一种方法是用`.pop()`函数。
+Another way to change the data in an array is with the `.pop()` function.
-`.pop()`函数用来“抛出”一个数组末尾的值。我们可以把这个“抛出”的值赋给一个变量存储起来。换句话说就是`.pop()`函数移除数组末尾的元素并返回这个元素。
+`.pop()` is used to "pop" a value off of the end of an array. We can store this "popped off" value by assigning it to a variable. In other words, `.pop()` removes the last element from an array and returns that element.
-数组中任何类型的元素（数值，字符串，甚至是数组）可以被“抛出来” 。
+Any type of entry can be "popped" off of an array - numbers, strings, even nested arrays.
```js
var threeArr = [1, 4, 6];
@@ -24,11 +24,11 @@ console.log(threeArr); // Returns [1, 4]
# --instructions--
-使用`.pop()`函数移除`myArray`中的最后一个元素，并且把“抛出”的值赋给`removedFromMyArray`。
+Use the `.pop()` function to remove the last item from `myArray`, assigning the "popped off" value to `removedFromMyArray`.
# --hints--
-`myArray`应该只包含`[["John", 23]]`。
+`myArray` should only contain `[["John", 23]]`.
```js
assert(
@@ -42,13 +42,13 @@ assert(
);
```
-对`myArray`使用`pop()`函数。
+You should use `pop()` on `myArray`.
```js
assert(/removedFromMyArray\s*=\s*myArray\s*.\s*pop\s*(\s*)/.test(code));
```
-`removedFromMyArray`应该只包含`["cat", 2]`。
+`removedFromMyArray` should only contain `["cat", 2]`.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulate-arrays-with-push.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulate-arrays-with-push.md
index c734a19d883..b6fcea06819 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulate-arrays-with-push.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulate-arrays-with-push.md
@@ -1,6 +1,6 @@
---
id: 56bbb991ad1ed5201cd392cb
-title: 使用 push() 操作数组
+title: Manipulate Arrays With push()
challengeType: 1
videoUrl: 'https://scrimba.com/c/cnqmVtJ'
forumTopicId: 18237
@@ -9,23 +9,29 @@ dashedName: manipulate-arrays-with-push
# --description--
-一个简单的方法将数据添加到一个数组的末尾是通过`push()`函数。
+An easy way to append data to the end of an array is via the `push()` function.
-`.push()`接受一个或多个参数，并把它“推”入到数组的末尾。
+`.push()` takes one or more <dfn>parameters</dfn> and "pushes" them onto the end of the array.
+
+Examples:
```js
-var arr = [1,2,3];
-arr.push(4);
-// arr is now [1,2,3,4]
+var arr1 = [1,2,3];
+arr1.push(4);
+// arr1 is now [1,2,3,4]
+
+var arr2 = ["Stimpson", "J", "cat"];
+arr2.push(["happy", "joy"]);
+// arr2 now equals ["Stimpson", "J", "cat", ["happy", "joy"]]
```
# --instructions--
-把`["dog", 3]`“推”入到`myArray`变量的末尾。
+Push `["dog", 3]` onto the end of the `myArray` variable.
# --hints--
-`myArray`应该等于`[["John", 23], ["cat", 2], ["dog", 3]]`。
+`myArray` should now equal `[["John", 23], ["cat", 2], ["dog", 3]]`.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulate-arrays-with-shift.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulate-arrays-with-shift.md
index 52da6a66795..4a68ea9135a 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulate-arrays-with-shift.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulate-arrays-with-shift.md
@@ -1,6 +1,6 @@
---
id: 56bbb991ad1ed5201cd392cd
-title: 使用 shift() 操作数组
+title: Manipulate Arrays With shift()
challengeType: 1
videoUrl: 'https://scrimba.com/c/cRbVETW'
forumTopicId: 18238
@@ -9,17 +9,25 @@ dashedName: manipulate-arrays-with-shift
# --description--
-`pop()`函数用来移出数组中最后一个元素。如果想要移出第一个元素要怎么办呢？
+`pop()` always removes the last element of an array. What if you want to remove the first?
-这就是`.shift()`的用武之地。它的工作原理就像`.pop()`，但它移除的是第一个元素，而不是最后一个。
+That's where `.shift()` comes in. It works just like `.pop()`, except it removes the first element instead of the last.
+
+Example:
+
+```js
+var ourArray = ["Stimpson", "J", ["cat"]];
+var removedFromOurArray = ourArray.shift();
+// removedFromOurArray now equals "Stimpson" and ourArray now equals ["J", ["cat"]].
+```
# --instructions--
-使用`.shift()`函数移出`myArray`中的第一项，并把“移出”的值赋给`removedFromMyArray`。
+Use the `.shift()` function to remove the first item from `myArray`, assigning the "shifted off" value to `removedFromMyArray`.
# --hints--
-`myArray`应该等于`[["dog", 3]]`。
+`myArray` should now equal `[["dog", 3]]`.
```js
assert(
@@ -33,7 +41,7 @@ assert(
);
```
-`removedFromMyArray`应该包含`["John", 23]`。
+`removedFromMyArray` should contain `["John", 23]`.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulate-arrays-with-unshift.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulate-arrays-with-unshift.md
index 99b1fbe8bb6..39d4b3c002d 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulate-arrays-with-unshift.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulate-arrays-with-unshift.md
@@ -1,6 +1,6 @@
---
id: 56bbb991ad1ed5201cd392ce
-title: 使用 unshift() 操作数组
+title: Manipulate Arrays With unshift()
challengeType: 1
videoUrl: 'https://scrimba.com/c/ckNDESv'
forumTopicId: 18239
@@ -9,17 +9,26 @@ dashedName: manipulate-arrays-with-unshift
# --description--
-你不仅可以`shift`（移出）数组中的第一个元素，你也可以`unshift`（移入）一个元素到数组的头部。
+Not only can you `shift` elements off of the beginning of an array, you can also `unshift` elements to the beginning of an array i.e. add elements in front of the array.
-`.unshift()`函数用起来就像`.push()`函数一样, 但不是在数组的末尾添加元素，而是在数组的头部添加元素。
+`.unshift()` works exactly like `.push()`, but instead of adding the element at the end of the array, `unshift()` adds the element at the beginning of the array.
+
+Example:
+
+```js
+var ourArray = ["Stimpson", "J", "cat"];
+ourArray.shift(); // ourArray now equals ["J", "cat"]
+ourArray.unshift("Happy");
+// ourArray now equals ["Happy", "J", "cat"]
+```
# --instructions--
-使用`unshift()`函数把`["Paul",35]`加入到`myArray`的头部。
+Add `["Paul",35]` to the beginning of the `myArray` variable using `unshift()`.
# --hints--
-`myArray`应该包含\[["Paul", 35], ["dog", 3]]。
+`myArray` should now have \[["Paul", 35], ["dog", 3]].
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulating-complex-objects.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulating-complex-objects.md
index 1dbca196df5..61d0af1cdae 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulating-complex-objects.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/manipulating-complex-objects.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244cb
-title: 操作复杂对象
+title: Manipulating Complex Objects
challengeType: 1
videoUrl: 'https://scrimba.com/c/c9yNMfR'
forumTopicId: 18208
@@ -9,9 +9,9 @@ dashedName: manipulating-complex-objects
# --description--
-有时你可能希望将数据存储在灵活的<dfn>数据结构</dfn>中。JavaScript 对象是处理灵活数据的一种方法。它可以储存<dfn>字符串</dfn>，<dfn>数字</dfn>，<dfn>布尔值</dfn>，<dfn>函数</dfn>，和<dfn>对象</dfn>以及这些值的任意组合。
+Sometimes you may want to store data in a flexible <dfn>Data Structure</dfn>. A JavaScript object is one way to handle flexible data. They allow for arbitrary combinations of <dfn>strings</dfn>, <dfn>numbers</dfn>, <dfn>booleans</dfn>, <dfn>arrays</dfn>, <dfn>functions</dfn>, and <dfn>objects</dfn>.
-这是一个复杂数据结构的示例：
+Here's an example of a complex data structure:
```js
var ourMusic = [
@@ -29,7 +29,7 @@ var ourMusic = [
];
```
-这是一个对象数组，并且对象有各种关于专辑的 <dfn>详细信息</dfn>。它也有一个嵌套的`formats`的数组。附加专辑记录可以被添加到数组的最上层。 对象将数据以一种键-值对的形式保存。在上面的示例中，`"artist": "Daft Punk"`是一个具有`"artist"`键和`"Daft Punk"`值的属性。 [JavaScript Object Notation](http://www.json.org/) 简称`JSON`是用于存储数据的相关数据交换格式。
+This is an array which contains one object inside. The object has various pieces of <dfn>metadata</dfn> about an album. It also has a nested `"formats"` array. If you want to add more album records, you can do this by adding records to the top level array. Objects hold data in a property, which has a key-value format. In the example above, `"artist": "Daft Punk"` is a property that has a key of `"artist"` and a value of `"Daft Punk"`. [JavaScript Object Notation](http://www.json.org/) or `JSON` is a related data interchange format used to store data.
```json
{
@@ -45,40 +45,40 @@ var ourMusic = [
}
```
-**提示**
-数组中有多个 JSON 对象的时候，对象与对象之间要用逗号隔开。
+**Note**
+You will need to place a comma after every object in the array, unless it is the last object in the array.
# --instructions--
-添加一个新专辑到`myMusic`的JSON对象。添加`artist`和`title`字符串，`release_year`数字和`formats`字符串数组。
+Add a new album to the `myMusic` array. Add `artist` and `title` strings, `release_year` number, and a `formats` array of strings.
# --hints--
-`myMusic`应该是一个数组。
+`myMusic` should be an array
```js
assert(Array.isArray(myMusic));
```
-`myMusic`应该至少包含两个元素。
+`myMusic` should have at least two elements
```js
assert(myMusic.length > 1);
```
-`myMusic[1]`应该是一个对象。
+`myMusic[1]` should be an object
```js
assert(typeof myMusic[1] === 'object');
```
-`myMusic[1]`至少要包含四个属性。
+`myMusic[1]` should have at least 4 properties
```js
assert(Object.keys(myMusic[1]).length > 3);
```
-`myMusic[1]`应该包含一个类型为字符串的`artist`的属性。
+`myMusic[1]` should contain an `artist` property which is a string
```js
assert(
@@ -86,7 +86,7 @@ assert(
);
```
-`myMusic[1]`应该包含一个类型为字符串的`title`的属性。
+`myMusic[1]` should contain a `title` property which is a string
```js
assert(
@@ -94,7 +94,7 @@ assert(
);
```
-`myMusic[1]`应该包含一个类型为数字的`release_year` 应该包含一个类型为数字的属性。
+`myMusic[1]` should contain a `release_year` property which is a number
```js
assert(
@@ -103,7 +103,7 @@ assert(
);
```
-`myMusic[1]`应该包含一个类型为数组的`formats`属性。
+`myMusic[1]` should contain a `formats` property which is an array
```js
assert(
@@ -111,7 +111,7 @@ assert(
);
```
-`formats`应该是一个至少包含两个字符串元素的数组。
+`formats` should be an array of strings with at least two elements
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/modify-array-data-with-indexes.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/modify-array-data-with-indexes.md
index e6dd8e8d7ce..86fa3ee5ba9 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/modify-array-data-with-indexes.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/modify-array-data-with-indexes.md
@@ -1,6 +1,6 @@
---
id: cf1111c1c11feddfaeb8bdef
-title: 通过索引修改数组中的数据
+title: Modify Array Data With Indexes
challengeType: 1
videoUrl: 'https://scrimba.com/c/czQM4A8'
forumTopicId: 18241
@@ -9,25 +9,25 @@ dashedName: modify-array-data-with-indexes
# --description--
-与字符串的数据不可变不同，数组的数据是可变的，并且可以自由地改变。
+Unlike strings, the entries of arrays are <dfn>mutable</dfn> and can be changed freely.
-**示例**
+**Example**
```js
var ourArray = [50,40,30];
ourArray[0] = 15; // equals [15,40,30]
```
-**提示**
-数组名称和方括号之间不应有任何空格，如`array [0]`尽管 JavaScript 能够正确处理，但可能会让看你代码的其他程序员感到困惑。
+**Note**
+There shouldn't be any spaces between the array name and the square brackets, like `array [0]`. Although JavaScript is able to process this correctly, this may confuse other programmers reading your code.
# --instructions--
-修改数组`myArray`中索引0上的值为`45`。
+Modify the data stored at index `0` of `myArray` to a value of `45`.
# --hints--
-`myArray`的值应该 [45,64,99]。
+`myArray` should now be [45,64,99].
```js
assert(
@@ -46,7 +46,7 @@ assert(
);
```
-你应该使用正确的索引修改`myArray`的值。
+You should be using correct index to modify the value in `myArray`.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/multiple-identical-options-in-switch-statements.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/multiple-identical-options-in-switch-statements.md
index f18b5dfbbd1..72b18e49284 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/multiple-identical-options-in-switch-statements.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/multiple-identical-options-in-switch-statements.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244df
-title: 在 Switch 语句添加多个相同选项
+title: Multiple Identical Options in Switch Statements
challengeType: 1
videoUrl: 'https://scrimba.com/c/cdBKWCV'
forumTopicId: 18242
@@ -9,9 +9,10 @@ dashedName: multiple-identical-options-in-switch-statements
# --description--
-如果你忘了给`switch`的每一条`case`添加`break`，那么直到遇见`break`为止，后续的`case`会一直执行。如果你想为多个不同的输入设置相同的结果，可以这样写：
+If the `break` statement is omitted from a `switch` statement's `case`, the following `case` statement(s) are executed until a `break` is encountered. If you have multiple inputs with the same output, you can represent them in a `switch` statement like this:
```js
+var result = "";
switch(val) {
case 1:
case 2:
@@ -23,81 +24,81 @@ switch(val) {
}
```
-这样，1、2、3 都会有相同的结果。
+Cases for 1, 2, and 3 will all produce the same result.
# --instructions--
-请写一个`switch`语句，根据输入的`val`的范围得出对应的`answer`：
+Write a switch statement to set `answer` for the following ranges:
`1-3` - "Low"
`4-6` - "Mid"
`7-9` - "High"
-**提示：**
-你的`case`应基于范围中的每一个数字编写。
+**Note**
+You will need to have a `case` statement for each number in the range.
# --hints--
-`sequentialSizes(1)`应该返回 "Low"。
+`sequentialSizes(1)` should return "Low"
```js
assert(sequentialSizes(1) === 'Low');
```
-`sequentialSizes(2)`应该返回 "Low"。
+`sequentialSizes(2)` should return "Low"
```js
assert(sequentialSizes(2) === 'Low');
```
-`sequentialSizes(3)`应该返回 "Low"。
+`sequentialSizes(3)` should return "Low"
```js
assert(sequentialSizes(3) === 'Low');
```
-`sequentialSizes(4)`应该返回 "Mid"。
+`sequentialSizes(4)` should return "Mid"
```js
assert(sequentialSizes(4) === 'Mid');
```
-`sequentialSizes(5)`应该返回 "Mid"。
+`sequentialSizes(5)` should return "Mid"
```js
assert(sequentialSizes(5) === 'Mid');
```
-`sequentialSizes(6)`应该返回 "Mid"。
+`sequentialSizes(6)` should return "Mid"
```js
assert(sequentialSizes(6) === 'Mid');
```
-`sequentialSizes(7)`应该返回 "High"。
+`sequentialSizes(7)` should return "High"
```js
assert(sequentialSizes(7) === 'High');
```
-`sequentialSizes(8)`应该返回 "High"。
+`sequentialSizes(8)` should return "High"
```js
assert(sequentialSizes(8) === 'High');
```
-`sequentialSizes(9)`应该返回 "High"。
+`sequentialSizes(9)` should return "High"
```js
assert(sequentialSizes(9) === 'High');
```
-你不应使用`if`或`else`语句。
+You should not use any `if` or `else` statements
```js
assert(!/else/g.test(code) || !/if/g.test(code));
```
-你应该编写 9 个`case`语句。
+You should have nine `case` statements
```js
assert(code.match(/case/g).length === 9);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/multiply-two-decimals-with-javascript.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/multiply-two-decimals-with-javascript.md
index efe7850d8c9..2a502314bb8 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/multiply-two-decimals-with-javascript.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/multiply-two-decimals-with-javascript.md
@@ -1,6 +1,6 @@
---
id: bd7993c9c69feddfaeb7bdef
-title: 两个小数相乘
+title: Multiply Two Decimals with JavaScript
challengeType: 1
videoUrl: 'https://scrimba.com/c/ce2GeHq'
forumTopicId: 301173
@@ -9,23 +9,23 @@ dashedName: multiply-two-decimals-with-javascript
# --description--
-在 JavaScript 中，你也可以用小数进行计算，就像整数一样。
+In JavaScript, you can also perform calculations with decimal numbers, just like whole numbers.
-把两个小数相乘，并得到它们乘积。
+Let's multiply two decimals together to get their product.
# --instructions--
-改变`0.0`的数值让变量`product`的值等于`5.0`。
+Change the `0.0` so that product will equal `5.0`.
# --hints--
-变量`product`应该等于`5.0`。
+The variable `product` should equal `5.0`.
```js
assert(product === 5.0);
```
-要使用`*`运算符。
+You should use the `*` operator
```js
assert(/\*/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/multiply-two-numbers-with-javascript.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/multiply-two-numbers-with-javascript.md
index 857970d428a..5be738a56fb 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/multiply-two-numbers-with-javascript.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/multiply-two-numbers-with-javascript.md
@@ -1,6 +1,6 @@
---
id: cf1231c1c11feddfaeb5bdef
-title: 乘法运算
+title: Multiply Two Numbers with JavaScript
challengeType: 1
videoUrl: 'https://scrimba.com/c/cP3y3Aq'
forumTopicId: 18243
@@ -9,11 +9,11 @@ dashedName: multiply-two-numbers-with-javascript
# --description--
-我们也可在 JavaScript 中使用乘法运算。
+We can also multiply one number by another.
-JavaScript 使用`*`符号表示两数相乘。
+JavaScript uses the `*` symbol for multiplication of two numbers.
-**示例**
+**Example**
```js
myVar = 13 * 13; // assigned 169
@@ -21,17 +21,17 @@ myVar = 13 * 13; // assigned 169
# --instructions--
-改变数值`0`来让变量 product 的值等于`80`。
+Change the `0` so that product will equal `80`.
# --hints--
-要使`product`的值等于 80。
+The variable `product` should be equal to 80.
```js
assert(product === 80);
```
-使用`*`运算符。
+You should use the `*` operator.
```js
assert(/\*/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/nest-one-array-within-another-array.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/nest-one-array-within-another-array.md
index 6214869efbf..de2c0bbe8d8 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/nest-one-array-within-another-array.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/nest-one-array-within-another-array.md
@@ -1,6 +1,6 @@
---
id: cf1111c1c11feddfaeb7bdef
-title: 将一个数组嵌套在另一个数组中
+title: Nest one Array within Another Array
challengeType: 1
videoUrl: 'https://scrimba.com/c/crZQZf8'
forumTopicId: 18247
@@ -9,15 +9,21 @@ dashedName: nest-one-array-within-another-array
# --description--
-你也可以在数组中包含其他数组，例如：`[["Bulls", 23], ["White Sox", 45]]`。这被称为一个<dfn>多维数组<dfn>。</dfn></dfn>
+You can also nest arrays within other arrays, like below:
+
+```js
+[["Bulls", 23], ["White Sox", 45]]
+```
+
+This is also called a <dfn>multi-dimensional array<dfn>.</dfn></dfn>
# --instructions--
-创建一个名为`myArray`的多维数组。
+Create a nested array called `myArray`.
# --hints--
-应该包含至少一个嵌入的数组。
+`myArray` should have at least one array nested within another array.
```js
assert(Array.isArray(myArray) && myArray.some(Array.isArray));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/nesting-for-loops.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/nesting-for-loops.md
index 15e4ce8fc85..8b9042e06fc 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/nesting-for-loops.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/nesting-for-loops.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244e1
-title: 循环嵌套
+title: Nesting For Loops
challengeType: 1
videoUrl: 'https://scrimba.com/c/cRn6GHM'
forumTopicId: 18248
@@ -9,7 +9,7 @@ dashedName: nesting-for-loops
# --description--
-如果你有一个二维数组，可以使用相同的逻辑，先遍历外面的数组，再遍历里面的子数组。下面是一个例子：
+If you have a multi-dimensional array, you can use the same logic as the prior waypoint to loop through both the array and any sub-arrays. Here is an example:
```js
var arr = [
@@ -22,21 +22,21 @@ for (var i=0; i < arr.length; i++) {
}
```
-一次输出`arr`中的每个子元素。提示，对于内部循环，我们可以通过`arr[i]`的`.length`来获得子数组的长度，因为`arr[i]`的本身就是一个数组。
+This outputs each sub-element in `arr` one at a time. Note that for the inner loop, we are checking the `.length` of `arr[i]`, since `arr[i]` is itself an array.
# --instructions--
-修改函数`multiplyAll`，获得`arr`内部数组的每个数字相乘的结果`product`。
+Modify function `multiplyAll` so that it returns the product of all the numbers in the sub-arrays of `arr`.
# --hints--
-`multiplyAll([[1],[2],[3]])`应该返回 `6`。
+`multiplyAll([[1],[2],[3]])` should return `6`
```js
assert(multiplyAll([[1], [2], [3]]) === 6);
```
-`multiplyAll([[1,2],[3,4],[5,6,7]])`应该返回 `5040`。
+`multiplyAll([[1,2],[3,4],[5,6,7]])` should return `5040`
```js
assert(
@@ -48,7 +48,7 @@ assert(
);
```
-`multiplyAll([[5,1],[0.2, 4, 0.5],[3, 9]])`应该返回 `54`。
+`multiplyAll([[5,1],[0.2, 4, 0.5],[3, 9]])` should return `54`
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/passing-values-to-functions-with-arguments.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/passing-values-to-functions-with-arguments.md
index 3ed59773029..9c38b815fca 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/passing-values-to-functions-with-arguments.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/passing-values-to-functions-with-arguments.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244bd
-title: 将值传递给带有参数的函数
+title: Passing Values to Functions with Arguments
challengeType: 1
videoUrl: 'https://scrimba.com/c/cy8rahW'
forumTopicId: 18254
@@ -9,9 +9,9 @@ dashedName: passing-values-to-functions-with-arguments
# --description--
-函数的参数`parameters`在函数中充当占位符(也叫形参)的作用，参数可以为一个或多个。调用一个函数时所传入的参数为实参，实参决定着形参真正的值。简单理解：形参即形式、实参即内容。
+<dfn>Parameters</dfn> are variables that act as placeholders for the values that are to be input to a function when it is called. When a function is defined, it is typically defined along with one or more parameters. The actual values that are input (or <dfn>"passed"</dfn>) into a function when it is called are known as <dfn>arguments</dfn>.
-这是带有两个参数的函数，`param1`和`param2`：
+Here is a function with two parameters, `param1` and `param2`:
```js
function testFun(param1, param2) {
@@ -19,21 +19,21 @@ function testFun(param1, param2) {
}
```
-接着我们调用`testFun`： `testFun("Hello", "World");` 我们传递了两个参数，`"Hello"`和`"World"`。在函数内部，`param1`等于“Hello”，`param2`等于“World”。请注意，`testFun`函数可以多次调用，每次调用时传递的参数会决定形参的实际值。
+Then we can call `testFun`: `testFun("Hello", "World");` We have passed two arguments, `"Hello"` and `"World"`. Inside the function, `param1` will equal "Hello" and `param2` will equal "World". Note that you could call `testFun` again with different arguments and the parameters would take on the value of the new arguments.
# --instructions--
-<ol><li>创建一个名为<code>functionWithArgs</code>的函数，它可以接收两个参数，计算参数的和，将结果输出到控制台。</li><li>调用这个函数。</li></ol>
+<ol><li>Create a function called <code>functionWithArgs</code> that accepts two arguments and outputs their sum to the dev console.</li><li>Call the function with two numbers as arguments.</li></ol>
# --hints--
-`functionWithArgs`应该是一个函数。
+`functionWithArgs` should be a function.
```js
assert(typeof functionWithArgs === 'function');
```
-`functionWithArgs(1,2)`应该输出`3`。
+`functionWithArgs(1,2)` should output `3`.
```js
if (typeof functionWithArgs === 'function') {
@@ -44,7 +44,7 @@ if (typeof functionWithArgs === 'function') {
assert(logOutput == 3);
```
-`functionWithArgs(7,9)`应该输出`16`。
+`functionWithArgs(7,9)` should output `16`.
```js
if (typeof functionWithArgs === 'function') {
@@ -55,10 +55,14 @@ if (typeof functionWithArgs === 'function') {
assert(logOutput == 16);
```
-在你定义`functionWithArgs`之后记得调用它。
+You should call `functionWithArgs` with two numbers after you define it.
```js
-assert(/^\s*functionWithArgs\s*\(\s*\d+\s*,\s*\d+\s*\)\s*;?/m.test(code));
+assert(
+ /functionWithArgs\([-+]?\d*\.?\d*,[-+]?\d*\.?\d*\)/.test(
+ code.replace(/\s/g, '')
+ )
+);
```
# --seed--
@@ -101,7 +105,6 @@ if (typeof functionWithArgs !== "function") {
```
## --seed-contents--
-
```js
```
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/practice-comparing-different-values.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/practice-comparing-different-values.md
index 6a03384773a..c97f6c7eb9d 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/practice-comparing-different-values.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/practice-comparing-different-values.md
@@ -1,6 +1,6 @@
---
id: 599a789b454f2bbd91a3ff4d
-title: 比较不同值
+title: Practice comparing different values
challengeType: 1
videoUrl: 'https://scrimba.com/c/cm8PqCa'
forumTopicId: 301174
@@ -9,21 +9,19 @@ dashedName: practice-comparing-different-values
# --description--
-在上两个挑战中，我们学习了相等运算符 (`==`) 和严格相等运算符 (`===`)。现在让我们快速回顾并实践一下。
+In the last two challenges, we learned about the equality operator (`==`) and the strict equality operator (`===`). Let's do a quick review and practice using these operators some more.
-如果要比较的值不是同一类型，相等运算符会先执行数据类型转换，然后比较值。而严格相等运算符只比较值，不会进行数据类型转换。
+If the values being compared are not of the same type, the equality operator will perform a type conversion, and then evaluate the values. However, the strict equality operator will compare both the data type and value as-is, without converting one type to the other.
-由此可见，相等运算符和严格相等运算符的区别是：前者会执行隐式类型转换，后者不会。
-
-**示例**
+**Examples**
```js
3 == '3' // returns true because JavaScript performs type conversion from string to number
3 === '3' // returns false because the types are different and type conversion is not performed
```
-**提示**
-在JavaScript中，你可以使用`typeof`运算符确定变量的类型或值，如下所示：
+**Note**
+In JavaScript, you can determine the type of a variable or a value with the `typeof` operator, as follows:
```js
typeof 3 // returns 'number'
@@ -32,23 +30,23 @@ typeof '3' // returns 'string'
# --instructions--
-编辑器中的`compareEquality`函数使用相等运算符比较两个值。修改函数，使其仅在值严格相等时返回 "Equal" 。
+The `compareEquality` function in the editor compares two values using the equality operator. Modify the function so that it returns "Equal" only when the values are strictly equal.
# --hints--
-`compareEquality(10, "10")`应该返回 "Not Equal"。
+`compareEquality(10, "10")` should return "Not Equal"
```js
assert(compareEquality(10, '10') === 'Not Equal');
```
-`compareEquality("20", 20)`应该返回 "Not Equal"。
+`compareEquality("20", 20)` should return "Not Equal"
```js
assert(compareEquality('20', 20) === 'Not Equal');
```
-你应该使用`===`运算符。
+You should use the `===` operator
```js
assert(code.match(/===/g));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/profile-lookup.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/profile-lookup.md
index ad35fdd2710..4f2c51e24c0 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/profile-lookup.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/profile-lookup.md
@@ -1,6 +1,6 @@
---
id: 5688e62ea601b2482ff8422b
-title: 资料查找
+title: Profile Lookup
challengeType: 1
videoUrl: 'https://scrimba.com/c/cDqW2Cg'
forumTopicId: 18259
@@ -9,27 +9,27 @@ dashedName: profile-lookup
# --description--
-我们有一个对象数组，里面存储着通讯录。
+We have an array of objects representing different people in our contacts lists.
-函数`lookUp`有两个预定义参数：`firstName`值和`prop`属性 。
+A `lookUpProfile` function that takes `name` and a property (`prop`) as arguments has been pre-written for you.
-函数将会检查通讯录中是否存在一个与传入的`firstName`相同的联系人。如果存在，那么还需要检查对应的联系人中是否存在`prop`属性。
+The function should check if `name` is an actual contact's `firstName` and the given property (`prop`) is a property of that contact.
-如果它们都存在，函数返回`prop`属性对应的值。
+If both are true, then return the "value" of that property.
-如果`firstName`值不存在，返回`"No such contact"`。
+If `name` does not correspond to any contacts then return `"No such contact"`.
-如果`prop`属性不存在，返回`"No such property"`。
+If `prop` does not correspond to any valid properties of a contact found to match `name` then return `"No such property"`.
# --hints--
-`"Kristian", "lastName"`应该返回 `"Vos"`。
+`lookUpProfile("Kristian", "lastName")` should return `"Vos"`
```js
assert(lookUpProfile('Kristian', 'lastName') === 'Vos');
```
-`"Sherlock", "likes"`应该返回 `["Intriguing Cases", "Violin"]`。
+`lookUpProfile("Sherlock", "likes")` should return `["Intriguing Cases", "Violin"]`
```js
assert.deepEqual(lookUpProfile('Sherlock', 'likes'), [
@@ -38,25 +38,25 @@ assert.deepEqual(lookUpProfile('Sherlock', 'likes'), [
]);
```
-`"Harry","likes"`应该返回 an array。
+`lookUpProfile("Harry", "likes")` should return an array
```js
assert(typeof lookUpProfile('Harry', 'likes') === 'object');
```
-`"Bob", "number"`应该返回 "No such contact"。
+`lookUpProfile("Bob", "number")` should return "No such contact"
```js
assert(lookUpProfile('Bob', 'number') === 'No such contact');
```
-`"Bob", "potato"`应该返回 "No such contact"。
+`lookUpProfile("Bob", "potato")` should return "No such contact"
```js
assert(lookUpProfile('Bob', 'potato') === 'No such contact');
```
-`"Akira", "address"`应该返回 "No such property"。
+`lookUpProfile("Akira", "address")` should return "No such property"
```js
assert(lookUpProfile('Akira', 'address') === 'No such property');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/quoting-strings-with-single-quotes.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/quoting-strings-with-single-quotes.md
index 5a23743bbc1..f7a36cb16d0 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/quoting-strings-with-single-quotes.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/quoting-strings-with-single-quotes.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244b4
-title: 用单引号引用字符串
+title: Quoting Strings with Single Quotes
challengeType: 1
videoUrl: 'https://scrimba.com/c/cbQmnhM'
forumTopicId: 18260
@@ -9,38 +9,39 @@ dashedName: quoting-strings-with-single-quotes
# --description--
-JavaScript 中的<dfn>字符串</dfn>可以使用开始和结束都是同类型的单引号或双引号表示，与其他一些编程语言不同的是，单引号和双引号的功能在 JavaScript 中是相同的。
+<dfn>String</dfn> values in JavaScript may be written with single or double quotes, as long as you start and end with the same type of quote. Unlike some other programming languages, single and double quotes work the same in JavaScript.
```js
doubleQuoteStr = "This is a string";
singleQuoteStr = 'This is also a string';
```
-当你需要在一个字符串中使用多个引号的时候，你可以使用单引号包裹双引号或者相反。常见的场景比如在字符串中包含对话的句子需要用引号包裹。另外比如在一个包含有`<a>`标签的字符串中，`<a>`标签的属性值需要用引号包裹。
+The reason why you might want to use one type of quote over the other is if you want to use both in a string. This might happen if you want to save a conversation in a string and have the conversation in quotes. Another use for it would be saving an `<a>` tag with various attributes in quotes, all within a string.
```js
conversation = 'Finn exclaims to Jake, "Algebraic!"';
```
-但是，如果你想在字符串中使用与最外层相同的引号，会有一些问题。要知道，字符串在开头和结尾都有相同的引号，如果在中间使用了相同的引号，字符串会提前中止并抛出错误。
+However, this becomes a problem if you need to use the outermost quotes within it. Remember, a string has the same kind of quote at the beginning and end. But if you have that same quote somewhere in the middle, the string will stop early and throw an error.
```js
goodStr = 'Jake asks Finn, "Hey, let\'s go on an adventure?"';
badStr = 'Finn responds, "Let's go!"'; // Throws an error
```
-在上面的`goodStr`中，通过使用反斜杠`\`转义字符可以安全地使用两种引号 **提示**
-不要把反斜杠`\`和斜杠`/`搞混，它们不是一回事。
+In the <dfn>goodStr</dfn> above, you can use both quotes safely by using the backslash <code>\\</code> as an escape character.
+
+**Note:** The backslash <code>\\</code> should not be confused with the forward slash `/`. They do not do the same thing.
# --instructions--
-把字符串更改为开头和结尾使用单引号的字符串，并且不包含转义字符。
+Change the provided string to a string with single quotes at the beginning and end and no escape characters.
-这样字符串中的`<a>`标签里面任何地方都可以使用双引号。你需要将最外层引号更改为单引号，以便删除转义字符。
+Right now, the `<a>` tag in the string uses double quotes everywhere. You will need to change the outer quotes to single quotes so you can remove the escape characters.
# --hints--
-删除所有`反斜杠` (`\`)。
+You should remove all the `backslashes` (<code>\\</code>).
```js
assert(
@@ -51,7 +52,7 @@ assert(
);
```
-应该要有两个单引号`'`和四个双引号`"`。
+You should have two single quotes `'` and four double quotes `"`.
```js
assert(code.match(/"/g).length === 4 && code.match(/'/g).length === 2);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/record-collection.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/record-collection.md
index 1924fc99bde..d3335ba646c 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/record-collection.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/record-collection.md
@@ -1,97 +1,88 @@
---
id: 56533eb9ac21ba0edf2244cf
-title: 记录集合
+title: Record Collection
challengeType: 1
-videoUrl: 'https://scrimba.com/c/c4mpysg'
forumTopicId: 18261
dashedName: record-collection
---
# --description--
-给定一个 JSON 对象，用来表示部分音乐专辑收藏。每张专辑都有几个属性和一个唯一的 id 号作为键值。并非所有专辑都有完整的信息。
+You are given a JSON object representing a part of your musical album collection. Each album has a unique id number as its key and several other properties. Not all albums have complete information.
-写一个函数，根据传入的`id`（如`2548`）、`prop`（属性，如`"artist"`或`"tracks"`）以及`value`（值，如`"Addicted to Love"`）来修改音乐专辑收藏的数据。
+You start with an `updateRecords` function that takes an object like `collection`, an `id`, a `prop` (like `artist` or `tracks`), and a `value`. Complete the function using the rules below to modify the object passed to the function.
-如果属性`prop`不是`"tracks"`且值`value`不为空（`""`），则更新或设置该专辑属性的值`value`。
+- Your function must always return the entire object.
+- If `prop` isn't `tracks` and `value` isn't an empty string, update or set that album's `prop` to `value`.
+- If `prop` is `tracks` but the album doesn't have a `tracks` property, create an empty array and add `value` to it.
+- If `prop` is `tracks` and `value` isn't an empty string, add `value` to the end of the album's existing `tracks` array.
+- If `value` is an empty string, delete the given `prop` property from the album.
-你的函数必须始终返回整个音乐专辑集合对象。
-
-处理不完整数据有几条规则：
-
-如果属性`prop`是`"tracks"`，但是专辑没有`"tracks"`属性，则在添加值之前先给`"tracks"`创建一个空数组。
-
-如果`prop`是`"tracks"`，并且值`value`不为空（`""`）， 把值`value`添加到`tracks`数组中。
-
-如果值`value`为空（`""`），则删除专辑的这一属性`prop`
-
-**提示：**
-[通过变量访问对象的属性](javascript-algorithms-and-data-structures/basic-javascript/accessing-object-properties-with-variables)时，应使用`中括号`。
-
-Push 是一个数组方法，详情请查看[Mozilla Developer Network](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/push).
-
-你可以参考[操作复杂对象](/javascript-algorithms-and-data-structures/basic-javascript/manipulating-complex-objects)这一节的内容复习相关知识。
+**Note:** A copy of the `collection` object is used for the tests.
# --hints--
-执行`updateRecords(5439, "artist", "ABBA")`后，`artist`属性值应该是`"ABBA"`。
+After `updateRecords(collection, 5439, "artist", "ABBA")`, `artist` should be `ABBA`
```js
assert(
- code.match(
- /var collection = {\s*2548: {\s*album: "Slippery When Wet",\s*artist: "Bon Jovi",\s*tracks: \[\s*"Let It Rock",\s*"You Give Love a Bad Name"\s*\]\s*},\s*2468: {\s*album: "1999",\s*artist: "Prince",\s*tracks: \[\s*"1999",\s*"Little Red Corvette"\s*\]\s*},\s*1245: {\s*artist: "Robert Palmer",\s*tracks: \[ \]\s*},\s*5439: {\s*album: "ABBA Gold"\s*}\s*};/g
- )
+ updateRecords(_recordCollection, 5439, 'artist', 'ABBA')[5439]['artist'] ===
+ 'ABBA'
);
```
-执行`updateRecords(5439, "artist", "ABBA")`后，`artist` 最后的元素应该是 `"ABBA"`。
-
-```js
-assert(updateRecords(5439, 'artist', 'ABBA')[5439]['artist'] === 'ABBA');
-```
-
-执行 `updateRecords(5439, "tracks", "Take a Chance on Me")` 后，`tracks` 最后的元素应该是 `"Take a Chance on Me"`。
+After `updateRecords(collection, 5439, "tracks", "Take a Chance on Me")`, `tracks` should have `Take a Chance on Me` as the last element.
```js
assert(
- updateRecords(5439, 'tracks', 'Take a Chance on Me')[5439]['tracks'].pop() ===
- 'Take a Chance on Me'
+ updateRecords(_recordCollection, 5439, 'tracks', 'Take a Chance on Me')[5439][
+ 'tracks'
+ ].pop() === 'Take a Chance on Me'
);
```
-执行`updateRecords(2548, "artist", "")`后，`artist`不应被创建。
+After `updateRecords(collection, 2548, "artist", "")`, `artist` should not be set
```js
-updateRecords(2548, 'artist', '');
-assert(!collection[2548].hasOwnProperty('artist'));
+updateRecords(_recordCollection, 2548, 'artist', '');
+assert(!_recordCollection[2548].hasOwnProperty('artist'));
```
-执行`updateRecords(1245, "tracks", "Addicted to Love")`后，`tracks`最后的元素应该是`"Addicted to Love"`。
+After `updateRecords(collection, 1245, "tracks", "Addicted to Love")`, `tracks` should have `Addicted to Love` as the last element.
```js
assert(
- updateRecords(1245, 'tracks', 'Addicted to Love')[1245]['tracks'].pop() ===
- 'Addicted to Love'
+ updateRecords(_recordCollection, 1245, 'tracks', 'Addicted to Love')[1245][
+ 'tracks'
+ ].pop() === 'Addicted to Love'
);
```
-执行`updateRecords(2468, "tracks", "Free")`后，`tracks`第一个元素应该是`"1999"`。
+After `updateRecords(collection, 2468, "tracks", "Free")`, `tracks` should have `1999` as the first element.
```js
-assert(updateRecords(2468, 'tracks', 'Free')[2468]['tracks'][0] === '1999');
+assert(
+ updateRecords(_recordCollection, 2468, 'tracks', 'Free')[2468][
+ 'tracks'
+ ][0] === '1999'
+);
```
-执行`updateRecords(2548, "tracks", "")`后，`tracks`不应被创建。
+After `updateRecords(collection, 2548, "tracks", "")`, `tracks` should not be set
```js
-updateRecords(2548, 'tracks', '');
-assert(!collection[2548].hasOwnProperty('tracks'));
+updateRecords(_recordCollection, 2548, 'tracks', '');
+assert(!_recordCollection[2548].hasOwnProperty('tracks'));
```
-执行`updateRecords(1245, "album", "Riptide")`后，`album`应该是`"Riptide"`。
+After `updateRecords(collection, 1245, "albumTitle", "Riptide")`, `albumTitle` should be `Riptide`
```js
-assert(updateRecords(1245, 'album', 'Riptide')[1245]['album'] === 'Riptide');
+assert(
+ updateRecords(_recordCollection, 1245, 'albumTitle', 'Riptide')[1245][
+ 'albumTitle'
+ ] === 'Riptide'
+);
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/replace-loops-using-recursion.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/replace-loops-using-recursion.md
index 078f52b99d8..0c060c2bcad 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/replace-loops-using-recursion.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/replace-loops-using-recursion.md
@@ -1,6 +1,6 @@
---
id: 5cfa3679138e7d9595b9d9d4
-title: 使用递归代替循环
+title: Replace Loops using Recursion
challengeType: 1
videoUrl: >-
https://www.freecodecamp.org/news/how-recursion-works-explained-with-flowcharts-and-a-video-de61f40cb7f9/
@@ -10,62 +10,74 @@ dashedName: replace-loops-using-recursion
# --description--
-函数调用自身的编程技巧称为递归。为了便于理解，有如下任务：计算数组内元素第 `0` 到第 `n` 的元素乘积，使用 `for` 循环， 可以这样做：
+Recursion is the concept that a function can be expressed in terms of itself. To help understand this, start by thinking about the following task: multiply the first `n` elements of an array to create the product of those elements. Using a `for` loop, you could do this:
```js
function multiply(arr, n) {
- var product = arr[0];
- for (var i = 1; i <= n; i++) {
+ var product = 1;
+ for (var i = 0; i < n; i++) {
product *= arr[i];
}
return product;
}
```
-下面是递归写法，注意代码里的 `multiply(arr, n) == multiply(arr, n - 1) * arr[n]`。这意味着可以重写 `multiply` 以调用自身而无需依赖循环。
+However, notice that `multiply(arr, n) == multiply(arr, n - 1) * arr[n - 1]`. That means you can rewrite `multiply` in terms of itself and never need to use a loop.
```js
function multiply(arr, n) {
if (n <= 0) {
- return arr[0];
+ return 1;
} else {
- return multiply(arr, n - 1) * arr[n];
+ return multiply(arr, n - 1) * arr[n - 1];
}
}
```
-递归版本的 `multiply` 详述如下。在 <dfn>base case</dfn> 里，也就是 `n <= 0` 时，返回结果，也就是 `arr[0]`。在 `n` 比 0 大的情况里，函数会调用自身，参数 n 的值为 `n - 1`。函数以相同的方式持续调用 `multiply`，直到 `n = 0` 为止。所以，所有函数都可以返回，原始的 `multiply` 返回结果。
+The recursive version of `multiply` breaks down like this. In the <dfn>base case</dfn>, where `n <= 0`, it returns 1. For larger values of `n`, it calls itself, but with `n - 1`. That function call is evaluated in the same way, calling `multiply` again until `n <= 0`. At this point, all the functions can return and the original `multiply` returns the answer.
-**注意：** 递归函数在没有函数调用时（在这个例子是，是当 `n <= 0` 时）必需有一个跳出结构，否则永远不会执行完毕。
+**Note:** Recursive functions must have a base case when they return without calling the function again (in this example, when `n <= 0`), otherwise they can never finish executing.
# --instructions--
-写一个递归函数，`sum(arr, n)`，返回递归调用数组 `arr` 从第 `0` 个到第 `n` 个元素和。
+Write a recursive function, `sum(arr, n)`, that returns the sum of the first `n` elements of an array `arr`.
# --hints--
-`sum([1], 0)` 应该返回 1 。
+`sum([1], 0)` should equal 0.
```js
-assert.equal(sum([1], 0), 1);
+assert.equal(sum([1], 0), 0);
```
-`sum([2, 3, 4], 1)` 应该返回 5 。
+`sum([2, 3, 4], 1)` should equal 2.
```js
-assert.equal(sum([2, 3, 4], 1), 5);
+assert.equal(sum([2, 3, 4], 1), 2);
```
-代码不应该包含任何形式的循环（`for` 或者 `while` 或者高阶函数比如 `forEach`，`map`，`filter`，或者 `reduce`)。
+`sum([2, 3, 4, 5], 3)` should equal 9.
```js
-assert(!removeJSComments(code).match(/for|while|forEach|map|filter|reduce/g));
+assert.equal(sum([2, 3, 4, 5], 3), 9);
```
-应该使用递归来解决这个问题。
+Your code should not rely on any kind of loops (`for` or `while` or higher order functions such as `forEach`, `map`, `filter`, or `reduce`.).
```js
-assert(removeJSComments(sum.toString()).match(/sum\(.*\)/g).length > 1);
+assert(
+ !__helpers
+ .removeJSComments(code)
+ .match(/for|while|forEach|map|filter|reduce/g)
+);
+```
+
+You should use recursion to solve this problem.
+
+```js
+assert(
+ __helpers.removeJSComments(sum.toString()).match(/sum\(.*\)/g).length > 1
+);
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/replacing-if-else-chains-with-switch.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/replacing-if-else-chains-with-switch.md
index 319544ad106..e724c8821c7 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/replacing-if-else-chains-with-switch.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/replacing-if-else-chains-with-switch.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244e0
-title: 用一个 Switch 语句来替代多个 if else 语句
+title: Replacing If Else Chains with Switch
challengeType: 1
videoUrl: 'https://scrimba.com/c/c3JE8fy'
forumTopicId: 18266
@@ -9,7 +9,7 @@ dashedName: replacing-if-else-chains-with-switch
# --description--
-如果你有多个选项需要选择，`switch`语句写起来会比多个串联的`if`/`if else`语句容易些，譬如:
+If you have many options to choose from, a `switch` statement can be easier to write than many chained `if`/`else if` statements. The following:
```js
if (val === 1) {
@@ -21,7 +21,7 @@ if (val === 1) {
}
```
-可以被下面替代：
+can be replaced with:
```js
switch(val) {
@@ -38,65 +38,65 @@ switch(val) {
# --instructions--
-把串联的`if`/`if else`语句改成`switch`语句。
+Change the chained `if`/`else if` statements into a `switch` statement.
# --hints--
-不要使用`else`表达式。
+You should not use any `else` statements anywhere in the editor
```js
assert(!/else/g.test(code));
```
-不要使用`if`表达式。
+You should not use any `if` statements anywhere in the editor
```js
assert(!/if/g.test(code));
```
-你应该有至少 4 个`break`表达式。
+You should have at least four `break` statements
```js
assert(code.match(/break/g).length >= 4);
```
-`chainToSwitch("bob")`应该为 "Marley"。
+`chainToSwitch("bob")` should be "Marley"
```js
assert(chainToSwitch('bob') === 'Marley');
```
-`chainToSwitch(42)`应该为 "The Answer"。
+`chainToSwitch(42)` should be "The Answer"
```js
assert(chainToSwitch(42) === 'The Answer');
```
-`chainToSwitch(1)`应该为 "There is no
+`chainToSwitch(1)` should be "There is no #1"
```js
assert(chainToSwitch(1) === 'There is no #1');
```
-`chainToSwitch(99)`应该为 "Missed me by this much!"。
+`chainToSwitch(99)` should be "Missed me by this much!"
```js
assert(chainToSwitch(99) === 'Missed me by this much!');
```
-`chainToSwitch(7)`应该为 "Ate Nine"。
+`chainToSwitch(7)` should be "Ate Nine"
```js
assert(chainToSwitch(7) === 'Ate Nine');
```
-`chainToSwitch("John")`应该为 "" (empty string)。
+`chainToSwitch("John")` should be "" (empty string)
```js
assert(chainToSwitch('John') === '');
```
-`chainToSwitch(156)`应该为 "" (empty string)。
+`chainToSwitch(156)` should be "" (empty string)
```js
assert(chainToSwitch(156) === '');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/return-a-value-from-a-function-with-return.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/return-a-value-from-a-function-with-return.md
index 575f4da440f..3feb8d7826a 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/return-a-value-from-a-function-with-return.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/return-a-value-from-a-function-with-return.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244c2
-title: 函数可以返回某个值
+title: Return a Value from a Function with Return
challengeType: 1
videoUrl: 'https://scrimba.com/c/cy87wue'
forumTopicId: 18271
@@ -9,9 +9,9 @@ dashedName: return-a-value-from-a-function-with-return
# --description--
-我们可以通过函数的<dfn>参数</dfn>把值传入函数，也可以使用`return`语句把数据从一个函数中传出来。
+We can pass values into a function with <dfn>arguments</dfn>. You can use a `return` statement to send a value back out of a function.
-**示例**
+**Example**
```js
function plusThree(num) {
@@ -20,33 +20,33 @@ function plusThree(num) {
var answer = plusThree(5); // 8
```
-`plusThree`带有一个`num`的<dfn>参数</dfn>并且返回（returns）一个等于`num + 3`的值。
+`plusThree` takes an <dfn>argument</dfn> for `num` and returns a value equal to `num + 3`.
# --instructions--
-创建一个函数`timesFive`接收一个参数, 把它乘以`5`之后返回，关于如何测试timesFive 函数，可以参考编辑器中最后一行的示例。
+Create a function `timesFive` that accepts one argument, multiplies it by `5`, and returns the new value. See the last line in the editor for an example of how you can test your `timesFive` function.
# --hints--
-`timesFive`应是一个函数。
+`timesFive` should be a function
```js
assert(typeof timesFive === 'function');
```
-`timesFive(5)`应该返回`25`。
+`timesFive(5)` should return `25`
```js
assert(timesFive(5) === 25);
```
-`timesFive(2)`应该返回`10`。
+`timesFive(2)` should return `10`
```js
assert(timesFive(2) === 10);
```
-`timesFive(0)`应该返回`0`。
+`timesFive(0)` should return `0`
```js
assert(timesFive(0) === 0);
@@ -55,7 +55,6 @@ assert(timesFive(0) === 0);
# --seed--
## --seed-contents--
-
```js
```
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/return-early-pattern-for-functions.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/return-early-pattern-for-functions.md
index 1807563f8eb..b51c7ebdb8c 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/return-early-pattern-for-functions.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/return-early-pattern-for-functions.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244c4
-title: 函数执行到 return 语句就结束
+title: Return Early Pattern for Functions
challengeType: 1
videoUrl: 'https://scrimba.com/c/cQe39Sq'
forumTopicId: 18272
@@ -9,9 +9,9 @@ dashedName: return-early-pattern-for-functions
# --description--
-当代码执行到 return 语句时，函数返回一个结果就结束运行了，return 后面的语句不会执行。
+When a `return` statement is reached, the execution of the current function stops and control returns to the calling location.
-**示例**
+**Example**
```js
function myFun() {
@@ -22,53 +22,59 @@ function myFun() {
myFun();
```
-上面的代码输出"Hello"到控制台、返回 "World"，但没有输出`"byebye"`，因为函数遇到 return 语句就退出了。
+The above outputs "Hello" to the console, returns "World", but `"byebye"` is never output, because the function exits at the `return` statement.
# --instructions--
-修改函数`abTest`当`a`或`b`小于0时，函数立即返回一个`undefined`并退出。
+Modify the function `abTest` so that if `a` or `b` are less than `0` the function will immediately exit with a value of `undefined`.
-**提示**
-记住[`undefined` 是一个关键字](/javascript-algorithms-and-data-structures/basic-javascript/understanding-uninitialized-variables)，而不是一个字符串。
+**Hint**
+Remember that [`undefined` is a keyword](https://www.freecodecamp.org/learn/javascript-algorithms-and-data-structures/basic-javascript/understanding-uninitialized-variables), not a string.
# --hints--
-`abTest(2,2)` 应该返回一个数字。
+`abTest(2,2)` should return a number
```js
assert(typeof abTest(2, 2) === 'number');
```
-`abTest(2,2)` 应该返回 `8`。
+`abTest(2,2)` should return `8`
```js
assert(abTest(2, 2) === 8);
```
-`abTest(-2,2)` 应该返回 `undefined`。
+`abTest(-2,2)` should return `undefined`
```js
assert(abTest(-2, 2) === undefined);
```
-`abTest(2,-2)` 应该返回 `undefined`。
+`abTest(2,-2)` should return `undefined`
```js
assert(abTest(2, -2) === undefined);
```
-`abTest(2,8)` 应该返回 `18`。
+`abTest(2,8)` should return `18`
```js
assert(abTest(2, 8) === 18);
```
-`abTest(3,3)` 应该返回 `12`。
+`abTest(3,3)` should return `12`
```js
assert(abTest(3, 3) === 12);
```
+`abTest(0,0)` should return `0`
+
+```js
+assert(abTest(0, 0) === 0);
+```
+
# --seed--
## --seed-contents--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/returning-boolean-values-from-functions.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/returning-boolean-values-from-functions.md
index a8bfa8b8260..1aabc48cbe4 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/returning-boolean-values-from-functions.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/returning-boolean-values-from-functions.md
@@ -1,6 +1,6 @@
---
id: 5679ceb97cbaa8c51670a16b
-title: 从函数返回布尔值
+title: Returning Boolean Values from Functions
challengeType: 1
videoUrl: 'https://scrimba.com/c/cp62qAQ'
forumTopicId: 18273
@@ -9,9 +9,9 @@ dashedName: returning-boolean-values-from-functions
# --description--
-你应该还记得[相等运算符](javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-equality-operator)这道挑战题。在那里我们提到，所有比较操作符都会返回 boolean：要么是`true`要么是`false`。
+You may recall from [Comparison with the Equality Operator](/learn/javascript-algorithms-and-data-structures/basic-javascript/comparison-with-the-equality-operator) that all comparison operators return a boolean `true` or `false` value.
-有时人们通过 if/else 语句来做比较然后返回`true`或`false`。
+Sometimes people use an if/else statement to do a comparison, like this:
```js
function isEqual(a,b) {
@@ -23,7 +23,7 @@ function isEqual(a,b) {
}
```
-有一个更好的方法，因为`===`总是返回`true`或`false`，所以我们可以直接返回比较的结果：
+But there's a better way to do this. Since `===` returns `true` or `false`, we can return the result of the comparison:
```js
function isEqual(a,b) {
@@ -33,23 +33,23 @@ function isEqual(a,b) {
# --instructions--
-移除`isLess`函数的`if/else`语句但不影响函数的功能。
+Fix the function `isLess` to remove the `if/else` statements.
# --hints--
-`isLess(10,15)`应该返回 `true`。
+`isLess(10,15)` should return `true`
```js
assert(isLess(10, 15) === true);
```
-`isLess(15,10)`应该返回 `false`。
+`isLess(15,10)` should return `false`
```js
assert(isLess(15, 10) === false);
```
-不应该使用 `if` 或者 `else` 语句。
+You should not use any `if` or `else` statements
```js
assert(!/if|else/g.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/selecting-from-many-options-with-switch-statements.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/selecting-from-many-options-with-switch-statements.md
index 61d954ff359..3e6c185dd09 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/selecting-from-many-options-with-switch-statements.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/selecting-from-many-options-with-switch-statements.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244dd
-title: 使用 Switch 语句从许多选项中进行选择
+title: Selecting from Many Options with Switch Statements
challengeType: 1
videoUrl: 'https://scrimba.com/c/c4mv4fm'
forumTopicId: 18277
@@ -9,9 +9,9 @@ dashedName: selecting-from-many-options-with-switch-statements
# --description--
-如果你有非常多的选项需要选择，可以使用 switch 语句。根据不同的参数值会匹配上不同的 case 分支，语句会从第一个匹配的 case 分支开始执行，直到碰到 break 就结束。
+If you have many options to choose from, use a <dfn>switch</dfn> statement. A `switch` statement tests a value and can have many <dfn>case</dfn> statements which define various possible values. Statements are executed from the first matched `case` value until a `break` is encountered.
-这是一个伪代码案例：
+Here is an example of a `switch` statement:
```js
switch(lowercaseLetter) {
@@ -24,49 +24,49 @@ switch(lowercaseLetter) {
}
```
-测试`case`值使用严格相等运算符进行比较，break 关键字告诉 JavaScript 停止执行语句。如果没有 break 关键字，下一个语句会继续执行。
+`case` values are tested with strict equality (`===`). The `break` tells JavaScript to stop executing statements. If the `break` is omitted, the next statement will be executed.
# --instructions--
-写一个测试`val`的 switch 语句，并且根据下面的条件来设置不同的`answer`：
-`1`- "alpha"
+Write a switch statement which tests `val` and sets `answer` for the following conditions:
+`1` - "alpha"
`2` - "beta"
-`3`- "gamma"
+`3` - "gamma"
`4` - "delta"
# --hints--
-`caseInSwitch(1)` 应该有一个值为 "alpha"。
+`caseInSwitch(1)` should have a value of "alpha"
```js
assert(caseInSwitch(1) === 'alpha');
```
-`caseInSwitch(2)` 应该有一个值为 "beta"。
+`caseInSwitch(2)` should have a value of "beta"
```js
assert(caseInSwitch(2) === 'beta');
```
-`caseInSwitch(3)` 应该有一个值为 "gamma"。
+`caseInSwitch(3)` should have a value of "gamma"
```js
assert(caseInSwitch(3) === 'gamma');
```
-`caseInSwitch(4)` 应该有一个值为 "delta"。
+`caseInSwitch(4)` should have a value of "delta"
```js
assert(caseInSwitch(4) === 'delta');
```
-不能使用任何`if`或`else`表达式。
+You should not use any `if` or `else` statements
```js
assert(!/else/g.test(code) || !/if/g.test(code));
```
-你应该有至少 3 个`break`表达式。
+You should have at least 3 `break` statements
```js
assert(code.match(/break/g).length > 2);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/shopping-list.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/shopping-list.md
index 3b69798a26d..84f22f2b423 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/shopping-list.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/shopping-list.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244bc
-title: 购物清单
+title: Shopping List
challengeType: 1
videoUrl: 'https://scrimba.com/c/c9MEKHZ'
forumTopicId: 18280
@@ -9,35 +9,35 @@ dashedName: shopping-list
# --description--
-创建一个名叫`myList`的购物清单，清单的数据格式就是多维数组。
+Create a shopping list in the variable `myList`. The list should be a multi-dimensional array containing several sub-arrays.
-每个子数组中的第一个元素应该是购买的物品名称，第二个元素应该是物品的数量，类似于：
+The first element in each sub-array should contain a string with the name of the item. The second element should be a number representing the quantity i.e.
`["Chocolate Bar", 15]`
-任务：你的购物清单至少应该有 5 个子数组。
+There should be at least 5 sub-arrays in the list.
# --hints--
-`myList`应该一个数组。
+`myList` should be an array.
```js
assert(isArray);
```
-你的每个子数组的第一个元素的类型都应该是字符串。
+The first elements in each of your sub-arrays should all be strings.
```js
assert(hasString);
```
-你的每个子数组的第二个元素的类型都应该是数字。
+The second elements in each of your sub-arrays should all be numbers.
```js
assert(hasNumber);
```
-你的列表中至少要包含 5 个元素。
+You should have at least 5 items in your list.
```js
assert(count > 4);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/stand-in-line.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/stand-in-line.md
index 9a2f141c4c9..1d882d80d63 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/stand-in-line.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/stand-in-line.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244c6
-title: 排队
+title: Stand in Line
challengeType: 1
videoUrl: 'https://scrimba.com/c/ca8Q8tP'
forumTopicId: 18307
@@ -9,41 +9,41 @@ dashedName: stand-in-line
# --description--
-在计算机科学中<dfn>队列</dfn>（queue）是一个抽象的数据结构，队列中的条目都是有秩序的。新的条目会被加到`队列`的末尾，旧的条目会从`队列`的头部被移出。
+In Computer Science a <dfn>queue</dfn> is an abstract <dfn>Data Structure</dfn> where items are kept in order. New items can be added at the back of the queue and old items are taken off from the front of the queue.
-写一个函数`nextInLine`，用一个数组(`arr`)和一个数字(`item`)作为参数。
+Write a function `nextInLine` which takes an array (`arr`) and a number (`item`) as arguments.
-把数字添加到数组的结尾，然后移出数组的第一个元素。
+Add the number to the end of the array, then remove the first element of the array.
-最后`nextInLine`函数应该返回被删除的元素。
+The `nextInLine` function should then return the element that was removed.
# --hints--
-`nextInLine([], 5)`应该返回一个数字。
+`nextInLine([], 5)` should return a number.
```js
assert.isNumber(nextInLine([], 5));
```
-`nextInLine([], 1)`应该返回`1`。
+`nextInLine([], 1)` should return `1`
```js
assert(nextInLine([], 1) === 1);
```
-`nextInLine([2], 1)`应该返回`2`。
+`nextInLine([2], 1)` should return `2`
```js
assert(nextInLine([2], 1) === 2);
```
-`nextInLine([5,6,7,8,9], 1)`应该返回`5`。
+`nextInLine([5,6,7,8,9], 1)` should return `5`
```js
assert(nextInLine([5, 6, 7, 8, 9], 1) === 5);
```
-在`nextInLine(testArr, 10)`执行后`testArr[4]`应该是`10`。
+After `nextInLine(testArr, 10)`, `testArr[4]` should be `10`
```js
nextInLine(testArr, 10);
@@ -90,10 +90,10 @@ testArr = [1,2,3,4,5];
```js
function nextInLine(arr, item) {
// Only change code below this line
-
+
return item;
// Only change code above this line
-
+
}
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/store-multiple-values-in-one-variable-using-javascript-arrays.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/store-multiple-values-in-one-variable-using-javascript-arrays.md
index 656c81fcb7d..ac96b059875 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/store-multiple-values-in-one-variable-using-javascript-arrays.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/store-multiple-values-in-one-variable-using-javascript-arrays.md
@@ -1,6 +1,6 @@
---
id: bd7993c9c69feddfaeb8bdef
-title: 使用 JavaScript 数组将多个值存储在一个变量中
+title: Store Multiple Values in one Variable using JavaScript Arrays
challengeType: 1
videoUrl: 'https://scrimba.com/c/crZQWAm'
forumTopicId: 18309
@@ -9,34 +9,34 @@ dashedName: store-multiple-values-in-one-variable-using-javascript-arrays
# --description--
-使用`数组`，我们可以在一个地方存储多个数据。
+With JavaScript `array` variables, we can store several pieces of data in one place.
-以左方括号`[`开始定义一个数组，以右方括号`]`结束，里面每个元素之间用逗号隔开，例如：
+You start an array declaration with an opening square bracket, end it with a closing square bracket, and put a comma between each entry, like this:
`var sandwich = ["peanut butter", "jelly", "bread"]`.
# --instructions--
-创建一个包含`字符串`和`数字`的数组`myArray`。
+Modify the new array `myArray` so that it contains both a `string` and a `number` (in that order).
-**提示**
-如果你遇到困难，请参考文本编辑器中的示例代码。
+**Hint**
+Refer to the example code in the text editor if you get stuck.
# --hints--
-`myArray`应该是一个`数组`。
+`myArray` should be an `array`.
```js
assert(typeof myArray == 'object');
```
-`myArray`数组的第一个元素应该是一个`字符串`。
+The first item in `myArray` should be a `string`.
```js
assert(typeof myArray[0] !== 'undefined' && typeof myArray[0] == 'string');
```
-`myArray`数组的第二个元素应该是一个`数字`。
+The second item in `myArray` should be a `number`.
```js
assert(typeof myArray[1] !== 'undefined' && typeof myArray[1] == 'number');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/storing-values-with-the-assignment-operator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/storing-values-with-the-assignment-operator.md
index fbbcabdf353..b60c20c2750 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/storing-values-with-the-assignment-operator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/storing-values-with-the-assignment-operator.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244a8
-title: 使用赋值运算符存储值
+title: Storing Values with the Assignment Operator
challengeType: 1
videoUrl: 'https://scrimba.com/c/cEanysE'
forumTopicId: 18310
@@ -9,53 +9,39 @@ dashedName: storing-values-with-the-assignment-operator
# --description--
-在 JavaScript 中，你可以使用赋值运算符将值存储在变量中。
+In JavaScript, you can store a value in a variable with the <dfn>assignment</dfn> operator (`=`).
`myVariable = 5;`
-这条语句把`Number`类型的值`5`赋给变量`myVariable`。
+This assigns the `Number` value `5` to `myVariable`.
-赋值过程是从右到左进行的。在将值分配给运算符左侧的变量之前，将解析`=`运算符右侧的所有内容。
+If there are any calculations to the right of the `=` operator, those are performed before the value is assigned to the variable on the left of the operator.
```js
+var myVar;
myVar = 5;
-myNum = myVar;
```
-数值`5`被赋给变量`myVar`中，然后再次将变量`myVar`解析为`5`并将其赋给`myNum`变量。
+First, this code creates a variable named `myVar`. Then, the code assigns `5` to `myVar`. Now, if `myVar` appears again in the code, the program will treat it as if it is `5`.
# --instructions--
-把数值`7`赋给变量 `a`。
-
-把变量`a`中的内容赋给变量`b`。
+Assign the value `7` to variable `a`.
# --hints--
-不要修改注释上方的代码。
+You should not change code above the specified comment.
```js
-assert(/var a;/.test(code) && /var b = 2;/.test(code));
+assert(/var a;/.test(code));
```
-`a`的值应该是 7。
+`a` should have a value of 7.
```js
assert(typeof a === 'number' && a === 7);
```
-`b`的值应该是 7。
-
-```js
-assert(typeof b === 'number' && b === 7);
-```
-
-你需要用`=`把`a`的值赋给`b`。
-
-```js
-assert(/b\s*=\s*a\s*;/g.test(code));
-```
-
# --seed--
## --before-user-code--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/subtract-one-number-from-another-with-javascript.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/subtract-one-number-from-another-with-javascript.md
index 3fbed35e9f6..e4b1e9c3469 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/subtract-one-number-from-another-with-javascript.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/subtract-one-number-from-another-with-javascript.md
@@ -1,6 +1,6 @@
---
id: cf1111c1c11feddfaeb4bdef
-title: 减法运算
+title: Subtract One Number from Another with JavaScript
challengeType: 1
videoUrl: 'https://scrimba.com/c/cP3yQtk'
forumTopicId: 18314
@@ -9,11 +9,11 @@ dashedName: subtract-one-number-from-another-with-javascript
# --description--
-我们也可以在 JavaScript 中进行减法运算。
+We can also subtract one number from another.
-JavaScript 中使用`-`来做减法运算。
+JavaScript uses the `-` symbol for subtraction.
-**示例**
+**Example**
```js
myVar = 12 - 6; // assigned 6
@@ -21,20 +21,20 @@ myVar = 12 - 6; // assigned 6
# --instructions--
-改变数字`0`让变量 difference 的值为`12`。
+Change the `0` so the difference is `12`.
# --hints--
-要使`difference`的值等于 12。
+The variable `difference` should be equal to 12.
```js
assert(difference === 12);
```
-只用 45 减去一个数。
+You should only subtract one number from 45.
```js
-assert(/var\s*difference\s*=\s*45\s*-\s*[0-9]*;(?!\s*[a-zA-Z0-9]+)/.test(code));
+assert(/difference=45-33;?/.test(__helpers.removeWhiteSpace(code)));
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/testing-objects-for-properties.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/testing-objects-for-properties.md
index 4fe47742125..d6c053d6018 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/testing-objects-for-properties.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/testing-objects-for-properties.md
@@ -1,17 +1,17 @@
---
id: 567af2437cbaa8c51670a16c
-title: 测试对象的属性
+title: Testing Objects for Properties
challengeType: 1
-videoUrl: 'https://scrimba.com/c/cm8Q7Ua'
+videoUrl: 'https://scrimba.com/c/c6Wz4ySr'
forumTopicId: 18324
dashedName: testing-objects-for-properties
---
# --description--
-有时检查一个对象属性是否存在是非常有用的，我们可以用`.hasOwnProperty(propname)`方法来检查对象是否有该属性。如果有返回`true`，反之返回`false`。
+Sometimes it is useful to check if the property of a given object exists or not. We can use the `.hasOwnProperty(propname)` method of objects to determine if that object has the given property name. `.hasOwnProperty()` returns `true` or `false` if the property is found or not.
-**示例**
+**Example**
```js
var myObj = {
@@ -24,26 +24,51 @@ myObj.hasOwnProperty("middle"); // false
# --instructions--
-修改函数`checkObj`检查`myObj`是否有`checkProp`属性，如果属性存在，返回属性对应的值，如果不存在，返回`"Not Found"`。
+Modify the function `checkObj` to test if an object passed to the function (`obj`) contains a specific property (`checkProp`). If the property is found, return that property's value. If not, return `"Not Found"`.
# --hints--
-`checkObj("gift")`应该返回`"pony"`。
+`checkObj({gift: "pony", pet: "kitten", bed: "sleigh"}, "gift")` should return `"pony"`.
```js
-assert(checkObj('gift') === 'pony');
+assert(
+ checkObj({ gift: 'pony', pet: 'kitten', bed: 'sleigh' }, 'gift') === 'pony'
+);
```
-`checkObj("pet")`应该返回`"kitten"`。
+`checkObj({gift: "pony", pet: "kitten", bed: "sleigh"}, "pet")` should return `"kitten"`.
```js
-assert(checkObj('pet') === 'kitten');
+assert(
+ checkObj({ gift: 'pony', pet: 'kitten', bed: 'sleigh' }, 'pet') === 'kitten'
+);
```
-`checkObj("house")`应该返回`"Not Found"`。
+`checkObj({gift: "pony", pet: "kitten", bed: "sleigh"}, "house")` should return `"Not Found"`.
```js
-assert(checkObj('house') === 'Not Found');
+assert(
+ checkObj({ gift: 'pony', pet: 'kitten', bed: 'sleigh' }, 'house') ===
+ 'Not Found'
+);
+```
+
+`checkObj({city: "Seattle"}, "city")` should return `"Seattle"`.
+
+```js
+assert(checkObj({ city: 'Seattle' }, 'city') === 'Seattle');
+```
+
+`checkObj({city: "Seattle"}, "district")` should return `"Not Found"`.
+
+```js
+assert(checkObj({ city: 'Seattle' }, 'district') === 'Not Found');
+```
+
+`checkObj({pet: "kitten", bed: "sleigh"}, "gift")` should return `"Not Found"`.
+
+```js
+assert(checkObj({ pet: 'kitten', bed: 'sleigh' }, 'gift') === 'Not Found');
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understand-string-immutability.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understand-string-immutability.md
index f51905e276a..8a78800648c 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understand-string-immutability.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understand-string-immutability.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244ba
-title: 了解字符串的不变性
+title: Understand String Immutability
challengeType: 1
videoUrl: 'https://scrimba.com/c/cWPVaUR'
forumTopicId: 18331
@@ -9,16 +9,16 @@ dashedName: understand-string-immutability
# --description--
-在 JavaScript 中，`字符串`的值是 <dfn>不可变的</dfn>，这意味着一旦字符串被创建就不能被改变。
+In JavaScript, `String` values are <dfn>immutable</dfn>, which means that they cannot be altered once created.
-例如，下面的代码：
+For example, the following code:
```js
var myStr = "Bob";
myStr[0] = "J";
```
-是不会把变量`myStr`的值改变成 "Job" 的，因为变量`myStr`是不可变的。注意，这*并不*意味着`myStr`永远不能被改变，只是字符串字面量 <dfn>string literal</dfn> 的各个字符不能被改变。改变`myStr`中的唯一方法是重新给它赋一个值，例如：
+cannot change the value of `myStr` to "Job", because the contents of `myStr` cannot be altered. Note that this does *not* mean that `myStr` cannot be changed, just that the individual characters of a <dfn>string literal</dfn> cannot be changed. The only way to change `myStr` would be to assign it with a new string, like this:
```js
var myStr = "Bob";
@@ -27,17 +27,17 @@ myStr = "Job";
# --instructions--
-把`myStr`的值改为`Hello World`。
+Correct the assignment to `myStr` so it contains the string value of `Hello World` using the approach shown in the example above.
# --hints--
-message:`myStr`的值应该是`Hello World`。
+`myStr` should have a value of `Hello World`.
```js
assert(myStr === 'Hello World');
```
-不要修改注释上面的代码。
+You should not change the code above the specified comment.
```js
assert(/myStr = "Jello World"/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understanding-boolean-values.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understanding-boolean-values.md
index bfaa50cb391..51208aa42a3 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understanding-boolean-values.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understanding-boolean-values.md
@@ -1,6 +1,6 @@
---
id: bd7123c9c441eddfaeb5bdef
-title: 理解布尔值
+title: Understanding Boolean Values
challengeType: 1
videoUrl: 'https://scrimba.com/c/c9Me8t4'
forumTopicId: 301176
@@ -9,24 +9,24 @@ dashedName: understanding-boolean-values
# --description--
-另一种数据类型是<dfn>布尔</dfn>（Boolean）。`布尔`值要么是`true`要么是`false`。它非常像电路开关，`true`是 “开”，`false`是 “关”。这两种状态是互斥的。
+Another data type is the <dfn>Boolean</dfn>. `Booleans` may only be one of two values: `true` or `false`. They are basically little on-off switches, where `true` is "on" and `false` is "off." These two states are mutually exclusive.
-**注意**
-`布尔值`是不带引号的，`"true"`和`"false"`是`字符串`而不是`布尔值`，在 JavaScript 中也没有特殊含义。
+**Note**
+`Boolean` values are never written with quotes. The `strings` `"true"` and `"false"` are not `Boolean` and have no special meaning in JavaScript.
# --instructions--
-修改`welcomeToBooleans`函数，让它返回`true`而不是`false`。
+Modify the `welcomeToBooleans` function so that it returns `true` instead of `false` when the run button is clicked.
# --hints--
-`welcomeToBooleans()`函数应该返回一个布尔值 (true/false)。
+The `welcomeToBooleans()` function should return a boolean (true/false) value.
```js
assert(typeof welcomeToBooleans() === 'boolean');
```
-`welcomeToBooleans()`应该返回 true。
+`welcomeToBooleans()` should return true.
```js
assert(welcomeToBooleans() === true);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understanding-case-sensitivity-in-variables.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understanding-case-sensitivity-in-variables.md
index 30799f09be4..e7fcb1247ba 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understanding-case-sensitivity-in-variables.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understanding-case-sensitivity-in-variables.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244ab
-title: 了解变量名区分大小写
+title: Understanding Case Sensitivity in Variables
challengeType: 1
videoUrl: 'https://scrimba.com/c/cd6GDcD'
forumTopicId: 18334
@@ -9,15 +9,15 @@ dashedName: understanding-case-sensitivity-in-variables
# --description--
-在 JavaScript 中所有的变量和函数名都是大小写敏感的。要区别对待大写字母和小写字母。
+In JavaScript all variables and function names are case sensitive. This means that capitalization matters.
-`MYVAR`与`MyVar`和`myvar`是截然不同的变量。这有可能导致出现多个相似名字的的变量。所以强烈地建议你，为了保持代码清晰不要使用这一特性。
+`MYVAR` is not the same as `MyVar` nor `myvar`. It is possible to have multiple distinct variables with the same name but different casing. It is strongly recommended that for the sake of clarity, you *do not* use this language feature.
-<h4>最佳实践</h4>
+**Best Practice**
-使用<dfn>驼峰命名法</dfn>来书写一个 Javascript 变量，在<dfn>驼峰命名法</dfn>中，变量名的第一个单词的首写字母小写，后面的单词的第一个字母大写。
+Write variable names in JavaScript in <dfn>camelCase</dfn>. In <dfn>camelCase</dfn>, multi-word variable names have the first word in lowercase and the first letter of each subsequent word is capitalized.
-**示例：**
+**Examples:**
```js
var someVariable;
@@ -27,17 +27,18 @@ var thisVariableNameIsSoLong;
# --instructions--
-修改已声明的变量，让它们的命名符合<dfn>驼峰命名法</dfn>的规范。
+Modify the existing declarations and assignments so their names use <dfn>camelCase</dfn>.
+Do not create any new variables.
# --hints--
-`studlyCapVar`应该被定义并且值为`10`。
+`studlyCapVar` should be defined and have a value of `10`.
```js
assert(typeof studlyCapVar !== 'undefined' && studlyCapVar === 10);
```
-`properCamelCase`应该被定义并且值为`"A String"`。
+`properCamelCase` should be defined and have a value of `"A String"`.
```js
assert(
@@ -45,25 +46,25 @@ assert(
);
```
-`titleCaseOver`应该被定义并且值为`9000`。
+`titleCaseOver` should be defined and have a value of `9000`.
```js
assert(typeof titleCaseOver !== 'undefined' && titleCaseOver === 9000);
```
-`studlyCapVar`在声明和赋值时都应该使用驼峰命名法。
+`studlyCapVar` should use camelCase in both declaration and assignment sections.
```js
assert(code.match(/studlyCapVar/g).length === 2);
```
-`properCamelCase` 在声明和赋值时都应该使用驼峰命名法。
+`properCamelCase` should use camelCase in both declaration and assignment sections.
```js
assert(code.match(/properCamelCase/g).length === 2);
```
-`titleCaseOver` 在声明和赋值时都应该使用驼峰命名法。
+`titleCaseOver` should use camelCase in both declaration and assignment sections.
```js
assert(code.match(/titleCaseOver/g).length === 2);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understanding-undefined-value-returned-from-a-function.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understanding-undefined-value-returned-from-a-function.md
index 8376c2ed0a5..94aa143e1ed 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understanding-undefined-value-returned-from-a-function.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understanding-undefined-value-returned-from-a-function.md
@@ -1,6 +1,6 @@
---
id: 598e8944f009e646fc236146
-title: 函数也可以返回 undefined
+title: Understanding Undefined Value returned from a Function
challengeType: 1
videoUrl: 'https://scrimba.com/c/ce2p7cL'
forumTopicId: 301177
@@ -9,9 +9,9 @@ dashedName: understanding-undefined-value-returned-from-a-function
# --description--
-函数一般用`return`语句来返回值，但这不是必须的。在函数没有`return`语句的情况下，当你调用它时，该函数会执行内部代码，返回的值是`undefined`。
+A function can include the `return` statement but it does not have to. In the case that the function doesn't have a `return` statement, when you call it, the function processes the inner code but the returned value is `undefined`.
-**示例**
+**Example**
```js
var sum = 0;
@@ -21,40 +21,37 @@ function addSum(num) {
addSum(3); // sum will be modified but returned value is undefined
```
-`addSum`是一个没有`return`语句的函数。该函数将更改全局变量`sum`，函数的返回值为`undefined`。
+`addSum` is a function without a `return` statement. The function will change the global `sum` variable but the returned value of the function is `undefined`.
# --instructions--
-创建一个没有任何参数的函数`addFive`。此函数使`sum`变量加 5，但其返回值是`undefined`。
+Create a function `addFive` without any arguments. This function adds 5 to the `sum` variable, but its returned value is `undefined`.
# --hints--
-`addFive`应该是一个函数。
+`addFive` should be a function.
```js
assert(typeof addFive === 'function');
```
-`sum`应该等于 8。
+Once both functions have ran, the `sum` should be equal to 8.
```js
assert(sum === 8);
```
-`addFive`的返回值应该是`undefined`。
+Returned value from `addFive` should be `undefined`.
```js
assert(addFive() === undefined);
```
-函数给变量 `sum` 加 5。
+Inside the `addFive` function, you should add `5` to the `sum` variable.
```js
assert(
- addFive
- .toString()
- .replace(/\s/g, '')
- .match(/sum=sum\+5|sum\+=5/)
+ __helpers.removeWhiteSpace(addFive.toString()).match(/sum=sum\+5|sum\+=5/)
);
```
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understanding-uninitialized-variables.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understanding-uninitialized-variables.md
index ab7ff51d912..37796550660 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understanding-uninitialized-variables.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/understanding-uninitialized-variables.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244aa
-title: 理解未初始化的变量
+title: Understanding Uninitialized Variables
challengeType: 1
videoUrl: 'https://scrimba.com/c/cBa2JAL'
forumTopicId: 18335
@@ -9,33 +9,33 @@ dashedName: understanding-uninitialized-variables
# --description--
-当 JavaScript 中的变量被声明的时候，程序内部会给它一个初始值`undefined`。当你对一个值为`undefined`的变量进行运算操作的时候，算出来的结果将会是`NaN`，`NaN`的意思是<dfn>"Not a Number"</dfn>。当你用一个值是`undefined`的变量来做字符串拼接操作的时候，它会输出字符串`"undefined"`。
+When JavaScript variables are declared, they have an initial value of `undefined`. If you do a mathematical operation on an `undefined` variable your result will be `NaN` which means <dfn>"Not a Number"</dfn>. If you concatenate a string with an `undefined` variable, you will get a literal <dfn>string</dfn> of `"undefined"`.
# --instructions--
-定义 3 个变量`a`、`b`、`c`，并且分别给他们赋值：`5`、`10`、`"I am a"`，这样它们值就不会是`undefined`了。
+Initialize the three variables `a`, `b`, and `c` with `5`, `10`, and `"I am a"` respectively so that they will not be `undefined`.
# --hints--
-`a`应该被定义，并且值为`6`。
+`a` should be defined and evaluated to have the value of `6`.
```js
assert(typeof a === 'number' && a === 6);
```
-`b`应该被定义，并且值为`15`。
+`b` should be defined and evaluated to have the value of `15`.
```js
assert(typeof b === 'number' && b === 15);
```
-`c`的值不能包含`undefined`，应该为 "I am a String!"。
+`c` should not contain `undefined` and should have a value of "I am a String!"
```js
assert(!/undefined/.test(c) && c === 'I am a String!');
```
-不要修改第二条注释下的代码。
+You should not change code below the specified comment.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/updating-object-properties.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/updating-object-properties.md
index 9512850d001..fade65082c4 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/updating-object-properties.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/updating-object-properties.md
@@ -1,6 +1,6 @@
---
id: 56bbb991ad1ed5201cd392d1
-title: 更新对象属性
+title: Updating Object Properties
challengeType: 1
videoUrl: 'https://scrimba.com/c/c9yEJT4'
forumTopicId: 18336
@@ -9,9 +9,9 @@ dashedName: updating-object-properties
# --description--
-当你创建了一个对象后，你可以用点操作符或中括号操作符来更新对象的属性。
+After you've created a JavaScript object, you can update its properties at any time just like you would update any other variable. You can use either dot or bracket notation to update.
-举个例子，让我们看看`ourDog`:
+For example, let's look at `ourDog`:
```js
var ourDog = {
@@ -22,21 +22,21 @@ var ourDog = {
};
```
-让我们更改它的名称为 "Happy Camper"，这有两种方式来更新对象的`name`属性： `ourDog.name = "Happy Camper";` 或 `ourDog["name"] = "Happy Camper";` 现在，`ourDog.name`的值就不再是 "Camper"，而是 "Happy Camper"。
+Since he's a particularly happy dog, let's change his name to "Happy Camper". Here's how we update his object's name property: `ourDog.name = "Happy Camper";` or `ourDog["name"] = "Happy Camper";` Now when we evaluate `ourDog.name`, instead of getting "Camper", we'll get his new name, "Happy Camper".
# --instructions--
-更新`myDog`对象的`name`属性，让它的名字从 "Coder" 变成 "Happy Coder"。
+Update the `myDog` object's name property. Let's change her name from "Coder" to "Happy Coder". You can use either dot or bracket notation.
# --hints--
-更新`myDog`的`"name"`属性, 使其等于 "Happy Coder"。
+You should update `myDog`'s `"name"` property to equal "Happy Coder".
```js
assert(/happy coder/gi.test(myDog.name));
```
-不要修改`myDog`的定义。
+You should not edit the `myDog` definition.
```js
assert(/"name": "Coder"/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-bracket-notation-to-find-the-first-character-in-a-string.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-bracket-notation-to-find-the-first-character-in-a-string.md
index 3269109dfad..78c7bc400d0 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-bracket-notation-to-find-the-first-character-in-a-string.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-bracket-notation-to-find-the-first-character-in-a-string.md
@@ -1,6 +1,6 @@
---
id: bd7123c9c549eddfaeb5bdef
-title: 使用方括号查找字符串中的第一个字符
+title: Use Bracket Notation to Find the First Character in a String
challengeType: 1
videoUrl: 'https://scrimba.com/c/ca8JwhW'
forumTopicId: 18341
@@ -9,28 +9,34 @@ dashedName: use-bracket-notation-to-find-the-first-character-in-a-string
# --description--
-方括号表示法是一种在字符串中的特定`index`（索引）处获取字符的方法。
+<dfn>Bracket notation</dfn> is a way to get a character at a specific `index` within a string.
-大多数现代编程语言，如JavaScript，不同于人类从 1 开始计数。它们是从 0 开始计数，这被称为 <dfn>基于零</dfn> 的索引。
+Most modern programming languages, like JavaScript, don't start counting at 1 like humans do. They start at 0. This is referred to as <dfn>Zero-based</dfn> indexing.
-例如, 在单词 "Charles" 中索引 0 上的字符为 "C"，所以在`var firstName = "Charles"`中，你可以使用`firstName[0]`来获得第一个位置上的字符。
+For example, the character at index 0 in the word "Charles" is "C". So if `var firstName = "Charles"`, you can get the value of the first letter of the string by using `firstName[0]`.
+
+Example:
+
+```js
+var firstName = "Charles";
+var firstLetter = firstName[0]; // firstLetter is "C"
+```
# --instructions--
-使用方括号获取变量`lastName`中的第一个字符，并赋给变量`firstLetterOfLastName`。
+Use bracket notation to find the first character in the `lastName` variable and assign it to `firstLetterOfLastName`.
-**提示**
-如果你遇到困难了，不妨看看变量`firstLetterOfFirstName`是如何赋值的。
+**Hint:** Try looking at the example above if you get stuck.
# --hints--
-`firstLetterOfLastName`的值应该是`L`。
+The `firstLetterOfLastName` variable should have the value of `L`.
```js
assert(firstLetterOfLastName === 'L');
```
-你应该使用中括号。
+You should use bracket notation.
```js
assert(code.match(/firstLetterOfLastName\s*?=\s*?lastName\[.*?\]/));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-bracket-notation-to-find-the-last-character-in-a-string.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-bracket-notation-to-find-the-last-character-in-a-string.md
index e0607f8b9f2..6643d6d469f 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-bracket-notation-to-find-the-last-character-in-a-string.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-bracket-notation-to-find-the-last-character-in-a-string.md
@@ -1,6 +1,6 @@
---
id: bd7123c9c451eddfaeb5bdef
-title: 使用方括号查找字符串中的最后一个字符
+title: Use Bracket Notation to Find the Last Character in a String
challengeType: 1
videoUrl: 'https://scrimba.com/c/cBZQGcv'
forumTopicId: 18342
@@ -9,26 +9,35 @@ dashedName: use-bracket-notation-to-find-the-last-character-in-a-string
# --description--
-要获取字符串的最后一个字符，可以用字符串的长度减 1 的索引值。
+In order to get the last letter of a string, you can subtract one from the string's length.
-例如，在`var firstName = "Charles"`中，你可以这样操作`firstName[firstName.length - 1]`来得到字符串的最后的一个字符。
+For example, if `var firstName = "Charles"`, you can get the value of the last letter of the string by using `firstName[firstName.length - 1]`.
+
+Example:
+
+```js
+var firstName = "Charles";
+var lastLetter = firstName[firstName.length - 1]; // lastLetter is "s"
+```
# --instructions--
-使用<dfn>方括号<dfn来取得<code>lastName变量中的最后一个字符。 <strong>提示</strong><br>如果你遇到困难了，不妨看看在<code>lastLetterOfFirstName</code>变量上是怎么做的。 </dfn来取得<code></code></code></dfn>
+Use <dfn>bracket notation</dfn> to find the last character in the `lastName` variable.
+
+**Hint:** Try looking at the example above if you get stuck.
# --hints--
-`lastLetterOfLastName`应该是"e"。
+`lastLetterOfLastName` should be "e".
```js
assert(lastLetterOfLastName === 'e');
```
-你需要使用`.length`获取最后一个字符。
+You should use `.length` to get the last letter.
```js
-assert(code.match(/\.length/g).length === 2);
+assert(code.match(/\.length/g).length > 0);
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-bracket-notation-to-find-the-nth-character-in-a-string.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-bracket-notation-to-find-the-nth-character-in-a-string.md
index 6232a29a72a..b3c183a9040 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-bracket-notation-to-find-the-nth-character-in-a-string.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-bracket-notation-to-find-the-nth-character-in-a-string.md
@@ -1,6 +1,6 @@
---
id: bd7123c9c450eddfaeb5bdef
-title: 使用方括号查找字符串中的第N个字符
+title: Use Bracket Notation to Find the Nth Character in a String
challengeType: 1
videoUrl: 'https://scrimba.com/c/cWPVJua'
forumTopicId: 18343
@@ -9,26 +9,32 @@ dashedName: use-bracket-notation-to-find-the-nth-character-in-a-string
# --description--
-你也可以使用方括号来获得一个字符串中的其他位置的字符。
+You can also use <dfn>bracket notation</dfn> to get the character at other positions within a string.
-请记住，程序是从`0`开始计数，所以获取第一个字符实际上是\[0]。
+Remember that computers start counting at `0`, so the first character is actually the zeroth character.
+
+Example:
+
+```js
+var firstName = "Ada";
+var secondLetterOfFirstName = firstName[1]; // secondLetterOfFirstName is "d"
+```
# --instructions--
-让我们使用方括号，把`lastName`变量的第三个字符赋值给`thirdLetterOfLastName`。
+Let's try to set `thirdLetterOfLastName` to equal the third letter of the `lastName` variable using bracket notation.
-**提示**
-如果你遇到困难了，看看`secondLetterOfFirstName`变量是如何做的。
+**Hint:** Try looking at the example above if you get stuck.
# --hints--
-`thirdLetterOfLastName`的值应该是`v`。
+The `thirdLetterOfLastName` variable should have the value of `v`.
```js
assert(thirdLetterOfLastName === 'v');
```
-你应该使用方括号。
+You should use bracket notation.
```js
assert(code.match(/thirdLetterOfLastName\s*?=\s*?lastName\[.*?\]/));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-bracket-notation-to-find-the-nth-to-last-character-in-a-string.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-bracket-notation-to-find-the-nth-to-last-character-in-a-string.md
index a27a7755739..0cb27719b04 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-bracket-notation-to-find-the-nth-to-last-character-in-a-string.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-bracket-notation-to-find-the-nth-to-last-character-in-a-string.md
@@ -1,6 +1,6 @@
---
id: bd7123c9c452eddfaeb5bdef
-title: 使用方括号查找字符串中的第N个字符到最后一个字符
+title: Use Bracket Notation to Find the Nth-to-Last Character in a String
challengeType: 1
videoUrl: 'https://scrimba.com/c/cw4vkh9'
forumTopicId: 18344
@@ -9,29 +9,35 @@ dashedName: use-bracket-notation-to-find-the-nth-to-last-character-in-a-string
# --description--
-我们既可以获取字符串的最后一个字符，也可以用获取字符串的倒数第N个字符。
+You can use the same principle we just used to retrieve the last character in a string to retrieve the Nth-to-last character.
-例如，你可以这样`firstName[firstName.length - 3]`操作来获得`var firstName = "Charles"`字符串中的倒数第三个字符。
+For example, you can get the value of the third-to-last letter of the `var firstName = "Charles"` string by using `firstName[firstName.length - 3]`
+
+Example:
+
+```js
+var firstName = "Charles";
+var thirdToLastLetter = firstName[firstName.length - 3]; // thirdToLastLetter is "l"
+```
# --instructions--
-使用<dfn>方括号</dfn>来获得`lastName`字符串中的倒数第二个字符。
+Use <dfn>bracket notation</dfn> to find the second-to-last character in the `lastName` string.
-**提示**
-如果你遇到困难了，不妨看看`thirdToLastLetterOfFirstName`变量是如何做到的。
+**Hint:** Try looking at the example above if you get stuck.
# --hints--
-`secondToLastLetterOfLastName`应该是"c"。
+`secondToLastLetterOfLastName` should be "c".
```js
assert(secondToLastLetterOfLastName === 'c');
```
-你需要使用`.length`获取倒数第二个字符。
+You should use `.length` to get the second last letter.
```js
-assert(code.match(/\.length/g).length === 2);
+assert(code.match(/\.length/g).length > 0);
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-conditional-logic-with-if-statements.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-conditional-logic-with-if-statements.md
index ab398d66d69..e90542e170d 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-conditional-logic-with-if-statements.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-conditional-logic-with-if-statements.md
@@ -1,6 +1,6 @@
---
id: cf1111c1c12feddfaeb3bdef
-title: 用 if 语句来表达条件逻辑
+title: Use Conditional Logic with If Statements
challengeType: 1
videoUrl: 'https://scrimba.com/c/cy87mf3'
forumTopicId: 18348
@@ -9,15 +9,15 @@ dashedName: use-conditional-logic-with-if-statements
# --description--
-`If`语句用于在代码中做条件判断。关键字`if`告诉 JavaScript 在小括号中的条件为真的情况下去执行定义在大括号里面的代码。这种条件被称为`Boolean`条件，因为他们只可能是`true`（真）或`false`（假）。
+`If` statements are used to make decisions in code. The keyword `if` tells JavaScript to execute the code in the curly braces under certain conditions, defined in the parentheses. These conditions are known as `Boolean` conditions and they may only be `true` or `false`.
-当条件的计算结果为`true`，程序执行大括号内的语句。当布尔条件的计算结果为`false`，大括号内的代码将不会执行。
+When the condition evaluates to `true`, the program executes the statement inside the curly braces. When the Boolean condition evaluates to `false`, the statement inside the curly braces will not execute.
-**伪代码**
+**Pseudocode**
-<blockquote>if(<i>条件为真</i>){<br> <i>语句被执行</i><br>}</blockquote>
+<blockquote>if (<i>condition is true</i>) {<br> <i>statement is executed</i><br>}</blockquote>
-**示例**
+**Example**
```js
function test (myCondition) {
@@ -30,39 +30,39 @@ test(true); // returns "It was true"
test(false); // returns "It was false"
```
-当`test`被调用，并且传递进来的参数值为`true`，`if`语句会计算`myCondition`的结果，看它是真还是假。如果条件为`true`，函数会返回`"It was true"`。当`test`被调用，并且传递进来的参数值为`false`，`myCondition`*不* 为`true`，并且不执行大括号后面的语句，函数返回`"It was false"`。
+When `test` is called with a value of `true`, the `if` statement evaluates `myCondition` to see if it is `true` or not. Since it is `true`, the function returns `"It was true"`. When we call `test` with a value of `false`, `myCondition` is *not* `true` and the statement in the curly braces is not executed and the function returns `"It was false"`.
# --instructions--
-在函数内部创建一个`if`语句，如果该参数`wasThatTrue`值为`true`，返回`"Yes, that was true"`，否则，并返回`"No, that was false"`。
+Create an `if` statement inside the function to return `"Yes, that was true"` if the parameter `wasThatTrue` is `true` and return `"No, that was false"` otherwise.
# --hints--
-`trueOrFalse`应该是一个函数。
+`trueOrFalse` should be a function
```js
assert(typeof trueOrFalse === 'function');
```
-`trueOrFalse(true)`应该返回一个字符串。
+`trueOrFalse(true)` should return a string
```js
assert(typeof trueOrFalse(true) === 'string');
```
-`trueOrFalse(false)`应该返回一个字符串。
+`trueOrFalse(false)` should return a string
```js
assert(typeof trueOrFalse(false) === 'string');
```
-`trueOrFalse(true)`应该返回 "Yes, that was true"。
+`trueOrFalse(true)` should return "Yes, that was true"
```js
assert(trueOrFalse(true) === 'Yes, that was true');
```
-`trueOrFalse(false)`应该返回 "No, that was false"。
+`trueOrFalse(false)` should return "No, that was false"
```js
assert(trueOrFalse(false) === 'No, that was false');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-multiple-conditional-ternary-operators.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-multiple-conditional-ternary-operators.md
index e59c5ca3bc2..73a2120e7f0 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-multiple-conditional-ternary-operators.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-multiple-conditional-ternary-operators.md
@@ -1,6 +1,6 @@
---
id: 587d7b7e367417b2b2512b21
-title: 使用多个三元运算符
+title: Use Multiple Conditional (Ternary) Operators
challengeType: 1
videoUrl: 'https://scrimba.com/c/cyWJBT4'
forumTopicId: 301179
@@ -9,9 +9,9 @@ dashedName: use-multiple-conditional-ternary-operators
# --description--
-在之前的挑战中，你使用了一个条件运算符。你也可以将多个运算符串联在一起以检查多种条件。
+In the previous challenge, you used a single conditional operator. You can also chain them together to check for multiple conditions.
-下面的函数使用 if，else if 和 else 语句来检查多个条件：
+The following function uses if, else if, and else statements to check multiple conditions:
```js
function findGreaterOrEqual(a, b) {
@@ -27,7 +27,7 @@ function findGreaterOrEqual(a, b) {
}
```
-上面的函数使用条件运算符写法如下：
+The above function can be re-written using multiple conditional operators:
```js
function findGreaterOrEqual(a, b) {
@@ -37,7 +37,7 @@ function findGreaterOrEqual(a, b) {
}
```
-即便如此，应谨慎使用多个三元运算符，因为在没有适当缩进的情况下使用多个三元运算符可能会使您的代码难以阅读。 例如：
+It is considered best practice to format multiple conditional operators such that each condition is on a separate line, as shown above. Using multiple conditional operators without proper indentation may make your code hard to read. For example:
```js
function findGreaterOrEqual(a, b) {
@@ -47,29 +47,29 @@ function findGreaterOrEqual(a, b) {
# --instructions--
-在 checkSign 函数中使用多个条件运算符来检查数字是正数 ("positive")、负数 ("negative") 或零 ("zero")。
+In the `checkSign` function, use multiple conditional operators - following the recommended format used in `findGreaterOrEqual` - to check if a number is positive, negative or zero. The function should return `"positive"`, `"negative"` or `"zero"`.
# --hints--
-`checkSign`应该使用多个条件运算符。
+`checkSign` should use multiple conditional operators
```js
assert(/.+?\s*?\?\s*?.+?\s*?:\s*?.+?\s*?\?\s*?.+?\s*?:\s*?.+?/gi.test(code));
```
-`checkSign(10)`应该返回 "positive" 注意，结果对大小写敏感。
+`checkSign(10)` should return "positive". Note that capitalization matters
```js
assert(checkSign(10) === 'positive');
```
-`checkSign(-12)`应该返回 "negative" 注意，结果对大小写敏感。
+`checkSign(-12)` should return "negative". Note that capitalization matters
```js
assert(checkSign(-12) === 'negative');
```
-`checkSign(0)`应该返回 "zero" 注意，结果对大小写敏感。
+`checkSign(0)` should return "zero". Note that capitalization matters
```js
assert(checkSign(0) === 'zero');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-recursion-to-create-a-countdown.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-recursion-to-create-a-countdown.md
index 7011c4089d0..44eace9821c 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-recursion-to-create-a-countdown.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-recursion-to-create-a-countdown.md
@@ -1,6 +1,6 @@
---
id: 5cd9a70215d3c4e65518328f
-title: 使用递归创建一个倒计时
+title: Use Recursion to Create a Countdown
challengeType: 1
forumTopicId: 305925
dashedName: use-recursion-to-create-a-countdown
@@ -8,11 +8,11 @@ dashedName: use-recursion-to-create-a-countdown
# --description--
-在上一个[挑战](/learn/javascript-algorithms-and-data-structures/basic-javascript/replace-loops-using-recursion)，学习了怎样用递归来代替循环。现在来学习一个更复杂的函数，函数返回一个从 `1` 到传递给函数的指定数字的连续数字数组。
+In a [previous challenge](/learn/javascript-algorithms-and-data-structures/basic-javascript/replace-loops-using-recursion), you learned how to use recursion to replace a for loop. Now, let's look at a more complex function that returns an array of consecutive integers starting with `1` through the number passed to the function.
-正如上一个挑战提到的，会有一个 <dfn>base case</dfn>。base case 告诉递归函数什么时候不在需要调用其自身。这是简单 情况，返回得到的值。还有 <dfn>recursive call</dfn>，继续用不同的参数调用自身。如果函数无误，一直执行直到 base case 为止。
+As mentioned in the previous challenge, there will be a <dfn>base case</dfn>. The base case tells the recursive function when it no longer needs to call itself. It is a simple case where the return value is already known. There will also be a <dfn>recursive call</dfn> which executes the original function with different arguments. If the function is written correctly, eventually the base case will be reached.
-比如，如果想写一个递归函数，返回一个数字 `1` 到 `n` 的连续数组。这个函数需要接收一个参数`n` 代表起始数字。然后会持续的调用自身，传入一个比 `n` 更小的值一直到传入的值是 `1` 为止。函数如下：
+For example, say you want to write a recursive function that returns an array containing the numbers `1` through `n`. This function will need to accept an argument, `n`, representing the final number. Then it will need to call itself with progressively smaller values of `n` until it reaches `1`. You could write the function as follows:
```javascript
function countup(n) {
@@ -27,42 +27,48 @@ function countup(n) {
console.log(countup(5)); // [ 1, 2, 3, 4, 5 ]
```
-起初，这似乎是违反直觉的，因为 n 的值*递减*，但是最终数组中的值却*递增*。 之所以发生这种情况，是因为在递归调用返回之后，才调用 push。 在将 `n` pushed 进数组时，`count(n - 1)` 已经调用赋值成功并返回了 `[1, 2, ..., n - 1]`。
+At first, this seems counterintuitive since the value of `n` *decreases*, but the values in the final array are *increasing*. This happens because the push happens last, after the recursive call has returned. At the point where `n` is pushed into the array, `countup(n - 1)` has already been evaluated and returned `[1, 2, ..., n - 1]`.
# --instructions--
-已经定义了一个函数 `countdown`，函数有一个参数（`n`）。函数应该基于参数 `n` 递归调用返回 `n` 到 `1` 的连续数字的数组。如果函数以小于 1 的参数调用，函数应该返回空数组。 比如，用 `n = 5` 调用函数应该返回数组 `[5, 4, 3, 2, 1]`。 函数必需使用递归函数调用自身，不能使用任何形式的循环。
+We have defined a function called `countdown` with one parameter (`n`). The function should use recursion to return an array containing the integers `n` through `1` based on the `n` parameter. If the function is called with a number less than 1, the function should return an empty array. For example, calling this function with `n = 5` should return the array `[5, 4, 3, 2, 1]`. Your function must use recursion by calling itself and must not use loops of any kind.
# --hints--
-`countdown(-1)` 应该返回一个空数组。
+`countdown(-1)` should return an empty array.
```js
assert.isEmpty(countdown(-1));
```
-`countdown(10)` 应该返回 `[10, 9, 8, 7, 6, 5, 4, 3, 2, 1]`。
+`countdown(10)` should return `[10, 9, 8, 7, 6, 5, 4, 3, 2, 1]`
```js
assert.deepStrictEqual(countdown(10), [10, 9, 8, 7, 6, 5, 4, 3, 2, 1]);
```
-`countdown(5)` 应该返回 `[5, 4, 3, 2, 1]`。
+`countdown(5)` should return `[5, 4, 3, 2, 1]`
```js
assert.deepStrictEqual(countdown(5), [5, 4, 3, 2, 1]);
```
-代码不能包含任意形式的循环（`for`、`while` 或者高阶函数如：`forEach`、`map`、`filter` 以及 `reduce`）。
+Your code should not rely on any kind of loops (`for`, `while` or higher order functions such as `forEach`, `map`, `filter`, and `reduce`).
```js
-assert(!removeJSComments(code).match(/for|while|forEach|map|filter|reduce/g));
+assert(
+ !__helpers
+ .removeJSComments(code)
+ .match(/for|while|forEach|map|filter|reduce/g)
+);
```
-应该用递归解决这个问题。
+You should use recursion to solve this problem.
```js
-assert(removeJSComments(countdown.toString()).match(/countdown\s*\(.+\)\;/));
+assert(
+ __helpers.removeJSComments(countdown.toString()).match(/countdown\s*\(.+\)/)
+);
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-recursion-to-create-a-range-of-numbers.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-recursion-to-create-a-range-of-numbers.md
index 4cd18381880..2e10a18bfb1 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-recursion-to-create-a-range-of-numbers.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-recursion-to-create-a-range-of-numbers.md
@@ -1,6 +1,6 @@
---
id: 5cc0bd7a49b71cb96132e54c
-title: 使用递归来创建一个数字序列
+title: Use Recursion to Create a Range of Numbers
challengeType: 1
forumTopicId: 301180
dashedName: use-recursion-to-create-a-range-of-numbers
@@ -8,47 +8,53 @@ dashedName: use-recursion-to-create-a-range-of-numbers
# --description--
-接着上一个挑战，有另外一个机会来创建递归函数解决问题。
+Continuing from the previous challenge, we provide you another opportunity to create a recursive function to solve a problem.
# --instructions--
-定义好了 `rangeOfNumbers` 函数，包含两个参数。函数应该返回一个连续数字数组，`startNum` 参数开始 `endNum` 参数截止。开始的数字小于或等于截止数字。函数必需递归调用自身，不能使用任意形式的循环。要考虑到 `startNum` 和 `endNum` 相同的情况。
+We have defined a function named `rangeOfNumbers` with two parameters. The function should return an array of integers which begins with a number represented by the `startNum` parameter and ends with a number represented by the `endNum` parameter. The starting number will always be less than or equal to the ending number. Your function must use recursion by calling itself and not use loops of any kind. It should also work for cases where both `startNum` and `endNum` are the same.
# --hints--
-函数应该返回一个数组。
+Your function should return an array.
```js
assert(Array.isArray(rangeOfNumbers(5, 10)));
```
-不能包含循环语句（`for` 或者 `while` 或者高阶函数比如 `forEach`、`map`、`filter` 或者 `reduce`）。
-
-```js
-assert(!removeJSComments(code).match(/for|while|forEach|map|filter|reduce/g));
-```
-
-`rangeOfNumbers` 应该使用递归函数（调用自身）来完成这个挑战。
+Your code should not use any loop syntax (`for` or `while` or higher order functions such as `forEach`, `map`, `filter`, or `reduce`).
```js
assert(
- removeJSComments(rangeOfNumbers.toString()).match(/rangeOfNumbers\s*\(.+\)/)
+ !__helpers
+ .removeJSComments(code)
+ .match(/for|while|forEach|map|filter|reduce/g)
);
```
-`rangeOfNumbers(1, 5)` 应该返回 `[1, 2, 3, 4, 5]`。
+`rangeOfNumbers` should use recursion (call itself) to solve this challenge.
+
+```js
+assert(
+ __helpers
+ .removeJSComments(rangeOfNumbers.toString())
+ .match(/rangeOfNumbers\s*\(.+\)/)
+);
+```
+
+`rangeOfNumbers(1, 5)` should return `[1, 2, 3, 4, 5]`.
```js
assert.deepStrictEqual(rangeOfNumbers(1, 5), [1, 2, 3, 4, 5]);
```
-`rangeOfNumbers(6, 9)` 应该返回 `[6, 7, 8, 9]`。
+`rangeOfNumbers(6, 9)` should return `[6, 7, 8, 9]`.
```js
assert.deepStrictEqual(rangeOfNumbers(6, 9), [6, 7, 8, 9]);
```
-`rangeOfNumbers(4, 4)` 应该返回 `[4]`。
+`rangeOfNumbers(4, 4)` should return `[4]`.
```js
assert.deepStrictEqual(rangeOfNumbers(4, 4), [4]);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-the-conditional-ternary-operator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-the-conditional-ternary-operator.md
index b4d311564d1..0a8808b0a8e 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-the-conditional-ternary-operator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-the-conditional-ternary-operator.md
@@ -1,21 +1,20 @@
---
id: 587d7b7e367417b2b2512b24
-title: 使用三元运算符
+title: Use the Conditional (Ternary) Operator
challengeType: 1
-videoUrl: 'https://scrimba.com/c/c3JRmSg'
forumTopicId: 301181
dashedName: use-the-conditional-ternary-operator
---
# --description--
-条件运算符（也称为三元运算符）的用处就像写成一行的 if-else 表达式
+The <dfn>conditional operator</dfn>, also called the <dfn>ternary operator</dfn>, can be used as a one line if-else expression.
-语法如下所示：
+The syntax is:
-`condition ? statement-if-true : statement-if-false;`
+`condition ? expression-if-true : expression-if-false;`
-以下函数使用 if-else 语句来检查条件：
+The following function uses an if-else statement to check a condition:
```js
function findGreater(a, b) {
@@ -28,7 +27,7 @@ function findGreater(a, b) {
}
```
-上面的函数使用条件运算符写法如下：
+This can be re-written using the `conditional operator`:
```js
function findGreater(a, b) {
@@ -38,29 +37,29 @@ function findGreater(a, b) {
# --instructions--
-在`checkEqual`函数中使用条件运算符检查两个数字是否相等，函数应该返回 "Equal" 或 "Not Equal"
+Use the `conditional operator` in the `checkEqual` function to check if two numbers are equal or not. The function should return either "Equal" or "Not Equal".
# --hints--
-`checkEqual`应该使用条件运算符。
+`checkEqual` should use the `conditional operator`
```js
assert(/.+?\s*?\?\s*?.+?\s*?:\s*?.+?/.test(code));
```
-`checkEqual(1, 2)`应该返回 "Not Equal"。
+`checkEqual(1, 2)` should return "Not Equal"
```js
assert(checkEqual(1, 2) === 'Not Equal');
```
-`checkEqual(1, 1)`应该返回 "Equal"。
+`checkEqual(1, 1)` should return "Equal"
```js
assert(checkEqual(1, 1) === 'Equal');
```
-`checkEqual(1, -1)`应该返回 "Not Equal"。
+`checkEqual(1, -1)` should return "Not Equal"
```js
assert(checkEqual(1, -1) === 'Not Equal');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-the-parseint-function-with-a-radix.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-the-parseint-function-with-a-radix.md
index c615f1cc46b..7d83a60eb96 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-the-parseint-function-with-a-radix.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-the-parseint-function-with-a-radix.md
@@ -1,6 +1,6 @@
---
id: 587d7b7e367417b2b2512b22
-title: 使用 parseInt 函数并传入一个基数
+title: Use the parseInt Function with a Radix
challengeType: 1
videoUrl: 'https://scrimba.com/c/c6K4Kh3'
forumTopicId: 301182
@@ -9,49 +9,49 @@ dashedName: use-the-parseint-function-with-a-radix
# --description--
-`parseInt()`函数解析一个字符串并返回一个整数。它同时可接受第二个参数，一个介于2和36之间的整数，表示字符串的基数。
+The `parseInt()` function parses a string and returns an integer. It takes a second argument for the radix, which specifies the base of the number in the string. The radix can be an integer between 2 and 36.
-函数调用如下所示：
+The function call looks like:
`parseInt(string, radix);`
-示例：
+And here's an example:
`var a = parseInt("11", 2);`
-参数 2 表示 "11" 使用二进制数值系统。此示例将字符串 "11" 转换为整数 3。
+The radix variable says that "11" is in the binary system, or base 2. This example converts the string "11" to an integer 3.
# --instructions--
-在`convertToIntegerparseInt()`函数把二进制数转换成十进制并返回。
+Use `parseInt()` in the `convertToInteger` function so it converts a binary number to an integer and returns it.
# --hints--
-`convertToInteger`中应该使用`parseInt()`函数。
+`convertToInteger` should use the `parseInt()` function
```js
assert(/parseInt/g.test(code));
```
-`convertToInteger("10011")`应该返回一个数字。
+`convertToInteger("10011")` should return a number
```js
assert(typeof convertToInteger('10011') === 'number');
```
-`convertToInteger("10011")`应该返回 19。
+`convertToInteger("10011")` should return 19
```js
assert(convertToInteger('10011') === 19);
```
-`convertToInteger("111001")`应该返回 57。
+`convertToInteger("111001")` should return 57
```js
assert(convertToInteger('111001') === 57);
```
-`convertToInteger("JamesBond")`应该返回 NaN。
+`convertToInteger("JamesBond")` should return NaN
```js
assert.isNaN(convertToInteger('JamesBond'));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-the-parseint-function.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-the-parseint-function.md
index d6538cb5b31..94d85f0bdda 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-the-parseint-function.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/use-the-parseint-function.md
@@ -1,6 +1,6 @@
---
id: 587d7b7e367417b2b2512b23
-title: 使用 parseInt 函数
+title: Use the parseInt Function
challengeType: 1
videoUrl: 'https://scrimba.com/c/cm83LSW'
forumTopicId: 301183
@@ -9,43 +9,43 @@ dashedName: use-the-parseint-function
# --description--
-`parseInt()`函数解析一个字符串返回一个整数下面是一个示例：
+The `parseInt()` function parses a string and returns an integer. Here's an example:
`var a = parseInt("007");`
-上面的函数把字符串 "007" 转换成数字 7。 如果字符串参数的第一个字符是字符串类型的，结果将不会转换成数字，而是返回`NaN`.
+The above function converts the string "007" to an integer 7. If the first character in the string can't be converted into a number, then it returns `NaN`.
# --instructions--
-在`convertToInteger`函数中使用`parseInt()`将字符串`str`转换为正数并返回。
+Use `parseInt()` in the `convertToInteger` function so it converts the input string `str` into an integer, and returns it.
# --hints--
-`convertToInteger`应该使用`parseInt()`函数。
+`convertToInteger` should use the `parseInt()` function
```js
assert(/parseInt/g.test(code));
```
-`convertToInteger("56")`应该返回一个数字。
+`convertToInteger("56")` should return a number
```js
assert(typeof convertToInteger('56') === 'number');
```
-`convertToInteger("56")`应该返回 56。
+`convertToInteger("56")` should return 56
```js
assert(convertToInteger('56') === 56);
```
-`convertToInteger("77")`应该返回 77。
+`convertToInteger("77")` should return 77
```js
assert(convertToInteger('77') === 77);
```
-`convertToInteger("JamesBond")`应该返回 NaN。
+`convertToInteger("JamesBond")` should return NaN
```js
assert.isNaN(convertToInteger('JamesBond'));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/using-objects-for-lookups.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/using-objects-for-lookups.md
index 15fed646d6c..1185e31c247 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/using-objects-for-lookups.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/using-objects-for-lookups.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244ca
-title: 使用对象进行查找
+title: Using Objects for Lookups
challengeType: 1
videoUrl: 'https://scrimba.com/c/cdBk8sM'
forumTopicId: 18373
@@ -9,9 +9,9 @@ dashedName: using-objects-for-lookups
# --description--
-对象和字典一样，可以用来存储键/值对。如果你的数据跟对象一样，你可以用对象来查找你想要的值，而不是使用switch或if/else语句。当你知道你的输入数据在某个范围时，这种查找方式极为有效。
+Objects can be thought of as a key/value storage, like a dictionary. If you have tabular data, you can use an object to "lookup" values rather than a `switch` statement or an `if/else` chain. This is most useful when you know that your input data is limited to a certain range.
-这是简单的反向字母表：
+Here is an example of a simple reverse alphabet lookup:
```js
var alpha = {
@@ -33,59 +33,59 @@ alpha[value]; // "Y"
# --instructions--
-把 switch 语句转化为`lookup`对象。使用它来查找`val`属性的值，并赋值给`result`变量。
+Convert the switch statement into an object called `lookup`. Use it to look up `val` and assign the associated string to the `result` variable.
# --hints--
-`phoneticLookup("alpha")`应该等于`"Adams"`。
+`phoneticLookup("alpha")` should equal `"Adams"`
```js
assert(phoneticLookup('alpha') === 'Adams');
```
-`phoneticLookup("bravo")`应该等于`"Boston"`。
+`phoneticLookup("bravo")` should equal `"Boston"`
```js
assert(phoneticLookup('bravo') === 'Boston');
```
-`phoneticLookup("charlie")`应该等于`"Chicago"`。
+`phoneticLookup("charlie")` should equal `"Chicago"`
```js
assert(phoneticLookup('charlie') === 'Chicago');
```
-`phoneticLookup("delta")`应该等于`"Denver"`。
+`phoneticLookup("delta")` should equal `"Denver"`
```js
assert(phoneticLookup('delta') === 'Denver');
```
-`phoneticLookup("echo")`应该等于`"Easy"`。
+`phoneticLookup("echo")` should equal `"Easy"`
```js
assert(phoneticLookup('echo') === 'Easy');
```
-`phoneticLookup("foxtrot")`应该等于`"Frank"`。
+`phoneticLookup("foxtrot")` should equal `"Frank"`
```js
assert(phoneticLookup('foxtrot') === 'Frank');
```
-`phoneticLookup("")`应该等于`undefined`。
+`phoneticLookup("")` should equal `undefined`
```js
assert(typeof phoneticLookup('') === 'undefined');
```
-请不要修改`return`语句。
+You should not modify the `return` statement
```js
assert(code.match(/return\sresult;/));
```
-请不要使用`case`，`switch`，或`if`语句。
+You should not use `case`, `switch`, or `if` statements
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/word-blanks.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/word-blanks.md
index 6da8a9b51c3..37ab2df30c7 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/word-blanks.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/word-blanks.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244bb
-title: 填词造句
+title: Word Blanks
challengeType: 1
videoUrl: 'https://scrimba.com/c/caqn8zuP'
forumTopicId: 18377
@@ -9,11 +9,11 @@ dashedName: word-blanks
# --description--
-现在，我们来用字符串的相关知识实现一个 "[Mad Libs](https://en.wikipedia.org/wiki/Mad_Libs)" 类的文字游戏，称为 "Word Blanks"。 你将创建一个（可选幽默的）“填空”样式句子。
+We will now use our knowledge of strings to build a "[Mad Libs](https://en.wikipedia.org/wiki/Mad_Libs)" style word game we're calling "Word Blanks". You will create an (optionally humorous) "Fill in the Blanks" style sentence.
-在 "Mad Libs" 游戏中，提供一个缺少一些单词的句子，缺少的单词包括名词，动词，形容词和副词等。然后，你选择一些单词填写句子缺失的地方，使句子完整并且有意义。
+In a "Mad Libs" game, you are provided sentences with some missing words, like nouns, verbs, adjectives and adverbs. You then fill in the missing pieces with words of your choice in a way that the completed sentence makes sense.
-思考一下这句话 - "It was really **\_\_\_\_**, and we **\_\_\_\_** ourselves **\_\_\_\_**"。这句话有三个缺失的部分 - 形容词，动词和副词，选择合适单词填入完成它。然后将完成的句子赋值给变量，如下所示：
+Consider this sentence - "It was really **\_\_\_\_**, and we **\_\_\_\_** ourselves **\_\_\_\_**". This sentence has three missing pieces- an adjective, a verb and an adverb, and we can add words of our choice to complete it. We can then assign the completed sentence to a variable as follows:
```js
var sentence = "It was really " + "hot" + ", and we " + "laughed" + " ourselves " + "silly" + ".";
@@ -21,21 +21,21 @@ var sentence = "It was really " + "hot" + ", and we " + "laughed" + " ourselves
# --instructions--
-在这个挑战中，我们为你提供名词，动词，形容词和副词。你需要使用合适单词以及我们提供的单词来形成完整的句子。
+In this challenge, we provide you with a noun, a verb, an adjective and an adverb. You need to form a complete sentence using words of your choice, along with the words we provide.
-你需要使用字符串连接运算符`+`来拼接字符串变量：`myNoun`，`myAdjective`，`myVerb`，和`myAdverb`来构建一个新字符串。然后，将新字符串赋给`result`变量。
+You will need to use the string concatenation operator `+` to build a new string, using the provided variables: `myNoun`, `myAdjective`, `myVerb`, and `myAdverb`. You will then assign the formed string to the `wordBlanks` variable. You should not change the words assigned to the variables.
-你还需要考虑字符串中的空格，确保句子的所有单词之间有空格。结果应该是一个完整的句子。
+You will also need to account for spaces in your string, so that the final sentence has spaces between all the words. The result should be a complete sentence.
# --hints--
-`wordBlanks("","","","")`应该返回一个字符串。
+`wordBlanks` should be a string.
```js
assert(typeof wordBlanks === 'string');
```
-不能改变 `myNoun`、`myVerb`、`myAdjective` 或者 `myAdverb` 的值。
+You should not change the values assigned to `myNoun`, `myVerb`, `myAdjective` or `myAdverb`.
```js
assert(
@@ -46,7 +46,7 @@ assert(
);
```
-不能直接使用 "dog"、"ran"、"big" 或者 "quickly" 来创建 `wordBlanks`。
+You should not directly use the values "dog", "ran", "big", or "quickly" to create `wordBlanks`.
```js
const newCode = removeAssignments(code);
@@ -58,7 +58,7 @@ assert(
);
```
-`wordBlanks` 应包含分配给变量 `myNoun`、`myVerb`、`myAdjective` 和 `myAdverb` 的所有单词，并用非单词字符（以及 madlib 中的其它单词）分隔。
+`wordBlanks` should contain all of the words assigned to the variables `myNoun`, `myVerb`, `myAdjective` and `myAdverb` separated by non-word characters (and any additional words in your madlib).
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/write-reusable-javascript-with-functions.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/write-reusable-javascript-with-functions.md
index c9a22920cfe..2fb3ca9cd80 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/write-reusable-javascript-with-functions.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/basic-javascript/write-reusable-javascript-with-functions.md
@@ -1,6 +1,6 @@
---
id: 56bbb991ad1ed5201cd392cf
-title: 用函数编写可重用代码
+title: Write Reusable JavaScript with Functions
challengeType: 1
videoUrl: 'https://scrimba.com/c/cL6dqfy'
forumTopicId: 18378
@@ -9,9 +9,9 @@ dashedName: write-reusable-javascript-with-functions
# --description--
-在 JavaScript 中，我们可以把代码的重复部分抽取出来，放到一个<dfn>函数</dfn>中。
+In JavaScript, we can divide up our code into reusable parts called <dfn>functions</dfn>.
-举个例子：
+Here's an example of a function:
```js
function functionName() {
@@ -19,27 +19,27 @@ function functionName() {
}
```
-你可以通过函数名`functionName`加上后面的小括号来调用这个函数，就像这样： `functionName();` 每次调用函数时，它都会在控制台上打印消息`"Hello World"`。每次调用函数时，大括号之间的所有代码都将被执行。
+You can call or <dfn>invoke</dfn> this function by using its name followed by parentheses, like this: `functionName();` Each time the function is called it will print out the message `"Hello World"` on the dev console. All of the code between the curly braces will be executed every time the function is called.
# --instructions--
-<ol><li>先创建一个名为<code>reusableFunction</code>的函数，这个函数可以打印<code>"Hi World"</code>到控制台上。</li><li>然后调用这个函数。</li></ol>
+<ol><li>Create a function called <code>reusableFunction</code> which prints <code>"Hi World"</code> to the dev console.</li><li>Call the function.</li></ol>
# --hints--
-`reusableFunction`应该是一个函数。
+`reusableFunction` should be a function.
```js
assert(typeof reusableFunction === 'function');
```
-`reusableFunction`应该在控制台中输出 "Hi World"。
+`reusableFunction` should output "Hi World" to the dev console.
```js
assert(hiWorldWasLogged);
```
-在你定义`reusableFunction`之后记得调用它。
+You should call `reusableFunction` after you define it.
```js
assert(/^\s*reusableFunction\(\)\s*/m.test(code));
@@ -87,7 +87,6 @@ if (typeof reusableFunction !== "function") {
```
## --seed-contents--
-
```js
```
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-arguments-passed-in-the-wrong-order-when-calling-a-function.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-arguments-passed-in-the-wrong-order-when-calling-a-function.md
index 6edfec092ed..7aea16b1e6b 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-arguments-passed-in-the-wrong-order-when-calling-a-function.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-arguments-passed-in-the-wrong-order-when-calling-a-function.md
@@ -1,6 +1,6 @@
---
id: 587d7b85367417b2b2512b3a
-title: 调用函数时，捕获以错误顺序传递的参数
+title: Catch Arguments Passed in the Wrong Order When Calling a Function
challengeType: 1
forumTopicId: 301184
dashedName: catch-arguments-passed-in-the-wrong-order-when-calling-a-function
@@ -8,21 +8,21 @@ dashedName: catch-arguments-passed-in-the-wrong-order-when-calling-a-function
# --description--
-继续讨论调用函数，需要注意的下一个 bug 是函数的参数传递顺序错误。 如果参数分别是不同的类型，例如接受数组和整数两个参数的函数，参数顺序传错就可能会引发运行时错误。对于接受相同类型参数的函数，传错参数也会导致逻辑错误或运行结果错误。确保以正确的顺序提供所有必需的参数以避免这些问题。
+Continuing the discussion on calling functions, the next bug to watch out for is when a function's arguments are supplied in the incorrect order. If the arguments are different types, such as a function expecting an array and an integer, this will likely throw a runtime error. If the arguments are the same type (all integers, for example), then the logic of the code won't make sense. Make sure to supply all required arguments, in the proper order to avoid these issues.
# --instructions--
-函数`raiseToPower`返回基数 (base) 的指数 (exponent) 次幂。不幸的是，它没有被正确调用 ———— 修改代码，使`power`的值为 8。
+The function `raiseToPower` raises a base to an exponent. Unfortunately, it's not called properly - fix the code so the value of `power` is the expected 8.
# --hints--
-你应修复变量`power`，使其等于 2 的 3 次方，而不是 3 的 2 次方。
+Your code should fix the variable `power` so it equals 2 raised to the 3rd power, not 3 raised to the 2nd power.
```js
assert(power == 8);
```
-你调用`raiseToPower`函数时，传递参数的顺序应正确。
+Your code should use the correct order of the arguments for the `raiseToPower` function call.
```js
assert(code.match(/raiseToPower\(\s*?base\s*?,\s*?exp\s*?\);/g));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-missing-open-and-closing-parenthesis-after-a-function-call.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-missing-open-and-closing-parenthesis-after-a-function-call.md
index 3a715dcd93a..5a4e0fc0601 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-missing-open-and-closing-parenthesis-after-a-function-call.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-missing-open-and-closing-parenthesis-after-a-function-call.md
@@ -1,6 +1,6 @@
---
id: 587d7b85367417b2b2512b39
-title: 捕捉函数调用后缺少的左括号和右括号
+title: Catch Missing Open and Closing Parenthesis After a Function Call
challengeType: 1
forumTopicId: 301185
dashedName: catch-missing-open-and-closing-parenthesis-after-a-function-call
@@ -8,9 +8,9 @@ dashedName: catch-missing-open-and-closing-parenthesis-after-a-function-call
# --description--
-当函数或方法不接受任何参数时，你可能忘记在调用它时加上空的左括号和右括号。通常，函数调用的结果会保存在变量中，供其他代码使用。可以通过将变量值（或其类型）打印到控制台，查看输出究竟是一个函数引用还是函数调用的返回值来检测这类错误。
+When a function or method doesn't take any arguments, you may forget to include the (empty) opening and closing parentheses when calling it. Often times the result of a function call is saved in a variable for other use in your code. This error can be detected by logging variable values (or their types) to the console and seeing that one is set to a function reference, instead of the expected value the function returns.
-下面示例中的两个变量是不同的:
+The variables in the following example are different:
```js
function myFunction() {
@@ -22,17 +22,17 @@ let varTwo = myFunction(); // set to equal the string "You rock!"
# --instructions--
-修复代码，把调用函数`getNine`的返回值赋给变量`result`。
+Fix the code so the variable `result` is set to the value returned from calling the function `getNine`.
# --hints--
-你应该修复变量`result`使其为函数`getNine`的返回值。
+Your code should fix the variable `result` so it is set to the number that the function `getNine` returns.
```js
assert(result == 9);
```
-你应该调用`getNine`函数。
+Your code should call the `getNine` function.
```js
assert(code.match(/getNine\(\)/g).length == 2);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-misspelled-variable-and-function-names.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-misspelled-variable-and-function-names.md
index 87c052c39d3..cdadad85349 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-misspelled-variable-and-function-names.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-misspelled-variable-and-function-names.md
@@ -1,6 +1,6 @@
---
id: 587d7b84367417b2b2512b35
-title: 捕获拼错的变量名和函数名
+title: Catch Misspelled Variable and Function Names
challengeType: 1
forumTopicId: 301186
dashedName: catch-misspelled-variable-and-function-names
@@ -8,41 +8,41 @@ dashedName: catch-misspelled-variable-and-function-names
# --description--
-`console.log()`和`typeof`方法是检查中间值和程序输出类型的两种主要方法。 现在是时候了解一下 bug 出现的常见的情形。一个语法级别的问题是打字太快带来的低级拼写错误。
+The `console.log()` and `typeof` methods are the two primary ways to check intermediate values and types of program output. Now it's time to get into the common forms that bugs take. One syntax-level issue that fast typers can commiserate with is the humble spelling error.
-变量或函数名的错写、漏写或大小写弄混都会让浏览器尝试查找并不存在的东西，并报出“引用错误”。JavaScript 变量和函数名称区分大小写。
+Transposed, missing, or mis-capitalized characters in a variable or function name will have the browser looking for an object that doesn't exist - and complain in the form of a reference error. JavaScript variable and function names are case-sensitive.
# --instructions--
-修复代码中的两个拼写错误，以便`netWorkingCapital`计算有效。
+Fix the two spelling errors in the code so the `netWorkingCapital` calculation works.
# --hints--
-检查计算 netWorkingCapital 值时使用的两个变量的拼写是否正确，控制台应该输出 "Net working capital is: 2"。
+Check the spelling of the two variables used in the netWorkingCapital calculation, the console output should show that "Net working capital is: 2".
```js
assert(netWorkingCapital === 2);
```
-代码中不应存在拼写错误的变量。
+There should be no instances of mis-spelled variables in the code.
```js
assert(!code.match(/recievables/g));
```
-应在代码中声明并正确使用`receivables`变量。
+The `receivables` variable should be declared and used properly in the code.
```js
assert(code.match(/receivables/g).length == 2);
```
-代码中不应存在拼写错误的变量。
+There should be no instances of mis-spelled variables in the code.
```js
assert(!code.match(/payable;/g));
```
-应在代码中声明并正确使用`payables`变量。
+The `payables` variable should be declared and used properly in the code.
```js
assert(code.match(/payables/g).length == 2);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-mixed-usage-of-single-and-double-quotes.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-mixed-usage-of-single-and-double-quotes.md
index 5b63e32d980..cde18db7989 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-mixed-usage-of-single-and-double-quotes.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-mixed-usage-of-single-and-double-quotes.md
@@ -1,6 +1,6 @@
---
id: 587d7b84367417b2b2512b37
-title: 捕捉单引号和双引号的混合用法
+title: Catch Mixed Usage of Single and Double Quotes
challengeType: 1
forumTopicId: 301188
dashedName: catch-mixed-usage-of-single-and-double-quotes
@@ -8,11 +8,11 @@ dashedName: catch-mixed-usage-of-single-and-double-quotes
# --description--
-JavaScript允许使用单引号('')和双引号("")声明字符串。决定使用哪一个通常看个人偏好，但有一些例外。
+JavaScript allows the use of both single (`'`) and double (`"`) quotes to declare a string. Deciding which one to use generally comes down to personal preference, with some exceptions.
-如果字符串中有缩写或存在一段带引号的文本，你就会明白为什么 JavaScript 允许两种引号了。请注意，不要提前用引号结束字符串，这会导致语法错误。
+Having two choices is great when a string has contractions or another piece of text that's in quotes. Just be careful that you don't close the string too early, which causes a syntax error.
-下面是混合使用引号的一些示例：
+Here are some examples of mixing quotes:
```js
// These are correct:
@@ -22,26 +22,26 @@ const quoteInString = "Groucho Marx once said 'Quote me as saying I was mis-quot
const uhOhGroucho = 'I've had a perfectly wonderful evening, but this wasn't it.';
```
-当然，只使用一种引号是可以的。你可以使用反斜杠 (`\`) 转义字符来转义字符串中的引号：
+Of course, it is okay to use only one style of quotes. You can escape the quotes inside the string by using the backslash (<code>\\</code>) escape character:
```js
-// 一种引号的正确使用方式
+// Correct use of same quotes:
const allSameQuotes = 'I\'ve had a perfectly wonderful evening, but this wasn\'t it.';
```
# --instructions--
-修复字符串，对`href`属性的值使用不同的引号，或者转义现有的引号。注意，整个字符串外面的双引号要保留。
+Fix the string so it either uses different quotes for the `href` value, or escape the existing ones. Keep the double quote marks around the entire string.
# --hints--
-你应通过更改或转义来修复`href`的值 '#Home' 周围的引号。
+Your code should fix the quotes around the `href` value "#Home" by either changing or escaping them.
```js
assert(code.match(/<a href=\s*?('|\\")#Home\1\s*?>/g));
```
-你应该在整个字符串外围保留双引号。
+Your code should keep the double quotes around the entire string.
```js
assert(code.match(/"<p>.*?<\/p>";/g));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-off-by-one-errors-when-using-indexing.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-off-by-one-errors-when-using-indexing.md
index afcf595eb73..d86487c484f 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-off-by-one-errors-when-using-indexing.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-off-by-one-errors-when-using-indexing.md
@@ -1,6 +1,6 @@
---
id: 587d7b86367417b2b2512b3b
-title: 捕获使用索引的时候出现的错误
+title: Catch Off By One Errors When Using Indexing
challengeType: 1
forumTopicId: 301189
dashedName: catch-off-by-one-errors-when-using-indexing
@@ -8,52 +8,52 @@ dashedName: catch-off-by-one-errors-when-using-indexing
# --description--
-当试图访问字符串或数组的特定索引（分割或访问一个片段）或循环索引时，有时会出现`Off by one errors`错误（有时称为 OBOE）。JavaScript 索引从`0`开始，而不是`1`，这意味着最后一个索引总会比字符串或数组的长度少 1。如果尝试访问等于长度的索引，程序可能会抛出“索引超出范围”引用错误或打印出`undefined`。
+<dfn>Off by one errors</dfn> (sometimes called OBOE) crop up when you're trying to target a specific index of a string or array (to slice or access a segment), or when looping over the indices of them. JavaScript indexing starts at zero, not one, which means the last index is always one less than the length of the item. If you try to access an index equal to the length, the program may throw an "index out of range" reference error or print `undefined`.
-当使用将索引范围作为参数的字符串或数组方法时，阅读相关的文档并了解参数中的索引的包含性（即是否考虑进返回值中）很重要。以下是一些错误的示例：
+When you use string or array methods that take index ranges as arguments, it helps to read the documentation and understand if they are inclusive (the item at the given index is part of what's returned) or not. Here are some examples of off by one errors:
```js
let alphabet = "abcdefghijklmnopqrstuvwxyz";
let len = alphabet.length;
for (let i = 0; i <= len; i++) {
- // 在最后多了一次循环
+ // loops one too many times at the end
console.log(alphabet[i]);
}
for (let j = 1; j < len; j++) {
- // 循环少了一次，漏掉了索引 0 处的字符
+ // loops one too few times and misses the first character at index 0
console.log(alphabet[j]);
}
for (let k = 0; k < len; k++) {
- // 不多不少，这才是正确的
+ // Goldilocks approves - this is just right
console.log(alphabet[k]);
}
```
# --instructions--
-修复以下函数中的两个索引错误，以便将 1 到 5 之间（包含 1 和 5）的所有数字打印到控制台。
+Fix the two indexing errors in the following function so all the numbers 1 through 5 are printed to the console.
# --hints--
-你应该设置循环的初始条件，使循环从第一个索引开始。
+Your code should set the initial condition of the loop so it starts at the first index.
```js
assert(code.match(/i\s*?=\s*?0\s*?;/g).length == 1);
```
-你应修复循环的初始条件，使循环从索引 0 开始。
+Your code should fix the initial condition of the loop so that the index starts at 0.
```js
assert(!code.match(/i\s?=\s*?1\s*?;/g));
```
-你应设置循环的终止条件，使循环在最后一个索引处停止。
+Your code should set the terminal condition of the loop so it stops at the last index.
```js
assert(code.match(/i\s*?<\s*?len\s*?;/g).length == 1);
```
-你应修复循环的终止条件，使循环在索引为字符串长度减 1 时停止。
+Your code should fix the terminal condition of the loop so that it stops at 1 before the length.
```js
assert(!code.match(/i\s*?<=\s*?len;/g));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-unclosed-parentheses-brackets-braces-and-quotes.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-unclosed-parentheses-brackets-braces-and-quotes.md
index 589425816ce..a55d812d8ac 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-unclosed-parentheses-brackets-braces-and-quotes.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-unclosed-parentheses-brackets-braces-and-quotes.md
@@ -1,6 +1,6 @@
---
id: 587d7b84367417b2b2512b36
-title: 捕获未闭合的括号、方括号、大括号和引号
+title: 'Catch Unclosed Parentheses, Brackets, Braces and Quotes'
challengeType: 1
forumTopicId: 301190
dashedName: catch-unclosed-parentheses-brackets-braces-and-quotes
@@ -8,23 +8,23 @@ dashedName: catch-unclosed-parentheses-brackets-braces-and-quotes
# --description--
-要注意的另一个语法错误是所有的小括号、方括号、花括号和引号都必须配对。当你编辑代码并插入新代码其中带有括号时，很容易忘记括号闭合。 此外，在将代码块嵌套到其他代码块时要小心，例如将回调函数作为参数添加到方法中。
+Another syntax error to be aware of is that all opening parentheses, brackets, curly braces, and quotes have a closing pair. Forgetting a piece tends to happen when you're editing existing code and inserting items with one of the pair types. Also, take care when nesting code blocks into others, such as adding a callback function as an argument to a method.
-避免这种错误的一种方法是，一次性输入完这些符号，然后将光标移回它们之间继续编写。好在，现在大部分编辑器都会帮你自动补全。
+One way to avoid this mistake is as soon as the opening character is typed, immediately include the closing match, then move the cursor back between them and continue coding. Fortunately, most modern code editors generate the second half of the pair automatically.
# --instructions--
-修复代码中的两个符号配对错误。
+Fix the two pair errors in the code.
# --hints--
-你应该修复数组缺少的部分。
+Your code should fix the missing piece of the array.
```js
assert(code.match(/myArray\s*?=\s*?\[\s*?1\s*?,\s*?2\s*?,\s*?3\s*?\];/g));
```
-你应该修复`.reduce()`方法缺少的部分。控制台应输出 "Sum of array values is: 6"。
+Your code should fix the missing piece of the `.reduce()` method. The console output should show that "Sum of array values is: 6".
```js
assert(arraySum === 6);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-use-of-assignment-operator-instead-of-equality-operator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-use-of-assignment-operator-instead-of-equality-operator.md
index 07e67f72d76..e164f373993 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-use-of-assignment-operator-instead-of-equality-operator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/catch-use-of-assignment-operator-instead-of-equality-operator.md
@@ -1,6 +1,6 @@
---
id: 587d7b85367417b2b2512b38
-title: 捕获使用赋值运算符而不是相等运算符
+title: Catch Use of Assignment Operator Instead of Equality Operator
challengeType: 1
forumTopicId: 301191
dashedName: catch-use-of-assignment-operator-instead-of-equality-operator
@@ -8,37 +8,37 @@ dashedName: catch-use-of-assignment-operator-instead-of-equality-operator
# --description--
-分支程序，即在满足某些条件时执行不同操作的程序，依赖于 JavaScript 中的`if`，`else if`、`else`语句。条件有时采取测试一个结果是否等于一个值的形式。
+Branching programs, i.e. ones that do different things if certain conditions are met, rely on `if`, `else if`, and `else` statements in JavaScript. The condition sometimes takes the form of testing whether a result is equal to a value.
-这种逻辑可以表述为“如果 x 等于 y ，则......”，听起来像是可以使用`=`（即赋值运算符）。然而，这会导致程序中流程出问题。
+This logic is spoken (in English, at least) as "if x equals y, then ..." which can literally translate into code using the `=`, or assignment operator. This leads to unexpected control flow in your program.
-如前面的挑战所述，JavaScript 中的赋值运算符 (`=`) 是用来为变量名赋值的。并且`==`和`===`运算符检查相等性（三等号`===`是用来测试是否严格相等的，严格相等的意思是值和类型都必须相同）。
+As covered in previous challenges, the assignment operator (`=`) in JavaScript assigns a value to a variable name. And the `==` and `===` operators check for equality (the triple `===` tests for strict equality, meaning both value and type are the same).
-下面的代码将`x`赋值为 2，表达式`x = y`会在执行后得到`true`。JavaScript 会把大部分的值都视为`true`，除了所谓的 "falsy" 值，即：`false`、`0`、`""`（空字符串）、`NaN`、`undefined` 和 `null`。
+The code below assigns `x` to be 2, which evaluates as `true`. Almost every value on its own in JavaScript evaluates to `true`, except what are known as the "falsy" values: `false`, `0`, `""` (an empty string), `NaN`, `undefined`, and `null`.
```js
let x = 1;
let y = 2;
if (x = y) {
- // 除了 "falsy" 值以外 y 为任意值时这个代码块都将执行
+ // this code block will run for any value of y (unless y were originally set as a falsy)
} else {
- // 按本例用意这个代码块应该执行（但其实不会）。
+ // this code block is what should run (but won't) in this example
}
```
# --instructions--
-修复条件语句，以便程序运行正确的分支，并给`result`赋上正确的值。
+Fix the condition so the program runs the right branch, and the appropriate value is assigned to `result`.
# --hints--
-你应该修复条件语句，使其判断是否相等,而不是赋值。
+Your code should fix the condition so it checks for equality, instead of using assignment.
```js
assert(result == 'Not equal!');
```
-条件语句可以使用`==`或`===`来测试是否相等。
+The condition should use either `==` or `===` to test for equality.
```js
assert(code.match(/x\s*?===?\s*?y/g));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/prevent-infinite-loops-with-a-valid-terminal-condition.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/prevent-infinite-loops-with-a-valid-terminal-condition.md
index d4b3142a8bd..9455e136450 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/prevent-infinite-loops-with-a-valid-terminal-condition.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/prevent-infinite-loops-with-a-valid-terminal-condition.md
@@ -1,6 +1,6 @@
---
id: 587d7b86367417b2b2512b3d
-title: 使用有效的终止条件防止无限循环
+title: Prevent Infinite Loops with a Valid Terminal Condition
challengeType: 1
forumTopicId: 301192
dashedName: prevent-infinite-loops-with-a-valid-terminal-condition
@@ -8,9 +8,9 @@ dashedName: prevent-infinite-loops-with-a-valid-terminal-condition
# --description--
-最后一个话题是可怕的无限循环。当需要程序运行代码块一定次数或满足条件时，循环是很好的工具，但是它们需要终止条件来结束循环。无限循环可能会使浏览器冻结或崩溃，并导致程序执行混乱，没有人想要这样的结果。
+The final topic is the dreaded infinite loop. Loops are great tools when you need your program to run a code block a certain number of times or until a condition is met, but they need a terminal condition that ends the looping. Infinite loops are likely to freeze or crash the browser, and cause general program execution mayhem, which no one wants.
-在本节的介绍中有一个无限循环的例子——它没有终止条件来摆脱`loopy()`内的`while`循环。不要调用这个函数！
+There was an example of an infinite loop in the introduction to this section - it had no terminal condition to break out of the `while` loop inside `loopy()`. Do NOT call this function!
```js
function loopy() {
@@ -20,21 +20,21 @@ function loopy() {
}
```
-程序员的工作是确保最终达到终止条件，该条件告诉程序何时跳出循环。有一种错误是从终端条件向错误方向递增或递减计数器变量。另一种是在循环代码中意外重置计数器或索引变量，而不是递增或递减它。
+It's the programmer's job to ensure that the terminal condition, which tells the program when to break out of the loop code, is eventually reached. One error is incrementing or decrementing a counter variable in the wrong direction from the terminal condition. Another one is accidentally resetting a counter or index variable within the loop code, instead of incrementing or decrementing it.
# --instructions--
-`myFunc()`函数包含一个无限循环，因为终止条件`i != 4`永远不会为`false`(并中断循环) -`i`将每次递增 2，然后跳过 4，因为`i`是从奇数开始递增。在终端条件中输入比较运算符，使循环仅在`i`小于或等于 4 的情况下运行。
+The `myFunc()` function contains an infinite loop because the terminal condition `i != 4` will never evaluate to `false` (and break the looping) - `i` will increment by 2 each pass, and jump right over 4 since `i` is odd to start. Fix the comparison operator in the terminal condition so the loop only runs for `i` less than or equal to 4.
# --hints--
-你应该在`for`循环的终止条件（中间部分）中更改比较运算符。
+Your code should change the comparison operator in the terminal condition (the middle part) of the `for` loop.
```js
assert(code.match(/i\s*?<=\s*?4;/g).length == 1);
```
-你应该修改比较运算符来避免出现死循环。
+Your code should fix the comparison operator in the terminal condition of the loop.
```js
assert(!code.match(/i\s*?!=\s*?4;/g));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/understanding-the-differences-between-the-freecodecamp-and-browser-console.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/understanding-the-differences-between-the-freecodecamp-and-browser-console.md
index 7d5f65b8d50..10f47a1581a 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/understanding-the-differences-between-the-freecodecamp-and-browser-console.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/understanding-the-differences-between-the-freecodecamp-and-browser-console.md
@@ -1,6 +1,6 @@
---
id: 587d7b83367417b2b2512b37
-title: 了解 freeCodeCamp 和浏览器控制台之间的差异
+title: Understanding the Differences between the freeCodeCamp and Browser Console
challengeType: 1
forumTopicId: 301193
dashedName: understanding-the-differences-between-the-freecodecamp-and-browser-console
@@ -8,37 +8,40 @@ dashedName: understanding-the-differences-between-the-freecodecamp-and-browser-c
# --description--
-你可能已经注意到一些 freeCodeCamp JavaScript 的挑战有自己的控制台。这些控制台的行为与上一次挑战中使用的浏览器控制台略有不同。
+You may have noticed that some freeCodeCamp JavaScript challenges include their own console. This console behaves a little differently than the browser console you used in the last challenge.
-以下挑战旨在强调 freeCodeCamp 控制台与浏览器控制台之间的一些差异。
+The following challenge is meant to highlight the main difference between the freeCodeCamp console and your browser console.
-对于浏览器控制台。当在浏览器中加载并运行 JavaScript 文件时，`console.log()`语句会在控制台中按照调用的次数准确地打印出要求的内容。然而，在 freeCodeCamp 的代码编辑器中使用`console.log()`会略有不同，一开始可能会让你感到困惑。
+When you run ordinary JavaScript, the browser's console will display your `console.log()` statements the exact number of times it is called.
-在 freeCodeCamp 代码编辑器中，传给`console.log()`的值会在每组测试执行的时候输出。另外，如果在代码中还手动调用过挑战题目的函数，调用几次就会增加几次传入值的输出。
+The freeCodeCamp console will print your `console.log()` statements a short time after the editor detects a change in the script, as well as during testing.
-这就产生了一些有趣的行为，并可能在一开始就让你感到困惑，因为你觉得只会输出一次的值可能会输出多次，具体次数取决于挑战题目本身测试的数量以及这些测试调用挑战函数的方式。
+The freeCodeCamp console is cleared before the tests are run and, to avoid spam, only prints the logs during the first test (see the note below for exceptions).
-如果你不打算执行挑战的测试，而只想查看自己调用`console.log()`的输出，可以使用`console.clear()`。
+If you would like to see every log for every test, run the tests, and open the browser console. If you prefer to use the browser console, and want it to mimic the freeCodeCamp console, place `console.clear()` before any other `console` calls, to clear the browser console.
+
+**Note:** `console.log`s inside functions are printed to the freeCodeCamp console whenever those functions are called, this can help debugging functions that are called during testing.
# --instructions--
-首先，使用 `console.clear()` 清空浏览器控制台。然后使用`console.log()`在代码中指定的位置打印 `output` 变量。
+First, use `console.log` to log the `output` variable. Then, use `console.clear` to clear the browser console.
# --hints--
-应该使用 `console.clear()` 来清空浏览器控制台。
+You should use `console.clear()` to clear the browser console.
```js
-const removeJSComments = code.replace(/\/\*[\s\S]*?\*\/|\/\/.*$/gm, '');
-const noSpaces = removeJSComments.replace(/\s/g, '');
-assert(noSpaces.match(/console.clear\(\)/));
+assert(
+ __helpers
+ .removeWhiteSpace(__helpers.removeJSComments(code))
+ .match(/console.clear\(\)/)
+);
```
-使用`console.log()`输出变量`output`的值。
+You should use `console.log()` to print the `output` variable.
```js
-const noSpaces = code.replace(/\s/g, '');
-assert(noSpaces.match(/console\.log\(output\)/));
+assert(__helpers.removeWhiteSpace(code).match(/console\.log\(output\)/));
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/use-caution-when-reinitializing-variables-inside-a-loop.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/use-caution-when-reinitializing-variables-inside-a-loop.md
index f19602dbfcc..19dd59b6056 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/use-caution-when-reinitializing-variables-inside-a-loop.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/use-caution-when-reinitializing-variables-inside-a-loop.md
@@ -1,6 +1,6 @@
---
id: 587d7b86367417b2b2512b3c
-title: 重新初始化循环中的变量时要小心
+title: Use Caution When Reinitializing Variables Inside a Loop
challengeType: 1
forumTopicId: 301194
dashedName: use-caution-when-reinitializing-variables-inside-a-loop
@@ -8,29 +8,29 @@ dashedName: use-caution-when-reinitializing-variables-inside-a-loop
# --description--
-有时需要在循环中保存信息以增加计数器或重置变量。一个潜在的问题是变量什么时候该重新初始化，什么时候不该重新初始化，反之亦然。如果你不小心重置了用于终止条件的变量，导致无限循环，这将特别危险。
+Sometimes it's necessary to save information, increment counters, or re-set variables within a loop. A potential issue is when variables either should be reinitialized, and aren't, or vice versa. This is particularly dangerous if you accidentally reset the variable being used for the terminal condition, causing an infinite loop.
-使用`console.log()`在每个循环中打印变量值可以发现与重置相关的错误或者重置变量失败。
+Printing variable values with each cycle of your loop by using `console.log()` can uncover buggy behavior related to resetting, or failing to reset a variable.
# --instructions--
-以下函数应该创建一个具有`m`行和`n`列“零”的二维数组。不幸的是，它没有产生预期的输出，因为`row`变量没有在外部循环中重新初始化（设置回空数组）。修改代码，使其正确地返回包含 3 行 2 列“零”的二维数组，即`[[0, 0], [0, 0], [0, 0]]`。
+The following function is supposed to create a two-dimensional array with `m` rows and `n` columns of zeroes. Unfortunately, it's not producing the expected output because the `row` variable isn't being reinitialized (set back to an empty array) in the outer loop. Fix the code so it returns a correct 3x2 array of zeroes, which looks like `[[0, 0], [0, 0], [0, 0]]`.
# --hints--
-你应将变量`matrix`设置为 3 行 2 列“零”的二维数组。
+Your code should set the `matrix` variable to an array holding 3 rows of 2 columns of zeroes each.
```js
assert(JSON.stringify(matrix) == '[[0,0],[0,0],[0,0]]');
```
-变量`matrix`应有 3 行。
+The `matrix` variable should have 3 rows.
```js
assert(matrix.length == 3);
```
-变量`matrix`每行应有 2 列。
+The `matrix` variable should have 2 columns in each row.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/use-the-javascript-console-to-check-the-value-of-a-variable.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/use-the-javascript-console-to-check-the-value-of-a-variable.md
index e6119f76929..eba22846147 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/use-the-javascript-console-to-check-the-value-of-a-variable.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/use-the-javascript-console-to-check-the-value-of-a-variable.md
@@ -1,6 +1,6 @@
---
id: 587d7b83367417b2b2512b33
-title: 使用控制台检查变量值
+title: Use the JavaScript Console to Check the Value of a Variable
challengeType: 1
forumTopicId: 18372
dashedName: use-the-javascript-console-to-check-the-value-of-a-variable
@@ -8,23 +8,23 @@ dashedName: use-the-javascript-console-to-check-the-value-of-a-variable
# --description--
-Chrome 和 Firefox 都有出色的 JavaScript 控制台（也称为 DevTools），可以用来调试 JavaScript 代码
+Both Chrome and Firefox have excellent JavaScript consoles, also known as DevTools, for debugging your JavaScript.
-可以在 Chrome 的菜单中找到“开发者工具”或 FireFox 的菜单中的 “Web 控制台”。如果你使用其他浏览器或手机，我们强烈建议你切换到桌面版 Firefox 或 Chrome。
+You can find Developer tools in your Chrome's menu or Web Console in Firefox's menu. If you're using a different browser, or a mobile phone, we strongly recommend switching to desktop Firefox or Chrome.
-`console.log()`方法可能是最有用的调试工具，它可以将括号中的内容输出到控制台，将它放在代码中的关键点可以显示变量在当时的值。在查看输出之前，最好先想清楚输出应该是什么。在代码的合适位置打点来查看变量状态有助于定位问题。
+The `console.log()` method, which "prints" the output of what's within its parentheses to the console, will likely be the most helpful debugging tool. Placing it at strategic points in your code can show you the intermediate values of variables. It's good practice to have an idea of what the output should be before looking at what it is. Having check points to see the status of your calculations throughout your code will help narrow down where the problem is.
-下面是输出 'Hello world!' 到控制台的示例：
+Here's an example to print 'Hello world!' to the console:
`console.log('Hello world!');`
# --instructions--
-请使用`console.log()`方法在代码中注明的地方输出变量`a`的值。
+Use the `console.log()` method to print the value of the variable `a` where noted in the code.
# --hints--
-你应使用`console.log()`来检查变量`a`的值。
+Your code should use `console.log()` to check the value of the variable `a`.
```js
assert(code.match(/console\.log\(a\)/g));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/use-typeof-to-check-the-type-of-a-variable.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/use-typeof-to-check-the-type-of-a-variable.md
index dbb96dfdeb9..e9d54bfb668 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/use-typeof-to-check-the-type-of-a-variable.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/debugging/use-typeof-to-check-the-type-of-a-variable.md
@@ -1,6 +1,6 @@
---
id: 587d7b84367417b2b2512b34
-title: 使用 type of 检查变量的类型
+title: Use typeof to Check the Type of a Variable
challengeType: 1
forumTopicId: 18374
dashedName: use-typeof-to-check-the-type-of-a-variable
@@ -8,38 +8,38 @@ dashedName: use-typeof-to-check-the-type-of-a-variable
# --description--
-可以使用`typeof`检查变量的数据结构或类型。在处理多种数据类型时，`typeof`会对调试很有帮助。如果想计算两数之和，但实际传入了一个字符串参数，则结果可能是错误的。类型错误可能隐藏在计算或函数调用中。当你以 JavaScript 对象（JSON）的形式访问和使用外部数据时尤其要小心。
+You can use `typeof` to check the data structure, or type, of a variable. This is useful in debugging when working with multiple data types. If you think you're adding two numbers, but one is actually a string, the results can be unexpected. Type errors can lurk in calculations or function calls. Be careful especially when you're accessing and working with external data in the form of a JavaScript Object Notation (JSON) object.
-下面是使用`typeof`的一些示例：
+Here are some examples using `typeof`:
```js
-console.log(typeof ""); // 输出 "string"
-console.log(typeof 0); // 输出 "number"
-console.log(typeof []); // 输出 "object"
-console.log(typeof {}); // 输出 "object"
+console.log(typeof ""); // outputs "string"
+console.log(typeof 0); // outputs "number"
+console.log(typeof []); // outputs "object"
+console.log(typeof {}); // outputs "object"
```
-JavaScript 有六种原始（不可变）数据类型：`Boolean`,`Null`,`Undefined`,`Number`,`String`, 和`Symbol`（ES6 新增）和一种可变的数据类型：`Object`。注意，在 JavaScript 中，数组在本质上是一种对象
+JavaScript recognizes six primitive (immutable) data types: `Boolean`, `Null`, `Undefined`, `Number`, `String`, and `Symbol` (new with ES6) and one type for mutable items: `Object`. Note that in JavaScript, arrays are technically a type of object.
# --instructions--
-添加两个`console.log()`语句来检查代码中的两个变量`seven`和`three`的 `typeof`值。
+Add two `console.log()` statements to check the `typeof` each of the two variables `seven` and `three` in the code.
# --hints--
-你应在两个`console.log()`语句中使用`typeof`来检查变量的类型。
+Your code should use `typeof` in two `console.log()` statements to check the type of the variables.
```js
assert(code.match(/console\.log\(typeof[\( ].*\)?\)/g).length == 2);
```
-你应使用`typeof`来检查变量`seven`的类型。
+Your code should use `typeof` to check the type of the variable `seven`.
```js
assert(code.match(/typeof[\( ]seven\)?/g));
```
-你应使用`typeof`来检查变量`three`的类型。
+Your code should use `typeof` to check the type of the variable `three`.
```js
assert(code.match(/typeof[\( ]three\)?/g));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/compare-scopes-of-the-var-and-let-keywords.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/compare-scopes-of-the-var-and-let-keywords.md
index 5478e0507d1..ba319ea8bce 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/compare-scopes-of-the-var-and-let-keywords.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/compare-scopes-of-the-var-and-let-keywords.md
@@ -1,6 +1,6 @@
---
id: 587d7b87367417b2b2512b40
-title: 比较 var 和 let 关键字的作用域
+title: Compare Scopes of the var and let Keywords
challengeType: 1
forumTopicId: 301195
dashedName: compare-scopes-of-the-var-and-let-keywords
@@ -8,11 +8,11 @@ dashedName: compare-scopes-of-the-var-and-let-keywords
# --description--
-当你使用`var`关键字来声明一个变量的时候，这个变量会被声明成全局变量，或是函数内的局部变量。
+When you declare a variable with the `var` keyword, it is declared globally, or locally if declared inside a function.
-`let`关键字的作用类似，但会有一些额外的特性。如果你在代码块、语句或表达式中使用关键字`let`声明变量，这个变量的作用域就被限制在当前的代码块，语句或表达式之中。
+The `let` keyword behaves similarly, but with some extra features. When you declare a variable with the `let` keyword inside a block, statement, or expression, its scope is limited to that block, statement, or expression.
-举个例子：
+For example:
```js
var numArray = [];
@@ -20,12 +20,12 @@ for (var i = 0; i < 3; i++) {
numArray.push(i);
}
console.log(numArray);
-// 返回 [0, 1, 2]
+// returns [0, 1, 2]
console.log(i);
-// 返回 3
+// returns 3
```
-当使用`var`关键字的时候，`i`会被声明成全局变量。当`i++`执行的时候，它会改变全局变量的值。这段代码可以看做下面这样:
+With the `var` keyword, `i` is declared globally. So when `i++` is executed, it updates the global variable. This code is similar to the following:
```js
var numArray = [];
@@ -34,12 +34,12 @@ for (i = 0; i < 3; i++) {
numArray.push(i);
}
console.log(numArray);
-// 返回 [0, 1, 2]
+// returns [0, 1, 2]
console.log(i);
-// 返回 3
+// returns 3
```
-如果你在`for`循环中创建了使用`i`变量的函数，那么在后续调用函数的时候，上面提到的这种行为就会出现问题。这是因为函数存储的值会因为全局变量`i`的变化而不断的改变。
+This behavior will cause problems if you were to create a function and store it for later use inside a for loop that uses the `i` variable. This is because the stored function will always refer to the value of the updated global `i` variable.
```js
var printNumTwo;
@@ -51,13 +51,12 @@ for (var i = 0; i < 3; i++) {
}
}
console.log(printNumTwo());
-// 返回 3
+// returns 3
```
-可以看到，`printNumTwo()`打印了 3 而不是 2。这是因为`i`发生了改变，并且函数`printNumTwo()`返回的是全局变量`i`的值，而不是`for`循环中创建函数时`i`的值。`let`关键字就不会有这种现象：
+As you can see, `printNumTwo()` prints 3 and not 2. This is because the value assigned to `i` was updated and the `printNumTwo()` returns the global `i` and not the value `i` had when the function was created in the for loop. The `let` keyword does not follow this behavior:
```js
-'use strict';
let printNumTwo;
for (let i = 0; i < 3; i++) {
if (i === 2) {
@@ -67,28 +66,28 @@ for (let i = 0; i < 3; i++) {
}
}
console.log(printNumTwo());
-// 返回 2
+// returns 2
console.log(i);
-// 返回 "i 未定义"
+// returns "i is not defined"
```
-`i`在全局作用域中没有声明，所以它没有被定义，它的声明只会发生在`for`循环内。在循环执行的时候，`let`关键字创建了三个不同的`i`变量，他们的值分别为 0、1 和 2，所以`printNumTwo()`返回了正确的值。
+`i` is not defined because it was not declared in the global scope. It is only declared within the for loop statement. `printNumTwo()` returned the correct value because three different `i` variables with unique values (0, 1, and 2) were created by the `let` keyword within the loop statement.
# --instructions--
-修改这段代码，使得在`if`语句中声明的`i`变量与在函数的第一行声明的`i`变量是彼此独立的。请注意不要在你的代码的任何地方使用`var`关键字。
+Fix the code so that `i` declared in the if statement is a separate variable than `i` declared in the first line of the function. Be certain not to use the `var` keyword anywhere in your code.
-这个练习说明了使用`var`与`let`关键字声明变量时，作用域之间的不同。当编写类似这个练习中的函数的时候，通常来说最好还是使用不同的变量名来避免误会。
+This exercise is designed to illustrate the difference between how `var` and `let` keywords assign scope to the declared variable. When programming a function similar to the one used in this exercise, it is often better to use different variable names to avoid confusion.
# --hints--
-`var`不应该在代码中存在。
+`var` should not exist in code.
```js
(getUserInput) => assert(!getUserInput('index').match(/var/g));
```
-在`if`语句中声明的`i`变量的值是 'block scope'。
+The variable `i` declared in the if statement should equal "block scope".
```js
(getUserInput) =>
@@ -97,7 +96,7 @@ console.log(i);
);
```
-`checkScope()`应当返回 'function scope'
+`checkScope()` should return "function scope"
```js
assert(checkScope() === 'function scope');
@@ -128,7 +127,7 @@ function checkScope() {
let i = 'block scope';
console.log('Block scope i is: ', i);
}
-
+
console.log('Function scope i is: ', i);
return i;
}
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/complete-a-promise-with-resolve-and-reject.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/complete-a-promise-with-resolve-and-reject.md
index ad4814573c8..171b4f549cd 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/complete-a-promise-with-resolve-and-reject.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/complete-a-promise-with-resolve-and-reject.md
@@ -1,6 +1,6 @@
---
id: 5cdafbc32913098997531680
-title: 通过 resolve 和 reject 完成 Promise
+title: Complete a Promise with resolve and reject
challengeType: 1
forumTopicId: 301196
dashedName: complete-a-promise-with-resolve-and-reject
@@ -8,7 +8,7 @@ dashedName: complete-a-promise-with-resolve-and-reject
# --description--
-promise 有三个状态：`pending`、`fulfilled` 和 `rejected`。上一个挑战里创建的 promise 一直阻塞在 `pending` 状态里，因为没有调用 promise 的完成方法。promise 提供的 `resolve` 和 `reject` 参数就是用来结束 promise 的。promise 成功时调用 `resolve`，promise 执行失败时调用 `reject`，这两个方法接收一个参数，如下所示。
+A promise has three states: `pending`, `fulfilled`, and `rejected`. The promise you created in the last challenge is forever stuck in the `pending` state because you did not add a way to complete the promise. The `resolve` and `reject` parameters given to the promise argument are used to do this. `resolve` is used when you want your promise to succeed, and `reject` is used when you want it to fail. These are methods that take an argument, as seen below.
```js
const myPromise = new Promise((resolve, reject) => {
@@ -20,31 +20,35 @@ const myPromise = new Promise((resolve, reject) => {
});
```
-上面的例子使用字符串做为函数的参数，也可以是任意类型，通常会是对象，里面可能是将要放到页面或其它地方的数据。
+The example above uses strings for the argument of these functions, but it can really be anything. Often, it might be an object, that you would use data from, to put on your website or elsewhere.
# --instructions--
-使 promise 可以处理成功和失败情况。如果 `responseFromServer` 是 `true`，调用 `resolve` 方法使 promise 成功。给 `resolve` 传递值为 `We got the data` 的字符串。如果 `responseFromServer` 是 `false`， 使用 `reject` 方法并传入值为 `Data not received` 的字符串。
+Make the promise handle success and failure. If `responseFromServer` is `true`, call the `resolve` method to successfully complete the promise. Pass `resolve` a string with the value `We got the data`. If `responseFromServer` is `false`, use the `reject` method instead and pass it the string: `Data not received`.
# --hints--
-当 `if` 条件是 `true` 时应该执行 `resolve`。
+`resolve` should be called with the expected string when the `if` condition is `true`.
```js
assert(
- removeJSComments(code).match(
- /if\s*\(\s*responseFromServer\s*\)\s*{\s*resolve\s*\(\s*('|"|`)We got the data\1\s*\)(\s*|\s*;\s*)}/g
- )
+ __helpers
+ .removeJSComments(code)
+ .match(
+ /if\s*\(\s*responseFromServer\s*\)\s*{\s*resolve\s*\(\s*('|"|`)We got the data\1\s*\)(\s*|\s*;\s*)}/g
+ )
);
```
-当 `if` 条件是 `false` 时应该执行 `reject`。
+`reject` should be called with the expected string when the `if` condition is `false`.
```js
assert(
- removeJSComments(code).match(
- /}\s*else\s*{\s*reject\s*\(\s*('|"|`)Data not received\1\s*\)(\s*|\s*;\s*)}/g
- )
+ __helpers
+ .removeJSComments(code)
+ .match(
+ /}\s*else\s*{\s*reject\s*\(\s*('|"|`)Data not received\1\s*\)(\s*|\s*;\s*)}/g
+ )
);
```
@@ -56,7 +60,7 @@ assert(
const makeServerRequest = new Promise((resolve, reject) => {
// responseFromServer represents a response from a server
let responseFromServer;
-
+
if(responseFromServer) {
// Change this line
} else {
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/create-a-javascript-promise.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/create-a-javascript-promise.md
index 9ac6bb9591a..5dbc786915d 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/create-a-javascript-promise.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/create-a-javascript-promise.md
@@ -1,6 +1,6 @@
---
id: 5cdafbb0291309899753167f
-title: 创建一个 JavaScript Promise
+title: Create a JavaScript Promise
challengeType: 1
forumTopicId: 301197
dashedName: create-a-javascript-promise
@@ -8,7 +8,7 @@ dashedName: create-a-javascript-promise
# --description--
-Promise 是异步编程的一种解决方案 - 它在未来的某时会生成一个值。任务完成，分执行成功和执行失败两种情况。`Promise` 是构造器函数，需要通过 `new` 关键字来创建。构造器参数是一个函数，该函数有两个参数 - `resolve` 和 `reject`。通过它们来判断 promise 的执行结果。用法如下：
+A promise in JavaScript is exactly what it sounds like - you use it to make a promise to do something, usually asynchronously. When the task completes, you either fulfill your promise or fail to do so. `Promise` is a constructor function, so you need to use the `new` keyword to create one. It takes a function, as its argument, with two parameters - `resolve` and `reject`. These are methods used to determine the outcome of the promise. The syntax looks like this:
```js
const myPromise = new Promise((resolve, reject) => {
@@ -18,17 +18,17 @@ const myPromise = new Promise((resolve, reject) => {
# --instructions--
-创建一个名为 `makeServerRequest` 的 promise。给构造器函数传入 `resolve` 和 `reject` 两个参数。
+Create a new promise called `makeServerRequest`. Pass in a function with `resolve` and `reject` parameters to the constructor.
# --hints--
-应该给名为 `makeServerRequest` 的变量指定一个 promise。
+You should assign a promise to a declared variable named `makeServerRequest`.
```js
assert(makeServerRequest instanceof Promise);
```
-promise 应该接收一个函数做为参数，该函数应该包含 `resolve` 和 `reject` 两个参数。
+Your promise should receive a function with `resolve` and `reject` as parameters.
```js
assert(
@@ -41,7 +41,6 @@ assert(
# --seed--
## --seed-contents--
-
```js
```
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/create-a-module-script.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/create-a-module-script.md
index 5a490bc55c8..507602f57dc 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/create-a-module-script.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/create-a-module-script.md
@@ -1,6 +1,6 @@
---
id: 5cddbfd622f1a59093ec611d
-title: 创建一个模块脚本
+title: Create a Module Script
challengeType: 6
forumTopicId: 301198
dashedName: create-a-module-script
@@ -8,27 +8,27 @@ dashedName: create-a-module-script
# --description--
-起初，JavaScript 几乎只在 HTML web 扮演一个很小的角色。今天，一切不同了，很多网站几乎全是用 JavaScript 所写。为了让 JavaScript 更模块化、更整洁以及更易于维护，ES6 引入了在多个 JavaScript 文件之间共享代码的机制。它可以导出文件的一部分供其它文件使用，然后在需要它的地方按需导入。为了使用这一功能， 需要在 HTML 文档里创建一个类型为 `module` 的脚本。例子如下：
+JavaScript started with a small role to play on an otherwise mostly HTML web. Today, it’s huge, and some websites are built almost entirely with JavaScript. In order to make JavaScript more modular, clean, and maintainable; ES6 introduced a way to easily share code among JavaScript files. This involves exporting parts of a file for use in one or more other files, and importing the parts you need, where you need them. In order to take advantage of this functionality, you need to create a script in your HTML document with a type of `module`. Here’s an example:
```html
<script type="module" src="filename.js"></script>
```
-使用了 `module` 类型的脚本后可以再接下来的挑战里使用 `import` 和 `export` 特性。
+A script that uses this `module` type can now use the `import` and `export` features you will learn about in the upcoming challenges.
# --instructions--
-给 HTML 文档添加 `module` 类型的脚本，指定源文件为 `index.js`。
+Add a script to the HTML document of type `module` and give it the source file of `index.js`
# --hints--
-应该创建一个 `script` 标签。
+You should create a `script` tag.
```js
assert(code.match(/<\s*script[^>]*>\s*<\/\s*script\s*>/g));
```
-`script` 标签的类型应该为 `module`。
+Your `script` tag should be of type `module`.
```js
assert(
@@ -38,7 +38,7 @@ assert(
);
```
-`script` 标签的 `src` 属性应该为 `index.js`。
+Your `script` tag should have a `src` of `index.js`.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/create-an-export-fallback-with-export-default.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/create-an-export-fallback-with-export-default.md
index fbb448a0757..ac32fc08485 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/create-an-export-fallback-with-export-default.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/create-an-export-fallback-with-export-default.md
@@ -1,6 +1,6 @@
---
id: 587d7b8c367417b2b2512b58
-title: 用 export default 创建一个默认导出
+title: Create an Export Fallback with export default
challengeType: 1
forumTopicId: 301199
dashedName: create-an-export-fallback-with-export-default
@@ -8,11 +8,11 @@ dashedName: create-an-export-fallback-with-export-default
# --description--
-在`export`的课程中，学习了<dfn>命名导出</dfn>语法。这可以在其他文件中引用一些函数或者变量。
+In the `export` lesson, you learned about the syntax referred to as a <dfn>named export</dfn>. This allowed you to make multiple functions and variables available for use in other files.
-还需要了解另外一种被称为<dfn>默认导出</dfn>的`export`的语法。在文件中只有一个值需要导出的时候，通常会使用这种语法。它也常常用于给文件或者模块创建返回值。
+There is another `export` syntax you need to know, known as <dfn>export default</dfn>. Usually you will use this syntax if only one value is being exported from a file. It is also used to create a fallback value for a file or module.
-下面是一个简单的`export default`例子：
+Below are examples using `export default`:
```js
// named function
@@ -26,15 +26,15 @@ export default function(x, y) {
}
```
-注意：当使用`export default`去声明一个文件或者模块的返回值时，在每个文件或者模块中应当只默认导出一个值。特别地，能将`export deafult`与`var`，`let`与`const`一起使用。
+Since `export default` is used to declare a fallback value for a module or file, you can only have one value be a default export in each module or file. Additionally, you cannot use `export default` with `var`, `let`, or `const`
# --instructions--
-下面的函数应该在这个模块中返回一个值。请添加需要的代码：
+The following function should be the fallback value for the module. Please add the necessary code to do so.
# --hints--
-正确的使用`export`进行返回。
+Your code should use `export` fallback.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/create-strings-using-template-literals.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/create-strings-using-template-literals.md
index 0f52b8a0bd4..cd5c70bd6f7 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/create-strings-using-template-literals.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/create-strings-using-template-literals.md
@@ -1,6 +1,6 @@
---
id: 587d7b8a367417b2b2512b4e
-title: 使用模板字面量创建字符串
+title: Create Strings using Template Literals
challengeType: 1
forumTopicId: 301200
dashedName: create-strings-using-template-literals
@@ -8,11 +8,11 @@ dashedName: create-strings-using-template-literals
# --description--
-模板字符串是 ES6 的另外一项新的功能。这是一种可以轻松构建复杂字符串的方法。
+A new feature of ES6 is the <dfn>template literal</dfn>. This is a special type of string that makes creating complex strings easier.
-模板字符串可以使用多行字符串和字符串插值功能。
+Template literals allow you to create multi-line strings and to use string interpolation features to create strings.
-请看以下代码：
+Consider the code below:
```js
const person = {
@@ -30,26 +30,33 @@ console.log(greeting); // prints
```
-这段代码有许多的不同： 首先，上面的例子使用了反引号（`` ` ``）而不是引号（`'` 或者 `"`）定义字符串。 其次，注意字符串是多行的，不管是代码还是输出。这是因为在字符串内插入了 `\n`。 上面使用的`${variable}`语法是一个占位符。这样一来，你将不再需要使用`+`运算符来连接字符串。当需要在字符串里增加变量的时候，你只需要在变量的外面括上`${`和`}`，并将其放在字符串里就可以了。 这个新的方式使你可以更灵活的创建复杂的字符串。
+A lot of things happened there. Firstly, the example uses backticks (`` ` ``), not quotes (`'` or `"`), to wrap the string. Secondly, notice that the string is multi-line, both in the code and the output. This saves inserting `\n` within strings. The `${variable}` syntax used above is a placeholder. Basically, you won't have to use concatenation with the `+` operator anymore. To add variables to strings, you just drop the variable in a template string and wrap it with `${` and `}`. Similarly, you can include other expressions in your string literal, for example `${a + b}`. This new way of creating strings gives you more flexibility to create robust strings.
# --instructions--
-使用模板字符串的反引号的语法来展示`result`对象的`failure`数组内的每个条目。每个条目应该括在带有`text-warning`类属性的`li`标签中，并赋值给`resultDisplayArray`。
+Use template literal syntax with backticks to create an array of list element (`li`) strings. Each list element's text should be one of the array elements from the `failure` property on the `result` object and have a `class` attribute with the value `text-warning`. The `makeList` function should return the array of list item strings.
-使用遍历方法（可以是任意形式的循环）输出指定值。
+Use an iterator method (any kind of loop) to get the desired output (shown below).
+
+```js
+[
+ '<li class="text-warning">no-var</li>',
+ '<li class="text-warning">var-on-top</li>',
+ '<li class="text-warning">linebreak</li>'
+]
+```
# --hints--
-`resultDisplayArray` 是一个包含了 `result failure` 内的消息的数组。
+`failuresList` should be an array containing `result failure` messages.
```js
assert(
- typeof makeList(result.failure) === 'object' &&
- resultDisplayArray.length === 3
+ typeof makeList(result.failure) === 'object' && failuresList.length === 3
);
```
-`resultDisplayArray` 要有正确的输出。
+`failuresList` should be equal to the specified output.
```js
assert(
@@ -61,13 +68,13 @@ assert(
);
```
-应使用模板字符串。
+Template strings and expression interpolation should be used.
```js
(getUserInput) => assert(getUserInput('index').match(/(`.*\${.*}.*`)/));
```
-应该遍历。
+An iterator should be used.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/declare-a-read-only-variable-with-the-const-keyword.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/declare-a-read-only-variable-with-the-const-keyword.md
index 919c81576bb..a869196788e 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/declare-a-read-only-variable-with-the-const-keyword.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/declare-a-read-only-variable-with-the-const-keyword.md
@@ -1,6 +1,6 @@
---
id: 587d7b87367417b2b2512b41
-title: 用 const 关键字声明只读变量
+title: Declare a Read-Only Variable with the const Keyword
challengeType: 1
forumTopicId: 301201
dashedName: declare-a-read-only-variable-with-the-const-keyword
@@ -8,45 +8,44 @@ dashedName: declare-a-read-only-variable-with-the-const-keyword
# --description--
-`let`并不是唯一的新的声明变量的方式。在 ES6里面，你还可以使用`const`关键字来声明变量。
+The keyword `let` is not the only new way to declare variables. In ES6, you can also declare variables using the `const` keyword.
-`const`拥有`let`的所有优点，所不同的是，通过`const`声明的变量是只读的。这意味着通过`const`声明的变量只能被赋值一次，而不能被再次赋值。
+`const` has all the awesome features that `let` has, with the added bonus that variables declared using `const` are read-only. They are a constant value, which means that once a variable is assigned with `const`, it cannot be reassigned.
```js
-"use strict";
const FAV_PET = "Cats";
FAV_PET = "Dogs"; // returns error
```
-可以看见，尝试给通过`const`声明的变量再次赋值会报错。你应该使用`const`关键字来对所有不打算再次赋值的变量进行声明。这有助于你避免给一个常量进行额外的再次赋值。一个最佳实践是对所有常量的命名采用全大写字母，并在单词之间使用下划线进行分隔。
+As you can see, trying to reassign a variable declared with `const` will throw an error. You should always name variables you don't want to reassign using the `const` keyword. This helps when you accidentally attempt to reassign a variable that is meant to stay constant. A common practice when naming constants is to use all uppercase letters, with words separated by an underscore.
-**注意：** 一般开发者会以大写做为常量标识符，小写字母或者驼峰命名做为变量（对象或数组）标识符。接下来的挑战里会涉及到小写变量标识符的数组。
+**Note:** It is common for developers to use uppercase variable identifiers for immutable values and lowercase or camelCase for mutable values (objects and arrays). In a later challenge you will see an example of a lowercase variable identifier being used for an array.
# --instructions--
-改变以下代码，使得所有的变量都使用`let`或`const`关键词来声明。当变量将会改变的时候使用`let`关键字，当变量要保持常量的时候使用`const`关键字。同时，对使用`const`声明的变量按照最佳实践重命名，变量名中的字母应该都是大写的。
+Change the code so that all variables are declared using `let` or `const`. Use `let` when you want the variable to change, and `const` when you want the variable to remain constant. Also, rename variables declared with `const` to conform to common practices, meaning constants should be in all caps.
# --hints--
-`var`在代码中不存在。
+`var` should not exist in your code.
```js
(getUserInput) => assert(!getUserInput('index').match(/var/g));
```
-`SENTENCE`应该是使用`const`声明的常量。
+`SENTENCE` should be a constant variable declared with `const`.
```js
(getUserInput) => assert(getUserInput('index').match(/(const SENTENCE)/g));
```
-`i`应该是使用`let`声明的变量。
+`i` should be declared with `let`.
```js
(getUserInput) => assert(getUserInput('index').match(/(let i)/g));
```
-`console.log`应该修改为用于打印`SENTENCE`变量。
+`console.log` should be changed to print the `SENTENCE` variable.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/explore-differences-between-the-var-and-let-keywords.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/explore-differences-between-the-var-and-let-keywords.md
index 72141e0748c..7792ee384bc 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/explore-differences-between-the-var-and-let-keywords.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/explore-differences-between-the-var-and-let-keywords.md
@@ -1,6 +1,6 @@
---
id: 587d7b87367417b2b2512b3f
-title: 探索 var 和 let 关键字之间的差异
+title: Explore Differences Between the var and let Keywords
challengeType: 1
forumTopicId: 301202
dashedName: explore-differences-between-the-var-and-let-keywords
@@ -8,7 +8,7 @@ dashedName: explore-differences-between-the-var-and-let-keywords
# --description--
-使用`var`关键字来声明变量，会出现重复声明导致变量被覆盖却不会报错的问题：
+One of the biggest problems with declaring variables with the `var` keyword is that you can overwrite variable declarations without an error.
```js
var camper = 'James';
@@ -17,15 +17,15 @@ console.log(camper);
// logs 'David'
```
-在上面的代码中，`camper`的初始值为`'James'`，然后又被覆盖成了`'David'`。 在小型的应用中，你可能不会遇到这样的问题，但是当你的代码规模变得更加庞大的时候，就可能会在不经意间覆盖了之前定义的变量。 这样的行为不会报错，导致了 debug 非常困难。
-在 ES6 中引入了新的关键字`let`来解决`var`关键字带来的潜在问题。 如果你在上面的代码中，使用了`let`关键字来代替`var`关键字，结果会是一个报错。
+As you can see in the code above, the `camper` variable is originally declared as `James` and then overridden to be `David`. In a small application, you might not run into this type of problem, but when your code becomes larger, you might accidentally overwrite a variable that you did not intend to overwrite. Because this behavior does not throw an error, searching and fixing bugs becomes more difficult.
+A new keyword called `let` was introduced in ES6 to solve this potential issue with the `var` keyword. If you were to replace `var` with `let` in the variable declarations of the code above, the result would be an error.
```js
let camper = 'James';
let camper = 'David'; // throws an error
```
-你可以在浏览器的控制台里看见这个错误。 与`var`不同的是，当使用`let`的时候，同一名字的变量只能被声明一次。 请注意`"use strict"`。这代表着开启了严格模式，用于检测常见的代码错误以及"不安全"的行为，例如：
+This error can be seen in the console of your browser. So unlike `var`, when using `let`, a variable with the same name can only be declared once. Note the `"use strict"`. This enables Strict Mode, which catches common coding mistakes and "unsafe" actions. For instance:
```js
"use strict";
@@ -34,23 +34,23 @@ x = 3.14; // throws an error because x is not declared
# --instructions--
-请更新这段代码，并且在其中只使用`let`关键字
+Update the code so it only uses the `let` keyword.
# --hints--
-在代码中不应存在`var`。
+`var` should not exist in the code.
```js
(getUserInput) => assert(!getUserInput('index').match(/var/g));
```
-`catName`变量的值应该为`'Oliver'`。
+`catName` should be `Oliver`.
```js
assert(catName === 'Oliver');
```
-`quote`变量的值应该为`'Oliver says Meow!'`
+`quote` should be `"Oliver says Meow!"`
```js
assert(quote === 'Oliver says Meow!');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/handle-a-fulfilled-promise-with-then.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/handle-a-fulfilled-promise-with-then.md
index 8eb422070b0..e39dd1da26a 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/handle-a-fulfilled-promise-with-then.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/handle-a-fulfilled-promise-with-then.md
@@ -1,6 +1,6 @@
---
id: 5cdafbd72913098997531681
-title: 在 then 中处理 Promise 完成的情况
+title: Handle a Fulfilled Promise with then
challengeType: 1
forumTopicId: 301203
dashedName: handle-a-fulfilled-promise-with-then
@@ -8,7 +8,7 @@ dashedName: handle-a-fulfilled-promise-with-then
# --description--
-当程序需要花费未知的时间才能完成时 Promise 很有用（比如，一些异步操作），一般是网络请求。网络请求会花费一些时间，当结束时需要根据服务器的响应执行一些操作。这可以用 `then` 方法来实现，当 promise 完成 `resolve` 时会触发 `then` 方法。例子如下：
+Promises are most useful when you have a process that takes an unknown amount of time in your code (i.e. something asynchronous), often a server request. When you make a server request it takes some amount of time, and after it completes you usually want to do something with the response from the server. This can be achieved by using the `then` method. The `then` method is executed immediately after your promise is fulfilled with `resolve`. Here’s an example:
```js
myPromise.then(result => {
@@ -16,32 +16,36 @@ myPromise.then(result => {
});
```
-`result` 即传入 `resolve` 方法的参数。
+`result` comes from the argument given to the `resolve` method.
# --instructions--
-给 promise 添加 `then` 方法。用 `result` 做为回调函数的参数并将 `result` 打印在控制台。
+Add the `then` method to your promise. Use `result` as the parameter of its callback function and log `result` to the console.
# --hints--
-应该给 promise 方法调用 `then` 方法。
+You should call the `then` method on the promise.
```js
-assert(codeWithoutSpaces.match(/(makeServerRequest|\))\.then\(/g));
+assert(
+ __helpers.removeWhiteSpace(code).match(/(makeServerRequest|\))\.then\(/g)
+);
```
-`then` 方法应该有一个回调函数，回调函数参数为 `result`。
+Your `then` method should have a callback function with `result` as its parameter.
```js
assert(resultIsParameter);
```
-应该打印 `result` 到控制台。
+You should log `result` to the console.
```js
assert(
resultIsParameter &&
- codeWithoutSpaces.match(/\.then\(.*?result.*?console.log\(result\).*?\)/)
+ __helpers
+ .removeWhiteSpace(code)
+ .match(/\.then\(.*?result.*?console.log\(result\).*?\)/)
);
```
@@ -59,7 +63,7 @@ const resultIsParameter = /\.then\((function\(result\){|result|\(result\)=>)/.te
const makeServerRequest = new Promise((resolve, reject) => {
// responseFromServer is set to true to represent a successful response from a server
let responseFromServer = true;
-
+
if(responseFromServer) {
resolve("We got the data");
} else {
@@ -74,7 +78,7 @@ const makeServerRequest = new Promise((resolve, reject) => {
const makeServerRequest = new Promise((resolve, reject) => {
// responseFromServer is set to true to represent a successful response from a server
let responseFromServer = true;
-
+
if(responseFromServer) {
resolve("We got the data");
} else {
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/handle-a-rejected-promise-with-catch.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/handle-a-rejected-promise-with-catch.md
index 1881a150b05..815c4b47862 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/handle-a-rejected-promise-with-catch.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/handle-a-rejected-promise-with-catch.md
@@ -1,6 +1,6 @@
---
id: 5cdafbe72913098997531682
-title: 在 catch 中处理 Promise 失败的情况
+title: Handle a Rejected Promise with catch
challengeType: 1
forumTopicId: 301204
dashedName: handle-a-rejected-promise-with-catch
@@ -8,7 +8,7 @@ dashedName: handle-a-rejected-promise-with-catch
# --description--
-当 promise 失败时会调用 `catch` 方法。当 promise 的 `reject` 方法执行时会直接调用。用法如下：
+`catch` is the method used when your promise has been rejected. It is executed immediately after a promise's `reject` method is called. Here’s the syntax:
```js
myPromise.catch(error => {
@@ -16,34 +16,36 @@ myPromise.catch(error => {
});
```
-`error` 是传入 `reject` 方法的参数。
-
-**注意：** `then` 和 `catch` 方法可以在 promise 后面链式调用。
+`error` is the argument passed in to the `reject` method.
# --instructions--
-给 promise 添加 `catch` 方法。用 `error` 做为回调函数的参数并把 `error` 打印到控制台。
+Add the `catch` method to your promise. Use `error` as the parameter of its callback function and log `error` to the console.
# --hints--
-应该在 promise 上调用 `catch` 方法。
+You should call the `catch` method on the promise.
```js
-assert(codeWithoutSpaces.match(/(makeServerRequest|\))\.catch\(/g));
+assert(
+ __helpers.removeWhiteSpace(code).match(/(makeServerRequest|\))\.catch\(/g)
+);
```
-`catch` 方法应该有一个回调函数，函数参数为`error`。
+Your `catch` method should have a callback function with `error` as its parameter.
```js
assert(errorIsParameter);
```
-应该打印`error`到控制台。
+You should log `error` to the console.
```js
assert(
errorIsParameter &&
- codeWithoutSpaces.match(/\.catch\(.*?error.*?console.log\(error\).*?\)/)
+ __helpers
+ .removeWhiteSpace(code)
+ .match(/\.catch\(.*?error.*?console.log\(error\).*?\)/)
);
```
@@ -61,7 +63,7 @@ const errorIsParameter = /\.catch\((function\(error\){|error|\(error\)=>)/.test(
const makeServerRequest = new Promise((resolve, reject) => {
// responseFromServer is set to false to represent an unsuccessful response from a server
let responseFromServer = false;
-
+
if(responseFromServer) {
resolve("We got the data");
} else {
@@ -80,7 +82,7 @@ makeServerRequest.then(result => {
const makeServerRequest = new Promise((resolve, reject) => {
// responseFromServer is set to false to represent an unsuccessful response from a server
let responseFromServer = false;
-
+
if(responseFromServer) {
resolve("We got the data");
} else {
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/import-a-default-export.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/import-a-default-export.md
index 8fa626e1459..d8a0a3d465e 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/import-a-default-export.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/import-a-default-export.md
@@ -1,6 +1,6 @@
---
id: 587d7b8d367417b2b2512b59
-title: 导入一个默认的导出
+title: Import a Default Export
challengeType: 1
forumTopicId: 301205
dashedName: import-a-default-export
@@ -8,23 +8,21 @@ dashedName: import-a-default-export
# --description--
-在上一个挑战里，学习了`export default`的用法。还需要一种`import`的语法来导入默认的导出。
-
-在下面的例子里有一个`add`函数, 它在`"math_functions"`文件里默认被导出。来看看来如何导入它：
+In the last challenge, you learned about `export default` and its uses. To import a default export, you need to use a different `import` syntax. In the following example, `add` is the default export of the `math_functions.js` file. Here is how to import it:
```js
import add from "./math_functions.js";
```
-这个语法只有一处不同的地方 —— 被导入的`add`值，并没有被花括号`{}`所包围。与导出值的方法不同，导入默认导出的写法仅仅只是简单的将变量名写在`import`之后。
+The syntax differs in one key place. The imported value, `add`, is not surrounded by curly braces (`{}`). `add` here is simply a variable name for whatever the default export of the `math_functions.js` file is. You can use any name here when importing a default.
# --instructions--
-在下面的代码中，请导入在同目录下的`"math_functions"`文件中默认导出的`subtract`值。
+In the following code, import the default export from the `math_functions.js` file, found in the same directory as this file. Give the import the name `subtract`.
# --hints--
-正确导入`export default`方法导出的值。
+You should properly import `subtract` from `math_functions.js`.
```js
assert(code.match(/import\s+subtract\s+from\s+('|")\.\/math_functions\.js\1/g));
@@ -35,7 +33,7 @@ assert(code.match(/import\s+subtract\s+from\s+('|")\.\/math_functions\.js\1/g));
## --seed-contents--
```js
-
+
// Only change code above this line
subtract(7,4);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/mutate-an-array-declared-with-const.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/mutate-an-array-declared-with-const.md
index 4824082aee5..90799c6bba5 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/mutate-an-array-declared-with-const.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/mutate-an-array-declared-with-const.md
@@ -1,6 +1,6 @@
---
id: 587d7b87367417b2b2512b42
-title: 改变一个用 const 声明的数组
+title: Mutate an Array Declared with const
challengeType: 1
forumTopicId: 301206
dashedName: mutate-an-array-declared-with-const
@@ -8,41 +8,40 @@ dashedName: mutate-an-array-declared-with-const
# --description--
-在现代的 JavaScript 里，`const`声明有很多用法。
+The `const` declaration has many use cases in modern JavaScript.
-一些开发者倾向默认使用`const`来声明所有变量，但如果它们打算在后续的代码中修改某个值，那在声明的时候就会用`let`。
+Some developers prefer to assign all their variables using `const` by default, unless they know they will need to reassign the value. Only in that case, they use `let`.
-然而，你要注意，对象（包括数组和函数）在使用`const`声明的时候依然是可变的。使用`const`来声明只会保证它的标识不会被重新赋值。
+However, it is important to understand that objects (including arrays and functions) assigned to a variable using `const` are still mutable. Using the `const` declaration only prevents reassignment of the variable identifier.
```js
-"use strict";
const s = [5, 6, 7];
s = [1, 2, 3]; // throws error, trying to assign a const
s[2] = 45; // works just as it would with an array declared with var or let
console.log(s); // returns [5, 6, 45]
```
-从以上代码看出，你可以改变`[5, 6, 7]`自身，所以`s`变量指向了改变后的数组`[5, 6, 45]`。和所有数组一样，数组`s`中的数组元素是可以被改变的，但是因为使用了`const`关键字，你不能使用赋值操作符将变量标识`s`指向另外一个数组。
+As you can see, you can mutate the object `[5, 6, 7]` itself and the variable `s` will still point to the altered array `[5, 6, 45]`. Like all arrays, the array elements in `s` are mutable, but because `const` was used, you cannot use the variable identifier `s` to point to a different array using the assignment operator.
# --instructions--
-这里有一个使用`const s = [5, 7, 2]`声明的数组。使用对各元素赋值的方法将数组改成`[2, 5, 7]`。
+An array is declared as `const s = [5, 7, 2]`. Change the array to `[2, 5, 7]` using various element assignments.
# --hints--
-不要替换`const`关键字。
+You should not replace `const` keyword.
```js
(getUserInput) => assert(getUserInput('index').match(/const/g));
```
-`s`应该为常量 (通过使用`const`)。
+`s` should be a constant variable (by using `const`).
```js
(getUserInput) => assert(getUserInput('index').match(/const\s+s/g));
```
-不要改变原数组的声明。
+You should not change the original array declaration.
```js
(getUserInput) =>
@@ -53,7 +52,7 @@ console.log(s); // returns [5, 6, 45]
);
```
-`s`应该等于`[2, 5, 7]`。
+`s` should be equal to `[2, 5, 7]`.
```js
assert.deepEqual(s, [2, 5, 7]);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/prevent-object-mutation.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/prevent-object-mutation.md
index ea76e46ce0c..fb97d403086 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/prevent-object-mutation.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/prevent-object-mutation.md
@@ -1,6 +1,6 @@
---
id: 598f48a36c8c40764b4e52b3
-title: 防止对象改变
+title: Prevent Object Mutation
challengeType: 1
forumTopicId: 301207
dashedName: prevent-object-mutation
@@ -8,9 +8,9 @@ dashedName: prevent-object-mutation
# --description--
-通过之前的挑战可以看出，`const`声明并不会真的保护你的数据不被改变。为了确保数据不被改变，JavaScript 提供了一个函数`Object.freeze`来防止数据改变。
+As seen in the previous challenge, `const` declaration alone doesn't really protect your data from mutation. To ensure your data doesn't change, JavaScript provides a function `Object.freeze` to prevent data mutation.
-当一个对象被冻结的时候，你不能再对它的属性再进行增、删、改的操作。任何试图改变对象的操作都会被阻止，却不会报错。
+Once the object is frozen, you can no longer add, update, or delete properties from it. Any attempt at changing the object will be rejected without an error.
```js
let obj = {
@@ -26,24 +26,24 @@ console.log(obj);
# --instructions--
-在这个挑战中，你将使用`Object.freeze`来防止数学常量被改变。你需要冻结`MATH_CONSTANTS`对象，使得没有人可以改变`PI`的值，抑或增加或删除属性。
+In this challenge you are going to use `Object.freeze` to prevent mathematical constants from changing. You need to freeze the `MATH_CONSTANTS` object so that no one is able to alter the value of `PI`, add, or delete properties.
# --hints--
-不要替换`const`关键字。
+You should not replace `const` keyword.
```js
(getUserInput) => assert(getUserInput('index').match(/const/g));
```
-`MATH_CONSTANTS`应该为一个常量 (使用`const`)。
+`MATH_CONSTANTS` should be a constant variable (by using `const`).
```js
(getUserInput) =>
assert(getUserInput('index').match(/const\s+MATH_CONSTANTS/g));
```
-不要改变原始的`MATH_CONSTANTS`。
+You should not change original `MATH_CONSTANTS`.
```js
(getUserInput) =>
@@ -54,7 +54,7 @@ console.log(obj);
);
```
-`PI`等于`3.14`。
+`PI` should equal `3.14`.
```js
assert(PI === 3.14);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/reuse-javascript-code-using-import.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/reuse-javascript-code-using-import.md
index b8942582427..ff3449dd2f8 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/reuse-javascript-code-using-import.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/reuse-javascript-code-using-import.md
@@ -1,6 +1,6 @@
---
id: 587d7b8c367417b2b2512b55
-title: 通过 import 复用 JavaScript 代码
+title: Reuse JavaScript Code Using import
challengeType: 1
forumTopicId: 301208
dashedName: reuse-javascript-code-using-import
@@ -8,15 +8,15 @@ dashedName: reuse-javascript-code-using-import
# --description--
-`import` 可以导入文件或模块的一部分。在之前的课程里，例子从 `math_functions.js` 文件里导出了 `add`，下面看一下如何在其它的文件导入它：
+`import` allows you to choose which parts of a file or module to load. In the previous lesson, the examples exported `add` from the `math_functions.js` file. Here's how you can import it to use in another file:
```js
import { add } from './math_functions.js';
```
-在这里，`import` 会在 `math_functions.js` 里找到 `add`，只导入这个函数，忽略剩余的部分。`./` 告诉程序在当前文件的相同目录寻找 `math_functions.js` 文件。用这种方式导入时，相对路径（`./`）和文件扩展名（`.js`）都是必需的。
+Here, `import` will find `add` in `math_functions.js`, import just that function for you to use, and ignore the rest. The `./` tells the import to look for the `math_functions.js` file in the same folder as the current file. The relative file path (`./`) and file extension (`.js`) are required when using import in this way.
-可以在导入语句里导入多个项目，如下：
+You can import more than one item from the file by adding them in the `import` statement like this:
```js
import { add, subtract } from './math_functions.js';
@@ -24,11 +24,11 @@ import { add, subtract } from './math_functions.js';
# --instructions--
-添加 `import` 语句，使当前文件可以使用你在之前课程里导出的 `uppercaseString` 和 `lowercaseString` 函数，函数在当前路径下的 `string_functions.js` 文件里。
+Add the appropriate `import` statement that will allow the current file to use the `uppercaseString` and `lowercaseString` functions you exported in the previous lesson. These functions are in a file called `string_functions.js`, which is in the same directory as the current file.
# --hints--
-应该导入 `uppercaseString`、
+You should properly import `uppercaseString`.
```js
assert(
@@ -38,7 +38,7 @@ assert(
);
```
-应该导入 `lowercaseString`、
+You should properly import `lowercaseString`.
```js
assert(
@@ -53,7 +53,7 @@ assert(
## --seed-contents--
```js
-
+
// Only change code above this line
uppercaseString("hello");
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/set-default-parameters-for-your-functions.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/set-default-parameters-for-your-functions.md
index f9c9b9bcefc..b804b67af1a 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/set-default-parameters-for-your-functions.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/set-default-parameters-for-your-functions.md
@@ -1,6 +1,6 @@
---
id: 587d7b88367417b2b2512b46
-title: 设置函数的默认参数
+title: Set Default Parameters for Your Functions
challengeType: 1
forumTopicId: 301209
dashedName: set-default-parameters-for-your-functions
@@ -8,9 +8,9 @@ dashedName: set-default-parameters-for-your-functions
# --description--
-ES6 里允许给函数传入<dfn>默认参数</dfn>，来构建更加灵活的函数。
+In order to help us create more flexible functions, ES6 introduces <dfn>default parameters</dfn> for functions.
-请看以下代码：
+Check out this code:
```js
const greeting = (name = "Anonymous") => "Hello " + name;
@@ -19,27 +19,27 @@ console.log(greeting("John")); // Hello John
console.log(greeting()); // Hello Anonymous
```
-默认参数会在参数没有被指定（值为 undefined ）的时候起作用。在上面的例子中，参数`name`会在没有得到新的值的时候，默认使用值 "Anonymous"。你还可以给多个参数赋予默认值。
+The default parameter kicks in when the argument is not specified (it is undefined). As you can see in the example above, the parameter `name` will receive its default value `"Anonymous"` when you do not provide a value for the parameter. You can add default values for as many parameters as you want.
# --instructions--
-给函数`increment`加上默认参数，使得在`value`没有被赋值的时候，默认给`number`加1。
+Modify the function `increment` by adding default parameters so that it will add 1 to `number` if `value` is not specified.
# --hints--
-`increment(5, 2)`的结果应该为`7`。
+The result of `increment(5, 2)` should be `7`.
```js
assert(increment(5, 2) === 7);
```
-`increment(5)`的结果应该为`6`。
+The result of `increment(5)` should be `6`.
```js
assert(increment(5) === 6);
```
-参数`value`的默认值应该为`1`。
+A default parameter value of `1` should be used for `value`.
```js
assert(code.match(/value\s*=\s*1/g));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use--to-import-everything-from-a-file.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use--to-import-everything-from-a-file.md
index 18fb113f972..a175f323833 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use--to-import-everything-from-a-file.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use--to-import-everything-from-a-file.md
@@ -1,6 +1,6 @@
---
id: 587d7b8c367417b2b2512b57
-title: 用 * 从文件中导入所有内容
+title: Use * to Import Everything from a File
challengeType: 1
forumTopicId: 301210
dashedName: use--to-import-everything-from-a-file
@@ -8,13 +8,13 @@ dashedName: use--to-import-everything-from-a-file
# --description--
-我们还可以用`import`语法从文件中导入所有的内容。下面是一个从同目录下的`"math_functions"`文件中导入所有内容的例子：
+Suppose you have a file and you wish to import all of its contents into the current file. This can be done with the `import * as` syntax. Here's an example where the contents of a file named `math_functions.js` are imported into a file in the same directory:
```js
import * as myMathModule from "./math_functions.js";
```
-上面的 `import` 语句会创建一个叫做 `myMathModule` 的对象。这只是一个变量名，可以随便命名。对象包含 `math_functions.js` 文件里的所有导出，可以像访问对象的属性那样访问里面的函数。下面是使用导入的 `add` 和 `subtract` 函数的例子：
+The above `import` statement will create an object called `myMathModule`. This is just a variable name, you can name it anything. The object will contain all of the exports from `math_functions.js` in it, so you can access the functions like you would any other object property. Here's how you can use the `add` and `subtract` functions that were imported:
```js
myMathModule.add(2,3);
@@ -23,11 +23,11 @@ myMathModule.subtract(5,3);
# --instructions--
-下面的代码需要从同目录下的`"string_functions"`文件中导入所有内容。使用提供的对象，在当前文件的顶部添加正确的`import *`语句
+The code in this file requires the contents of the file: `string_functions.js`, that is in the same directory as the current file. Use the `import * as` syntax to import everything from the file into an object called `stringFunctions`.
# --hints--
-正确使用`import * as`语法。
+Your code should properly use `import * as` syntax.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-arrow-functions-to-write-concise-anonymous-functions.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-arrow-functions-to-write-concise-anonymous-functions.md
index 24086615f84..b1e92a91d2f 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-arrow-functions-to-write-concise-anonymous-functions.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-arrow-functions-to-write-concise-anonymous-functions.md
@@ -1,6 +1,6 @@
---
id: 587d7b87367417b2b2512b43
-title: 使用箭头函数编写简洁的匿名函数
+title: Use Arrow Functions to Write Concise Anonymous Functions
challengeType: 1
forumTopicId: 301211
dashedName: use-arrow-functions-to-write-concise-anonymous-functions
@@ -8,9 +8,9 @@ dashedName: use-arrow-functions-to-write-concise-anonymous-functions
# --description--
-在 JavaScript 里，我们会经常遇到不需要给函数命名的情况，尤其是在需要将一个函数作为参数传给另外一个函数的时候。这时，我们会创建匿名函数。因为这些函数不会在其他地方复用，所以我们不需要给它们命名。
+In JavaScript, we often don't need to name our functions, especially when passing a function as an argument to another function. Instead, we create inline functions. We don't need to name these functions because we do not reuse them anywhere else.
-这种情况下，我们通常会使用以下语法：
+To achieve this, we often use the following syntax:
```js
const myFunc = function() {
@@ -19,7 +19,7 @@ const myFunc = function() {
}
```
-ES6 提供了其他写匿名函数的方式的语法糖。你可以使用箭头函数：
+ES6 provides us with the syntactic sugar to not have to write anonymous functions this way. Instead, you can use **arrow function syntax**:
```js
const myFunc = () => {
@@ -28,45 +28,45 @@ const myFunc = () => {
}
```
-当不需要函数体，只返回一个值的时候，箭头函数允许你省略`return`关键字和外面的大括号。这样就可以将一个简单的函数简化成一个单行语句。
+When there is no function body, and only a return value, arrow function syntax allows you to omit the keyword `return` as well as the brackets surrounding the code. This helps simplify smaller functions into one-line statements:
```js
const myFunc = () => "value";
```
-这段代码仍然会返回`value`。
+This code will still return the string `value` by default.
# --instructions--
-使用箭头函数的语法重写`magic`函数，使其返回一个新的`Date()`。同时不要用`var`关键字来定义任何变量。
+Rewrite the function assigned to the variable `magic` which returns a `new Date()` to use arrow function syntax. Also, make sure nothing is defined using the keyword `var`.
# --hints--
-替换掉`var`关键字。
+User should replace `var` keyword.
```js
(getUserInput) => assert(!getUserInput('index').match(/var/g));
```
-`magic`应该为一个常量 (使用`const`)。
+`magic` should be a constant variable (by using `const`).
```js
(getUserInput) => assert(getUserInput('index').match(/const\s+magic/g));
```
-`magic`是一个`function`。
+`magic` should be a `function`.
```js
assert(typeof magic === 'function');
```
-`magic()`返回正确的日期。
+`magic()` should return correct date.
```js
assert(magic().setHours(0, 0, 0, 0) === new Date().setHours(0, 0, 0, 0));
```
-不要使用`function`关键字。
+`function` keyword should not be used.
```js
(getUserInput) => assert(!getUserInput('index').match(/function/g));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-class-syntax-to-define-a-constructor-function.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-class-syntax-to-define-a-constructor-function.md
index 76c4bc644c1..9e27796ab3a 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-class-syntax-to-define-a-constructor-function.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-class-syntax-to-define-a-constructor-function.md
@@ -1,6 +1,6 @@
---
id: 587d7b8b367417b2b2512b53
-title: 使用 class 语法定义构造函数
+title: Use class Syntax to Define a Constructor Function
challengeType: 1
forumTopicId: 301212
dashedName: use-class-syntax-to-define-a-constructor-function
@@ -8,11 +8,11 @@ dashedName: use-class-syntax-to-define-a-constructor-function
# --description--
-ES6 提供了一个新的创建对象的语法，使用关键字`class`。
+ES6 provides a new syntax to create objects, using the <dfn>class</dfn> keyword.
-值得注意的是，`class`只是一个语法糖，它并不像 Java、Python 或者 Ruby 这一类的语言一样，严格履行了面向对象的开发规范。
+It should be noted that the `class` syntax is just syntax, and not a full-fledged class-based implementation of an object-oriented paradigm, unlike in languages such as Java, Python, Ruby, etc.
-在 ES5 里面，我们通常会定义一个构造函数，然后使用 `new` 关键字来实例化一个对象：
+In ES5, we usually define a constructor function and use the `new` keyword to instantiate an object.
```js
var SpaceShuttle = function(targetPlanet){
@@ -21,7 +21,7 @@ var SpaceShuttle = function(targetPlanet){
var zeus = new SpaceShuttle('Jupiter');
```
-`class`的语法只是简单地替换了构造函数的写法：
+The `class` syntax simply replaces the constructor function creation:
```js
class SpaceShuttle {
@@ -32,24 +32,21 @@ class SpaceShuttle {
const zeus = new SpaceShuttle('Jupiter');
```
-应该注意 `class` 关键字声明了一个函数，里面添加了一个构造器（constructor）。当调用 `new` 来创建一个新对象时构造器会被调用。
+It should be noted that the `class` keyword declares a new function, to which a constructor is added. This constructor is invoked when `new` is called to create a new object.
+**Notes:**
-**注意：**
-
--
-
-<li> 首字母大写驼峰命名法是 ES6 class 名的惯例，就像上面的 <code>SpaceShuttle</code>。</li>
-<li> 构造函数是一个特殊的函数，在用 class 创建时来创建和初始化对象。在 JavaScript 算法和数据结构证书的面向对象章节里会更深入介绍。</li></ul>
+- UpperCamelCase should be used by convention for ES6 class names, as in `SpaceShuttle` used above.
+- The constructor method is a special method for creating and initializing an object created with a class. You will learn more about it in the Object Oriented Programming section of the JavaScript Algorithms And Data Structures Certification.
# --instructions--
-使用`class`关键字，并写出正确的构造函数，来创建`Vegetable`这个类：
+Use the `class` keyword and write a constructor to create the `Vegetable` class.
-`Vegetable`这个类可以创建 vegetable 对象，这个对象拥有一个在构造函数中赋值的`name`属性。
+The `Vegetable` class allows you to create a vegetable object with a property `name` that gets passed to the constructor.
# --hints--
-`Vegetable` 应该是一个 `class`，并在其中定义了`constructor`方法。
+`Vegetable` should be a `class` with a defined `constructor` method.
```js
assert(
@@ -57,13 +54,13 @@ assert(
);
```
-使用了`class`关键字。
+`class` keyword should be used.
```js
assert(code.match(/class/g));
```
-`Vegetable`可以被实例化。
+`Vegetable` should be able to be instantiated.
```js
assert(() => {
@@ -72,7 +69,7 @@ assert(() => {
});
```
-`carrot.name` 应该返回 `carrot`.
+`carrot.name` should return `carrot`.
```js
assert(carrot.name == 'carrot');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-assign-variables-from-arrays.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-assign-variables-from-arrays.md
index 6d941fc9efb..1c13043c134 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-assign-variables-from-arrays.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-assign-variables-from-arrays.md
@@ -1,6 +1,6 @@
---
id: 587d7b89367417b2b2512b4b
-title: 使用解构赋值从数组中分配变量
+title: Use Destructuring Assignment to Assign Variables from Arrays
challengeType: 1
forumTopicId: 301213
dashedName: use-destructuring-assignment-to-assign-variables-from-arrays
@@ -8,18 +8,18 @@ dashedName: use-destructuring-assignment-to-assign-variables-from-arrays
# --description--
-在 ES6 里面，解构数组可以如同解构对象一样简单。
+ES6 makes destructuring arrays as easy as destructuring objects.
-与数组解构不同，数组的扩展运算会将数组里的所有内容分解成一个由逗号分隔的列表。所以，你不能选择哪个元素来给变量赋值。
+One key difference between the spread operator and array destructuring is that the spread operator unpacks all contents of an array into a comma-separated list. Consequently, you cannot pick or choose which elements you want to assign to variables.
-而对数组进行解构却可以让我们做到这一点：
+Destructuring an array lets us do exactly that:
```js
const [a, b] = [1, 2, 3, 4, 5, 6];
console.log(a, b); // 1, 2
```
-变量`a`以及`b`分别被数组的第一、第二个元素赋值。 我们甚至能在数组解构中使用逗号分隔符，来获取任意一个想要的值：
+The variable `a` is assigned the first value of the array, and `b` is assigned the second value of the array. We can also access the value at any index in an array with destructuring by using commas to reach the desired index:
```js
const [a, b,,, c] = [1, 2, 3, 4, 5, 6];
@@ -28,23 +28,23 @@ console.log(a, b, c); // 1, 2, 5
# --instructions--
-使用数组解构来交换变量`a`与`b`的值。使`a`、`b`能分别获得对方的值。
+Use destructuring assignment to swap the values of `a` and `b` so that `a` receives the value stored in `b`, and `b` receives the value stored in `a`.
# --hints--
-在交换后，`a`的值应该为6。
+Value of `a` should be 6, after swapping.
```js
assert(a === 6);
```
-在交换后，`b`的值应该为8。
+Value of `b` should be 8, after swapping.
```js
assert(b === 8);
```
-使用数组解构来交换`a`和`b`。
+You should use array destructuring to swap a and b.
```js
assert(/\[\s*(\w)\s*,\s*(\w)\s*\]\s*=\s*\[\s*\2\s*,\s*\1\s*\]/g.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-assign-variables-from-nested-objects.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-assign-variables-from-nested-objects.md
index 8b5e152951f..23f5ff9888e 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-assign-variables-from-nested-objects.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-assign-variables-from-nested-objects.md
@@ -1,6 +1,6 @@
---
id: 587d7b89367417b2b2512b4a
-title: 使用解构赋值从嵌套对象中分配变量
+title: Use Destructuring Assignment to Assign Variables from Nested Objects
challengeType: 1
forumTopicId: 301214
dashedName: use-destructuring-assignment-to-assign-variables-from-nested-objects
@@ -8,9 +8,9 @@ dashedName: use-destructuring-assignment-to-assign-variables-from-nested-objects
# --description--
-同样，我们可以将 *嵌套的对象*解构到变量中。
+You can use the same principles from the previous two lessons to destructure values from nested objects.
-请看以下代码：
+Using an object similar to previous examples:
```js
const user = {
@@ -21,13 +21,13 @@ const user = {
};
```
-这是解构对象的属性并赋值给相同名字的变量：
+Here's how to extract the values of object properties and assign them to variables with the same name:
```js
const { johnDoe: { age, email }} = user;
```
-这是将对象的属性值指定给一个不同的名字：
+And here's how you can assign an object properties' values to variables with different names:
```js
const { johnDoe: { age: userAge, email: userEmail }} = user;
@@ -35,11 +35,11 @@ const { johnDoe: { age: userAge, email: userEmail }} = user;
# --instructions--
-将两个赋值语句替换成等价的解构赋值。`lowToday` 和 `highToday` 应该为 `LOCAL_FORECAST` 中 `today.low` 和 `today.high` 的值。
+Replace the two assignments with an equivalent destructuring assignment. It should still assign the variables `lowToday` and `highToday` the values of `today.low` and `today.high` from the `LOCAL_FORECAST` object.
# --hints--
-不能使用 ES5 的赋值语句。
+You should remove the ES5 assignment syntax.
```js
assert(
@@ -48,7 +48,7 @@ assert(
);
```
-应该使用解构创建 `lowToday` 变量。
+You should use destructuring to create the `lowToday` variable.
```js
assert(
@@ -58,7 +58,7 @@ assert(
);
```
-应该使用解构创建 `highToday` 变量。
+You should use destructuring to create the `highToday` variable.
```js
assert(
@@ -68,6 +68,12 @@ assert(
);
```
+`lowToday` should be equal to `64` and `highToday` should be equal to `77`.
+
+```js
+assert(lowToday === 64 && highToday === 77);
+```
+
# --seed--
## --seed-contents--
@@ -80,7 +86,7 @@ const LOCAL_FORECAST = {
};
// Only change code below this line
-
+
const lowToday = LOCAL_FORECAST.today.low;
const highToday = LOCAL_FORECAST.today.high;
@@ -95,6 +101,6 @@ const LOCAL_FORECAST = {
today: { low: 64, high: 77 },
tomorrow: { low: 68, high: 80 }
};
-
+
const { today: { low: lowToday, high: highToday }} = LOCAL_FORECAST;
```
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-assign-variables-from-objects.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-assign-variables-from-objects.md
index bcee9750cf0..c0689cacb3a 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-assign-variables-from-objects.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-assign-variables-from-objects.md
@@ -1,6 +1,6 @@
---
id: 587d7b89367417b2b2512b49
-title: 使用解构赋值从对象中分配变量
+title: Use Destructuring Assignment to Assign Variables from Objects
challengeType: 1
forumTopicId: 301215
dashedName: use-destructuring-assignment-to-assign-variables-from-objects
@@ -8,30 +8,30 @@ dashedName: use-destructuring-assignment-to-assign-variables-from-objects
# --description--
-可以在解构的属性后添加冒号和新的变量名来给解构的值赋予一个新的变量名。
+Destructuring allows you to assign a new variable name when extracting values. You can do this by putting the new name after a colon when assigning the value.
-还是以上个例子的对象来举例：
+Using the same object from the last example:
```js
const user = { name: 'John Doe', age: 34 };
```
-这是指定新的变量名的例子：
+Here's how you can give new variable names in the assignment:
```js
const { name: userName, age: userAge } = user;
// userName = 'John Doe', userAge = 34
```
-获取到了 `user.name` 的值并赋值给名为 `userName` 的变量。
+You may read it as "get the value of `user.name` and assign it to a new variable named `userName`" and so on.
# --instructions--
-使用解构赋值语句替换两个赋值语句。确保 `HIGH_TEMPERATURES` 的 `today` 和 `tomorrow` 属性赋值给 `highToday` 和 `highTomorrow`。
+Replace the two assignments with an equivalent destructuring assignment. It should still assign the variables `highToday` and `highTomorrow` the values of `today` and `tomorrow` from the `HIGH_TEMPERATURES` object.
# --hints--
-应该移除 ES5 赋值语句。
+You should remove the ES5 assignment syntax.
```js
assert(
@@ -40,26 +40,32 @@ assert(
);
```
-应该使用解构赋值语句创建 `highToday` 变量。
+You should use destructuring to create the `highToday` variable.
```js
assert(
code.match(
- /(var|const|let)\s*{\s*(today:\s*highToday[^}]*|[^,]*,\s*today\s*:\s*highToday\s*)}\s*=\s*HIGH_TEMPERATURES(;|\s+|\/\/)/g
+ /(var|const|let)\s*{\s*(today\s*:\s*highToday[^}]*|[^,]*,\s*today\s*:\s*highToday\s*)}\s*=\s*HIGH_TEMPERATURES(;|\s+|\/\/)/g
)
);
```
-应该使用解构赋值语句创建 `highTomorrow` 变量。
+You should use destructuring to create the `highTomorrow` variable.
```js
assert(
code.match(
- /(var|const|let)\s*{\s*(tomorrow:\s*highTomorrow[^}]*|[^,]*,\s*tomorrow\s*:\s*highTomorrow\s*)}\s*=\s*HIGH_TEMPERATURES(;|\s+|\/\/)/g
+ /(var|const|let)\s*{\s*(tomorrow\s*:\s*highTomorrow[^}]*|[^,]*,\s*tomorrow\s*:\s*highTomorrow\s*)}\s*=\s*HIGH_TEMPERATURES(;|\s+|\/\/)/g
)
);
```
+`highToday` should be equal to `77` and `highTomorrow` should be equal to `80`.
+
+```js
+assert(highToday === 77 && highTomorrow === 80);
+```
+
# --seed--
## --seed-contents--
@@ -72,7 +78,7 @@ const HIGH_TEMPERATURES = {
};
// Only change code below this line
-
+
const highToday = HIGH_TEMPERATURES.today;
const highTomorrow = HIGH_TEMPERATURES.tomorrow;
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-extract-values-from-objects.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-extract-values-from-objects.md
index 561dee4e444..ec873a1fa22 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-extract-values-from-objects.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-extract-values-from-objects.md
@@ -1,6 +1,6 @@
---
id: 5cfa550e84205a357704ccb6
-title: 使用解构赋值来获取对象的值
+title: Use Destructuring Assignment to Extract Values from Objects
challengeType: 1
forumTopicId: 301216
dashedName: use-destructuring-assignment-to-extract-values-from-objects
@@ -8,9 +8,9 @@ dashedName: use-destructuring-assignment-to-extract-values-from-objects
# --description--
-<dfn>解构赋值</dfn> 是 ES6 引入的新语法，用来从数组和对象中提取值，并优雅的对变量进行赋值。
+<dfn>Destructuring assignment</dfn> is special syntax introduced in ES6, for neatly assigning values taken directly from an object.
-有如下 ES5 代码：
+Consider the following ES5 code:
```js
const user = { name: 'John Doe', age: 34 };
@@ -19,52 +19,63 @@ const name = user.name; // name = 'John Doe'
const age = user.age; // age = 34
```
-下面是使用 ES6 解构赋值的等价语句：
+Here's an equivalent assignment statement using the ES6 destructuring syntax:
```js
const { name, age } = user;
// name = 'John Doe', age = 34
```
-在这里，`name` 和 `age` 被自动创建并赋予 `user` 对象相应属性的值。一目了然。
+Here, the `name` and `age` variables will be created and assigned the values of their respective values from the `user` object. You can see how much cleaner this is.
-解构赋值的参数数量可以任意。
+You can extract as many or few values from the object as you want.
# --instructions--
-把两个赋值语句替换成等价的解构赋值。`today` 和 `tomorrow` 的值应该还为 `HIGH_TEMPERATURES` 对象的 `today` 和 `tomorrow` 属性的值。
+Replace the two assignments with an equivalent destructuring assignment. It should still assign the variables `today` and `tomorrow` the values of `today` and `tomorrow` from the `HIGH_TEMPERATURES` object.
# --hints--
-应该移除 ES5 赋值语句。
+You should remove the ES5 assignment syntax.
```js
assert(
- !code.match(/today = HIGH_TEMPERATURES\.today/g) &&
- !code.match(/tomorrow = HIGH_TEMPERATURES\.tomorrow/g)
+ !__helpers
+ .removeJSComments(code)
+ .match(/today\s*=\s*HIGH_TEMPERATURES\.(today|tomorrow)/g)
);
```
-应该解构创建 `today` 变量。
+You should use destructuring to create the `today` variable.
```js
assert(
- code.match(
- /(var|let|const)\s*{\s*(today[^}]*|[^,]*,\s*today)\s*}\s*=\s*HIGH_TEMPERATURES(;|\s+|\/\/)/g
- )
+ __helpers
+ .removeJSComments(code)
+ .match(
+ /(var|let|const)\s*{\s*(today[^}]*|[^,]*,\s*today)\s*}\s*=\s*HIGH_TEMPERATURES(;|\s+|\/\/)/g
+ )
);
```
-应该解构创建 `tomorrow` 变量。
+You should use destructuring to create the `tomorrow` variable.
```js
assert(
- code.match(
- /(var|let|const)\s*{\s*(tomorrow[^}]*|[^,]*,\s*tomorrow)\s*}\s*=\s*HIGH_TEMPERATURES(;|\s+|\/\/)/g
- )
+ __helpers
+ .removeJSComments(code)
+ .match(
+ /(var|let|const)\s*{\s*(tomorrow[^}]*|[^,]*,\s*tomorrow)\s*}\s*=\s*HIGH_TEMPERATURES(;|\s+|\/\/)/g
+ )
);
```
+`today` should be equal to `77` and `tomorrow` should be equal to `80`.
+
+```js
+assert(today === 77 && tomorrow === 80);
+```
+
# --seed--
## --seed-contents--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-pass-an-object-as-a-functions-parameters.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-pass-an-object-as-a-functions-parameters.md
index f732b9f950b..b142b79a75c 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-pass-an-object-as-a-functions-parameters.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-to-pass-an-object-as-a-functions-parameters.md
@@ -1,6 +1,6 @@
---
id: 587d7b8a367417b2b2512b4d
-title: 使用解构赋值将对象作为函数的参数传递
+title: Use Destructuring Assignment to Pass an Object as a Function's Parameters
challengeType: 1
forumTopicId: 301217
dashedName: use-destructuring-assignment-to-pass-an-object-as-a-functions-parameters
@@ -8,9 +8,9 @@ dashedName: use-destructuring-assignment-to-pass-an-object-as-a-functions-parame
# --description--
-在某些情况下，你可以在函数的参数里直接解构对象。
+In some cases, you can destructure the object in a function argument itself.
-请看以下代码：
+Consider the code below:
```js
const profileUpdate = (profileData) => {
@@ -19,7 +19,7 @@ const profileUpdate = (profileData) => {
}
```
-上面的操作解构了传给函数的对象。这样的操作也可以直接在参数里完成：
+This effectively destructures the object sent into the function. This can also be done in-place:
```js
const profileUpdate = ({ name, age, nationality, location }) => {
@@ -27,33 +27,33 @@ const profileUpdate = ({ name, age, nationality, location }) => {
}
```
-这样的操作去除了多余的代码，使代码更加整洁。 这样做还有个额外的好处：函数不需要再去操作整个对象，而仅仅是操作复制到函数作用域内部的参数。
+When `profileData` is passed to the above function, the values are destructured from the function parameter for use within the function.
# --instructions--
-对`half`的参数进行解构赋值，使得仅仅将`max`与`min`的值传进函数。
+Use destructuring assignment within the argument to the function `half` to send only `max` and `min` inside the function.
# --hints--
-`stats`的类型应该是一个`object`。
+`stats` should be an `object`.
```js
assert(typeof stats === 'object');
```
-`half(stats)`应该等于`28.015`
+`half(stats)` should be `28.015`
```js
assert(half(stats) === 28.015);
```
-应该使用解构赋值。
+Destructuring should be used.
```js
-assert(code.replace(/\s/g, '').match(/half=\({\w+,\w+}\)/));
+assert(__helpers.removeWhiteSpace(code).match(/half=\({\w+,\w+}\)/));
```
-应该使用解构参数。
+Destructured parameter should be used.
```js
assert(!code.match(/stats\.max|stats\.min/));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-with-the-rest-parameter-to-reassign-array-elements.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-with-the-rest-parameter-to-reassign-array-elements.md
index dab982a160a..47b00256aaa 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-with-the-rest-parameter-to-reassign-array-elements.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-destructuring-assignment-with-the-rest-parameter-to-reassign-array-elements.md
@@ -1,6 +1,7 @@
---
id: 587d7b8a367417b2b2512b4c
-title: 使用解构赋值配合 rest 操作符来重新分配数组元素
+title: >-
+ Use Destructuring Assignment with the Rest Parameter to Reassign Array Elements
challengeType: 1
forumTopicId: 301218
dashedName: >-
@@ -9,9 +10,9 @@ dashedName: >-
# --description--
-在解构数组的某些情况下，我们可能希望将剩下的元素放进另一个数组里面。
+In some situations involving array destructuring, we might want to collect the rest of the elements into a separate array.
-以下代码的结果与使用`Array.prototype.slice()`相同：
+The result is similar to `Array.prototype.slice()`, as shown below:
```js
const [a, b, ...arr] = [1, 2, 3, 4, 5, 7];
@@ -19,31 +20,39 @@ console.log(a, b); // 1, 2
console.log(arr); // [3, 4, 5, 7]
```
-变量`a`与`b`分别获取了数组的前两个元素的值。之后，因为`rest`操作符的存在，`arr`获取了原数组剩余的元素的值，并构成了一个新的数组。 `rest`操作只能对数组列表最后的元素起作用。这意味着你不能使用`rest`操作符来截取原数组中间元素的子数组。
+Variables `a` and `b` take the first and second values from the array. After that, because of the rest parameter's presence, `arr` gets the rest of the values in the form of an array. The rest element only works correctly as the last variable in the list. As in, you cannot use the rest parameter to catch a subarray that leaves out the last element of the original array.
# --instructions--
-使用解构赋值以及`rest`操作符来进行一个`Array.prototype.slice`相同的操作。使得`arr`是原数组`source`除开前两个元素的子数组。
+Use destructuring assignment with the rest parameter to perform an effective `Array.prototype.slice()` so that `arr` is a sub-array of the original array `source` with the first two elements omitted.
# --hints--
-`arr`应该为`[3,4,5,6,7,8,9,10]`
+`arr` should be `[3,4,5,6,7,8,9,10]`
```js
assert(arr.every((v, i) => v === i + 3) && arr.length === 8);
```
-没有使用`Array.slice()`。
+`source` should be `[1,2,3,4,5,6,7,8,9,10]`
+
+```js
+assert(source.every((v, i) => v === i + 1) && source.length === 10);
+```
+
+`Array.slice()` should not be used.
```js
(getUserInput) => assert(!getUserInput('index').match(/slice/g));
```
-使用了解构赋值。
+Destructuring on `list` should be used.
```js
assert(
- code.replace(/\s/g, '').match(/\[(([_$a-z]\w*)?,){1,}\.\.\.arr\]=list/i)
+ __helpers
+ .removeWhiteSpace(code)
+ .match(/\[(([_$a-z]\w*)?,){1,}\.\.\.arr\]=list/i)
);
```
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-export-to-share-a-code-block.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-export-to-share-a-code-block.md
index e13465d4a8d..cffcbbe642a 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-export-to-share-a-code-block.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-export-to-share-a-code-block.md
@@ -1,6 +1,6 @@
---
id: 587d7b8c367417b2b2512b56
-title: 用 export 来重用代码块
+title: Use export to Share a Code Block
challengeType: 1
forumTopicId: 301219
dashedName: use-export-to-share-a-code-block
@@ -8,7 +8,7 @@ dashedName: use-export-to-share-a-code-block
# --description--
-假设有一个文件 `math_functions.js`，该文件包含了数学运算相关的一些函数。其中一个存储在变量 `add` 里，该函数接受两个数字做为参数返回它们的和。如果想在其它不同的 JavaScript 文件里使用这个函数，就需要 `export` 它。
+Imagine a file called `math_functions.js` that contains several functions related to mathematical operations. One of them is stored in a variable, `add`, that takes in two numbers and returns their sum. You want to use this function in several different JavaScript files. In order to share it with these other files, you first need to `export` it.
```js
export const add = (x, y) => {
@@ -16,7 +16,7 @@ export const add = (x, y) => {
}
```
-上面是导出单个函数常用方法，还可以这样导出：
+The above is a common way to export a single function, but you can achieve the same thing like this:
```js
const add = (x, y) => {
@@ -26,7 +26,7 @@ const add = (x, y) => {
export { add };
```
-导出变量和函数后，就可以在其它文件里导入使用从而避免了代码冗余。重复第一个例子的代码可以导出多个对象或函数，在第二个例子里面的导出语句中添加更多值也可以导出多项，例子如下：
+When you export a variable or function, you can import it in another file and use it without having to rewrite the code. You can export multiple things by repeating the first example for each thing you want to export, or by placing them all in the export statement of the second example, like this:
```js
export { add, subtract };
@@ -34,11 +34,11 @@ export { add, subtract };
# --instructions--
-下面有两个变量需要在别的文件中可以使用。利用刚才展示的第一种方式，导出两个变量。
+There are two string-related functions in the editor. Export both of them using the method of your choice.
# --hints--
-应该导出`uppercaseString`变量。
+You should properly export `uppercaseString`.
```js
assert(
@@ -48,7 +48,7 @@ assert(
);
```
-应该导出`lowercaseString`变量。
+You should properly export `lowercaseString`.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-getters-and-setters-to-control-access-to-an-object.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-getters-and-setters-to-control-access-to-an-object.md
index 383a9a66858..04e7ec825a4 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-getters-and-setters-to-control-access-to-an-object.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-getters-and-setters-to-control-access-to-an-object.md
@@ -1,6 +1,6 @@
---
id: 587d7b8c367417b2b2512b54
-title: 使用 getter 和 setter 来控制对象的访问
+title: Use getters and setters to Control Access to an Object
challengeType: 1
forumTopicId: 301220
dashedName: use-getters-and-setters-to-control-access-to-an-object
@@ -8,13 +8,13 @@ dashedName: use-getters-and-setters-to-control-access-to-an-object
# --description--
-你可以从对象中获得一个值，也可以给对象的属性赋值。
+You can obtain values from an object and set the value of a property within an object.
-这些通常行为被称为 <dfn>getters</dfn> 以及 <dfn>setters</dfn>。
+These are classically called <dfn>getters</dfn> and <dfn>setters</dfn>.
-Getter 函数的作用是可以让对象返回一个私有变量，而不需要直接去访问私有变量。
+Getter functions are meant to simply return (get) the value of an object's private variable to the user without the user directly accessing the private variable.
-Setter 函数的作用是可以基于传进的参数来修改对象中私有变量。这些修改可以是计算，或者是直接替换之前的值。
+Setter functions are meant to modify (set) the value of an object's private variable based on the value passed into the setter function. This change could involve calculations, or even overwriting the previous value completely.
```js
class Book {
@@ -30,33 +30,31 @@ class Book {
this._author = updatedAuthor;
}
}
-const lol = new Book('anonymous');
-console.log(lol.writer); // anonymous
-lol.writer = 'wut';
-console.log(lol.writer); // wut
+const novel = new Book('anonymous');
+console.log(novel.writer); // anonymous
+novel.writer = 'newAuthor';
+console.log(novel.writer); // newAuthor
```
-注意我们调用 getter 和 setter 的语法，它们看起来并不像一个函数调用。 Getter 和 Setter 非常重要，因为它们隐藏了内部的实现细节。
-
-**注意：** 通常会在私有变量前添加下划线（`_`）。这里并没有真正意义上让变量私有。
+Notice the syntax used to invoke the getter and setter. They do not even look like functions. Getters and setters are important because they hide internal implementation details. **Note:** It is convention to precede the name of a private variable with an underscore (`_`). However, the practice itself does not make a variable private.
# --instructions--
-使用`class`关键字来创建`Thermostat`类，它的构造函数应该可以接收 fahrenheit（华氏温度）作为参数。
+Use the `class` keyword to create a Thermostat class. The constructor accepts a Fahrenheit temperature.
-在类中创建 temperature 的 `getter`和`setter`，将温度转换成摄氏温度。
+In the class, create a `getter` to obtain the temperature in Celsius and a `setter` to set the temperature in Celsius.
-温度转换的公式是`C = 5/9 * (F - 32)`以及`F = C * 9.0 / 5 + 32`，F 代表华氏温度，C 代表摄氏温度。
+Remember that `C = 5/9 * (F - 32)` and `F = C * 9.0 / 5 + 32`, where `F` is the value of temperature in Fahrenheit, and `C` is the value of the same temperature in Celsius.
-**注意：** 当你实现这个作业的时候，你应当在类中使用一个温度标准，无论是华氏温度还是摄氏温度。
+**Note:** When you implement this, you will track the temperature inside the class in one scale, either Fahrenheit or Celsius.
-是时候展现 getter 和 setter 的威力了——无论你的 API 内部使用的是哪种温度标准，用户都能得到正确的结果。
+This is the power of a getter and a setter. You are creating an API for another user, who can get the correct result regardless of which one you track.
-或者说，你从用户需求中抽象出了实现细节。
+In other words, you are abstracting implementation details from the user.
# --hints--
-`Thermostat`应该是一个`class`，并且在其中定义了`constructor`方法。
+`Thermostat` should be a `class` with a defined `constructor` method.
```js
assert(
@@ -65,24 +63,35 @@ assert(
);
```
-应该使用 `class` 关键字。
+`class` keyword should be used.
```js
assert(code.match(/class/g));
```
-`Thermostat`应该可以被实例化。
+`Thermostat` should be able to be instantiated.
```js
assert(
(() => {
- const t = new Thermostat(32);
- return typeof t === 'object' && t.temperature === 0;
+ const t = new Thermostat(122);
+ return typeof t === 'object';
})()
);
```
-应该定义一个 `getter`。
+When instantiated with a Fahrenheit value, `Thermostat` should set the correct temperature.
+
+```js
+assert(
+ (() => {
+ const t = new Thermostat(122);
+ return t.temperature === 50;
+ })()
+);
+```
+
+A `getter` should be defined.
```js
assert(
@@ -96,7 +105,7 @@ assert(
);
```
-应该定义一个 `setter`。
+A `setter` should be defined.
```js
assert(
@@ -110,14 +119,16 @@ assert(
);
```
-调用 `setter` 应该设置 temperature。
+Calling the `setter` with a Celsius value should set the temperature.
```js
assert(
(() => {
const t = new Thermostat(32);
t.temperature = 26;
- return t.temperature !== 0;
+ const u = new Thermostat(32);
+ u.temperature = 50;
+ return t.temperature === 26 && u.temperature === 50;
})()
);
```
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-the-rest-parameter-with-function-parameters.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-the-rest-parameter-with-function-parameters.md
index 9320c9b2d74..888e68da6ba 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-the-rest-parameter-with-function-parameters.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-the-rest-parameter-with-function-parameters.md
@@ -1,6 +1,6 @@
---
id: 587d7b88367417b2b2512b47
-title: 将 rest 操作符与函数参数一起使用
+title: Use the Rest Parameter with Function Parameters
challengeType: 1
forumTopicId: 301221
dashedName: use-the-rest-parameter-with-function-parameters
@@ -8,9 +8,9 @@ dashedName: use-the-rest-parameter-with-function-parameters
# --description--
-ES6 推出了用于函数参数的<dfn>rest 操作符</dfn>帮助我们创建更加灵活的函数。在`rest`操作符的帮助下，你可以创建有一个变量来接受多个参数的函数。这些参数被储存在一个可以在函数内部读取的数组中。
+In order to help us create more flexible functions, ES6 introduces the <dfn>rest parameter</dfn> for function parameters. With the rest parameter, you can create functions that take a variable number of arguments. These arguments are stored in an array that can be accessed later from inside the function.
-请看以下代码：
+Check out this code:
```js
function howMany(...args) {
@@ -20,42 +20,42 @@ console.log(howMany(0, 1, 2)); // You have passed 3 arguments.
console.log(howMany("string", null, [1, 2, 3], { })); // You have passed 4 arguments.
```
-`rest`操作符可以避免查看`args`数组的需求，并且允许我们在参数数组上使用`map()`、`fiter()` 和 `reduce()`。
+The rest parameter eliminates the need to check the `args` array and allows us to apply `map()`, `filter()` and `reduce()` on the parameters array.
# --instructions--
-修改`sum`函数，来让它使用`rest`操作符，并且它可以在有任何数量的参数时以相同的形式工作。
+Modify the function `sum` using the rest parameter in such a way that the function `sum` is able to take any number of arguments and return their sum.
# --hints--
-`sum(0,1,2)`的返回结果应该为3。
+The result of `sum(0,1,2)` should be 3
```js
assert(sum(0, 1, 2) === 3);
```
-`sum(1,2,3,4)`的返回结果应该为10。
+The result of `sum(1,2,3,4)` should be 10
```js
assert(sum(1, 2, 3, 4) === 10);
```
-`sum(5)`的返回结果应该为5。
+The result of `sum(5)` should be 5
```js
assert(sum(5) === 5);
```
-`sum()`的返回结果应该为 0。
+The result of `sum()` should be 0
```js
assert(sum() === 0);
```
-对`sum`函数的`args`参数使用了`...`展开操作符。
+`sum` should be an arrow function which uses the rest parameter syntax (`...`) on the `args` parameter.
```js
-assert(code.replace(/\s/g, '').match(/sum=\(\.\.\.args\)=>/));
+assert(__helpers.removeWhiteSpace(code).match(/sum=\(\.\.\.args\)=>/));
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-the-spread-operator-to-evaluate-arrays-in-place.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-the-spread-operator-to-evaluate-arrays-in-place.md
index c5140ce21e0..9fa3b6b1b58 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-the-spread-operator-to-evaluate-arrays-in-place.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/use-the-spread-operator-to-evaluate-arrays-in-place.md
@@ -1,6 +1,6 @@
---
id: 587d7b89367417b2b2512b48
-title: 使用 spread 运算符展开数组项
+title: Use the Spread Operator to Evaluate Arrays In-Place
challengeType: 1
forumTopicId: 301222
dashedName: use-the-spread-operator-to-evaluate-arrays-in-place
@@ -8,23 +8,23 @@ dashedName: use-the-spread-operator-to-evaluate-arrays-in-place
# --description--
-ES6 允许我们使用 <dfn>展开操作符</dfn> 来展开数组，以及需要多个参数或元素的表达式。
+ES6 introduces the <dfn>spread operator</dfn>, which allows us to expand arrays and other expressions in places where multiple parameters or elements are expected.
-下面的 ES5 代码使用了`apply()`来计算数组的最大值：
+The ES5 code below uses `apply()` to compute the maximum value in an array:
```js
var arr = [6, 89, 3, 45];
var maximus = Math.max.apply(null, arr); // returns 89
```
-我们必须使用`Math.max.apply(null,arr)`，是因为直接调用`Math.max(arr)`会返回`NaN`。`Math.max()`函数需要传入的是一系列由逗号分隔的参数，而不是一个数组。 展开操作符可以提升代码的可读性，这对后续的代码维护是有积极作用的。
+We had to use `Math.max.apply(null, arr)` because `Math.max(arr)` returns `NaN`. `Math.max()` expects comma-separated arguments, but not an array. The spread operator makes this syntax much better to read and maintain.
```js
const arr = [6, 89, 3, 45];
const maximus = Math.max(...arr); // returns 89
```
-`...arr`返回了一个“打开”的数组。或者说它 *展开* 了数组。 然而，展开操作符只能够在函数的参数中，或者数组之中使用。下面的代码将会报错：
+`...arr` returns an unpacked array. In other words, it *spreads* the array. However, the spread operator only works in-place, like in an argument to a function or in an array literal. The following code will not work:
```js
const spreaded = ...arr; // will throw a syntax error
@@ -32,23 +32,23 @@ const spreaded = ...arr; // will throw a syntax error
# --instructions--
-使用展开操作符将`arr1`中的内容都赋值到`arr2`中去。
+Copy all contents of `arr1` into another array `arr2` using the spread operator.
# --hints--
-`arr2`的值是由`arr1`拷贝而来的。
+`arr2` should be correct copy of `arr1`.
```js
-assert(arr2.every((v, i) => v === arr1[i]));
+assert(arr2.every((v, i) => v === arr1[i]) && arr2.length);
```
-用`...`展开操作符来赋值`arr1`。
+`...` spread operator should be used to duplicate `arr1`.
```js
assert(code.match(/Array\(\s*\.\.\.arr1\s*\)|\[\s*\.\.\.arr1\s*\]/));
```
-当`arr1`改变的时候，`arr2`不会改变。
+`arr2` should remain unchanged when `arr1` is changed.
```js
assert((arr1, arr2) => {
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/write-arrow-functions-with-parameters.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/write-arrow-functions-with-parameters.md
index 13ab4e77c2f..390cb0933cc 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/write-arrow-functions-with-parameters.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/write-arrow-functions-with-parameters.md
@@ -1,6 +1,6 @@
---
id: 587d7b88367417b2b2512b44
-title: 编写带参数的箭头函数
+title: Write Arrow Functions with Parameters
challengeType: 1
forumTopicId: 301223
dashedName: write-arrow-functions-with-parameters
@@ -8,61 +8,63 @@ dashedName: write-arrow-functions-with-parameters
# --description--
-和一般的函数一样，你也可以给箭头函数传递参数。
+Just like a regular function, you can pass arguments into an arrow function.
```js
-// 给传入的数值乘以 2 并返回结果
+// doubles input value and returns it
const doubler = (item) => item * 2;
+doubler(4); // returns 8
```
-如果箭头函数只有一个参数，则可以省略包含该参数的括号。
+If an arrow function has a single parameter, the parentheses enclosing the parameter may be omitted.
```js
-// the same function, without the argument parentheses
+// the same function, without the parameter parentheses
const doubler = item => item * 2;
```
-可以将多个参数传递到箭头函数中。
+It is possible to pass more than one argument into an arrow function.
```js
// multiplies the first input value by the second and returns it
const multiplier = (item, multi) => item * multi;
+multiplier(4, 2); // returns 8
```
# --instructions--
-使用箭头函数的语法重写`myConcat`函数，使其可以将`arr2`的内容填充在`arr1`里。
+Rewrite the `myConcat` function which appends contents of `arr2` to `arr1` so that the function uses arrow function syntax.
# --hints--
-替换掉所有的`var`关键字。
+You should replace the `var` keyword.
```js
(getUserInput) => assert(!getUserInput('index').match(/var/g));
```
-`myConcat`应该是一个常量 (使用`const`)。
+`myConcat` should be a constant variable (by using `const`).
```js
(getUserInput) => assert(getUserInput('index').match(/const\s+myConcat/g));
```
-`myConcat`应该是一个函数。
+`myConcat` should be an arrow function with two parameters
```js
-assert(typeof myConcat === 'function');
+assert(
+ /myConcat=\(\w+,\w+\)=>/.test(code.replace(/\s/g, '')) &&
+ typeof myConcat === 'function'
+);
```
-`myConcat()` 应该返回 `[1, 2, 3, 4, 5]`。
+`myConcat()` should return `[1, 2, 3, 4, 5]`.
```js
-assert(() => {
- const a = myConcat([1], [2]);
- return a[0] == 1 && a[1] == 2;
-});
+assert.deepEqual(myConcat([1, 2], [3, 4, 5]), [1, 2, 3, 4, 5]);
```
-不要使用`function`关键字。
+`function` keyword should not be used.
```js
(getUserInput) => assert(!getUserInput('index').match(/function/g));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/write-concise-declarative-functions-with-es6.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/write-concise-declarative-functions-with-es6.md
index 44d6f7ea243..b7bd2904d89 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/write-concise-declarative-functions-with-es6.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/write-concise-declarative-functions-with-es6.md
@@ -1,6 +1,6 @@
---
id: 587d7b8b367417b2b2512b50
-title: 用 ES6 编写简洁的函数声明
+title: Write Concise Declarative Functions with ES6
challengeType: 1
forumTopicId: 301224
dashedName: write-concise-declarative-functions-with-es6
@@ -8,7 +8,7 @@ dashedName: write-concise-declarative-functions-with-es6
# --description--
-在 ES5 中，当我们需要在对象中定义一个函数的时候，我们必须如下面这般使用`function`关键字：
+When defining functions within objects in ES5, we have to use the keyword `function` as follows:
```js
const person = {
@@ -19,7 +19,7 @@ const person = {
};
```
-在 ES6 语法的对象中定义函数的时候，你可以完全删除`function`关键字和冒号。请看以下例子：
+With ES6, You can remove the `function` keyword and colon altogether when defining functions in objects. Here's an example of this syntax:
```js
const person = {
@@ -32,17 +32,17 @@ const person = {
# --instructions--
-使用以上这种简短的语法，重构在`bicycle`对象中的`setGear`函数。
+Refactor the function `setGear` inside the object `bicycle` to use the shorthand syntax described above.
# --hints--
-不应使用`function`关键字定义方法。
+Traditional function expression should not be used.
```js
-(getUserInput) => assert(!removeJSComments(code).match(/function/));
+(getUserInput) => assert(!__helpers.removeJSComments(code).match(/function/));
```
-`setGear`应是一个函数。
+`setGear` should be a declarative function.
```js
assert(
@@ -50,7 +50,7 @@ assert(
);
```
-执行`bicycle.setGear(48)`应可以让`gear`的值变为 48。
+`bicycle.setGear(48)` should change the `gear` value to 48.
```js
assert(new bicycle.setGear(48).gear === 48);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/write-concise-object-literal-declarations-using-object-property-shorthand.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/write-concise-object-literal-declarations-using-object-property-shorthand.md
index e6210607b94..33dabc5d3a5 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/write-concise-object-literal-declarations-using-object-property-shorthand.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/es6/write-concise-object-literal-declarations-using-object-property-shorthand.md
@@ -1,6 +1,6 @@
---
id: 587d7b8a367417b2b2512b4f
-title: 使用简单字段编写简洁的对象字面量声明
+title: Write Concise Object Literal Declarations Using Object Property Shorthand
challengeType: 1
forumTopicId: 301225
dashedName: write-concise-object-literal-declarations-using-object-property-shorthand
@@ -8,9 +8,9 @@ dashedName: write-concise-object-literal-declarations-using-object-property-shor
# --description--
-ES6 添加了一些很棒的功能，以便于更方便地定义对象。
+ES6 adds some nice support for easily defining object literals.
-请看以下代码：
+Consider the following code:
```js
const getMousePosition = (x, y) => ({
@@ -19,7 +19,7 @@ const getMousePosition = (x, y) => ({
});
```
-`getMousePosition`是一个返回了拥有2个属性的对象的简单函数。 ES6 提供了一个语法糖，消除了类似`x: x`这种冗余的写法.你可以仅仅只写一次`x`，解释器会自动将其转换成`x: x`。 下面是使用这种语法重写的同样的函数：
+`getMousePosition` is a simple function that returns an object containing two properties. ES6 provides the syntactic sugar to eliminate the redundancy of having to write `x: x`. You can simply write `x` once, and it will be converted to`x: x` (or something equivalent) under the hood. Here is the same function from above rewritten to use this new syntax:
```js
const getMousePosition = (x, y) => ({ x, y });
@@ -27,11 +27,11 @@ const getMousePosition = (x, y) => ({ x, y });
# --instructions--
-请使用简单属性对象的语法来创建并返回一个`Person`对象。
+Use object property shorthand with object literals to create and return an object with `name`, `age` and `gender` properties.
# --hints--
-输出是`{name: "Zodiac Hasbro", age: 56, gender: "male"}`。
+`createPerson("Zodiac Hasbro", 56, "male")` should return `{name: "Zodiac Hasbro", age: 56, gender: "male"}`.
```js
assert.deepEqual(
@@ -40,7 +40,7 @@ assert.deepEqual(
);
```
-不要使用`key:value`。
+Your code should not use `key:value`.
```js
(getUserInput) => assert(!getUserInput('index').match(/:/g));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/add-elements-to-the-end-of-an-array-using-concat-instead-of-push.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/add-elements-to-the-end-of-an-array-using-concat-instead-of-push.md
index 340f66b5c52..2f689c3b728 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/add-elements-to-the-end-of-an-array-using-concat-instead-of-push.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/add-elements-to-the-end-of-an-array-using-concat-instead-of-push.md
@@ -1,6 +1,6 @@
---
id: 587d7da9367417b2b2512b67
-title: 使用 concat 而不是 push 将元素添加到数组的末尾
+title: Add Elements to the End of an Array Using concat Instead of push
challengeType: 1
forumTopicId: 301226
dashedName: add-elements-to-the-end-of-an-array-using-concat-instead-of-push
@@ -8,9 +8,9 @@ dashedName: add-elements-to-the-end-of-an-array-using-concat-instead-of-push
# --description--
-函数式编程就是创建和使用具有不变性的函数。
+Functional programming is all about creating and using non-mutating functions.
-上一个挑战介绍了`concat`方法，这是一种在不改变原始数组的前提下，将数组组合成新数组的方法。将`concat`方法与`push`方法做比较，`Push`将元素添加到调用它的数组的末尾，这样会改变该数组。举个例子：
+The last challenge introduced the `concat` method as a way to combine arrays into a new one without mutating the original arrays. Compare `concat` to the `push` method. `Push` adds an item to the end of the same array it is called on, which mutates that array. Here's an example:
```js
var arr = [1, 2, 3];
@@ -19,39 +19,39 @@ arr.push([4, 5, 6]);
// Not the functional programming way
```
-`Concat`方法可以将新项目添加到数组末尾，而不产生任何变化。
+`Concat` offers a way to add new items to the end of an array without any mutating side effects.
# --instructions--
-修改`nonMutatingPush`函数，用`concat`替代`push`将`newItem`添加到`original`末尾，该函数应返回一个数组。
+Change the `nonMutatingPush` function so it uses `concat` to add `newItem` to the end of `original` instead of `push`. The function should return an array.
# --hints--
-应该使用`concat`方法。
+Your code should use the `concat` method.
```js
assert(code.match(/\.concat/g));
```
-不能使用`push`方法。
+Your code should not use the `push` method.
```js
-assert(!code.match(/\.push/g));
+assert(!code.match(/\.?[\s\S]*?push/g));
```
-不能改变`first`数组。
+The `first` array should not change.
```js
assert(JSON.stringify(first) === JSON.stringify([1, 2, 3]));
```
-不能改变`second`数组。
+The `second` array should not change.
```js
assert(JSON.stringify(second) === JSON.stringify([4, 5]));
```
-`nonMutatingPush([1, 2, 3], [4, 5])`应返回`[1, 2, 3, 4, 5]`。
+`nonMutatingPush([1, 2, 3], [4, 5])` should return `[1, 2, 3, 4, 5]`.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/apply-functional-programming-to-convert-strings-to-url-slugs.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/apply-functional-programming-to-convert-strings-to-url-slugs.md
index ed507170a1d..bec46bed96e 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/apply-functional-programming-to-convert-strings-to-url-slugs.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/apply-functional-programming-to-convert-strings-to-url-slugs.md
@@ -1,6 +1,6 @@
---
id: 587d7dab367417b2b2512b6d
-title: 应用函数式编程将字符串转换为URL片段
+title: Apply Functional Programming to Convert Strings to URL Slugs
challengeType: 1
forumTopicId: 301227
dashedName: apply-functional-programming-to-convert-strings-to-url-slugs
@@ -8,51 +8,45 @@ dashedName: apply-functional-programming-to-convert-strings-to-url-slugs
# --description--
-最后几个挑战中涵盖了许多符合函数式编程原则并在处理数组和字符串中非常有用的方法。我们还学习了强大的、可以将问题简化为更简单形式的`reduce`方法，从计算平均值到排序，任何数组操作都可以用它来实现。回想一下，`map`和`filter`方法都是`reduce`的特殊实现。
+The last several challenges covered a number of useful array and string methods that follow functional programming principles. We've also learned about `reduce`, which is a powerful method used to reduce problems to simpler forms. From computing averages to sorting, any array operation can be achieved by applying it. Recall that `map` and `filter` are special cases of `reduce`.
-让我们把学到的知识结合起来解决一个实际问题。
+Let's combine what we've learned to solve a practical problem.
-许多内容管理站点（CMS）为了让添加书签更简单，会将帖子的标题添加到 URL 上。举个例子，如果你写了一篇标题为 "Stop Using Reduce" 的帖子，URL很可能会包含标题字符串的某种形式 (如：".../stop-using-reduce")，你可能已经在 freeCodeCamp 网站上注意到了这一点。
+Many content management sites (CMS) have the titles of a post added to part of the URL for simple bookmarking purposes. For example, if you write a Medium post titled "Stop Using Reduce", it's likely the URL would have some form of the title string in it (".../stop-using-reduce"). You may have already noticed this on the freeCodeCamp site.
# --instructions--
-填写`urlSlug`函数，将字符串`title`转换成带有连字符号的 URL。您可以使用本节中介绍的任何方法，但不要用`replace`方法。以下是本次挑战的要求：
+Fill in the `urlSlug` function so it converts a string `title` and returns the hyphenated version for the URL. You can use any of the methods covered in this section, and don't use `replace`. Here are the requirements:
-输入包含空格和标题大小写单词的字符串
+The input is a string with spaces and title-cased words
-输出字符串，单词之间的空格用连字符(`-`)替换
+The output is a string with the spaces between words replaced by a hyphen (`-`)
-输出应该是小写字母
+The output should be all lower-cased letters
-输出不应有任何空格
+The output should not have any spaces
# --hints--
-`globalTitle`变量应保持不变。
+Your code should not use the `replace` method for this challenge.
```js
-assert(globalTitle === 'Winter Is Coming');
+assert(!code.match(/\.?[\s\S]*?replace/g));
```
-在此挑战中不能使用`replace`方法。
-
-```js
-assert(!code.match(/\.replace/g));
-```
-
-`urlSlug('Winter Is Coming')`应返回`'winter-is-coming'`。
+`urlSlug("Winter Is Coming")` should return `"winter-is-coming"`.
```js
assert(urlSlug('Winter Is Coming') === 'winter-is-coming');
```
-`urlSlug(' Winter Is Coming')`应返回`'winter-is-coming'`。
+`urlSlug(" Winter Is Coming")` should return `"winter-is-coming"`.
```js
assert(urlSlug(' Winter Is Coming') === 'winter-is-coming');
```
-`urlSlug('A Mind Needs Books Like A Sword Needs A Whetstone')`应返回`'a-mind-needs-books-like-a-sword-needs-a-whetstone'`。
+`urlSlug("A Mind Needs Books Like A Sword Needs A Whetstone")` should return `"a-mind-needs-books-like-a-sword-needs-a-whetstone"`.
```js
assert(
@@ -61,7 +55,7 @@ assert(
);
```
-`urlSlug('Hold The Door')`应返回`'hold-the-door'`。
+`urlSlug("Hold The Door")` should return `"hold-the-door"`.
```js
assert(urlSlug('Hold The Door') === 'hold-the-door');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/avoid-mutations-and-side-effects-using-functional-programming.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/avoid-mutations-and-side-effects-using-functional-programming.md
index beb8aafa9d7..0f401beeab6 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/avoid-mutations-and-side-effects-using-functional-programming.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/avoid-mutations-and-side-effects-using-functional-programming.md
@@ -1,6 +1,6 @@
---
id: 587d7b8e367417b2b2512b5e
-title: 使用函数式编程避免变化和副作用
+title: Avoid Mutations and Side Effects Using Functional Programming
challengeType: 1
forumTopicId: 301228
dashedName: avoid-mutations-and-side-effects-using-functional-programming
@@ -8,34 +8,47 @@ dashedName: avoid-mutations-and-side-effects-using-functional-programming
# --description--
-如果你还没想通，上一个挑战的问题出在`tabClose()`函数里的`splice`。不幸的是，`splice`修改了调用它的原始数组，所以第二次调用它时是基于修改后的数组，才给出了意料之外的结果。
+If you haven't already figured it out, the issue in the previous challenge was with the `splice` call in the `tabClose()` function. Unfortunately, `splice` changes the original array it is called on, so the second call to it used a modified array, and gave unexpected results.
-这是一个小例子，还有更广义的定义——在变量，数组或对象上调用一个函数，这个函数会改变对象中的变量或其他东西。
+This is a small example of a much larger pattern - you call a function on a variable, array, or an object, and the function changes the variable or something in the object.
-函数式编程的核心原则之一是不改变任何东西。变化会导致错误。如果一个函数不改变传入的参数、全局变量等数据，那么它造成问题的可能性就会小很多。
+One of the core principles of functional programming is to not change things. Changes lead to bugs. It's easier to prevent bugs knowing that your functions don't change anything, including the function arguments or any global variable.
-前面的例子没有任何复杂的操作，但是`splice`方法改变了原始数组，导致 bug 产生。
+The previous example didn't have any complicated operations but the `splice` method changed the original array, and resulted in a bug.
-回想一下，在函数式编程中，改变或变更叫做`mutation`，这种改变的结果叫做“副作用”（`side effect`）。理想情况下，函数应该是不会产生任何副作用的`纯函数`。
+Recall that in functional programming, changing or altering things is called <dfn>mutation</dfn>, and the outcome is called a <dfn>side effect</dfn>. A function, ideally, should be a <dfn>pure function</dfn>, meaning that it does not cause any side effects.
-让我们尝试掌握这个原则：不要改变代码中的任何变量或对象。
+Let's try to master this discipline and not alter any variable or object in our code.
# --instructions--
-填写`incrementer`函数的代码，使其返回全局变量`fixedValue`的值增加 1。
+Fill in the code for the function `incrementer` so it returns the value of the global variable `fixedValue` increased by one.
# --hints--
-`incrementer`函数不能改变`fixedValue`的值。
+Your function `incrementer` should not change the value of `fixedValue` (which is `4`).
```js
+incrementer();
assert(fixedValue === 4);
```
-`incrementer`函数应返回比`fixedValue`变量更大的值。
+Your `incrementer` function should return a value that is one larger than the `fixedValue` value.
```js
-assert(newValue === 5);
+const __newValue = incrementer();
+assert(__newValue === 5);
+```
+
+Your `incrementer` function should return a value based on the global `fixedValue` variable value.
+
+```js
+(function () {
+ fixedValue = 10;
+ const newValue = incrementer();
+ assert(fixedValue === 10 && newValue === 11);
+ fixedValue = 4;
+})();
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/combine-an-array-into-a-string-using-the-join-method.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/combine-an-array-into-a-string-using-the-join-method.md
index 46338dea576..c1b662432f1 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/combine-an-array-into-a-string-using-the-join-method.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/combine-an-array-into-a-string-using-the-join-method.md
@@ -1,6 +1,6 @@
---
id: 587d7daa367417b2b2512b6c
-title: 使用 join 方法将数组组合成字符串
+title: Combine an Array into a String Using the join Method
challengeType: 1
forumTopicId: 18221
dashedName: combine-an-array-into-a-string-using-the-join-method
@@ -8,9 +8,9 @@ dashedName: combine-an-array-into-a-string-using-the-join-method
# --description--
-`join`方法用来把数组中的所有元素放入一个字符串，并通过指定的分隔符参数进行分隔。
+The `join` method is used to join the elements of an array together to create a string. It takes an argument for the delimiter that is used to separate the array elements in the string.
-举个例子：
+Here's an example:
```js
var arr = ["Hello", "World"];
@@ -20,35 +20,35 @@ var str = arr.join(" ");
# --instructions--
-在函数`sentensify`内用`join`方法（及其他方法）用字符串`str`中的单词造句，这个函数应返回一个字符串。举个例子，"I-like-Star-Wars" 会被转换成 "I like Star Wars"。在此挑战中请勿使用`replace`方法。
+Use the `join` method (among others) inside the `sentensify` function to make a sentence from the words in the string `str`. The function should return a string. For example, "I-like-Star-Wars" would be converted to "I like Star Wars". For this challenge, do not use the `replace` method.
# --hints--
-应使用`join`方法。
+Your code should use the `join` method.
```js
assert(code.match(/\.join/g));
```
-不能使用`replace`方法。
+Your code should not use the `replace` method.
```js
-assert(!code.match(/\.replace/g));
+assert(!code.match(/\.?[\s\S]*?replace/g));
```
-`sentensify('May-the-force-be-with-you')`应返回一个字符串
+`sentensify("May-the-force-be-with-you")` should return a string.
```js
assert(typeof sentensify('May-the-force-be-with-you') === 'string');
```
-`sentensify('May-the-force-be-with-you')`应返回`'May the force be with you'`。
+`sentensify("May-the-force-be-with-you")` should return `"May the force be with you"`.
```js
assert(sentensify('May-the-force-be-with-you') === 'May the force be with you');
```
-`sentensify('The.force.is.strong.with.this.one')`应返回`'The force is strong with this one'`。
+`sentensify("The.force.is.strong.with.this.one")` should return `"The force is strong with this one"`.
```js
assert(
@@ -57,7 +57,7 @@ assert(
);
```
-`sentensify('There,has,been,an,awakening')`应返回`'There has been an awakening'`。
+`sentensify("There,has,been,an,awakening")` should return `"There has been an awakening"`.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/combine-two-arrays-using-the-concat-method.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/combine-two-arrays-using-the-concat-method.md
index fbf018dd17c..4bfd0ef5c31 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/combine-two-arrays-using-the-concat-method.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/combine-two-arrays-using-the-concat-method.md
@@ -1,6 +1,6 @@
---
id: 587d7da9367417b2b2512b66
-title: 使用 concat 方法组合两个数组
+title: Combine Two Arrays Using the concat Method
challengeType: 1
forumTopicId: 301229
dashedName: combine-two-arrays-using-the-concat-method
@@ -8,38 +8,38 @@ dashedName: combine-two-arrays-using-the-concat-method
# --description--
-`Concatenation`意思是将元素连接到尾部。同理，JavaScript 为字符串和数组提供了`concat`方法。对数组来说，在一个数组上调用`concat`方法，然后提供另一个数组作为参数添加到第一个数组末尾，返回一个新数组，不会改变任何一个原始数组。举个例子：
+<dfn>Concatenation</dfn> means to join items end to end. JavaScript offers the `concat` method for both strings and arrays that work in the same way. For arrays, the method is called on one, then another array is provided as the argument to `concat`, which is added to the end of the first array. It returns a new array and does not mutate either of the original arrays. Here's an example:
```js
[1, 2, 3].concat([4, 5, 6]);
-// 返回新数组 [1, 2, 3, 4, 5, 6]
+// Returns a new array [1, 2, 3, 4, 5, 6]
```
# --instructions--
-在`nonMutatingConcat`函数里使用`concat`，将`attach`拼接到`original`尾部，返回拼接后的数组。
+Use the `concat` method in the `nonMutatingConcat` function to concatenate `attach` to the end of `original`. The function should return the concatenated array.
# --hints--
-应该使用`concat`方法。
+Your code should use the `concat` method.
```js
assert(code.match(/\.concat/g));
```
-不能改变`first`数组。
+The `first` array should not change.
```js
assert(JSON.stringify(first) === JSON.stringify([1, 2, 3]));
```
-不能改变`second`数组。
+The `second` array should not change.
```js
assert(JSON.stringify(second) === JSON.stringify([4, 5]));
```
-`nonMutatingConcat([1, 2, 3], [4, 5])`应返回`[1, 2, 3, 4, 5]`。
+`nonMutatingConcat([1, 2, 3], [4, 5])` should return `[1, 2, 3, 4, 5]`.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/implement-map-on-a-prototype.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/implement-map-on-a-prototype.md
index db109e9271e..b1098b2b28a 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/implement-map-on-a-prototype.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/implement-map-on-a-prototype.md
@@ -1,6 +1,6 @@
---
id: 587d7b8f367417b2b2512b62
-title: 在原型上实现 map 方法
+title: Implement map on a Prototype
challengeType: 1
forumTopicId: 301230
dashedName: implement-map-on-a-prototype
@@ -8,30 +8,28 @@ dashedName: implement-map-on-a-prototype
# --description--
-我们之前用`map`方法（即`Array.prototype.map()`）返回一个与调用它的数组长度相同的数组。只要它的回调函数不改变原始数组，它就不会改变原始数组。
+As you have seen from applying `Array.prototype.map()`, or simply `map()` earlier, the `map` method returns an array of the same length as the one it was called on. It also doesn't alter the original array, as long as its callback function doesn't.
-换句话说，`map`是一个纯函数，它的输出仅取决于输入的数组和作为参数传入的回调函数。
+In other words, `map` is a pure function, and its output depends solely on its inputs. Plus, it takes another function as its argument.
-为了加深对`map`方法的理解，现在我们来用`for`或`Array.prototype.forEach()`自己实现一下这个方法。
-
-注意：纯函数可以改变其作用域内定义的局部变量，但我们最好不要这样做。
+You might learn a lot about the `map` method if you implement your own version of it. It is recommended you use a `for` loop or `Array.prototype.forEach()`.
# --instructions--
-写一个和`Array.prototype.map()`一样的`Array.prototype.myMap()`。你可以用`for`循环或者`forEach`方法。
+Write your own `Array.prototype.myMap()`, which should behave exactly like `Array.prototype.map()`. You should not use the built-in `map` method. The `Array` instance can be accessed in the `myMap` method using `this`.
# --hints--
-`new_s`应等于`[46, 130, 196, 10]`。
+`new_s` should equal `[46, 130, 196, 10]`.
```js
assert(JSON.stringify(new_s) === JSON.stringify([46, 130, 196, 10]));
```
-不能使用`map`方法。
+Your code should not use the `map` method.
```js
-assert(!code.match(/\.map/g));
+assert(!code.match(/\.?[\s\S]*?map/g));
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/implement-the-filter-method-on-a-prototype.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/implement-the-filter-method-on-a-prototype.md
index 1da28069fe0..a9469de9772 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/implement-the-filter-method-on-a-prototype.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/implement-the-filter-method-on-a-prototype.md
@@ -1,6 +1,6 @@
---
id: 587d7b8f367417b2b2512b64
-title: 在原型上实现 filter 方法
+title: Implement the filter Method on a Prototype
challengeType: 1
forumTopicId: 301231
dashedName: implement-the-filter-method-on-a-prototype
@@ -8,26 +8,24 @@ dashedName: implement-the-filter-method-on-a-prototype
# --description--
-为了加深对`filter`的理解，现在我们来自己实现一下`Array.prototype.filter()`方法。可以用`for`循环或`Array.prototype.forEach()`。
-
-请注意：纯函数可以改变其作用域内定义的局部变量，但我们最好不要这样做。
+You might learn a lot about the `filter` method if you implement your own version of it. It is recommended you use a `for` loop or `Array.prototype.forEach()`.
# --instructions--
-编写一个和`Array.prototype.filter()`功能一样的`Array.prototype.myFilter()`方法。你可以用`for`循环或`Array.prototype.forEach()`方法。
+Write your own `Array.prototype.myFilter()`, which should behave exactly like `Array.prototype.filter()`. You should not use the built-in `filter` method. The `Array` instance can be accessed in the `myFilter` method using `this`.
# --hints--
-`new_s`应等于`[23, 65, 5]`。
+`new_s` should equal `[23, 65, 5]`.
```js
assert(JSON.stringify(new_s) === JSON.stringify([23, 65, 5]));
```
-不能使用`filter`方法。
+Your code should not use the `filter` method.
```js
-assert(!code.match(/\.filter/g));
+assert(!code.match(/\.?[\s\S]*?filter/g));
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/introduction-to-currying-and-partial-application.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/introduction-to-currying-and-partial-application.md
index 7494df5df42..1d3651620f2 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/introduction-to-currying-and-partial-application.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/introduction-to-currying-and-partial-application.md
@@ -1,6 +1,6 @@
---
id: 587d7dab367417b2b2512b70
-title: 函数柯里化
+title: Introduction to Currying and Partial Application
challengeType: 1
forumTopicId: 301232
dashedName: introduction-to-currying-and-partial-application
@@ -8,11 +8,11 @@ dashedName: introduction-to-currying-and-partial-application
# --description--
-`arity`是函数所需的形参的数量。函数`柯里化`意思是把接受多个`arity`的函数变换成接受单一`arity`的函数。
+The <dfn>arity</dfn> of a function is the number of arguments it requires. <dfn>Currying</dfn> a function means to convert a function of N arity into N functions of arity 1.
-换句话说，就是重构函数让它接收一个参数，然后返回接收下一个参数的函数，依此类推。
+In other words, it restructures a function so it takes one argument, then returns another function that takes the next argument, and so on.
-举个例子：
+Here's an example:
```js
//Un-curried function
@@ -20,7 +20,7 @@ function unCurried(x, y) {
return x + y;
}
-//柯里化函数
+//Curried function
function curried(x) {
return function(y) {
return x + y;
@@ -29,10 +29,10 @@ function curried(x) {
//Alternative using ES6
const curried = x => y => x + y
-curried(1)(2) // 返回 3
+curried(1)(2) // Returns 3
```
-柯里化在不能一次为函数提供所有参数情况下很有用。因为它可以将每个函数的调用保存到一个变量中，该变量将保存返回的函数引用，该引用在下一个参数可用时接受该参数。下面是使用`柯里化`函数的例子：
+This is useful in your program if you can't supply all the arguments to a function at one time. You can save each function call into a variable, which will hold the returned function reference that takes the next argument when it's available. Here's an example using the curried function in the example above:
```js
// Call a curried function in parts:
@@ -40,7 +40,7 @@ var funcForY = curried(1);
console.log(funcForY(2)); // Prints 3
```
-类似地，`局部应用`的意思是一次对一个函数应用几个参数，然后返回另一个应用更多参数的函数。 举个例子：
+Similarly, <dfn>partial application</dfn> can be described as applying a few arguments to a function at a time and returning another function that is applied to more arguments. Here's an example:
```js
//Impartial function
@@ -53,29 +53,29 @@ partialFn(10); // Returns 13
# --instructions--
-填写`add`函数主体部分，用柯里化添加参数`x`，`y`和`z`.
+Fill in the body of the `add` function so it uses currying to add parameters `x`, `y`, and `z`.
# --hints--
-`add(10)(20)(30)`应返回`60`。
+`add(10)(20)(30)` should return `60`.
```js
assert(add(10)(20)(30) === 60);
```
-`add(1)(2)(3)`应返回`6`。
+`add(1)(2)(3)` should return `6`.
```js
assert(add(1)(2)(3) === 6);
```
-`add(11)(22)(33)`应返回`66`。
+`add(11)(22)(33)` should return `66`.
```js
assert(add(11)(22)(33) === 66);
```
-应返回`x + y + z`的最终结果。
+Your code should include a final statement that returns `x + y + z`.
```js
assert(code.match(/[xyz]\s*?\+\s*?[xyz]\s*?\+\s*?[xyz]/g));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/learn-about-functional-programming.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/learn-about-functional-programming.md
index 94b0ddeb270..a0cc4d99a36 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/learn-about-functional-programming.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/learn-about-functional-programming.md
@@ -1,6 +1,6 @@
---
id: 587d7b8d367417b2b2512b5b
-title: 学习函数式编程
+title: Learn About Functional Programming
challengeType: 1
forumTopicId: 301233
dashedName: learn-about-functional-programming
@@ -8,33 +8,33 @@ dashedName: learn-about-functional-programming
# --description--
-函数式编程是一种方案简单、功能独立、对作用域外没有任何副作用的编程范式。
+Functional programming is a style of programming where solutions are simple, isolated functions, without any side effects outside of the function scope.
`INPUT -> PROCESS -> OUTPUT`
-函数式编程：
+Functional programming is about:
-1）功能独立——不依赖于程序的状态（比如可能发生变化的全局变量）；
+1) Isolated functions - there is no dependence on the state of the program, which includes global variables that are subject to change
-2）纯函数——同一个输入永远能得到同一个输出；
+2) Pure functions - the same input always gives the same output
-3）有限的副作用——可以严格地限制函数外部对状态的更改。
+3) Functions with limited side effects - any changes, or mutations, to the state of the program outside the function are carefully controlled
# --instructions--
-freeCodeCamp 成员在 love tea 的故事。
+The members of freeCodeCamp happen to love tea.
-在代码编辑器中，已经为你定义好了`prepareTea`和`getTea`函数。调用`getTea`函数为团队准备 40 杯茶，并将它们存储在`tea4TeamFCC`变量里。
+In the code editor, the `prepareTea` and `getTea` functions are already defined for you. Call the `getTea` function to get 40 cups of tea for the team, and store them in the `tea4TeamFCC` variable.
# --hints--
-`tea4TeamFCC`变量里应有 40 杯为团队准备的茶。
+The `tea4TeamFCC` variable should hold 40 cups of tea for the team.
```js
assert(tea4TeamFCC.length === 40);
```
-`tea4TeamFCC`变量里应有 greenTea。
+The `tea4TeamFCC` variable should hold cups of green tea.
```js
assert(tea4TeamFCC[0] === 'greenTea');
@@ -75,7 +75,7 @@ const prepareTea = () => 'greenTea';
const getTea = (numOfCups) => {
const teaCups = [];
-
+
for(let cups = 1; cups <= numOfCups; cups += 1) {
const teaCup = prepareTea();
teaCups.push(teaCup);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/pass-arguments-to-avoid-external-dependence-in-a-function.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/pass-arguments-to-avoid-external-dependence-in-a-function.md
index cac392ad489..19d7d355e92 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/pass-arguments-to-avoid-external-dependence-in-a-function.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/pass-arguments-to-avoid-external-dependence-in-a-function.md
@@ -1,6 +1,6 @@
---
id: 587d7b8e367417b2b2512b5f
-title: 传递参数以避免函数中的外部依赖
+title: Pass Arguments to Avoid External Dependence in a Function
challengeType: 1
forumTopicId: 301234
dashedName: pass-arguments-to-avoid-external-dependence-in-a-function
@@ -8,42 +8,43 @@ dashedName: pass-arguments-to-avoid-external-dependence-in-a-function
# --description--
-上一个挑战是更接近函数式编程原则的挑战，但是仍然缺少一些东西。
+The last challenge was a step closer to functional programming principles, but there is still something missing.
-虽然我们没有改变全局变量值，但在没有全局变量`fixedValue`情况下，`incrementer`函数将不起作用。
+We didn't alter the global variable value, but the function `incrementer` would not work without the global variable `fixedValue` being there.
-函数式编程的另一个原则是：总是显式声明依赖关系。如果函数依赖于一个变量或对象，那么将该变量或对象作为参数直接传递到函数中。
+Another principle of functional programming is to always declare your dependencies explicitly. This means if a function depends on a variable or object being present, then pass that variable or object directly into the function as an argument.
-这样做会有很多好处，其中一点是让函数更容易测试，因为你确切地知道参数是什么，并且这个参数也不依赖于程序中的任何其他内容。
+There are several good consequences from this principle. The function is easier to test, you know exactly what input it takes, and it won't depend on anything else in your program.
-其次，这样做可以让你更加自信地更改，删除或添加新代码。因为你很清楚哪些是可以改的，哪些是不可以改的，这样你就知道哪里可能会有潜在的陷阱。
+This can give you more confidence when you alter, remove, or add new code. You would know what you can or cannot change and you can see where the potential traps are.
-最后，无论代码的哪一部分执行它，函数总是会为同一组输入生成相同的输出。
+Finally, the function would always produce the same output for the same set of inputs, no matter what part of the code executes it.
# --instructions--
-更新`incrementer`函数，明确声明其依赖项。
+Let's update the `incrementer` function to clearly declare its dependencies.
-编写`incrementer`函数，获取它的参数，然后将值增加 1。
+Write the `incrementer` function so it takes an argument, and then returns a result after increasing the value by one.
# --hints--
-`incrementer`函数不能修改`fixedValue`的值。
+Your function `incrementer` should not change the value of `fixedValue` (which is `4`).
```js
assert(fixedValue === 4);
```
-`incrementer`函数应该接收一个参数。
+Your `incrementer` function should take an argument.
```js
assert(incrementer.length === 1);
```
-`incrementer`函数应返回比`fixedValue`更大的值。
+Your `incrementer` function should return a value that is one larger than the `fixedValue` value.
```js
-assert(newValue === 5);
+const __newValue = incrementer(fixedValue);
+assert(__newValue === 5);
```
# --seed--
@@ -75,5 +76,5 @@ function incrementer (fixedValue) {
// Only change code above this line
}
-
+
```
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/refactor-global-variables-out-of-functions.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/refactor-global-variables-out-of-functions.md
index 3c43bc16b2f..0eea537b775 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/refactor-global-variables-out-of-functions.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/refactor-global-variables-out-of-functions.md
@@ -1,6 +1,6 @@
---
id: 587d7b8f367417b2b2512b60
-title: 在函数中重构全局变量
+title: Refactor Global Variables Out of Functions
challengeType: 1
forumTopicId: 301235
dashedName: refactor-global-variables-out-of-functions
@@ -8,21 +8,23 @@ dashedName: refactor-global-variables-out-of-functions
# --description--
-目前为止，我们已经看到了函数式编程的两个原则：
+So far, we have seen two distinct principles for functional programming:
-1) 不要更改变量或对象——创建新变量和对象，并在需要时从函数返回它们。
+1) Don't alter a variable or object - create new variables and objects and return them if need be from a function. Hint: using something like `var newArr = arrVar`, where `arrVar` is an array will simply create a reference to the existing variable and not a copy. So changing a value in `newArr` would change the value in `arrVar`.
-2) 声明函数参数——函数内的任何计算仅取决于参数，而不取决于任何全局对象或变量。
+2) Declare function parameters - any computation inside a function depends only on the arguments passed to the function, and not on any global object or variable.
-给数字增加 1 不够刺激，我们可以在处理数组或更复杂的对象时应用这些原则。
+Adding one to a number is not very exciting, but we can apply these principles when working with arrays or more complex objects.
# --instructions--
-重构代码，使全局数组`bookList`在函数内部不会被改变。`add`函数可以将指定的`bookName`增加到数组末尾。`remove`函数可以从数组中移除指定`bookName`。两个函数都返回数组，并且任何参数都应该添加到`bookName`前面。
+Rewrite the code so the global array `bookList` is not changed inside either function. The `add` function should add the given `bookName` to the end of the array passed to it and return a new array (list). The `remove` function should remove the given `bookName` from the array passed to it.
+
+**Note:** Both functions should return an array, and any new parameters should be added before the `bookName` parameter.
# --hints--
-`bookList`不应该改变，应等于`['The Hound of the Baskervilles', 'On The Electrodynamics of Moving Bodies', 'Philosophiæ Naturalis Principia Mathematica', 'Disquisitiones Arithmeticae']`.
+`bookList` should not change and still equal `["The Hound of the Baskervilles", "On The Electrodynamics of Moving Bodies", "Philosophiæ Naturalis Principia Mathematica", "Disquisitiones Arithmeticae"]`.
```js
assert(
@@ -36,7 +38,7 @@ assert(
);
```
-`newBookList`应等于`['The Hound of the Baskervilles', 'On The Electrodynamics of Moving Bodies', 'Philosophiæ Naturalis Principia Mathematica', 'Disquisitiones Arithmeticae', 'A Brief History of Time']`.
+`newBookList` should equal `["The Hound of the Baskervilles", "On The Electrodynamics of Moving Bodies", "Philosophiæ Naturalis Principia Mathematica", "Disquisitiones Arithmeticae", "A Brief History of Time"]`.
```js
assert(
@@ -51,7 +53,7 @@ assert(
);
```
-`newerBookList`应等于`['The Hound of the Baskervilles', 'Philosophiæ Naturalis Principia Mathematica', 'Disquisitiones Arithmeticae']`.
+`newerBookList` should equal `["The Hound of the Baskervilles", "Philosophiæ Naturalis Principia Mathematica", "Disquisitiones Arithmeticae"]`.
```js
assert(
@@ -64,7 +66,7 @@ assert(
);
```
-`newestBookList`应等于`['The Hound of the Baskervilles', 'Philosophiæ Naturalis Principia Mathematica', 'Disquisitiones Arithmeticae', 'A Brief History of Time']`.
+`newestBookList` should equal `["The Hound of the Baskervilles", "Philosophiæ Naturalis Principia Mathematica", "Disquisitiones Arithmeticae", "A Brief History of Time"]`.
```js
assert(
@@ -91,7 +93,7 @@ function add (bookName) {
bookList.push(bookName);
return bookList;
-
+
// Change code above this line
}
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/remove-elements-from-an-array-using-slice-instead-of-splice.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/remove-elements-from-an-array-using-slice-instead-of-splice.md
index 4367439e893..082deb8650f 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/remove-elements-from-an-array-using-slice-instead-of-splice.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/remove-elements-from-an-array-using-slice-instead-of-splice.md
@@ -1,6 +1,6 @@
---
id: 9d7123c8c441eeafaeb5bdef
-title: 使用 slice 而不是 splice 从数组中移除元素
+title: Remove Elements from an Array Using slice Instead of splice
challengeType: 1
forumTopicId: 301236
dashedName: remove-elements-from-an-array-using-slice-instead-of-splice
@@ -8,7 +8,7 @@ dashedName: remove-elements-from-an-array-using-slice-instead-of-splice
# --description--
-使用数组时经常遇到要删除一些元素并保留数组剩余部分的情况。为此，JavaScript 提供了`splice`方法，它接收两个参数：从哪里开始删除项目的索引，和要删除的项目数。如果没有提供第二个参数，默认情况下是移除到结尾的元素。但`splice`方法会改变调用它的原始数组。举个例子：
+A common pattern while working with arrays is when you want to remove items and keep the rest of the array. JavaScript offers the `splice` method for this, which takes arguments for the index of where to start removing items, then the number of items to remove. If the second argument is not provided, the default is to remove items through the end. However, the `splice` method mutates the original array it is called on. Here's an example:
```js
var cities = ["Chicago", "Delhi", "Islamabad", "London", "Berlin"];
@@ -16,29 +16,29 @@ cities.splice(3, 1); // Returns "London" and deletes it from the cities array
// cities is now ["Chicago", "Delhi", "Islamabad", "Berlin"]
```
-正如我们在上一次挑战中看到的那样，`slice`方法不会改变原始数组，而是返回一个可以保存到变量中的新数组。回想一下，`slice`方法接收两个参数，从开始索引开始选取到结束（不包括该元素），并在新数组中返回这些元素。使用`slice`方法替代`splice`有助于避免数组变化产生的副作用。
+As we saw in the last challenge, the `slice` method does not mutate the original array, but returns a new one which can be saved into a variable. Recall that the `slice` method takes two arguments for the indices to begin and end the slice (the end is non-inclusive), and returns those items in a new array. Using the `slice` method instead of `splice` helps to avoid any array-mutating side effects.
# --instructions--
-用`slice`代替`splice`重写`nonMutatingSplice`函数。将`cities`数组长度限制为3，并返回一个仅包含前 3 项的新数组。
+Rewrite the function `nonMutatingSplice` by using `slice` instead of `splice`. It should limit the provided `cities` array to a length of 3, and return a new array with only the first three items.
-不要改变提供给函数的原始数组。
+Do not mutate the original array provided to the function.
# --hints--
-应该使用`slice`方法。
+Your code should use the `slice` method.
```js
assert(code.match(/\.slice/g));
```
-不能使用`splice`方法。
+Your code should not use the `splice` method.
```js
-assert(!code.match(/\.splice/g));
+assert(!code.match(/\.?[\s\S]*?splice/g));
```
-不能改变`inputCities`数组。
+The `inputCities` array should not change.
```js
assert(
@@ -47,7 +47,7 @@ assert(
);
```
-`nonMutatingSplice(['Chicago', 'Delhi', 'Islamabad', 'London', 'Berlin'])`应返回`['Chicago', 'Delhi', 'Islamabad']`。
+`nonMutatingSplice(["Chicago", "Delhi", "Islamabad", "London", "Berlin"])` should return `["Chicago", "Delhi", "Islamabad"]`.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/return-a-sorted-array-without-changing-the-original-array.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/return-a-sorted-array-without-changing-the-original-array.md
index 5b551553721..3d77603ca7a 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/return-a-sorted-array-without-changing-the-original-array.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/return-a-sorted-array-without-changing-the-original-array.md
@@ -1,6 +1,6 @@
---
id: 587d7da9367417b2b2512b6a
-title: 在不更改原始数组的前提下返回排序后的数组
+title: Return a Sorted Array Without Changing the Original Array
challengeType: 1
forumTopicId: 301237
dashedName: return-a-sorted-array-without-changing-the-original-array
@@ -8,27 +8,27 @@ dashedName: return-a-sorted-array-without-changing-the-original-array
# --description--
-`sort`方法会产生改变原始数组中元素顺序的副作用。换句话说，它会改变数组的位置。避免这种情况的一种方法是先将空数组连接到正在排序的数组上（记住`concat`返回一个新数组），再用`sort`方法。
+A side effect of the `sort` method is that it changes the order of the elements in the original array. In other words, it mutates the array in place. One way to avoid this is to first concatenate an empty array to the one being sorted (remember that `slice` and `concat` return a new array), then run the `sort` method.
# --instructions--
-在`nonMutatingSort`函数中使用`sort`方法对数组中的元素按升序进行排列。函数不能改变`globalArray`变量，应返回一个新数组。
+Use the `sort` method in the `nonMutatingSort` function to sort the elements of an array in ascending order. The function should return a new array, and not mutate the `globalArray` variable.
# --hints--
-应该使用`sort`方法。
+Your code should use the `sort` method.
```js
assert(nonMutatingSort.toString().match(/\.sort/g));
```
-应该使用`concat`方法。
+The `globalArray` variable should not change.
```js
assert(JSON.stringify(globalArray) === JSON.stringify([5, 6, 3, 2, 9]));
```
-`globalArray`variable 应保持不变。
+`nonMutatingSort(globalArray)` should return `[2, 3, 5, 6, 9]`.
```js
assert(
@@ -37,12 +37,18 @@ assert(
);
```
-`nonMutatingSort(globalArray)`应返回`[2, 3, 5, 6, 9]`。
+`nonMutatingSort(globalArray)` should not be hard coded.
```js
assert(!nonMutatingSort.toString().match(/[23569]/g));
```
+The function should return a new array, not the array passed to it.
+
+```js
+assert(nonMutatingSort(globalArray) !== globalArray);
+```
+
# --seed--
## --seed-contents--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/return-part-of-an-array-using-the-slice-method.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/return-part-of-an-array-using-the-slice-method.md
index a8a5a0323d3..75362aa5016 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/return-part-of-an-array-using-the-slice-method.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/return-part-of-an-array-using-the-slice-method.md
@@ -1,6 +1,6 @@
---
id: 587d7b90367417b2b2512b65
-title: 使用 slice 方法返回数组的一部分
+title: Return Part of an Array Using the slice Method
challengeType: 1
forumTopicId: 301239
dashedName: return-part-of-an-array-using-the-slice-method
@@ -8,9 +8,9 @@ dashedName: return-part-of-an-array-using-the-slice-method
# --description--
-`slice`方法可以从已有数组中返回指定元素。它接受两个参数，第一个规定从何处开始选取，第二个规定从何处结束选取（不包括该元素）。如果没有传参，则默认为从数组的开头开始到结尾结束，这是复制整个数组的简单方式。`slice`返回一个新数组，不会修改原始数组。
+The `slice` method returns a copy of certain elements of an array. It can take two arguments, the first gives the index of where to begin the slice, the second is the index for where to end the slice (and it's non-inclusive). If the arguments are not provided, the default is to start at the beginning of the array through the end, which is an easy way to make a copy of the entire array. The `slice` method does not mutate the original array, but returns a new one.
-举个例子：
+Here's an example:
```js
var arr = ["Cat", "Dog", "Tiger", "Zebra"];
@@ -20,17 +20,17 @@ var newArray = arr.slice(1, 3);
# --instructions--
-在`sliceArray`函数中使用`slice`方法，给出`beginSlice`和`endSlice`索引，返回`anim`数组的一部分，这个函数应返回一个数组。
+Use the `slice` method in the `sliceArray` function to return part of the `anim` array given the provided `beginSlice` and `endSlice` indices. The function should return an array.
# --hints--
-你的代码中应使用`slice`方法。
+Your code should use the `slice` method.
```js
assert(code.match(/\.slice/g));
```
-不能改变`inputAnim`变量。
+The `inputAnim` variable should not change.
```js
assert(
@@ -39,7 +39,7 @@ assert(
);
```
-`sliceArray(['Cat', 'Dog', 'Tiger', 'Zebra', 'Ant'], 1, 3)`应返回`['Dog', 'Tiger']`。
+`sliceArray(["Cat", "Dog", "Tiger", "Zebra", "Ant"], 1, 3)` should return `["Dog", "Tiger"]`.
```js
assert(
@@ -48,7 +48,7 @@ assert(
);
```
-`sliceArray(['Cat', 'Dog', 'Tiger', 'Zebra', 'Ant'], 0, 1)`应返回`['Cat']`。
+`sliceArray(["Cat", "Dog", "Tiger", "Zebra", "Ant"], 0, 1)` should return `["Cat"]`.
```js
assert(
@@ -57,7 +57,7 @@ assert(
);
```
-`sliceArray(['Cat', 'Dog', 'Tiger', 'Zebra', 'Ant'], 1, 4)`应返回`['Dog', 'Tiger', 'Zebra']`。
+`sliceArray(["Cat", "Dog", "Tiger", "Zebra", "Ant"], 1, 4)` should return `["Dog", "Tiger", "Zebra"]`.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/sort-an-array-alphabetically-using-the-sort-method.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/sort-an-array-alphabetically-using-the-sort-method.md
index ce9ce69b268..359ea528a91 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/sort-an-array-alphabetically-using-the-sort-method.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/sort-an-array-alphabetically-using-the-sort-method.md
@@ -1,6 +1,6 @@
---
id: 587d7da9367417b2b2512b69
-title: 使用 sort 方法按字母顺序给数组排序
+title: Sort an Array Alphabetically using the sort Method
challengeType: 1
forumTopicId: 18303
dashedName: sort-an-array-alphabetically-using-the-sort-method
@@ -8,9 +8,9 @@ dashedName: sort-an-array-alphabetically-using-the-sort-method
# --description--
-`sort`方法可以根据回调函数对数组元素进行排序。
+The `sort` method sorts the elements of an array according to the callback function.
-举个例子：
+For example:
```js
function ascendingOrder(arr) {
@@ -30,21 +30,21 @@ reverseAlpha(['l', 'h', 'z', 'b', 's']);
// Returns ['z', 's', 'l', 'h', 'b']
```
-JavaScript 的默认排序方法是 Unicode 值顺序排序，有时可能会得到意想不到的结果。所以，建议传入一个回调函数指定数组项目的排序方式，这个回调函数通常叫做 `compareFunction`，它根据 `compareFunction` 的返回值决定数组元素的排序方式： 如果两个元素 `a` 和 `b`，`compareFunction(a,b)` 返回一个比 0 小的值，那么 `a` 会在 `b` 的前面。 如果两个元素 `a` 和 `b`，`compareFunction(a,b)` 返回一个比 0 大的值，那么 `b` 会在 `a` 的前面。 如果两个元素 `a` 和 `b`，`compareFunction(a,b)` 返回等于 0 的值，那么 `a` 和 `b` 的位置保持不变。
+JavaScript's default sorting method is by string Unicode point value, which may return unexpected results. Therefore, it is encouraged to provide a callback function to specify how to sort the array items. When such a callback function, normally called `compareFunction`, is supplied, the array elements are sorted according to the return value of the `compareFunction`: If `compareFunction(a,b)` returns a value less than 0 for two elements `a` and `b`, then `a` will come before `b`. If `compareFunction(a,b)` returns a value greater than 0 for two elements `a` and `b`, then `b` will come before `a`. If `compareFunction(a,b)` returns a value equal to 0 for two elements `a` and `b`, then `a` and `b` will remain unchanged.
# --instructions--
-在`alphabeticalOrder`函数中使用`sort`方法对`arr`中的元素按照字母顺序排列。
+Use the `sort` method in the `alphabeticalOrder` function to sort the elements of `arr` in alphabetical order.
# --hints--
-应该使用`sort`方法。
+Your code should use the `sort` method.
```js
assert(code.match(/\.sort/g));
```
-`alphabeticalOrder(['a', 'd', 'c', 'a', 'z', 'g'])`应返回`['a', 'a', 'c', 'd', 'g', 'z']`。
+`alphabeticalOrder(["a", "d", "c", "a", "z", "g"])` should return `["a", "a", "c", "d", "g", "z"]`.
```js
assert(
@@ -53,7 +53,7 @@ assert(
);
```
-`alphabeticalOrder(['x', 'h', 'a', 'm', 'n', 'm'])`应返回`['a', 'h', 'm', 'm', 'n', 'x']`。
+`alphabeticalOrder(["x", "h", "a", "m", "n", "m"])` should return `["a", "h", "m", "m", "n", "x"]`.
```js
assert(
@@ -62,7 +62,7 @@ assert(
);
```
-`alphabeticalOrder(['a', 'a', 'a', 'a', 'x', 't'])`应返回`['a', 'a', 'a', 'a', 't', 'x']`。
+`alphabeticalOrder(["a", "a", "a", "a", "x", "t"])` should return `["a", "a", "a", "a", "t", "x"]`.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/split-a-string-into-an-array-using-the-split-method.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/split-a-string-into-an-array-using-the-split-method.md
index f6cf7d1ec82..9f702386e01 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/split-a-string-into-an-array-using-the-split-method.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/split-a-string-into-an-array-using-the-split-method.md
@@ -1,6 +1,6 @@
---
id: 587d7daa367417b2b2512b6b
-title: 使用 split 方法将字符串拆分成数组
+title: Split a String into an Array Using the split Method
challengeType: 1
forumTopicId: 18305
dashedName: split-a-string-into-an-array-using-the-split-method
@@ -8,9 +8,9 @@ dashedName: split-a-string-into-an-array-using-the-split-method
# --description--
-`split`方法用于把字符串分割成字符串数组，接收一个分隔符参数，分隔符可以是用于分解字符串或正则表达式的字符。举个例子，如果分隔符是空格，你会得到一个单词数组；如果分隔符是空字符串，你会得到一个由字符串中每个字符组成的数组。
+The `split` method splits a string into an array of strings. It takes an argument for the delimiter, which can be a character to use to break up the string or a regular expression. For example, if the delimiter is a space, you get an array of words, and if the delimiter is an empty string, you get an array of each character in the string.
-下面是两个用空格分隔一个字符串的例子，另一个是用数字的正则表达式分隔：
+Here are two examples that split one string by spaces, then another by digits using a regular expression:
```js
var str = "Hello World";
@@ -22,21 +22,21 @@ var byDigits = otherString.split(/\d/);
// Sets byDigits to ["How", "are", "you", "today"]
```
-因为字符串是固定的，`split`方法可以更简单的操作它们。
+Since strings are immutable, the `split` method makes it easier to work with them.
# --instructions--
-在`splitify`函数中用`split`方法将`str`分割成单词数组，这个方法应该返回一个数组。单词不一定都是用空格分隔，所以数组中不应包含标点符号。
+Use the `split` method inside the `splitify` function to split `str` into an array of words. The function should return the array. Note that the words are not always separated by spaces, and the array should not contain punctuation.
# --hints--
-应该使用`split`方法。
+Your code should use the `split` method.
```js
assert(code.match(/\.split/g));
```
-`splitify('Hello World,I-am code')`应返回`['Hello', 'World', 'I', 'am', 'code']`。
+`splitify("Hello World,I-am code")` should return `["Hello", "World", "I", "am", "code"]`.
```js
assert(
@@ -45,7 +45,7 @@ assert(
);
```
-`splitify('Earth-is-our home')`应返回`['Earth', 'is', 'our', 'home']`。
+`splitify("Earth-is-our home")` should return `["Earth", "is", "our", "home"]`.
```js
assert(
@@ -54,7 +54,7 @@ assert(
);
```
-`splitify('This.is.a-sentence')`应返回`['This', 'is', 'a', 'sentence']`。
+`splitify("This.is.a-sentence")` should return `["This", "is", "a", "sentence"]`.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/understand-functional-programming-terminology.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/understand-functional-programming-terminology.md
index 09f387a25f5..ee3572d6aec 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/understand-functional-programming-terminology.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/understand-functional-programming-terminology.md
@@ -1,6 +1,6 @@
---
id: 587d7b8e367417b2b2512b5c
-title: 了解函数式编程术语
+title: Understand Functional Programming Terminology
challengeType: 1
forumTopicId: 301240
dashedName: understand-functional-programming-terminology
@@ -8,47 +8,47 @@ dashedName: understand-functional-programming-terminology
# --description--
-FCC 团队需求有变更，现在想要两种茶：绿茶（green tea）和红茶（black tea）。事实证明，用户需求变更是很常见的。
+The FCC Team had a mood swing and now wants two types of tea: green tea and black tea. General Fact: Client mood swings are pretty common.
-基于以上信息，我们需要重构上一节挑战中的`getTea`函数来处理多种茶的请求。我们可以修改`getTea`接受一个函数作为参数，使它能够修改茶的类型。这让`getTea`更灵活，也使需求变更时为程序员提供更多控制权。
+With that information, we'll need to revisit the `getTea` function from last challenge to handle various tea requests. We can modify `getTea` to accept a function as a parameter to be able to change the type of tea it prepares. This makes `getTea` more flexible, and gives the programmer more control when client requests change.
-首先，我们将介绍一些术语：
+But first, let's cover some functional terminology:
-`Callbacks`是被传递到另一个函数中调用的函数。你应该已经在其他函数中看过这个写法，例如在`filter`中，回调函数告诉 JavaScript 以什么规则过滤数组。
+<dfn>Callbacks</dfn> are the functions that are slipped or passed into another function to decide the invocation of that function. You may have seen them passed to other methods, for example in `filter`, the callback function tells JavaScript the criteria for how to filter an array.
-函数就像其他正常值一样，可以赋值给变量、传递给另一个函数，或从其它函数返回，这种函数叫做`头等`函数。在 JavaScript 中，所有函数都是`头等`函数。
+Functions that can be assigned to a variable, passed into another function, or returned from another function just like any other normal value, are called <dfn>first class</dfn> functions. In JavaScript, all functions are first class functions.
-将函数为参数或返回值的函数叫做`高阶`函数。
+The functions that take a function as an argument, or return a function as a return value are called <dfn>higher order</dfn> functions.
-当函数传递给另一个函数或从另一个函数返回时，那些传入或返回的函数可以叫做`lambda`。
+When the functions are passed in to another function or returned from another function, then those functions which gets passed in or returned can be called a <dfn>lambda</dfn>.
# --instructions--
-准备 27 杯绿茶和 13 杯红茶，分别存入`tea4GreenTeamFCC`和`tea4BlackTeamFCC`变量。请注意，`getTea`函数已经变了，现在它接收一个函数作为第一个参数。
+Prepare 27 cups of green tea and 13 cups of black tea and store them in `tea4GreenTeamFCC` and `tea4BlackTeamFCC` variables, respectively. Note that the `getTea` function has been modified so it now takes a function as the first argument.
-注意：数据（茶的数量）作为最后一个参数。我们将在后面的课程中对此进行更多讨论。
+Note: The data (the number of cups of tea) is supplied as the last argument. We'll discuss this more in later lessons.
# --hints--
-`tea4GreenTeamFCC`变量应存有为团队准备的 27 杯茶。
+The `tea4GreenTeamFCC` variable should hold 27 cups of green tea for the team.
```js
assert(tea4GreenTeamFCC.length === 27);
```
-`tea4GreenTeamFCC`变量应存有绿茶。
+The `tea4GreenTeamFCC` variable should hold cups of green tea.
```js
assert(tea4GreenTeamFCC[0] === 'greenTea');
```
-`tea4BlackTeamFCC`变量应存有 13 杯红茶。
+The `tea4BlackTeamFCC` variable should hold 13 cups of black tea.
```js
assert(tea4BlackTeamFCC.length === 13);
```
-`tea4BlackTeamFCC`变量应存有红茶。
+The `tea4BlackTeamFCC` variable should hold cups of black tea.
```js
assert(tea4BlackTeamFCC[0] === 'blackTea');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/understand-the-hazards-of-using-imperative-code.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/understand-the-hazards-of-using-imperative-code.md
index e99b149713c..f717be6c1e4 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/understand-the-hazards-of-using-imperative-code.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/understand-the-hazards-of-using-imperative-code.md
@@ -1,6 +1,6 @@
---
id: 587d7b8e367417b2b2512b5d
-title: 了解使用命令式编程的危害
+title: Understand the Hazards of Using Imperative Code
challengeType: 1
forumTopicId: 301241
dashedName: understand-the-hazards-of-using-imperative-code
@@ -8,31 +8,31 @@ dashedName: understand-the-hazards-of-using-imperative-code
# --description--
-函数式编程是一种好习惯，它能让代码管理更简单，不受隐藏 bug 影响。在我们开始函数式编程之前，为了更好的突显可能遇到的问题，我们先看看命令式编程。
+Functional programming is a good habit. It keeps your code easy to manage, and saves you from sneaky bugs. But before we get there, let's look at an imperative approach to programming to highlight where you may have issues.
-类似在英语（和许多其他语言）中，命令式时态用于给出命令，编程中的命令式是给计算机一组语句来执行任务。
+In English (and many other languages), the imperative tense is used to give commands. Similarly, an imperative style in programming is one that gives the computer a set of statements to perform a task.
-这些语句通常会改变程序的状态，例如更新全局变量，典型的例子就是写一个 `for` 循环，它给出了迭代数组索引的精确方向。
+Often the statements change the state of the program, like updating global variables. A classic example is writing a `for` loop that gives exact directions to iterate over the indices of an array.
-相反，函数式编程是声明式编程的一种形式，通过调用方法或函数来告诉计算机要做什么。
+In contrast, functional programming is a form of declarative programming. You tell the computer what you want done by calling a method or function.
-JavaScript 提供了许多处理常见任务的方法，所以你无需写出计算机应如何执行它们。例如，你可以用 `map` 函数替代上面提到的 `for` 循环来处理数组迭代。这有助于避免语义错误，如调试章节介绍的"Off By One Errors"。
+JavaScript offers many predefined methods that handle common tasks so you don't need to write out how the computer should perform them. For example, instead of using the `for` loop mentioned above, you could call the `map` method which handles the details of iterating over an array. This helps to avoid semantic errors, like the "Off By One Errors" that were covered in the Debugging section.
-考虑这样的场景：你正在浏览器中浏览网页，并想操作你打开的标签。下面我们来试试用面向对象的思路来描述这种情景。
+Consider the scenario: you are browsing the web in your browser, and want to track the tabs you have opened. Let's try to model this using some simple object-oriented code.
-窗口对象由选项卡组成，通常会打开多个窗口。窗口对象中每个打开网站的标题都保存在一个数组中。在对浏览器进行了如打开新标签、合并窗口、关闭标签之类的操作后，你需要输出所有打开的标签。关掉的标签将从数组中删除，新打开的标签（为简单起见）则添加到数组的末尾。
+A Window object is made up of tabs, and you usually have more than one Window open. The titles of each open site in each Window object is held in an array. After working in the browser (opening new tabs, merging windows, and closing tabs), you want to print the tabs that are still open. Closed tabs are removed from the array and new tabs (for simplicity) get added to the end of it.
-代码编辑器中显示了此功能的实现，其中包含 `tabOpen()`，`tabClose()`，和 `join()` 函数。tabs数组是窗口对象的一部分用于储存打开页面的名称。
+The code editor shows an implementation of this functionality with functions for `tabOpen()`, `tabClose()`, and `join()`. The array `tabs` is part of the Window object that stores the name of the open pages.
# --instructions--
-在编辑器中运行代码。它使用了有副作用的方法，导致输出错误。打开标签的最终列表应该是 `['FB', 'Gitter', 'Reddit', 'Twitter', 'Medium', 'new tab', 'Netflix', 'YouTube', 'Vine', 'GMail', 'Work mail', 'Docs', 'freeCodeCamp', 'new tab']` 但输出会略有不同。
+Examine the code in the editor. It's using a method that has side effects in the program, causing incorrect behaviour. The final list of open tabs, stored in `finalTabs.tabs`, should be `['FB', 'Gitter', 'Reddit', 'Twitter', 'Medium', 'new tab', 'Netflix', 'YouTube', 'Vine', 'GMail', 'Work mail', 'Docs', 'freeCodeCamp', 'new tab']` but the list produced by the code is slightly different.
-修改 `Window.prototype.tabClose` 使其删除正确的标签。
+Change `Window.prototype.tabClose` so that it removes the correct tab.
# --hints--
-`finalTabs.tabs` 应该是 `['FB', 'Gitter', 'Reddit', 'Twitter', 'Medium', 'new tab', 'Netflix', 'YouTube', 'Vine', 'GMail', 'Work mail', 'Docs', 'freeCodeCamp', 'new tab']`
+`finalTabs.tabs` should be `['FB', 'Gitter', 'Reddit', 'Twitter', 'Medium', 'new tab', 'Netflix', 'YouTube', 'Vine', 'GMail', 'Work mail', 'Docs', 'freeCodeCamp', 'new tab']`
```js
assert.deepEqual(finalTabs.tabs, [
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-higher-order-functions-map-filter-or-reduce-to-solve-a-complex-problem.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-higher-order-functions-map-filter-or-reduce-to-solve-a-complex-problem.md
index dc9f28e58d0..bc3fb9f25e2 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-higher-order-functions-map-filter-or-reduce-to-solve-a-complex-problem.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-higher-order-functions-map-filter-or-reduce-to-solve-a-complex-problem.md
@@ -1,6 +1,6 @@
---
id: 587d7b88367417b2b2512b45
-title: 使用高阶函数 map、filter 或者 reduce 来解决复杂问题
+title: 'Use Higher-Order Functions map, filter, or reduce to Solve a Complex Problem'
challengeType: 1
forumTopicId: 301311
dashedName: use-higher-order-functions-map-filter-or-reduce-to-solve-a-complex-problem
@@ -8,42 +8,46 @@ dashedName: use-higher-order-functions-map-filter-or-reduce-to-solve-a-complex-p
# --description--
-已经接触了高阶函数如 `map()`、 `filter()` 和 `reduce()`的使用，是时候用它们来完成一些复杂的挑战了。
+Now that you have worked through a few challenges using higher-order functions like `map()`, `filter()`, and `reduce()`, you now get to apply them to solve a more complex challenge.
# --instructions--
-已经定义了一个函数 `squareList`。任意组合 `map()`、`filter()` 和 `reduce()` 来完成函数，满足以下条件，当传入实数数组（如，`[-3, 4.8, 5, 3, -3.2]`）时，返回*仅*包含正整数的平方的新数组。
+We have defined a function named `squareList`. You need to complete the code for the `squareList` function using any combination of `map()`, `filter()`, and `reduce()` so that it returns a new array containing only the square of *only* the positive integers (decimal numbers are not integers) when an array of real numbers is passed to it. An example of an array containing only real numbers is `[-3, 4.8, 5, 3, -3.2]`.
-**注意：** 函数不应该包含任何形式的 `for` 或者 `while` 循环或者 `forEach()` 函数。
+**Note:** Your function should not use any kind of `for` or `while` loops or the `forEach()` function.
# --hints--
-`squareList` 应该是一个 `function`。
+`squareList` should be a `function`.
```js
assert.typeOf(squareList, 'function'),
'<code>squareList</code> should be a <code>function</code>';
```
-不应该使用 for 或者 while 循环或者 forEach。
+`for`, `while`, and `forEach` should not be used.
```js
-assert(!removeJSComments(code).match(/for|while|forEach/g));
+assert(!__helpers.removeJSComments(code).match(/for|while|forEach/g));
```
-应该使用 `map`、`filter` 或者 `reduce`。
+`map`, `filter`, or `reduce` should be used.
```js
-assert(removeJSComments(code).match(/\.(map|filter|reduce)\s*\(/g));
+assert(
+ __helpers
+ .removeWhiteSpace(__helpers.removeJSComments(code))
+ .match(/\.(map|filter|reduce)\(/g)
+);
```
-函数应该返回 `array`。
+The function should return an `array`.
```js
assert(Array.isArray(squareList([4, 5.6, -9.8, 3.14, 42, 6, 8.34, -2])));
```
-`squareList([4, 5.6, -9.8, 3.14, 42, 6, 8.34, -2])` 应该返回 `[16, 1764, 36]`。
+`squareList([4, 5.6, -9.8, 3.14, 42, 6, 8.34, -2])` should return `[16, 1764, 36]`.
```js
assert.deepStrictEqual(squareList([4, 5.6, -9.8, 3.14, 42, 6, 8.34, -2]), [
@@ -53,7 +57,7 @@ assert.deepStrictEqual(squareList([4, 5.6, -9.8, 3.14, 42, 6, 8.34, -2]), [
]);
```
-`squareList([-3.7, -5, 3, 10, 12.5, 7, -4.5, -17, 0.3])` 应该返回 `[9, 100, 49]`。
+`squareList([-3.7, -5, 3, 10, 12.5, 7, -4.5, -17, 0.3])` should return `[9, 100, 49]`.
```js
assert.deepStrictEqual(squareList([-3.7, -5, 3, 10, 12.5, 7, -4.5, -17, 0.3]), [
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-every-method-to-check-that-every-element-in-an-array-meets-a-criteria.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-every-method-to-check-that-every-element-in-an-array-meets-a-criteria.md
index 827b0dc8dec..4e63ab1df30 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-every-method-to-check-that-every-element-in-an-array-meets-a-criteria.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-every-method-to-check-that-every-element-in-an-array-meets-a-criteria.md
@@ -1,6 +1,6 @@
---
id: 587d7dab367417b2b2512b6e
-title: 使用 every 方法检查数组中的每个元素是否符合条件
+title: Use the every Method to Check that Every Element in an Array Meets a Criteria
challengeType: 1
forumTopicId: 301312
dashedName: use-the-every-method-to-check-that-every-element-in-an-array-meets-a-criteria
@@ -8,9 +8,9 @@ dashedName: use-the-every-method-to-check-that-every-element-in-an-array-meets-a
# --description--
-`every`方法用于检测数组中*所有*元素是否都符合指定条件。如果所有元素满足条件，返回布尔值`true`，反之返回`false`。
+The `every` method works with arrays to check if *every* element passes a particular test. It returns a Boolean value - `true` if all values meet the criteria, `false` if not.
-举个例子，下面的代码检测数组`numbers`的所有元素是否都小于 10：
+For example, the following code would check if every element in the `numbers` array is less than 10:
```js
var numbers = [1, 5, 8, 0, 10, 11];
@@ -22,29 +22,29 @@ numbers.every(function(currentValue) {
# --instructions--
-在`checkPositive`函数中使用`every`方法检查`arr`中是否所有元素都是正数，函数应返回一个布尔值。
+Use the `every` method inside the `checkPositive` function to check if every element in `arr` is positive. The function should return a Boolean value.
# --hints--
-应使用`every`方法。
+Your code should use the `every` method.
```js
assert(code.match(/\.every/g));
```
-`checkPositive([1, 2, 3, -4, 5])`应返回`false`。
+`checkPositive([1, 2, 3, -4, 5])` should return `false`.
```js
assert.isFalse(checkPositive([1, 2, 3, -4, 5]));
```
-`checkPositive([1, 2, 3, 4, 5])`应返回`true`。
+`checkPositive([1, 2, 3, 4, 5])` should return `true`.
```js
assert.isTrue(checkPositive([1, 2, 3, 4, 5]));
```
-`checkPositive([1, -2, 3, -4, 5])`应返回`false`。
+`checkPositive([1, -2, 3, -4, 5])` should return `false`.
```js
assert.isFalse(checkPositive([1, -2, 3, -4, 5]));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-filter-method-to-extract-data-from-an-array.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-filter-method-to-extract-data-from-an-array.md
index f720f42431e..0942be39242 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-filter-method-to-extract-data-from-an-array.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-filter-method-to-extract-data-from-an-array.md
@@ -1,6 +1,6 @@
---
id: 587d7b8f367417b2b2512b63
-title: 使用 filter 方法从数组中提取数据
+title: Use the filter Method to Extract Data from an Array
challengeType: 1
forumTopicId: 18179
dashedName: use-the-filter-method-to-extract-data-from-an-array
@@ -8,13 +8,13 @@ dashedName: use-the-filter-method-to-extract-data-from-an-array
# --description--
-另一个有用的数组方法是`filter()`（即`Array.prototype.filter()`）。`filter`方法会返回一个长度不大于原始数组的新数组。
+Another useful array function is `Array.prototype.filter()`, or simply `filter()`.
-和`map`一样，`Filter`不会改变原始数组，它接收一个回调函数，将回调内的逻辑应用于数组的每个元素。新数组包含根据回调函数内条件返回 true 的元素。
+`filter` calls a function on each element of an array and returns a new array containing only the elements for which that function returns `true`. In other words, it filters the array, based on the function passed to it. Like `map`, it does this without needing to modify the original array.
-回调函数接收三个参数。第一个参数是当前正在被处理的元素，第二个参数是元素的索引，第三个参数是在其上调用 `filter` 方法的数组。
+The callback function accepts three arguments. The first argument is the current element being processed. The second is the index of that element and the third is the array upon which the `filter` method was called.
-看下在 `users` 上使用 `filter` 方法的例子，返回了一个新数组包含了 30 岁以下的用户。为了简化，例子里只使用了回调函数的第一个参数。
+See below for an example using the `filter` method on the `users` array to return a new array containing only the users under the age of 30. For simplicity, the example only uses the first argument of the callback.
```js
const users = [
@@ -29,11 +29,11 @@ console.log(usersUnder30); // [ { name: 'Amy', age: 20 }, { name: 'camperCat', a
# --instructions--
-`watchList`是包含一些电影信息的对象。结合`filter`和`map`返回一个只包含`title`和`rating`属性的新数组，并且`imdbRating`值大于或等于 8.0。请注意，评级值在对象中保存为字符串，你可能需要将它转换成数字来执行运算。
+The variable `watchList` holds an array of objects with information on several movies. Use a combination of `filter` and `map` on `watchList` to assign a new array of objects with only `title` and `rating` keys. The new array should only include objects where `imdbRating` is greater than or equal to 8.0. Note that the rating values are saved as strings in the object and you may need to convert them into numbers to perform mathematical operations on them.
# --hints--
-`watchList`应保持不变。
+The `watchList` variable should not change.
```js
assert(
@@ -41,19 +41,19 @@ assert(
);
```
-应使用`filter`方法。
+Your code should use the `filter` method.
```js
assert(code.match(/\.filter/g));
```
-不能使用`for`循环。
+Your code should not use a `for` loop.
```js
-assert(!code.match(/for\s*?\(.+?\)/g));
+assert(!code.match(/for\s*?\([\s\S]*?\)/g));
```
-`filteredList`应等于`[{"title": "Inception","rating": "8.8"},{"title": "Interstellar","rating": "8.6"},{"title": "The Dark Knight","rating": "9.0"},{"title": "Batman Begins","rating": "8.3"}]`。
+`filteredList` should equal `[{"title": "Inception","rating": "8.8"},{"title": "Interstellar","rating": "8.6"},{"title": "The Dark Knight","rating": "9.0"},{"title": "Batman Begins","rating": "8.3"}]`.
```js
assert.deepEqual(filteredList, [
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-map-method-to-extract-data-from-an-array.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-map-method-to-extract-data-from-an-array.md
index ac0400c2e79..0ddd1c3d56c 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-map-method-to-extract-data-from-an-array.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-map-method-to-extract-data-from-an-array.md
@@ -1,6 +1,6 @@
---
id: 587d7b8f367417b2b2512b61
-title: 使用 map 方法从数组中提取数据
+title: Use the map Method to Extract Data from an Array
challengeType: 1
forumTopicId: 18214
dashedName: use-the-map-method-to-extract-data-from-an-array
@@ -8,19 +8,19 @@ dashedName: use-the-map-method-to-extract-data-from-an-array
# --description--
-目前为止，我们已经学会了使用纯函数来避免程序中的副作用。此外，我们已经看到函数的值仅取决于其输入参数。
+So far we have learned to use pure functions to avoid side effects in a program. Also, we have seen the value in having a function only depend on its input arguments.
-这仅仅是个开始。顾名思义，函数式编程以函数理论为中心。
+This is only the beginning. As its name suggests, functional programming is centered around a theory of functions.
-能够将它们作为参数传递给其他函数，和从另一个函数返回一个函数是有意义的。函数在 JavaScript 中被视为`First Class Objects`，它们可以像任何其他对象一样使用。它们可以保存在变量中，存储在对象中，也可以作为函数参数传递。
+It would make sense to be able to pass them as arguments to other functions, and return a function from another function. Functions are considered <dfn>first class objects</dfn> in JavaScript, which means they can be used like any other object. They can be saved in variables, stored in an object, or passed as function arguments.
-让我们从一些简单的数组函数开始，这些函数是数组对象原型上的方法。在本练习中，我们来了解下数组的`map`方法（即`Array.prototype.map()`）。
+Let's start with some simple array functions, which are methods on the array object prototype. In this exercise we are looking at `Array.prototype.map()`, or more simply `map`.
-请记住，`map`方法是迭代数组中每一项的方式之一。在对每个元素应用回调函数后，它会创建一个新数组(不改变原来的数组)。
+The `map` method iterates over each item in an array and returns a new array containing the results of calling the callback function on each element. It does this without mutating the original array.
-当调用回调函数时，传入了三个参数。第一个参数是当前正在处理的数组项，第二个参数是当前数组项的索引值，第三个参数是在其上调用 `map` 方法的数组。
+When the callback is used, it is passed three arguments. The first argument is the current element being processed. The second is the index of that element and the third is the array upon which the `map` method was called.
-看下在 `users` 上使用 `map` 方法的例子，返回了一个新数组只包含了用户的名字。为了简化，例子里只使用了回调函数的第一个参数。
+See below for an example using the `map` method on the `users` array to return a new array containing only the names of the users as elements. For simplicity, the example only uses the first argument of the callback.
```js
const users = [
@@ -35,11 +35,11 @@ console.log(names); // [ 'John', 'Amy', 'camperCat' ]
# --instructions--
-`watchList`数组保存了包含一些电影信息的对象。使用`map`从`watchList`中提取标题（`title`）和评分（`rating`），并将新数组保存在`rating`变量里。目前编辑器中的代码是使用`for`循环实现，使用`map`表达式替换循环功能。
+The `watchList` array holds objects with information on several movies. Use `map` on `watchList` to assign a new array of objects with only `title` and `rating` keys to the `ratings` variable. The code in the editor currently uses a `for` loop to do this, so you should replace the loop functionality with your `map` expression.
# --hints--
-`watchList`不能被改变
+The `watchList` variable should not change.
```js
assert(
@@ -47,31 +47,28 @@ assert(
);
```
-你的代码不能使用`for`循环。
+Your code should not use a `for` loop.
```js
-assert(!removeJSComments(code).match(/for\s*?\(.*?\)/));
+assert(!__helpers.removeJSComments(code).match(/for\s*?\([\s\S]*?\)/));
```
-你的代码应使用`map`方法。
+Your code should use the `map` method.
```js
assert(code.match(/\.map/g));
```
-`rating`应等于`[{"title":"Inception","rating":"8.8"},{"title":"Interstellar","rating":"8.6"},{"title":"The Dark Knight","rating":"9.0"},{"title":"Batman Begins","rating":"8.3"},{"title":"Avatar","rating":"7.9"}]`.
+`ratings` should equal `[{"title":"Inception","rating":"8.8"},{"title":"Interstellar","rating":"8.6"},{"title":"The Dark Knight","rating":"9.0"},{"title":"Batman Begins","rating":"8.3"},{"title":"Avatar","rating":"7.9"}]`.
```js
-assert(
- JSON.stringify(ratings) ===
- JSON.stringify([
- { title: 'Inception', rating: '8.8' },
- { title: 'Interstellar', rating: '8.6' },
- { title: 'The Dark Knight', rating: '9.0' },
- { title: 'Batman Begins', rating: '8.3' },
- { title: 'Avatar', rating: '7.9' }
- ])
-);
+assert.deepEqual(ratings, [
+ { title: 'Inception', rating: '8.8' },
+ { title: 'Interstellar', rating: '8.6' },
+ { title: 'The Dark Knight', rating: '9.0' },
+ { title: 'Batman Begins', rating: '8.3' },
+ { title: 'Avatar', rating: '7.9' }
+]);
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-reduce-method-to-analyze-data.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-reduce-method-to-analyze-data.md
index 2a3a370071c..f9a80004a40 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-reduce-method-to-analyze-data.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-reduce-method-to-analyze-data.md
@@ -1,6 +1,6 @@
---
id: 587d7da9367417b2b2512b68
-title: 使用 reduce 方法分析数据
+title: Use the reduce Method to Analyze Data
challengeType: 1
forumTopicId: 301313
dashedName: use-the-reduce-method-to-analyze-data
@@ -8,15 +8,15 @@ dashedName: use-the-reduce-method-to-analyze-data
# --description--
-`reduce()`（即`Array.prototype.reduce()`），是 JavaScript 所有数组操作中最常用的方法。几乎可以用`reduce`方法解决所有数组处理问题。
+`Array.prototype.reduce()`, or simply `reduce()`, is the most general of all array operations in JavaScript. You can solve almost any array processing problem using the `reduce` method.
-`reduce`方法是处理数组更通用的方式，而且`filter`和`map`方法都可以当作是`reduce`的特殊实现。 `reduce`方法遍历数组中的每个项目并返回单个值（即字符串、数字、对象、数组）。 这是通过在每次迭代中调用一个回调函数来实现的。
+The `reduce` method allows for more general forms of array processing, and it's possible to show that both `filter` and `map` can be derived as special applications of `reduce`. The `reduce` method iterates over each item in an array and returns a single value (i.e. string, number, object, array). This is achieved via a callback function that is called on each iteration.
-回调函数接受四个参数。第一个参数称为叠加器，它是上一次迭代中回调函数的返回值，第二个参数是当前正在处理的数组元素，第三个参数是该参数的索引，第四个参数是在其上调用 `reduce` 方法的数组。
+The callback function accepts four arguments. The first argument is known as the accumulator, which gets assigned the return value of the callback function from the previous iteration, the second is the current element being processed, the third is the index of that element and the fourth is the array upon which `reduce` is called.
-除了回调函数，`reduce` 还有一个额外的参数做为叠加器的初始值。如果没有第二个参数，会跳过第一次迭代，第二次迭代给叠加器传入数组的第一个元素。
+In addition to the callback function, `reduce` has an additional parameter which takes an initial value for the accumulator. If this second parameter is not used, then the first iteration is skipped and the second iteration gets passed the first element of the array as the accumulator.
-见下面的例子，给 `users` 数组使用 `reduce` 方法，返回所有用户数组的和。为了简化，例子仅使用了回调函数的第一个参数和第二个参数。
+See below for an example using `reduce` on the `users` array to return the sum of all the users' ages. For simplicity, the example only uses the first and second arguments.
```js
const users = [
@@ -29,7 +29,7 @@ const sumOfAges = users.reduce((sum, user) => sum + user.age, 0);
console.log(sumOfAges); // 64
```
-在另一个例子里，查看如何返回一个包含用户名称做为属性，其年龄做为值的对象。
+In another example, see how an object can be returned containing the names of the users as properties with their ages as values.
```js
const users = [
@@ -47,11 +47,11 @@ console.log(usersObj); // { John: 34, Amy: 20, camperCat: 10 }
# --instructions--
-`watchList`变量中包含一组存有多部电影信息对象。使用`reduce`查找由**Christopher Nolan 导演**的电影的 IMDB 评级平均值。回想一下之前的挑战，如何`filter`数据，以及使用`map`来获取你想要的数据。你可能需要创建一些变量，但是请将最后的平均值保存到`averageRating`变量中。请注意，评级在对象中是字符串，需要将其转换为数字再用于数学运算。
+The variable `watchList` holds an array of objects with information on several movies. Use `reduce` to find the average IMDB rating of the movies **directed by Christopher Nolan**. Recall from prior challenges how to `filter` data and `map` over it to pull what you need. You may need to create other variables, and return the average rating from `getRating` function. Note that the rating values are saved as strings in the object and need to be converted into numbers before they are used in any mathematical operations.
# --hints--
-`watchList`应保持不变。
+The `watchList` variable should not change.
```js
assert(
@@ -59,25 +59,25 @@ assert(
);
```
-应该使用`reduce`方法。
+Your code should use the `reduce` method.
```js
assert(code.match(/\.reduce/g));
```
-The`averageRating`应等于 8.675。
+The `getRating(watchList)` should equal 8.675.
```js
assert(getRating(watchList) === 8.675);
```
-不能使用`for`循环。
+Your code should not use a `for` loop.
```js
-assert(!code.match(/for\s*?\(.*\)/g));
+assert(!code.match(/for\s*?\([\s\S]*?\)/g));
```
-在修改 `watchList` 对象后应该返回正确的输出。
+Your code should return correct output after modifying the `watchList` object.
```js
assert(getRating(watchList.filter((_, i) => i < 1 || i > 2)) === 8.55);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-some-method-to-check-that-any-elements-in-an-array-meet-a-criteria.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-some-method-to-check-that-any-elements-in-an-array-meet-a-criteria.md
index 759b1162a8f..3aa85c4dab0 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-some-method-to-check-that-any-elements-in-an-array-meet-a-criteria.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/functional-programming/use-the-some-method-to-check-that-any-elements-in-an-array-meet-a-criteria.md
@@ -1,6 +1,6 @@
---
id: 587d7dab367417b2b2512b6f
-title: 使用 some 方法检查数组中是否有元素是否符合条件
+title: Use the some Method to Check that Any Elements in an Array Meet a Criteria
challengeType: 1
forumTopicId: 301314
dashedName: use-the-some-method-to-check-that-any-elements-in-an-array-meet-a-criteria
@@ -8,9 +8,9 @@ dashedName: use-the-some-method-to-check-that-any-elements-in-an-array-meet-a-cr
# --description--
-`some`方法用于检测数组中*任何*元素是否满足指定条件。如果有一个元素满足条件，返回布尔值`true`，反之返回`false`。
+The `some` method works with arrays to check if *any* element passes a particular test. It returns a Boolean value - `true` if any of the values meet the criteria, `false` if not.
-举个例子，下面的代码检测数组`numbers`中是否有元素小于10：
+For example, the following code would check if any element in the `numbers` array is less than 10:
```js
var numbers = [10, 50, 8, 220, 110, 11];
@@ -22,29 +22,29 @@ numbers.some(function(currentValue) {
# --instructions--
-在`checkPositive`函数值中使用`some`检查`arr`中是否有元素为正数，函数应返回一个布尔值。
+Use the `some` method inside the `checkPositive` function to check if any element in `arr` is positive. The function should return a Boolean value.
# --hints--
-应该使用`some`method.
+Your code should use the `some` method.
```js
assert(code.match(/\.some/g));
```
-`checkPositive([1, 2, 3, -4, 5])`应返回`true`。
+`checkPositive([1, 2, 3, -4, 5])` should return `true`.
```js
assert(checkPositive([1, 2, 3, -4, 5]));
```
-`checkPositive([1, 2, 3, 4, 5])`应返回`true`。
+`checkPositive([1, 2, 3, 4, 5])` should return `true`.
```js
assert(checkPositive([1, 2, 3, 4, 5]));
```
-`checkPositive([-1, -2, -3, -4, -5])`应返回`false`。
+`checkPositive([-1, -2, -3, -4, -5])` should return `false`.
```js
assert(!checkPositive([-1, -2, -3, -4, -5]));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/arguments-optional.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/arguments-optional.md
index 2d34d1a8b06..c3c51c294de 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/arguments-optional.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/arguments-optional.md
@@ -1,6 +1,6 @@
---
id: a97fd23d9b809dac9921074f
-title: 可选参数
+title: Arguments Optional
challengeType: 5
forumTopicId: 14271
dashedName: arguments-optional
@@ -8,51 +8,51 @@ dashedName: arguments-optional
# --description--
-创建一个将两个参数相加的函数。如果调用时只传入了一个参数，则应返回一个接收新的参数的函数。待传入下一个参数后，再返回与之前传入的参数之和。
+Create a function that sums two arguments together. If only one argument is provided, then return a function that expects one argument and returns the sum.
-比如，`addTogether(2, 3)` 应该返回 `5`。而 `addTogether(2)` 应该返回一个函数。
+For example, `addTogether(2, 3)` should return `5`, and `addTogether(2)` should return a function.
-调用这个返回的函数，为它传入一个值，然后再返回总和：
+Calling this returned function with a single argument will then return the sum:
`var sumTwoAnd = addTogether(2);`
-`sumTwoAnd(3)` 此时应返回 `5`。
+`sumTwoAnd(3)` returns `5`.
-任何时候，只要任一传入的参数不是数字，就应返回 `undefined`。
+If either argument isn't a valid number, return undefined.
# --hints--
-`addTogether(2, 3)` 应返回 5。
+`addTogether(2, 3)` should return 5.
```js
assert.deepEqual(addTogether(2, 3), 5);
```
-`addTogether(23, 30)` 应返回 53。
+`addTogether(23, 30)` should return 53.
```js
assert.deepEqual(addTogether(23, 30), 53);
```
-`addTogether(5)(7)` 应返回 12。
+`addTogether(5)(7)` should return 12.
```js
assert.deepEqual(addTogether(5)(7), 12);
```
-`addTogether("http://bit.ly/IqT6zt")` 应返回 undefined。
+`addTogether("http://bit.ly/IqT6zt")` should return undefined.
```js
assert.isUndefined(addTogether('http://bit.ly/IqT6zt'));
```
-`addTogether(2, "3")` 应返回 undefined。
+`addTogether(2, "3")` should return undefined.
```js
assert.isUndefined(addTogether(2, '3'));
```
-`addTogether(2)([3])` 应返回 undefined。
+`addTogether(2)([3])` should return undefined.
```js
assert.isUndefined(addTogether(2)([3]));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/binary-agents.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/binary-agents.md
index ff0a0cadc47..954865c4281 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/binary-agents.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/binary-agents.md
@@ -1,6 +1,6 @@
---
id: a8d97bd4c764e91f9d2bda01
-title: 翻译二进制字符串
+title: Binary Agents
challengeType: 5
forumTopicId: 14273
dashedName: binary-agents
@@ -8,13 +8,13 @@ dashedName: binary-agents
# --description--
-请实现一个函数，把传入的二进制字符串转换成英文句子。
+Return an English translated sentence of the passed binary string.
-二进制字符串会以空格分隔。
+The binary string will be space separated.
# --hints--
-`binaryAgent('01000001 01110010 01100101 01101110 00100111 01110100 00100000 01100010 01101111 01101110 01100110 01101001 01110010 01100101 01110011 00100000 01100110 01110101 01101110 00100001 00111111')` 应返回 "Aren't bonfires fun!?"。
+`binaryAgent("01000001 01110010 01100101 01101110 00100111 01110100 00100000 01100010 01101111 01101110 01100110 01101001 01110010 01100101 01110011 00100000 01100110 01110101 01101110 00100001 00111111")` should return "Aren't bonfires fun!?"
```js
assert.deepEqual(
@@ -25,7 +25,7 @@ assert.deepEqual(
);
```
-`binaryAgent('01001001 00100000 01101100 01101111 01110110 01100101 00100000 01000110 01110010 01100101 01100101 01000011 01101111 01100100 01100101 01000011 01100001 01101101 01110000 00100001')` 应返回 "I love FreeCodeCamp!"。
+`binaryAgent("01001001 00100000 01101100 01101111 01110110 01100101 00100000 01000110 01110010 01100101 01100101 01000011 01101111 01100100 01100101 01000011 01100001 01101101 01110000 00100001")` should return "I love FreeCodeCamp!"
```js
assert.deepEqual(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/convert-html-entities.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/convert-html-entities.md
index 2cd10ea406c..707290ef0c8 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/convert-html-entities.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/convert-html-entities.md
@@ -1,6 +1,6 @@
---
id: a6b0bb188d873cb2c8729495
-title: 转换 HTML 实体字符
+title: Convert HTML Entities
challengeType: 5
forumTopicId: 16007
dashedName: convert-html-entities
@@ -8,17 +8,17 @@ dashedName: convert-html-entities
# --description--
-请将字符串中的 `&`、`<`、`>`、`"`（双引号）和 `'`（单引号）转换为 [HTML 实体字符](https://developer.mozilla.org/zh-CN/docs/Glossary/Entity)，并返回转换之后的字符串。
+Convert the characters `&`, `<`, `>`, `"` (double quote), and `'` (apostrophe), in a string to their corresponding HTML entities.
# --hints--
-`convertHTML("Dolce & Gabbana")` 应返回 `"Dolce & Gabbana"`。
+`convertHTML("Dolce & Gabbana")` should return `"Dolce & Gabbana"`.
```js
assert.match(convertHTML('Dolce & Gabbana'), /Dolce & Gabbana/);
```
-`convertHTML("Hamburgers < Pizza < Tacos")` 应返回 `"Hamburgers < Pizza < Tacos"`。
+`convertHTML("Hamburgers < Pizza < Tacos")` should return `"Hamburgers < Pizza < Tacos"`.
```js
assert.match(
@@ -27,13 +27,13 @@ assert.match(
);
```
-`convertHTML("Sixty > twelve")` 应返回 `"Sixty > twelve"`。
+`convertHTML("Sixty > twelve")` should return `"Sixty > twelve"`.
```js
assert.match(convertHTML('Sixty > twelve'), /Sixty > twelve/);
```
-`convertHTML('Stuff in "quotation marks"')` 应返回 `"Stuff in "quotation marks""`。
+`convertHTML('Stuff in "quotation marks"')` should return `"Stuff in "quotation marks""`.
```js
assert.match(
@@ -42,19 +42,19 @@ assert.match(
);
```
-`convertHTML("Schindler's List")` 应返回 `"Schindler's List"`。
+`convertHTML("Schindler's List")` should return `"Schindler's List"`.
```js
assert.match(convertHTML("Schindler's List"), /Schindler's List/);
```
-`convertHTML("<>")` 应返回 `"<>"`。
+`convertHTML("<>")` should return `"<>"`.
```js
assert.match(convertHTML('<>'), /<>/);
```
-`convertHTML("abc")` 应返回 `"abc"`。
+`convertHTML("abc")` should return `"abc"`.
```js
assert.strictEqual(convertHTML('abc'), 'abc');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/diff-two-arrays.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/diff-two-arrays.md
index d631b6be00b..228d55cf80b 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/diff-two-arrays.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/diff-two-arrays.md
@@ -1,6 +1,6 @@
---
id: a5de63ebea8dbee56860f4f2
-title: 数组的对称差
+title: Diff Two Arrays
challengeType: 5
forumTopicId: 16008
dashedName: diff-two-arrays
@@ -8,20 +8,20 @@ dashedName: diff-two-arrays
# --description--
-在这道题目中，我们需要实现一个函数，它可以比较两个输入数组并返回一个新数组；返回的新数组需包含传入的两个数组中，仅在一个数组里出现的元素。如果某个元素同时出现在两个数组中，则不应包含在返回的数组里。换言之，我们需要返回两个数组的对称差。
+Compare two arrays and return a new array with any items only found in one of the two given arrays, but not both. In other words, return the symmetric difference of the two arrays.
-**注意：**
-返回数组中的元素顺序不会影响挑战是否通过。
+**Note**
+You can return the array with its elements in any order.
# --hints--
-`diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5])` 应返回一个数组。
+`diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5])` should return an array.
```js
assert(typeof diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]) === 'object');
```
-`["diorite", "andesite", "grass", "dirt", "pink wool", "dead shrub"], ["diorite", "andesite", "grass", "dirt", "dead shrub"]` 应返回 `["pink wool"]`。
+`["diorite", "andesite", "grass", "dirt", "pink wool", "dead shrub"], ["diorite", "andesite", "grass", "dirt", "dead shrub"]` should return `["pink wool"]`.
```js
assert.sameMembers(
@@ -33,7 +33,7 @@ assert.sameMembers(
);
```
-`["diorite", "andesite", "grass", "dirt", "pink wool", "dead shrub"], ["diorite", "andesite", "grass", "dirt", "dead shrub"]` 应返回一个长度为 1 的数组。
+`["diorite", "andesite", "grass", "dirt", "pink wool", "dead shrub"], ["diorite", "andesite", "grass", "dirt", "dead shrub"]` should return an array with one item.
```js
assert(
@@ -44,7 +44,7 @@ assert(
);
```
-`["andesite", "grass", "dirt", "pink wool", "dead shrub"], ["diorite", "andesite", "grass", "dirt", "dead shrub"]` 应返回 `["diorite", "pink wool"]`。
+`["andesite", "grass", "dirt", "pink wool", "dead shrub"], ["diorite", "andesite", "grass", "dirt", "dead shrub"]` should return `["diorite", "pink wool"]`.
```js
assert.sameMembers(
@@ -56,7 +56,7 @@ assert.sameMembers(
);
```
-`["andesite", "grass", "dirt", "pink wool", "dead shrub"], ["diorite", "andesite", "grass", "dirt", "dead shrub"]` 应返回一个长度为 2 的数组。
+`["andesite", "grass", "dirt", "pink wool", "dead shrub"], ["diorite", "andesite", "grass", "dirt", "dead shrub"]` should return an array with two items.
```js
assert(
@@ -67,7 +67,7 @@ assert(
);
```
-`["andesite", "grass", "dirt", "dead shrub"], ["andesite", "grass", "dirt", "dead shrub"]` 应返回 `[]`。
+`["andesite", "grass", "dirt", "dead shrub"], ["andesite", "grass", "dirt", "dead shrub"]` should return `[]`.
```js
assert.sameMembers(
@@ -79,7 +79,7 @@ assert.sameMembers(
);
```
-`["andesite", "grass", "dirt", "dead shrub"], ["andesite", "grass", "dirt", "dead shrub"]` 应返回一个空数组。
+`["andesite", "grass", "dirt", "dead shrub"], ["andesite", "grass", "dirt", "dead shrub"]` should return an empty array.
```js
assert(
@@ -90,19 +90,19 @@ assert(
);
```
-`[1, 2, 3, 5], [1, 2, 3, 4, 5]` 应返回 `[4]`。
+`[1, 2, 3, 5], [1, 2, 3, 4, 5]` should return `[4]`.
```js
assert.sameMembers(diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]), [4]);
```
-`[1, 2, 3, 5], [1, 2, 3, 4, 5]` 应返回一个长度为 1 的数组。
+`[1, 2, 3, 5], [1, 2, 3, 4, 5]` should return an array with one item.
```js
assert(diffArray([1, 2, 3, 5], [1, 2, 3, 4, 5]).length === 1);
```
-`[1, "calf", 3, "piglet"], [1, "calf", 3, 4]` 应返回 `["piglet", 4]`。
+`[1, "calf", 3, "piglet"], [1, "calf", 3, 4]` should return `["piglet", 4]`.
```js
assert.sameMembers(diffArray([1, 'calf', 3, 'piglet'], [1, 'calf', 3, 4]), [
@@ -111,13 +111,13 @@ assert.sameMembers(diffArray([1, 'calf', 3, 'piglet'], [1, 'calf', 3, 4]), [
]);
```
-`[1, "calf", 3, "piglet"], [1, "calf", 3, 4]` 应返回一个长度为 2 的数组。
+`[1, "calf", 3, "piglet"], [1, "calf", 3, 4]` should return an array with two items.
```js
assert(diffArray([1, 'calf', 3, 'piglet'], [1, 'calf', 3, 4]).length === 2);
```
-`[], ["snuffleupagus", "cookie monster", "elmo"]` 应返回 `["snuffleupagus", "cookie monster", "elmo"]`。
+`[], ["snuffleupagus", "cookie monster", "elmo"]` should return `["snuffleupagus", "cookie monster", "elmo"]`.
```js
assert.sameMembers(diffArray([], ['snuffleupagus', 'cookie monster', 'elmo']), [
@@ -127,13 +127,13 @@ assert.sameMembers(diffArray([], ['snuffleupagus', 'cookie monster', 'elmo']), [
]);
```
-`[], ["snuffleupagus", "cookie monster", "elmo"]` 应返回一个长度为 3 的数组。
+`[], ["snuffleupagus", "cookie monster", "elmo"]` should return an array with three items.
```js
assert(diffArray([], ['snuffleupagus', 'cookie monster', 'elmo']).length === 3);
```
-`[1, "calf", 3, "piglet"], [7, "filly"]` 应返回 `[1, "calf", 3, "piglet", 7, "filly"]`。
+`[1, "calf", 3, "piglet"], [7, "filly"]` should return `[1, "calf", 3, "piglet", 7, "filly"]`.
```js
assert.sameMembers(diffArray([1, 'calf', 3, 'piglet'], [7, 'filly']), [
@@ -146,7 +146,7 @@ assert.sameMembers(diffArray([1, 'calf', 3, 'piglet'], [7, 'filly']), [
]);
```
-`[1, "calf", 3, "piglet"], [7, "filly"]` 应返回一个长度为 6 的数组。
+`[1, "calf", 3, "piglet"], [7, "filly"]` should return an array with six items.
```js
assert(diffArray([1, 'calf', 3, 'piglet'], [7, 'filly']).length === 6);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/dna-pairing.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/dna-pairing.md
index 1676fa960aa..d9f36cd2814 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/dna-pairing.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/dna-pairing.md
@@ -1,6 +1,6 @@
---
id: afd15382cdfb22c9efe8b7de
-title: DNA 配对
+title: DNA Pairing
challengeType: 5
forumTopicId: 16009
dashedName: dna-pairing
@@ -8,19 +8,19 @@ dashedName: dna-pairing
# --description--
-给出的 DNA 链上缺少配对元素。请基于每个字符，获取与其配对的元素，并将结果作为二维数组返回。
+The DNA strand is missing the pairing element. Take each character, get its pair, and return the results as a 2d array.
-DNA 的[碱基对](http://en.wikipedia.org/wiki/Base_pair) 有两种形式：一种是 A 与 T，一种是 C 与 G。请为参数中给出的每个字符配对相应的碱基。
+[Base pairs](http://en.wikipedia.org/wiki/Base_pair) are a pair of AT and CG. Match the missing element to the provided character.
-注意，参数中给出的字符应作为每个子数组中的第一个元素返回。
+Return the provided character as the first element in each array.
-例如，传入 GCG 时，应返回 \[\["G", "C"], \["C", "G"], \["G", "C"]]。
+For example, for the input GCG, return \[\["G", "C"], \["C","G"],\["G", "C"]]
-参数中的字符及与其配对的碱基应存在于一个数组中，代表碱基对。再将每个配对完成的碱基对数组按顺序放到一个数组中，作为最终的返回结果。
+The character and its pair are paired up in an array, and all the arrays are grouped into one encapsulating array.
# --hints--
-`pairElement("ATCGA")` 应返回 `[["A","T"],["T","A"],["C","G"],["G","C"],["A","T"]]`。
+`pairElement("ATCGA")` should return `[["A","T"],["T","A"],["C","G"],["G","C"],["A","T"]]`.
```js
assert.deepEqual(pairElement('ATCGA'), [
@@ -32,7 +32,7 @@ assert.deepEqual(pairElement('ATCGA'), [
]);
```
-`pairElement("TTGAG")` 应返回 `[["T","A"],["T","A"],["G","C"],["A","T"],["G","C"]]`。
+`pairElement("TTGAG")` should return `[["T","A"],["T","A"],["G","C"],["A","T"],["G","C"]]`.
```js
assert.deepEqual(pairElement('TTGAG'), [
@@ -44,7 +44,7 @@ assert.deepEqual(pairElement('TTGAG'), [
]);
```
-`pairElement("CTCTA")` 应返回 `[["C","G"],["T","A"],["C","G"],["T","A"],["A","T"]]`。
+`pairElement("CTCTA")` should return `[["C","G"],["T","A"],["C","G"],["T","A"],["A","T"]]`.
```js
assert.deepEqual(pairElement('CTCTA'), [
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/drop-it.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/drop-it.md
index b7d2cc6d87e..b2d97fca2fa 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/drop-it.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/drop-it.md
@@ -1,6 +1,6 @@
---
id: a5deed1811a43193f9f1c841
-title: 根据参数删除数组元素
+title: Drop it
challengeType: 5
forumTopicId: 16010
dashedName: drop-it
@@ -8,13 +8,13 @@ dashedName: drop-it
# --description--
-给定数组 `arr`，从数组的第一个元素开始，用函数 `func` 来检查数组的每个元素是否返回 `true`。如果返回 `false`，就把这个元素删除。持续执行删除操作，直到某个元素传入 `func` 时返回 `true` 为止。换言之，函数的最终返回值也应是一个数组，它由原数组中第一个使得 `func` 为 `true` 的元素及其之后的所有元素组成。
+Given the array `arr`, iterate through and remove each element starting from the first element (the 0 index) until the function `func` returns `true` when the iterated element is passed through it.
-如果数组中的所有元素都不能让 `func` 为 `true`，则返回空数组 `[]`。
+Then return the rest of the array once the condition is satisfied, otherwise, `arr` should be returned as an empty array.
# --hints--
-`dropElements([1, 2, 3, 4], function(n) {return n >= 3;})` 应返回 `[3, 4]`。
+`dropElements([1, 2, 3, 4], function(n) {return n >= 3;})` should return `[3, 4]`.
```js
assert.deepEqual(
@@ -25,7 +25,7 @@ assert.deepEqual(
);
```
-`dropElements([0, 1, 0, 1], function(n) {return n === 1;})` 应返回 `[1, 0, 1]`。
+`dropElements([0, 1, 0, 1], function(n) {return n === 1;})` should return `[1, 0, 1]`.
```js
assert.deepEqual(
@@ -36,7 +36,7 @@ assert.deepEqual(
);
```
-`dropElements([1, 2, 3], function(n) {return n > 0;})` 应返回 `[1, 2, 3]`。
+`dropElements([1, 2, 3], function(n) {return n > 0;})` should return `[1, 2, 3]`.
```js
assert.deepEqual(
@@ -47,7 +47,7 @@ assert.deepEqual(
);
```
-`dropElements([1, 2, 3, 4], function(n) {return n > 5;})` 应返回 `[]`。
+`dropElements([1, 2, 3, 4], function(n) {return n > 5;})` should return `[]`.
```js
assert.deepEqual(
@@ -58,7 +58,7 @@ assert.deepEqual(
);
```
-`dropElements([1, 2, 3, 7, 4], function(n) {return n > 3;})` 应返回 `[7, 4]`。
+`dropElements([1, 2, 3, 7, 4], function(n) {return n > 3;})` should return `[7, 4]`.
```js
assert.deepEqual(
@@ -69,7 +69,7 @@ assert.deepEqual(
);
```
-`dropElements([1, 2, 3, 9, 2], function(n) {return n > 2;})` 应返回 `[3, 9, 2]`。
+`dropElements([1, 2, 3, 9, 2], function(n) {return n > 2;})` should return `[3, 9, 2]`.
```js
assert.deepEqual(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/everything-be-true.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/everything-be-true.md
index 901376159e8..7a93f8ae2af 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/everything-be-true.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/everything-be-true.md
@@ -1,6 +1,6 @@
---
id: a10d2431ad0c6a099a4b8b52
-title: 检查对象属性
+title: Everything Be True
challengeType: 5
forumTopicId: 16011
dashedName: everything-be-true
@@ -8,17 +8,17 @@ dashedName: everything-be-true
# --description--
-对于第一个参数（对象数组）中的每个元素，检查第二个参数（字符串）所对应的属性值是否都为 <dfn>truthy</dfn>。
+Check if the predicate (second argument) is <dfn>truthy</dfn> on all elements of a collection (first argument).
-换句话说，函数的第一个参数是一个对象数组，第二个参数是一个字符串 `pre`，表示对象的属性。如果数组中的每个对象里，`pre` 对应属性值均为 `truthy`，则返回 `true`；否则返回 `false`。
+In other words, you are given an array collection of objects. The predicate `pre` will be an object property and you need to return `true` if its value is `truthy`. Otherwise, return `false`.
-JavaScript 中，如果一个值在 Boolean 的上下文中（比如 `if` 语句）的执行结果为 `true`，那么我们称这个值是 `truthy` 的。
+In JavaScript, `truthy` values are values that translate to `true` when evaluated in a Boolean context.
-别忘了，你可以使用点号表示法（`.`）或方括号表示法（`[]`）来访问对象的属性。
+Remember, you can access object properties through either dot notation or `[]` notation.
# --hints--
-`truthCheck([{"user": "Tinky-Winky", "sex": "male"}, {"user": "Dipsy", "sex": "male"}, {"user": "Laa-Laa", "sex": "female"}, {"user": "Po", "sex": "female"}], "sex")` 应返回 true。
+`truthCheck([{"user": "Tinky-Winky", "sex": "male"}, {"user": "Dipsy", "sex": "male"}, {"user": "Laa-Laa", "sex": "female"}, {"user": "Po", "sex": "female"}], "sex")` should return true.
```js
assert.strictEqual(
@@ -35,7 +35,7 @@ assert.strictEqual(
);
```
-`truthCheck([{"user": "Tinky-Winky", "sex": "male"}, {"user": "Dipsy"}, {"user": "Laa-Laa", "sex": "female"}, {"user": "Po", "sex": "female"}], "sex")` 应返回 false。
+`truthCheck([{"user": "Tinky-Winky", "sex": "male"}, {"user": "Dipsy"}, {"user": "Laa-Laa", "sex": "female"}, {"user": "Po", "sex": "female"}], "sex")` should return false.
```js
assert.strictEqual(
@@ -52,7 +52,7 @@ assert.strictEqual(
);
```
-`truthCheck([{"user": "Tinky-Winky", "sex": "male", "age": 0}, {"user": "Dipsy", "sex": "male", "age": 3}, {"user": "Laa-Laa", "sex": "female", "age": 5}, {"user": "Po", "sex": "female", "age": 4}], "age")` 应返回 false。
+`truthCheck([{"user": "Tinky-Winky", "sex": "male", "age": 0}, {"user": "Dipsy", "sex": "male", "age": 3}, {"user": "Laa-Laa", "sex": "female", "age": 5}, {"user": "Po", "sex": "female", "age": 4}], "age")` should return false.
```js
assert.strictEqual(
@@ -69,7 +69,7 @@ assert.strictEqual(
);
```
-`truthCheck([{"name": "Pete", "onBoat": true}, {"name": "Repeat", "onBoat": true}, {"name": "FastForward", "onBoat": null}], "onBoat")` 应返回 false。
+`truthCheck([{"name": "Pete", "onBoat": true}, {"name": "Repeat", "onBoat": true}, {"name": "FastForward", "onBoat": null}], "onBoat")` should return false
```js
assert.strictEqual(
@@ -85,7 +85,7 @@ assert.strictEqual(
);
```
-`truthCheck([{"name": "Pete", "onBoat": true}, {"name": "Repeat", "onBoat": true, "alias": "Repete"}, {"name": "FastForward", "onBoat": true}], "onBoat")` 应返回 true。
+`truthCheck([{"name": "Pete", "onBoat": true}, {"name": "Repeat", "onBoat": true, "alias": "Repete"}, {"name": "FastForward", "onBoat": true}], "onBoat")` should return true
```js
assert.strictEqual(
@@ -101,13 +101,13 @@ assert.strictEqual(
);
```
-`truthCheck([{"single": "yes"}], "single")` 应返回 true。
+`truthCheck([{"single": "yes"}], "single")` should return true
```js
assert.strictEqual(truthCheck([{ single: 'yes' }], 'single'), true);
```
-`truthCheck([{"single": ""}, {"single": "double"}], "single")` 应返回 false。
+`truthCheck([{"single": ""}, {"single": "double"}], "single")` should return false
```js
assert.strictEqual(
@@ -116,7 +116,7 @@ assert.strictEqual(
);
```
-`truthCheck([{"single": "double"}, {"single": undefined}], "single")` 应返回 false。
+`truthCheck([{"single": "double"}, {"single": undefined}], "single")` should return false
```js
assert.strictEqual(
@@ -125,7 +125,7 @@ assert.strictEqual(
);
```
-`truthCheck([{"single": "double"}, {"single": NaN}], "single")` 应返回 false。
+`truthCheck([{"single": "double"}, {"single": NaN}], "single")` should return false
```js
assert.strictEqual(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/make-a-person.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/make-a-person.md
index fc48498fd51..b9458749e7e 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/make-a-person.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/make-a-person.md
@@ -1,6 +1,6 @@
---
id: a2f1d72d9b908d0bd72bb9f6
-title: 构建 Person 类
+title: Make a Person
challengeType: 5
forumTopicId: 16020
dashedName: make-a-person
@@ -8,7 +8,7 @@ dashedName: make-a-person
# --description--
-在这道题目中，我们需要写一个构造器（constructor）函数。它只接收一个字符串参数 `firstAndLast`，这个参数代表一个英文名的全名（包含姓和名）。通过这个构造函数创建出的实例需要具有以下方法：
+Fill in the object constructor with the following methods below:
```js
getFirstName()
@@ -19,53 +19,53 @@ setLastName(last)
setFullName(firstAndLast)
```
-你可以点击“运行测试”，然后就可以在底下的控制台中看到每个测试用例执行的情况。以上方法中，接收一个参数的方法只会接收一个字符串作为参数。同时，我们只会通过以上定义的这些方法与实例进行交互。
+Run the tests to see the expected output for each method. The methods that take an argument must accept only one argument and it has to be a string. These methods must be the only available means of interacting with the object.
# --hints--
-`Object.keys(bob).length` 应返回 6。
+`Object.keys(bob).length` should return 6.
```js
assert.deepEqual(Object.keys(bob).length, 6);
```
-`bob instanceof Person` 应返回 true。
+`bob instanceof Person` should return true.
```js
assert.deepEqual(bob instanceof Person, true);
```
-`bob.firstName` 应返回 undefined。
+`bob.firstName` should return undefined.
```js
assert.deepEqual(bob.firstName, undefined);
```
-`bob.lastName` 应返回 undefined。
+`bob.lastName` should return undefined.
```js
assert.deepEqual(bob.lastName, undefined);
```
-`bob.getFirstName()` 应返回 "Bob"。
+`bob.getFirstName()` should return "Bob".
```js
assert.deepEqual(bob.getFirstName(), 'Bob');
```
-`bob.getLastName()` 应返回 "Ross"。
+`bob.getLastName()` should return "Ross".
```js
assert.deepEqual(bob.getLastName(), 'Ross');
```
-`bob.getFullName()` 应返回 "Bob Ross"。
+`bob.getFullName()` should return "Bob Ross".
```js
assert.deepEqual(bob.getFullName(), 'Bob Ross');
```
-在执行 `bob.setFirstName("Haskell")` 后，`bob.getFullName()` 应返回 "Haskell Ross"。
+`bob.getFullName()` should return "Haskell Ross" after `bob.setFirstName("Haskell")`.
```js
assert.strictEqual(
@@ -77,7 +77,7 @@ assert.strictEqual(
);
```
-在执行 `bob.setLastName("Curry")` 后，`bob.getFullName()` 应返回 "Haskell Curry"。
+`bob.getFullName()` should return "Haskell Curry" after `bob.setLastName("Curry")`.
```js
assert.strictEqual(
@@ -90,7 +90,7 @@ assert.strictEqual(
);
```
-在执行 `bob.setFullName("Haskell Curry")` 后，`bob.getFullName()` 应返回 "Haskell Curry"。
+`bob.getFullName()` should return "Haskell Curry" after `bob.setFullName("Haskell Curry")`.
```js
assert.strictEqual(
@@ -102,7 +102,7 @@ assert.strictEqual(
);
```
-在执行 `bob.setFullName("Haskell Curry")` 后，`bob.getFirstName()` 应返回 "Haskell"。
+`bob.getFirstName()` should return "Haskell" after `bob.setFullName("Haskell Curry")`.
```js
assert.strictEqual(
@@ -114,7 +114,7 @@ assert.strictEqual(
);
```
-在执行 `bob.setFullName("Haskell Curry")` 后，`bob.getLastName()` 应返回 "Curry"。
+`bob.getLastName()` should return "Curry" after `bob.setFullName("Haskell Curry")`.
```js
assert.strictEqual(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/map-the-debris.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/map-the-debris.md
index fbeffd5cdc9..74118356a40 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/map-the-debris.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/map-the-debris.md
@@ -1,6 +1,6 @@
---
id: af4afb223120f7348cdfc9fd
-title: 计算轨道周期
+title: Map the Debris
challengeType: 5
forumTopicId: 16021
dashedName: map-the-debris
@@ -8,25 +8,19 @@ dashedName: map-the-debris
# --description--
-在这道题目中，我们需要写一个计算天体轨道周期（单位是秒）的函数。
+Return a new array that transforms the elements' average altitude into their orbital periods (in seconds).
-它接收一个对象数组参数 `arr`，对象中包含表示天体名称的 `name` 属性，及表示天体表面平均海拔的 `avgAlt` 属性。就像这样：`{name: 'name', avgAlt: avgAlt}`。
+The array will contain objects in the format `{name: 'name', avgAlt: avgAlt}`.
-这个函数的返回值也是一个对象数组，应保留原对象中的 `name` 属性及其属性值，然后根据 `avgAlt` 的属性值求出轨道周期（单位是秒），并赋值给 `orbitalPeriod` 属性。返回值中不应保留原数据中的 `avgAlt` 属性。
+You can read about orbital periods [on Wikipedia](http://en.wikipedia.org/wiki/Orbital_period).
-你可以在这条[维基百科](http://en.wikipedia.org/wiki/Orbital_period)的链接中找到轨道周期的计算公式：
+The values should be rounded to the nearest whole number. The body being orbited is Earth.
-```
-T = 2π*(a^3/GM)^(1/2)
-```
-
-轨道周期的计算以地球为基准（即环绕地球的轨道周期），因此公式中的 `a` 为数组对象中的 `avgAlt` 加上地球半径。最终的计算结果应取整到最接近的整数。
-
-地球的半径是 6367.4447 千米，地球的 GM 值为 398600.4418 km<sup>3</sup>s<sup>-2</sup>。
+The radius of the earth is 6367.4447 kilometers, and the GM value of earth is 398600.4418 km<sup>3</sup>s<sup>-2</sup>.
# --hints--
-`orbitalPeriod([{name : "sputnik", avgAlt : 35873.5553}])` 应返回 `[{name: "sputnik", orbitalPeriod: 86400}]`。
+`orbitalPeriod([{name : "sputnik", avgAlt : 35873.5553}])` should return `[{name: "sputnik", orbitalPeriod: 86400}]`.
```js
assert.deepEqual(orbitalPeriod([{ name: 'sputnik', avgAlt: 35873.5553 }]), [
@@ -34,7 +28,7 @@ assert.deepEqual(orbitalPeriod([{ name: 'sputnik', avgAlt: 35873.5553 }]), [
]);
```
-`orbitalPeriod([{name: "iss", avgAlt: 413.6}, {name: "hubble", avgAlt: 556.7}, {name: "moon", avgAlt: 378632.553}])` 应返回 `[{name : "iss", orbitalPeriod: 5557}, {name: "hubble", orbitalPeriod: 5734}, {name: "moon", orbitalPeriod: 2377399}]`。
+`orbitalPeriod([{name: "iss", avgAlt: 413.6}, {name: "hubble", avgAlt: 556.7}, {name: "moon", avgAlt: 378632.553}])` should return `[{name : "iss", orbitalPeriod: 5557}, {name: "hubble", orbitalPeriod: 5734}, {name: "moon", orbitalPeriod: 2377399}]`.
```js
assert.deepEqual(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/missing-letters.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/missing-letters.md
index 96e1bdd025d..c5e807a7a41 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/missing-letters.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/missing-letters.md
@@ -1,43 +1,44 @@
---
id: af7588ade1100bde429baf20
-title: 寻找缺失的字母
+title: Missing letters
challengeType: 5
+forumTopicId: 16023
dashedName: missing-letters
---
# --description--
-在这道题目中，我们需要写一个函数，找出传入的字符串里缺失的字母并返回它。
+Find the missing letter in the passed letter range and return it.
-判断缺失的依据是字母顺序。对于没有缺失的情况，请返回 `undefined`。
+If all letters are present in the range, return undefined.
# --hints--
-`fearNotLetter("abce")` 应返回 "d"。
+`fearNotLetter("abce")` should return "d".
```js
assert.deepEqual(fearNotLetter('abce'), 'd');
```
-`fearNotLetter("abcdefghjklmno")` 应返回 "i"。
+`fearNotLetter("abcdefghjklmno")` should return "i".
```js
assert.deepEqual(fearNotLetter('abcdefghjklmno'), 'i');
```
-`fearNotLetter("stvwx")` 应返回 "u"。
+`fearNotLetter("stvwx")` should return "u".
```js
assert.deepEqual(fearNotLetter('stvwx'), 'u');
```
-`fearNotLetter("bcdf")` 应返回 "e"。
+`fearNotLetter("bcdf")` should return "e".
```js
assert.deepEqual(fearNotLetter('bcdf'), 'e');
```
-`fearNotLetter("abcdefghijklmnopqrstuvwxyz")` 应返回 undefined。
+`fearNotLetter("abcdefghijklmnopqrstuvwxyz")` should return undefined.
```js
assert.isUndefined(fearNotLetter('abcdefghijklmnopqrstuvwxyz'));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/pig-latin.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/pig-latin.md
index 103a226e5a5..af40b695bf0 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/pig-latin.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/pig-latin.md
@@ -1,6 +1,6 @@
---
id: aa7697ea2477d1316795783b
-title: 儿童黑话
+title: Pig Latin
challengeType: 5
forumTopicId: 16039
dashedName: pig-latin
@@ -8,57 +8,55 @@ dashedName: pig-latin
# --description--
-[儿童黑话](http://en.wikipedia.org/wiki/Pig_Latin)，也叫 Pig Latin，是一种英语语言游戏。规则如下：
+Pig Latin is a way of altering English Words. The rules are as follows:
-\- 如果单词以辅音开头，就把第一个辅音字母或第一组辅音簇移到单词的结尾，并在后面加上 "ay"。
+\- If a word begins with a consonant, take the first consonant or consonant cluster, move it to the end of the word, and add "ay" to it.
-\- 如果单词以元音开头，只需要在结尾加上 "way"。
-
-在英语中，字母 a、e、i、o、u 为元音，其余的字母均为辅音。辅音簇的意思是连续的多个辅音字母。
+\- If a word begins with a vowel, just add "way" at the end.
# --instructions--
-请把传入的字符串根据上述规则翻译成儿童黑话并返回结果。输入的字符串一定是一个小写的英文单词。
+Translate the provided string to Pig Latin. Input strings are guaranteed to be English words in all lowercase.
# --hints--
-`translatePigLatin("california")` 应返回 "aliforniacay"。
+`translatePigLatin("california")` should return "aliforniacay".
```js
assert.deepEqual(translatePigLatin('california'), 'aliforniacay');
```
-`translatePigLatin("paragraphs")` 应返回 "aragraphspay"。
+`translatePigLatin("paragraphs")` should return "aragraphspay".
```js
assert.deepEqual(translatePigLatin('paragraphs'), 'aragraphspay');
```
-`translatePigLatin("glove")` 应返回 "oveglay"。
+`translatePigLatin("glove")` should return "oveglay".
```js
assert.deepEqual(translatePigLatin('glove'), 'oveglay');
```
-`translatePigLatin("algorithm")` 应返回 "algorithmway"。
+`translatePigLatin("algorithm")` should return "algorithmway".
```js
assert.deepEqual(translatePigLatin('algorithm'), 'algorithmway');
```
-`translatePigLatin("eight")` 应返回 "eightway"。
+`translatePigLatin("eight")` should return "eightway".
```js
assert.deepEqual(translatePigLatin('eight'), 'eightway');
```
-Should handle words where the first vowel comes in the middle of the word. `translatePigLatin("schwartz")` 应返回 "artzschway"。
+Should handle words where the first vowel comes in the middle of the word. `translatePigLatin("schwartz")` should return "artzschway".
```js
assert.deepEqual(translatePigLatin('schwartz'), 'artzschway');
```
-应可以处理不含元音的单词，`translatePigLatin("rhythm")` 应返回 "rhythmay"。
+Should handle words without vowels. `translatePigLatin("rhythm")` should return "rhythmay".
```js
assert.deepEqual(translatePigLatin('rhythm'), 'rhythmay');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/search-and-replace.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/search-and-replace.md
index 1465c7748c1..f4ca7362bd0 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/search-and-replace.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/search-and-replace.md
@@ -1,6 +1,6 @@
---
id: a0b5010f579e69b815e7c5d6
-title: 搜索与替换
+title: Search and Replace
challengeType: 5
forumTopicId: 16045
dashedName: search-and-replace
@@ -8,20 +8,20 @@ dashedName: search-and-replace
# --description--
-在这道题目中，我们需要写一个字符串的搜索与替换函数，它的返回值为完成替换后的新字符串。
+Perform a search and replace on the sentence using the arguments provided and return the new sentence.
-这个函数接收的第一个参数为待替换的句子。
+First argument is the sentence to perform the search and replace on.
-第二个参数为句中需要被替换的单词。
+Second argument is the word that you will be replacing (before).
-第三个参数为替换后的单词。
+Third argument is what you will be replacing the second argument with (after).
-**注意：**
-你需要保留被替换单词首字母的大小写格式。即如果传入的第二个参数为 "Book"，第三个参数为 "dog"，那么替换后的结果应为 "Dog"
+**Note**
+Preserve the case of the first character in the original word when you are replacing it. For example if you mean to replace the word "Book" with the word "dog", it should be replaced as "Dog"
# --hints--
-`myReplace("Let us go to the store", "store", "mall")` 应返回 "Let us go to the mall"。
+`myReplace("Let us go to the store", "store", "mall")` should return "Let us go to the mall".
```js
assert.deepEqual(
@@ -30,7 +30,7 @@ assert.deepEqual(
);
```
-`myReplace("He is Sleeping on the couch", "Sleeping", "sitting")` 应返回 "He is Sitting on the couch"。
+`myReplace("He is Sleeping on the couch", "Sleeping", "sitting")` should return "He is Sitting on the couch".
```js
assert.deepEqual(
@@ -39,7 +39,7 @@ assert.deepEqual(
);
```
-`myReplace("I think we should look up there", "up", "Down")` 应返回 "I think we should look down there"。
+`myReplace("I think we should look up there", "up", "Down")` should return "I think we should look down there".
```js
assert.deepEqual(
@@ -48,7 +48,7 @@ assert.deepEqual(
);
```
-`myReplace("This has a spellngi error", "spellngi", "spelling")` 应返回 "This has a spelling error"。
+`myReplace("This has a spellngi error", "spellngi", "spelling")` should return "This has a spelling error".
```js
assert.deepEqual(
@@ -57,7 +57,7 @@ assert.deepEqual(
);
```
-`myReplace("His name is Tom", "Tom", "john")` 应返回 "His name is John"。
+`myReplace("His name is Tom", "Tom", "john")` should return "His name is John".
```js
assert.deepEqual(
@@ -66,7 +66,7 @@ assert.deepEqual(
);
```
-`myReplace("Let us get back to more Coding", "Coding", "algorithms")` 应返回 "Let us get back to more Algorithms"。
+`myReplace("Let us get back to more Coding", "Coding", "algorithms")` should return "Let us get back to more Algorithms".
```js
assert.deepEqual(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/seek-and-destroy.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/seek-and-destroy.md
index 9af518070bf..5fbf7d5aa82 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/seek-and-destroy.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/seek-and-destroy.md
@@ -1,6 +1,6 @@
---
id: a39963a4c10bc8b4d4f06d7e
-title: 过滤数组元素
+title: Seek and Destroy
challengeType: 5
forumTopicId: 16046
dashedName: seek-and-destroy
@@ -8,38 +8,38 @@ dashedName: seek-and-destroy
# --description--
-在这道题目中，我们要写一个叫 `destroyer` 的函数。传入的第一个参数是一个数组，我们称他为初始数组。后续的参数数量是不确定的，可能有一个或多个。你需要做的是，从初始数组中移除所有与后续参数相等的元素，并返回移除元素后的数组。
+You will be provided with an initial array (the first argument in the destroyer function), followed by one or more arguments. Remove all elements from the initial array that are of the same value as these arguments.
-**注意：**
-你可以使用 `arguments` 对象，也可以使用 `...`，即“剩余参数”（Rest Parameters）语法。
+**Note**
+You have to use the `arguments` object.
# --hints--
-`destroyer([1, 2, 3, 1, 2, 3], 2, 3)` 应返回 `[1, 1]`。
+`destroyer([1, 2, 3, 1, 2, 3], 2, 3)` should return `[1, 1]`.
```js
assert.deepEqual(destroyer([1, 2, 3, 1, 2, 3], 2, 3), [1, 1]);
```
-`destroyer([1, 2, 3, 5, 1, 2, 3], 2, 3)` 应返回 `[1, 5, 1]`。
+`destroyer([1, 2, 3, 5, 1, 2, 3], 2, 3)` should return `[1, 5, 1]`.
```js
assert.deepEqual(destroyer([1, 2, 3, 5, 1, 2, 3], 2, 3), [1, 5, 1]);
```
-`destroyer([3, 5, 1, 2, 2], 2, 3, 5)` 应返回 `[1]`。
+`destroyer([3, 5, 1, 2, 2], 2, 3, 5)` should return `[1]`.
```js
assert.deepEqual(destroyer([3, 5, 1, 2, 2], 2, 3, 5), [1]);
```
-`destroyer([2, 3, 2, 3], 2, 3)` 应返回 `[]`。
+`destroyer([2, 3, 2, 3], 2, 3)` should return `[]`.
```js
assert.deepEqual(destroyer([2, 3, 2, 3], 2, 3), []);
```
-`destroyer(["tree", "hamburger", 53], "tree", 53)` 应返回 `["hamburger"]`。
+`destroyer(["tree", "hamburger", 53], "tree", 53)` should return `["hamburger"]`.
```js
assert.deepEqual(destroyer(['tree', 'hamburger', 53], 'tree', 53), [
@@ -47,7 +47,7 @@ assert.deepEqual(destroyer(['tree', 'hamburger', 53], 'tree', 53), [
]);
```
-`destroyer(["possum", "trollo", 12, "safari", "hotdog", 92, 65, "grandma", "bugati", "trojan", "yacht"], "yacht", "possum", "trollo", "safari", "hotdog", "grandma", "bugati", "trojan")` 应返回 `[12,92,65]`。
+`destroyer(["possum", "trollo", 12, "safari", "hotdog", 92, 65, "grandma", "bugati", "trojan", "yacht"], "yacht", "possum", "trollo", "safari", "hotdog", "grandma", "bugati", "trojan")` should return `[12,92,65]`.
```js
assert.deepEqual(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/smallest-common-multiple.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/smallest-common-multiple.md
index 2b82cc17e2e..e6c39f5ee60 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/smallest-common-multiple.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/smallest-common-multiple.md
@@ -1,6 +1,6 @@
---
id: ae9defd7acaf69703ab432ea
-title: 找出数字范围内的最小公倍数
+title: Smallest Common Multiple
challengeType: 5
forumTopicId: 16075
dashedName: smallest-common-multiple
@@ -8,45 +8,45 @@ dashedName: smallest-common-multiple
# --description--
-在这道题目中，我们需要写一个函数，它接收一个包含两个数字的数组参数 `arr`；它的返回值为这两个数字范围内所有数字（包含这两个数字）的最小公倍数。
+Find the smallest common multiple of the provided parameters that can be evenly divided by both, as well as by all sequential numbers in the range between these parameters.
-注意，较小数不一定总是出现在数组的第一个元素。
+The range will be an array of two numbers that will not necessarily be in numerical order.
-比如，传入 `[1, 3]`，那么函数的返回结果应为 1、2、3 的最小公倍数，即为 6。
+For example, if given 1 and 3, find the smallest common multiple of both 1 and 3 that is also evenly divisible by all numbers *between* 1 and 3. The answer here would be 6.
# --hints--
-`smallestCommons([1, 5])` 应返回 a number。
+`smallestCommons([1, 5])` should return a number.
```js
assert.deepEqual(typeof smallestCommons([1, 5]), 'number');
```
-`smallestCommons([1, 5])` 应返回 60。
+`smallestCommons([1, 5])` should return 60.
```js
assert.deepEqual(smallestCommons([1, 5]), 60);
```
-`smallestCommons([5, 1])` 应返回 60。
+`smallestCommons([5, 1])` should return 60.
```js
assert.deepEqual(smallestCommons([5, 1]), 60);
```
-`smallestCommons([2, 10])` 应返回 2520。
+`smallestCommons([2, 10])` should return 2520.
```js
assert.deepEqual(smallestCommons([2, 10]), 2520);
```
-`smallestCommons([1, 13])` 应返回 360360。
+`smallestCommons([1, 13])` should return 360360.
```js
assert.deepEqual(smallestCommons([1, 13]), 360360);
```
-`smallestCommons([23, 18])` 应返回 6056820。
+`smallestCommons([23, 18])` should return 6056820.
```js
assert.deepEqual(smallestCommons([23, 18]), 6056820);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sorted-union.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sorted-union.md
index 93fafeafa55..cbc9cd794a1 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sorted-union.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sorted-union.md
@@ -1,6 +1,6 @@
---
id: a105e963526e7de52b219be9
-title: 集合排序
+title: Sorted Union
challengeType: 5
forumTopicId: 16077
dashedName: sorted-union
@@ -8,29 +8,29 @@ dashedName: sorted-union
# --description--
-在这道题目中，我们需要写一个函数，它接收两个或多个数组为参数。我们需要首先求出所有数组的并集，然后进行去除重复元素的处理，并以数组的形式返回去重结果。
+Write a function that takes two or more arrays and returns a new array of unique values in the order of the original provided arrays.
-换句话说，所有数组中出现的所有值都应按其原始顺序包括在内，但最终数组中不得重复。
+In other words, all values present from all arrays should be included in their original order, but with no duplicates in the final array.
-去重后的数字应按其出现在参数中的原始顺序排序，最终数组不应按数字大小进行排序。
+The unique numbers should be sorted by their original order, but the final array should not be sorted in numerical order.
-如有疑问，请先浏览下方的测试用例。
+Check the assertion tests for examples.
# --hints--
-`uniteUnique([1, 3, 2], [5, 2, 1, 4], [2, 1])` 应返回 `[1, 3, 2, 5, 4]`。
+`uniteUnique([1, 3, 2], [5, 2, 1, 4], [2, 1])` should return `[1, 3, 2, 5, 4]`.
```js
assert.deepEqual(uniteUnique([1, 3, 2], [5, 2, 1, 4], [2, 1]), [1, 3, 2, 5, 4]);
```
-`uniteUnique([1, 2, 3], [5, 2, 1])` 应返回 `[1, 2, 3, 5]`。
+`uniteUnique([1, 2, 3], [5, 2, 1])` should return `[1, 2, 3, 5]`.
```js
assert.deepEqual(uniteUnique([1, 2, 3], [5, 2, 1]), [1, 2, 3, 5]);
```
-`uniteUnique([1, 2, 3], [5, 2, 1, 4], [2, 1], [6, 7, 8])` 应返回 `[1, 2, 3, 5, 4, 6, 7, 8]`。
+`uniteUnique([1, 2, 3], [5, 2, 1, 4], [2, 1], [6, 7, 8])` should return `[1, 2, 3, 5, 4, 6, 7, 8]`.
```js
assert.deepEqual(uniteUnique([1, 2, 3], [5, 2, 1, 4], [2, 1], [6, 7, 8]), [
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/spinal-tap-case.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/spinal-tap-case.md
index c0cb11076fa..c68938ce893 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/spinal-tap-case.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/spinal-tap-case.md
@@ -1,6 +1,6 @@
---
id: a103376db3ba46b2d50db289
-title: 短线连接格式
+title: Spinal Tap Case
challengeType: 5
forumTopicId: 16078
dashedName: spinal-tap-case
@@ -8,23 +8,23 @@ dashedName: spinal-tap-case
# --description--
-在这道题目中，我们需要写一个函数，把一个字符串转换为“短线连接格式”。短线连接格式的意思是，所有字母都是小写，且用 `-` 连接。比如，`Hello World` 的短线连接格式为 `hello-world`；`I love_Javascript-VeryMuch` 的短线连接格式为 `i-love-javascript-very-much`。
+Convert a string to spinal case. Spinal case is all-lowercase-words-joined-by-dashes.
# --hints--
-`spinalCase("This Is Spinal Tap")` 应返回 `"this-is-spinal-tap"`。
+`spinalCase("This Is Spinal Tap")` should return `"this-is-spinal-tap"`.
```js
assert.deepEqual(spinalCase('This Is Spinal Tap'), 'this-is-spinal-tap');
```
-`spinalCase("thisIsSpinalTap")` 应返回 `"this-is-spinal-tap"`。
+`spinalCase("thisIsSpinalTap")` should return `"this-is-spinal-tap"`.
```js
assert.strictEqual(spinalCase('thisIsSpinalTap'), 'this-is-spinal-tap');
```
-`spinalCase("The_Andy_Griffith_Show")` 应返回 `"the-andy-griffith-show"`。
+`spinalCase("The_Andy_Griffith_Show")` should return `"the-andy-griffith-show"`.
```js
assert.strictEqual(
@@ -33,7 +33,7 @@ assert.strictEqual(
);
```
-`spinalCase("Teletubbies say Eh-oh")` 应返回 `"teletubbies-say-eh-oh"`。
+`spinalCase("Teletubbies say Eh-oh")` should return `"teletubbies-say-eh-oh"`.
```js
assert.strictEqual(
@@ -42,7 +42,7 @@ assert.strictEqual(
);
```
-`spinalCase("AllThe-small Things")` 应返回 `"all-the-small-things"`。
+`spinalCase("AllThe-small Things")` should return `"all-the-small-things"`.
```js
assert.strictEqual(spinalCase('AllThe-small Things'), 'all-the-small-things');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/steamroller.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/steamroller.md
index fd3ab165ca0..5744ab7e207 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/steamroller.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/steamroller.md
@@ -1,6 +1,6 @@
---
id: ab306dbdcc907c7ddfc30830
-title: 数组扁平化
+title: Steamroller
challengeType: 5
forumTopicId: 16079
dashedName: steamroller
@@ -8,35 +8,35 @@ dashedName: steamroller
# --description--
-在这道题目中，我们需要写一个数组扁平化的函数。请注意考虑多层数组嵌套的情景。
+Flatten a nested array. You must account for varying levels of nesting.
# --hints--
-`steamrollArray([[["a"]], [["b"]]])` 应返回 `["a", "b"]`。
+`steamrollArray([[["a"]], [["b"]]])` should return `["a", "b"]`.
```js
assert.deepEqual(steamrollArray([[['a']], [['b']]]), ['a', 'b']);
```
-`steamrollArray([1, [2], [3, [[4]]]])` 应返回 `[1, 2, 3, 4]`。
+`steamrollArray([1, [2], [3, [[4]]]])` should return `[1, 2, 3, 4]`.
```js
assert.deepEqual(steamrollArray([1, [2], [3, [[4]]]]), [1, 2, 3, 4]);
```
-`steamrollArray([1, [], [3, [[4]]]])` 应返回 `[1, 3, 4]`。
+`steamrollArray([1, [], [3, [[4]]]])` should return `[1, 3, 4]`.
```js
assert.deepEqual(steamrollArray([1, [], [3, [[4]]]]), [1, 3, 4]);
```
-`steamrollArray([1, {}, [3, [[4]]]])` 应返回 `[1, {}, 3, 4]`。
+`steamrollArray([1, {}, [3, [[4]]]])` should return `[1, {}, 3, 4]`.
```js
assert.deepEqual(steamrollArray([1, {}, [3, [[4]]]]), [1, {}, 3, 4]);
```
-代码中不应使用 `Array.prototype.flat()` 或 `Array.prototype.flatMap()` 方法。
+Your solution should not use the `Array.prototype.flat()` or `Array.prototype.flatMap()` methods.
```js
assert(!code.match(/\.\s*flat\s*\(/) && !code.match(/\.\s*flatMap\s*\(/));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sum-all-numbers-in-a-range.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sum-all-numbers-in-a-range.md
index bc2dd9d1879..a7dbdaeecc9 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sum-all-numbers-in-a-range.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sum-all-numbers-in-a-range.md
@@ -1,6 +1,6 @@
---
id: a3566b1109230028080c9345
-title: 范围内的数字求和
+title: Sum All Numbers in a Range
challengeType: 5
forumTopicId: 16083
dashedName: sum-all-numbers-in-a-range
@@ -8,37 +8,37 @@ dashedName: sum-all-numbers-in-a-range
# --description--
-给出一个含有两个数字的数组，我们需要写一个函数，让它返回这两个数字间所有数字（包含这两个数字）的总和。注意，较小数不一定总是出现在数组的第一个元素。
+We'll pass you an array of two numbers. Return the sum of those two numbers plus the sum of all the numbers between them. The lowest number will not always come first.
-例如，`sumAll([4,1])` 应返回 `10`，因为从 1 到 4（包含 1、4）的所有数字的和是 `10`。
+For example, `sumAll([4,1])` should return `10` because sum of all the numbers between 1 and 4 (both inclusive) is `10`.
# --hints--
-`sumAll([1, 4])` 应返回一个数字。
+`sumAll([1, 4])` should return a number.
```js
assert(typeof sumAll([1, 4]) === 'number');
```
-`sumAll([1, 4])` 应返回 10。
+`sumAll([1, 4])` should return 10.
```js
assert.deepEqual(sumAll([1, 4]), 10);
```
-`sumAll([4, 1])` 应返回 10。
+`sumAll([4, 1])` should return 10.
```js
assert.deepEqual(sumAll([4, 1]), 10);
```
-`sumAll([5, 10])` 应返回 45。
+`sumAll([5, 10])` should return 45.
```js
assert.deepEqual(sumAll([5, 10]), 45);
```
-`sumAll([10, 5])` 应返回 45。
+`sumAll([10, 5])` should return 45.
```js
assert.deepEqual(sumAll([10, 5]), 45);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sum-all-odd-fibonacci-numbers.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sum-all-odd-fibonacci-numbers.md
index 7827ca93bf7..0c5b625e71b 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sum-all-odd-fibonacci-numbers.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sum-all-odd-fibonacci-numbers.md
@@ -1,6 +1,6 @@
---
id: a5229172f011153519423690
-title: 求斐波那契数列中的奇数之和
+title: Sum All Odd Fibonacci Numbers
challengeType: 5
forumTopicId: 16084
dashedName: sum-all-odd-fibonacci-numbers
@@ -8,45 +8,45 @@ dashedName: sum-all-odd-fibonacci-numbers
# --description--
-在这道题目中，我们需要写一个函数，参数为一个正整数 `num`，返回值为斐波那契数列中，小于或等于 `num` 的奇数之和。
+Given a positive integer `num`, return the sum of all odd Fibonacci numbers that are less than or equal to `num`.
-斐波那契数列中，第一和第二个数字都是 1，后面的每个数字由之前两数相加得出。斐波那契数列的前六个数字分别为：1、1、2、3、5、8。
+The first two numbers in the Fibonacci sequence are 1 and 1. Every additional number in the sequence is the sum of the two previous numbers. The first six numbers of the Fibonacci sequence are 1, 1, 2, 3, 5 and 8.
-比如，`sumFibs(10)` 应该返回 `10`。因为斐波那契数列中，比 `10` 小的数字只有 1、1、3、5。
+For example, `sumFibs(10)` should return `10` because all odd Fibonacci numbers less than or equal to `10` are 1, 1, 3, and 5.
# --hints--
-`sumFibs(1)` 应返回一个数字。
+`sumFibs(1)` should return a number.
```js
assert(typeof sumFibs(1) === 'number');
```
-`sumFibs(1000)` 应返回 1785。
+`sumFibs(1000)` should return 1785.
```js
assert(sumFibs(1000) === 1785);
```
-`sumFibs(4000000)` 应返回 4613732。
+`sumFibs(4000000)` should return 4613732.
```js
assert(sumFibs(4000000) === 4613732);
```
-`sumFibs(4)` 应返回 5。
+`sumFibs(4)` should return 5.
```js
assert(sumFibs(4) === 5);
```
-`sumFibs(75024)` 应返回 60696。
+`sumFibs(75024)` should return 60696.
```js
assert(sumFibs(75024) === 60696);
```
-`sumFibs(75025)` 应返回 135721。
+`sumFibs(75025)` should return 135721.
```js
assert(sumFibs(75025) === 135721);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sum-all-primes.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sum-all-primes.md
index 5c71f40187f..57295dd11e5 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sum-all-primes.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/sum-all-primes.md
@@ -1,6 +1,6 @@
---
id: a3bfc1673c0526e06d3ac698
-title: 质数求和
+title: Sum All Primes
challengeType: 5
forumTopicId: 16085
dashedName: sum-all-primes
@@ -8,25 +8,25 @@ dashedName: sum-all-primes
# --description--
-<dfn>质数</dfn>是大于 1 且仅可以被 1 和自己整除的数。比如，2 就是一个质数，因为它只可以被 1 和 2（它本身）整除。相反，4 不是质数，因为它可以被 1, 2 和 4 整除。
+A <dfn>prime number</dfn> is a whole number greater than 1 with exactly two divisors: 1 and itself. For example, 2 is a prime number because it is only divisible by 1 and 2. In contrast, 4 is not prime since it is divisible by 1, 2 and 4.
-请完成 `sumPrimes` 方法，使其返回小于等于传入参数数字的所有质数之和。
+Rewrite `sumPrimes` so it returns the sum of all prime numbers that are less than or equal to num.
# --hints--
-`sumPrimes(10)` 应返回一个数字。
+`sumPrimes(10)` should return a number.
```js
assert.deepEqual(typeof sumPrimes(10), 'number');
```
-`sumPrimes(10)` 应返回 17。
+`sumPrimes(10)` should return 17.
```js
assert.deepEqual(sumPrimes(10), 17);
```
-`sumPrimes(977)` 应返回 73156。
+`sumPrimes(977)` should return 73156.
```js
assert.deepEqual(sumPrimes(977), 73156);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/wherefore-art-thou.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/wherefore-art-thou.md
index 09054fabf38..b0a565be943 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/wherefore-art-thou.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/wherefore-art-thou.md
@@ -1,6 +1,6 @@
---
id: a8e512fbe388ac2f9198f0fa
-title: 找出包含特定键值对的对象
+title: Wherefore art thou
challengeType: 5
forumTopicId: 16092
dashedName: wherefore-art-thou
@@ -8,13 +8,13 @@ dashedName: wherefore-art-thou
# --description--
-在这道题目中，我们要写一个函数，它接收两个参数：第一个参数是对象数组，第二个参数是一个对象。我们需要从对象数组中找出与第二个参数的属性和属性值均相等或包含第二个参数的所有对象，并以对象数组的形式返回。其中，相等的意思是原数组中的对象与第二个参数中对象的所有键值对完全相等；包含的意思是只要第二个参数中对象的所有键存在于原数组对象中，且它们对应的值相同即可。
+Make a function that looks through an array of objects (first argument) and returns an array of all objects that have matching name and value pairs (second argument). Each name and value pair of the source object has to be present in the object from the collection if it is to be included in the returned array.
-比如，如果第一个参数是 `[{ first: "Romeo", last: "Montague" }, { first: "Mercutio", last: null }, { first: "Tybalt", last: "Capulet" }]`，第二个参数是 `{ last: "Capulet" }`。那么你需要以对象数组的形式返回第一个参数中的第三个元素，因为它包含第二个参数中定义的键 `last`，且对应的值 `"Capulet"` 相同
+For example, if the first argument is `[{ first: "Romeo", last: "Montague" }, { first: "Mercutio", last: null }, { first: "Tybalt", last: "Capulet" }]`, and the second argument is `{ last: "Capulet" }`, then you must return the third object from the array (the first argument), because it contains the name and its value, that was passed on as the second argument.
# --hints--
-`whatIsInAName([{ first: "Romeo", last: "Montague" }, { first: "Mercutio", last: null }, { first: "Tybalt", last: "Capulet" }], { last: "Capulet" })` 应返回 `[{ first: "Tybalt", last: "Capulet" }]`。
+`whatIsInAName([{ first: "Romeo", last: "Montague" }, { first: "Mercutio", last: null }, { first: "Tybalt", last: "Capulet" }], { last: "Capulet" })` should return `[{ first: "Tybalt", last: "Capulet" }]`.
```js
assert.deepEqual(
@@ -30,7 +30,7 @@ assert.deepEqual(
);
```
-`whatIsInAName([{ "apple": 1 }, { "apple": 1 }, { "apple": 1, "bat": 2 }], { "apple": 1 })` 应返回 `[{ "apple": 1 }, { "apple": 1 }, { "apple": 1, "bat": 2 }]`。
+`whatIsInAName([{ "apple": 1 }, { "apple": 1 }, { "apple": 1, "bat": 2 }], { "apple": 1 })` should return `[{ "apple": 1 }, { "apple": 1 }, { "apple": 1, "bat": 2 }]`.
```js
assert.deepEqual(
@@ -41,7 +41,7 @@ assert.deepEqual(
);
```
-`whatIsInAName([{ "apple": 1, "bat": 2 }, { "bat": 2 }, { "apple": 1, "bat": 2, "cookie": 2 }], { "apple": 1, "bat": 2 })` 应返回 `[{ "apple": 1, "bat": 2 }, { "apple": 1, "bat": 2, "cookie": 2 }]`。
+`whatIsInAName([{ "apple": 1, "bat": 2 }, { "bat": 2 }, { "apple": 1, "bat": 2, "cookie": 2 }], { "apple": 1, "bat": 2 })` should return `[{ "apple": 1, "bat": 2 }, { "apple": 1, "bat": 2, "cookie": 2 }]`.
```js
assert.deepEqual(
@@ -56,7 +56,7 @@ assert.deepEqual(
);
```
-`whatIsInAName([{ "apple": 1, "bat": 2 }, { "apple": 1 }, { "apple": 1, "bat": 2, "cookie": 2 }], { "apple": 1, "cookie": 2 })` 应返回 `[{ "apple": 1, "bat": 2, "cookie": 2 }]`。
+`whatIsInAName([{ "apple": 1, "bat": 2 }, { "apple": 1 }, { "apple": 1, "bat": 2, "cookie": 2 }], { "apple": 1, "cookie": 2 })` should return `[{ "apple": 1, "bat": 2, "cookie": 2 }]`.
```js
assert.deepEqual(
@@ -68,7 +68,7 @@ assert.deepEqual(
);
```
-`whatIsInAName([{ "apple": 1, "bat": 2 }, { "apple": 1 }, { "apple": 1, "bat": 2, "cookie": 2 }, { "bat":2 }], { "apple": 1, "bat": 2 })` 应返回 `[{ "apple": 1, "bat": 2 }, { "apple": 1, "bat": 2, "cookie":2 }]`。
+`whatIsInAName([{ "apple": 1, "bat": 2 }, { "apple": 1 }, { "apple": 1, "bat": 2, "cookie": 2 }, { "bat":2 }], { "apple": 1, "bat": 2 })` should return `[{ "apple": 1, "bat": 2 }, { "apple": 1, "bat": 2, "cookie":2 }]`.
```js
assert.deepEqual(
@@ -88,7 +88,7 @@ assert.deepEqual(
);
```
-`whatIsInAName([{"a": 1, "b": 2, "c": 3}], {"a": 1, "b": 9999, "c": 3})` 应返回 `[]`。
+`whatIsInAName([{"a": 1, "b": 2, "c": 3}], {"a": 1, "b": 9999, "c": 3})` should return `[]`
```js
assert.deepEqual(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/caesars-cipher.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/caesars-cipher.md
index e01f6ad62d0..7e0a8c9f362 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/caesars-cipher.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/caesars-cipher.md
@@ -1,6 +1,6 @@
---
id: 56533eb9ac21ba0edf2244e2
-title: 凯撒密码
+title: Caesars Cipher
challengeType: 5
forumTopicId: 16003
dashedName: caesars-cipher
@@ -8,35 +8,35 @@ dashedName: caesars-cipher
# --description--
-<dfn>凯撒密码</dfn>是最简单且最广为人知的<dfn>密码</dfn>，也被称为<dfn>移位密码</dfn>。在移位密码中，明文中的字母通过按照一个固定数目进行偏移后被替换成新的字母。
+One of the simplest and most widely known <dfn>ciphers</dfn> is a <dfn>Caesar cipher</dfn>, also known as a <dfn>shift cipher</dfn>. In a shift cipher the meanings of the letters are shifted by some set amount.
-[ROT13](https://en.wikipedia.org/wiki/ROT13) 是一个被广泛使用的加密技术，明文中的所有字母都被移动 13 位。因此有：'A' ↔ 'N'、'B' ↔ 'O' 等等。
+A common modern use is the [ROT13](https://en.wikipedia.org/wiki/ROT13) cipher, where the values of the letters are shifted by 13 places. Thus 'A' ↔ 'N', 'B' ↔ 'O' and so on.
-请编写一个函数，用于解码一个被 [ROT13](https://en.wikipedia.org/wiki/ROT13) 编码的字符串，然后返回解码后的结果。
+Write a function which takes a [ROT13](https://en.wikipedia.org/wiki/ROT13) encoded string as input and returns a decoded string.
-所有解码后的字母都必须为字母大写。请不要解码非字母的字符（例如，空格、标点符号），但你需要在结果中保留它们。
+All letters will be uppercase. Do not transform any non-alphabetic character (i.e. spaces, punctuation), but do pass them on.
# --hints--
-`rot13("SERR PBQR PNZC")` 应解码为 `FREE CODE CAMP`。
+`rot13("SERR PBQR PNZC")` should decode to `FREE CODE CAMP`
```js
assert(rot13('SERR PBQR PNZC') === 'FREE CODE CAMP');
```
-`rot13("SERR CVMMN!")` 应解码为 `FREE PIZZA!`。
+`rot13("SERR CVMMN!")` should decode to `FREE PIZZA!`
```js
assert(rot13('SERR CVMMN!') === 'FREE PIZZA!');
```
-`rot13("SERR YBIR?")` 应解码为 `FREE LOVE?`。
+`rot13("SERR YBIR?")` should decode to `FREE LOVE?`
```js
assert(rot13('SERR YBIR?') === 'FREE LOVE?');
```
-`rot13("GUR DHVPX OEBJA SBK WHZCF BIRE GUR YNML QBT.")` 应解码为 `THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG.`。
+`rot13("GUR DHVPX OEBJA SBK WHZCF BIRE GUR YNML QBT.")` should decode to `THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG.`
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/cash-register.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/cash-register.md
index 8a4ca1dac89..49336bf3738 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/cash-register.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/cash-register.md
@@ -1,6 +1,6 @@
---
id: aa2e6f85cab2ab736c9a9b24
-title: 计算找零
+title: Cash Register
challengeType: 5
forumTopicId: 16012
dashedName: cash-register
@@ -8,21 +8,21 @@ dashedName: cash-register
# --description--
-请编写一个用于收银机的函数 `checkCashRegister()`：它的第一个参数为售价 `price`、第二个参数为支付金额 `cash`、第三个参数为收银机內的金额 `cid`。
+Design a cash register drawer function `checkCashRegister()` that accepts purchase price as the first argument (`price`), payment as the second argument (`cash`), and cash-in-drawer (`cid`) as the third argument.
-`cid` 是包含货币面值的二维数组。
+`cid` is a 2D array listing available currency.
-函数 `checkCashRegister()` 应返回含有 `status` 属性和 `change` 属性的对象。
+The `checkCashRegister()` function should always return an object with a `status` key and a `change` key.
-如果收银机內的金额少于应找回的零钱数，或者你无法返回确切的数目时，返回 `{status: "INSUFFICIENT_FUNDS", change: []}`。
+Return `{status: "INSUFFICIENT_FUNDS", change: []}` if cash-in-drawer is less than the change due, or if you cannot return the exact change.
-如果收银机內的金额恰好等于应找回的零钱数，返回 `{status: "CLOSED", change: [...]}`，其中 `change` 的属性值就是收银机內的金额。
+Return `{status: "CLOSED", change: [...]}` with cash-in-drawer as the value for the key `change` if it is equal to the change due.
-否则，返回 `{status: "OPEN", change: [...]}`，其中 `change` 键值是应找回的零钱数，并将找零的面值由高到低排序。
+Otherwise, return `{status: "OPEN", change: [...]}`, with the change due in coins and bills, sorted in highest to lowest order, as the value of the `change` key.
-<table class='table table-striped'><tbody><tr><th>货币单位 Unit</th><th>面值</th></tr><tr><td>Penny</td><td>$0.01 (PENNY)</td></tr><tr><td>Nickel</td><td>$0.05 (NICKEL)</td></tr><tr><td>Dime</td><td>$0.1 (DIME)</td></tr><tr><td>Quarter</td><td>$0.25 (QUARTER)</td></tr><tr><td>Dollar</td><td>$1 (ONE)</td></tr><tr><td>Five Dollars</td><td>$5 (FIVE)</td></tr><tr><td>Ten Dollars</td><td>$10 (TEN)</td></tr><tr><td>Twenty Dollars</td><td>$20 (TWENTY)</td></tr><tr><td>One-hundred Dollars</td><td>$100 (ONE HUNDRED)</td></tr></tbody></table>
+<table class='table table-striped'><tbody><tr><th>Currency Unit</th><th>Amount</th></tr><tr><td>Penny</td><td>$0.01 (PENNY)</td></tr><tr><td>Nickel</td><td>$0.05 (NICKEL)</td></tr><tr><td>Dime</td><td>$0.1 (DIME)</td></tr><tr><td>Quarter</td><td>$0.25 (QUARTER)</td></tr><tr><td>Dollar</td><td>$1 (ONE)</td></tr><tr><td>Five Dollars</td><td>$5 (FIVE)</td></tr><tr><td>Ten Dollars</td><td>$10 (TEN)</td></tr><tr><td>Twenty Dollars</td><td>$20 (TWENTY)</td></tr><tr><td>One-hundred Dollars</td><td>$100 (ONE HUNDRED)</td></tr></tbody></table>
-以下为 `cid` 参数的示例：
+See below for an example of a cash-in-drawer array:
```js
[
@@ -40,7 +40,7 @@ dashedName: cash-register
# --hints--
-`checkCashRegister(19.5, 20, [["PENNY", 1.01], ["NICKEL", 2.05], ["DIME", 3.1], ["QUARTER", 4.25], ["ONE", 90], ["FIVE", 55], ["TEN", 20], ["TWENTY", 60], ["ONE HUNDRED", 100]])` 应返回一个对象。
+`checkCashRegister(19.5, 20, [["PENNY", 1.01], ["NICKEL", 2.05], ["DIME", 3.1], ["QUARTER", 4.25], ["ONE", 90], ["FIVE", 55], ["TEN", 20], ["TWENTY", 60], ["ONE HUNDRED", 100]])` should return an object.
```js
assert.deepEqual(
@@ -61,7 +61,7 @@ assert.deepEqual(
);
```
-`checkCashRegister(19.5, 20, [["PENNY", 1.01], ["NICKEL", 2.05], ["DIME", 3.1], ["QUARTER", 4.25], ["ONE", 90], ["FIVE", 55], ["TEN", 20], ["TWENTY", 60], ["ONE HUNDRED", 100]])` 应返回 `{status: "OPEN", change: [["QUARTER", 0.5]]}`。
+`checkCashRegister(19.5, 20, [["PENNY", 1.01], ["NICKEL", 2.05], ["DIME", 3.1], ["QUARTER", 4.25], ["ONE", 90], ["FIVE", 55], ["TEN", 20], ["TWENTY", 60], ["ONE HUNDRED", 100]])` should return `{status: "OPEN", change: [["QUARTER", 0.5]]}`.
```js
assert.deepEqual(
@@ -80,7 +80,7 @@ assert.deepEqual(
);
```
-`checkCashRegister(3.26, 100, [["PENNY", 1.01], ["NICKEL", 2.05], ["DIME", 3.1], ["QUARTER", 4.25], ["ONE", 90], ["FIVE", 55], ["TEN", 20], ["TWENTY", 60], ["ONE HUNDRED", 100]])` 应返回 `{status: "OPEN", change: [["TWENTY", 60], ["TEN", 20], ["FIVE", 15], ["ONE", 1], ["QUARTER", 0.5], ["DIME", 0.2], ["PENNY", 0.04]]}`。
+`checkCashRegister(3.26, 100, [["PENNY", 1.01], ["NICKEL", 2.05], ["DIME", 3.1], ["QUARTER", 4.25], ["ONE", 90], ["FIVE", 55], ["TEN", 20], ["TWENTY", 60], ["ONE HUNDRED", 100]])` should return `{status: "OPEN", change: [["TWENTY", 60], ["TEN", 20], ["FIVE", 15], ["ONE", 1], ["QUARTER", 0.5], ["DIME", 0.2], ["PENNY", 0.04]]}`.
```js
assert.deepEqual(
@@ -110,7 +110,7 @@ assert.deepEqual(
);
```
-`checkCashRegister(19.5, 20, [["PENNY", 0.01], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 0], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]])` 应返回 `{status: "INSUFFICIENT_FUNDS", change: []}`。
+`checkCashRegister(19.5, 20, [["PENNY", 0.01], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 0], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]])` should return `{status: "INSUFFICIENT_FUNDS", change: []}`.
```js
assert.deepEqual(
@@ -129,7 +129,7 @@ assert.deepEqual(
);
```
-`checkCashRegister(19.5, 20, [["PENNY", 0.01], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 1], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]])` 应返回 `{status: "INSUFFICIENT_FUNDS", change: []}`。
+`checkCashRegister(19.5, 20, [["PENNY", 0.01], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 1], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]])` should return `{status: "INSUFFICIENT_FUNDS", change: []}`.
```js
assert.deepEqual(
@@ -148,7 +148,7 @@ assert.deepEqual(
);
```
-`checkCashRegister(19.5, 20, [["PENNY", 0.5], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 0], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]])` 应返回 `{status: "CLOSED", change: [["PENNY", 0.5], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 0], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]]}`。
+`checkCashRegister(19.5, 20, [["PENNY", 0.5], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 0], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]])` should return `{status: "CLOSED", change: [["PENNY", 0.5], ["NICKEL", 0], ["DIME", 0], ["QUARTER", 0], ["ONE", 0], ["FIVE", 0], ["TEN", 0], ["TWENTY", 0], ["ONE HUNDRED", 0]]}`.
```js
assert.deepEqual(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/palindrome-checker.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/palindrome-checker.md
index c365e939447..49ec05e0e88 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/palindrome-checker.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/palindrome-checker.md
@@ -1,6 +1,6 @@
---
id: aaa48de84e1ecc7c742e1124
-title: 回文检查器
+title: Palindrome Checker
challengeType: 5
forumTopicId: 16004
dashedName: palindrome-checker
@@ -8,91 +8,91 @@ dashedName: palindrome-checker
# --description--
-如果传入的字符串是回文字符串，则返回 `true`，否则返回 `false`。
+Return `true` if the given string is a palindrome. Otherwise, return `false`.
-<dfn>Palindrome（回文）</dfn>，指在忽略标点符号、大小写和空格的前提下，正着读和反着读一模一样。
+A <dfn>palindrome</dfn> is a word or sentence that's spelled the same way both forward and backward, ignoring punctuation, case, and spacing.
-**注意：**检查回文时，你需要先去除**所有非字母数字的字符**（标点、空格和符号），并将所有字母都转换成大写或都转换成小写。
+**Note:** You'll need to remove **all non-alphanumeric characters** (punctuation, spaces and symbols) and turn everything into the same case (lower or upper case) in order to check for palindromes.
-我们会传入不同格式的字符串，例如：`"racecar"`、`"RaceCar"`、`"race CAR"` 等等。
+We'll pass strings with varying formats, such as `"racecar"`, `"RaceCar"`, and `"race CAR"` among others.
-我们也会传入一些包含特殊符号的字符串，例如 `"2A3*3a2"`、`"2A3 3a2"`、`"2_A3*3#A2"`。
+We'll also pass strings with special symbols, such as `"2A3*3a2"`, `"2A3 3a2"`, and `"2_A3*3#A2"`.
# --hints--
-`palindrome("eye")` 应返回一个布尔值。
+`palindrome("eye")` should return a boolean.
```js
assert(typeof palindrome('eye') === 'boolean');
```
-`palindrome("eye")` 应返回 true。
+`palindrome("eye")` should return true.
```js
assert(palindrome('eye') === true);
```
-`palindrome("_eye")` 应返回 true。
+`palindrome("_eye")` should return true.
```js
assert(palindrome('_eye') === true);
```
-`palindrome("race car")` 应返回 true。
+`palindrome("race car")` should return true.
```js
assert(palindrome('race car') === true);
```
-`palindrome("not a palindrome")` 应返回 false。
+`palindrome("not a palindrome")` should return false.
```js
assert(palindrome('not a palindrome') === false);
```
-`palindrome("A man, a plan, a canal. Panama")` 应返回 true。
+`palindrome("A man, a plan, a canal. Panama")` should return true.
```js
assert(palindrome('A man, a plan, a canal. Panama') === true);
```
-`palindrome("never odd or even")` 应返回 true。
+`palindrome("never odd or even")` should return true.
```js
assert(palindrome('never odd or even') === true);
```
-`palindrome("nope")` 应返回 false。
+`palindrome("nope")` should return false.
```js
assert(palindrome('nope') === false);
```
-`palindrome("almostomla")` 应返回 false。
+`palindrome("almostomla")` should return false.
```js
assert(palindrome('almostomla') === false);
```
-`palindrome("My age is 0, 0 si ega ym.")` 应返回 true。
+`palindrome("My age is 0, 0 si ega ym.")` should return true.
```js
assert(palindrome('My age is 0, 0 si ega ym.') === true);
```
-`palindrome("1 eye for of 1 eye.")` 应返回 false。
+`palindrome("1 eye for of 1 eye.")` should return false.
```js
assert(palindrome('1 eye for of 1 eye.') === false);
```
-`palindrome("0_0 (: /-\ :) 0-0")` 应返回 true。
+`palindrome("0_0 (: /-\ :) 0-0")` should return true.
```js
assert(palindrome('0_0 (: /- :) 0-0') === true);
```
-`palindrome("five|\_/|four")` 应返回 false。
+`palindrome("five|\_/|four")` should return false.
```js
assert(palindrome('five|_/|four') === false);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/roman-numeral-converter.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/roman-numeral-converter.md
index 466e0cbc0ab..9f750e7aed5 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/roman-numeral-converter.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/roman-numeral-converter.md
@@ -1,6 +1,6 @@
---
id: a7f4d8f2483413a6ce226cac
-title: 罗马数字转换器
+title: Roman Numeral Converter
challengeType: 5
forumTopicId: 16044
dashedName: roman-numeral-converter
@@ -8,163 +8,163 @@ dashedName: roman-numeral-converter
# --description--
-把传入的数字转为罗马数字。
+Convert the given number into a roman numeral.
-转换后的[罗马数字](http://www.mathsisfun.com/roman-numerals.html)字母必须都是大写。
+All [roman numerals](http://www.mathsisfun.com/roman-numerals.html) answers should be provided in upper-case.
# --hints--
-`convertToRoman(2)` 应返回 "II"。
+`convertToRoman(2)` should return "II".
```js
assert.deepEqual(convertToRoman(2), 'II');
```
-`convertToRoman(3)` 应返回 "III"。
+`convertToRoman(3)` should return "III".
```js
assert.deepEqual(convertToRoman(3), 'III');
```
-`convertToRoman(4)` 应返回 "IV"。
+`convertToRoman(4)` should return "IV".
```js
assert.deepEqual(convertToRoman(4), 'IV');
```
-`convertToRoman(5)` 应返回 "V"。
+`convertToRoman(5)` should return "V".
```js
assert.deepEqual(convertToRoman(5), 'V');
```
-`convertToRoman(9)` 应返回 "IX"。
+`convertToRoman(9)` should return "IX".
```js
assert.deepEqual(convertToRoman(9), 'IX');
```
-`convertToRoman(12)` 应返回 "XII"。
+`convertToRoman(12)` should return "XII".
```js
assert.deepEqual(convertToRoman(12), 'XII');
```
-`convertToRoman(16)` 应返回 "XVI"。
+`convertToRoman(16)` should return "XVI".
```js
assert.deepEqual(convertToRoman(16), 'XVI');
```
-`convertToRoman(29)` 应返回 "XXIX"。
+`convertToRoman(29)` should return "XXIX".
```js
assert.deepEqual(convertToRoman(29), 'XXIX');
```
-`convertToRoman(44)` 应返回 "XLIV"。
+`convertToRoman(44)` should return "XLIV".
```js
assert.deepEqual(convertToRoman(44), 'XLIV');
```
-`convertToRoman(45)` 应返回 "XLV"。
+`convertToRoman(45)` should return "XLV"
```js
assert.deepEqual(convertToRoman(45), 'XLV');
```
-`convertToRoman(68)` 应返回 "LXVIII"。
+`convertToRoman(68)` should return "LXVIII"
```js
assert.deepEqual(convertToRoman(68), 'LXVIII');
```
-`convertToRoman(83)` 应返回 "LXXXIII"。
+`convertToRoman(83)` should return "LXXXIII"
```js
assert.deepEqual(convertToRoman(83), 'LXXXIII');
```
-`convertToRoman(97)` 应返回 "XCVII"。
+`convertToRoman(97)` should return "XCVII"
```js
assert.deepEqual(convertToRoman(97), 'XCVII');
```
-`convertToRoman(99)` 应返回 "XCIX"。
+`convertToRoman(99)` should return "XCIX"
```js
assert.deepEqual(convertToRoman(99), 'XCIX');
```
-`convertToRoman(400)` 应返回 "CD"。
+`convertToRoman(400)` should return "CD"
```js
assert.deepEqual(convertToRoman(400), 'CD');
```
-`convertToRoman(500)` 应返回 "D"。
+`convertToRoman(500)` should return "D"
```js
assert.deepEqual(convertToRoman(500), 'D');
```
-`convertToRoman(501)` 应返回 "DI"。
+`convertToRoman(501)` should return "DI"
```js
assert.deepEqual(convertToRoman(501), 'DI');
```
-`convertToRoman(649)` 应返回 "DCXLIX"。
+`convertToRoman(649)` should return "DCXLIX"
```js
assert.deepEqual(convertToRoman(649), 'DCXLIX');
```
-`convertToRoman(798)` 应返回 "DCCXCVIII"。
+`convertToRoman(798)` should return "DCCXCVIII"
```js
assert.deepEqual(convertToRoman(798), 'DCCXCVIII');
```
-`convertToRoman(891)` 应返回 "DCCCXCI"。
+`convertToRoman(891)` should return "DCCCXCI"
```js
assert.deepEqual(convertToRoman(891), 'DCCCXCI');
```
-`convertToRoman(1000)` 应返回 "M"。
+`convertToRoman(1000)` should return "M"
```js
assert.deepEqual(convertToRoman(1000), 'M');
```
-`convertToRoman(1004)` 应返回 "MIV"。
+`convertToRoman(1004)` should return "MIV"
```js
assert.deepEqual(convertToRoman(1004), 'MIV');
```
-`convertToRoman(1006)` 应返回 "MVI"。
+`convertToRoman(1006)` should return "MVI"
```js
assert.deepEqual(convertToRoman(1006), 'MVI');
```
-`convertToRoman(1023)` 应返回 "MXXIII"。
+`convertToRoman(1023)` should return "MXXIII"
```js
assert.deepEqual(convertToRoman(1023), 'MXXIII');
```
-`convertToRoman(2014)` 应返回 "MMXIV"。
+`convertToRoman(2014)` should return "MMXIV"
```js
assert.deepEqual(convertToRoman(2014), 'MMXIV');
```
-`convertToRoman(3999)` 应返回 "MMMCMXCIX"。
+`convertToRoman(3999)` should return "MMMCMXCIX"
```js
assert.deepEqual(convertToRoman(3999), 'MMMCMXCIX');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/telephone-number-validator.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/telephone-number-validator.md
index 7aab53bbff9..67d1515d0f2 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/telephone-number-validator.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/javascript-algorithms-and-data-structures-projects/telephone-number-validator.md
@@ -1,6 +1,6 @@
---
id: aff0395860f5d3034dc0bfc9
-title: 电话号码检查器
+title: Telephone Number Validator
challengeType: 5
forumTopicId: 16090
dashedName: telephone-number-validator
@@ -8,173 +8,173 @@ dashedName: telephone-number-validator
# --description--
-如果传入的字符串是一个有效的美国电话号码格式，则返回`true`。
+Return `true` if the passed string looks like a valid US phone number.
-只要是有效的美国电话号码的格式，用户可以按照他们的方式填写表单中的电话号码。以下是一些正确的例子（其他格式变形请参考以下例子）：
+The user may fill out the form field any way they choose as long as it has the format of a valid US number. The following are examples of valid formats for US numbers (refer to the tests below for other variants):
<blockquote>555-555-5555<br>(555)555-5555<br>(555) 555-5555<br>555 555 5555<br>5555555555<br>1 555 555 5555</blockquote>
-在这个挑战中，参数可能是 `800-692-7753` 或者 `8oo-six427676;laskdjf` 的号码。你的任务是根据上面不同的格式组合，判断它是否为有效的电话号码。其中，地区码（电话号码中的前三位）是必须的。如果提供国家代码，则国家代码只能为 `1`。如果传入的参数是有效的美国电话号码就返回 `true`，否则返回 `false`。
+For this challenge you will be presented with a string such as `800-692-7753` or `8oo-six427676;laskdjf`. Your job is to validate or reject the US phone number based on any combination of the formats provided above. The area code is required. If the country code is provided, you must confirm that the country code is `1`. Return `true` if the string is a valid US phone number; otherwise return `false`.
# --hints--
-`telephoneCheck("555-555-5555")` 应返回一个布尔值。
+`telephoneCheck("555-555-5555")` should return a boolean.
```js
assert(typeof telephoneCheck('555-555-5555') === 'boolean');
```
-`telephoneCheck("1 555-555-5555")` 应返回 true。
+`telephoneCheck("1 555-555-5555")` should return true.
```js
assert(telephoneCheck('1 555-555-5555') === true);
```
-`telephoneCheck("1 (555) 555-5555")` 应返回 true。
+`telephoneCheck("1 (555) 555-5555")` should return true.
```js
assert(telephoneCheck('1 (555) 555-5555') === true);
```
-`telephoneCheck("5555555555")` 应返回 true。
+`telephoneCheck("5555555555")` should return true.
```js
assert(telephoneCheck('5555555555') === true);
```
-`telephoneCheck("555-555-5555")` 应返回 true。
+`telephoneCheck("555-555-5555")` should return true.
```js
assert(telephoneCheck('555-555-5555') === true);
```
-`telephoneCheck("(555)555-5555")` 应返回 true。
+`telephoneCheck("(555)555-5555")` should return true.
```js
assert(telephoneCheck('(555)555-5555') === true);
```
-`telephoneCheck("1(555)555-5555")` 应返回 true。
+`telephoneCheck("1(555)555-5555")` should return true.
```js
assert(telephoneCheck('1(555)555-5555') === true);
```
-`telephoneCheck("555-5555")` 应返回 false。
+`telephoneCheck("555-5555")` should return false.
```js
assert(telephoneCheck('555-5555') === false);
```
-`telephoneCheck("5555555")` 应返回 false。
+`telephoneCheck("5555555")` should return false.
```js
assert(telephoneCheck('5555555') === false);
```
-`telephoneCheck("1 555)555-5555")` 应返回 false。
+`telephoneCheck("1 555)555-5555")` should return false.
```js
assert(telephoneCheck('1 555)555-5555') === false);
```
-`telephoneCheck("1 555 555 5555")` 应返回 true。
+`telephoneCheck("1 555 555 5555")` should return true.
```js
assert(telephoneCheck('1 555 555 5555') === true);
```
-`telephoneCheck("1 456 789 4444")` 应返回 true。
+`telephoneCheck("1 456 789 4444")` should return true.
```js
assert(telephoneCheck('1 456 789 4444') === true);
```
-`telephoneCheck("123**&!!asdf#")` 应返回 false。
+`telephoneCheck("123**&!!asdf#")` should return false.
```js
assert(telephoneCheck('123**&!!asdf#') === false);
```
-`telephoneCheck("55555555")` 应返回 false。
+`telephoneCheck("55555555")` should return false.
```js
assert(telephoneCheck('55555555') === false);
```
-`telephoneCheck("(6054756961)")` 应返回 false。
+`telephoneCheck("(6054756961)")` should return false
```js
assert(telephoneCheck('(6054756961)') === false);
```
-`telephoneCheck("2 (757) 622-7382")` 应返回 false。
+`telephoneCheck("2 (757) 622-7382")` should return false.
```js
assert(telephoneCheck('2 (757) 622-7382') === false);
```
-`telephoneCheck("0 (757) 622-7382")` 应返回 false。
+`telephoneCheck("0 (757) 622-7382")` should return false.
```js
assert(telephoneCheck('0 (757) 622-7382') === false);
```
-`telephoneCheck("-1 (757) 622-7382")` 应返回 false。
+`telephoneCheck("-1 (757) 622-7382")` should return false
```js
assert(telephoneCheck('-1 (757) 622-7382') === false);
```
-`telephoneCheck("2 757 622-7382")` 应返回 false。
+`telephoneCheck("2 757 622-7382")` should return false.
```js
assert(telephoneCheck('2 757 622-7382') === false);
```
-`telephoneCheck("10 (757) 622-7382")` 应返回 false。
+`telephoneCheck("10 (757) 622-7382")` should return false.
```js
assert(telephoneCheck('10 (757) 622-7382') === false);
```
-`telephoneCheck("27576227382")` 应返回 false。
+`telephoneCheck("27576227382")` should return false.
```js
assert(telephoneCheck('27576227382') === false);
```
-`telephoneCheck("(275)76227382")` 应返回 false。
+`telephoneCheck("(275)76227382")` should return false.
```js
assert(telephoneCheck('(275)76227382') === false);
```
-`telephoneCheck("2(757)6227382")` 应返回 false。
+`telephoneCheck("2(757)6227382")` should return false.
```js
assert(telephoneCheck('2(757)6227382') === false);
```
-`telephoneCheck("2(757)622-7382")` 应返回 false。
+`telephoneCheck("2(757)622-7382")` should return false.
```js
assert(telephoneCheck('2(757)622-7382') === false);
```
-`telephoneCheck("555)-555-5555")` 应返回 false。
+`telephoneCheck("555)-555-5555")` should return false.
```js
assert(telephoneCheck('555)-555-5555') === false);
```
-`telephoneCheck("(555-555-5555")` 应返回 false。
+`telephoneCheck("(555-555-5555")` should return false.
```js
assert(telephoneCheck('(555-555-5555') === false);
```
-`telephoneCheck("(555)5(55?)-5555")` 应返回 false。
+`telephoneCheck("(555)5(55?)-5555")` should return false.
```js
assert(telephoneCheck('(555)5(55?)-5555') === false);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/add-methods-after-inheritance.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/add-methods-after-inheritance.md
index f1abd71c610..4475a902b1e 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/add-methods-after-inheritance.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/add-methods-after-inheritance.md
@@ -1,6 +1,6 @@
---
id: 587d7db1367417b2b2512b87
-title: 继承后添加方法
+title: Add Methods After Inheritance
challengeType: 1
forumTopicId: 301315
dashedName: add-methods-after-inheritance
@@ -8,9 +8,9 @@ dashedName: add-methods-after-inheritance
# --description--
-从`父类`继承其`原型`对象的构造函数除了继承的方法之外，还可以有自己的方法。
+A constructor function that inherits its `prototype` object from a supertype constructor function can still have its own methods in addition to inherited methods.
-请看举例：`Bird`是一个构造函数，它继承了`Animal`构造函数的`原型`：
+For example, `Bird` is a constructor that inherits its `prototype` from `Animal`:
```js
function Animal() { }
@@ -22,7 +22,7 @@ Bird.prototype = Object.create(Animal.prototype);
Bird.prototype.constructor = Bird;
```
-除了从`Animal`构造函数继承的行为之外，还需要给`Bird`对象添加它独有的行为。这里，我们给`Bird`对象添加一个`fly()`函数。函数会以一种与其他构造函数相同的方式添加到`Bird`的`原型`中：
+In addition to what is inherited from `Animal`, you want to add behavior that is unique to `Bird` objects. Here, `Bird` will get a `fly()` function. Functions are added to `Bird's` `prototype` the same way as any constructor function:
```js
Bird.prototype.fly = function() {
@@ -30,7 +30,7 @@ Bird.prototype.fly = function() {
};
```
-现在`Bird`的实例中就有了`eat()`和`fly()`这两个方法：
+Now instances of `Bird` will have both `eat()` and `fly()` methods:
```js
let duck = new Bird();
@@ -40,40 +40,58 @@ duck.fly(); // prints "I'm flying!"
# --instructions--
-添加必要的代码，使得`Dog`对象继承`Animal`构造函数，并且把`Dog 原型`上的 constructor 属性设置为 Dog。然后给`Dog`对象添加一个`bark()`方法，这样的话，`beagle`将同时拥有`eat()`和`bark()`这两个方法。`bark()`方法中应该输出 "Woof!" 到控制台。
+Add all necessary code so the `Dog` object inherits from `Animal` and the `Dog's` `prototype` constructor is set to Dog. Then add a `bark()` method to the `Dog` object so that `beagle` can both `eat()` and `bark()`. The `bark()` method should print "Woof!" to the console.
# --hints--
-`Animal`应该没有`bark()`方法。
+`Animal` should not respond to the `bark()` method.
```js
assert(typeof Animal.prototype.bark == 'undefined');
```
-`Dog`应该继承了`Animal`构造函数的`eat()`方法。
+`Dog` should inherit the `eat()` method from `Animal`.
```js
assert(typeof Dog.prototype.eat == 'function');
```
-`Dog`应该有一个`bark()`方法作为`自身`属性。
+`Dog` should have the `bark()` method as an `own` property.
```js
assert(Dog.prototype.hasOwnProperty('bark'));
```
-`beagle`应该是`Animal`的一个`instanceof`。
+`beagle` should be an `instanceof` `Animal`.
```js
assert(beagle instanceof Animal);
```
-`beagle`的 constructor 属性应该被设置为`Dog`。
+The constructor for `beagle` should be set to `Dog`.
```js
assert(beagle.constructor === Dog);
```
+`beagle.eat()` should log `"nom nom nom"`
+
+```js
+console.log = function (msg) {
+ throw msg;
+};
+assert.throws(() => beagle.eat(), 'nom nom nom');
+```
+
+`beagle.bark()` should log `"Woof!"`
+
+```js
+console.log = function (msg) {
+ throw msg;
+};
+assert.throws(() => beagle.bark(), 'Woof!');
+```
+
# --seed--
## --seed-contents--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/change-the-prototype-to-a-new-object.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/change-the-prototype-to-a-new-object.md
index 61c2b5d6c40..5313994f242 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/change-the-prototype-to-a-new-object.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/change-the-prototype-to-a-new-object.md
@@ -1,6 +1,6 @@
---
id: 587d7daf367417b2b2512b7f
-title: 将原型更改为新对象
+title: Change the Prototype to a New Object
challengeType: 1
forumTopicId: 301316
dashedName: change-the-prototype-to-a-new-object
@@ -8,13 +8,13 @@ dashedName: change-the-prototype-to-a-new-object
# --description--
-到目前为止，你已经可以单独给`prototype`添加属性了：
+Up until now you have been adding properties to the `prototype` individually:
```js
Bird.prototype.numLegs = 2;
```
-这将在添加许多属性的时候变得单调乏味。
+This becomes tedious after more than a few properties.
```js
Bird.prototype.eat = function() {
@@ -26,7 +26,7 @@ Bird.prototype.describe = function() {
}
```
-一种更有效的方法就是给对象的`prototype`设置为一个已经包含了属性的新对象。这样一来，所有属性都可以一次性添加进来：
+A more efficient way is to set the `prototype` to a new object that already contains the properties. This way, the properties are added all at once:
```js
Bird.prototype = {
@@ -42,29 +42,29 @@ Bird.prototype = {
# --instructions--
-通过给`prototype`设置为新对象的方法，在`Dog`构造函数的`原型`上添加一个属性`numLegs`以及两个方法：`eat()`和`describe()`。
+Add the property `numLegs` and the two methods `eat()` and `describe()` to the `prototype` of `Dog` by setting the `prototype` to a new object.
# --hints--
-`Dog.prototype`应该被设置为一个新对象。
+`Dog.prototype` should be set to a new object.
```js
assert(/Dog\.prototype\s*?=\s*?{/.test(code));
```
-`Dog.prototype`应该拥有属性`numLegs`。
+`Dog.prototype` should have the property `numLegs`.
```js
assert(Dog.prototype.numLegs !== undefined);
```
-`Dog.prototype`应该拥有方法`eat()`。
+`Dog.prototype` should have the method `eat()`.
```js
assert(typeof Dog.prototype.eat === 'function');
```
-`Dog.prototype`应该拥有方法`describe()`。
+`Dog.prototype` should have the method `describe()`.
```js
assert(typeof Dog.prototype.describe === 'function');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/create-a-basic-javascript-object.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/create-a-basic-javascript-object.md
index 72ae7285a90..c71fb6cd852 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/create-a-basic-javascript-object.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/create-a-basic-javascript-object.md
@@ -1,6 +1,6 @@
---
id: 587d7dac367417b2b2512b73
-title: 创建一个基本的 JavaScript 对象
+title: Create a Basic JavaScript Object
challengeType: 1
forumTopicId: 301317
dashedName: create-a-basic-javascript-object
@@ -8,13 +8,13 @@ dashedName: create-a-basic-javascript-object
# --description--
-想想我们在生活中每天都可见到的事物：比如汽车、商店以及小鸟等。它们都是`物体`：即人们可以观察和与之互动的实体事物。
+Think about things people see every day, like cars, shops, and birds. These are all <dfn>objects</dfn>: tangible things people can observe and interact with.
-那么这些`物体`都有哪些特征呢？比如汽车的特征是有轮子，商店是用来出售商品的，而小鸟的特征是有翅膀。
+What are some qualities of these objects? A car has wheels. Shops sell items. Birds have wings.
-这些特征，或者说是`属性`定义了一个`物体`由什么构成的。需要注意的是：那些相似的`物体`可以拥有相同的`属性`，但是这些`属性`可能会有不同的值。举个例子：所有的汽车都有轮子，但并不是所有汽车的轮子个数都是一样的。
+These qualities, or <dfn>properties</dfn>, define what makes up an object. Note that similar objects share the same properties, but may have different values for those properties. For example, all cars have wheels, but not all cars have the same number of wheels.
-JavaScript 中的`对象`可以用来描述现实世界中的物体，并赋予他们`属性`和行为，就像它们在现实世界中的对应物一样。下面是使用这些概念来创建一个`duck 对象`的示例：
+Objects in JavaScript are used to model real-world objects, giving them properties and behavior just like their real-world counterparts. Here's an example using these concepts to create a `duck` object:
```js
let duck = {
@@ -23,27 +23,27 @@ let duck = {
};
```
-这个`duck 对象` 有两组键值对：一个是`name`属性，它的值是 "Aflac"；另一个是`numLegs`属性，它的值是 2。
+This `duck` object has two property/value pairs: a `name` of "Aflac" and a `numLegs` of 2.
# --instructions--
-创建一个`dog 对象`，并给这个对象添加两个属性：`name`和`numLegs`，同时把这两个属性的值分别设为字符串和数字。
+Create a `dog` object with `name` and `numLegs` properties, and set them to a string and a number, respectively.
# --hints--
-`dog`应该是一个`object`。
+`dog` should be an object.
```js
assert(typeof dog === 'object');
```
-`dog`应该有一个`name`属性，且它的值是一个<code>字符串<code>。
+`dog` should have a `name` property set to a `string`.
```js
assert(typeof dog.name === 'string');
```
-`dog`应该有一个`numLegs`属性，且它的值是一个`数字`。
+`dog` should have a `numLegs` property set to a `number`.
```js
assert(typeof dog.numLegs === 'number');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/create-a-method-on-an-object.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/create-a-method-on-an-object.md
index 6631aec77e7..0399c52c151 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/create-a-method-on-an-object.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/create-a-method-on-an-object.md
@@ -1,6 +1,6 @@
---
id: 587d7dad367417b2b2512b75
-title: 在对象上创建方法
+title: Create a Method on an Object
challengeType: 1
forumTopicId: 301318
dashedName: create-a-method-on-an-object
@@ -8,9 +8,9 @@ dashedName: create-a-method-on-an-object
# --description--
-`对象`可以有一个叫做`方法`的特殊`属性`。
+Objects can have a special type of property, called a <dfn>method</dfn>.
-`方法`其实是一个值为函数的`属性`，它可以为一个`对象`添加不同的行为。以下就是一个带有方法属性的`duck`示例：
+Methods are properties that are functions. This adds different behavior to an object. Here is the `duck` example with a method:
```js
let duck = {
@@ -22,21 +22,21 @@ duck.sayName();
// Returns "The name of this duck is Aflac."
```
-这个例子给`duck`对象添加了一个`sayName 方法`，这个方法返回一个包含`duck`名字的句子。 注意：这个`方法`在返回语句中使用`duck.name`的方式来获取`name`的属性值。在下一个挑战中我们将会使用另外一种方法来实现。
+The example adds the `sayName` method, which is a function that returns a sentence giving the name of the `duck`. Notice that the method accessed the `name` property in the return statement using `duck.name`. The next challenge will cover another way to do this.
# --instructions--
-给`dog 对象`设置一个名为`sayLegs`的方法，并让它返回 "This dog has 4 legs." 这句话。
+Using the `dog` object, give it a method called `sayLegs`. The method should return the sentence "This dog has 4 legs."
# --hints--
-`dog.sayLegs()`应该是一个函数。
+`dog.sayLegs()` should be a function.
```js
assert(typeof dog.sayLegs === 'function');
```
-`dog.sayLegs()`应该返回给定的字符串，需要注意标点和间距的问题。
+`dog.sayLegs()` should return the given string - note that punctuation and spacing matter.
```js
assert(dog.sayLegs() === 'This dog has 4 legs.');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/define-a-constructor-function.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/define-a-constructor-function.md
index f3ebf6250fa..ffc215f1f61 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/define-a-constructor-function.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/define-a-constructor-function.md
@@ -1,6 +1,6 @@
---
id: 587d7dad367417b2b2512b77
-title: 定义构造函数
+title: Define a Constructor Function
challengeType: 1
forumTopicId: 16804
dashedName: define-a-constructor-function
@@ -8,9 +8,9 @@ dashedName: define-a-constructor-function
# --description--
-`构造函数`用以创建一个新对象，并给这个新对象定义属性和行为。因此这是创建新对象的一个最基本的方式。
+<dfn>Constructors</dfn> are functions that create new objects. They define properties and behaviors that will belong to the new object. Think of them as a blueprint for the creation of new objects.
-以下就是一个`构造函数`的示例：
+Here is an example of a constructor:
```js
function Bird() {
@@ -20,29 +20,29 @@ function Bird() {
}
```
-这个`构造函数`定义了一个`Bird`对象，其属性`name`、`color`和`numLegs`的值分别被设置为`Albert`、`blue`和 2。 `构造函数`遵循一些惯例规则：
+This constructor defines a `Bird` object with properties `name`, `color`, and `numLegs` set to Albert, blue, and 2, respectively. Constructors follow a few conventions:
-<ul><li><code>构造函数</code>函数名的首字母最好大写，这是为了方便我们区分<code>构造函数</code>和其他非构造函数。</li><li><code>构造函数</code>使用<code>this</code>关键字来给它将创建的这个对象设置新的属性。在<code>构造函数</code>里面，<code>this</code>指向的就是它新创建的这个对象。</li><li><code>构造函数</code>定义了属性和行为就可创建对象，而不是像其他函数一样需要设置返回值。</li></ul>
+<ul><li>Constructors are defined with a capitalized name to distinguish them from other functions that are not <code>constructors</code>.</li><li>Constructors use the keyword <code>this</code> to set properties of the object they will create. Inside the constructor, <code>this</code> refers to the new object it will create.</li><li>Constructors define properties and behaviors instead of returning a value as other functions might.</li></ul>
# --instructions--
-创建一个`构造函数`：`Dog`。给其添加`name`，`color`和`numLegs`属性并分别给它们设置为：字符串，字符串和数字。
+Create a constructor, `Dog`, with properties `name`, `color`, and `numLegs` that are set to a string, a string, and a number, respectively.
# --hints--
-`Dog`应该有一个`name`属性且它的值是一个字符串。
+`Dog` should have a `name` property set to a string.
```js
assert(typeof new Dog().name === 'string');
```
-`Dog`应该有一个`color`属性且它的值是一个字符串。
+`Dog` should have a `color` property set to a string.
```js
assert(typeof new Dog().color === 'string');
```
-`Dog`应该有一个`numLegs`属性且它的值是一个数字。
+`Dog` should have a `numLegs` property set to a number.
```js
assert(typeof new Dog().numLegs === 'number');
@@ -51,7 +51,6 @@ assert(typeof new Dog().numLegs === 'number');
# --seed--
## --seed-contents--
-
```js
```
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/extend-constructors-to-receive-arguments.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/extend-constructors-to-receive-arguments.md
index 52ccabddee0..99a679937ab 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/extend-constructors-to-receive-arguments.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/extend-constructors-to-receive-arguments.md
@@ -1,6 +1,6 @@
---
id: 587d7dae367417b2b2512b79
-title: 扩展构造函数以接收参数
+title: Extend Constructors to Receive Arguments
challengeType: 1
forumTopicId: 18235
dashedName: extend-constructors-to-receive-arguments
@@ -8,7 +8,7 @@ dashedName: extend-constructors-to-receive-arguments
# --description--
-上一个挑战中`Bird`和`Dog`构造函数运行得不错。但是，注意到没有：所有通过`Bird`构造函数创建出来的实例`Birds`都自动的取名为 Albert，颜色都是蓝色，还都有两条腿。如果你想要新创建出来的小鸟们拥有不同的名字和颜色要怎么办呢？当然，手动的去修改每一个小鸟实例自己的属性也是可以实现的，只是会增加很多无谓的工作量：
+The `Bird` and `Dog` constructors from last challenge worked well. However, notice that all `Birds` that are created with the `Bird` constructor are automatically named Albert, are blue in color, and have two legs. What if you want birds with different values for name and color? It's possible to change the properties of each bird manually but that would be a lot of work:
```js
let swan = new Bird();
@@ -16,7 +16,7 @@ swan.name = "Carlos";
swan.color = "white";
```
-假如你写了一个程序来追踪一个鸟舍里面的几百只甚至几千只不同的小鸟，你将会花费很多时间去创建所有的小鸟实例并给它们的属性一一修改为不同的值。 为了减轻创建不同`Bird`对象的工作量，你可以给你的`Bird`设置为可以接收参数的构造函数：
+Suppose you were writing a program to keep track of hundreds or even thousands of different birds in an aviary. It would take a lot of time to create all the birds, then change the properties to different values for every one. To more easily create different `Bird` objects, you can design your Bird constructor to accept parameters:
```js
function Bird(name, color) {
@@ -26,7 +26,7 @@ function Bird(name, color) {
}
```
-然后将值通过参数的方式传递给`Bird`构造函数来定义每一个唯一的小鸟实例： `let cardinal = new Bird("Bruce", "red");` 这给`Bird`的名字和颜色属性分别赋值为 Bruce 和红色提供了另外一种方法。但`numLegs`属性被默认赋值为 2。 `cardinal`有以下这些属性：
+Then pass in the values as arguments to define each unique bird into the `Bird` constructor: `let cardinal = new Bird("Bruce", "red");` This gives a new instance of `Bird` with name and color properties set to Bruce and red, respectively. The `numLegs` property is still set to 2. The `cardinal` has these properties:
```js
cardinal.name // => Bruce
@@ -34,33 +34,33 @@ cardinal.color // => red
cardinal.numLegs // => 2
```
-这样一来构造函数就变得很灵活了。现在可以直接定义每个`Bird`实例在创建时的属性，这是 JavaScript 构造函数非常实用的用法之一。它们根据共同或相似的属性和行为将对象归纳为一组，并能够自动的创建各自实例。
+The constructor is more flexible. It's now possible to define the properties for each `Bird` at the time it is created, which is one way that JavaScript constructors are so useful. They group objects together based on shared characteristics and behavior and define a blueprint that automates their creation.
# --instructions--
-创建另一个`Dog`构造函数。这一次，给它设置两个参数：`name`和`color`，同时给`numLegs`赋值为 4。然后创建一个新`Dog`实例保存为变量名：`terrier`，再将两个字符串通过参数的形式传入`name`和`color`属性。
+Create another `Dog` constructor. This time, set it up to take the parameters `name` and `color`, and have the property `numLegs` fixed at 4. Then create a new `Dog` saved in a variable `terrier`. Pass it two strings as arguments for the `name` and `color` properties.
# --hints--
-`Dog`应该接收一个`name`参数。
+`Dog` should receive an argument for `name`.
```js
assert(new Dog('Clifford').name === 'Clifford');
```
-`Dog`应该接收一个`color`参数。
+`Dog` should receive an argument for `color`.
```js
assert(new Dog('Clifford', 'yellow').color === 'yellow');
```
-`Dog`应该有一个`numLegs`属性且值为 4。
+`Dog` should have property `numLegs` set to 4.
```js
assert(new Dog('Clifford').numLegs === 4);
```
-`terrier`应该是通过`Dog`构造函数创建的。
+`terrier` should be created using the `Dog` constructor.
```js
assert(terrier instanceof Dog);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/inherit-behaviors-from-a-supertype.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/inherit-behaviors-from-a-supertype.md
index cb61823f971..357fdeab57e 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/inherit-behaviors-from-a-supertype.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/inherit-behaviors-from-a-supertype.md
@@ -1,6 +1,6 @@
---
id: 587d7db0367417b2b2512b84
-title: 从超类继承行为
+title: Inherit Behaviors from a Supertype
challengeType: 1
forumTopicId: 301319
dashedName: inherit-behaviors-from-a-supertype
@@ -8,7 +8,7 @@ dashedName: inherit-behaviors-from-a-supertype
# --description--
-在上一个挑战中，我们创建了一个`Animal 超类`，用来定义所有动物共有的行为：
+In the previous challenge, you created a `supertype` called `Animal` that defined behaviors shared by all animals:
```js
function Animal() { }
@@ -17,19 +17,19 @@ Animal.prototype.eat = function() {
};
```
-在这一节以及下一节挑战中我们将学习如何给`Bird`、`Dog`重写`Animal`中的方法，而无需重新定义它们。这里我们会用到构造函数的`继承`特性。 这一节挑战中我们学习第一步：创建一个`超类`（或者叫父类）的实例。 你已经学会了一种创建`Animal`实例的方法，即使用`new`操作符：
+This and the next challenge will cover how to reuse `Animal's` methods inside `Bird` and `Dog` without defining them again. It uses a technique called inheritance. This challenge covers the first step: make an instance of the `supertype` (or parent). You already know one way to create an instance of `Animal` using the `new` operator:
```js
let animal = new Animal();
```
-此语法用于`继承`时会存在一些缺点，这些缺点对于当前我们这个挑战来说太复杂了。相反，我们学习另外一种没有这些缺点的方法来替代`new`操作：
+There are some disadvantages when using this syntax for inheritance, which are too complex for the scope of this challenge. Instead, here's an alternative approach without those disadvantages:
```js
let animal = Object.create(Animal.prototype);
```
-`Object.create(obj)`创建了一个新对象，并指定了`obj`作为新对象的`原型`。回忆一下，我们之前说过`原型`就像是创建对象的“配方”。如果我们把`animal`的`原型`设置为与`Animal`构造函数的`原型`一样，那么就相当于让`animal`这个实例的配方与`Animal`其他实例的配方一样了。
+`Object.create(obj)` creates a new object, and sets `obj` as the new object's `prototype`. Recall that the `prototype` is like the "recipe" for creating an object. By setting the `prototype` of `animal` to be `Animal's` `prototype`, you are effectively giving the `animal` instance the same "recipe" as any other instance of `Animal`.
```js
animal.eat(); // prints "nom nom nom"
@@ -38,23 +38,23 @@ animal instanceof Animal; // => true
# --instructions--
-使用`Object.create`方法给`Animal`创建两个实例：`duck`和`beagle`。
+Use `Object.create` to make two instances of `Animal` named `duck` and `beagle`.
# --hints--
-应该定义一个`duck`变量。
+The `duck` variable should be defined.
```js
assert(typeof duck !== 'undefined');
```
-应该定义一个`beagle`变量。
+The `beagle` variable should be defined.
```js
assert(typeof beagle !== 'undefined');
```
-`duck` 变量应该通过 `Object.create` 初始化。
+The `duck` variable should be initialised with `Object.create`.
```js
assert(
@@ -64,7 +64,7 @@ assert(
);
```
-`beagle` 变量应该通过 `Object.create` 初始化。
+The `beagle` variable should be initialised with `Object.create`.
```js
assert(
@@ -74,13 +74,13 @@ assert(
);
```
-`duck`的原型应该被设置为`Animal`构造函数的`原型`。
+`duck` should have a `prototype` of `Animal`.
```js
assert(duck instanceof Animal);
```
-`beagle`的原型应该被设置为`Animal`构造函数的`原型`。
+`beagle` should have a `prototype` of `Animal`.
```js
assert(beagle instanceof Animal);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/iterate-over-all-properties.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/iterate-over-all-properties.md
index 132f7517d71..bc6aa77d267 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/iterate-over-all-properties.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/iterate-over-all-properties.md
@@ -1,6 +1,6 @@
---
id: 587d7daf367417b2b2512b7d
-title: 迭代所有属性
+title: Iterate Over All Properties
challengeType: 1
forumTopicId: 301320
dashedName: iterate-over-all-properties
@@ -8,19 +8,19 @@ dashedName: iterate-over-all-properties
# --description--
-现在你已经了解了两种属性: `自身`属性和`原型`属性。`自身`属性是直接在对象上定义的。而`原型`属性是定义在`prototype`上的：
+You have now seen two kinds of properties: `own` properties and `prototype` properties. `Own` properties are defined directly on the object instance itself. And `prototype` properties are defined on the `prototype`.
```js
function Bird(name) {
- this.name = name; // 自身属性
+ this.name = name; //own property
}
-Bird.prototype.numLegs = 2; // 原型属性
+Bird.prototype.numLegs = 2; // prototype property
let duck = new Bird("Donald");
```
-这个示例会告诉你如何将`duck`的`自身`属性和`原型`属性分别添加到`ownProps`数组和`prototypeProps`数组里面：
+Here is how you add `duck`'s `own` properties to the array `ownProps` and `prototype` properties to the array `prototypeProps`:
```js
let ownProps = [];
@@ -40,23 +40,23 @@ console.log(prototypeProps); // prints ["numLegs"]
# --instructions--
-将`beagle`的自身属性都添加到`ownProps`数组里面去。将`Dog`的所有`原型`属性添加到`prototypeProps`数组中。
+Add all of the `own` properties of `beagle` to the array `ownProps`. Add all of the `prototype` properties of `Dog` to the array `prototypeProps`.
# --hints--
-这个`ownProps`数组应该包含`'name'`这个值。
+The `ownProps` array should only contain `"name"`.
```js
-assert(ownProps.indexOf('name') !== -1);
+assert.deepEqual(ownProps, ['name']);
```
-这个`prototypeProps`数组应该包含`'numLegs'`这个值。
+The `prototypeProps` array should only contain `"numLegs"`.
```js
-assert(prototypeProps.indexOf('numLegs') !== -1);
+assert.deepEqual(prototypeProps, ['numLegs']);
```
-在不使用内置方法`Object.keys()`的情况下完成这个挑战。
+You should solve this challenge without using the built in method `Object.keys()`.
```js
assert(!/\Object.keys/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/make-code-more-reusable-with-the-this-keyword.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/make-code-more-reusable-with-the-this-keyword.md
index 045e5af8f2c..fc3324f4797 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/make-code-more-reusable-with-the-this-keyword.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/make-code-more-reusable-with-the-this-keyword.md
@@ -1,6 +1,6 @@
---
id: 587d7dad367417b2b2512b76
-title: 使用 this 关键字使代码更加可重用
+title: Make Code More Reusable with the this Keyword
challengeType: 1
forumTopicId: 301321
dashedName: make-code-more-reusable-with-the-this-keyword
@@ -8,13 +8,13 @@ dashedName: make-code-more-reusable-with-the-this-keyword
# --description--
-在上一个挑战中我们了解了该如何给`duck`对象设置一个`方法`属性。然后在 `return` 语句里，我们通过使用 “点号表示法” `duck.name`来获取`name`的属性值：
+The last challenge introduced a method to the `duck` object. It used `duck.name` dot notation to access the value for the `name` property within the return statement:
`sayName: function() {return "The name of this duck is " + duck.name + ".";}`
-虽然这是访问对象属性的有效方法，但是这里有一个陷阱。如果变量名发生了改变，那么引用了原始名称的任何代码都需要更新。在一个简短的对象定义中这并不是问题，但是如果对象有很多对其属性的引用，那么发生错误的可能性就更大了。
+While this is a valid way to access the object's property, there is a pitfall here. If the variable name changes, any code referencing the original name would need to be updated as well. In a short object definition, it isn't a problem, but if an object has many references to its properties there is a greater chance for error.
-我们可以使用`this`关键字这个方法来避免这一问题：
+A way to avoid these issues is with the `this` keyword:
```js
let duck = {
@@ -24,21 +24,21 @@ let duck = {
};
```
-`this`是一个很复杂的知识点，而上面那个例子也只是使用`this`的一种方法而已。在当前的上下文环境中，`this`指向的就是与这个方法有关联的`duck`对象。 如果把对象的变量名改为`mallard`，那使用`this`就没有必要在代码中找到所有指向`duck`的部分，这样可以使得代码更具有可读性和复用性。
+`this` is a deep topic, and the above example is only one way to use it. In the current context, `this` refers to the object that the method is associated with: `duck`. If the object's name is changed to `mallard`, it is not necessary to find all the references to `duck` in the code. It makes the code reusable and easier to read.
# --instructions--
-修改`dog.sayLegs`方法以将所有直接对`dog`的引用删除。可以参考上面的例子。
+Modify the `dog.sayLegs` method to remove any references to `dog`. Use the `duck` example for guidance.
# --hints--
-`dog.sayLegs()`应该返回一个指定的字符串。
+`dog.sayLegs()` should return the given string.
```js
assert(dog.sayLegs() === 'This dog has 4 legs.');
```
-你的代码应该使用`this`关键字来访问`dog`对象的`numLegs`属性值。
+Your code should use the `this` keyword to access the `numLegs` property of `dog`.
```js
assert(code.match(/this\.numLegs/g));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/override-inherited-methods.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/override-inherited-methods.md
index 04b05570c64..24ee9ad5134 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/override-inherited-methods.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/override-inherited-methods.md
@@ -1,6 +1,6 @@
---
id: 587d7db1367417b2b2512b88
-title: 重写继承的方法
+title: Override Inherited Methods
challengeType: 1
forumTopicId: 301322
dashedName: override-inherited-methods
@@ -8,19 +8,19 @@ dashedName: override-inherited-methods
# --description--
-在上一个挑战中，我们学习了一个对象可以通过复制另一个对象的`原型`来继承其属性和行为（或方法）：
+In previous lessons, you learned that an object can inherit its behavior (methods) from another object by referencing its `prototype` object:
```js
ChildObject.prototype = Object.create(ParentObject.prototype);
```
-然后，`ChildObject`将自己的方法链接到它的`原型`中：
+Then the `ChildObject` received its own methods by chaining them onto its `prototype`:
```js
ChildObject.prototype.methodName = function() {...};
```
-我们还可以重写继承的方法。以同样的方式——通过使用一个与需要重写的方法相同的方法名，向`ChildObject.prototype`中添加方法。 请看下面的举例：`Bird`重写了从`Animal`继承来的`eat()`方法：
+It's possible to override an inherited method. It's done the same way - by adding a method to `ChildObject.prototype` using the same method name as the one to override. Here's an example of `Bird` overriding the `eat()` method inherited from `Animal`:
```js
function Animal() { }
@@ -29,35 +29,35 @@ Animal.prototype.eat = function() {
};
function Bird() { }
-// 继承了 Animal 的所有方法
+// Inherit all methods from Animal
Bird.prototype = Object.create(Animal.prototype);
-// Bird.eat() 重写了 Animal.eat() 方法
+// Bird.eat() overrides Animal.eat()
Bird.prototype.eat = function() {
return "peck peck peck";
};
```
-如果你有一个实例：`let duck = new Bird();`，然后你调用了`duck.eat()`，以下就是 JavaScript 在`duck`的`原型`链上寻找方法的过程：
+If you have an instance `let duck = new Bird();` and you call `duck.eat()`, this is how JavaScript looks for the method on `duck’s` `prototype` chain:
-1. duck => 这里定义了 eat() 方法吗？没有。
-2. Bird => 这里定义了 eat() 方法吗？=> 是的。执行它并停止往上搜索。
-3. Animal => 这里也定义了 eat() 方法，但是 JavaScript 在到达这层原型链之前已停止了搜索。
-4. Object => JavaScript 在到达这层原型链之前也已经停止了搜索。
+1. duck => Is eat() defined here? No.
+2. Bird => Is eat() defined here? => Yes. Execute it and stop searching.
+3. Animal => eat() is also defined, but JavaScript stopped searching before reaching this level.
+4. Object => JavaScript stopped searching before reaching this level.
# --instructions--
-重写`Penguin`的`fly()`方法，使其返回 "Alas, this is a flightless bird."
+Override the `fly()` method for `Penguin` so that it returns "Alas, this is a flightless bird."
# --hints--
-`penguin.fly()`方法应该返回字符串：'Alas, this is a flightless bird.'
+`penguin.fly()` should return the string "Alas, this is a flightless bird."
```js
assert(penguin.fly() === 'Alas, this is a flightless bird.');
```
-The `bird.fly()`方法应该返回 'I am flying!'
+The `bird.fly()` method should return "I am flying!"
```js
assert(new Bird().fly() === 'I am flying!');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/remember-to-set-the-constructor-property-when-changing-the-prototype.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/remember-to-set-the-constructor-property-when-changing-the-prototype.md
index 45b9639d757..76874fedecf 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/remember-to-set-the-constructor-property-when-changing-the-prototype.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/remember-to-set-the-constructor-property-when-changing-the-prototype.md
@@ -1,6 +1,6 @@
---
id: 587d7daf367417b2b2512b80
-title: 更改原型时，记得设置构造函数属性
+title: Remember to Set the Constructor Property when Changing the Prototype
challengeType: 1
forumTopicId: 301323
dashedName: remember-to-set-the-constructor-property-when-changing-the-prototype
@@ -8,15 +8,15 @@ dashedName: remember-to-set-the-constructor-property-when-changing-the-prototype
# --description--
-手动给新对象重新设置`原型`对象，会产生一个重要的副作用：删除了`constructor`属性！我们来看一下，上一个挑战中`duck`的`constructor`属性输出到控制台的结果：
+There is one crucial side effect of manually setting the prototype to a new object. It erases the `constructor` property! This property can be used to check which constructor function created the instance, but since the property has been overwritten, it now gives false results:
```js
-duck.constructor === Bird; // false -- 啊哦
-duck.constructor === Object; // true, 所有的对象都继承自 Object.prototype
-duck instanceof Bird; // true, 依然生效
+duck.constructor === Bird; // false -- Oops
+duck.constructor === Object; // true, all objects inherit from Object.prototype
+duck instanceof Bird; // true, still works
```
-为了解决这个问题，凡是手动给新对象重新设置过原型对象的，都别忘记在原型对象中定义一个`constructor`属性：
+To fix this, whenever a prototype is manually set to a new object, remember to define the `constructor` property:
```js
Bird.prototype = {
@@ -33,11 +33,11 @@ Bird.prototype = {
# --instructions--
-给`Dog 的原型`对象定义一个`constructor`属性。
+Define the `constructor` property on the `Dog` `prototype`.
# --hints--
-`Dog.prototype`应该定义一个`constructor`属性。
+`Dog.prototype` should set the `constructor` property.
```js
assert(Dog.prototype.constructor === Dog);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/reset-an-inherited-constructor-property.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/reset-an-inherited-constructor-property.md
index 4b9ef2c5b02..ce051f75a23 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/reset-an-inherited-constructor-property.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/reset-an-inherited-constructor-property.md
@@ -1,6 +1,6 @@
---
id: 587d7db1367417b2b2512b86
-title: 重置一个继承的构造函数属性
+title: Reset an Inherited Constructor Property
challengeType: 1
forumTopicId: 301324
dashedName: reset-an-inherited-constructor-property
@@ -8,9 +8,9 @@ dashedName: reset-an-inherited-constructor-property
# --description--
-当一个对象从另一个对象那里继承了其`原型`，那它也继承了`父类`的 constructor 属性。
+When an object inherits its `prototype` from another object, it also inherits the supertype's constructor property.
-请看下面的举例：
+Here's an example:
```js
function Bird() { }
@@ -19,7 +19,7 @@ let duck = new Bird();
duck.constructor // function Animal(){...}
```
-但是`duck`和其他所有`Bird`的实例都应该表明它们是由`Bird`创建的，而不是由`Animal`创建的。为此，你可以手动把`Bird`的 constructor 属性设置为`Bird`对象：
+But `duck` and all instances of `Bird` should show that they were constructed by `Bird` and not `Animal`. To do so, you can manually set `Bird's` constructor property to the `Bird` object:
```js
Bird.prototype.constructor = Bird;
@@ -28,29 +28,29 @@ duck.constructor // function Bird(){...}
# --instructions--
-修改你的代码，使得`duck.constructor`和`beagle.constructor`返回各自的构造函数。
+Fix the code so `duck.constructor` and `beagle.constructor` return their respective constructors.
# --hints--
-`Bird.prototype`应该是`Animal`的一个实例。
+`Bird.prototype` should be an instance of `Animal`.
```js
assert(Animal.prototype.isPrototypeOf(Bird.prototype));
```
-`duck.constructor`应该返回`Bird`。
+`duck.constructor` should return `Bird`.
```js
assert(duck.constructor === Bird);
```
-`Dog.prototype`应该是`Animal`的一个实例。
+`Dog.prototype` should be an instance of `Animal`.
```js
assert(Animal.prototype.isPrototypeOf(Dog.prototype));
```
-`beagle.constructor`应该返回`Dog`。
+`beagle.constructor` should return `Dog`.
```js
assert(beagle.constructor === Dog);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/set-the-childs-prototype-to-an-instance-of-the-parent.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/set-the-childs-prototype-to-an-instance-of-the-parent.md
index 0c4ed3d5a0f..6e450e1bfe8 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/set-the-childs-prototype-to-an-instance-of-the-parent.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/set-the-childs-prototype-to-an-instance-of-the-parent.md
@@ -1,6 +1,6 @@
---
id: 587d7db1367417b2b2512b85
-title: 将子辈的原型设置为父辈的实例
+title: Set the Child's Prototype to an Instance of the Parent
challengeType: 1
forumTopicId: 301325
dashedName: set-the-childs-prototype-to-an-instance-of-the-parent
@@ -8,30 +8,30 @@ dashedName: set-the-childs-prototype-to-an-instance-of-the-parent
# --description--
-在上一个挑战中，我们学习了从`超类（或者叫父类） Animal`继承其行为的第一个步骤：创建一个`Animal`的实例。
+In the previous challenge you saw the first step for inheriting behavior from the supertype (or parent) `Animal`: making a new instance of `Animal`.
-这一节挑战我们将学习第二个步骤：给`子类型（或者子类）`设置`原型`。这样一来，`Bird`就是`Animal`的一个实例了。
+This challenge covers the next step: set the `prototype` of the subtype (or child)—in this case, `Bird`—to be an instance of `Animal`.
```js
Bird.prototype = Object.create(Animal.prototype);
```
-请记住，`原型`类似于创建对象的“配方”。从某种意义上来说，`Bird`对象的配方包含了`Animal`构造函数的所有关键“成分”。
+Remember that the `prototype` is like the "recipe" for creating an object. In a way, the recipe for `Bird` now includes all the key "ingredients" from `Animal`.
```js
let duck = new Bird("Donald");
duck.eat(); // prints "nom nom nom"
```
-`duck`继承了`Animal`构造函数的所有属性，其中包括了`eat`方法。
+`duck` inherits all of `Animal`'s properties, including the `eat` method.
# --instructions--
-修改你的代码，实现一个继承自`Animal`的`Dog`构造函数。
+Modify the code so that instances of `Dog` inherit from `Animal`.
# --hints--
-`Dog.prototype`应该是`Animal`的一个实例。
+`Dog.prototype` should be an instance of `Animal`.
```js
assert(Animal.prototype.isPrototypeOf(Dog.prototype));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-own-properties.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-own-properties.md
index e541309734c..f4f369581c8 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-own-properties.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-own-properties.md
@@ -1,6 +1,6 @@
---
id: 587d7dae367417b2b2512b7b
-title: 了解自己的属性
+title: Understand Own Properties
challengeType: 1
forumTopicId: 301326
dashedName: understand-own-properties
@@ -8,7 +8,7 @@ dashedName: understand-own-properties
# --description--
-请看下面的实例，`Bird`构造函数定义了两个属性：`name`和`numLegs`：
+In the following example, the `Bird` constructor defines two properties: `name` and `numLegs`:
```js
function Bird(name) {
@@ -20,7 +20,7 @@ let duck = new Bird("Donald");
let canary = new Bird("Tweety");
```
-`name`和`numLegs`被叫做`自身`属性，因为他们是直接在实例对象上定义的。这就意味着`duck`和`canary`这两个对象分别拥有这些属性的独立副本。 事实上，`Bird`的这些实例都将拥有这些属性的独立副本。 以下的代码将`duck`里面所有的`自身`属性都存到一个叫`ownProps`的数组里面：
+`name` and `numLegs` are called `own` properties, because they are defined directly on the instance object. That means that `duck` and `canary` each has its own separate copy of these properties. In fact every instance of `Bird` will have its own copy of these properties. The following code adds all of the `own` properties of `duck` to the array `ownProps`:
```js
let ownProps = [];
@@ -36,17 +36,17 @@ console.log(ownProps); // prints [ "name", "numLegs" ]
# --instructions--
-将`canary`对象里面的`自身`属性添加到`ownProps`数组里面。
+Add the `own` properties of `canary` to the array `ownProps`.
# --hints--
-`ownProps`应该包含`'numLegs'`和`'name'`两个属性的值。
+`ownProps` should include the values `"numLegs"` and `"name"`.
```js
assert(ownProps.indexOf('name') !== -1 && ownProps.indexOf('numLegs') !== -1);
```
-在不使用内置方法`Object.keys()`的情况下完成这个挑战。
+You should solve this challenge without using the built in method `Object.keys()`.
```js
assert(!/Object(\.keys|\[(['"`])keys\2\])/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-the-constructor-property.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-the-constructor-property.md
index ef6b7b564ea..6d1fd731db9 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-the-constructor-property.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-the-constructor-property.md
@@ -1,6 +1,6 @@
---
id: 587d7daf367417b2b2512b7e
-title: 了解构造函数属性
+title: Understand the Constructor Property
challengeType: 1
forumTopicId: 301327
dashedName: understand-the-constructor-property
@@ -8,7 +8,7 @@ dashedName: understand-the-constructor-property
# --description--
-在上一个挑战中创建的实例对象`duck`和`beagle`都有一个特殊的`constructor`属性：
+There is a special `constructor` property located on the object instances `duck` and `beagle` that were created in the previous challenges:
```js
let duck = new Bird();
@@ -18,7 +18,7 @@ console.log(duck.constructor === Bird); //prints true
console.log(beagle.constructor === Dog); //prints true
```
-需要注意到的是这个`constructor`属性是对创建这个实例的构造函数的一个引用。 `constructor`属性存在的一个优势是，我们可以通过检查这个属性来找出它是一个什么样的对象。下面是一个例子，来看看是怎么使用的：
+Note that the `constructor` property is a reference to the constructor function that created the instance. The advantage of the `constructor` property is that it's possible to check for this property to find out what kind of object it is. Here's an example of how this could be used:
```js
function joinBirdFraternity(candidate) {
@@ -30,28 +30,28 @@ function joinBirdFraternity(candidate) {
}
```
-**注意：**
-由于`constructor`属性可以被重写（在下面两节挑战中将会遇到），所以使用`instanceof`方法来检查对象的类型会更好。
+**Note**
+Since the `constructor` property can be overwritten (which will be covered in the next two challenges) it’s generally better to use the `instanceof` method to check the type of an object.
# --instructions--
-写一个`joinDogFraternity`函数，传入一个`candidate`参数并使用`constructor`属性来判断传入的 candidate 是不是`Dog`创建的对象实例，如果是，就返回`true`，否则返回`false`。
+Write a `joinDogFraternity` function that takes a `candidate` parameter and, using the `constructor` property, return `true` if the candidate is a `Dog`, otherwise return `false`.
# --hints--
-`joinDogFraternity`应该被定义为一个函数。
+`joinDogFraternity` should be defined as a function.
```js
assert(typeof joinDogFraternity === 'function');
```
-如果`candidate`是`Dog`的一个对象实例，则`joinDogFraternity`函数应该返回`true`。
+`joinDogFraternity` should return true if`candidate` is an instance of `Dog`.
```js
assert(joinDogFraternity(new Dog('')) === true);
```
-`joinDogFraternity`中应该用到`constructor`属性。
+`joinDogFraternity` should use the `constructor` property.
```js
assert(/\.constructor/.test(code) && !/instanceof/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-the-immediately-invoked-function-expression-iife.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-the-immediately-invoked-function-expression-iife.md
index 85e0822128f..cc33fc1883d 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-the-immediately-invoked-function-expression-iife.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-the-immediately-invoked-function-expression-iife.md
@@ -1,6 +1,6 @@
---
id: 587d7db2367417b2b2512b8b
-title: 了解立即调用函数表达（IIFE）
+title: Understand the Immediately Invoked Function Expression (IIFE)
challengeType: 1
forumTopicId: 301328
dashedName: understand-the-immediately-invoked-function-expression-iife
@@ -8,33 +8,33 @@ dashedName: understand-the-immediately-invoked-function-expression-iife
# --description--
-JavaScript 中的一个常见模式就是，函数在声明后立刻执行：
+A common pattern in JavaScript is to execute a function as soon as it is declared:
```js
(function () {
console.log("Chirp, chirp!");
-})(); // 这是一个立即执行的匿名函数表达式
-// 立即输出 "Chirp, chirp!"
+})(); // this is an anonymous function expression that executes right away
+// Outputs "Chirp, chirp!" immediately
```
-请注意，函数没有名称，也不存储在变量中。函数表达式末尾的两个括号（）导致它被立即执行或调用。这种模式被叫做`自执行函数表达式`或者`IIFE`。
+Note that the function has no name and is not stored in a variable. The two parentheses () at the end of the function expression cause it to be immediately executed or invoked. This pattern is known as an <dfn>immediately invoked function expression</dfn> or <dfn>IIFE</dfn>.
# --instructions--
-重写函数`makeNest`，并删除它的调用，取而代之是一个匿名的`自执行函数表达式`（`IIFE`）。
+Rewrite the function `makeNest` and remove its call so instead it's an anonymous immediately invoked function expression (IIFE).
# --hints--
-该函数应该是匿名的。
+The function should be anonymous.
```js
-assert(/\((function|\(\))(=>|\(\)){/.test(code.replace(/\s/g, '')));
+assert(/\((function|\(\))(=>|\(\)){?/.test(code.replace(/\s/g, '')));
```
-函数应该在表达式的末尾有括号，以便立即调用它。
+Your function should have parentheses at the end of the expression to call it immediately.
```js
-assert(/}\)\(\)/.test(code.replace(/\s/g, '')));
+assert(/\(.*(\)\(|\}\(\))\)/.test(code.replace(/[\s;]/g, '')));
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-the-prototype-chain.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-the-prototype-chain.md
index a3ec0e1f5f5..7632ddb4b34 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-the-prototype-chain.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-the-prototype-chain.md
@@ -1,6 +1,6 @@
---
id: 587d7db0367417b2b2512b82
-title: 了解原型链
+title: Understand the Prototype Chain
challengeType: 1
forumTopicId: 301329
dashedName: understand-the-prototype-chain
@@ -8,39 +8,38 @@ dashedName: understand-the-prototype-chain
# --description--
-JavaScript 中所有的对象（除了少数例外）都有自己的`原型`。而且，对象的`原型`本身也是一个对象。
+All objects in JavaScript (with a few exceptions) have a `prototype`. Also, an object’s `prototype` itself is an object.
```js
function Bird(name) {
this.name = name;
}
-typeof Bird.prototype; // => object
+typeof Bird.prototype; // yields 'object'
```
-正因为`原型`是一个对象，所以`原型`对象也有它自己的`原型`！这样看来的话，`Bird.prototype`的`原型`就是`Object.prototype`：
+Because a `prototype` is an object, a `prototype` can have its own `prototype`! In this case, the `prototype` of `Bird.prototype` is `Object.prototype`:
```js
-Object.prototype.isPrototypeOf(Bird.prototype);
-// 返回 true
+Object.prototype.isPrototypeOf(Bird.prototype); // returns true
```
-这有什么作用呢？你可能还记得我们在上一个挑战中学到的`hasOwnProperty`方法：
+How is this useful? You may recall the `hasOwnProperty` method from a previous challenge:
```js
let duck = new Bird("Donald");
-duck.hasOwnProperty("name"); // => true
+duck.hasOwnProperty("name"); // yields true
```
-`hasOwnProperty`是定义在`Object.prototype`上的一个方法，尽管在`Bird.prototype`和`duck`上并没有定义该方法，但是我们依然可以在这两个对象上访问到。这就是一个`原型`链。 在这个`原型`链中，`Bird`构造函数是`父级`，`duck`是`子级`。`Object`则是`Bird`构造函数和`duck`实例共同的`父级`。 `Object`是 JavaScript 中所有对象的`父级`，也就是原型链的最顶层。因此，所有对象都可以访问`hasOwnProperty`方法。
+The `hasOwnProperty` method is defined in `Object.prototype`, which can be accessed by `Bird.prototype`, which can then be accessed by `duck`. This is an example of the `prototype` chain. In this `prototype` chain, `Bird` is the `supertype` for `duck`, while `duck` is the `subtype`. `Object` is a `supertype` for both `Bird` and `duck`. `Object` is a `supertype` for all objects in JavaScript. Therefore, any object can use the `hasOwnProperty` method.
# --instructions--
-修改以下代码使其展示出正确的原型链。
+Modify the code to show the correct prototype chain.
# --hints--
-你的代码应该展示`Object.prototype`是`Dog.prototype`的原型。
+Your code should show that `Object.prototype` is the prototype of `Dog.prototype`
```js
assert(/Object\.prototype\.isPrototypeOf/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-where-an-objects-prototype-comes-from.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-where-an-objects-prototype-comes-from.md
index 6afeead7d4a..5644bfa873f 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-where-an-objects-prototype-comes-from.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/understand-where-an-objects-prototype-comes-from.md
@@ -1,6 +1,6 @@
---
id: 587d7db0367417b2b2512b81
-title: 了解对象的原型来自哪里
+title: Understand Where an Object’s Prototype Comes From
challengeType: 1
forumTopicId: 301330
dashedName: understand-where-an-objects-prototype-comes-from
@@ -8,7 +8,7 @@ dashedName: understand-where-an-objects-prototype-comes-from
# --description--
-就像人们从父母那里继承基因一样，对象也可直接从创建它的构造函数那里继承其`原型`。请看下面的例子：`Bird`构造函数创建了一个`duck`对象：
+Just like people inherit genes from their parents, an object inherits its `prototype` directly from the constructor function that created it. For example, here the `Bird` constructor creates the `duck` object:
```js
function Bird(name) {
@@ -18,20 +18,20 @@ function Bird(name) {
let duck = new Bird("Donald");
```
-`duck`从`Bird`构造函数那里继承了它的`原型`，你可以使用`isPrototypeOf`方法来验证他们之间的关系：
+`duck` inherits its `prototype` from the `Bird` constructor function. You can show this relationship with the `isPrototypeOf` method:
```js
Bird.prototype.isPrototypeOf(duck);
-// 返回 true
+// returns true
```
# --instructions--
-使用`isPrototypeOf`方法验证`beagle`是否继承了`Dog`构造函数的`原型`。
+Use `isPrototypeOf` to check the `prototype` of `beagle`.
# --hints--
-`Dog.prototype`应该是`beagle`的`原型`。
+You should show that `Dog.prototype` is the `prototype` of `beagle`
```js
assert(/Dog\.prototype\.isPrototypeOf\(beagle\)/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-a-constructor-to-create-objects.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-a-constructor-to-create-objects.md
index 8c88082fbd6..d26baffb1a8 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-a-constructor-to-create-objects.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-a-constructor-to-create-objects.md
@@ -1,6 +1,6 @@
---
id: 587d7dad367417b2b2512b78
-title: 使用构造函数创建对象
+title: Use a Constructor to Create Objects
challengeType: 1
forumTopicId: 18233
dashedName: use-a-constructor-to-create-objects
@@ -8,20 +8,20 @@ dashedName: use-a-constructor-to-create-objects
# --description--
-在上一个挑战中，我们用所学到的知识创建了一个`Bird`构造函数：
+Here's the `Bird` constructor from the previous challenge:
```js
function Bird() {
this.name = "Albert";
this.color = "blue";
this.numLegs = 2;
- // 构造器内的 "this" 一直指向创建的对象
+ // "this" inside the constructor always refers to the object being created
}
let blueBird = new Bird();
```
-注意：通过构造函数创建对象的时候要使用`new`操作符。因为只有这样，JavaScript 才知道要给`Bird`这个构造函数创建一个新的`实例`：`blueBird`。如果不使用`new`操作符来新建对象，那么构造函数里面的`this`就无法指向新创建的这个对象实例，从而产生不可预见的错误。 现在`blueBird`这个实例就继承了`Bird`这个构造函数的所有属性，如下：
+Notice that the `new` operator is used when calling a constructor. This tells JavaScript to create a new instance of `Bird` called `blueBird`. Without the `new` operator, `this` inside the constructor would not point to the newly created object, giving unexpected results. Now `blueBird` has all the properties defined inside the `Bird` constructor:
```js
blueBird.name; // => Albert
@@ -29,7 +29,7 @@ blueBird.color; // => blue
blueBird.numLegs; // => 2
```
-由构造函数创建的实例也和其他对象一样，它的属性可以被访问和修改：
+Just like any other object, its properties can be accessed and modified:
```js
blueBird.name = 'Elvira';
@@ -38,17 +38,17 @@ blueBird.name; // => Elvira
# --instructions--
-使用上一个课时中的`Dog`构造函数创建一个`Dog`的新实例，并把它赋值给变量`hound`。
+Use the `Dog` constructor from the last lesson to create a new instance of `Dog`, assigning it to a variable `hound`.
# --hints--
-`hound`应该是通过`Dog`构造函数来创建的。
+`hound` should be created using the `Dog` constructor.
```js
assert(hound instanceof Dog);
```
-你的代码中应该使用`new`操作符来创建`Dog`构造函数的新`实例`。
+Your code should use the `new` operator to create an instance of `Dog`.
```js
assert(code.match(/new/g));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-a-mixin-to-add-common-behavior-between-unrelated-objects.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-a-mixin-to-add-common-behavior-between-unrelated-objects.md
index 3bf7e445369..5664c1e414b 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-a-mixin-to-add-common-behavior-between-unrelated-objects.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-a-mixin-to-add-common-behavior-between-unrelated-objects.md
@@ -1,6 +1,6 @@
---
id: 587d7db2367417b2b2512b89
-title: 使用 Mixin 在不相关对象之间添加共同行为
+title: Use a Mixin to Add Common Behavior Between Unrelated Objects
challengeType: 1
forumTopicId: 301331
dashedName: use-a-mixin-to-add-common-behavior-between-unrelated-objects
@@ -8,9 +8,9 @@ dashedName: use-a-mixin-to-add-common-behavior-between-unrelated-objects
# --description--
-正如你所见，行为是可以通过继承来共享的。然而，在有些情况下，继承不是最好的解决方案。继承不适用于不相关的对象，比如`Bird`和`Airplane`。虽然它们都可以飞行，但是`Bird`并不是一种`Airplane`，反之亦然。
+As you have seen, behavior is shared through inheritance. However, there are cases when inheritance is not the best solution. Inheritance does not work well for unrelated objects like `Bird` and `Airplane`. They can both fly, but a `Bird` is not a type of `Airplane` and vice versa.
-对于不相关的对象，更好的方法是使用`mixins`。`mixin`允许其他对象使用函数集合。
+For unrelated objects, it's better to use <dfn>mixins</dfn>. A mixin allows other objects to use a collection of functions.
```js
let flyMixin = function(obj) {
@@ -20,7 +20,7 @@ let flyMixin = function(obj) {
};
```
-`flyMixin`能接受任何对象，并为其提供`fly`方法。
+The `flyMixin` takes any object and gives it the `fly` method.
```js
let bird = {
@@ -37,34 +37,34 @@ flyMixin(bird);
flyMixin(plane);
```
-这里的`flyMixin`接收了`bird`和`plane`对象，然后将`fly`方法分配给了每一个对象。现在`bird`和`plane`都可以飞行了：
+Here `bird` and `plane` are passed into `flyMixin`, which then assigns the `fly` function to each object. Now `bird` and `plane` can both fly:
```js
bird.fly(); // prints "Flying, wooosh!"
plane.fly(); // prints "Flying, wooosh!"
```
-注意观察`mixin`是如何允许相同的`fly`方法被不相关的对象`bird`和`plane`重用的。
+Note how the mixin allows for the same `fly` method to be reused by unrelated objects `bird` and `plane`.
# --instructions--
-创建一个名为`glideMixin`的`mixin`，并定义一个`glide`方法。然后使用`glideMixin`来给`bird`和`boat`赋予滑行（glide）的能力。
+Create a mixin named `glideMixin` that defines a method named `glide`. Then use the `glideMixin` to give both `bird` and `boat` the ability to glide.
# --hints--
-你应该声明一个变量名为`glideMixin`的函数。
+Your code should declare a `glideMixin` variable that is a function.
```js
assert(typeof glideMixin === 'function');
```
-你应该在`bird`上使用`glideMixin`，以提供`glide`方法。
+Your code should use the `glideMixin` on the `bird` object to give it the `glide` method.
```js
assert(typeof bird.glide === 'function');
```
-你应该在`boat`上使用`glideMixin`，以提供`glide`方法。
+Your code should use the `glideMixin` on the `boat` object to give it the `glide` method.
```js
assert(typeof boat.glide === 'function');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-an-iife-to-create-a-module.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-an-iife-to-create-a-module.md
index 17ab12999e5..6f5a69f157e 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-an-iife-to-create-a-module.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-an-iife-to-create-a-module.md
@@ -1,6 +1,6 @@
---
id: 587d7db2367417b2b2512b8c
-title: 使用 IIFE 创建一个模块
+title: Use an IIFE to Create a Module
challengeType: 1
forumTopicId: 301332
dashedName: use-an-iife-to-create-a-module
@@ -8,7 +8,7 @@ dashedName: use-an-iife-to-create-a-module
# --description--
-一个`自执行函数表达式`（`IIFE`）通常用于将相关功能分组到单个对象或者是`模块`中。例如，先前的挑战中定义了一个混合类：
+An immediately invoked function expression (IIFE) is often used to group related functionality into a single object or <dfn>module</dfn>. For example, an earlier challenge defined two mixins:
```js
function glideMixin(obj) {
@@ -23,7 +23,7 @@ function flyMixin(obj) {
}
```
-我们可以将这些`mixins`分成以下模块：
+We can group these mixins into a module as follows:
```js
let motionModule = (function () {
@@ -39,10 +39,10 @@ let motionModule = (function () {
};
}
}
-})(); // 末尾的两个括号导致函数被立即调用
+})(); // The two parentheses cause the function to be immediately invoked
```
-注意：一个`自执行函数表达式`（`IIFE`）返回了一个`motionModule`对象。返回的这个对象包含了作为对象属性的所有`mixin`行为。 `模块`模式的优点是，所有的运动行为都可以打包成一个对象，然后由代码的其他部分使用。下面是一个使用它的例子：
+Note that you have an immediately invoked function expression (IIFE) that returns an object `motionModule`. This returned object contains all of the mixin behaviors as properties of the object. The advantage of the module pattern is that all of the motion behaviors can be packaged into a single object that can then be used by other parts of your code. Here is an example using it:
```js
motionModule.glideMixin(duck);
@@ -51,23 +51,23 @@ duck.glide();
# --instructions--
-创建一个名为`funModule`的`模块`，将这两个`mixins`：`isCuteMixin`和`singMixin`包装起来。`funModule`应该返回一个对象。
+Create a module named `funModule` to wrap the two mixins `isCuteMixin` and `singMixin`. `funModule` should return an object.
# --hints--
-`funModule`应该被定义并返回一个对象。
+`funModule` should be defined and return an object.
```js
assert(typeof funModule === 'object');
```
-`funModule.isCuteMixin`应该访问一个函数。
+`funModule.isCuteMixin` should access a function.
```js
assert(typeof funModule.isCuteMixin === 'function');
```
-`funModule.singMixin`应该访问一个函数。
+`funModule.singMixin` should access a function.
```js
assert(typeof funModule.singMixin === 'function');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-closure-to-protect-properties-within-an-object-from-being-modified-externally.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-closure-to-protect-properties-within-an-object-from-being-modified-externally.md
index 53629ff9296..cdbb9b51c44 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-closure-to-protect-properties-within-an-object-from-being-modified-externally.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-closure-to-protect-properties-within-an-object-from-being-modified-externally.md
@@ -1,6 +1,7 @@
---
id: 587d7db2367417b2b2512b8a
-title: 使用闭包保护对象内的属性不被外部修改
+title: >-
+ Use Closure to Protect Properties Within an Object from Being Modified Externally
challengeType: 1
forumTopicId: 18234
dashedName: >-
@@ -9,50 +10,47 @@ dashedName: >-
# --description--
-在上一次挑战中，`bird`有一个公共属性`name`。公共属性的定义就是：它可以在`bird`的定义范围之外被访问和更改。
+In the previous challenge, `bird` had a public property `name`. It is considered public because it can be accessed and changed outside of `bird`'s definition.
```js
bird.name = "Duffy";
```
-因此，代码的任何地方都可以轻松地将`bird`的 name 属性更改为任意值。想想密码和银行账户之类的东西，如果代码库的任何部分都可以轻易改变，那么将会引起很多问题。
+Therefore, any part of your code can easily change the name of `bird` to any value. Think about things like passwords and bank accounts being easily changeable by any part of your codebase. That could cause a lot of issues.
-使属性私有化最简单的方法就是在构造函数中创建变量。可以将该变量范围限定在构造函数中，而不是全局可用。这样，属性只能由构造函数中的方法访问和更改。
+The simplest way to make this public property private is by creating a variable within the constructor function. This changes the scope of that variable to be within the constructor function versus available globally. This way, the variable can only be accessed and changed by methods also within the constructor function.
```js
function Bird() {
- let hatchedEgg = 10; // 私有变量
+ let hatchedEgg = 10; // private variable
- /* bird 对象可用的公共方法 */
+ /* publicly available method that a bird object can use */
this.getHatchedEggCount = function() {
return hatchedEgg;
};
}
let ducky = new Bird();
-ducky.getHatchedEggCount(); // 返回 10
+ducky.getHatchedEggCount(); // returns 10
```
-这里的`getHachedEggCount`是一种特权方法，因为它可以访问私有属性`hatchedEgg`。这是因为`hatchedEgg`是在与`getHachedEggCount`相同的上下文中声明的。在 JavaScript 中，函数总是可以访问创建它的上下文。这就叫做`闭包`。
+Here `getHatchedEggCount` is a privileged method, because it has access to the private variable `hatchedEgg`. This is possible because `hatchedEgg` is declared in the same context as `getHatchedEggCount`. In JavaScript, a function always has access to the context in which it was created. This is called `closure`.
# --instructions--
-更改在`Bird`函数中声明的`weight`方法，使其成为私有变量。然后，创建一个返回`weight`值的`getWeight`方法。
+Change how `weight` is declared in the `Bird` function so it is a private variable. Then, create a method `getWeight` that returns the value of `weight` 15.
# --hints--
-`weight`属性应该是一个私有变量，值应该是 `15`。
+The `weight` property should be a private variable and should be assigned the value of `15`.
```js
assert(code.match(/(var|let|const)\s+weight\s*\=\s*15\;?/g));
```
-你的代码应该在`Bird`中创建一个名为`getWeight`方法，该方法返回`weight`值。
+Your code should create a method in `Bird` called `getWeight` that returns the value of the private variable `weight`.
```js
-assert(
- new Bird().getWeight() === 15,
- '你的代码应该在<code>Bird</code>中创建一个名为<code>getWeight</code>方法，该方法返回<code>weight</code>值。'
-);
+assert(new Bird().getWeight() === 15);
```
Your `getWeight` function should return the private variable `weight`.
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-dot-notation-to-access-the-properties-of-an-object.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-dot-notation-to-access-the-properties-of-an-object.md
index 119c84e1e2b..b4e34f42f97 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-dot-notation-to-access-the-properties-of-an-object.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-dot-notation-to-access-the-properties-of-an-object.md
@@ -1,6 +1,6 @@
---
id: 587d7dac367417b2b2512b74
-title: 使用点符号来访问对象的属性
+title: Use Dot Notation to Access the Properties of an Object
challengeType: 1
forumTopicId: 301333
dashedName: use-dot-notation-to-access-the-properties-of-an-object
@@ -8,7 +8,7 @@ dashedName: use-dot-notation-to-access-the-properties-of-an-object
# --description--
-在上一个挑战中，我们创建了一个拥有不同`属性`的`对象`，现在我们来看看该如何访问这些`属性`：
+The last challenge created an object with various properties. Now you'll see how to access the values of those properties. Here's an example:
```js
let duck = {
@@ -16,24 +16,24 @@ let duck = {
numLegs: 2
};
console.log(duck.name);
-// 这段代码会在控制台中输出 "Aflac"
+// This prints "Aflac" to the console
```
-我们可以用“点号表示法”来访问对象的属性，只需要在`对象`名称后面加上`.`以及`属性`名即可。比如，`duck.name`就可以访问到 "Aflac"。
+Dot notation is used on the object name, `duck`, followed by the name of the property, `name`, to access the value of "Aflac".
# --instructions--
-请在控制台里面输出`dog`对象中两个`属性`对应的值。
+Print both properties of the `dog` object to your console.
# --hints--
-你应该使用`console.log`来将`dog`对象的`name`属性值输出到控制台。
+Your code should use `console.log` to print the value for the `name` property of the `dog` object.
```js
assert(/console.log\(.*dog\.name.*\)/g.test(code));
```
-你应该使用`console.log`来将`dog`对象的`numLegs`属性值输出到控制台。
+Your code should use `console.log` to print the value for the `numLegs` property of the `dog` object.
```js
assert(/console.log\(.*dog\.numLegs.*\)/g.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-inheritance-so-you-dont-repeat-yourself.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-inheritance-so-you-dont-repeat-yourself.md
index fa99c94b6ec..4411c1a12e4 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-inheritance-so-you-dont-repeat-yourself.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-inheritance-so-you-dont-repeat-yourself.md
@@ -1,6 +1,6 @@
---
id: 587d7db0367417b2b2512b83
-title: 使用继承避免重复
+title: Use Inheritance So You Don't Repeat Yourself
challengeType: 1
forumTopicId: 301334
dashedName: use-inheritance-so-you-dont-repeat-yourself
@@ -8,9 +8,9 @@ dashedName: use-inheritance-so-you-dont-repeat-yourself
# --description--
-有一条原则叫做：`Don't Repeat Yourself`，常以缩写形式`DRY`出现，意思是“不要自己重复”。编写重复代码会产生的问题是：任何改变都需要去多个地方修复所有重复的代码。这通常意味着我们需要做更多的工作，会产生更高的出错率。
+There's a principle in programming called <dfn>Don't Repeat Yourself (DRY)</dfn>. The reason repeated code is a problem is because any change requires fixing code in multiple places. This usually means more work for programmers and more room for errors.
-请观察下面的示例，`Bird`和`Dog`共享`describe`方法：
+Notice in the example below that the `describe` method is shared by `Bird` and `Dog`:
```js
Bird.prototype = {
@@ -28,7 +28,7 @@ Dog.prototype = {
};
```
-我们可以看到`describe`方法在两个地方重复定义了。根据以上所说的`DRY`原则，我们可以通过创建一个`Animal 超类（或者父类）`来重写这段代码：
+The `describe` method is repeated in two places. The code can be edited to follow the DRY principle by creating a `supertype` (or parent) called `Animal`:
```js
function Animal() { };
@@ -41,7 +41,7 @@ Animal.prototype = {
};
```
-`Animal`构造函数中定义了`describe`方法，可将`Bird`和`Dog`这两个构造函数的方法删除掉：
+Since `Animal` includes the `describe` method, you can remove it from `Bird` and `Dog`:
```js
Bird.prototype = {
@@ -55,23 +55,23 @@ Dog.prototype = {
# --instructions--
-`Cat`和`Bear`重复定义了`eat`方法。本着`DRY`的原则，通过将`eat`方法移动到`Animal 超类`中来重写你的代码。
+The `eat` method is repeated in both `Cat` and `Bear`. Edit the code in the spirit of DRY by moving the `eat` method to the `Animal` `supertype`.
# --hints--
-`Animal.prototype`应该有`eat`属性。
+`Animal.prototype` should have the `eat` property.
```js
assert(Animal.prototype.hasOwnProperty('eat'));
```
-`Bear.prototype`不应该有`eat`属性。
+`Bear.prototype` should not have the `eat` property.
```js
assert(!Bear.prototype.hasOwnProperty('eat'));
```
-`Cat.prototype`不应该有`eat`属性。
+`Cat.prototype` should not have the `eat` property.
```js
assert(!Cat.prototype.hasOwnProperty('eat'));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-prototype-properties-to-reduce-duplicate-code.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-prototype-properties-to-reduce-duplicate-code.md
index bd02b2dc923..2f0875f8c3e 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-prototype-properties-to-reduce-duplicate-code.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/use-prototype-properties-to-reduce-duplicate-code.md
@@ -1,6 +1,6 @@
---
id: 587d7dae367417b2b2512b7c
-title: 使用原型属性来减少重复代码
+title: Use Prototype Properties to Reduce Duplicate Code
challengeType: 1
forumTopicId: 301336
dashedName: use-prototype-properties-to-reduce-duplicate-code
@@ -8,44 +8,44 @@ dashedName: use-prototype-properties-to-reduce-duplicate-code
# --description--
-所有`Bird`实例可能会有相同的`numLegs`值，所以在每一个`Bird`的实例中本质上都有一个重复的变量`numLegs`。
+Since `numLegs` will probably have the same value for all instances of `Bird`, you essentially have a duplicated variable `numLegs` inside each `Bird` instance.
-当只有两个实例时可能并不是什么问题，但想象一下如果有数百万个实例，这将会产生许许多多重复的变量。
+This may not be an issue when there are only two instances, but imagine if there are millions of instances. That would be a lot of duplicated variables.
-这里有一个更好的方法可以解决上述问题，那就是使用`Bird`的`原型`。`原型`是一个可以在所有`Bird`实例之间共享的对象。以下是一个在`Bird prototype`中添加`numLegs`属性的示例：
+A better way is to use `Bird’s` `prototype`. Properties in the `prototype` are shared among ALL instances of `Bird`. Here's how to add `numLegs` to the `Bird prototype`:
```js
Bird.prototype.numLegs = 2;
```
-现在所有的`Bird`实例都拥有了共同的`numLegs`属性值。
+Now all instances of `Bird` have the `numLegs` property.
```js
console.log(duck.numLegs); // prints 2
console.log(canary.numLegs); // prints 2
```
-由于所有的实例都可以继承`原型`上的属性，所以可以把`原型`看作是创建对象的 "配方"。 请注意：`duck`和`canary`的`原型`属于`Bird`的构造函数，即 Bird 的原型 `Bird.prototype`。JavaScript 中几乎所有的对象都有一个`原型`属性，这个属性是属于它所在的构造函数。
+Since all instances automatically have the properties on the `prototype`, think of a `prototype` as a "recipe" for creating objects. Note that the `prototype` for `duck` and `canary` is part of the `Bird` constructor as `Bird.prototype`. Nearly every object in JavaScript has a `prototype` property which is part of the constructor function that created it.
# --instructions--
-给`Dog`的`原型`添加一个`numLegs`属性。
+Add a `numLegs` property to the `prototype` of `Dog`
# --hints--
-`beagle`应该有一个`numLegs`属性。
+`beagle` should have a `numLegs` property.
```js
assert(beagle.numLegs !== undefined);
```
-`beagle.numLegs`应该是一个数字。
+`beagle.numLegs` should be a number.
```js
assert(typeof beagle.numLegs === 'number');
```
-`numLegs`应该是一个`原型`属性而不是一个`自身`属性。
+`numLegs` should be a `prototype` property not an `own` property.
```js
assert(beagle.hasOwnProperty('numLegs') === false);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/verify-an-objects-constructor-with-instanceof.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/verify-an-objects-constructor-with-instanceof.md
index ec33e645065..5c4258a1b42 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/verify-an-objects-constructor-with-instanceof.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/object-oriented-programming/verify-an-objects-constructor-with-instanceof.md
@@ -1,6 +1,6 @@
---
id: 587d7dae367417b2b2512b7a
-title: 使用 instance of 验证对象的构造函数
+title: Verify an Object's Constructor with instanceof
challengeType: 1
forumTopicId: 301337
dashedName: verify-an-objects-constructor-with-instanceof
@@ -8,7 +8,7 @@ dashedName: verify-an-objects-constructor-with-instanceof
# --description--
-凡是通过构造函数创建出的新对象，都叫做这个构造函数的`实例`。JavaScript 提供了一种很简便的方法来验证这个事实，那就是通过`instanceof`操作符。`instanceof`允许你将对象与构造函数之间进行比较，根据对象是否由这个构造函数创建的返回`true`或者`false`。以下是一个示例：
+Anytime a constructor function creates a new object, that object is said to be an <dfn>instance</dfn> of its constructor. JavaScript gives a convenient way to verify this with the `instanceof` operator. `instanceof` allows you to compare an object to a constructor, returning `true` or `false` based on whether or not that object was created with the constructor. Here's an example:
```js
let Bird = function(name, color) {
@@ -22,7 +22,7 @@ let crow = new Bird("Alexis", "black");
crow instanceof Bird; // => true
```
-如果一个对象不是使用构造函数创建的，那么`instanceof`将会验证这个对象不是构造函数的实例：
+If an object is created without using a constructor, `instanceof` will verify that it is not an instance of that constructor:
```js
let canary = {
@@ -36,17 +36,17 @@ canary instanceof Bird; // => false
# --instructions--
-给`House`构造函数创建一个新实例，取名为`myHouse`并且传递一个数字给`bedrooms`参数。然后使用`instanceof`操作符验证这个对象是否为`House`的实例。
+Create a new instance of the `House` constructor, calling it `myHouse` and passing a number of bedrooms. Then, use `instanceof` to verify that it is an instance of `House`.
# --hints--
-`myHouse`应该有一个`numBedrooms`属性被赋值为一个数字。
+`myHouse` should have a `numBedrooms` attribute set to a number.
```js
assert(typeof myHouse.numBedrooms === 'number');
```
-请务必使用`instanceof`操作符验证`myHouse`这个对象是`House`构造函数的一个实例。
+You should verify that `myHouse` is an instance of `House` using the `instanceof` operator.
```js
assert(/myHouse\s*instanceof\s*House/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/check-for-all-or-none.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/check-for-all-or-none.md
index 5aa2cdc080c..2b8ee673e9c 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/check-for-all-or-none.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/check-for-all-or-none.md
@@ -1,6 +1,6 @@
---
id: 587d7dba367417b2b2512ba8
-title: 检查全部或无
+title: Check for All or None
challengeType: 1
forumTopicId: 301338
dashedName: check-for-all-or-none
@@ -8,11 +8,11 @@ dashedName: check-for-all-or-none
# --description--
-有时，想要搜寻的匹配模式可能有不确定是否存在的部分。尽管如此，还是想检查它们。
+Sometimes the patterns you want to search for may have parts of it that may or may not exist. However, it may be important to check for them nonetheless.
-为此，可以使用问号`?`指定可能存在的元素。这将检查前面的零个或一个元素。可以将此符号视为前面的元素是可选的。
+You can specify the possible existence of an element with a question mark, `?`. This checks for zero or one of the preceding element. You can think of this symbol as saying the previous element is optional.
-例如，美式英语和英式英语略有不同，可以使用问号来匹配两种拼写。
+For example, there are slight differences in American and British English and you can use the question mark to match both spellings.
```js
let american = "color";
@@ -24,31 +24,35 @@ rainbowRegex.test(british); // Returns true
# --instructions--
-修改正则表达式`favRegex`以匹配美式英语（favorite）和英式英语（favourite）的单词版本。
+Change the regex `favRegex` to match both the American English (favorite) and the British English (favourite) version of the word.
# --hints--
-你的正则表达式应该使用可选符号`?`。
+Your regex should use the optional symbol, `?`.
```js
+favRegex.lastIndex = 0;
assert(favRegex.source.match(/\?/).length > 0);
```
-你的正则表达式应该匹配`'favorite'`。
+Your regex should match `"favorite"`
```js
+favRegex.lastIndex = 0;
assert(favRegex.test('favorite'));
```
-你的正则表达式应该匹配`'favourite'`。
+Your regex should match `"favourite"`
```js
+favRegex.lastIndex = 0;
assert(favRegex.test('favourite'));
```
-你的正则表达式不应该匹配`'fav'`。
+Your regex should not match `"fav"`
```js
+favRegex.lastIndex = 0;
assert(!favRegex.test('fav'));
```
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/check-for-mixed-grouping-of-characters.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/check-for-mixed-grouping-of-characters.md
index f84f9bde3a8..1b47e0c90ec 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/check-for-mixed-grouping-of-characters.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/check-for-mixed-grouping-of-characters.md
@@ -1,6 +1,6 @@
---
id: 5c3dda8b4d8df89bea71600f
-title: 检查混合字符组
+title: Check For Mixed Grouping of Characters
challengeType: 1
forumTopicId: 301339
dashedName: check-for-mixed-grouping-of-characters
@@ -8,55 +8,62 @@ dashedName: check-for-mixed-grouping-of-characters
# --description--
-有时候我们想使用正则表达式里的括号 `()` 来检查字符组。
+Sometimes we want to check for groups of characters using a Regular Expression and to achieve that we use parentheses `()`.
-如果想在字符串找到 `Penguin` 或 `Pumpkin`，可以这个正则表达式：`/P(engu|umpk)in/g`。
+If you want to find either `Penguin` or `Pumpkin` in a string, you can use the following Regular Expression: `/P(engu|umpk)in/g`
-然后使用 `test()` 方法检查 test 字符串里面是否包含字符组。
+Then check whether the desired string groups are in the test string by using the `test()` method.
```js
let testStr = "Pumpkin";
-let testRegex = /P(engu|umpk)in/g;
+let testRegex = /P(engu|umpk)in/;
testRegex.test(testStr);
// Returns true
```
# --instructions--
-完善正则表达式，使其以区分大小写的方式检查 `Franklin Roosevelt` 或 `Eleanor Roosevelt` 的名字，并且应该忽略 middle names。
+Fix the regex so that it checks for the names of `Franklin Roosevelt` or `Eleanor Roosevelt` in a case sensitive manner and it should make concessions for middle names.
-然后完善代码，使创建的正则检查 `myString`，根据正则是否匹配返回 `true` 或 `false`。
+Then fix the code so that the regex that you have created is checked against `myString` and either `true` or `false` is returned depending on whether the regex matches.
# --hints--
-正则 `myRegex` 测试 `Franklin D. Roosevelt` 应该返回 `true`。
+Your regex `myRegex` should return `true` for the string `Franklin D. Roosevelt`
```js
myRegex.lastIndex = 0;
assert(myRegex.test('Franklin D. Roosevelt'));
```
-正则 `myRegex` 测试 `Eleanor Roosevelt` 应该返回 `true`。
+Your regex `myRegex` should return `true` for the string `Eleanor Roosevelt`
```js
myRegex.lastIndex = 0;
assert(myRegex.test('Eleanor Roosevelt'));
```
-正则 `myRegex` 测试 `Franklin Rosevelt` 应该返回 `false`。
+Your regex `myRegex` should return `false` for the string `Franklin Rosevelt`
```js
myRegex.lastIndex = 0;
assert(!myRegex.test('Franklin Rosevelt'));
```
-应该使用 `.test()` 来测试正则。
+Your regex `myRegex` should return `false` for the string `Frank Roosevelt`
+
+```js
+myRegex.lastIndex = 0;
+assert(!myRegex.test('Frank Roosevelt'));
+```
+
+You should use `.test()` to test the regex.
```js
assert(code.match(/myRegex.test\(\s*myString\s*\)/));
```
-result 应该返回 `true`。
+Your result should return `true`.
```js
assert(result === true);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/extract-matches.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/extract-matches.md
index fdb8ca1a087..293eb6da977 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/extract-matches.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/extract-matches.md
@@ -1,6 +1,6 @@
---
id: 587d7db4367417b2b2512b92
-title: 提取匹配项
+title: Extract Matches
challengeType: 1
forumTopicId: 301340
dashedName: extract-matches
@@ -8,9 +8,11 @@ dashedName: extract-matches
# --description--
-到目前为止，只是检查了一个匹配模式是否存在于字符串中。还可以使用`.match()`方法来提取找到的实际匹配项。
+So far, you have only been checking if a pattern exists or not within a string. You can also extract the actual matches you found with the `.match()` method.
-可以使用字符串来调用`.match()`方法，并在括号内传入正则表达式。以下是一个示例：
+To use the `.match()` method, apply the method on a string and pass in the regex inside the parentheses.
+
+Here's an example:
```js
"Hello, World!".match(/Hello/);
@@ -21,25 +23,32 @@ ourStr.match(ourRegex);
// Returns ["expressions"]
```
+Note that the `.match` syntax is the "opposite" of the `.test` method you have been using thus far:
+
+```js
+'string'.match(/regex/);
+/regex/.test('string');
+```
+
# --instructions--
-利用`.match()`方法提取单词`coding`。
+Apply the `.match()` method to extract the word `coding`.
# --hints--
-`结果`应该包含单词`coding`。
+The `result` should have the word `coding`
```js
assert(result.join() === 'coding');
```
-你的正则表达式`codingRegex`应该搜寻`coding`。
+Your regex `codingRegex` should search for `coding`
```js
assert(codingRegex.source === 'coding');
```
-你应该使用`.match()`方法。
+You should use the `.match()` method.
```js
assert(code.match(/\.match\(.*\)/));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/find-characters-with-lazy-matching.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/find-characters-with-lazy-matching.md
index 78b7b7a4b6f..df05cd9717d 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/find-characters-with-lazy-matching.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/find-characters-with-lazy-matching.md
@@ -1,6 +1,6 @@
---
id: 587d7db6367417b2b2512b9b
-title: 用惰性匹配来查找字符
+title: Find Characters with Lazy Matching
challengeType: 1
forumTopicId: 301341
dashedName: find-characters-with-lazy-matching
@@ -8,29 +8,41 @@ dashedName: find-characters-with-lazy-matching
# --description--
-在正则表达式中，`贪婪`匹配会匹配到符合正则表达式匹配模式的字符串的最长可能部分，并将其作为匹配项返回。另一种方案称为`懒惰`匹配，它会匹配到满足正则表达式的字符串的最小可能部分。
+In regular expressions, a <dfn>greedy</dfn> match finds the longest possible part of a string that fits the regex pattern and returns it as a match. The alternative is called a <dfn>lazy</dfn> match, which finds the smallest possible part of the string that satisfies the regex pattern.
-可以将正则表达式`/t[a-z]*i/`应用于字符串`"titanic"`。这个正则表达式是一个以`t`开始，以`i`结束，并且中间有一些字母的匹配模式。
+You can apply the regex `/t[a-z]*i/` to the string `"titanic"`. This regex is basically a pattern that starts with `t`, ends with `i`, and has some letters in between.
-正则表达式默认是`贪婪`匹配，因此匹配返回为`["titani"]`。它会匹配到适合该匹配模式的最大子字符串。
+Regular expressions are by default greedy, so the match would return `["titani"]`. It finds the largest sub-string possible to fit the pattern.
-但是，你可以使用`?`字符来将其变成`懒惰`匹配。调整后的正则表达式`/t[a-z]*?i/`匹配字符串`"titanic"`返回`["ti"]`。
+However, you can use the `?` character to change it to lazy matching. `"titanic"` matched against the adjusted regex of `/t[a-z]*?i/` returns `["ti"]`.
-**注意**
-应该避免使用正则表达式解析 HTML，但是可以用正则表达式匹配 HTML 字符串。
+**Note**
+Parsing HTML with regular expressions should be avoided, but pattern matching an HTML string with regular expressions is completely fine.
# --instructions--
-修复正则表达式`/<.*>/`，让它返回 HTML 标签`<h1>`，而不是文本`"<h1>Winter is coming</h1>"`。请记得在正则表达式中使用通配符`.`来匹配任意字符。
+Fix the regex `/<.*>/` to return the HTML tag `<h1>` and not the text `"<h1>Winter is coming</h1>"`. Remember the wildcard `.` in a regular expression matches any character.
# --hints--
-`结果`变量应该是一个包含`<h1>`的数组。
+The `result` variable should be an array with `<h1>` in it
```js
assert(result[0] == '<h1>');
```
+`myRegex` should use lazy matching
+
+```js
+assert(/\?/g.test(myRegex));
+```
+
+`myRegex` should not include the string 'h1'
+
+```js
+assert(!myRegex.source.match('h1'));
+```
+
# --seed--
## --seed-contents--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/find-more-than-the-first-match.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/find-more-than-the-first-match.md
index 2d366816107..cb76eff45d9 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/find-more-than-the-first-match.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/find-more-than-the-first-match.md
@@ -1,6 +1,6 @@
---
id: 587d7db4367417b2b2512b93
-title: 全局匹配
+title: Find More Than the First Match
challengeType: 1
forumTopicId: 301342
dashedName: find-more-than-the-first-match
@@ -8,7 +8,7 @@ dashedName: find-more-than-the-first-match
# --description--
-到目前为止，只能提取或搜寻一次模式匹配。
+So far, you have only been able to extract or search a pattern once.
```js
let testStr = "Repeat, Repeat, Repeat";
@@ -17,7 +17,7 @@ testStr.match(ourRegex);
// Returns ["Repeat"]
```
-若要多次搜寻或提取模式匹配，可以使用`g`标志。
+To search or extract a pattern more than once, you can use the `g` flag.
```js
let repeatRegex = /Repeat/g;
@@ -27,26 +27,26 @@ testStr.match(repeatRegex);
# --instructions--
-使用正则表达式`starRegex`，从字符串`twinkleStar`中匹配到所有的`"Twinkle"`单词并提取出来。
+Using the regex `starRegex`, find and extract both `"Twinkle"` words from the string `twinkleStar`.
-**注意：**
-在正则表达式上可以有多个标志，比如`/search/gi`。
+**Note**
+You can have multiple flags on your regex like `/search/gi`
# --hints--
-你的正则表达式`starRegex`应该使用全局标志`g`。
+Your regex `starRegex` should use the global flag `g`
```js
assert(starRegex.flags.match(/g/).length == 1);
```
-你的正则表达式`starRegex`应该使用忽略大小写标志`i`。
+Your regex `starRegex` should use the case insensitive flag `i`
```js
assert(starRegex.flags.match(/i/).length == 1);
```
-你的匹配应该匹配单词`'Twinkle'`的两个匹配项。
+Your match should match both occurrences of the word `"Twinkle"`
```js
assert(
@@ -58,7 +58,7 @@ assert(
);
```
-你的匹配`结果`应该包含两个元素。
+Your match `result` should have two elements in it.
```js
assert(result.length == 2);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/find-one-or-more-criminals-in-a-hunt.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/find-one-or-more-criminals-in-a-hunt.md
index d84ac46c748..e94ab1943cd 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/find-one-or-more-criminals-in-a-hunt.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/find-one-or-more-criminals-in-a-hunt.md
@@ -1,6 +1,6 @@
---
id: 587d7db7367417b2b2512b9c
-title: 在狩猎中找到一个或多个罪犯
+title: Find One or More Criminals in a Hunt
challengeType: 1
forumTopicId: 301343
dashedName: find-one-or-more-criminals-in-a-hunt
@@ -8,11 +8,11 @@ dashedName: find-one-or-more-criminals-in-a-hunt
# --description--
-是时候暂停和测试你的新正则表达式写作技巧了。一群罪犯逃出监狱逃跑，但你不知道有多少人。但是，你知道他们和其他人在一起时会保持紧密联系。你有责任立刻找到所有的罪犯。
+Time to pause and test your new regex writing skills. A group of criminals escaped from jail and ran away, but you don't know how many. However, you do know that they stay close together when they are around other people. You are responsible for finding all of the criminals at once.
-这里有一个示例来回顾如何做到这一点：
+Here's an example to review how to do this:
-当字母`z`在一行中出现一次或连续多次时，正则表达式`/z+/`会匹配到它。它会在以下所有字符串中找到匹配项：
+The regex `/z+/` matches the letter `z` when it appears one or more times in a row. It would find matches in all of the following strings:
```js
"z"
@@ -22,7 +22,7 @@ dashedName: find-one-or-more-criminals-in-a-hunt
"abczzzzzzzzzzzzzzzzzzzzzabc"
```
-但是它不会在以下字符串中找到匹配项，因为它们中没有字母`z`：
+But it does not find matches in the following strings since there are no letter `z` characters:
```js
""
@@ -32,23 +32,23 @@ dashedName: find-one-or-more-criminals-in-a-hunt
# --instructions--
-编写一个`贪婪`正则表达式，在一组其他人中匹配到一个或多个罪犯。罪犯由大写字母`C`表示。
+Write a greedy regex that finds one or more criminals within a group of other people. A criminal is represented by the capital letter `C`.
# --hints--
-你的正则表达式应该匹配`'C'`中的 <em>一个</em> 罪犯（'`C`'）。
+Your regex should match one criminal (`C`) in `"C"`
```js
assert('C'.match(reCriminals) && 'C'.match(reCriminals)[0] == 'C');
```
-你的正则表达式应该匹配`'CC'`中的 <em>两个</em> 罪犯（'`CC`'）。
+Your regex should match two criminals (`CC`) in `"CC"`
```js
assert('CC'.match(reCriminals) && 'CC'.match(reCriminals)[0] == 'CC');
```
-你的正则表达式应该匹配`'P1P5P4CCCP2P6P3'`中的 <em>三个</em> 罪犯（'`CCC`'）。
+Your regex should match three criminals (`CCC`) in `"P1P5P4CCCP2P6P3"`
```js
assert(
@@ -57,7 +57,7 @@ assert(
);
```
-你的正则表达式应该匹配`'P6P2P7P4P5CCCCCP3P1'`中的 <em>五个</em> 罪犯（'`CCCCC`'）。
+Your regex should match five criminals (`CCCCC`) in `"P6P2P7P4P5CCCCCP3P1"`
```js
assert(
@@ -66,19 +66,19 @@ assert(
);
```
-你的正则表达式在`''`中不应该匹配到任何罪犯。
+Your regex should not match any criminals in `""`
```js
assert(!reCriminals.test(''));
```
-你的正则表达式在`'P1P2P3'`中不应该匹配到任何罪犯。
+Your regex should not match any criminals in `"P1P2P3"`
```js
assert(!reCriminals.test('P1P2P3'));
```
-你的正则表达式应该匹配`'P2P1P5P4CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCP3'`中的 <em>五十个</em> 罪犯（'`CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC`'）。
+Your regex should match fifty criminals (`CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC`) in `"P2P1P5P4CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCP3"`.
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/ignore-case-while-matching.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/ignore-case-while-matching.md
index d3012ca65f2..8ea7f4218ec 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/ignore-case-while-matching.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/ignore-case-while-matching.md
@@ -1,6 +1,6 @@
---
id: 587d7db4367417b2b2512b91
-title: 匹配时忽略大小写
+title: Ignore Case While Matching
challengeType: 1
forumTopicId: 301344
dashedName: ignore-case-while-matching
@@ -8,73 +8,73 @@ dashedName: ignore-case-while-matching
# --description--
-到目前为止，已经了解了如何用正则表达式来执行字符串的匹配。但有时候，并不关注匹配字母的大小写。
+Up until now, you've looked at regexes to do literal matches of strings. But sometimes, you might want to also match case differences.
-大小写即大写字母和小写字母。大写字母如`"A"`、`"B"`和`"C"`。小写字母如`"a"`、`"b"`和`"c"`。
+Case (or sometimes letter case) is the difference between uppercase letters and lowercase letters. Examples of uppercase are `"A"`, `"B"`, and `"C"`. Examples of lowercase are `"a"`, `"b"`, and `"c"`.
-可以使用标志（flag）来匹配这两种情况。标志有很多，不过这里我们只关注忽略大小写的标志——`i`。可以通过将它附加到正则表达式之后来使用它。这里给出使用该标志的一个实例`/ignorecase/i`。这个字符串可以匹配字符串`"ignorecase"`、`"igNoreCase"`和`"IgnoreCase"`。
+You can match both cases using what is called a flag. There are other flags but here you'll focus on the flag that ignores case - the `i` flag. You can use it by appending it to the regex. An example of using this flag is `/ignorecase/i`. This regex can match the strings `"ignorecase"`, `"igNoreCase"`, and `"IgnoreCase"`.
# --instructions--
-编写正则表达式`fccRegex`以匹配`"freeCodeCamp"`，忽略大小写。正则表达式不应与任何缩写或带有空格的变体匹配。
+Write a regex `fccRegex` to match `"freeCodeCamp"`, no matter its case. Your regex should not match any abbreviations or variations with spaces.
# --hints--
-你的正则表达式应该匹配`freeCodeCamp`。
+Your regex should match `freeCodeCamp`
```js
assert(fccRegex.test('freeCodeCamp'));
```
-你的正则表达式应该匹配`FreeCodeCamp`。
+Your regex should match `FreeCodeCamp`
```js
assert(fccRegex.test('FreeCodeCamp'));
```
-你的正则表达式应该匹配`FreecodeCamp`。
+Your regex should match `FreecodeCamp`
```js
assert(fccRegex.test('FreecodeCamp'));
```
-你的正则表达式应该匹配`FreeCodecamp`。
+Your regex should match `FreeCodecamp`
```js
assert(fccRegex.test('FreeCodecamp'));
```
-你的正则表达式不应该匹配`Free Code Camp`。
+Your regex should not match `Free Code Camp`
```js
assert(!fccRegex.test('Free Code Camp'));
```
-Your regex should match`FreeCOdeCamp`。
+Your regex should match `FreeCOdeCamp`
```js
assert(fccRegex.test('FreeCOdeCamp'));
```
-你的正则表达式不应该匹配`FCC`。
+Your regex should not match `FCC`
```js
assert(!fccRegex.test('FCC'));
```
-你的正则表达式应该匹配`FrEeCoDeCamp`。
+Your regex should match `FrEeCoDeCamp`
```js
assert(fccRegex.test('FrEeCoDeCamp'));
```
-你的正则表达式应该匹配`FrEeCodECamp`。
+Your regex should match `FrEeCodECamp`
```js
assert(fccRegex.test('FrEeCodECamp'));
```
-你的正则表达式应该匹配`FReeCodeCAmp`。
+Your regex should match `FReeCodeCAmp`
```js
assert(fccRegex.test('FReeCodeCAmp'));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-a-literal-string-with-different-possibilities.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-a-literal-string-with-different-possibilities.md
index 825ebd8c6cc..c98f45b8519 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-a-literal-string-with-different-possibilities.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-a-literal-string-with-different-possibilities.md
@@ -1,6 +1,6 @@
---
id: 587d7db4367417b2b2512b90
-title: 同时用多种模式匹配文字字符串
+title: Match a Literal String with Different Possibilities
challengeType: 1
forumTopicId: 301345
dashedName: match-a-literal-string-with-different-possibilities
@@ -8,57 +8,57 @@ dashedName: match-a-literal-string-with-different-possibilities
# --description--
-使用正则表达式`/coding/`，你可以在其他字符串中查找`"coding"`。
+Using regexes like `/coding/`, you can look for the pattern `"coding"` in another string.
-这对于搜寻单个字符串非常有用，但仅限于一种匹配模式。你可以使用`|`操作符来匹配多个规则。
+This is powerful to search single strings, but it's limited to only one pattern. You can search for multiple patterns using the `alternation` or `OR` operator: `|`.
-此操作符匹配操作符前面或后面的字符。例如，如果你想匹配`"yes"`或`"no"`，你需要的正则表达式是`/yes|no/`。
+This operator matches patterns either before or after it. For example, if you wanted to match `"yes"` or `"no"`, the regex you want is `/yes|no/`.
-你还可以匹配多个规则，这可以通过添加更多的匹配模式来实现。这些匹配模式将包含更多的`|`操作符来分隔它们，比如`/yes|no|maybe/`。
+You can also search for more than just two patterns. You can do this by adding more patterns with more `OR` operators separating them, like `/yes|no|maybe/`.
# --instructions--
-完成正则表达式`petRegex`以匹配`"dog"`、`"cat"`、`"bird"`或者`"fish"`。
+Complete the regex `petRegex` to match the pets `"dog"`, `"cat"`, `"bird"`, or `"fish"`.
# --hints--
-对于字符串`'John has a pet dog.'`，你的正则表达式`petRegex`的`test`方法应该返回`true`。
+Your regex `petRegex` should return `true` for the string `"John has a pet dog."`
```js
assert(petRegex.test('John has a pet dog.'));
```
-对于字符串`'Emma has a pet rock.'`，你的正则表达式`petRegex`的`test`方法应该返回`false`。
+Your regex `petRegex` should return `false` for the string `"Emma has a pet rock."`
```js
assert(!petRegex.test('Emma has a pet rock.'));
```
-对于字符串`'Emma has a pet bird.'`，你的正则表达式`petRegex`的`test`方法应该返回`true`。
+Your regex `petRegex` should return `true` for the string `"Emma has a pet bird."`
```js
assert(petRegex.test('Emma has a pet bird.'));
```
-对于字符串`'Liz has a pet cat.'`，你的正则表达式`petRegex`的`test`方法应该返回`true`。
+Your regex `petRegex` should return `true` for the string `"Liz has a pet cat."`
```js
assert(petRegex.test('Liz has a pet cat.'));
```
-对于字符串`'Kara has a pet dolphin.'`，你的正则表达式`petRegex`的`test`方法应该返回`false`。
+Your regex `petRegex` should return `false` for the string `"Kara has a pet dolphin."`
```js
assert(!petRegex.test('Kara has a pet dolphin.'));
```
-对于字符串`'Alice has a pet fish.'`，你的正则表达式`petRegex`的`test`方法应该返回`true`。
+Your regex `petRegex` should return `true` for the string `"Alice has a pet fish."`
```js
assert(petRegex.test('Alice has a pet fish.'));
```
-对于字符串`'Jimmy has a pet computer.'`，你的正则表达式`petRegex`的`test`方法应该返回`false`。
+Your regex `petRegex` should return `false` for the string `"Jimmy has a pet computer."`
```js
assert(!petRegex.test('Jimmy has a pet computer.'));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-all-letters-and-numbers.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-all-letters-and-numbers.md
index 2f72db40cfd..70cb70395a3 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-all-letters-and-numbers.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-all-letters-and-numbers.md
@@ -1,6 +1,6 @@
---
id: 587d7db7367417b2b2512b9f
-title: 匹配所有的字母和数字
+title: Match All Letters and Numbers
challengeType: 1
forumTopicId: 301346
dashedName: match-all-letters-and-numbers
@@ -8,9 +8,9 @@ dashedName: match-all-letters-and-numbers
# --description--
-使用元字符，可以使用`[a-z]`搜寻字母表中的所有字母。这种元字符是很常见的，它有一个缩写，但这个缩写也包含额外的字符。
+Using character classes, you were able to search for all letters of the alphabet with `[a-z]`. This kind of character class is common enough that there is a shortcut for it, although it includes a few extra characters as well.
-JavaScript 中与字母表匹配的最接近的元字符是`\w`，这个缩写等同于`[A-Za-z0-9_]`。它不仅可以匹配大小写字母和数字，注意，它还会匹配下划线字符（`_`）。
+The closest character class in JavaScript to match the alphabet is `\w`. This shortcut is equal to `[A-Za-z0-9_]`. This character class matches upper and lowercase letters plus numbers. Note, this character class also includes the underscore character (`_`).
```js
let longHand = /[A-Za-z0-9_]+/;
@@ -23,25 +23,27 @@ longHand.test(varNames); // Returns true
shortHand.test(varNames); // Returns true
```
+These shortcut character classes are also known as <dfn>shorthand character classes</dfn>.
+
# --instructions--
-使用缩写`\w`来计算所有引号中字母和数字字符的数量。
+Use the shorthand character class `\w` to count the number of alphanumeric characters in various quotes and strings.
# --hints--
-你的正则表达式应该使用全局状态修正符。
+Your regex should use the global flag.
```js
assert(alphabetRegexV2.global);
```
-正则表达式应该使用元字符 `\w` 匹配字母表里的所有字符。
+Your regex should use the shorthand character `\w` to match all characters which are alphanumeric.
```js
assert(/\\w/.test(alphabetRegexV2.source));
```
-你的正则表达式应该在`'The five boxing wizards jump quickly.'`中匹配到 31 个字母数字字符。
+Your regex should find 31 alphanumeric characters in `"The five boxing wizards jump quickly."`
```js
assert(
@@ -49,7 +51,7 @@ assert(
);
```
-你的正则表达式应该在`'Pack my box with five dozen liquor jugs.'`中匹配到 32 个字母数字字符。
+Your regex should find 32 alphanumeric characters in `"Pack my box with five dozen liquor jugs."`
```js
assert(
@@ -58,7 +60,7 @@ assert(
);
```
-你的正则表达式应该在`'How vexingly quick daft zebras jump!'`中匹配到 30 个字母数字字符。
+Your regex should find 30 alphanumeric characters in `"How vexingly quick daft zebras jump!"`
```js
assert(
@@ -66,7 +68,7 @@ assert(
);
```
-你的正则表达式应该在`'123 456 7890 ABC def GHI jkl MNO pqr STU vwx YZ.'`中匹配到 36 个字母数字字符。
+Your regex should find 36 alphanumeric characters in `"123 456 7890 ABC def GHI jkl MNO pqr STU vwx YZ."`
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-all-non-numbers.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-all-non-numbers.md
index efffcc6ab33..2290a226d99 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-all-non-numbers.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-all-non-numbers.md
@@ -1,6 +1,6 @@
---
id: 587d7db8367417b2b2512ba1
-title: 匹配所有非数字
+title: Match All Non-Numbers
challengeType: 1
forumTopicId: 301347
dashedName: match-all-non-numbers
@@ -8,59 +8,59 @@ dashedName: match-all-non-numbers
# --description--
-上一项挑战中展示了如何使用带有小写`d`的缩写`\d`来搜寻数字。也可以使用类似的缩写来搜寻非数字，该缩写使用大写的`D`。
+The last challenge showed how to search for digits using the shortcut `\d` with a lowercase `d`. You can also search for non-digits using a similar shortcut that uses an uppercase `D` instead.
-查找非数字字符的缩写是`\D`。这等同于字符串`[^0-9]`，它查找不是 0 - 9 之间数字的单个字符。
+The shortcut to look for non-digit characters is `\D`. This is equal to the character class `[^0-9]`, which looks for a single character that is not a number between zero and nine.
# --instructions--
-使用非数字缩写`\D`来计算电影标题中有多少非数字。
+Use the shorthand character class for non-digits `\D` to count how many non-digits are in movie titles.
# --hints--
-你的正则表达式应该使用缩写来匹配非数字字符。
+Your regex should use the shortcut character to match non-digit characters
```js
assert(/\\D/.test(noNumRegex.source));
```
-你的正则表达式应该使用全局状态修正符。
+Your regex should use the global flag.
```js
assert(noNumRegex.global);
```
-你的正则表达式在`'9'`中应该匹配不到非数字。
+Your regex should find no non-digits in `"9"`.
```js
assert('9'.match(noNumRegex) == null);
```
-你的正则表达式应该在`'Catch 22'`中匹配到 6 个非数字。
+Your regex should find 6 non-digits in `"Catch 22"`.
```js
assert('Catch 22'.match(noNumRegex).length == 6);
```
-你的正则表达式应该在`'101 Dalmatians'`中匹配到 11 个非数字。
+Your regex should find 11 non-digits in `"101 Dalmatians"`.
```js
assert('101 Dalmatians'.match(noNumRegex).length == 11);
```
-你的正则表达式应该在`'One, Two, Three'`中匹配到 15 个非数字。
+Your regex should find 15 non-digits in `"One, Two, Three"`.
```js
assert('One, Two, Three'.match(noNumRegex).length == 15);
```
-你的正则表达式应该在`'21 Jump Street'`中匹配到 12 个非数字。
+Your regex should find 12 non-digits in `"21 Jump Street"`.
```js
assert('21 Jump Street'.match(noNumRegex).length == 12);
```
-你的正则表达式应该在`"2001: A Space Odyssey"`中匹配到 17 个非数字。
+Your regex should find 17 non-digits in `"2001: A Space Odyssey"`.
```js
assert('2001: A Space Odyssey'.match(noNumRegex).length == 17);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-all-numbers.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-all-numbers.md
index 371fa1a3fa7..86c84f015a7 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-all-numbers.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-all-numbers.md
@@ -1,6 +1,6 @@
---
id: 5d712346c441eddfaeb5bdef
-title: 匹配所有数字
+title: Match All Numbers
challengeType: 1
forumTopicId: 18181
dashedName: match-all-numbers
@@ -8,59 +8,59 @@ dashedName: match-all-numbers
# --description--
-已经了解了常见字符串匹配模式的元字符，如字母数字。另一个常见的匹配模式是只寻找数字。
+You've learned shortcuts for common string patterns like alphanumerics. Another common pattern is looking for just digits or numbers.
-查找数字字符的缩写是`\d`，注意是小写的`d`。这等同于元字符`[0-9]`，它查找 0 到 9 之间任意数字的单个字符。
+The shortcut to look for digit characters is `\d`, with a lowercase `d`. This is equal to the character class `[0-9]`, which looks for a single character of any number between zero and nine.
# --instructions--
-使用缩写`\d`来计算电影标题中有多少个数字。书面数字（"six" 而不是 6）不计算在内。
+Use the shorthand character class `\d` to count how many digits are in movie titles. Written out numbers ("six" instead of 6) do not count.
# --hints--
-你的正则表达式应该使用缩写来匹配数字字符。
+Your regex should use the shortcut character to match digit characters
```js
assert(/\\d/.test(numRegex.source));
```
-你的正则表达式应该使用全局状态修正符。
+Your regex should use the global flag.
```js
assert(numRegex.global);
```
-你的正则表达式应该在`'9'`中匹配到 1 个数字。
+Your regex should find 1 digit in `"9"`.
```js
assert('9'.match(numRegex).length == 1);
```
-你的正则表达式应该在`'Catch 22'`中匹配到 2 个数字。
+Your regex should find 2 digits in `"Catch 22"`.
```js
assert('Catch 22'.match(numRegex).length == 2);
```
-你的正则表达式应该在`'101 Dalmatians'`中匹配到 3 个数字。
+Your regex should find 3 digits in `"101 Dalmatians"`.
```js
assert('101 Dalmatians'.match(numRegex).length == 3);
```
-你的正则表达式在`'One, Two, Three'`中应该匹配不到数字。
+Your regex should find no digits in `"One, Two, Three"`.
```js
assert('One, Two, Three'.match(numRegex) == null);
```
-你的正则表达式应该在`'21 Jump Street'`中匹配到 2 个数字。
+Your regex should find 2 digits in `"21 Jump Street"`.
```js
assert('21 Jump Street'.match(numRegex).length == 2);
```
-你的正则表达式应该在`"2001: A Space Odyssey"`中匹配到 4 个数字。
+Your regex should find 4 digits in `"2001: A Space Odyssey"`.
```js
assert('2001: A Space Odyssey'.match(numRegex).length == 4);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-anything-with-wildcard-period.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-anything-with-wildcard-period.md
index 550f4f2ec95..64c17e62a69 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-anything-with-wildcard-period.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-anything-with-wildcard-period.md
@@ -1,6 +1,6 @@
---
id: 587d7db5367417b2b2512b94
-title: 用通配符.匹配任何内容
+title: Match Anything with Wildcard Period
challengeType: 1
forumTopicId: 301348
dashedName: match-anything-with-wildcard-period
@@ -8,9 +8,9 @@ dashedName: match-anything-with-wildcard-period
# --description--
-有时不（或不需要）知道匹配模式中的确切字符。如果要精确匹配到完整的单词，那出现一个拼写错误就会匹配不到。幸运的是，可以使用通配符`.`来处理这种情况。
+Sometimes you won't (or don't need to) know the exact characters in your patterns. Thinking of all words that match, say, a misspelling would take a long time. Luckily, you can save time using the wildcard character: `.`
-通配符`.`将匹配任何一个字符。通配符也叫`dot`或`period`。可以像使用正则表达式中任何其他字符一样使用通配符。例如，如果想匹配`"hug"`、`"huh"`、`"hut"`和`"hum"`，可以使用正则表达式`/hu./`匹配以上四个单词。
+The wildcard character `.` will match any one character. The wildcard is also called `dot` and `period`. You can use the wildcard character just like any other character in the regex. For example, if you wanted to match `"hug"`, `"huh"`, `"hut"`, and `"hum"`, you can use the regex `/hu./` to match all four words.
```js
let humStr = "I'll hum a song";
@@ -22,55 +22,61 @@ huRegex.test(hugStr); // Returns true
# --instructions--
-完成正则表达式`unRegex`以匹配字符串`"run"`、`"sun"`、`"fun"`、`"pun"`、`"nun"`和`"bun"`。正则表达式中应该使用通配符。
+Complete the regex `unRegex` so that it matches the strings `"run"`, `"sun"`, `"fun"`, `"pun"`, `"nun"`, and `"bun"`. Your regex should use the wildcard character.
# --hints--
-你应该使用`.test()`方法。
+You should use the `.test()` method.
```js
assert(code.match(/\.test\(.*\)/));
```
-你应该在你的正则表达式`unRegex`中使用通配符。
+You should use the wildcard character in your regex `unRegex`
```js
assert(/\./.test(unRegex.source));
```
-你的正则表达式`unRegex`应该在字符串`'Let us go on a run.'`中匹配到`'run'`单词。
+Your regex `unRegex` should match `"run"` in `"Let us go on a run."`
```js
+unRegex.lastIndex = 0;
assert(unRegex.test('Let us go on a run.'));
```
-你的正则表达式`unRegex`应该在字符串`'The sun is out today.'`中匹配到`'sun'`单词。
+Your regex `unRegex` should match `"sun"` in `"The sun is out today."`
```js
+unRegex.lastIndex = 0;
assert(unRegex.test('The sun is out today.'));
```
-你的正则表达式`unRegex`应该在字符串`'Coding is a lot of fun.'`中匹配到`'fun'`单词。
+Your regex `unRegex` should match `"fun"` in `"Coding is a lot of fun."`
```js
+unRegex.lastIndex = 0;
assert(unRegex.test('Coding is a lot of fun.'));
```
-你的正则表达式`unRegex`应该在字符串`'Seven days without a pun makes one weak.'`中匹配到`'pun'`单词。
+Your regex `unRegex` should match `"pun"` in `"Seven days without a pun makes one weak."`
```js
+unRegex.lastIndex = 0;
assert(unRegex.test('Seven days without a pun makes one weak.'));
```
-你的正则表达式`unRegex`应该在字符串`'One takes a vow to be a nun.'`中匹配到`'nun'`单词。
+Your regex `unRegex` should match `"nun"` in `"One takes a vow to be a nun."`
```js
+unRegex.lastIndex = 0;
assert(unRegex.test('One takes a vow to be a nun.'));
```
-你的正则表达式`unRegex`应该在字符串`'She got fired from the hot dog stand for putting her hair in a bun.'`中匹配到`'bun'`单词。
+Your regex `unRegex` should match `"bun"` in `"She got fired from the hot dog stand for putting her hair in a bun."`
```js
+unRegex.lastIndex = 0;
assert(
unRegex.test(
'She got fired from the hot dog stand for putting her hair in a bun.'
@@ -78,16 +84,18 @@ assert(
);
```
-你的正则表达式`unRegex`不应该匹配`'There is a bug in my code.'`。
+Your regex `unRegex` should not match `"There is a bug in my code."`
```js
+unRegex.lastIndex = 0;
assert(!unRegex.test('There is a bug in my code.'));
```
-你的正则表达式`unRegex`不应该匹配`'Catch me if you can.'`。
+Your regex `unRegex` should not match `"Catch me if you can."`
```js
-assert(!unRegex.test('Can me if you can.'));
+unRegex.lastIndex = 0;
+assert(!unRegex.test('Catch me if you can.'));
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-beginning-string-patterns.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-beginning-string-patterns.md
index b7addcee015..099238eebf0 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-beginning-string-patterns.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-beginning-string-patterns.md
@@ -1,6 +1,6 @@
---
id: 587d7db7367417b2b2512b9d
-title: 匹配字符串的开头
+title: Match Beginning String Patterns
challengeType: 1
forumTopicId: 301349
dashedName: match-beginning-string-patterns
@@ -8,9 +8,9 @@ dashedName: match-beginning-string-patterns
# --description--
-回顾一下之前的挑战，正则表达式可以用于查找多项匹配。还可以查询字符串中符合指定匹配模式的字符。
+Prior challenges showed that regular expressions can be used to look for a number of matches. They are also used to search for patterns in specific positions in strings.
-在之前的挑战中，使用`字符集`中的`插入`符号（`^`）来创建一个`否定字符集`，形如`[^thingsThatWillNotBeMatched]`。在`字符集`之外，`插入`符号用于字符串的开头搜寻匹配模式。
+In an earlier challenge, you used the caret character (`^`) inside a character set to create a negated character set in the form `[^thingsThatWillNotBeMatched]`. Outside of a character set, the caret is used to search for patterns at the beginning of strings.
```js
let firstString = "Ricky is first and can be found.";
@@ -24,29 +24,29 @@ firstRegex.test(notFirst);
# --instructions--
-在正则表达式中使用`^`符号，以匹配仅在字符串`rickyAndCal`的开头出现的`"Cal"`。
+Use the caret character in a regex to find `"Cal"` only in the beginning of the string `rickyAndCal`.
# --hints--
-你的正则表达式应该搜寻有一个大写字母的`'Cal'`。
+Your regex should search for `"Cal"` with a capital letter.
```js
assert(calRegex.source == '^Cal');
```
-你的正则表达式不应该使用任何标志。
+Your regex should not use any flags.
```js
assert(calRegex.flags == '');
```
-你的正则表达式应该匹配字符串开头的`'Cal'`。
+Your regex should match `"Cal"` at the beginning of the string.
```js
assert(calRegex.test('Cal and Ricky both like racing.'));
```
-你的正则表达式不应该匹配字符串中间的`'Cal'`。
+Your regex should not match `"Cal"` in the middle of a string.
```js
assert(!calRegex.test('Ricky and Cal both like racing.'));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-characters-that-occur-one-or-more-times.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-characters-that-occur-one-or-more-times.md
index 68325201b41..cf54e942ffa 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-characters-that-occur-one-or-more-times.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-characters-that-occur-one-or-more-times.md
@@ -1,6 +1,6 @@
---
id: 587d7db6367417b2b2512b99
-title: 匹配出现一次或多次的字符
+title: Match Characters that Occur One or More Times
challengeType: 1
forumTopicId: 301350
dashedName: match-characters-that-occur-one-or-more-times
@@ -8,33 +8,33 @@ dashedName: match-characters-that-occur-one-or-more-times
# --description--
-有时，需要匹配出现一次或者连续多次的的字符（或字符组）。这意味着它至少出现一次，并且可能重复出现。
+Sometimes, you need to match a character (or group of characters) that appears one or more times in a row. This means it occurs at least once, and may be repeated.
-可以使用`+`符号来检查情况是否如此。记住，字符或匹配模式必须一个接一个地连续出现。
+You can use the `+` character to check if that is the case. Remember, the character or pattern has to be present consecutively. That is, the character has to repeat one after the other.
-例如，`/a+/g`会在`"abc"`中匹配到一个匹配项，并且返回`["a"]`。因为`+`的存在，它也会在`"aabc"`中匹配到一个匹配项，然后返回`["aa"]`。
+For example, `/a+/g` would find one match in `"abc"` and return `["a"]`. Because of the `+`, it would also find a single match in `"aabc"` and return `["aa"]`.
-如果它是检查字符串`"abab"`，它将匹配到两个匹配项并且返回`["a", "a"]`，因为`a`字符不连续，在它们之间有一个`b`字符。最后，因为在字符串`"bcd"`中没有`"a"`，因此找不到匹配项。
+If it were instead checking the string `"abab"`, it would find two matches and return `["a", "a"]` because the `a` characters are not in a row - there is a `b` between them. Finally, since there is no `"a"` in the string `"bcd"`, it wouldn't find a match.
# --instructions--
-想要在字符串`"Mississippi"`中匹配到出现一次或多次的字母`s`的匹配项。编写一个使用`+`符号的正则表达式。
+You want to find matches when the letter `s` occurs one or more times in `"Mississippi"`. Write a regex that uses the `+` sign.
# --hints--
-你的正则表达式`myRegex`应该使用`+`符号来匹配一个或多个`s`字符。
+Your regex `myRegex` should use the `+` sign to match one or more `s` characters.
```js
assert(/\+/.test(myRegex.source));
```
-你的正则表达式`myRegex`应该匹配两项。
+Your regex `myRegex` should match 2 items.
```js
assert(result.length == 2);
```
-`结果`变量应该是一个包含两个`'ss'`匹配项的数组。
+The `result` variable should be an array with two matches of `"ss"`
```js
assert(result[0] == 'ss' && result[1] == 'ss');
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-characters-that-occur-zero-or-more-times.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-characters-that-occur-zero-or-more-times.md
index 8e2a70ee45b..745202e5fe9 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-characters-that-occur-zero-or-more-times.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-characters-that-occur-zero-or-more-times.md
@@ -1,6 +1,6 @@
---
id: 587d7db6367417b2b2512b9a
-title: 匹配出现零次或多次的字符
+title: Match Characters that Occur Zero or More Times
challengeType: 1
forumTopicId: 301351
dashedName: match-characters-that-occur-zero-or-more-times
@@ -8,9 +8,9 @@ dashedName: match-characters-that-occur-zero-or-more-times
# --description--
-上一次的挑战中使用了加号`+`来查找出现一次或多次的字符。还有一个选项可以匹配出现零次或多次的字符。
+The last challenge used the plus `+` sign to look for characters that occur one or more times. There's also an option that matches characters that occur zero or more times.
-执行该操作的字符叫做`asterisk`或`star`，即`*`。
+The character to do this is the asterisk or star: `*`.
```js
let soccerWord = "gooooooooal!";
@@ -24,35 +24,35 @@ oPhrase.match(goRegex); // Returns null
# --instructions--
-在这个挑战里，`chewieQuote` 已经被初始化为 "Aaaaaaaaaaaaaaaarrrgh!"。创建一个变量为`chewieRegex`的正则表达式，使用`*`符号在`chewieQuote`中匹配`"A"`及其之后出现的零个或多个`"a"`。你的正则表达式不需要使用修饰符，也不需要匹配引号。
+For this challenge, `chewieQuote` has been initialized as "Aaaaaaaaaaaaaaaarrrgh!" behind the scenes. Create a regex `chewieRegex` that uses the `*` character to match an uppercase `"A"` character immediately followed by zero or more lowercase `"a"` characters in `chewieQuote`. Your regex does not need flags or character classes, and it should not match any of the other quotes.
# --hints--
-你的正则表达式`chewieRegex`应该使用`*`符号匹配`'A'`之后出现的零个或多个`'a'`字符。
+Your regex `chewieRegex` should use the `*` character to match zero or more `a` characters.
```js
assert(/\*/.test(chewieRegex.source));
```
-正则表达式应当匹配 `chewieQuote` 里的 `"A"`。
+Your regex should match `"A"` in `chewieQuote`.
```js
assert(result[0][0] === 'A');
```
-你的正则表达式应该匹配`'Aaaaaaaaaaaaaaaa'`。
+Your regex should match `"Aaaaaaaaaaaaaaaa"` in `chewieQuote`.
```js
assert(result[0] === 'Aaaaaaaaaaaaaaaa');
```
-你的正则表达式`chewieRegex`应该匹配 16 个字符。
+Your regex `chewieRegex` should match 16 characters in `chewieQuote`.
```js
assert(result[0].length === 16);
```
-你的正则表达式在`'He made a fair move. Screaming about it can't help you.'`中不应该匹配任何字符。
+Your regex should not match any characters in "He made a fair move. Screaming about it can't help you."
```js
assert(
@@ -60,7 +60,7 @@ assert(
);
```
-你的正则表达式在`'Let him have it. It's not wise to upset a Wookiee.'`中不应该匹配任何字符。
+Your regex should not match any characters in "Let him have it. It's not wise to upset a Wookiee."
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-ending-string-patterns.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-ending-string-patterns.md
index efcb8500cef..60e1638ce7d 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-ending-string-patterns.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-ending-string-patterns.md
@@ -1,6 +1,6 @@
---
id: 587d7db7367417b2b2512b9e
-title: 匹配字符串的末尾
+title: Match Ending String Patterns
challengeType: 1
forumTopicId: 301352
dashedName: match-ending-string-patterns
@@ -8,9 +8,9 @@ dashedName: match-ending-string-patterns
# --description--
-在上一个挑战中，学习了使用`^`符号来搜寻字符串开头的匹配模式。还有一种方法可以搜寻字符串末尾的匹配模式。
+In the last challenge, you learned to use the caret character to search for patterns at the beginning of strings. There is also a way to search for patterns at the end of strings.
-可以使用正则表达式的`美元`符号`$`来搜寻字符串的结尾。
+You can search the end of strings using the dollar sign character `$` at the end of the regex.
```js
let theEnding = "This is a never ending story";
@@ -25,23 +25,23 @@ storyRegex.test(noEnding);
# --instructions--
-使用`$`在字符串`caboose`的末尾匹配`"caboose"`。
+Use the anchor character (`$`) to match the string `"caboose"` at the end of the string `caboose`.
# --hints--
-你应该在正则表达式使用美元符号`$`来搜寻`'caboose'`。
+You should search for `"caboose"` with the dollar sign `$` anchor in your regex.
```js
assert(lastRegex.source == 'caboose$');
```
-你的正则表达式不应该使用任何标志。
+Your regex should not use any flags.
```js
assert(lastRegex.flags == '');
```
-你应该在字符串`'The last car on a train is the caboose'`的末尾匹配`'caboose'`。
+You should match `"caboose"` at the end of the string `"The last car on a train is the caboose"`
```js
assert(lastRegex.test('The last car on a train is the caboose'));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-everything-but-letters-and-numbers.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-everything-but-letters-and-numbers.md
index bd1fd72dba6..541817a6632 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-everything-but-letters-and-numbers.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-everything-but-letters-and-numbers.md
@@ -1,6 +1,6 @@
---
id: 587d7db8367417b2b2512ba0
-title: 匹配除了字母和数字的所有符号
+title: Match Everything But Letters and Numbers
challengeType: 1
forumTopicId: 301353
dashedName: match-everything-but-letters-and-numbers
@@ -8,9 +8,9 @@ dashedName: match-everything-but-letters-and-numbers
# --description--
-已经了解到可以使用缩写`\w`来匹配字母和数字`[A-Za-z0-9_]`。不过，有可能想要搜寻的匹配模式是非字母数字字符。
+You've learned that you can use a shortcut to match alphanumerics `[A-Za-z0-9_]` using `\w`. A natural pattern you might want to search for is the opposite of alphanumerics.
-可以使用`\W`搜寻和`\w`相反的匹配模式。注意，相反匹配模式使用大写字母。此缩写与`[^A-Za-z0-9_]`是一样的。
+You can search for the opposite of the `\w` with `\W`. Note, the opposite pattern uses a capital letter. This shortcut is the same as `[^A-Za-z0-9_]`.
```js
let shortHand = /\W/;
@@ -22,17 +22,17 @@ sentence.match(shortHand); // Returns ["!"]
# --instructions--
-使用缩写`\W`来计算不同引号和字符串中非字母数字字符的数量。
+Use the shorthand character class `\W` to count the number of non-alphanumeric characters in various quotes and strings.
# --hints--
-你的正则表达式应该使用全局状态修正符。
+Your regex should use the global flag.
```js
assert(nonAlphabetRegex.global);
```
-你的正则表达式应该在`'The five boxing wizards jump quickly.'`中匹配到 6 个非字母数字字符。
+Your regex should find 6 non-alphanumeric characters in `"The five boxing wizards jump quickly."`.
```js
assert(
@@ -40,13 +40,13 @@ assert(
);
```
-正则表达式应该使用元字符来匹配非字母字符。
+Your regex should use the shorthand character to match characters which are non-alphanumeric.
```js
assert(/\\W/.test(nonAlphabetRegex.source));
```
-你的正则表达式应该在`'Pack my box with five dozen liquor jugs.'`中匹配到 8 个非字母数字字符。
+Your regex should find 8 non-alphanumeric characters in `"Pack my box with five dozen liquor jugs."`
```js
assert(
@@ -54,7 +54,7 @@ assert(
);
```
-你的正则表达式应该在`'How vexingly quick daft zebras jump!'`中匹配到 6 个非字母数字字符。
+Your regex should find 6 non-alphanumeric characters in `"How vexingly quick daft zebras jump!"`
```js
assert(
@@ -62,7 +62,7 @@ assert(
);
```
-你的正则表达式应该在`'123 456 7890 ABC def GHI jkl MNO pqr STU vwx YZ.'`中匹配到 12 个非字母数字字符。
+Your regex should find 12 non-alphanumeric characters in `"123 456 7890 ABC def GHI jkl MNO pqr STU vwx YZ."`
```js
assert(
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-letters-of-the-alphabet.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-letters-of-the-alphabet.md
index 7c6dd790f94..7b48c2b89df 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-letters-of-the-alphabet.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-letters-of-the-alphabet.md
@@ -1,6 +1,6 @@
---
id: 587d7db5367417b2b2512b96
-title: 匹配字母表中的字母
+title: Match Letters of the Alphabet
challengeType: 1
forumTopicId: 301354
dashedName: match-letters-of-the-alphabet
@@ -8,11 +8,11 @@ dashedName: match-letters-of-the-alphabet
# --description--
-了解了如何使用`字符集`来指定要匹配的一组字符串，但是当需要匹配大量字符（例如，字母表中的每个字母）时，有一种写法可以让实现这个功能变得简短。
+You saw how you can use <dfn>character sets</dfn> to specify a group of characters to match, but that's a lot of typing when you need to match a large range of characters (for example, every letter in the alphabet). Fortunately, there is a built-in feature that makes this short and simple.
-在`字符集`中，可以使用`连字符`（`-`）来定义要匹配的字符范围。
+Inside a character set, you can define a range of characters to match using a hyphen character: `-`.
-例如，要匹配小写字母`a`到`e`，你可以使用`[a-e]`。
+For example, to match lowercase letters `a` through `e` you would use `[a-e]`.
```js
let catStr = "cat";
@@ -26,26 +26,25 @@ matStr.match(bgRegex); // Returns null
# --instructions--
-匹配字符串`quoteSample`中的所有字母。
+Match all the letters in the string `quoteSample`.
-**注意：**
-一定要同时匹配大小写**字母**。\*\*\*\*
+**Note**: Be sure to match both uppercase and lowercase letters.
# --hints--
-你的正则表达式`alphabetRegex`应该匹配 35 项。
+Your regex `alphabetRegex` should match 35 items.
```js
assert(result.length == 35);
```
-你的正则表达式`alphabetRegex`应该使用全局标志。
+Your regex `alphabetRegex` should use the global flag.
```js
assert(alphabetRegex.flags.match(/g/).length == 1);
```
-你的正则表达式`alphabetRegex`应该使用忽略大小写标志。
+Your regex `alphabetRegex` should use the case insensitive flag.
```js
assert(alphabetRegex.flags.match(/i/).length == 1);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-literal-strings.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-literal-strings.md
index b7dd2fa1e57..132c48dbe45 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-literal-strings.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-literal-strings.md
@@ -1,6 +1,6 @@
---
id: 587d7db3367417b2b2512b8f
-title: 匹配文字字符串
+title: Match Literal Strings
challengeType: 1
forumTopicId: 301355
dashedName: match-literal-strings
@@ -8,7 +8,7 @@ dashedName: match-literal-strings
# --description--
-在上一个挑战中，使用正则表达式`/Hello/`搜索到了字符串`"Hello"`。那个正则表达式在字符串中搜寻`"Hello"`的文字匹配。下面是另一个在字符串中搜寻`"Kevin"`的示例：
+In the last challenge, you searched for the word `"Hello"` using the regular expression `/Hello/`. That regex searched for a literal match of the string `"Hello"`. Here's another example searching for a literal match of the string `"Kevin"`:
```js
let testStr = "Hello, my name is Kevin.";
@@ -17,7 +17,7 @@ testRegex.test(testStr);
// Returns true
```
-任何其他形式的`"Kevin"`都不会被匹配。例如，正则表达式`/Kevin/`不会匹配`"kevin"`或者`"KEVIN"`。
+Any other forms of `"Kevin"` will not match. For example, the regex `/Kevin/` will not match `"kevin"` or `"KEVIN"`.
```js
let wrongRegex = /kevin/;
@@ -25,27 +25,27 @@ wrongRegex.test(testStr);
// Returns false
```
-后续的挑战将为你展示如何匹配其他形式的字符串。
+A future challenge will show how to match those other forms as well.
# --instructions--
-完成正则表达式`waldoRegex`，在字符串`waldoIsHiding`中匹配到文本`"Waldo"`。
+Complete the regex `waldoRegex` to find `"Waldo"` in the string `waldoIsHiding` with a literal match.
# --hints--
-你的正则表达式`waldoRegex`应该匹配到`'Waldo'`。
+Your regex `waldoRegex` should find `"Waldo"`
```js
assert(waldoRegex.test(waldoIsHiding));
```
-你的正则表达式`waldoRegex`不应该搜寻其他的任何内容。
+Your regex `waldoRegex` should not search for anything else.
```js
assert(!waldoRegex.test('Somewhere is hiding in this text.'));
```
-你应该使用你的正则表达式对字符串执行文字匹配。
+You should perform a literal string match with your regex.
```js
assert(!/\/.*\/i/.test(code));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-non-whitespace-characters.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-non-whitespace-characters.md
index 680f77a68d3..9607d58cc93 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-non-whitespace-characters.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-non-whitespace-characters.md
@@ -1,6 +1,6 @@
---
id: 587d7db9367417b2b2512ba4
-title: 匹配非空白字符
+title: Match Non-Whitespace Characters
challengeType: 1
forumTopicId: 18210
dashedName: match-non-whitespace-characters
@@ -8,9 +8,9 @@ dashedName: match-non-whitespace-characters
# --description--
-已经学会了如何使用带有小写`s`的缩写`\s`来搜寻空白字符。还可以搜寻除了空格之外的所有内容。
+You learned about searching for whitespace using `\s`, with a lowercase `s`. You can also search for everything except whitespace.
-使用`\S`搜寻非空白字符，其中`S`是大写。此匹配模式将不匹配空格、回车符、制表符、换页符和换行符。可以认为这类似于元字符`[^\r\t\f\n\v]`。
+Search for non-whitespace using `\S`, which is an uppercase `s`. This pattern will not match whitespace, carriage return, tab, form feed, and new line characters. You can think of it being similar to the character class `[^ \r\t\f\n\v]`.
```js
let whiteSpace = "Whitespace. Whitespace everywhere!"
@@ -20,23 +20,23 @@ whiteSpace.match(nonSpaceRegex).length; // Returns 32
# --instructions--
-修改正则表达式`countNonWhiteSpace`以查找字符串中的多个非空字符。
+Change the regex `countNonWhiteSpace` to look for multiple non-whitespace characters in a string.
# --hints--
-你的正则表达式应该使用全局状态修正符。
+Your regex should use the global flag.
```js
assert(countNonWhiteSpace.global);
```
-正则表达式应该使用元字符 <code>\\S/code> 来匹配所有的非空格字符。
+Your regex should use the shorthand character `\S` to match all non-whitespace characters.
```js
assert(/\\S/.test(countNonWhiteSpace.source));
```
-你的正则表达式应该在`'Men are from Mars and women are from Venus.'`中匹配到 35 个非空白字符。
+Your regex should find 35 non-spaces in `"Men are from Mars and women are from Venus."`
```js
assert(
@@ -45,13 +45,13 @@ assert(
);
```
-你的正则表达式应该在`"Space: the final frontier."`中匹配到 23 个非空白字符。
+Your regex should find 23 non-spaces in `"Space: the final frontier."`
```js
assert('Space: the final frontier.'.match(countNonWhiteSpace).length == 23);
```
-你的正则表达式应该在`'MindYourPersonalSpace'`中匹配到 21 个非空白字符。
+Your regex should find 21 non-spaces in `"MindYourPersonalSpace"`
```js
assert('MindYourPersonalSpace'.match(countNonWhiteSpace).length == 21);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-numbers-and-letters-of-the-alphabet.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-numbers-and-letters-of-the-alphabet.md
index de5c72f12e7..1702acbad8f 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-numbers-and-letters-of-the-alphabet.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-numbers-and-letters-of-the-alphabet.md
@@ -1,6 +1,6 @@
---
id: 587d7db5367417b2b2512b97
-title: 匹配字母表中的数字和字母
+title: Match Numbers and Letters of the Alphabet
challengeType: 1
forumTopicId: 301356
dashedName: match-numbers-and-letters-of-the-alphabet
@@ -8,11 +8,11 @@ dashedName: match-numbers-and-letters-of-the-alphabet
# --description--
-使用连字符（`-`）匹配字符范围并不仅限于字母。它还可以匹配一系列数字。
+Using the hyphen (`-`) to match a range of characters is not limited to letters. It also works to match a range of numbers.
-例如，`/[0-5]/`匹配`0`和`5`之间的任意数字，包含`0`和`5`。
+For example, `/[0-5]/` matches any number between `0` and `5`, including the `0` and `5`.
-此外，还可以在单个字符集中组合一系列字母和数字。
+Also, it is possible to combine a range of letters and numbers in a single character set.
```js
let jennyStr = "Jenny8675309";
@@ -23,23 +23,23 @@ jennyStr.match(myRegex);
# --instructions--
-创建一个正则表达式，使其可以匹配`h`和`s`之间的一系列字母，以及`2`和`6`之间的一系列数字。请记得在正则表达式中包含恰当的标志。
+Create a single regex that matches a range of letters between `h` and `s`, and a range of numbers between `2` and `6`. Remember to include the appropriate flags in the regex.
# --hints--
-你的正则表达式`myRegex`应该匹配 17 项。
+Your regex `myRegex` should match 17 items.
```js
assert(result.length == 17);
```
-你的正则表达式`myRegex`应该使用全局标志。
+Your regex `myRegex` should use the global flag.
```js
assert(myRegex.flags.match(/g/).length == 1);
```
-你的正则表达式`myRegex`应该使用忽略大小写的标志。
+Your regex `myRegex` should use the case insensitive flag.
```js
assert(myRegex.flags.match(/i/).length == 1);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-single-character-with-multiple-possibilities.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-single-character-with-multiple-possibilities.md
index f93bec3b6fa..097ded3a042 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-single-character-with-multiple-possibilities.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-single-character-with-multiple-possibilities.md
@@ -1,6 +1,6 @@
---
id: 587d7db5367417b2b2512b95
-title: 将单个字符与多种可能性匹配
+title: Match Single Character with Multiple Possibilities
challengeType: 1
forumTopicId: 301357
dashedName: match-single-character-with-multiple-possibilities
@@ -8,11 +8,11 @@ dashedName: match-single-character-with-multiple-possibilities
# --description--
-已经了解了文字匹配模式（`/literal/`）和通配符（`/./`）。这是正则表达式的两种极端情况，一种是精确匹配，而另一种则是匹配所有。在这两种极端情况之间有一个平衡选项。
+You learned how to match literal patterns (`/literal/`) and wildcard character (`/./`). Those are the extremes of regular expressions, where one finds exact matches and the other matches everything. There are options that are a balance between the two extremes.
-可以使用`字符集`搜寻具有一定灵活性的文字匹配模式。可以把字符集放在方括号（`[`和`]`）之间来定义一组需要匹配的字符串。
+You can search for a literal pattern with some flexibility with <dfn>character classes</dfn>. Character classes allow you to define a group of characters you wish to match by placing them inside square (`[` and `]`) brackets.
-例如，如果想要匹配`"bag"`、`"big"`和`"bug"`，但是不想匹配`"bog"`。可以创建正则表达式`/b[aiu]g/`来执行此操作。`[aiu]`是只匹配字符`"a"`、`"i"`或者`"u"`的字符集。
+For example, you want to match `"bag"`, `"big"`, and `"bug"` but not `"bog"`. You can create the regex `/b[aiu]g/` to do this. The `[aiu]` is the character class that will only match the characters `"a"`, `"i"`, or `"u"`.
```js
let bigStr = "big";
@@ -28,38 +28,38 @@ bogStr.match(bgRegex); // Returns null
# --instructions--
-使用元音字符集（`a`、`e`、`i`、`o`、`u`）在正则表达式`vowelRegex`中匹配到字符串`quoteSample`中的所有元音。
+Use a character class with vowels (`a`, `e`, `i`, `o`, `u`) in your regex `vowelRegex` to find all the vowels in the string `quoteSample`.
-**注意**
-一定要同时匹配大小写元音。
+**Note**
+Be sure to match both upper- and lowercase vowels.
# --hints--
-你应该匹配到所有25个元音。
+You should find all 25 vowels.
```js
assert(result.length == 25);
```
-你的正则表达式`vowelRegex`应该使用字符集。
+Your regex `vowelRegex` should use a character class.
```js
assert(/\[.*\]/.test(vowelRegex.source));
```
-你的正则表达式`vowelRegex`应该使用全局标志。
+Your regex `vowelRegex` should use the global flag.
```js
assert(vowelRegex.flags.match(/g/).length == 1);
```
-你的正则表达式`vowelRegex`应该使用忽略大小写标志。
+Your regex `vowelRegex` should use the case insensitive flag.
```js
assert(vowelRegex.flags.match(/i/).length == 1);
```
-你的正则表达式不应该匹配任何辅音。
+Your regex should not match any consonants.
```js
assert(!/[b-df-hj-np-tv-z]/gi.test(result.join()));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-single-characters-not-specified.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-single-characters-not-specified.md
index bb16f286914..a0f6672834e 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-single-characters-not-specified.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-single-characters-not-specified.md
@@ -1,6 +1,6 @@
---
id: 587d7db6367417b2b2512b98
-title: 匹配单个未指定的字符
+title: Match Single Characters Not Specified
challengeType: 1
forumTopicId: 301358
dashedName: match-single-characters-not-specified
@@ -8,31 +8,31 @@ dashedName: match-single-characters-not-specified
# --description--
-到目前为止，已经创建了一个想要匹配的字符集合，但也可以创建一个不想匹配的字符集合。这些类型的字符集称为`否定字符集`。
+So far, you have created a set of characters that you want to match, but you could also create a set of characters that you do not want to match. These types of character sets are called <dfn>negated character sets</dfn>.
-要创建`否定字符集`，需要在开始括号后面和不想匹配的字符前面放置`插入字符`（即`^`）。
+To create a negated character set, you place a caret character (`^`) after the opening bracket and before the characters you do not want to match.
-例如，`/[^aeiou]/gi`匹配所有非元音字符。注意，字符`.`、`!`、`[`、`@`、`/`和空白字符等也会被匹配，该否定字符集仅排除元音字符。
+For example, `/[^aeiou]/gi` matches all characters that are not a vowel. Note that characters like `.`, `!`, `[`, `@`, `/` and white space are matched - the negated vowel character set only excludes the vowel characters.
# --instructions--
-创建一个匹配所有非数字或元音字符的正则表达式。请记得在正则表达式中包含恰当的标志。
+Create a single regex that matches all characters that are not a number or a vowel. Remember to include the appropriate flags in the regex.
# --hints--
-你的正则表达式`myRegex`应该匹配 9 项。
+Your regex `myRegex` should match 9 items.
```js
assert(result.length == 9);
```
-你的正则表达式`myRegex`应该使用全局标志。
+Your regex `myRegex` should use the global flag.
```js
assert(myRegex.flags.match(/g/).length == 1);
```
-你的正则表达式`myRegex`应该使用忽略大小写标志。
+Your regex `myRegex` should use the case insensitive flag.
```js
assert(myRegex.flags.match(/i/).length == 1);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-whitespace.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-whitespace.md
index e9509f25e35..b59d501509c 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-whitespace.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/match-whitespace.md
@@ -1,6 +1,6 @@
---
id: 587d7db8367417b2b2512ba3
-title: 匹配空白字符
+title: Match Whitespace
challengeType: 1
forumTopicId: 301359
dashedName: match-whitespace
@@ -8,9 +8,9 @@ dashedName: match-whitespace
# --description--
-迄今为止的挑战包括匹配的字母和数字。还可以匹配字母之间的空格。
+The challenges so far have covered matching letters of the alphabet and numbers. You can also match the whitespace or spaces between letters.
-可以使用`\s`搜寻空格，其中`s`是小写。此匹配模式不仅匹配空格，还匹配回车符、制表符、换页符和换行符，可以将其视为与`[\r\t\f\n\v]`类似。
+You can search for whitespace using `\s`, which is a lowercase `s`. This pattern not only matches whitespace, but also carriage return, tab, form feed, and new line characters. You can think of it as similar to the character class `[ \r\t\f\n\v]`.
```js
let whiteSpace = "Whitespace. Whitespace everywhere!"
@@ -21,23 +21,23 @@ whiteSpace.match(spaceRegex);
# --instructions--
-修改正则表达式`countWhiteSpace`查找字符串中的多个空白字符。
+Change the regex `countWhiteSpace` to look for multiple whitespace characters in a string.
# --hints--
-你的正则表达式应该使用全局状态修正符。
+Your regex should use the global flag.
```js
assert(countWhiteSpace.global);
```
-正则表达式应该使用元字符 `\s` 匹配所有的空白。
+Your regex should use the shorthand character `\s` to match all whitespace characters.
```js
assert(/\\s/.test(countWhiteSpace.source));
```
-你的正则表达式应该在`'Men are from Mars and women are from Venus.'`中匹配到 8 个空白字符。
+Your regex should find eight spaces in `"Men are from Mars and women are from Venus."`
```js
assert(
@@ -46,13 +46,13 @@ assert(
);
```
-你的正则表达式应该在`"Space: the final frontier."`中匹配到 3 个空白字符。
+Your regex should find three spaces in `"Space: the final frontier."`
```js
assert('Space: the final frontier.'.match(countWhiteSpace).length == 3);
```
-你的正则表达式在`'MindYourPersonalSpace'`中应该匹配不到空白字符。
+Your regex should find no spaces in `"MindYourPersonalSpace"`
```js
assert('MindYourPersonalSpace'.match(countWhiteSpace) == null);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/positive-and-negative-lookahead.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/positive-and-negative-lookahead.md
index 04cdcb33ae0..e7b48147260 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/positive-and-negative-lookahead.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/positive-and-negative-lookahead.md
@@ -1,6 +1,6 @@
---
id: 587d7dba367417b2b2512ba9
-title: 正向先行断言和负向先行断言
+title: Positive and Negative Lookahead
challengeType: 1
forumTopicId: 301360
dashedName: positive-and-negative-lookahead
@@ -8,15 +8,15 @@ dashedName: positive-and-negative-lookahead
# --description--
-`先行断言`是告诉 JavaScript 在字符串中向前查找的匹配模式。当想要在同一个字符串上搜寻多个匹配模式时，这可能非常有用。
+<dfn>Lookaheads</dfn> are patterns that tell JavaScript to look-ahead in your string to check for patterns further along. This can be useful when you want to search for multiple patterns over the same string.
-有两种`先行断言`：`正向先行断言`和`负向先行断言`。
+There are two kinds of lookaheads: <dfn>positive lookahead</dfn> and <dfn>negative lookahead</dfn>.
-`正向先行断言`会查看并确保搜索匹配模式中的元素存在，但实际上并不匹配。正向先行断言的用法是`(?=...)`，其中`...`就是需要存在但不会被匹配的部分。
+A positive lookahead will look to make sure the element in the search pattern is there, but won't actually match it. A positive lookahead is used as `(?=...)` where the `...` is the required part that is not matched.
-另一方面，`负向先行断言`会查看并确保搜索匹配模式中的元素不存在。负向先行断言的用法是`(?!...)`，其中`...`是希望不存在的匹配模式。如果负向先行断言部分不存在，将返回匹配模式的其余部分。
+On the other hand, a negative lookahead will look to make sure the element in the search pattern is not there. A negative lookahead is used as `(?!...)` where the `...` is the pattern that you do not want to be there. The rest of the pattern is returned if the negative lookahead part is not present.
-尽管先行断言有点儿令人困惑，但是这些示例会有所帮助。
+Lookaheads are a bit confusing but some examples will help.
```js
let quit = "qu";
@@ -27,7 +27,7 @@ quit.match(quRegex); // Returns ["q"]
noquit.match(qRegex); // Returns ["q"]
```
-`先行断言`的更实际用途是检查一个字符串中的两个或更多匹配模式。这里有一个简单的密码检查器，密码规则是 3 到 6 个字符且至少包含一个数字：
+A more practical use of lookaheads is to check two or more patterns in one string. Here is a (naively) simple password checker that looks for between 3 and 6 characters and at least one number:
```js
let password = "abc123";
@@ -37,68 +37,62 @@ checkPass.test(password); // Returns true
# --instructions--
-在正则表达式`pwRegex`中使用`先行断言`以匹配大于5个字符且有两个连续数字的密码，并且不能以数字开头。
+Use lookaheads in the `pwRegex` to match passwords that are greater than 5 characters long, and have two consecutive digits.
# --hints--
-你的正则表达式应该使用两个正向`先行断言`。
+Your regex should use two positive `lookaheads`.
```js
assert(pwRegex.source.match(/\(\?=.*?\)\(\?=.*?\)/) !== null);
```
-你的正则表达式不应该匹配`'astronaut'`。
+Your regex should not match `"astronaut"`
```js
assert(!pwRegex.test('astronaut'));
```
-你的正则表达式不应该匹配`'airplanes'`。
-
-```js
-assert(!pwRegex.test('airplanes'));
-```
-
-正则不应该匹配 `"banan1"`
+Your regex should not match `"banan1"`
```js
assert(!pwRegex.test('banan1'));
```
-你的正则表达式应该匹配`'bana12'`。
+Your regex should match `"bana12"`
```js
assert(pwRegex.test('bana12'));
```
-你的正则表达式应该匹配`'abc123'`。
+Your regex should match `"abc123"`
```js
assert(pwRegex.test('abc123'));
```
-你的正则表达式不应该匹配`'123'`。
+Your regex should not match `"12345"`
```js
-assert(!pwRegex.test('123'));
+assert(!pwRegex.test('12345'));
```
-你的正则表达式不应该匹配`'1234'`。
+Your regex should match `"8pass99"`
```js
-assert(!pwRegex.test('1234'));
+assert(pwRegex.test('8pass99'));
```
-正则不应该匹配 `"8pass99"`
+Your regex should not match `"1a2bcde"`
```js
-assert(!pwRegex.test('8pass99'));
+assert(!pwRegex.test('1a2bcde'));
```
-正则不应该匹配 `"12abcde"`
+Your regex should match `"astr1on11aut"`
```js
-assert(!pwRegex.test('12abcde'));
+assert(pwRegex.test('astr1on11aut'));
```
# --seed--
@@ -114,5 +108,5 @@ let result = pwRegex.test(sampleWord);
# --solutions--
```js
-var pwRegex = /^\D(?=\w{5})(?=\w*\d{2})/;
+let pwRegex = /(?=\w{6})(?=\w*\d{2})/;
```
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/remove-whitespace-from-start-and-end.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/remove-whitespace-from-start-and-end.md
index 5b4fe0c53c1..63496d77334 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/remove-whitespace-from-start-and-end.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/remove-whitespace-from-start-and-end.md
@@ -1,6 +1,6 @@
---
id: 587d7dbb367417b2b2512bac
-title: 删除开头和结尾的空白
+title: Remove Whitespace from Start and End
challengeType: 1
forumTopicId: 301362
dashedName: remove-whitespace-from-start-and-end
@@ -8,30 +8,29 @@ dashedName: remove-whitespace-from-start-and-end
# --description--
-有时字符串周围存在的空白字符并不是必需的。字符串的典型处理是删除字符串开头和结尾处的空格。
+Sometimes whitespace characters around strings are not wanted but are there. Typical processing of strings is to remove the whitespace at the start and end of it.
# --instructions--
-编写一个正则表达式并使用适当的字符串方法删除字符串开头和结尾的空格。
+Write a regex and use the appropriate string methods to remove whitespace at the beginning and end of strings.
-**注意：**
-`.trim()`方法在这里也可以实现同样的效果，但是你需要使用正则表达式来完成此项挑战。
+**Note:** The `String.prototype.trim()` method would work here, but you'll need to complete this challenge using regular expressions.
# --hints--
-`结果`应该等于`'Hello, World!'`。
+`result` should equal to `"Hello, World!"`
```js
assert(result == 'Hello, World!');
```
-你不应该使用`.trim()`方法。
+Your solution should not use the `String.prototype.trim()` method.
```js
-assert(!code.match(/\.trim\(.*?\)/));
+assert(!code.match(/\.?[\s\S]*?trim/));
```
-`结果`变量不应该设置为等于字符串。
+The `result` variable should not be set equal to a string.
```js
assert(!code.match(/result\s*=\s*".*?"/));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/restrict-possible-usernames.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/restrict-possible-usernames.md
index 415f06e92c3..59da9fb3608 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/restrict-possible-usernames.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/restrict-possible-usernames.md
@@ -1,6 +1,6 @@
---
id: 587d7db8367417b2b2512ba2
-title: 限制可能的用户名
+title: Restrict Possible Usernames
challengeType: 1
forumTopicId: 301363
dashedName: restrict-possible-usernames
@@ -8,90 +8,96 @@ dashedName: restrict-possible-usernames
# --description--
-用户名在互联网上随处可见。它们是用户在自己喜欢的网站上的唯一身份。
+Usernames are used everywhere on the internet. They are what give users a unique identity on their favorite sites.
-需要检索数据库中的所有用户名。以下是用户在创建用户名时必须遵守的一些简单规则。
+You need to check all the usernames in a database. Here are some simple rules that users have to follow when creating their username.
-1) 用户名只能是数字字母字符。
+1) Usernames can only use alpha-numeric characters.
-2) 用户名中的数字必须在最后，且数字可以有零个或多个。
+2) The only numbers in the username have to be at the end. There can be zero or more of them at the end. Username cannot start with the number.
-3) 用户名字母可以是小写字母和大写字母。
+3) Username letters can be lowercase and uppercase.
-4) 用户名长度必须至少为两个字符。两位用户名只能使用字母。
+4) Usernames have to be at least two characters long. A two-character username can only use alphabet letters as characters.
# --instructions--
-修改正则表达式`userCheck`以适合上面列出的约束。
+Change the regex `userCheck` to fit the constraints listed above.
# --hints--
-你的正则表达式应该匹配`JACK`。
+Your regex should match `JACK`
```js
assert(userCheck.test('JACK'));
```
-你的正则表达式不应该匹配`J`。
+Your regex should not match `J`
```js
assert(!userCheck.test('J'));
```
-正则表达式应该匹配 `Jo`。
+Your regex should match `Jo`
```js
assert(userCheck.test('Jo'));
```
-你的正则表达式应该匹配`Oceans11`。
+Your regex should match `Oceans11`
```js
assert(userCheck.test('Oceans11'));
```
-你的正则表达式应该匹配`RegexGuru`。
+Your regex should match `RegexGuru`
```js
assert(userCheck.test('RegexGuru'));
```
-你的正则表达式不应该匹配`007`。
+Your regex should not match `007`
```js
assert(!userCheck.test('007'));
```
-你的正则表达式不应该匹配`9`。
+Your regex should not match `9`
```js
assert(!userCheck.test('9'));
```
-正则表达式不应该匹配 `A1`。
+Your regex should not match `A1`
```js
assert(!userCheck.test('A1'));
```
-正则表达式不应该匹配 `BadUs3rnam3`。
+Your regex should not match `BadUs3rnam3`
```js
assert(!userCheck.test('BadUs3rnam3'));
```
-正则表达式应该匹配 `Z97`。
+Your regex should match `Z97`
```js
assert(userCheck.test('Z97'));
```
-正则表达式不应该匹配 `c57bT3`。
+Your regex should not match `c57bT3`
```js
assert(!userCheck.test('c57bT3'));
```
+Your regex should match `AB1`
+
+```js
+assert(userCheck.test('AB1'));
+```
+
# --seed--
## --seed-contents--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/reuse-patterns-using-capture-groups.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/reuse-patterns-using-capture-groups.md
index ae630a2b997..0c639d63a26 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/reuse-patterns-using-capture-groups.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/reuse-patterns-using-capture-groups.md
@@ -27,7 +27,7 @@ repeatStr.match(repeatRegex); // Returns ["regex regex", "regex"]
# --instructions--
-在正则表达式 `reRegex` 中使用捕获组，以匹配在字符串中被空格隔开的三个相同数字。
+Use capture groups in `reRegex` to match a string that consists of only the same number repeated exactly three times separated by single spaces.
# --hints--
@@ -43,52 +43,43 @@ assert(reRegex.source.match(/\\d/));
assert(reRegex.source.match(/\\1|\\2/g).length >= 2);
```
-你的正则表达式应该有两个空格分隔三个数字。
-
-```js
-assert(
- reRegex.source.match(/ |\\s/g).length === 2 ||
- reRegex.source.match(/\(\\s\)(?=.*\\(1|2))/g)
-);
-```
-
-你的正则表达式应该匹配 `"42 42 42"`。
+Your regex should match `"42 42 42"`.
```js
assert(reRegex.test('42 42 42'));
```
-你的正则表达式应该匹配 `"100 100 100"`。
+Your regex should match `"100 100 100"`.
```js
assert(reRegex.test('100 100 100'));
```
-你的正则表达式不应该匹配 `"42 42 42 42"`。
+Your regex should not match `"42 42 42 42"`.
```js
assert.equal('42 42 42 42'.match(reRegex.source), null);
```
-你的正则表达式不应该匹配 `"42 42"`。
+Your regex should not match `"42 42"`.
```js
assert.equal('42 42'.match(reRegex.source), null);
```
-你的正则表达式不应该匹配 `"101 102 103"`。
+Your regex should not match `"101 102 103"`.
```js
assert(!reRegex.test('101 102 103'));
```
-你的正则表达式不应该匹配 `"1 2 3"`。
+Your regex should not match `"1 2 3"`.
```js
assert(!reRegex.test('1 2 3'));
```
-你的正则表达式应该匹配 `"10 10 10"`。
+Your regex should match `"10 10 10"`.
```js
assert(reRegex.test('10 10 10'));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/specify-exact-number-of-matches.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/specify-exact-number-of-matches.md
index 3f195fe2138..8f7a68cc7b5 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/specify-exact-number-of-matches.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/specify-exact-number-of-matches.md
@@ -1,6 +1,6 @@
---
id: 587d7db9367417b2b2512ba7
-title: 指定匹配的确切数量
+title: Specify Exact Number of Matches
challengeType: 1
forumTopicId: 301365
dashedName: specify-exact-number-of-matches
@@ -8,11 +8,11 @@ dashedName: specify-exact-number-of-matches
# --description--
-可以使用带有花括号的`数量说明符`来指定匹配模式的上下限。但有时只需要特定数量的匹配。
+You can specify the lower and upper number of patterns with quantity specifiers using curly brackets. Sometimes you only want a specific number of matches.
-要指定一定数量的匹配模式，只需在大括号之间放置一个数字。
+To specify a certain number of patterns, just have that one number between the curly brackets.
-例如，要只匹配字母`a`出现`3`次的单词`"hah"`，正则表达式应为`/ha{3}h/`。
+For example, to match only the word `"hah"` with the letter `a` `3` times, your regex would be `/ha{3}h/`.
```js
let A4 = "haaaah";
@@ -26,43 +26,48 @@ multipleHA.test(A100); // Returns false
# --instructions--
-修改正则表达式`timRegex`，以匹配仅有四个字母单词`m`的单词`"Timber"`。
+Change the regex `timRegex` to match the word `"Timber"` only when it has four letter `m`'s.
# --hints--
-你的正则表达式应该使用花括号。
+Your regex should use curly brackets.
```js
assert(timRegex.source.match(/{.*?}/).length > 0);
```
-你的正则表达式不应该匹配`'Timber'`。
+Your regex should not match `"Timber"`
```js
+timRegex.lastIndex = 0;
assert(!timRegex.test('Timber'));
```
-你的正则表达式不应该匹配`'Timmber'`。
+Your regex should not match `"Timmber"`
```js
+timRegex.lastIndex = 0;
assert(!timRegex.test('Timmber'));
```
-你的正则表达式不应该匹配`'Timmmber'`。
+Your regex should not match `"Timmmber"`
```js
+timRegex.lastIndex = 0;
assert(!timRegex.test('Timmmber'));
```
-你的正则表达式应该匹配`'Timmmmber'`。
+Your regex should match `"Timmmmber"`
```js
+timRegex.lastIndex = 0;
assert(timRegex.test('Timmmmber'));
```
-你的正则表达式不应该匹配包含 30 个字母`m`的`'Timber'`。
+Your regex should not match `"Timber"` with 30 `m`'s in it.
```js
+timRegex.lastIndex = 0;
assert(!timRegex.test('Ti' + 'm'.repeat(30) + 'ber'));
```
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/specify-only-the-lower-number-of-matches.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/specify-only-the-lower-number-of-matches.md
index 271cc82b8df..f75994614f2 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/specify-only-the-lower-number-of-matches.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/specify-only-the-lower-number-of-matches.md
@@ -1,6 +1,6 @@
---
id: 587d7db9367417b2b2512ba6
-title: 只指定匹配的下限
+title: Specify Only the Lower Number of Matches
challengeType: 1
forumTopicId: 301366
dashedName: specify-only-the-lower-number-of-matches
@@ -8,11 +8,11 @@ dashedName: specify-only-the-lower-number-of-matches
# --description--
-可以使用带有花括号的`数量说明符`来指定匹配模式的上下限。但有时候只想指定匹配模式的下限而不需要指定上限。
+You can specify the lower and upper number of patterns with quantity specifiers using curly brackets. Sometimes you only want to specify the lower number of patterns with no upper limit.
-为此，在第一个数字后面跟一个逗号即可。
+To only specify the lower number of patterns, keep the first number followed by a comma.
-例如，要匹配至少出现`3`次字母`a`的字符串`"hah"`，正则表达式应该是`/ha{3,}h/`。
+For example, to match only the string `"hah"` with the letter `a` appearing at least `3` times, your regex would be `/ha{3,}h/`.
```js
let A4 = "haaaah";
@@ -26,47 +26,47 @@ multipleA.test(A100); // Returns true
# --instructions--
-修改正则表达式`haRegex`，匹配包含四个或更多字母`z`的单词`"Hazzah"`。
+Change the regex `haRegex` to match the word `"Hazzah"` only when it has four or more letter `z`'s.
# --hints--
-你的正则表达式应该使用花括号。
+Your regex should use curly brackets.
```js
assert(haRegex.source.match(/{.*?}/).length > 0);
```
-你的正则表达式不应该匹配`'Hazzah'`。
+Your regex should not match `"Hazzah"`
```js
assert(!haRegex.test('Hazzah'));
```
-你的正则表达式不应该匹配`'Hazzzah'`。
+Your regex should not match `"Hazzzah"`
```js
assert(!haRegex.test('Hazzzah'));
```
-正则表达式应该匹配 `"Hazzzzah"`
+Your regex should match `"Hazzzzah"`
```js
assert('Hazzzzah'.match(haRegex)[0].length === 8);
```
-你的正则表达式应该匹配`'Hazzzzah'`。
+Your regex should match `"Hazzzzzah"`
```js
assert('Hazzzzzah'.match(haRegex)[0].length === 9);
```
-正则表达式应该匹配 `"Hazzzzzzah"`
+Your regex should match `"Hazzzzzzah"`
```js
assert('Hazzzzzzah'.match(haRegex)[0].length === 10);
```
-正则表达式应该匹配 `"Hazzah"` with 30 `z`'s in it.
+Your regex should match `"Hazzah"` with 30 `z`'s in it.
```js
assert('Hazzzzzzzzzzzzzzzzzzzzzzzzzzzzzzah'.match(haRegex)[0].length === 34);
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/specify-upper-and-lower-number-of-matches.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/specify-upper-and-lower-number-of-matches.md
index eaf5c083d9c..83ed14cecce 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/specify-upper-and-lower-number-of-matches.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/specify-upper-and-lower-number-of-matches.md
@@ -1,6 +1,6 @@
---
id: 587d7db9367417b2b2512ba5
-title: 指定匹配的上限和下限
+title: Specify Upper and Lower Number of Matches
challengeType: 1
forumTopicId: 301367
dashedName: specify-upper-and-lower-number-of-matches
@@ -8,11 +8,11 @@ dashedName: specify-upper-and-lower-number-of-matches
# --description--
-回想一下，使用加号`+`查找一个或多个字符，使用星号`*`查找零个或多个字符。这些都很方便，但有时需要匹配一定范围的匹配模式。
+Recall that you use the plus sign `+` to look for one or more characters and the asterisk `*` to look for zero or more characters. These are convenient but sometimes you want to match a certain range of patterns.
-可以使用`数量说明符`指定匹配模式的上下限。数量说明符与花括号（`{`和`}`）一起使用。可以在花括号之间放两个数字，这两个数字代表匹配模式的上限和下限。
+You can specify the lower and upper number of patterns with <dfn>quantity specifiers</dfn>. Quantity specifiers are used with curly brackets (`{` and `}`). You put two numbers between the curly brackets - for the lower and upper number of patterns.
-例如，要在字符串`"ah"`中匹配仅出现`3`到`5`次的字母`a`，正则表达式应为`/a{3,5}h/`。
+For example, to match only the letter `a` appearing between `3` and `5` times in the string `"ah"`, your regex would be `/a{3,5}h/`.
```js
let A4 = "aaaah";
@@ -24,47 +24,47 @@ multipleA.test(A2); // Returns false
# --instructions--
-修改正则表达式`ohRegex`以匹配在`"Oh no"`中仅出现`3`到`6`次的字母`h`。
+Change the regex `ohRegex` to match the entire phrase `"Oh no"` only when it has `3` to `6` letter `h`'s.
# --hints--
-你的正则表达式应该使用花括号。
+Your regex should use curly brackets.
```js
assert(ohRegex.source.match(/{.*?}/).length > 0);
```
-你的正则表达式不应该匹配`'Ohh no'`。
+Your regex should not match `"Ohh no"`
```js
assert(!ohRegex.test('Ohh no'));
```
-你的正则表达式应该匹配`'Ohhh no'`。
+Your regex should match `"Ohhh no"`
```js
assert('Ohhh no'.match(ohRegex)[0].length === 7);
```
-正则表达式应该匹配 `"Ohhhh no"`。
+Your regex should match `"Ohhhh no"`
```js
assert('Ohhhh no'.match(ohRegex)[0].length === 8);
```
-你的正则表达式应该匹配`'Ohhhhh no'`。
+Your regex should match `"Ohhhhh no"`
```js
assert('Ohhhhh no'.match(ohRegex)[0].length === 9);
```
-你的正则表达式应该匹配`'Ohhhhhh no'`。
+Your regex should match `"Ohhhhhh no"`
```js
assert('Ohhhhhh no'.match(ohRegex)[0].length === 10);
```
-你的正则表达式不应该匹配`'Ohhhhhhh no'`。
+Your regex should not match `"Ohhhhhhh no"`
```js
assert(!ohRegex.test('Ohhhhhhh no'));
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/use-capture-groups-to-search-and-replace.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/use-capture-groups-to-search-and-replace.md
index dd4c9705651..d7d2771dae7 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/use-capture-groups-to-search-and-replace.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/use-capture-groups-to-search-and-replace.md
@@ -1,6 +1,6 @@
---
id: 587d7dbb367417b2b2512bab
-title: 使用捕获组搜索和替换
+title: Use Capture Groups to Search and Replace
challengeType: 1
forumTopicId: 301368
dashedName: use-capture-groups-to-search-and-replace
@@ -8,9 +8,9 @@ dashedName: use-capture-groups-to-search-and-replace
# --description--
-搜索功能是很有用的。但是，当搜索同时也执行更改（或替换）匹配文本的操作时，搜索功能就会显得更加强大。
+Searching is useful. However, you can make searching even more powerful when it also changes (or replaces) the text you match.
-可以使用字符串上`.replace()`方法来搜索并替换字符串中的文本。`.replace()`的输入首先是想要搜索的正则表达式匹配模式，第二个参数是用于替换匹配的字符串或用于执行某些操作的函数。
+You can search and replace text in a string using `.replace()` on a string. The inputs for `.replace()` is first the regex pattern you want to search for. The second parameter is the string to replace the match or a function to do something.
```js
let wrongText = "The sky is silver.";
@@ -19,7 +19,7 @@ wrongText.replace(silverRegex, "blue");
// Returns "The sky is blue."
```
-你还可以使用美元符号（`$`）访问替换字符串中的捕获组。
+You can also access capture groups in the replacement string with dollar signs (`$`).
```js
"Code Camp".replace(/(\w+)\s(\w+)/, '$2 $1');
@@ -28,28 +28,41 @@ wrongText.replace(silverRegex, "blue");
# --instructions--
-编写一个正则表达式，以搜索字符串`"good"`。然后更新变量`replaceText`，用字符串`"okey-dokey"`替换`"good"`。
+Write a regex `fixRegex` using three capture groups that will search for each word in the string "one two three". Then update the `replaceText` variable to replace "one two three" with the string "three two one" and assign the result to the `result` variable. Make sure you are utilizing capture groups in the replacement string using the dollar sign (`$`) syntax.
# --hints--
-你应该使用`.replace()`搜索并替换。
+You should use `.replace()` to search and replace.
```js
assert(code.match(/\.replace\(.*\)/));
```
-你的正则表达式应该把`'This sandwich is good.'`变成`'This sandwich is okey-dokey.'`。
+Your regex should change `"one two three"` to `"three two one"`
```js
-assert(
- result == 'This sandwich is okey-dokey.' && replaceText === 'okey-dokey'
-);
+assert(result === 'three two one');
```
-你不应该改变最后一行。
+You should not change the last line.
```js
-assert(code.match(/result\s*=\s*huhText\.replace\(.*?\)/));
+assert(code.match(/result\s*=\s*str\.replace\(.*?\)/));
+```
+
+`fixRegex` should use at least three capture groups.
+
+```js
+assert(new RegExp(fixRegex.source + '|').exec('').length - 1 >= 3);
+```
+
+`replaceText` should use parenthesized submatch string(s) (i.e. the nth parenthesized submatch string, $n, corresponds to the nth capture group).
+
+```js
+{
+ const re = /(\$\d{1,2})+(?:[\D]|\b)/g;
+ assert(replaceText.match(re).length >= 3);
+}
```
# --seed--
diff --git a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/using-the-test-method.md b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/using-the-test-method.md
index f823f6522a2..e39467f661d 100644
--- a/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/using-the-test-method.md
+++ b/curriculum/challenges/chinese/02-javascript-algorithms-and-data-structures/regular-expressions/using-the-test-method.md
@@ -1,6 +1,6 @@
---
id: 587d7db3367417b2b2512b8e
-title: 使用测试方法
+title: Using the Test Method
challengeType: 1
forumTopicId: 301369
dashedName: using-the-test-method
@@ -8,11 +8,11 @@ dashedName: using-the-test-method
# --description--
-在编程语言中，正则表达式用于匹配指定的字符串。通过正则表达式创建匹配模式（规则）可以帮你完成指定匹配。
+Regular expressions are used in programming languages to match parts of strings. You create patterns to help you do that matching.
-如果想要在字符串`"The dog chased the cat"`中匹配到`"the"`这个单词，可以使用如下正则表达式：`/the/`。注意，正则表达式中不需要引号。
+If you want to find the word `"the"` in the string `"The dog chased the cat"`, you could use the following regular expression: `/the/`. Notice that quote marks are not required within the regular expression.
-JavaScript 中有多种使用正则表达式的方法。测试正则表达式的一种方法是使用`.test()`方法。`.test()`方法会把编写的正则表达式和字符串（即括号内的内容）匹配，如果成功匹配到字符，则返回`true`，反之，返回`false`。
+JavaScript has multiple ways to use regexes. One way to test a regex is using the `.test()` method. The `.test()` method takes the regex, applies it to a string (which is placed inside the parentheses), and returns `true` or `false` if your pattern finds something or not.
```js
let testStr = "freeCodeCamp";
@@ -23,17 +23,17 @@ testRegex.test(testStr);
# --instructions--
-使用`.test()`方法，检测字符串`myString`是否符合正则表达式`myRegex`定义的规则。
+Apply the regex `myRegex` on the string `myString` using the `.test()` method.
# --hints--
-你应该使用`.test()`方法来检测正则表达式。
+You should use `.test()` to test the regex.
```js
assert(code.match(/myRegex.test\(\s*myString\s*\)/));
```
-你的返回结果应该为`true`。
+Your result should return `true`.
```js
assert(result === true);
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/add-elements-within-your-bootstrap-wells.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/add-elements-within-your-bootstrap-wells.md
index 69ce23cf4ea..2326bddd3cc 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/add-elements-within-your-bootstrap-wells.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/add-elements-within-your-bootstrap-wells.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aec908849
-title: 在 Bootstrap Wells 中添加元素
+title: Add Elements within Your Bootstrap Wells
challengeType: 0
forumTopicId: 16636
dashedName: add-elements-within-your-bootstrap-wells
@@ -8,13 +8,13 @@ dashedName: add-elements-within-your-bootstrap-wells
# --description--
-现在我们的每一行的每一列都已经有了 `div` 元素。这已经足够了，现在让我们添加 `button` 元素吧。
+Now we're several `div` elements deep on each column of our row. This is as deep as we'll need to go. Now we can add our `button` elements.
-在每一个 `well` `div` 元素下放置三个 `button` 元素。
+Nest three `button` elements within each of your `div` elements having the class name `well`.
# --hints--
-在每一个 class 属性为 `well` 的 `div` 元素内分别放置三个 `button` 元素。
+Three `button` elements should be nested within each of your `div` elements with class `well`.
```js
assert(
@@ -23,13 +23,13 @@ assert(
);
```
-总共有 6 个 `button` 元素。
+You should have a total of 6 `button` elements.
```js
assert($('button') && $('button').length > 5);
```
-确保 `button` 元素都有一个闭合标签。
+All of your `button` elements should have closing tags.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/add-font-awesome-icons-to-all-of-our-buttons.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/add-font-awesome-icons-to-all-of-our-buttons.md
index 1357a9b0f24..1cfb689b06d 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/add-font-awesome-icons-to-all-of-our-buttons.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/add-font-awesome-icons-to-all-of-our-buttons.md
@@ -1,26 +1,28 @@
---
id: bad87fee1348bd9aedc08845
-title: 将字体图标添加到我们所有的按钮上
+title: Add Font Awesome Icons to all of our Buttons
challengeType: 0
+forumTopicId: 16637
required:
- - link: 'https://use.fontawesome.com/releases/v5.8.1/css/all.css'
+ -
+ link: 'https://use.fontawesome.com/releases/v5.8.1/css/all.css'
raw: true
dashedName: add-font-awesome-icons-to-all-of-our-buttons
---
# --description--
-Font Awesome 是一个非常便利的图标库。这些图标都是被保存在 `.svg` 的文件格式中的矢量图。这些图标就和字体一样，不仅能通过像素单位指定他们的大小，它们也同样会继承父级 HTML 元素的字体大小。
+Font Awesome is a convenient library of icons. These icons can be web fonts or vector graphics. These icons are treated just like fonts. You can specify their size using pixels, and they will assume the font size of their parent HTML elements.
# --instructions--
-使用 Font Awesome 分别为你的 info 按钮添加 `info-circle` 图标，为你的 delete 按钮添加 `trash` 图标。
+Use Font Awesome to add an `info-circle` icon to your info button and a `trash` icon to your delete button.
-**记住：** 下述要求中的 `i` 元素可以用 `span` 元素代替。
+**Note:** The `span` element is an acceptable alternative to the `i` element for the directions below.
# --hints--
-添加一个 `<i class='fas fa-info-circle'></i>` 图标到你的 info 按钮元素中。
+You should add a `<i class="fas fa-info-circle"></i>` within your info button element.
```js
assert(
@@ -29,7 +31,7 @@ assert(
);
```
-添加一个 `<i class='fas fa-trash'></i>` 图标到你的 delete 按钮元素中。
+You should add a `<i class="fas fa-trash"></i>` within your delete button element.
```js
assert(
@@ -38,7 +40,7 @@ assert(
);
```
-确保每一个 `i` 元素都有一个闭合标签并且 `<i class='fa fa-thumbs-up'></i>` 在 like 按钮元素中。
+Each of your `i` elements should have a closing tag and `<i class="fas fa-thumbs-up"></i>` is in your like button element.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/add-font-awesome-icons-to-our-buttons.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/add-font-awesome-icons-to-our-buttons.md
index 113915b79ec..4cec89bf4c4 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/add-font-awesome-icons-to-our-buttons.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/add-font-awesome-icons-to-our-buttons.md
@@ -1,43 +1,44 @@
---
id: bad87fee1348bd9aedd08845
-title: 将字体图标添加到我们的按钮中
+title: Add Font Awesome Icons to our Buttons
challengeType: 0
forumTopicId: 16638
required:
- - link: 'https://use.fontawesome.com/releases/v5.8.1/css/all.css'
+ -
+ link: 'https://use.fontawesome.com/releases/v5.8.1/css/all.css'
raw: true
dashedName: add-font-awesome-icons-to-our-buttons
---
# --description--
-Font Awesome 是一个非常便利的图标库。这些图标都是被保存在 `.svg` 的文件格式中的矢量图。这些图标就和字体一样，不仅能通过像素单位指定他们的大小，它们也同样会继承父级 HTML 元素的字体大小。
+Font Awesome is a convenient library of icons. These icons can be webfonts or vector graphics. These icons are treated just like fonts. You can specify their size using pixels, and they will assume the font size of their parent HTML elements.
-你可以将 Font Awesome 图标库添加至任何一个 app 中，方法很简单，只需要在你的 HTML 头部增加下列代码即可：
+You can include Font Awesome in any app by adding the following code to the top of your HTML:
`<link rel="stylesheet" href="https://use.fontawesome.com/releases/v5.8.1/css/all.css" integrity="sha384-50oBUHEmvpQ+1lW4y57PTFmhCaXp0ML5d60M1M7uH2+nqUivzIebhndOJK28anvf" crossorigin="anonymous">`
-不过在这里，我们已经预先为此页面添加了上述代码。
+In this case, we've already added it for you to this page behind the scenes.
-`i` 元素起初用于让其它元素具有斜体（italic）的效果，不过现在一般用于指代图标。你可以把 Font Awesome 中的 class 属性添加到 i 元素中，让它变成一个图标，比如：
+The `i` element was originally used to make other elements italic, but is now commonly used for icons. You can add the Font Awesome classes to the `i` element to turn it into an icon, for example:
`<i class="fas fa-info-circle"></i>`
-记住 `span` 元素也一样可以用于指代图标。
+Note that the `span` element is also acceptable for use with icons.
# --instructions--
-通过 Font Awesome 库增加一个 `thumbs-up` 图标到你的 like 按钮上，具体方法是给 i 元素添加 class 属性 `fas` 和 `fa-thumbs-up`；确保你的 "Like" 文本在图标旁边。
+Use Font Awesome to add a `thumbs-up` icon to your like button by giving it an `i` element with the classes `fas` and `fa-thumbs-up`. Make sure to keep the text "Like" next to the icon.
# --hints--
-增加一个 class 为 `fas` 和 `fa-thumbs-up` 的 `i` 元素。
+You should add an `i` element with the classes `fas` and `fa-thumbs-up`.
```js
assert($('i').is('.fas.fa-thumbs-up') || $('span').is('.fas.fa-thumbs-up'));
```
-`fa-thumbs-up` 图标应该放在 Like 按钮中。
+Your `fa-thumbs-up` icon should be located within the Like button.
```js
assert(
@@ -48,7 +49,7 @@ assert(
);
```
-将 `i` 元素放置在你的 `button` 元素中。
+Your `i` element should be nested within your `button` element.
```js
assert(
@@ -57,7 +58,7 @@ assert(
);
```
-确保你的图标元素有一个闭合标签。
+Your icon element should have a closing tag.
```js
assert(code.match(/<\/i>|<\/span>/g));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/add-id-attributes-to-bootstrap-elements.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/add-id-attributes-to-bootstrap-elements.md
index aa12b3c62cc..f2354d36db5 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/add-id-attributes-to-bootstrap-elements.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/add-id-attributes-to-bootstrap-elements.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aec908853
-title: 给 Bootstrap 元素添加id属性
+title: Add id Attributes to Bootstrap Elements
challengeType: 0
forumTopicId: 16639
dashedName: add-id-attributes-to-bootstrap-elements
@@ -8,21 +8,21 @@ dashedName: add-id-attributes-to-bootstrap-elements
# --description--
-回忆一下，除了可以给元素添加 class 属性，我们还可以给元素设置 `id` 属性。
+Recall that in addition to class attributes, you can give each of your elements an `id` attribute.
-每个指定元素的 id 都是唯一的，并且在每个页面中都只能使用一次。
+Each id must be unique to a specific element and used only once per page.
-让我们为每个 class 为 `well` 的 `div` 元素添加一个唯一的 id。
+Let's give a unique id to each of our `div` elements of class `well`.
-记住，你可以这样给一个元素设置 ID。
+Remember that you can give an element an id like this:
`<div class="well" id="center-well">`
-把左边 well 的 ID 设置为 `left-well`。右边的 well 的 ID 设置为 `right-well`。
+Give the well on the left the id of `left-well`. Give the well on the right the id of `right-well`.
# --hints--
-把左边的 `well` 的 ID 设置为 `left-well`。
+Your left `well` should have the id of `left-well`.
```js
assert(
@@ -31,7 +31,7 @@ assert(
);
```
-把右边的 `well` 的 ID 设置为 `right-well`。
+Your right `well` should have the id of `right-well`.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/apply-the-default-bootstrap-button-style.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/apply-the-default-bootstrap-button-style.md
index 28c303667a9..4ec8c688fca 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/apply-the-default-bootstrap-button-style.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/apply-the-default-bootstrap-button-style.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aec908850
-title: 给 Bootstrap 按钮添加默认样式
+title: Apply the Default Bootstrap Button Style
challengeType: 0
forumTopicId: 16657
dashedName: apply-the-default-bootstrap-button-style
@@ -8,19 +8,19 @@ dashedName: apply-the-default-bootstrap-button-style
# --description--
-Bootstrap 还有另外一种属于按钮的 class 属性叫做 `btn-default`。
+Bootstrap has another button class called `btn-default`.
-为 `button` 元素增加两个 class： `btn` 和 `btn-default`。
+Apply both the `btn` and `btn-default` classes to each of your `button` elements.
# --hints--
-将 `btn` class 添加到所有的 `button` 元素中。
+You should apply the `btn` class to each of your `button` elements.
```js
assert($('.btn').length > 5);
```
-将 `btn-default` class 添加到每一个 `button` 元素中。
+You should apply the `btn-default` class to each of your `button` elements.
```js
assert($('.btn-default').length > 5);
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/call-out-optional-actions-with-btn-info.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/call-out-optional-actions-with-btn-info.md
index dea10e4c1fe..e0661892708 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/call-out-optional-actions-with-btn-info.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/call-out-optional-actions-with-btn-info.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348cd8acef08813
-title: 使用 btn-info 调出可选操作
+title: Call out Optional Actions with btn-info
challengeType: 0
forumTopicId: 16770
dashedName: call-out-optional-actions-with-btn-info
@@ -8,33 +8,33 @@ dashedName: call-out-optional-actions-with-btn-info
# --description--
-Bootstrap 有着丰富的预定义按钮颜色。浅蓝色 `btn-info` class 通常用在用户可能采取的操作上。
+Bootstrap comes with several pre-defined colors for buttons. The `btn-info` class is used to call attention to optional actions that the user can take.
-在你的 "Like" 按钮下方创建包含文本 "Info" 的块级 Bootstrap 按钮， 然后为该按钮添加 Bootstrap 的 `btn-info` 和 `btn-block` class。
+Create a new block-level Bootstrap button below your "Like" button with the text "Info", and add Bootstrap's `btn-info` and `btn-block` classes to it.
-记住：你的这些按钮仍然需要 `btn` 和 `btn-block` class。
+Note that these buttons still need the `btn` and `btn-block` classes.
# --hints--
-创建包含文本 'Info' 的 `button` 元素。
+You should create a new `button` element with the text "Info".
```js
assert(new RegExp('info', 'gi').test($('button').text()));
```
-两个按钮的 class 属性应该仍然具有 `btn` 和 `btn-block`。
+Both of your Bootstrap buttons should have the `btn` and `btn-block` classes.
```js
assert($('button.btn-block.btn').length > 1);
```
-新按钮的 class 属性应该含有 `btn-info`。
+Your new button should have the class `btn-info`.
```js
assert($('button').hasClass('btn-info'));
```
-确保你所有的 `button` 元素都有一个闭合标签。
+All of your `button` elements should have closing tags.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/center-text-with-bootstrap.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/center-text-with-bootstrap.md
index 765f0a23f88..bcf6be7ef84 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/center-text-with-bootstrap.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/center-text-with-bootstrap.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd8acde08812
-title: 用 Bootstrap 居中文本
+title: Center Text with Bootstrap
challengeType: 0
forumTopicId: 16771
dashedName: center-text-with-bootstrap
@@ -8,21 +8,21 @@ dashedName: center-text-with-bootstrap
# --description--
-我们可以使用 Bootstrap 将顶部的元素居中来美化页面。只需要将 `h2` 元素的 class 属性设置为 `text-center` 就可以实现。
+Now that we're using Bootstrap, we can center our heading element to make it look better. All we need to do is add the class `text-center` to our `h2` element.
-记住：我们通过空格分隔不同的 class 可以为一个元素添加多个 class ，就像这样：
+Remember that you can add several classes to the same element by separating each of them with a space, like this:
`<h2 class="red-text text-center">your text</h2>`
# --hints--
-你的 `h2` 元素应该居中且有一个 class 为 `text-center`
+Your `h2` element should be centered by applying the class `text-center`
```js
assert($('h2').hasClass('text-center'));
```
-你的 `h2` 元素应该还有另一个 class 为 `red-text`
+Your `h2` element should still have the class `red-text`
```js
assert($('h2').hasClass('red-text'));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-block-element-bootstrap-button.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-block-element-bootstrap-button.md
index 67a423220ab..784250c7e57 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-block-element-bootstrap-button.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-block-element-bootstrap-button.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348cd8acef08812
-title: 创建一个 Bootstrap 块级元素
+title: Create a Block Element Bootstrap Button
challengeType: 0
forumTopicId: 16810
dashedName: create-a-block-element-bootstrap-button
@@ -8,41 +8,41 @@ dashedName: create-a-block-element-bootstrap-button
# --description--
-一般情况下，具有 `btn` 和 `btn-default` 两个 class 的 `button` 元素宽度与它包含的文本相同，举个例子：
+Normally, your `button` elements with the `btn` and `btn-default` classes are only as wide as the text that they contain. For example:
`<button class="btn btn-default">Submit</button>`
-这个按钮的宽度应该和文本 "Submit" 相同
+This button would only be as wide as the word "Submit".
<button class='btn btn-default'>Submit</button>
-通过为按钮添加 class 属性 `btn-block` 使其成为块级元素，按钮会伸展并填满页面整个水平空间，后续的元素会流到这个块级元素的下方，即 "新开一行"。
+By making them block elements with the additional class of `btn-block`, your button will stretch to fill your page's entire horizontal space and any elements following it will flow onto a "new line" below the block.
`<button class="btn btn-default btn-block">Submit</button>`
-这个按钮会 100% 占满所有的可用宽度。
+This button would take up 100% of the available width.
<button class='btn btn-default btn-block'>Submit</button>
-记住这些按钮仍然需要 `btn` 这个 class。
+Note that these buttons still need the `btn` class.
-添加 Bootstrap 的 `btn-block` class 到你 Bootstrap 按钮上吧。
+Add Bootstrap's `btn-block` class to your Bootstrap button.
# --hints--
-按钮的 class 属性应该仍然具有 `btn` 和 `btn-default`。
+Your button should still have the `btn` and `btn-default` classes.
```js
assert($('button').hasClass('btn') && $('button').hasClass('btn-default'));
```
-按钮的 class 属性应该包含 `btn-block`。
+Your button should have the class `btn-block`.
```js
assert($('button').hasClass('btn-block'));
```
-确保所有的 `button` 元素都有一个闭合标签。
+All of your `button` elements should have closing tags.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-bootstrap-button.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-bootstrap-button.md
index 8ec5d8dcf16..2cf8ce1aff6 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-bootstrap-button.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-bootstrap-button.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348cd8acdf08812
-title: 创建一个 Bootstrap 按钮
+title: Create a Bootstrap Button
challengeType: 0
forumTopicId: 16811
dashedName: create-a-bootstrap-button
@@ -8,13 +8,13 @@ dashedName: create-a-bootstrap-button
# --description--
-Bootstrap 的 `button` 元素有着独有的、比默认 HTML 按钮更好的样式风格。
+Bootstrap has its own styles for `button` elements, which look much better than the plain HTML ones.
-在较大的小猫图片下方创建新的 `button` 元素。 为它添加 `btn` 与 `btn-default` 两个 class 和 "Like" 文本。
+Create a new `button` element below your large kitten photo. Give it the `btn` and `btn-default` classes, as well as the text of "Like".
# --hints--
-创建新的 `button` 元素，设置文本为 'Like'。
+You should create a new `button` element with the text "Like".
```js
assert(
@@ -23,13 +23,13 @@ assert(
);
```
-新的按钮元素应该有两个 class : `btn` 和 `btn-default`。
+Your new button should have two classes: `btn` and `btn-default`.
```js
assert($('button').hasClass('btn') && $('button').hasClass('btn-default'));
```
-保证所有 `button` 元素都有一个闭合标签。
+All of your `button` elements should have closing tags.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-bootstrap-headline.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-bootstrap-headline.md
index a751aa9d149..074aa2fc926 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-bootstrap-headline.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-bootstrap-headline.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aec908846
-title: 创建一个 Bootstrap 标题
+title: Create a Bootstrap Headline
challengeType: 0
forumTopicId: 16812
dashedName: create-a-bootstrap-headline
@@ -8,23 +8,23 @@ dashedName: create-a-bootstrap-headline
# --description--
-现在，让我们运用 HTML，CSS 和 Bootstrap 从零开始做点东西。
+Now let's build something from scratch to practice our HTML, CSS and Bootstrap skills.
-我们将会搭建一个 jQuery playground，以便在后续的 jQuery 课程中使用它。
+We'll build a jQuery playground, which we'll soon put to use in our jQuery challenges.
-首先，创建一个包含 `jQuery Playground` 文本内容的 `h3` 元素。
+To start with, create an `h3` element, with the text `jQuery Playground`.
-通过给你的 `h3` 元素设置 Bootstrap 的 class 属性 `text-primary` 来为其上色，然后添加 Bootstrap 的 class 属性 `text-center` 使其文本居中显示。
+Color your `h3` element with the `text-primary` Bootstrap class, and center it with the `text-center` Bootstrap class.
# --hints--
-为你的页面添加一个 `h3` 元素。
+You should add an `h3` element to your page.
```js
assert($('h3') && $('h3').length > 0);
```
-确保你的 `h3` 元素有一个闭合标签。
+Your `h3` element should have a closing tag.
```js
assert(
@@ -34,19 +34,19 @@ assert(
);
```
-为了确保成功上色，`h3` 元素应该具有 `text-primary` class。
+Your `h3` element should be colored by applying the class `text-primary`
```js
assert($('h3').hasClass('text-primary'));
```
-为了确保文本居中显示，你的 `h3` 元素应该具有 `text-center` class。
+Your `h3` element should be centered by applying the class `text-center`
```js
assert($('h3').hasClass('text-center'));
```
-`h3` 元素文本为 `jQuery Playground`。
+Your `h3` element should have the text `jQuery Playground`.
```js
assert.isTrue(/jquery(\s)+playground/gi.test($('h3').text()));
@@ -55,7 +55,6 @@ assert.isTrue(/jquery(\s)+playground/gi.test($('h3').text()));
# --seed--
## --seed-contents--
-
```html
```
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-bootstrap-row.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-bootstrap-row.md
index fc9a440e462..3a5ff315bbf 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-bootstrap-row.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-bootstrap-row.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9bec908846
-title: 创建一个 Bootstrap Row
+title: Create a Bootstrap Row
challengeType: 0
forumTopicId: 16813
dashedName: create-a-bootstrap-row
@@ -8,13 +8,13 @@ dashedName: create-a-bootstrap-row
# --description--
-这次让我们为内联元素创建一个 Bootstrap 栅格系统的 Row（行）。
+Now we'll create a Bootstrap row for our inline elements.
-在 `h3` 标签下方创建一个 class 属性为 `row` 的 `div` 元素。
+Create a `div` element below the `h3` tag, with a class of `row`.
# --hints--
-`h3` 元素下增加一个 `div` 元素。
+You should add a `div` element below your `h3` element.
```js
assert(
@@ -25,19 +25,19 @@ assert(
);
```
-`div` 元素的 class 属性应为 `row`。
+Your `div` element should have the class `row`
```js
assert($('div').hasClass('row'));
```
-`row div` 应该内嵌于 `container-fluid div`。
+Your `row div` should be nested inside the `container-fluid div`
```js
assert($('div.container-fluid div.row').length > 0);
```
-确保所有 `div` 元素都有一个闭合标签。
+Your `div` element should have a closing tag.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-class-to-target-with-jquery-selectors.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-class-to-target-with-jquery-selectors.md
index 8b2bf71139b..eaea9337356 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-class-to-target-with-jquery-selectors.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-class-to-target-with-jquery-selectors.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aec908852
-title: 分别给每个目标元素的 class 属性添加一个 target 值
+title: Create a Class to Target with jQuery Selectors
challengeType: 0
forumTopicId: 16815
dashedName: create-a-class-to-target-with-jquery-selectors
@@ -8,13 +8,13 @@ dashedName: create-a-class-to-target-with-jquery-selectors
# --description--
-并不是所有 class 属性都需要有对应的 CSS 样式。有时候我们设置 class 只是为了更方便地在 jQuery 中选中这些元素。
+Not every class needs to have corresponding CSS. Sometimes we create classes just for the purpose of selecting these elements more easily using jQuery.
-为每一个 `button` 元素添加 `target` class。
+Give each of your `button` elements the class `target`.
# --hints--
-给你的每一个 `button` 元素设置 `target` class。
+You should apply the `target` class to each of your `button` elements.
```js
assert($('.target').length > 5);
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-custom-heading.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-custom-heading.md
index d28dd786bc9..646afcbe0fe 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-custom-heading.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-a-custom-heading.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aede08845
-title: 创建自定义标题
+title: Create a Custom Heading
challengeType: 0
forumTopicId: 16816
dashedName: create-a-custom-heading
@@ -8,23 +8,23 @@ dashedName: create-a-custom-heading
# --description--
-让我们来为 Cat Photo App 做一个导航吧，把标题和惬意的猫咪图片放在同一行。
+We will make a simple heading for our Cat Photo App by putting the title and relaxing cat image in the same row.
-记住，由于 Bootstrap 使用了响应式栅格系统，使得我们可以很方便的放置元素并为元素指定相对的宽度。大部分的 Bootstrap 的 class 都能用在 `div` 元素上。
+Remember, Bootstrap uses a responsive grid system, which makes it easy to put elements into rows and specify each element's relative width. Most of Bootstrap's classes can be applied to a `div` element.
-把我们的第一张图片和 `h2` 元素用一个简单的 `<div class="row">` 元素包裹起来。再用 `<div class="col-xs-8">` 包裹我们的 `h2` 元素，用 `<div class="col-xs-4">` 包裹我们的图片，这样它们就能位于同一行了。
+Nest your first image and your `h2` element within a single `<div class="row">` element. Nest your `h2` element within a `<div class="col-xs-8">` and your image in a `<div class="col-xs-4">` so that they are on the same line.
-注意现在图片是否与文字大小一致呢？
+Notice how the image is now just the right size to fit along the text?
# --hints--
-`h2` 元素和最上方的 `img` 元素应该一起内嵌于具有 `row` class 的 `div` 元素内。
+Your `h2` element and topmost `img` element should both be nested together within a `div` element with the class `row`.
```js
assert($('div.row:has(h2)').length > 0 && $('div.row:has(img)').length > 0);
```
-最上方的 `img` 元素应该内嵌于含有 `col-xs-4` class 的 `div` 元素中。
+Your topmost `img` element should be nested within a `div` with the class `col-xs-4`.
```js
assert(
@@ -33,7 +33,7 @@ assert(
);
```
-`h2` 元素应该内嵌于含有 `col-xs-8` class的 `div` 元素中。
+Your `h2` element should be nested within a `div` with the class `col-xs-8`.
```js
assert(
@@ -42,7 +42,7 @@ assert(
);
```
-确保每一个 `div` 元素都有一个闭合标签。
+All of your `div` elements should have closing tags.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-bootstrap-wells.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-bootstrap-wells.md
index 1e07118d2fb..6e79e8f12e2 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-bootstrap-wells.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/create-bootstrap-wells.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aec908848
-title: 创建 Bootstrap Wells
+title: Create Bootstrap Wells
challengeType: 0
forumTopicId: 16825
dashedName: create-bootstrap-wells
@@ -8,25 +8,25 @@ dashedName: create-bootstrap-wells
# --description--
-Bootstrap 有一个叫做 `well` 的 class，作用是赋予列一种视觉上的深度感（视觉上的效果）。
+Bootstrap has a class called `well` that can create a visual sense of depth for your columns.
-在每一个 class 属性为 `col-xs-6` 的 `div` 元素中都嵌入一个带有 `well` 的 `div` 元素。
+Nest one `div` element with the class `well` within each of your `col-xs-6` `div` elements.
# --hints--
-在每一个 class 属性为 `'col-xs-6'` 的 `div` 元素中都嵌入一个带有 `well` 的 `div` 元素。
+You should add a `div` element with the class `well` inside each of your `div` elements with the class `"col-xs-6"`
```js
assert($('div.col-xs-6').not(':has(>div.well)').length < 1);
```
-将你的两个 class 属性为 `'col-xs-6'` 的 `div` 元素都内嵌入一个带有 `'row'` 的 `div` 元素中。
+Both of your `div` elements with the class `"col-xs-6"` should be nested within your `div` element with the class `"row"`.
```js
assert($('div.row > div.col-xs-6').length > 1);
```
-确保你的 `div` 元素都有一个闭合标签。
+All your `div` elements should have closing tags.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/ditch-custom-css-for-bootstrap.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/ditch-custom-css-for-bootstrap.md
index 76055ec11e4..543e90d7a25 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/ditch-custom-css-for-bootstrap.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/ditch-custom-css-for-bootstrap.md
@@ -1,6 +1,6 @@
---
id: bad87fee1347bd9aedf08845
-title: 用 Bootstrap 来取代我们之前的自定义样式
+title: Ditch Custom CSS for Bootstrap
challengeType: 0
forumTopicId: 17565
dashedName: ditch-custom-css-for-bootstrap
@@ -8,31 +8,31 @@ dashedName: ditch-custom-css-for-bootstrap
# --description--
-现在我们可以清理一下之前代码，用 Bootstrap 的内置样式来替换我们之前定义的样式，这样会让我们的 Cat Photo App 看起来更简洁些。
+We can clean up our code and make our Cat Photo App look more conventional by using Bootstrap's built-in styles instead of the custom styles we created earlier.
-别担心————以后我们会有大把时间来自定义我们的 CSS 样式的。
+Don't worry - there will be plenty of time to customize our CSS later.
-删除 `style` 元素里的 `.red-text`, `p`, 和 `.smaller-image` CSS 声明，使 `style` 元素留下的声明只有 `h2` 和 `thick-green-border`.
+Delete the `.red-text`, `p`, and `.smaller-image` CSS declarations from your `style` element so that the only declarations left in your `style` element are `h2` and `thick-green-border`.
-删除包含死链接的 `p` 元素。 然后将 `h2` 的 `red-text` class 替换为 Bootstrap 的 `text-primary` class.
+Then delete the `p` element that contains a dead link. Then remove the `red-text` class from your `h2` element and replace it with the `text-primary` Bootstrap class.
-最后, 将你第一个 `img` 元素的 "smaller-image" class 替换为 `img-responsive` class.
+Finally, remove the "smaller-image" class from your first `img` element and replace it with the `img-responsive` class.
# --hints--
-h2 元素的 class 不应为 `red-text`。
+Your h2 element should no longer have the class `red-text`.
```js
assert(!$('h2').hasClass('red-text'));
```
-h2 元素的 class 应为 `text-primary`。
+Your h2 element should now have the class `text-primary`.
```js
assert($('h2').hasClass('text-primary'));
```
-你的段落元素（p）应该不再使用 `Monospace` 字体。
+Your paragraph elements should no longer use the font `Monospace`.
```js
assert(
@@ -42,13 +42,13 @@ assert(
);
```
-移除你第一张图片的 class 属性 `smaller-image`。
+The `smaller-image` class should be removed from your top image.
```js
assert(!$('img').hasClass('smaller-image'));
```
-给你的第一张图片添加 class 属性 `img-responsive`。
+You should add the `img-responsive` class to your top image.
```js
assert($('.img-responsive').length > 1);
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/give-each-element-a-unique-id.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/give-each-element-a-unique-id.md
index 8844df755f0..290766d634e 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/give-each-element-a-unique-id.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/give-each-element-a-unique-id.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aec908855
-title: 给每个元素一个唯一的 id
+title: Give Each Element a Unique id
challengeType: 0
forumTopicId: 18191
dashedName: give-each-element-a-unique-id
@@ -8,15 +8,15 @@ dashedName: give-each-element-a-unique-id
# --description--
-我们也可以通过 jQuery 根据每个按钮唯一的 id 来标识出它们。
+We will also want to be able to use jQuery to target each button by its unique id.
-给你的每一个按钮设置唯一的 id，以 `target1` 开始，`target6` 结束。
+Give each of your buttons a unique id, starting with `target1` and ending with `target6`.
-确保 `target1` 到 `target3` 在 `#left-well` 之中，`target4` 到 `target6` 在 `#right-well` 之中。
+Make sure that `target1` to `target3` are in `#left-well`, and `target4` to `target6` are in `#right-well`.
# --hints--
-其中一个 `button` 元素应该有 id `target1`。
+One `button` element should have the id `target1`.
```js
assert(
@@ -25,7 +25,7 @@ assert(
);
```
-其中一个 `button` 元素应该有 id `target2`。
+One `button` element should have the id `target2`.
```js
assert(
@@ -34,7 +34,7 @@ assert(
);
```
-其中一个 `button` 元素应该有 id `target3`。
+One `button` element should have the id `target3`.
```js
assert(
@@ -43,7 +43,7 @@ assert(
);
```
-其中一个 `button` 元素应该有 id `target4`。
+One `button` element should have the id `target4`.
```js
assert(
@@ -52,7 +52,7 @@ assert(
);
```
-其中一个 `button` 元素应该有 id `target5`。
+One `button` element should have the id `target5`.
```js
assert(
@@ -61,7 +61,7 @@ assert(
);
```
-其中一个 `button` 元素应该有 id `target6`。
+One `button` element should have the id `target6`.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/house-our-page-within-a-bootstrap-container-fluid-div.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/house-our-page-within-a-bootstrap-container-fluid-div.md
index bc5cb829512..452f52ab470 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/house-our-page-within-a-bootstrap-container-fluid-div.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/house-our-page-within-a-bootstrap-container-fluid-div.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aec908746
-title: 将我们的页面放在 Fluid 容器中
+title: House our page within a Bootstrap container-fluid div
challengeType: 0
forumTopicId: 18198
dashedName: house-our-page-within-a-bootstrap-container-fluid-div
@@ -8,19 +8,19 @@ dashedName: house-our-page-within-a-bootstrap-container-fluid-div
# --description--
-现在让我们确保页面所有内容应该都是响应式的。
+Now let's make sure all the content on your page is mobile-responsive.
-将我们的 `h3` 元素内嵌进一个具有 `container-fluid` class 的`div` 元素中。
+Let's nest your `h3` element within a `div` element with the class `container-fluid`.
# --hints--
-`div` 元素 class 属性应该为 `container-fluid`。
+Your `div` element should have the class `container-fluid`.
```js
assert($('div').hasClass('container-fluid'));
```
-确保每一个 `div` 元素都有一个闭合标签。
+Each of your `div` elements should have closing tags.
```js
assert(
@@ -30,7 +30,7 @@ assert(
);
```
-`h3` 元素应该内嵌于 `div` 元素。
+Your `h3` element should be nested inside a `div` element.
```js
assert($('div').children('h3').length > 0);
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/label-bootstrap-buttons.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/label-bootstrap-buttons.md
index 0b5e0c004c4..06e1b6f63b4 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/label-bootstrap-buttons.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/label-bootstrap-buttons.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aec908856
-title: 给 Bootstrap 按钮贴标签
+title: Label Bootstrap Buttons
challengeType: 0
forumTopicId: 18222
dashedName: label-bootstrap-buttons
@@ -8,43 +8,43 @@ dashedName: label-bootstrap-buttons
# --description--
-正如我们标注了每个 wells 一样，我们同样想要标注每一个按钮。
+Just like we labeled our wells, we want to label our buttons.
-为每一个 `button` 元素设置与其 `id` 选择器相同的文本。
+Give each of your `button` elements text that corresponds to its `id`'s selector.
# --hints--
-给 id 为 `target1` 的 `button` 元素设置文本 `#target1`。
+Your `button` element with the id `target1` should have the text `#target1`.
```js
assert(new RegExp('#target1', 'gi').test($('#target1').text()));
```
-给 id 为 `target2` 的 `button` 元素设置文本 `#target2`。
+Your `button` element with the id `target2` should have the text `#target2`.
```js
assert(new RegExp('#target2', 'gi').test($('#target2').text()));
```
-给 id 为 `target3` 的 `button` 元素设置文本 `#target3`。
+Your `button` element with the id `target3` should have the text `#target3`.
```js
assert(new RegExp('#target3', 'gi').test($('#target3').text()));
```
-给 id 为 `target4` 的 `button` 元素设置文本 `#target4`。
+Your `button` element with the id `target4` should have the text `#target4`.
```js
assert(new RegExp('#target4', 'gi').test($('#target4').text()));
```
-给 id 为 `target5` 的 `button` 元素设置文本 `#target5`。
+Your `button` element with the id `target5` should have the text `#target5`.
```js
assert(new RegExp('#target5', 'gi').test($('#target5').text()));
```
-给 id 为 `target6` 的 `button` 元素设置文本 `#target6`。
+Your `button` element with the id `target6` should have the text `#target6`.
```js
assert(new RegExp('#target6', 'gi').test($('#target6').text()));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/label-bootstrap-wells.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/label-bootstrap-wells.md
index 79479e17a57..cae4c4837f2 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/label-bootstrap-wells.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/label-bootstrap-wells.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aec908854
-title: 给 Bootstrap Wells 贴标签
+title: Label Bootstrap Wells
challengeType: 0
forumTopicId: 18223
dashedName: label-bootstrap-wells
@@ -8,15 +8,15 @@ dashedName: label-bootstrap-wells
# --description--
-为了让我们页面逻辑更清晰，让我们为 wells 都标上它们的 id 吧。
+For the sake of clarity, let's label both of our wells with their ids.
-在 left-well 的上一层，class 属性为 `col-xs-6` 的 `div` 元素里面，增加一个文本为 `#left-well` 的 `h4` 元素。
+Above your left-well, inside its `col-xs-6` `div` element, add a `h4` element with the text `#left-well`.
-在 right-well 的上一层，class 属性为 `col-xs-6` 的 `div` 元素里面，增加一个文本为 `#right-well` 的 `h4` 元素。
+Above your right-well, inside its `col-xs-6` `div` element, add a `h4` element with the text `#right-well`.
# --hints--
-为每个 `<div class='col-xs-6'>` 元素添加一个 `h4` 元素。
+You should add an `h4` element to each of your `<div class="col-xs-6">` elements.
```js
assert(
@@ -24,19 +24,19 @@ assert(
);
```
-其中一个 `h4` 元素应该含有文本内容 `#left-well`。
+One `h4` element should have the text `#left-well`.
```js
assert(new RegExp('#left-well', 'gi').test($('h4').text()));
```
-其中一个 `h4` 元素应该含有文本内容 `#right-well`。
+One `h4` element should have the text `#right-well`.
```js
assert(new RegExp('#right-well', 'gi').test($('h4').text()));
```
-确保每个 `h4` 元素都有一个闭合标签。
+All of your `h4` elements should have closing tags.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/line-up-form-elements-responsively-with-bootstrap.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/line-up-form-elements-responsively-with-bootstrap.md
index ef31b90a030..b6037f83336 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/line-up-form-elements-responsively-with-bootstrap.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/line-up-form-elements-responsively-with-bootstrap.md
@@ -1,10 +1,11 @@
---
id: bad87fee1348bd9aec908845
-title: 使用 Bootstrap 响应式排列表单元素
+title: Line up Form Elements Responsively with Bootstrap
challengeType: 0
forumTopicId: 18225
required:
- - link: >-
+ -
+ link: >-
https://cdnjs.cloudflare.com/ajax/libs/font-awesome/4.2.0/css/font-awesome.css
raw: true
dashedName: line-up-form-elements-responsively-with-bootstrap
@@ -12,15 +13,15 @@ dashedName: line-up-form-elements-responsively-with-bootstrap
# --description--
-现在让我们把你表单里的 `input` 元素和 submit `button` （提交按钮）放在同一行。跟着之前的做法来就行：准备好一个具有 `row` class 的 `div` 元素还有几个具有 `col-xs-*` class 的 `div` 元素。
+Now let's get your form `input` and your submission `button` on the same line. We'll do this the same way we have previously: by using a `div` element with the class `row`, and other `div` elements within it using the `col-xs-*` class.
-先把你表单的 text `input` （文本输入框）和 submit `button` （提交按钮）放进具有 `row` class 的 `div` 中。再用 `col-xs-7` class 的 div 包裹表单的 text `input` （文本输入框），`col-xs-5` class 的 div 包裹表单的 submit `button` （提交按钮）。
+Nest both your form's text `input` and submit `button` within a `div` with the class `row`. Nest your form's text `input` within a div with the class of `col-xs-7`. Nest your form's submit `button` in a `div` with the class `col-xs-5`.
-这是我们到目前为止 Cat Photo App 的最后一个挑战了。我希望你能喜欢 Font Awesome， Bootstrap，和响应式设计！
+This is the last challenge we'll do for our Cat Photo App for now. We hope you've enjoyed learning Font Awesome, Bootstrap, and responsive design!
# --hints--
-确保提交按钮和文本输入框都在同一个具有 class 属性 `row` 的 div 元素中。
+Your form submission button and text input should be nested in a div with class `row`.
```js
assert(
@@ -29,19 +30,19 @@ assert(
);
```
-表单的文本输入框应该嵌入到具有 class 属性 `col-xs-7` 的 div 内。
+Your form text input should be nested in a div with the class `col-xs-7`.
```js
assert($('div.col-xs-7:has(input[type="text"])').length > 0);
```
-表单的提交按钮应该嵌入到具有 class 属性 `col-xs-5` 的 div 内。
+Your form submission button should be nested in a div with the class `col-xs-5`.
```js
assert($('div.col-xs-5:has(button[type="submit"])').length > 0);
```
-确保每一个 `div` 元素都有一个闭合标签。
+All of your `div` elements should have closing tags.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/make-images-mobile-responsive.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/make-images-mobile-responsive.md
index 94eb27049b2..54c1ec8e26b 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/make-images-mobile-responsive.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/make-images-mobile-responsive.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9acde08812
-title: 使图片自适应移动端
+title: Make Images Mobile Responsive
challengeType: 0
forumTopicId: 18232
dashedName: make-images-mobile-responsive
@@ -8,39 +8,39 @@ dashedName: make-images-mobile-responsive
# --description--
-首先，在已有的图片下面添加一张新的图片. 设置 `src` 属性为 `https://bit.ly/fcc-running-cats`。
+First, add a new image below the existing one. Set its `src` attribute to `https://bit.ly/fcc-running-cats`.
-如果图片的大小恰恰和我们手机屏幕尺寸大小一样自然是最好的。
+It would be great if this image could be exactly the width of our phone's screen.
-幸运的是现在通过 Bootstrap，我们仅仅只需要为 image 标签上设置 class 属性为 `img-responsive` 就可以让它完美的适应你页面的宽度了。
+Fortunately, with Bootstrap, all we need to do is add the `img-responsive` class to your image. Do this, and the image should perfectly fit the width of your page.
# --hints--
-该页面拥有总计两个图片。
+You should have a total of two images.
```js
assert($('img').length === 2);
```
-新的图片应该在旧的图片下面并且含有 class 属性 `img-responsive`。
+Your new image should be below your old one and have the class `img-responsive`.
```js
assert($('img:eq(1)').hasClass('img-responsive'));
```
-新的图片不应该含有 class 属性 `smaller-image`。
+Your new image should not have the class `smaller-image`.
```js
assert(!$('img:eq(1)').hasClass('smaller-image'));
```
-新图片的 `src` 属性应该为 `https://bit.ly/fcc-running-cats`。
+Your new image should have a `src` of `https://bit.ly/fcc-running-cats`.
```js
assert($('img:eq(1)').attr('src') === 'https://bit.ly/fcc-running-cats');
```
-确保新的 `img` 元素有一个闭合的右尖括号 “/>”。
+Your new `img` element should have a closing angle bracket.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/responsively-style-checkboxes.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/responsively-style-checkboxes.md
index 369fcf72351..8fcf0774c88 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/responsively-style-checkboxes.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/responsively-style-checkboxes.md
@@ -1,10 +1,11 @@
---
id: bad87fee1348bd9aeda08845
-title: 响应式风格的复选框
+title: Responsively Style Checkboxes
challengeType: 0
forumTopicId: 18269
required:
- - link: >-
+ -
+ link: >-
https://cdnjs.cloudflare.com/ajax/libs/font-awesome/4.2.0/css/font-awesome.css
raw: true
dashedName: responsively-style-checkboxes
@@ -12,27 +13,27 @@ dashedName: responsively-style-checkboxes
# --description--
-你还可以将 Bootstrap 的 `col-xs-*` 用在 `form` 元素上!这样我们就可以在不关心屏幕大小的情况下，将我们的复选框均匀的放在页面上了。
+Since Bootstrap's `col-xs-*` classes are applicable to all `form` elements, you can use them on your checkboxes too! This way, the checkboxes will be evenly spread out across the page, regardless of how wide the screen resolution is.
# --instructions--
-将所有复选框都放置于一个 `<div class="row">` 元素中。然后分别把每个复选框都放置于一个 `<div class="col-xs-4">` 元素中。
+Nest all three of your checkboxes in a `<div class="row">` element. Then nest each of them in a `<div class="col-xs-4">` element.
# --hints--
-将所有的复选框嵌入一个含有 `row` class 的 `div` 元素中。
+All of your checkboxes should be nested inside one `div` with the class `row`.
```js
assert($('div.row:has(input[type="checkbox"])').length > 0);
```
-每一个复选框应该嵌套于自己的 `div` 元素中，每个 `div` 元素都具有 `col-xs-4` class。
+Each of your checkboxes should be nested inside its own `div` with the class `col-xs-4`.
```js
assert($('div.col-xs-4:has(input[type="checkbox"])').length > 2);
```
-确保所有 `div` 元素都有一个闭合标签。
+All of your `div` elements should have closing tags.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/responsively-style-radio-buttons.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/responsively-style-radio-buttons.md
index 09adc2f41c6..3541d7fead3 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/responsively-style-radio-buttons.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/responsively-style-radio-buttons.md
@@ -1,10 +1,11 @@
---
id: bad87fee1348bd9aedb08845
-title: 响应式风格的单选按钮
+title: Responsively Style Radio Buttons
challengeType: 0
forumTopicId: 18270
required:
- - link: >-
+ -
+ link: >-
https://cdnjs.cloudflare.com/ajax/libs/font-awesome/4.2.0/css/font-awesome.css
raw: true
dashedName: responsively-style-radio-buttons
@@ -12,27 +13,27 @@ dashedName: responsively-style-radio-buttons
# --description--
-你还可以将 Bootstrap 的 `col-xs-*` class 用在 `form` 元素上！这样我们就可以在不关心屏幕大小的情况下，将我们的单选按钮均匀的平铺在页面上。
+You can use Bootstrap's `col-xs-*` classes on `form` elements, too! This way, our radio buttons will be evenly spread out across the page, regardless of how wide the screen resolution is.
-将你的所有单选按钮放入 `<div class="row">` 元素。再用 `<div class="col-xs-6">` 元素包裹每一个单选按钮。
+Nest both your radio buttons within a `<div class="row">` element. Then nest each of them within a `<div class="col-xs-6">` element.
-**记住：** 提醒一句，单选按钮是 type 为 `radio` 的 `input` 元素。
+**Note:** As a reminder, radio buttons are `input` elements of type `radio`.
# --hints--
-把所有的单选按钮放置于具有 `row` class 的 `div` 元素中。
+All of your radio buttons should be nested inside one `div` with the class `row`.
```js
assert($('div.row:has(input[type="radio"])').length > 0);
```
-每一个单选按钮应该嵌套于 `div` 元素之中，该 `div` 元素的 class 属性为 `col-xs-6`。
+Each of your radio buttons should be nested inside its own `div` with the class `col-xs-6`.
```js
assert($('div.col-xs-6:has(input[type="radio"])').length > 1);
```
-确保所有 `div` 元素都有一个闭合标签。
+All of your `div` elements should have closing tags.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/split-your-bootstrap-row.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/split-your-bootstrap-row.md
index ec47f8315d1..440c9a7740b 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/split-your-bootstrap-row.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/split-your-bootstrap-row.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aec908847
-title: 分割你的 Bootstrap Row
+title: Split Your Bootstrap Row
challengeType: 0
forumTopicId: 18306
dashedName: split-your-bootstrap-row
@@ -8,19 +8,19 @@ dashedName: split-your-bootstrap-row
# --description--
-现在我们已经有了一个 Bootstrap Row，让我们把它分成两列来放置我们的元素。
+Now that we have a Bootstrap Row, let's split it into two columns to house our elements.
-在行内创建两个 class 属性为 `col-xs-6` 的 `div` 元素。
+Create two `div` elements within your row, both with the class `col-xs-6`.
# --hints--
-将两个 `div class='col-xs-6'` 元素内嵌入你的 `div class='row'` 元素中。
+Two `div class="col-xs-6"` elements should be nested within your `div class="row"` element.
```js
assert($('div.row > div.col-xs-6').length > 1);
```
-确保你的 `div` 元素都有一个闭合标签。
+All your `div` elements should have closing tags.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/style-text-inputs-as-form-controls.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/style-text-inputs-as-form-controls.md
index 182ff04dd63..513456f26c0 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/style-text-inputs-as-form-controls.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/style-text-inputs-as-form-controls.md
@@ -1,10 +1,11 @@
---
id: bad87fee1348bd9aed908845
-title: 给表单控件的输入框添加样式
+title: Style Text Inputs as Form Controls
challengeType: 0
forumTopicId: 18312
required:
- - link: >-
+ -
+ link: >-
https://cdnjs.cloudflare.com/ajax/libs/font-awesome/4.2.0/css/font-awesome.css
raw: true
dashedName: style-text-inputs-as-form-controls
@@ -12,33 +13,33 @@ dashedName: style-text-inputs-as-form-controls
# --description--
-你可以通过在 submit `button` 内加上 `<i class="fa fa-paper-plane"></i>` 来添加 Font Awesome 的 `fa-paper-plane` 图标。
+You can add the `fa-paper-plane` Font Awesome icon by adding `<i class="fa fa-paper-plane"></i>` within your submit `button` element.
-为表单的文本输入框（text input）设置 `form-control` class。为表单的提交（submit）按钮设置 `btn btn-primary` class，并为它加上 Font Awesome 的 `fa-paper-plane` 图标。
+Give your form's text input field a class of `form-control`. Give your form's submit button the classes `btn btn-primary`. Also give this button the Font Awesome icon of `fa-paper-plane`.
-所有文本输入类的元素如 `<input>`，`<textarea>` 和 `<select>` 只要设置 `.form-control` class 就会占满100%的宽度。
+All textual `<input>`, `<textarea>`, and `<select>` elements with the class `.form-control` have a width of 100%.
# --hints--
-给你的 submit 按钮添加 `btn btn-primary` class。
+The submit button in your form should have the classes `btn btn-primary`.
```js
assert($('button[type="submit"]').hasClass('btn btn-primary'));
```
-在你的 submit `button` 元素嵌入 `<i class='fa fa-paper-plane'></i>`。
+You should add a `<i class="fa fa-paper-plane"></i>` within your submit `button` element.
```js
assert($('button[type="submit"]:has(i.fa.fa-paper-plane)').length > 0);
```
-给表单中的 text `input` 元素添加 `form-control` class。
+The text `input` in your form should have the class `form-control`.
```js
assert($('input[type="text"]').hasClass('form-control'));
```
-确保每一个 `i` 元素都有一个闭合标签。
+Each of your `i` elements should have a closing tag.
```js
assert(code.match(/<\/i>/g) && code.match(/<\/i/g).length > 3);
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/taste-the-bootstrap-button-color-rainbow.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/taste-the-bootstrap-button-color-rainbow.md
index 24a16703f6a..a3e9e1b9ec0 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/taste-the-bootstrap-button-color-rainbow.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/taste-the-bootstrap-button-color-rainbow.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348cd8acef08811
-title: 体验 Bootstrap 彩虹色的按钮
+title: Taste the Bootstrap Button Color Rainbow
challengeType: 0
forumTopicId: 18323
dashedName: taste-the-bootstrap-button-color-rainbow
@@ -8,27 +8,27 @@ dashedName: taste-the-bootstrap-button-color-rainbow
# --description--
-`btn-primary` class 的颜色是你在应用中的主题色。这样 “突出显示” 是引导用户按步就班进行操作的有效办法。
+The `btn-primary` class is the main color you'll use in your app. It is useful for highlighting actions you want your user to take.
-将按钮的 class 从 Bootstrap 的 `btn-default` 替换为 `btn-primary`。
+Replace Bootstrap's `btn-default` class with `btn-primary` in your button.
-记住：你的按钮仍然需要 `btn` 和 `btn-block` class。
+Note that this button will still need the `btn` and `btn-block` classes.
# --hints--
-按钮的 class 属性应该有 `btn-primary`。
+Your button should have the class `btn-primary`.
```js
assert($('button').hasClass('btn-primary'));
```
-按钮的 class 属性应该仍有 `btn` 和 `btn-block`。
+Your button should still have the `btn` and `btn-block` classes.
```js
assert($('button').hasClass('btn-block') && $('button').hasClass('btn'));
```
-确保所有 `button` 元素都有一个闭合标签。
+All your `button` elements should have closing tags.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/use-a-span-to-target-inline-elements.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/use-a-span-to-target-inline-elements.md
index 43c733ef99e..dfb243376e3 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/use-a-span-to-target-inline-elements.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/use-a-span-to-target-inline-elements.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aedf08845
-title: 使用 span 创建行内元素
+title: Use a span to Target Inline Elements
challengeType: 0
forumTopicId: 18370
dashedName: use-a-span-to-target-inline-elements
@@ -8,31 +8,31 @@ dashedName: use-a-span-to-target-inline-elements
# --description--
-你可以使用 span 标签来创建行内元素。还记得我们怎么使用 `btn-block` class 来创建填满整行的按钮吗？
+You can use spans to create inline elements. Remember when we used the `btn-block` class to make the button fill the entire row?
<button class='btn' style='background-color: rgb(0, 100, 0); color: rgb(255, 255, 255);'>normal button</button>
<button class='btn btn-block' style='background-color: rgb(0, 100, 0); color: rgb(255, 255, 255);'>btn-block button</button>
-上面的例子就是 "inline" （行内）元素和 "block" （块级）元素的区别。
+That illustrates the difference between an "inline" element and a "block" element.
-通过使用行内元素 `span`，你可以把不同的元素放在同一行，甚至能为一个元素的不同部分指定样式。
+By using the inline `span` element, you can put several elements on the same line, and even style different parts of the same line differently.
-把 "Things cats love" 中的 "love" 放入 `span` 标签。然后为其添加 `text-danger` class 来使其文字变成红色。
+Nest the word "love" in your "Things cats love" element below within a `span` element. Then give that `span` the class `text-danger` to make the text red.
-"Top 3 things cats hate" 元素的写法如下：
+Here's how you would do this with the "Top 3 things cats hate" element:
`<p>Top 3 things cats <span class="text-danger">hate:</span></p>`
# --hints--
-`span` 元素应该在 `p` 元素内。
+Your `span` element should be inside your `p` element.
```js
assert($('p span') && $('p span').length > 0);
```
-`span` 元素应该只含有文本 `love`。
+Your `span` element should have just the text `love`.
```js
assert(
@@ -44,13 +44,13 @@ assert(
);
```
-`span` 元素应该含有 class `text-danger`。
+Your `span` element should have class `text-danger`.
```js
assert($('span').hasClass('text-danger'));
```
-确保你的 `span` 元素有一个闭合标签。
+Your `span` element should have a closing tag.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/use-comments-to-clarify-code.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/use-comments-to-clarify-code.md
index 08d96522515..d53b7685098 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/use-comments-to-clarify-code.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/use-comments-to-clarify-code.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aec908857
-title: 使用注释来说明代码
+title: Use Comments to Clarify Code
challengeType: 0
forumTopicId: 18347
dashedName: use-comments-to-clarify-code
@@ -8,35 +8,35 @@ dashedName: use-comments-to-clarify-code
# --description--
-当我们开始使用 jQuery，我们将修改HTML元素，但是实际上我们并不是直接在 HTML 文本中修改。
+When we start using jQuery, we will modify HTML elements without needing to actually change them in HTML.
-我们必须确保让每个人都知道，他们不应该直接修改此页面上这些代码。
+Let's make sure that everyone knows they shouldn't actually modify any of this code directly.
-记住，你可以在 `<!--` 为开始，`-->` 为结束的地方进行评论注释。
+Remember that you can start a comment with `<!--` and end a comment with `-->`
-请你在你的 HTML 顶部加如下一段注释： `Code below this line should not be changed` 。
+Add a comment at the top of your HTML that says `Code below this line should not be changed`
# --hints--
-在添加注释前，在你的 HTML 顶部增加此代码 `<!--`。
+You should start a comment with `<!--` at the top of your HTML.
```js
assert(code.match(/^\s*<!--/));
```
-注释内容应该为 `Code below this line should not be changed`。
+Your comment should have the text `Code below this line should not be changed`.
```js
assert(code.match(/<!--(?!(>|->|.*-->.*this line))\s*.*this line.*\s*-->/gi));
```
-注释应该用 `-->` 进行闭合。
+You should close your comment with `-->`.
```js
assert(code.match(/-->.*\n+.+/g));
```
-注意，注释的开始标签和闭合标签数量应该一一对应，保持数量一致。
+You should have the same number of comment openers and closers.
```js
assert(code.match(/<!--/g).length === code.match(/-->/g).length);
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/use-responsive-design-with-bootstrap-fluid-containers.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/use-responsive-design-with-bootstrap-fluid-containers.md
index d591ae5041a..7de266b0995 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/use-responsive-design-with-bootstrap-fluid-containers.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/use-responsive-design-with-bootstrap-fluid-containers.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9acde08712
-title: 使用 Fluid 容器实现响应式设计
+title: Use Responsive Design with Bootstrap Fluid Containers
challengeType: 0
forumTopicId: 18362
dashedName: use-responsive-design-with-bootstrap-fluid-containers
@@ -8,29 +8,29 @@ dashedName: use-responsive-design-with-bootstrap-fluid-containers
# --description--
-之前，在 freeCodeCamp 的 HTML5 和 CSS 章节中我们构建了一个 Cat Photo App。这次我们将会使用最受欢迎的响应式 CSS 框架 Bootstrap 来美化它。
+In the HTML5 and CSS section of freeCodeCamp we built a Cat Photo App. Now let's go back to it. This time, we'll style it using the popular Bootstrap responsive CSS framework.
-Bootstrap 会根据你的屏幕大小来调节 HTML 元素的大小————因此称为 `Responsive Design`（响应式设计）。
+Bootstrap will figure out how wide your screen is and respond by resizing your HTML elements - hence the name <dfn>responsive design</dfn>.
-通过响应式设计，我们将无需额外设计一个手机版的网页，因为它在任何尺寸的屏幕上看起来都还不错。
+With responsive design, there is no need to design a mobile version of your website. It will look good on devices with screens of any width.
-无论开发什么应用，你都可以通过将以下代码添加到你的 HTML 顶部来引入 Bootstrap 。
+You can add Bootstrap to any app by adding the following code to the top of your HTML:
`<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css" integrity="sha384-BVYiiSIFeK1dGmJRAkycuHAHRg32OmUcww7on3RYdg4Va+PmSTsz/K68vbdEjh4u" crossorigin="anonymous"/>`
-在该案例中，我们已经帮你把相应代码添加到页面中。记住使用 `>` 和 `/>` 闭合 `link` 标签来保证引入成功。
+In this case, we've already added it for you to this page behind the scenes. Note that using either `>` or `/>` to close the `link` tag is acceptable.
-首先，我们应该把所有 HTML 标签放在 class 为 `container-fluid` 的 `div` 元素下（除了 `link` 标签和 `style` 标签）。
+To get started, we should nest all of our HTML (except the `link` tag and the `style` element) in a `div` element with the class `container-fluid`.
# --hints--
-`div` 元素的 class 属性应该为 `container-fluid`。
+Your `div` element should have the class `container-fluid`.
```js
assert($('div').hasClass('container-fluid'));
```
-确保你的 `div` 元素有闭合标签.
+Your `div` element should have a closing tag.
```js
assert(
@@ -40,7 +40,7 @@ assert(
);
```
-确保你已经将所有 HTML 元素内嵌在闭合的 `style` 标签后的 `.container-fluid` 元素中。
+All HTML elements after the closing `style` tag should be nested in `.container-fluid`.
```js
assert($('.container-fluid').children().length >= 8);
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/use-the-bootstrap-grid-to-put-elements-side-by-side.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/use-the-bootstrap-grid-to-put-elements-side-by-side.md
index 1c7d18b7f01..17f0f4c1465 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/use-the-bootstrap-grid-to-put-elements-side-by-side.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/use-the-bootstrap-grid-to-put-elements-side-by-side.md
@@ -1,6 +1,6 @@
---
id: bad88fee1348ce8acef08815
-title: 使用 Bootstrap 网格并排放置元素
+title: Use the Bootstrap Grid to Put Elements Side By Side
challengeType: 0
forumTopicId: 18371
dashedName: use-the-bootstrap-grid-to-put-elements-side-by-side
@@ -8,33 +8,33 @@ dashedName: use-the-bootstrap-grid-to-put-elements-side-by-side
# --description--
-Bootstrap 具有一套基于 12 列的响应式栅格系统————可以轻松实现将多个元素放入一行并指定它们的相对宽度。 Bootstrap 的大部分 class 属性都可以应用在 `div` 元素上。
+Bootstrap uses a responsive 12-column grid system, which makes it easy to put elements into rows and specify each element's relative width. Most of Bootstrap's classes can be applied to a `div` element.
-Bootstrap 的列宽度属性取决于用户的屏幕宽度。 比如，手机有着窄屏幕而笔记本电脑有者更大的屏幕.
+Bootstrap has different column width attributes that it uses depending on how wide the user's screen is. For example, phones have narrow screens, and laptops have wider screens.
-就拿 Bootstrap 的 `col-md-*` class 来说。 在这里， `md` 表示 medium （中等的）， 而 `*` 是一个数字，说明了这个元素占有多少个列宽度。这个例子就是指定了中等大小屏幕（例如笔记本电脑）下元素所占的列宽度。
+Take for example Bootstrap's `col-md-*` class. Here, `md` means medium, and `*` is a number specifying how many columns wide the element should be. In this case, the column width of an element on a medium-sized screen, such as a laptop, is being specified.
-在我们创建的 Cat Photo App 中，我们将使用 `col-xs-*` ， 其中 `xs` 是 extra small 的缩写 (应用于较小的屏幕，比如手机屏幕)， `*` 是你填写的数字，代表一行中，你的元素该占多少列宽。
+In the Cat Photo App that we're building, we'll use `col-xs-*`, where `xs` means extra small (like an extra-small mobile phone screen), and `*` is the number of columns specifying how many columns wide the element should be.
-将我们的 `Like`, `Info` 和 `Delete` 三个按钮并排放入一个 `<div class="row">` 元素中，然后每个按钮都各用一个 `<div class="col-xs-4">` 元素包裹起来。
+Put the `Like`, `Info` and `Delete` buttons side-by-side by nesting all three of them within one `<div class="row">` element, then each of them within a `<div class="col-xs-4">` element.
-当 `div` 元素设置了 `row` class 之后，那几个按钮便会嵌入其中了。
+The `row` class is applied to a `div`, and the buttons themselves can be nested within it.
# --hints--
-所有按钮都需要嵌入到同一个 `div` 元素中， 并且该元素包含 class 属性 `row`。
+Your buttons should all be nested within the same `div` element with the class `row`.
```js
assert($('div.row:has(button)').length > 0);
```
-每个 Bootstrap 按钮都需要嵌入各自的 `div` 元素，并且该元素包含 class 属性 `col-xs-4`。
+Each of your Bootstrap buttons should be nested within its own `div` element with the class `col-xs-4`.
```js
assert($('div.col-xs-4:has(button)').length > 2);
```
-确保每一个 `button` 元素都有一个闭合标签。
+Each of your `button` elements should have a closing tag.
```js
assert(
@@ -44,7 +44,7 @@ assert(
);
```
-确保每一个 `div` 元素都有一个闭合标签。
+Each of your `div` elements should have a closing tag.
```js
assert(
@@ -170,7 +170,7 @@ assert(
<button class="btn btn-block btn-danger">Delete</button>
</div>
</div>
-
+
<p>Things cats love:</p>
<ul>
<li>cat nip</li>
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/warn-your-users-of-a-dangerous-action-with-btn-danger.md b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/warn-your-users-of-a-dangerous-action-with-btn-danger.md
index 0bfa55ed765..fa35d89a563 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/warn-your-users-of-a-dangerous-action-with-btn-danger.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/bootstrap/warn-your-users-of-a-dangerous-action-with-btn-danger.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348ce8acef08814
-title: 使用 btn-danger 提示危险操作
+title: Warn Your Users of a Dangerous Action with btn-danger
challengeType: 0
forumTopicId: 18375
dashedName: warn-your-users-of-a-dangerous-action-with-btn-danger
@@ -8,33 +8,33 @@ dashedName: warn-your-users-of-a-dangerous-action-with-btn-danger
# --description--
-Bootstrap 有着丰富的预定义按钮颜色。 红色 `btn-danger` class 用来提醒用户此行为具有破坏性，比如删除一张猫的图片。
+Bootstrap comes with several pre-defined colors for buttons. The `btn-danger` class is the button color you'll use to notify users that the button performs a destructive action, such as deleting a cat photo.
-创建一个包含文本 "Delete" 的按钮并为它设置 class 为 `btn-danger`。
+Create a button with the text "Delete" and give it the class `btn-danger`.
-记住：你的这些按钮仍然需要 `btn` 和 `btn-block` class。
+Note that these buttons still need the `btn` and `btn-block` classes.
# --hints--
-创建一个包含文本 'Delete' 的 `button` 元素。
+You should create a new `button` element with the text "Delete".
```js
assert(new RegExp('Delete', 'gi').test($('button').text()));
```
-所有 Bootstrap 按钮的 class 属性都应该有 `btn` 和 `btn-block`。
+All of your Bootstrap buttons should have the `btn` and `btn-block` classes.
```js
assert($('button.btn-block.btn').length > 2);
```
-新创建按钮的 class 属性应该有 `btn-danger`。
+Your new button should have the class `btn-danger`.
```js
assert($('button').hasClass('btn-danger'));
```
-确保你所有的 `button` 元素都有一个闭合标签。
+All of your `button` elements should have closing tags.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-25-5-clock.md b/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-25-5-clock.md
index 97a98da8b51..0a332ee2b3b 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-25-5-clock.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-25-5-clock.md
@@ -1,6 +1,6 @@
---
id: bd7158d8c442eddfaeb5bd0f
-title: 构建一个番茄时钟
+title: Build a 25 + 5 Clock
challengeType: 3
forumTopicId: 301373
dashedName: build-a-25--5-clock
@@ -8,71 +8,71 @@ dashedName: build-a-25--5-clock
# --description--
-**目标：** 在 [CodePen.io](https://codepen.io) 上实现一个功能类似 <https://codepen.io/freeCodeCamp/full/XpKrrW> 的 App。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/XpKrrW>.
-在满足以下[需求](https://en.wikipedia.org/wiki/User_story)并能通过所有测试的前提下，你可以根据自己的喜好来美化你的 app。
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story) and get all of the tests to pass. Give it your own personal style.
-你可以使用 HTML、JavaScript、CSS、Bootstrap、SASS、React、Redux、jQuery 来完成这个挑战。但鉴于这个章节的学习内容与前端框架相关，推荐使用一款前端框架（比如 React）来完成这个挑战；不推荐你使用前面没有提到的技术，否则风险自负。我们有计划新增其他前端框架课程，例如 Angular 和 Vue，不过目前还没有这些内容。我们会接受并尽力处理你在使用建议的技术栈过程中遇到的问题。编码愉快！
+You can use any mix of HTML, JavaScript, CSS, Bootstrap, SASS, React, Redux, and jQuery to complete this project. You should use a frontend framework (like React for example) because this section is about learning frontend frameworks. Additional technologies not listed above are not recommended and using them is at your own risk. We are looking at supporting other frontend frameworks like Angular and Vue, but they are not currently supported. We will accept and try to fix all issue reports that use the suggested technology stack for this project. Happy coding!
-**需求 1：** 我应该能看到一个具有`id="break-label"`属性的元素，这个元素的内容应该是一个字符串(例如："Break Length")。
+**User Story #1:** I can see an element with `id="break-label"` that contains a string (e.g. "Break Length").
-**需求 2：** 我应该能看到一个具有`id="session-label"`属性的元素，这个元素的内容应该是一个字符串(例如："Session Length")。
+**User Story #2:** I can see an element with `id="session-label"` that contains a string (e.g. "Session Length").
-**需求 3：** 我应该能看到两个可以点击的元素，他们分别具有如下 id：`id="break-decrement"`和`id="session-decrement"`。
+**User Story #3:** I can see two clickable elements with corresponding IDs: `id="break-decrement"` and `id="session-decrement"`.
-**需求 4：** 我应该能看到两个可以点击的元素，他们分别具有如下 id：`id="break-increment"`和`id="session-increment"`。
+**User Story #4:** I can see two clickable elements with corresponding IDs: `id="break-increment"` and `id="session-increment"`.
-**需求 5：** 我应该能看到一个具有`id="break-length"`属性的元素，这个元素默认展示数值应该为 5（加载后）。
+**User Story #5:** I can see an element with a corresponding `id="break-length"`, which by default (on load) displays a value of 5.
-**需求 6：** 我应该能看到一个具有`id="session-length"`属性的元素，这个元素默认展示数值应该为 25（加载后）。
+**User Story #6:** I can see an element with a corresponding `id="session-length"`, which by default displays a value of 25.
-**需求 7：** 我应该能看到一个具有`id="timer-label"`属性的元素，这个元素包含一个表示当前状态的字符串（例如："Session"）。
+**User Story #7:** I can see an element with a corresponding `id="timer-label"`, that contains a string indicating a session is initialized (e.g. "Session").
-**需求 8：** 我应该能看到一个具有`id="time-left"`属性的元素。注意：暂停或者运行时，该元素的内容应始终以`mm:ss`格式显示（如 25:00）。
+**User Story #8:** I can see an element with corresponding `id="time-left"`. NOTE: Paused or running, the value in this field should always be displayed in `mm:ss` format (i.e. 25:00).
-**需求 9：** 我应该能看到一个具有`id="start_stop"`属性的可点击的元素。
+**User Story #9:** I can see a clickable element with a corresponding `id="start_stop"`.
-**需求 10：** 我应该能看到一个具有`id="reset"`属性的可点击的元素。
+**User Story #10:** I can see a clickable element with a corresponding `id="reset"`.
-**需求 11：** 当我点击 id 属性为`reset`的元素时，应该停止任何运行中的计时器，`id="break-length"`元素中的值应该重新回到`5`，`id="session-length"`元素中的值应该重新回到 25，且`id="time-left"`元素应该重置为默认状态。
+**User Story #11:** When I click the element with the id of `reset`, any running timer should be stopped, the value within `id="break-length"` should return to `5`, the value within `id="session-length"` should return to 25, and the element with `id="time-left"` should reset to it's default state.
-**需求 12：** 当我点击 id 属性为`break-decrement`的元素时，`id="break-length"`元素的值应该减去 1，且我应该能看到更新后的值。
+**User Story #12:** When I click the element with the id of `break-decrement`, the value within `id="break-length"` decrements by a value of 1, and I can see the updated value.
-**需求 13：** 当我点击 id 属性为`break-increment` 的元素时，`id="break-length"`元素的值应该增加 1，且我应该能看到更新后的值。
+**User Story #13:** When I click the element with the id of `break-increment`, the value within `id="break-length"` increments by a value of 1, and I can see the updated value.
-**需求 14：** 当我点击 id 属性为`session-decrement` 的元素时，`id="session-length"`元素的值应该减去 1，且我应该能看到更新后的值。
+**User Story #14:** When I click the element with the id of `session-decrement`, the value within `id="session-length"` decrements by a value of 1, and I can see the updated value.
-**需求 15：** 当我点击 id 属性为`session-increment` 的元素时，`id="session-length"`元素的值应该增加 1，且我应该能看到更新后的值。
+**User Story #15:** When I click the element with the id of `session-increment`, the value within `id="session-length"` increments by a value of 1, and I can see the updated value.
-**需求 16：** 工作或者休息长度不应该可以设置为 <= 0 的值。
+**User Story #16:** I should not be able to set a session or break length to <= 0.
-**需求 17：** 工作或者休息长度比应该可以设置为 > 60 的值。
+**User Story #17:** I should not be able to set a session or break length to > 60.
-**需求 18：** 当我首次点击具有`id="start_stop"`属性的元素时，计时器应该根据`id="session-length"`元素当前显示的值开始运行，即使该值已从原始值 25 递增过或递减过。
+**User Story #18:** When I first click the element with `id="start_stop"`, the timer should begin running from the value currently displayed in `id="session-length"`, even if the value has been incremented or decremented from the original value of 25.
-**需求 19：** 如果计时器正在运行，id 属性为`time-left`的元素应该以`mm:ss`的格式展示剩余的时间（按1递减并且每秒更新一次显示的值）。
+**User Story #19:** If the timer is running, the element with the id of `time-left` should display the remaining time in `mm:ss` format (decrementing by a value of 1 and updating the display every 1000ms).
-**需求 20：** 如果计时器正在运行，当我点击`id="start_stop"`元素时，倒计时应该暂停。
+**User Story #20:** If the timer is running and I click the element with `id="start_stop"`, the countdown should pause.
-**需求 21：** 如果计时器已经暂停，当我点击`id="start_stop"`元素时，倒计时应该从暂停的时间点恢复运行。
+**User Story #21:** If the timer is paused and I click the element with `id="start_stop"`, the countdown should resume running from the point at which it was paused.
-**需求 22：** 当一个工作倒计时结束（注意：计时器必须达到00:00），并且新的倒计时开始运行时，id 属性为`timer-label`的元素应该显示一个表示已经开始休息的字符串。
+**User Story #22:** When a session countdown reaches zero (NOTE: timer MUST reach 00:00), and a new countdown begins, the element with the id of `timer-label` should display a string indicating a break has begun.
-**需求 23：** 当一个工作倒计时结束（注意：计时器必须达到00:00），应该开始一个新的休息倒计时，时间应该从`id="break-length"`元素中当前显示的值开始计算。
+**User Story #23:** When a session countdown reaches zero (NOTE: timer MUST reach 00:00), a new break countdown should begin, counting down from the value currently displayed in the `id="break-length"` element.
-**需求 24：** 当一个休息倒计时结束（注意：计时器必须达到00:00），并且新的倒计时开始运行时，id 属性为`timer-label`的元素应该显示一个表示已经开始工作的字符串。
+**User Story #24:** When a break countdown reaches zero (NOTE: timer MUST reach 00:00), and a new countdown begins, the element with the id of `timer-label` should display a string indicating a session has begun.
-**需求 25：** 当一个休息倒计时结束（注意：计时器必须达到00:00），应该开始一个新的工作倒计时，时间应该从`id="session-length"`元素中当前显示的值开始计算。
+**User Story #25:** When a break countdown reaches zero (NOTE: timer MUST reach 00:00), a new session countdown should begin, counting down from the value currently displayed in the `id="session-length"` element.
-**需求 26：** 当一个倒计时结束（注意：计时器必须达到00:00），应该播放一个表示时间到了的声音提示。这个提示音应该使用 HTML5 的`audio`标签并有一个`id="beep"`属性。
+**User Story #26:** When a countdown reaches zero (NOTE: timer MUST reach 00:00), a sound indicating that time is up should play. This should utilize an HTML5 `audio` tag and have a corresponding `id="beep"`.
-**需求 27：** 具有`id="beep"`属性的音频元素时长应该至少有一秒。
+**User Story #27:** The audio element with `id="beep"` must be 1 second or longer.
-**需求 28：** 当点击 id 属性为`reset`的元素时，id 属性为`beep`的音频元素必须停止播放并回到开头。
+**User Story #28:** The audio element with id of `beep` must stop playing and be rewound to the beginning when the element with the id of `reset` is clicked.
-你可以 fork [这个 CodePen pen 项目](http://codepen.io/freeCodeCamp/pen/MJjpwO) 来构建你的项目。 或者你可以在任何你喜欢的环境中使用以下 CDN 链接来运行测试： `https://gitcdn.link/repo/freeCodeCamp/testable-projects-fcc/master/build/bundle.js`
+You can build your project by forking [this CodePen pen](https://codepen.io/freeCodeCamp/pen/MJjpwO). Or you can use this CDN link to run the tests in any environment you like: `https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`
-一旦你完成了本项目并且该项目所有测试运行通过， 请提交项目的 URL。
+Once you're done, submit the URL to your working project with all its tests passing.
# --solutions--
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-drum-machine.md b/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-drum-machine.md
index ba07bae119d..add89ca96a1 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-drum-machine.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-drum-machine.md
@@ -1,6 +1,6 @@
---
id: 587d7dbc367417b2b2512bae
-title: 构建一台鼓式机器
+title: Build a Drum Machine
challengeType: 3
forumTopicId: 301370
dashedName: build-a-drum-machine
@@ -8,29 +8,29 @@ dashedName: build-a-drum-machine
# --description--
-**目标：** 在 [CodePen.io](https://codepen.io) 上实现一个功能类似 <https://codepen.io/freeCodeCamp/full/MJyNMd> 的 App。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/MJyNMd>.
-在满足以下[需求](https://en.wikipedia.org/wiki/User_story)并能通过所有测试的前提下，你可以根据自己的喜好来美化你的 app。
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story) and get all of the tests to pass. Give it your own personal style.
-你可以使用 HTML、JavaScript、CSS、Bootstrap、SASS、React、Redux、jQuery 来完成这个挑战。但鉴于这个章节的学习内容与前端框架相关，推荐使用一款前端框架（比如 React）来完成这个挑战；不推荐你使用前面没有提到的技术，否则风险自负。我们有计划新增其他前端框架课程，例如 Angular 和 Vue，不过目前还没有这些内容。我们会接受并尽力处理你在使用建议的技术栈过程中遇到的问题。编码愉快！
+You can use any mix of HTML, JavaScript, CSS, Bootstrap, SASS, React, Redux, and jQuery to complete this project. You should use a frontend framework (like React for example) because this section is about learning frontend frameworks. Additional technologies not listed above are not recommended and using them is at your own risk. We are looking at supporting other frontend frameworks like Angular and Vue, but they are not currently supported. We will accept and try to fix all issue reports that use the suggested technology stack for this project. Happy coding!
-**需求 1：** 我应该可以看到一个具有`id="drum-machine"`属性的外层容器，该容器包含了其他所有元素。
+**User Story #1:** I should be able to see an outer container with a corresponding `id="drum-machine"` that contains all other elements.
-**需求 2：** 在具有`#drum-machine`属性的元素内，我应该能看到一个具有 `id="display"` 属性的元素。
+**User Story #2:** Within `#drum-machine` I can see an element with a corresponding `id="display"`.
-**需求 3：** 在具有`#drum-machine`属性的元素内，我应该能看到 9 个可以点击的鼓垫元素，且每个鼓垫元素都应该有一个值为`drum-pad`的 class 属性。一个用于描述触发鼓垫音频片段的特殊 id，以及以下键值之一的文本内容：Q、W、E、A、S、D、Z、X、C。这些鼓垫必须按照以上顺序排列。
+**User Story #3:** Within `#drum-machine` I can see 9 clickable drum pad elements, each with a class name of `drum-pad`, a unique id that describes the audio clip the drum pad will be set up to trigger, and an inner text that corresponds to one of the following keys on the keyboard: Q, W, E, A, S, D, Z, X, C. The drum pads MUST be in this order.
-**需求 4：** 在每一个具有`.drum-pad`属性的元素内，应该有一个具有指向音频片段地址的`src`属性的 HTML5`audio`元素，一个值为`clip`的 class 属性，以及一个 id 属性，它的值应该是其父元素`.drum-pad`的文本内容（例如`id="Q"`、`id="W"`、`id="E"` 等等）。
+**User Story #4:** Within each `.drum-pad`, there should be an HTML5 `audio` element which has a `src` attribute pointing to an audio clip, a class name of `clip`, and an id corresponding to the inner text of its parent `.drum-pad` (e.g. `id="Q"`, `id="W"`, `id="E"` etc.).
-**需求 5：** 当我点击一个具有`.drum-pad`属性的元素时，应该触发它的子元素`audio`包含的音频片段。
+**User Story #5:** When I click on a `.drum-pad` element, the audio clip contained in its child `audio` element should be triggered.
-**需求 6：** 当我按下每一个`.drum-pad`元素的关联键时，应该触发其子元素`audio`包含的音频片段（例如：按下 Q 键应该触发包含字符串 "Q" 的鼓垫，按下 W 键应该触发包含字符串 "W" 的鼓垫等等）。
+**User Story #6:** When I press the trigger key associated with each `.drum-pad`, the audio clip contained in its child `audio` element should be triggered (e.g. pressing the Q key should trigger the drum pad which contains the string "Q", pressing the W key should trigger the drum pad which contains the string "W", etc.).
-**需求 7：** 当触发一个具有`.drum-pad`属性的元素时，`#display`元素内应该展示这个触发元素关联音频片段的描述字符串（每一个字符串都应该是独一无二的）。
+**User Story #7:** When a `.drum-pad` is triggered, a string describing the associated audio clip is displayed as the inner text of the `#display` element (each string must be unique).
-你可以 fork [这个 CodePen pen 项目](http://codepen.io/freeCodeCamp/pen/MJjpwO) 来构建你的项目。或者你可以在任何你喜欢的环境中使用以下 CDN 链接来运行测试：`https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`。
+You can build your project by forking [this CodePen pen](https://codepen.io/freeCodeCamp/pen/MJjpwO). Or you can use this CDN link to run the tests in any environment you like: `https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`
-一旦你完成了本项目并且该项目所有测试运行通过，请提交项目的 URL。
+Once you're done, submit the URL to your working project with all its tests passing.
# --solutions--
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-javascript-calculator.md b/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-javascript-calculator.md
index 672f3fd140c..07a5bf85e8d 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-javascript-calculator.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-javascript-calculator.md
@@ -1,6 +1,6 @@
---
id: bd7158d8c442eddfaeb5bd17
-title: 构建一个 JavaScript 计算器
+title: Build a JavaScript Calculator
challengeType: 3
forumTopicId: 301371
dashedName: build-a-javascript-calculator
@@ -8,52 +8,52 @@ dashedName: build-a-javascript-calculator
# --description--
-**目标：** 在 [CodePen.io](https://codepen.io) 上实现一个功能类似 <https://codepen.io/freeCodeCamp/full/wgGVVX> 的 App。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/wgGVVX>.
-在满足以下[需求](https://en.wikipedia.org/wiki/User_story)并能通过所有测试的前提下，你可以根据自己的喜好来美化你的 app。
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story) and get all of the tests to pass. Give it your own personal style.
-你可以使用 HTML、JavaScript、CSS、Bootstrap、SASS、React、Redux、jQuery 来完成这个挑战。但鉴于这个章节的学习内容与前端框架相关，推荐使用一款前端框架（比如 React）来完成这个挑战；不推荐你使用前面没有提到的技术，否则风险自负。我们有计划新增其他前端框架课程，例如 Angular 和 Vue，不过目前还没有这些内容。我们会接受并尽力处理你在使用建议的技术栈过程中遇到的问题。编码愉快！
+You can use any mix of HTML, JavaScript, CSS, Bootstrap, SASS, React, Redux, and jQuery to complete this project. You should use a frontend framework (like React for example) because this section is about learning frontend frameworks. Additional technologies not listed above are not recommended and using them is at your own risk. We are looking at supporting other frontend frameworks like Angular and Vue, but they are not currently supported. We will accept and try to fix all issue reports that use the suggested technology stack for this project. Happy coding!
-**需求 1：** 我的计算器应该包含一个具有`id="equals"`属性的可以点击的元素，元素的文本内容为`=`（等于符号）。
+**User Story #1:** My calculator should contain a clickable element containing an `=` (equal sign) with a corresponding `id="equals"`.
-**需求 2：** 我的计算器应该包含 10 个具有如下 id 属性的可以点击的元素，每个元素的文本内容对应 0-9 的数字：`id="zero"`、`id="one"`、`id="two"`、`id="three"`、`id="four"`、`id="five"`、`id="six"`、`id="seven"`、`id="eight"`、`id="nine"`。
+**User Story #2:** My calculator should contain 10 clickable elements containing one number each from 0-9, with the following corresponding IDs: `id="zero"`, `id="one"`, `id="two"`, `id="three"`, `id="four"`, `id="five"`, `id="six"`, `id="seven"`, `id="eight"`, and `id="nine"`.
-**需求 3：** 我的计算器应该包含四个可以点击的元素，文本内容对应4个主要数学运算符，且应具有如下 id 属性： `id="add"`, `id="subtract"`, `id="multiply"`, `id="divide"`。
+**User Story #3:** My calculator should contain 4 clickable elements each containing one of the 4 primary mathematical operators with the following corresponding IDs: `id="add"`, `id="subtract"`, `id="multiply"`, `id="divide"`.
-**需求 4：** 计算器应该包含一个可点击的 `.`（小数点）符号，对应的`id="decimal"`。
+**User Story #4:** My calculator should contain a clickable element containing a `.` (decimal point) symbol with a corresponding `id="decimal"`.
-**需求 5：** 我的计算器应该包含一个具有`id="clear"`属性的可以点击的元素。
+**User Story #5:** My calculator should contain a clickable element with an `id="clear"`.
-**需求 6：** 我的计算器应该包含一个用于展示数值的元素，这个元素具有`id="display"`属性。
+**User Story #6:** My calculator should contain an element to display values with a corresponding `id="display"`.
-**需求 7：** 在任何时候按下清除键，都会清空输入和输出的数值并且使计算器回到初始状态；此时在 id 为`display`的元素中应该显示数字 0。
+**User Story #7:** At any time, pressing the clear button clears the input and output values, and returns the calculator to its initialized state; 0 should be shown in the element with the id of `display`.
-**需求 8：** 在我输入数字的同时，我应该能看到我输入的数字展示在 id 为`display`的元素中。
+**User Story #8:** As I input numbers, I should be able to see my input in the element with the id of `display`.
-**需求 9：** 我应该可以在任意顺序下对一串任意长度的数字执行加、减、乘、除操作，并且当我按下`=`时，正确答案应该显示在 id 为`display`的元素中。
+**User Story #9:** In any order, I should be able to add, subtract, multiply and divide a chain of numbers of any length, and when I hit `=`, the correct result should be shown in the element with the id of `display`.
-**需求 10：** 在我输入数字的同时，我的计算器应该不允许一个数字以多个零开头。
+**User Story #10:** When inputting numbers, my calculator should not allow a number to begin with multiple zeros.
-**需求 11：** 当点击小数点元素时，当前展示的数值后面应该添加一个`.`符号；数字中不允许出现两个`.`符号。
+**User Story #11:** When the decimal element is clicked, a `.` should append to the currently displayed value; two `.` in one number should not be accepted.
-**需求 12：** 我可以对包含小数点的数字执行任何四则运算（+、-、\*、/）。
+**User Story #12:** I should be able to perform any operation (+, -, \*, /) on numbers containing decimal points.
-**需求 13：** 如果连续键入了两个及以上的运算符，应该执行最后一次键入的运算符（负数（-）运算符除外）。例如，如果输入 `5 + * 7 = `，结果应该是 `35` （等同于 5 \* 7）；如果输入 `5 * - 5 =`，结果应该是 `-25`（等同于 5 x (-5)）。
+**User Story #13:** If 2 or more operators are entered consecutively, the operation performed should be the last operator entered (excluding the negative (-) sign). For example, if `5 + * 7 =` is entered, the result should be `35` (i.e. 5 \* 7); if `5 * - 5 =` is entered, the result should be `-25` (i.e. 5 x (-5)).
-**需求 14：** 如果在按下`=`符号后继续按一个运算符，则应该在上一次计算结果的基础上进行新的计算。
+**User Story #14:** Pressing an operator immediately following `=` should start a new calculation that operates on the result of the previous evaluation.
-**需求 15：** 当需要四舍五入时，我的计算器应该可以处理结果的精度（注意：处理到多少位没有一个准确的标准，但是你应该能处理类似`2 / 7`这样的计算，使之至少有 4 位小数的精度）。
+**User Story #15:** My calculator should have several decimal places of precision when it comes to rounding (note that there is no exact standard, but you should be able to handle calculations like `2 / 7` with reasonable precision to at least 4 decimal places).
-**注意计算器的逻辑：** 应当注意的是，计算器输入逻辑主要有两种思路：<dfn>立即执行逻辑</dfn>和<dfn>公式逻辑</dfn>。我们的示例使用公式逻辑并遵守运算优先顺序，而立即执行则不然。两者都是可以接受的，但请注意，根据你的选择，你的计算器对于某些算式可能会得到与我们不同的计算结果（参见下面的示例）。只要你的数学计算可以通过其他现实中计算器的验证，那么你的代码就是合理的。
+**Note On Calculator Logic:** It should be noted that there are two main schools of thought on calculator input logic: <dfn>immediate execution logic</dfn> and <dfn>formula logic</dfn>. Our example utilizes formula logic and observes order of operation precedence, immediate execution does not. Either is acceptable, but please note that depending on which you choose, your calculator may yield different results than ours for certain equations (see below example). As long as your math can be verified by another production calculator, please do not consider this a bug.
-**示例：** `3 + 5 x 6 - 2 / 4 =`
+**EXAMPLE:** `3 + 5 x 6 - 2 / 4 =`
-- **立即执行逻辑：** `11.5`
-- **公式／表达式逻辑：** `32.5`
+- **Immediate Execution Logic:** `11.5`
+- **Formula/Expression Logic:** `32.5`
-你可以 fork [这个 CodePen pen 项目](http://codepen.io/freeCodeCamp/pen/MJjpwO) 来构建你的项目。或者你可以在任何你喜欢的环境中使用以下 CDN 链接来运行测试：`https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`。
+You can build your project by forking [this CodePen pen](https://codepen.io/freeCodeCamp/pen/MJjpwO). Or you can use this CDN link to run the tests in any environment you like: `https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`
-一旦你完成了本项目并且该项目所有测试运行通过，请提交项目的 URL。
+Once you're done, submit the URL to your working project with all its tests passing.
# --solutions--
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-markdown-previewer.md b/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-markdown-previewer.md
index ebec5953203..431e4ce92cc 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-markdown-previewer.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-markdown-previewer.md
@@ -1,6 +1,6 @@
---
id: bd7157d8c242eddfaeb5bd13
-title: 构建一个 Markdown 文件预览器
+title: Build a Markdown Previewer
challengeType: 3
forumTopicId: 301372
dashedName: build-a-markdown-previewer
@@ -8,29 +8,29 @@ dashedName: build-a-markdown-previewer
# --description--
-**目标：** 在 [CodePen.io](https://codepen.io) 上实现一个功能类似 <https://codepen.io/freeCodeCamp/full/GrZVVO> 的 App。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/GrZVVO>.
-在满足以下[需求](https://en.wikipedia.org/wiki/User_story)并能通过所有测试的前提下，你可以根据自己的喜好来美化你的 app。
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story) and get all of the tests to pass. Give it your own personal style.
-你可以使用 HTML、JavaScript、CSS、Bootstrap、SASS、React、Redux、jQuery 来完成这个挑战。但鉴于这个章节的学习内容与前端框架相关，推荐使用一款前端框架（比如 React）来完成这个挑战；不推荐你使用前面没有提到的技术，否则风险自负。我们有计划新增其他前端框架课程，例如 Angular 和 Vue，不过目前还没有这些内容。我们会接受并尽力处理你在使用建议的技术栈过程中遇到的问题。编码愉快！
+You can use any mix of HTML, JavaScript, CSS, Bootstrap, SASS, React, Redux, and jQuery to complete this project. You should use a frontend framework (like React for example) because this section is about learning frontend frameworks. Additional technologies not listed above are not recommended and using them is at your own risk. We are looking at supporting other frontend frameworks like Angular and Vue, but they are not currently supported. We will accept and try to fix all issue reports that use the suggested technology stack for this project. Happy coding!
-**需求 1：** 我应该能看到一个具有`id="editor"`属性的`textarea`元素。
+**User Story #1:** I can see a `textarea` element with a corresponding `id="editor"`.
-**需求 2：** 我应该能看到一个具有`id="preview"`属性的元素。
+**User Story #2:** I can see an element with a corresponding `id="preview"`.
-**需求 3：** 当我在具有`#editor`属性的元素内输入文本时，具有`#preview`属性的元素应该同步更新展示我键入的内容。
+**User Story #3:** When I enter text into the `#editor` element, the `#preview` element is updated as I type to display the content of the textarea.
-**需求 4：** 当我在具有`#editor`属性的元素内输入 GitHub 风格的 markdown 内容时，文本应该以 HTML 的形式，把我所键入的内容渲染在具有`#preview`属性的元素中（提示：你不需要自己解析 Markdown——你可以引入一个叫做 Marked 的库来完成这项工作：<https://cdnjs.com/libraries/marked>）。
+**User Story #4:** When I enter GitHub flavored markdown into the `#editor` element, the text is rendered as HTML in the `#preview` element as I type (HINT: You don't need to parse Markdown yourself - you can import the Marked library for this: <https://cdnjs.com/libraries/marked>).
-**需求 5：** 当我的 Markdown 预览器首次加载时，具有`#editor`属性的元素内的默认内容应该包含以下每个种类的至少一段有效的 Markdown 代码：标题（H1 标签）、次级标题（H2 标签）、链接、行内代码、代码块、列表、引用块、图片、加粗文本。
+**User Story #5:** When my markdown previewer first loads, the default text in the `#editor` field should contain valid markdown that represents at least one of each of the following elements: a header (H1 size), a sub header (H2 size), a link, inline code, a code block, a list item, a blockquote, an image, and bolded text.
-**需求 6：** 当我的 Markdown 预览器首次加载时，具有`#editor`属性的元素内容应该以 HTML 的形式渲染在具有`#preview`属性的元素中。
+**User Story #6:** When my markdown previewer first loads, the default markdown in the `#editor` field should be rendered as HTML in the `#preview` element.
-**可选需求（你无需通过这项测试）：** 我的 Markdown 预览器能够解析回车符并且将他们以`br`（换行）元素的形式渲染出来。
+**Optional Bonus (you do not need to make this test pass):** My markdown previewer interprets carriage returns and renders them as `br` (line break) elements.
-你可以 fork [这个 CodePen pen 项目](http://codepen.io/freeCodeCamp/pen/MJjpwO) 来构建你的项目。或者你可以在任何你喜欢的环境中使用以下 CDN 链接来运行测试：`https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`。
+You can build your project by forking [this CodePen pen](https://codepen.io/freeCodeCamp/pen/MJjpwO). Or you can use this CDN link to run the tests in any environment you like: `https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`
-一旦你完成了本项目并且该项目所有测试运行通过，请提交项目的 URL。
+Once you're done, submit the URL to your working project with all its tests passing.
# --solutions--
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-random-quote-machine.md b/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-random-quote-machine.md
index bb5784996b7..82093285ea6 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-random-quote-machine.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/front-end-libraries-projects/build-a-random-quote-machine.md
@@ -1,6 +1,6 @@
---
id: bd7158d8c442eddfaeb5bd13
-title: 构建一个随机引语生成器
+title: Build a Random Quote Machine
challengeType: 3
forumTopicId: 301374
dashedName: build-a-random-quote-machine
@@ -8,39 +8,39 @@ dashedName: build-a-random-quote-machine
# --description--
-**目标：** 在 [CodePen.io](https://codepen.io) 上实现一个功能类似 <https://codepen.io/freeCodeCamp/full/qRZeGZ> 的 App。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/qRZeGZ>.
-在满足以下[需求](https://en.wikipedia.org/wiki/User_story)并能通过所有测试的前提下，你可以根据自己的喜好来美化你的 app。
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story) and get all of the tests to pass. Give it your own personal style.
-你可以使用 HTML、JavaScript、CSS、Bootstrap、SASS、React、Redux、jQuery 来完成这个挑战。但鉴于这个章节的学习内容与前端框架相关，推荐使用一款前端框架（比如 React）来完成这个挑战；不推荐你使用前面没有提到的技术，否则风险自担。我们有计划新增其他前端框架课程，例如 Angular 和 Vue，不过目前还没有这些内容。我们会接受并尽力处理你在使用建议的技术栈过程中遇到的问题。编码愉快！
+You can use any mix of HTML, JavaScript, CSS, Bootstrap, SASS, React, Redux, and jQuery to complete this project. You should use a frontend framework (like React for example) because this section is about learning frontend frameworks. Additional technologies not listed above are not recommended and using them is at your own risk. We are looking at supporting other frontend frameworks like Angular and Vue, but they are not currently supported. We will accept and try to fix all issue reports that use the suggested technology stack for this project. Happy coding!
-**需求 1：** 我应该能看到一个具有`id="quote-box"`属性的包裹元素。
+**User Story #1:** I can see a wrapper element with a corresponding `id="quote-box"`.
-**需求 2：** 在`#quote-box`元素内，我应该能看到一个具有`id="text"`属性的元素。
+**User Story #2:** Within `#quote-box`, I can see an element with a corresponding `id="text"`.
-**需求 3：** 在`#quote-box`元素内，我应该能看到一个具有`id="author"`属性的元素。
+**User Story #3:** Within `#quote-box`, I can see an element with a corresponding `id="author"`.
-**需求 4：** 在`#quote-box`元素内，我应该能看到一个具有`id="new-quote"`属性的可点击元素。
+**User Story #4:** Within `#quote-box`, I can see a clickable element with a corresponding `id="new-quote"`.
-**需求 5：** 在`#quote-box`元素内，我应该能看到一个具有`id="tweet-quote"`属性的可点击 `a` 元素。
+**User Story #5:** Within `#quote-box`, I can see a clickable `a` element with a corresponding `id="tweet-quote"`.
-**需求 6：** 首次加载时，我的 App 应该在具有`id="text"`属性的元素内展示一条随机引语。
+**User Story #6:** On first load, my quote machine displays a random quote in the element with `id="text"`.
-**需求 7：** 首次加载时，我的 App 应该在具有`id="author"`属性的元素内展示该条随机引语的作者。
+**User Story #7:** On first load, my quote machine displays the random quote's author in the element with `id="author"`.
-**需求 8：** 当点击具有`#new-quote`属性的按钮时，我的 App 应该得到一条新的引语并在具有`#text`属性的元素内展示出来。
+**User Story #8:** When the `#new-quote` button is clicked, my quote machine should fetch a new quote and display it in the `#text` element.
-**需求 9：** 当点击具有`#new-quote`属性的按钮时，我的 App 应该得到新引语的作者并在具有`#author`属性的元素内展示出来。
+**User Story #9:** My quote machine should fetch the new quote's author when the `#new-quote` button is clicked and display it in the `#author` element.
-**需求 10：** 我应该可以通过点击具有`#tweet-quote`属性的`a`标签将当前展示的引语发送推特。该`a`标签的`href`属性应该是`"twitter.com/intent/tweet"`以便成功发送。
+**User Story #10:** I can tweet the current quote by clicking on the `#tweet-quote``a` element. This `a` element should include the `"twitter.com/intent/tweet"` path in its `href` attribute to tweet the current quote.
-**需求 11：** 具有`#quote-box`属性的包裹元素应该水平居中。请在浏览器缩放尺寸为 100% 且页面窗口最大化时运行测试。
+**User Story #11:** The `#quote-box` wrapper element should be horizontally centered. Please run tests with browser's zoom level at 100% and page maximized.
-你可以 fork [这个 CodePen pen 项目](http://codepen.io/freeCodeCamp/pen/MJjpwO)来构建你的项目。或者你可以在任何你喜欢的环境中使用以下 CDN 链接来运行测试：`https://gitcdn.link/repo/freeCodeCamp/testable-projects-fcc/master/build/bundle.js`。
+You can build your project by forking [this CodePen pen](https://codepen.io/freeCodeCamp/pen/MJjpwO). Or you can use this CDN link to run the tests in any environment you like: `https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`
-一旦你完成了本项目并且该项目所有测试运行通过，请提交项目的 URL。
+Once you're done, submit the URL to your working project with all its tests passing.
-**注意：** Twitter 不允许在 iframe 里加载链接。如果你的 tweet 不能加载，尝试在 `#tweet-quote` 元素上使用 `target="_blank"` 或者 `target="_top"` 属性。`target="_top"` 会替换当前 tab 页的内容，所以确保当前内容已经保存了。
+**Note:** Twitter does not allow links to be loaded in an iframe. Try using the `target="_blank"` or `target="_top"` attribute on the `#tweet-quote` element if your tweet won't load. `target="_top"` will replace the current tab so make sure your work is saved.
# --solutions--
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/jquery/change-text-inside-an-element-using-jquery.md b/curriculum/challenges/chinese/03-front-end-libraries/jquery/change-text-inside-an-element-using-jquery.md
index 900a40ecec2..2e02864f610 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/jquery/change-text-inside-an-element-using-jquery.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/jquery/change-text-inside-an-element-using-jquery.md
@@ -1,6 +1,6 @@
---
id: 564944c91be2204b269d51e3
-title: 使用 jQuery 更改元素内部的文本
+title: Change Text Inside an Element Using jQuery
challengeType: 6
forumTopicId: 16773
dashedName: change-text-inside-an-element-using-jquery
@@ -8,25 +8,25 @@ dashedName: change-text-inside-an-element-using-jquery
# --description--
-你能用 jQuery 改变起始标签和结束标签之间的文本，甚至改变 HTML 标签。
+Using jQuery, you can change the text between the start and end tags of an element. You can even change HTML markup.
-jQuery 有一个`.html()`函数，你能用其在标签里添加 HTML 标签和文本，函数提供的内容将完全替换之前标签的内容。
+jQuery has a function called `.html()` that lets you add HTML tags and text within an element. Any content previously within the element will be completely replaced with the content you provide using this function.
-下面的代码效果是重写并强调标题文本：
+Here's how you would rewrite and emphasize the text of our heading:
`$("h3").html("<em>jQuery Playground</em>");`
-类似的，jQuery 有一个`.text()`函数，其改变文本而不增加标签。换而言之，该函数会忽略所有传递过来的 HTML 标签，并将其视为用来替换原文本的文本。
+jQuery also has a similar function called `.text()` that only alters text without adding tags. In other words, this function will not evaluate any HTML tags passed to it, but will instead treat it as the text you want to replace the existing content with.
-请强调 id 为`target4`的按钮的文本。
+Change the button with id `target4` by emphasizing its text.
-请查看此[链接](https://www.freecodecamp.org/news/html-elements-explained-what-are-html-tags/#em-elemen)来了解更多`<i>`和`<em>`的区别和用法。
+[View our news article for <em>](https://www.freecodecamp.org/news/html-elements-explained-what-are-html-tags/#em-element) to learn the difference between `<i>` and `<em>` and their uses.
-注意，`<i>`标签虽然传统上用来强调文本，但此后常用作图标的标签；`<em>`标签作为强调标签现在已被广泛接受。两者都可以应对本次挑战。
+Note that while the `<i>` tag has traditionally been used to emphasize text, it has since been adopted for use as a tag for icons. The `<em>` tag is now widely accepted as the tag for emphasis. Either will work for this challenge.
# --hints--
-通过添加 HTML 标签强调`target4`按钮中的文本。
+You should emphasize the text in your `target4` button by adding HTML tags.
```js
assert.isTrue(
@@ -34,25 +34,25 @@ assert.isTrue(
);
```
-确保文本不乱。
+The text should otherwise remain unchanged.
```js
assert($('#target4') && $('#target4').text().trim() === '#target4');
```
-不改变其他任何文本内容。
+You should not alter any other text.
```js
assert.isFalse(/<em>|<i>/gi.test($('h3').html()));
```
-确保使用`.html()`方法而不是`.text()`方法。
+You should be using `.html()` and not `.text()`.
```js
assert(code.match(/\.html\(/g));
```
-确保用 jQuery 选取`button id='target4'`。
+You should select `button id="target4"` with jQuery.
```js
assert(code.match(/\$\(\s*?(\"|\')#target4(\"|\')\s*?\)\.html\(/));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/jquery/change-the-css-of-an-element-using-jquery.md b/curriculum/challenges/chinese/03-front-end-libraries/jquery/change-the-css-of-an-element-using-jquery.md
index 686009e71c8..aac990940e2 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/jquery/change-the-css-of-an-element-using-jquery.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/jquery/change-the-css-of-an-element-using-jquery.md
@@ -1,38 +1,39 @@
---
id: bad87fee1348bd9aed908826
-title: 使用 jQuery 更改元素的 CSS
+title: Change the CSS of an Element Using jQuery
challengeType: 6
forumTopicId: 16776
required:
- - link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
+ -
+ link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
dashedName: change-the-css-of-an-element-using-jquery
---
# --description--
-我们也能用 jQuery 直接改变 HTML 标签的 CSS。
+We can also change the CSS of an HTML element directly with jQuery.
-jQuery 有一个`.css()`方法，其能改变标签的 CSS。
+jQuery has a function called `.css()` that allows you to change the CSS of an element.
-下面的代码效果是把颜色变蓝：
+Here's how we would change its color to blue:
`$("#target1").css("color", "blue");`
-这与通常的 CSS 声明略有不同，因为这个 CSS 属性和它的值在英文引号里，并且它们用逗号而不是冒号间隔开。
+This is slightly different from a normal CSS declaration, because the CSS property and its value are in quotes, and separated with a comma instead of a colon.
-删除你的jQuery选择器，并留下空的`document ready function`。
+Delete your jQuery selectors, leaving an empty `document ready function`.
-请选择`target1`并将其颜色变为红色（red）。
+Select `target1` and change its color to red.
# --hints--
-`target1`标签应该有红色文本。
+Your `target1` element should have red text.
```js
assert($('#target1').css('color') === 'rgb(255, 0, 0)');
```
-仅用 jQuery 给标签添加类。
+You should only use jQuery to add these classes to the element.
```js
assert(!code.match(/class.*animated/g));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/jquery/clone-an-element-using-jquery.md b/curriculum/challenges/chinese/03-front-end-libraries/jquery/clone-an-element-using-jquery.md
index ce6446bf556..73307848d45 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/jquery/clone-an-element-using-jquery.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/jquery/clone-an-element-using-jquery.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aed508826
-title: 使用 jQuery 克隆元素
+title: Clone an Element Using jQuery
challengeType: 6
forumTopicId: 16780
dashedName: clone-an-element-using-jquery
@@ -8,33 +8,33 @@ dashedName: clone-an-element-using-jquery
# --description--
-除了移动标签，也可以把元素从一个地方复制到另一地方。
+In addition to moving elements, you can also copy them from one place to another.
-jQuery 有一个`clone()`方法，可以复制标签。
+jQuery has a function called `clone()` that makes a copy of an element.
-例如，如果想把`target2`从`left-well`复制到`right-well`，可以设置如下：
+For example, if we wanted to copy `target2` from our `left-well` to our `right-well`, we would use:
`$("#target2").clone().appendTo("#right-well");`
-你是否注意到这两个 jQuery 方法连在一起了吗？这被称为`链式调用（function chaining）`，是一种用 jQuery 实现效果的简便方法。
+Did you notice this involves sticking two jQuery functions together? This is called <dfn>function chaining</dfn> and it's a convenient way to get things done with jQuery.
-克隆`target5`标签并附加到`left-well`。
+Clone your `target5` element and append it to your `left-well`.
# --hints--
-`target5`标签应该在`right-well`内。
+Your `target5` element should be inside your `right-well`.
```js
assert($('#right-well').children('#target5').length > 0);
```
-克隆`target5`标签并放在`left-well`内。
+A copy of your `target5` element should also be inside your `left-well`.
```js
assert($('#left-well').children('#target5').length > 0);
```
-仅用 jQuery 移动这些标签。
+You should only use jQuery to move these elements.
```js
assert(!code.match(/class.*animated/g));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/jquery/delete-your-jquery-functions.md b/curriculum/challenges/chinese/03-front-end-libraries/jquery/delete-your-jquery-functions.md
index f977e807a7a..b1ed6c05f29 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/jquery/delete-your-jquery-functions.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/jquery/delete-your-jquery-functions.md
@@ -1,46 +1,47 @@
---
id: bad87fee1348bd9aeda08726
-title: 删除 jQuery 函数
+title: Delete Your jQuery Functions
challengeType: 6
forumTopicId: 17561
required:
- - link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
+ -
+ link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
dashedName: delete-your-jquery-functions
---
# --description--
-这些动画开始看起来很酷，但是有时可能会让用户分心。
+These animations were cool at first, but now they're getting kind of distracting.
-请删除`document ready function`内的三个 jQuery 函数，但保留`document ready function`本身。
+Delete all three of these jQuery functions from your `document ready function`, but leave your `document ready function` itself intact.
# --hints--
-删除`document ready function`中的三个 jQuery 函数。
+All three of your jQuery functions should be deleted from your `document ready function`.
```js
assert(code.match(/\{\s*\}\);/g));
```
-保持`script`标签不变。
+You should leave your `script` element intact.
```js
assert(code.match(/<script>/g));
```
-保持`$(document).ready(function() {`在`script`标签的开头不变。
+You should leave your `$(document).ready(function() {` at the beginning of your `script` element.
```js
assert(code.match(/\$\(document\)\.ready\(function\(\)\s?\{/g));
```
-保持 'document ready function' 用`})`闭合。
+You should leave your "document ready function" closing `})` intact.
```js
assert(code.match(/.*\s*\}\);/g));
```
-保持`script`标签闭合。
+You should leave your `script` element closing tag intact.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/jquery/disable-an-element-using-jquery.md b/curriculum/challenges/chinese/03-front-end-libraries/jquery/disable-an-element-using-jquery.md
index dd5f52bfd4e..bdfefa4578e 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/jquery/disable-an-element-using-jquery.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/jquery/disable-an-element-using-jquery.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aed808826
-title: 使用 jQuery 禁用元素
+title: Disable an Element Using jQuery
challengeType: 6
forumTopicId: 17563
dashedName: disable-an-element-using-jquery
@@ -8,21 +8,21 @@ dashedName: disable-an-element-using-jquery
# --description--
-你还能用 jQuery 改变 HTML 标签的非 CSS 属性,例如:能禁用按钮。
+You can also change the non-CSS properties of HTML elements with jQuery. For example, you can disable buttons.
-当你禁用按钮时，它将变成灰色并无法点击。
+When you disable a button, it will become grayed-out and can no longer be clicked.
-jQuery 有一个`.prop()`方法，你可以用其调整标签的属性。
+jQuery has a function called `.prop()` that allows you to adjust the properties of elements.
-下面的代码效果是禁用所有的按钮：
+Here's how you would disable all buttons:
`$("button").prop("disabled", true);`
-请仅禁用`target1`按钮。
+Disable only the `target1` button.
# --hints--
-禁用`target1`按钮。
+Your `target1` button should be disabled.
```js
assert(
@@ -32,13 +32,13 @@ assert(
);
```
-不禁用其他的按钮。
+No other buttons should be disabled.
```js
assert($('#target2') && !$('#target2').prop('disabled'));
```
-仅用 jQuery 给标签添加类。
+You should only use jQuery to add these classes to the element.
```js
assert(!code.match(/disabled[^<]*>/g));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/jquery/learn-how-script-tags-and-document-ready-work.md b/curriculum/challenges/chinese/03-front-end-libraries/jquery/learn-how-script-tags-and-document-ready-work.md
index be8c6d1b537..3835dfeca8f 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/jquery/learn-how-script-tags-and-document-ready-work.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/jquery/learn-how-script-tags-and-document-ready-work.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9acdd08826
-title: 了解 script 和 document.ready 是如何工作的
+title: Learn How Script Tags and Document Ready Work
challengeType: 6
forumTopicId: 18224
dashedName: learn-how-script-tags-and-document-ready-work
@@ -8,23 +8,23 @@ dashedName: learn-how-script-tags-and-document-ready-work
# --description--
-现在我们已经准备好学习有史以来最受欢迎的 JavaScript 框架——jQuery 了。
+Now we're ready to learn jQuery, the most popular JavaScript tool of all time.
-在使用 jQuery 之前，我们需要在 HTML 页面中添加一些东西。
+Before we can start using jQuery, we need to add some things to our HTML.
-首先，在页面顶部添加`script`标签，记得在后面为`script`标签添加结束标签。
+First, add a `script` element at the top of your page. Be sure to close it on the following line.
-浏览器在`script`标签中运行所有的 JavaScript 脚本包括 jQuery。
+Your browser will run any JavaScript inside a `script` element, including jQuery.
-在`script`标签中添加代码`$(document).ready(function() {`。然后在后面（仍在该`script`标签内）用`});`闭合它。
+Inside your `script` element, add this code: `$(document).ready(function() {` to your `script`. Then close it on the following line (still inside your `script` element) with: `});`
-稍后我们将详细介绍`functions`，现在需要知道的是，只要浏览器加载页面，`function`中放入的代码就会运行。
+We'll learn more about `functions` later. The important thing to know is that code you put inside this `function` will run as soon as your browser has loaded your page.
-有一点很重要，如果没有`document ready function`，你的代码将在 HTML 页面呈现之前运行，这将导致错误。
+This is important because without your `document ready function`, your code may run before your HTML is rendered, which would cause bugs.
# --hints--
-创建一个`script`标签，确保其有效并具有闭合标签。
+You should create a `script` element making sure it is valid and has a closing tag.
```js
assert(
@@ -39,7 +39,7 @@ assert(
);
```
-在`script`的开头添加`$(document).ready(function() {`。
+You should add `$(document).ready(function() {` to the beginning of your `script` element.
```js
assert(
@@ -49,7 +49,7 @@ assert(
);
```
-用`});`闭合`$(document).ready(function() {`函数。
+You should close your `$(document).ready(function() {` function with `});`
```js
assert(code.match(/\n*?\s*?\}\s*?\);/g));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/jquery/remove-an-element-using-jquery.md b/curriculum/challenges/chinese/03-front-end-libraries/jquery/remove-an-element-using-jquery.md
index de5ff768e74..f4517e844f9 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/jquery/remove-an-element-using-jquery.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/jquery/remove-an-element-using-jquery.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aed708826
-title: 使用 jQuery 删除元素
+title: Remove an Element Using jQuery
challengeType: 6
forumTopicId: 18262
dashedName: remove-an-element-using-jquery
@@ -8,15 +8,15 @@ dashedName: remove-an-element-using-jquery
# --description--
-现在学习用 jQuery 从页面移除 HTML 标签。
+Now let's remove an HTML element from your page using jQuery.
-jQuery 有一个`.remove()`方法，能完全移除 HTML 标签。
+jQuery has a function called `.remove()` that will remove an HTML element entirely
-请用`.remove()`方法从页面移除`target4`标签。
+Remove element `target4` from the page by using the `.remove()` function.
# --hints--
-用 jQuery 从页面中移除`target4`标签。
+You should use jQuery to remove your `target4` element from your page.
```js
assert(
@@ -24,7 +24,7 @@ assert(
);
```
-仅用 jQuery 移除该标签。
+You should only use jQuery to remove this element.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/jquery/remove-classes-from-an-element-with-jquery.md b/curriculum/challenges/chinese/03-front-end-libraries/jquery/remove-classes-from-an-element-with-jquery.md
index 4424c770b0e..aeaa523467e 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/jquery/remove-classes-from-an-element-with-jquery.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/jquery/remove-classes-from-an-element-with-jquery.md
@@ -1,38 +1,39 @@
---
id: bad87fee1348bd9aed918626
-title: 使用 jQuery 从元素中移除 class
+title: Remove Classes from an Element with jQuery
challengeType: 6
forumTopicId: 18264
required:
- - link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
+ -
+ link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
dashedName: remove-classes-from-an-element-with-jquery
---
# --description--
-和用 jQuery 的`addClass()`方法给标签添加类一样，也可以利用 jQuery 的`removeClass()`方法移除他们。
+In the same way you can add classes to an element with jQuery's `addClass()` function, you can remove them with jQuery's `removeClass()` function.
-下面的代码效果是为特定的按钮执行上面的操作：
+Here's how you would do this for a specific button:
`$("#target2").removeClass("btn-default");`
-请把所有`button`标签的`btn-default`类移除。
+Let's remove the `btn-default` class from all of our `button` elements.
# --hints--
-移除所有的`button`标签的`btn-default`属性。
+The `btn-default` class should be removed from all of your `button` elements.
```js
assert($('.btn-default').length === 0);
```
-仅用 jQuery 从标签中移除类。
+You should only use jQuery to remove this class from the element.
```js
assert(code.match(/btn btn-default/g));
```
-仅移除`btn-default`类。
+You should only remove the `btn-default` class.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-a-specific-child-of-an-element-using-jquery.md b/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-a-specific-child-of-an-element-using-jquery.md
index ee8b974d07a..72721a290f8 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-a-specific-child-of-an-element-using-jquery.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-a-specific-child-of-an-element-using-jquery.md
@@ -1,30 +1,31 @@
---
id: bad87fee1348bd9aed108826
-title: 使用 jQuery 选择元素的特定子元素
+title: Target a Specific Child of an Element Using jQuery
challengeType: 6
forumTopicId: 18315
required:
- - link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
+ -
+ link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
dashedName: target-a-specific-child-of-an-element-using-jquery
---
# --description--
-你已经看到了为什么 id 属性对于 jQuery 选择器选取标签非常方便，但这并不适用于所有情景。
+You've seen why id attributes are so convenient for targeting with jQuery selectors. But you won't always have such neat ids to work with.
-幸运的是，jQuery 有一些其他的方法可以选取正确的标签。
+Fortunately, jQuery has some other tricks for targeting the right elements.
-jQuery 可以用`CSS 选择器（CSS Selectors）`选取标签。`target:nth-child(n)`CSS 选择器可以选取所有的第 n 个标签并设置目标属性和目标样式。
+jQuery uses CSS Selectors to target elements. The `target:nth-child(n)` CSS selector allows you to select all the nth elements with the target class or element type.
-下面的代码展示了给每个区域（well）的第 3 个标签设置`bounce`类：
+Here's how you would give the third element in each well the bounce class:
`$(".target:nth-child(3)").addClass("animated bounce");`
-请给每个区域（well）的第 2 个标签设置`bounce`类，必须用`target`类选取标签。
+Make the second child in each of your well elements bounce. You must select the elements' children with the `target` class.
# --hints--
-`target`标签中的第二个标签应该有弹性的动画效果。
+The second element in your `target` elements should bounce.
```js
assert(
@@ -33,19 +34,19 @@ assert(
);
```
-应该仅两个标签有弹性的动画效果。
+Only two elements should bounce.
```js
assert($('.animated.bounce').length === 2);
```
-应该用`:nth-child()`选择器修改这些标签。
+You should use the `:nth-child()` selector to modify these elements.
```js
assert(code.match(/\:nth-child\(/g));
```
-仅用 jQuery 给标签添加类。
+You should only use jQuery to add these classes to the element.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-elements-by-class-using-jquery.md b/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-elements-by-class-using-jquery.md
index c72c4f88d81..caf2a759898 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-elements-by-class-using-jquery.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-elements-by-class-using-jquery.md
@@ -1,38 +1,39 @@
---
id: bad87fee1348bd9aedc08826
-title: 使用 jQuery 配合 class 选择器选择元素
+title: Target Elements by Class Using jQuery
challengeType: 6
forumTopicId: 18316
required:
- - link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
+ -
+ link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
dashedName: target-elements-by-class-using-jquery
---
# --description--
-我们如何使所有的`button`标签有弹性的动画效果？我们用`$("button")`选取所有的`button`标签，并用`.addClass("animated bounce");`给其添加一些 CSS 属性。
+You see how we made all of your `button` elements bounce? We selected them with `$("button")`, then we added some CSS classes to them with `.addClass("animated bounce");`.
-jQuery 的`.addClass()`方法用来给标签添加类。
+You just used jQuery's `.addClass()` function, which allows you to add classes to elements.
-首先，我们使用`$(".well")`选取类为`well`的`div`标签。
+First, let's target your `div` elements with the class `well` by using the `$(".well")` selector.
-值得注意的是，和 CSS 声明一样，在类名前需要添加`.`。
+Note that, just like with CSS declarations, you type a `.` before the class's name.
-然后，用 jQuery 的`.addClass()`方法添加`animated`和`shake`类。
+Then use jQuery's `.addClass()` function to add the classes `animated` and `shake`.
-例如，在`document ready function`中添加下面的代码，能使所有类为`text-primary`的标签抖动：
+For example, you could make all the elements with the class `text-primary` shake by adding the following to your `document ready function`:
`$(".text-primary").addClass("animated shake");`
# --hints--
-用 jQuery 的`addClass()`方法给所有类为`well`的标签添加`animated`和`shake`类。
+You should use the jQuery `addClass()` function to give the classes `animated` and `shake` to all your elements with the class `well`.
```js
assert($('.well').hasClass('animated') && $('.well').hasClass('shake'));
```
-仅用 jQuery 给标签添加类。
+You should only use jQuery to add these classes to the element.
```js
assert(!code.match(/class\.\*animated/g));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-elements-by-id-using-jquery.md b/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-elements-by-id-using-jquery.md
index f987183c9fa..ff20eab9476 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-elements-by-id-using-jquery.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-elements-by-id-using-jquery.md
@@ -1,36 +1,37 @@
---
id: bad87fee1348bd9aeda08826
-title: 使用 jQuery 配合 id 选择器选择元素
+title: Target Elements by id Using jQuery
challengeType: 6
forumTopicId: 18317
required:
- - link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
+ -
+ link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
dashedName: target-elements-by-id-using-jquery
---
# --description--
-你也能通过 id 属性选取标签。
+You can also target elements by their id attributes.
-首先，用`$("#target3")`选择器选取 id 为`target3`的`button`标签。
+First target your `button` element with the id `target3` by using the `$("#target3")` selector.
-注意，和 CSS 属性一样，在 id 名前需要添加`#`。
+Note that, just like with CSS declarations, you type a `#` before the id's name.
-然后，用 jQuery 的`.addClass()`方法添加`animated`和`fadeOut`类。
+Then use jQuery's `.addClass()` function to add the classes `animated` and `fadeOut`.
-下面的代码的效果是使 id 为`target6`的`button`标签淡出：
+Here's how you'd make the `button` element with the id `target6` fade out:
`$("#target6").addClass("animated fadeOut")`.
# --hints--
-用 jQuery 的`addClass()`方法给`id`为`target3`的`button`标签添加`animated`类。
+You should select the `button` element with the `id` of `target3` and use the jQuery `addClass()` function to give it the class of `animated`.
```js
assert($('#target3').hasClass('animated'));
```
-用 jQuery 的`addClass()`方法给`id`为`target3`的标签的类添加`fadeOut`类。
+You should target the element with the id `target3` and use the jQuery `addClass()` function to give it the class `fadeOut`.
```js
assert(
@@ -39,7 +40,7 @@ assert(
);
```
-仅用 jQuery 给标签设置类。
+You should only use jQuery to add these classes to the element.
```js
assert(!code.match(/class.*animated/g));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-even-elements-using-jquery.md b/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-even-elements-using-jquery.md
index 1320c9b1173..71d4b9738ee 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-even-elements-using-jquery.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-even-elements-using-jquery.md
@@ -1,28 +1,29 @@
---
id: bad87fee1348bd9aed008826
-title: 使用 jQuery 选择偶数元素
+title: Target Even Elements Using jQuery
challengeType: 6
forumTopicId: 18318
required:
- - link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
+ -
+ link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
dashedName: target-even-elements-using-jquery
---
# --description--
-你也可以用基于位置的`:odd`和`:even`选择器选取标签。
+You can also target elements based on their positions using `:odd` or `:even` selectors.
-注意，jQuery 是`零索引（zero-indexed）`的，这意味着第 1 个标签的位置编号是`0`。这有点混乱和反常——`:odd`表示选择第 2 个标签（位置编号 1），第 4 个标签（位置编号 3）……等等，以此类推。
+Note that jQuery is zero-indexed which means the first element in a selection has a position of 0. This can be a little confusing as, counter-intuitively, `:odd` selects the second element (position 1), fourth element (position 3), and so on.
-下面的代码展示了选取所有的奇标签并设置类：
+Here's how you would target all the odd elements with class `target` and give them classes:
`$(".target:odd").addClass("animated shake");`
-请尝试选取所有`target`标签的偶标签并给他们设置`animated`和`shake`类。要考虑到**偶（even）**指的是标签位置编号基于`0`的系统。
+Try selecting all the even `target` elements and giving them the classes of `animated` and `shake`. Remember that **even** refers to the position of elements with a zero-based system in mind.
# --hints--
-所有的`target`标签应该抖动。
+All of the `target` elements that jQuery considers to be even should shake.
```js
assert(
@@ -30,13 +31,13 @@ assert(
);
```
-应该用`:even`选择器修改这些标签。
+You should use the `:even` selector to modify these elements.
```js
assert(code.match(/\:even/g));
```
-仅用 jQuery 给标签添加类。
+You should only use jQuery to add these classes to the element.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-html-elements-with-selectors-using-jquery.md b/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-html-elements-with-selectors-using-jquery.md
index 25c5e4ae4dc..bd06f7a7b88 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-html-elements-with-selectors-using-jquery.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-html-elements-with-selectors-using-jquery.md
@@ -1,48 +1,47 @@
---
id: bad87fee1348bd9bedc08826
-title: 使用 jQuery 配合元素选择器选择元素
+title: Target HTML Elements with Selectors Using jQuery
challengeType: 6
forumTopicId: 18319
required:
- - link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
+ -
+ link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
dashedName: target-html-elements-with-selectors-using-jquery
---
# --description--
-接下来我们学习`document ready function`。
+Now we have a `document ready function`.
-首先，我们完成第一个 jQuery 语句。所有的 jQuery 函数以`$`开头，这个符号通常被称为`美元符号（dollar sign operator）`或`bling`。
+Now let's write our first jQuery statement. All jQuery functions start with a `$`, usually referred to as a dollar sign operator, or as bling.
-jQuery 通常选取并操作带有`选择器（selector）`的 HTML 标签。
+jQuery often selects an HTML element with a <dfn>selector</dfn>, then does something to that element.
-例如，如果要所有`button`有弹性的动画效果，只需在`document ready function`中添加如下代码即可：
+For example, let's make all of your `button` elements bounce. Just add this code inside your document ready function:
`$("button").addClass("animated bounce");`
-请注意，为了能在编辑器里直接使用，我们已经为你在后台引入了 jQuery 库和 Animate.css 库。因此，你只需要通过 jQuery 给`button`元素添加`bounce`类就可以了。
+Note that we've already included both the jQuery library and the Animate.css library in the background so that you can use them in the editor. So you are using jQuery to apply the Animate.css `bounce` class to your `button` elements.
# --hints--
-用 jQuery 的`addClass()`方法给`button`标签添加`animated`和`bounce`类。
+You should use the jQuery `addClass()` function to give the classes `animated` and `bounce` to your `button` elements.
```js
assert($('button').hasClass('animated') && $('button').hasClass('bounce'));
```
-仅用 jQuery 给标签添加颜色。
+You should only use jQuery to add these classes to the element.
```js
assert(!code.match(/class.*animated/g));
```
-jQuery 代码应该放在`$(document).ready();`函数里。
+Your jQuery code should be within the `$(document).ready();` function.
```js
assert(
- code.match(
- /\$\(document\)\.ready\(function.*(\s|\n)*.*button.*.addClass.*\);/g
- )
+ code.replace(/\s/g, '').match(/\$\(document\)\.ready\(function\(\)\{\$/g)
);
```
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-the-children-of-an-element-using-jquery.md b/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-the-children-of-an-element-using-jquery.md
index a4caec86a69..cd7add5572d 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-the-children-of-an-element-using-jquery.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-the-children-of-an-element-using-jquery.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aed208826
-title: 使用 jQuery 选择元素的子元素
+title: Target the Children of an Element Using jQuery
challengeType: 6
forumTopicId: 18320
dashedName: target-the-children-of-an-element-using-jquery
@@ -8,33 +8,33 @@ dashedName: target-the-children-of-an-element-using-jquery
# --description--
-把 HTML 标签放到另一个级别的标签里，这些 HTML 标签被称为该标签的`子标签（children element）`。例如，本次挑战中文本为 "#target1"、"#target2" 和 "#target3" 的按钮都是`<div class="well" id="left-well">`标签的`子标签`。
+When HTML elements are placed one level below another they are called <dfn>children</dfn> of that element. For example, the button elements in this challenge with the text "#target1", "#target2", and "#target3" are all children of the `<div class="well" id="left-well">` element.
-jQuery 有一个`children()`方法，可以访问被选取标签的子标签。
+jQuery has a function called `children()` that allows you to access the children of whichever element you've selected.
-下面的代码展示了用`children()`方法把`left-well`标签的子标签的颜色设置成`蓝色（blue）`：
+Here's an example of how you would use the `children()` function to give the children of your `left-well` element the color `blue`:
`$("#left-well").children().css("color", "blue")`
# --instructions--
-请把`right-well`标签的所有子标签颜色设置成`橙色（orange）`。
+Give all the children of your `right-well` element the color orange.
# --hints--
-`#right-well`所有的子标签文本应该是橙色。
+All children of `#right-well` should have orange text.
```js
assert($('#right-well').children().css('color') === 'rgb(255, 165, 0)');
```
-应该用`children()`方法修改标签。
+You should use the `children()` function to modify these elements.
```js
assert(code.match(/\.children\(\)\.css/g));
```
-仅用 jQuery 给标签添加类。
+You should only use jQuery to add these classes to the element.
```js
assert(code.match(/<div class="well" id="right-well">/g));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-the-parent-of-an-element-using-jquery.md b/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-the-parent-of-an-element-using-jquery.md
index e511981a014..7e74abd49f7 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-the-parent-of-an-element-using-jquery.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-the-parent-of-an-element-using-jquery.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aed308826
-title: 使用 jQuery 选择元素的父元素
+title: Target the Parent of an Element Using jQuery
challengeType: 6
forumTopicId: 18321
dashedName: target-the-parent-of-an-element-using-jquery
@@ -8,21 +8,21 @@ dashedName: target-the-parent-of-an-element-using-jquery
# --description--
-每个 HTML 标签都默认`继承（inherits）`其`父标签（parent element）`的 CSS 属性。
+Every HTML element has a `parent` element from which it `inherits` properties.
-例如，`h3`标签`jQuery Playground`的父标签是`<div class="container-fluid">`，`<div class="container-fluid">`的父标签是`body`。
+For example, your `jQuery Playground` `h3` element has the parent element of `<div class="container-fluid">`, which itself has the parent `body`.
-jQuery 有一个`parent()`方法，可以访问被选取标签的父标签。
+jQuery has a function called `parent()` that allows you to access the parent of whichever element you've selected.
-下面的代码展示了使用`parent()`方法把`left-well`标签的父标签背景色设置成`蓝色（blue）`的方式：
+Here's an example of how you would use the `parent()` function if you wanted to give the parent element of the `left-well` element a background color of blue:
`$("#left-well").parent().css("background-color", "blue")`
-请把`#target1`标签的父标签背景色设置成`红色（red）`。
+Give the parent of the `#target1` element a background-color of red.
# --hints--
-`left-well`标签应该有红色的背景。
+Your `left-well` element should have a red background.
```js
assert(
@@ -33,13 +33,13 @@ assert(
);
```
-应该用`.parent()`方法修改该标签。
+You should use the `.parent()` function to modify this element.
```js
assert(code.match(/\.parent\s*\(\s*\)\s*\.css/g));
```
-应该在`#target1`标签上调用`.parent()`方法。
+The `.parent()` method should be called on the `#target1` element.
```js
assert(
@@ -47,7 +47,7 @@ assert(
);
```
-仅用 jQuery 给标签添加类。
+You should only use jQuery to add these classes to the element.
```js
assert(code.match(/<div class="well" id="left-well">/g));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-the-same-element-with-multiple-jquery-selectors.md b/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-the-same-element-with-multiple-jquery-selectors.md
index 92fc91cdce8..65750f36e70 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-the-same-element-with-multiple-jquery-selectors.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/jquery/target-the-same-element-with-multiple-jquery-selectors.md
@@ -1,51 +1,52 @@
---
id: bad87fee1348bd9aed908626
-title: 用多个 jQuery 选择器选择同一个元素
+title: Target the Same Element with Multiple jQuery Selectors
challengeType: 6
forumTopicId: 18322
required:
- - link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
+ -
+ link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
dashedName: target-the-same-element-with-multiple-jquery-selectors
---
# --description--
-现在你知道三种选取标签的方法：用标签选择器，如`$("button")`；用类选择器，`$(".btn")`以及用 id 选择器，`$("#target1")`。
+Now you know three ways of targeting elements: by type: `$("button")`, by class: `$(".btn")`, and by id `$("#target1")`.
-虽然可以在单个`.addClass()`内添加多个类，但是我们可以用*三种不同的方式*给一种标签添加类。
+Although it is possible to add multiple classes in a single `.addClass()` call, let's add them to the same element in *three separate ways*.
-以三种不同的方式用`.addClass()`方法每次只给一种标签添加一个类：
+Using `.addClass()`, add only one class at a time to the same element, three different ways:
-给所有的`button`标签添加`animated`类。
+Add the `animated` class to all elements with type `button`.
-给所有类为`.btn`的`button`标签添加`shake`类。
+Add the `shake` class to all the buttons with class `.btn`.
-给所有 id 为`#target1`的`button`标签添加`btn-primary`类。
+Add the `btn-primary` class to the button with id `#target1`.
-**注意：**
-虽然三个选择器最终给 id 为`#target1`的`button`标签添加`shake`、`animated`和`btn-primary`等三个类，但是你需要用且仅用三种不同的选择器给三种标签各添加一个类（译者注：所谓的“一种标签”是指他们有某种共同的特点，如包含同一个 class）。
+**Note**
+You should only be targeting one element and adding only one class at a time. Altogether, your three individual selectors will end up adding the three classes `shake`, `animated`, and `btn-primary` to `#target1`.
# --hints--
-用`$('button')`选择标签。
+Your code should use the `$("button")` selector.
```js
assert(code.match(/\$\s*?\(\s*?(?:'|")\s*?button\s*?(?:'|")/gi));
```
-用`$('.btn')`选择标签。
+Your code should use the `$(".btn")` selector.
```js
assert(code.match(/\$\s*?\(\s*?(?:'|")\s*?\.btn\s*?(?:'|")/gi));
```
-用`$('#target1')`选择标签。
+Your code should use the `$("#target1")` selector.
```js
assert(code.match(/\$\s*?\(\s*?(?:'|")\s*?#target1\s*?(?:'|")/gi));
```
-三个选择器每个只添加一个类。
+You should only add one class with each of your three selectors.
```js
assert(
@@ -54,7 +55,7 @@ assert(
);
```
-`#target1`标签应具有`animated`、`shake`和`btn-primary`三个类。
+Your `#target1` element should have the classes `animated`‚ `shake` and `btn-primary`.
```js
assert(
@@ -64,7 +65,7 @@ assert(
);
```
-仅用 jQuery 给标签添加类。
+You should only use jQuery to add these classes to the element.
```js
assert(!code.match(/class.*animated/g));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/jquery/use-appendto-to-move-elements-with-jquery.md b/curriculum/challenges/chinese/03-front-end-libraries/jquery/use-appendto-to-move-elements-with-jquery.md
index 613036bae01..cac332f2f59 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/jquery/use-appendto-to-move-elements-with-jquery.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/jquery/use-appendto-to-move-elements-with-jquery.md
@@ -1,6 +1,6 @@
---
id: bad87fee1348bd9aed608826
-title: 使用 jQuery 的 appendTo 方法移动元素
+title: Use appendTo to Move Elements with jQuery
challengeType: 6
forumTopicId: 18340
dashedName: use-appendto-to-move-elements-with-jquery
@@ -8,31 +8,31 @@ dashedName: use-appendto-to-move-elements-with-jquery
# --description--
-现在我们学习把标签从一个`div`移动到另一个`div`。
+Now let's try moving elements from one `div` to another.
-jQuery 有一个`appendTo()`方法，可以选取 HTML 标签并将其添加到另一个标签里面。
+jQuery has a function called `appendTo()` that allows you to select HTML elements and append them to another element.
-例如，如果要把`target4`从右框移到左框，可以设置如下：
+For example, if we wanted to move `target4` from our right well to our left well, we would use:
`$("#target4").appendTo("#left-well");`
-请把`target2`标签从`left-well`移动到`right-well`。
+Move your `target2` element from your `left-well` to your `right-well`.
# --hints--
-`target2`标签应该不在`left-well`内。
+Your `target2` element should not be inside your `left-well`.
```js
assert($('#left-well').children('#target2').length === 0);
```
-`target2`标签应该在`right-well`内。
+Your `target2` element should be inside your `right-well`.
```js
assert($('#right-well').children('#target2').length > 0);
```
-仅用 jQuery 移动这些标签。
+You should only use jQuery to move these elements.
```js
assert(!code.match(/class.*animated/g));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/jquery/use-jquery-to-modify-the-entire-page.md b/curriculum/challenges/chinese/03-front-end-libraries/jquery/use-jquery-to-modify-the-entire-page.md
index 9bbe3aa3f00..540607dfe71 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/jquery/use-jquery-to-modify-the-entire-page.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/jquery/use-jquery-to-modify-the-entire-page.md
@@ -1,26 +1,27 @@
---
id: bad87fee1348bd9aecb08826
-title: 使用 jQuery 修改整个页面
+title: Use jQuery to Modify the Entire Page
challengeType: 6
forumTopicId: 18361
required:
- - link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
+ -
+ link: 'https://cdnjs.cloudflare.com/ajax/libs/animate.css/3.2.0/animate.css'
dashedName: use-jquery-to-modify-the-entire-page
---
# --description--
-jQuery 的学习到这里就告一段落了，现在我们来试一试让元素消失的特效。
+We're done playing with our jQuery playground. Let's tear it down!
-jQuery 也能选取`body`标签。
+jQuery can target the `body` element as well.
-后面的代码效果是使整个`body`标签淡出：`$("body").addClass("animated fadeOut");`
+Here's how we would make the entire body fade out: `$("body").addClass("animated fadeOut");`
-接下来我们做一些更有戏剧性的事：给`body`标签添加`animated`和`hinge`类。
+But let's do something more dramatic. Add the classes `animated` and `hinge` to your `body` element.
# --hints--
-给`body`标签添加`animated`和`hinge`类。
+You should add the classes `animated` and `hinge` to your `body` element.
```js
assert($('body').hasClass('animated') && $('body').hasClass('hinge'));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/connect-redux-to-react.md b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/connect-redux-to-react.md
index 5877d194161..731a2081822 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/connect-redux-to-react.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/connect-redux-to-react.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036147
-title: 连接 Redux 和 React
+title: Connect Redux to React
challengeType: 6
forumTopicId: 301426
dashedName: connect-redux-to-react
@@ -8,21 +8,21 @@ dashedName: connect-redux-to-react
# --description--
-既然写了`mapStateToProps()`、`mapDispatchToProps()`两个函数，现在你可以用它们来把`state`和`dispatch`映射到 React 组件的`props`了。React Redux 的`connect`方法可以完成这个任务。此方法有`mapStateToProps()`、`mapDispatchToProps()`两个可选参数，它们是可选的，原因是你的组件可能仅需要访问`状态`但不需要分发任何 actions，反之亦然。
+Now that you've written both the `mapStateToProps()` and the `mapDispatchToProps()` functions, you can use them to map `state` and `dispatch` to the `props` of one of your React components. The `connect` method from React Redux can handle this task. This method takes two optional arguments, `mapStateToProps()` and `mapDispatchToProps()`. They are optional because you may have a component that only needs access to `state` but doesn't need to dispatch any actions, or vice versa.
-为了使用此方法，需要传入函数参数并在调用时传入组件。这种语法有些不寻常，如下所示：
+To use this method, pass in the functions as arguments, and immediately call the result with your component. This syntax is a little unusual and looks like:
`connect(mapStateToProps, mapDispatchToProps)(MyComponent)`
-**注意：** 如果要省略`connect`方法中的某个参数，则应当用`null`替换这个参数。
+**Note:** If you want to omit one of the arguments to the `connect` method, you pass `null` in its place.
# --instructions--
-在编辑器上有两个函数：`mapStateToProps()`、`mapDispatchToProps()`，还有一个叫`Presentational`的 React 组件。用`ReactRedux`全局对象中的`connect`方法将此组件连接到 Redux，并立即在`Presentational`组件中调用，把结果赋值给一个名为`ConnectedComponent`的代表已连接组件的新常量。大功告成！你已成功把 React 连接到 Redux！尝试更改任何一个`connect`参数为`null`并观察测试结果。
+The code editor has the `mapStateToProps()` and `mapDispatchToProps()` functions and a new React component called `Presentational`. Connect this component to Redux with the `connect` method from the `ReactRedux` global object, and call it immediately on the `Presentational` component. Assign the result to a new `const` called `ConnectedComponent` that represents the connected component. That's it, now you're connected to Redux! Try changing either of `connect`'s arguments to `null` and observe the test results.
# --hints--
-应渲染`Presentational`组件。
+The `Presentational` component should render.
```js
assert(
@@ -33,7 +33,7 @@ assert(
);
```
-`Presentational`组件应通过`connect`接收一个`messages`属性。
+The `Presentational` component should receive a prop `messages` via `connect`.
```js
assert(
@@ -45,7 +45,7 @@ assert(
);
```
-`Presentational`组件应通过`connect`接收一个`submitNewMessage`属性。
+The `Presentational` component should receive a prop `submitNewMessage` via `connect`.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/connect-redux-to-the-messages-app.md b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/connect-redux-to-the-messages-app.md
index 78f056a99d1..ce25f89ba65 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/connect-redux-to-the-messages-app.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/connect-redux-to-the-messages-app.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036148
-title: 将 Redux 连接到 Messages App
+title: Connect Redux to the Messages App
challengeType: 6
forumTopicId: 301427
dashedName: connect-redux-to-the-messages-app
@@ -8,17 +8,17 @@ dashedName: connect-redux-to-the-messages-app
# --description--
-知道`connect`怎么实现 React 和 Redux 的连接后，我们可以在 React 组件中应用上面学到的内容。
+Now that you understand how to use `connect` to connect React to Redux, you can apply what you've learned to your React component that handles messages.
-在上一课，连接到 Redux 的组件命名为`Presentational`，这个命名不是任意的，这样的术语通常是指未直接连接到 Redux 的 React 组件，他们只负责执行接收 props 的函数来实现 UI 的呈现。与上一挑战相比，本挑战需要把容器组件连接到 Redux。这些组件通常负责把 actions 分派给 store，且经常给子组件传入 store state 属性。
+In the last lesson, the component you connected to Redux was named `Presentational`, and this wasn't arbitrary. This term *generally* refers to React components that are not directly connected to Redux. They are simply responsible for the presentation of UI and do this as a function of the props they receive. By contrast, container components are connected to Redux. These are typically responsible for dispatching actions to the store and often pass store state to child components as props.
# --instructions--
-到目前为止，我们的编辑器上已包含了整个章节的代码，唯一不同的是，React 组件被重新命名为`Presentational`，即展示层组件。创建一个新组件，保存在名为`Container`的常量中。这个常量用`connect`把`Presentational`组件和 Redux 连接起来。然后，在`AppWrapper`中渲染 React Redux 的`Provider`组件，给`Provider`传入 Redux`store`属性并渲染`Container`为子组件。完成这些，消息 app 应用会再次渲染页面。
+The code editor has all the code you've written in this section so far. The only change is that the React component is renamed to `Presentational`. Create a new component held in a constant called `Container` that uses `connect` to connect the `Presentational` component to Redux. Then, in the `AppWrapper`, render the React Redux `Provider` component. Pass `Provider` the Redux `store` as a prop and render `Container` as a child. Once everything is setup, you will see the messages app rendered to the page again.
# --hints--
-`AppWrapper`应渲染该页面。
+The `AppWrapper` should render to the page.
```js
assert(
@@ -29,7 +29,7 @@ assert(
);
```
-`Presentational` 组件应该渲染到页面上。
+The `Presentational` component should render to page.
```js
assert(
@@ -40,7 +40,7 @@ assert(
);
```
-`Presentational`组件应渲染`h2`、`input`、`button`、`ul`四个元素。
+The `Presentational` component should render an `h2`, `input`, `button`, and `ul` elements.
```js
assert(
@@ -57,7 +57,7 @@ assert(
);
```
-`Presentational`组件应接收 Redux store 的`消息`属性。
+The `Presentational` component should receive `messages` from the Redux store as a prop.
```js
assert(
@@ -70,7 +70,7 @@ assert(
);
```
-`Presentational`组件应接收创建 action 的函数`submitMessage`属性。
+The `Presentational` component should receive the `submitMessage` action creator as a prop.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/extract-local-state-into-redux.md b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/extract-local-state-into-redux.md
index b4833714e0c..b1c4581f2e1 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/extract-local-state-into-redux.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/extract-local-state-into-redux.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036149
-title: 将局部状态提取到 Redux 中
+title: Extract Local State into Redux
challengeType: 6
forumTopicId: 301428
dashedName: extract-local-state-into-redux
@@ -8,17 +8,17 @@ dashedName: extract-local-state-into-redux
# --description--
-胜利就在眼前了！请回顾一下为管理 React messages app 的状态写的 Redux 代码。现在有了连接好的 Redux，你还要从`Presentational`组件中提取状态管理到 Redux，在`Presentational`组件内处理本地状态。
+You're almost done! Recall that you wrote all the Redux code so that Redux could control the state management of your React messages app. Now that Redux is connected, you need to extract the state management out of the `Presentational` component and into Redux. Currently, you have Redux connected, but you are handling the state locally within the `Presentational` component.
# --instructions--
-在`Presentational`组件中，先删除本地`state`中的`messages`属性，被删的 messages 将由 Redux 管理。接着，修改`submitMessage()`方法，使该方法从`this.props`那里分发`submitNewMessage()`；从本地`state`中传入当前消息输入作为参数。因本地状态删除了`messages`属性，所以在调用`this.setState()`时也要删除该属性。最后，修改`render()`方法，使其所映射的消息是从`props`接收的，而不是`state`
+In the `Presentational` component, first, remove the `messages` property in the local `state`. These messages will be managed by Redux. Next, modify the `submitMessage()` method so that it dispatches `submitNewMessage()` from `this.props`, and pass in the current message input from local `state` as an argument. Because you removed `messages` from local state, remove the `messages` property from the call to `this.setState()` here as well. Finally, modify the `render()` method so that it maps over the messages received from `props` rather than `state`.
-完成这些更改后，我们的应用会实现 Redux 管理应用的状态，但它继续运行着相同的功能。此示例还阐明了组件获得本地状态的方式，即在自己的状态中继续跟踪用户本地输入。由此可见，Redux 为 React 提供了很有用的状态管理框架。先前，你仅使用 React 的本地状态也实现了相同的结果，这在应付简单的应用时通常是可行的。但是，随着应用变得越来越大，越来越复杂，应用的状态管理也变得非常困难，Redux 就是为解决这样的问题而诞生的。
+Once these changes are made, the app will continue to function the same, except Redux manages the state. This example also illustrates how a component may have local `state`: your component still tracks user input locally in its own `state`. You can see how Redux provides a useful state management framework on top of React. You achieved the same result using only React's local state at first, and this is usually possible with simple apps. However, as your apps become larger and more complex, so does your state management, and this is the problem Redux solves.
# --hints--
-`AppWrapper`应该渲染该到页面。
+The `AppWrapper` should render to the page.
```js
assert(
@@ -29,7 +29,7 @@ assert(
);
```
-`Presentational` 应该渲染到页面上.
+The `Presentational` component should render to page.
```js
assert(
@@ -40,7 +40,7 @@ assert(
);
```
-`Presentational`组件应渲染`h2`、`input`、`button`、`ul`四个元素。
+The `Presentational` component should render an `h2`, `input`, `button`, and `ul` elements.
```js
assert(
@@ -57,7 +57,7 @@ assert(
);
```
-`Presentational`组件应接收 Redux store 的`消息`属性。
+The `Presentational` component should receive `messages` from the Redux store as a prop.
```js
assert(
@@ -70,7 +70,7 @@ assert(
);
```
-`Presentational`组件应接收创建 action 的函数`submitMessage`属性。
+The `Presentational` component should receive the `submitMessage` action creator as a prop.
```js
assert(
@@ -83,7 +83,7 @@ assert(
);
```
-`Presentational`组件的状态应包含一个初始化为空字符串的input属性。
+The state of the `Presentational` component should contain one property, `input`, which is initialized to an empty string.
```js
assert(
@@ -100,7 +100,7 @@ assert(
);
```
-键入`input`元素应更新`Presentational`组件的状态。
+Typing in the `input` element should update the state of the `Presentational` component.
```js
async () => {
@@ -124,7 +124,7 @@ async () => {
};
```
-在`Presentational`组件上 dispatch `submitMessage`应更新 Redux store 并清除本地状态中的输入。
+Dispatching the `submitMessage` on the `Presentational` component should update Redux store and clear the input in local state.
```js
async () => {
@@ -156,7 +156,7 @@ async () => {
};
```
-`Presentational`组件应渲染 Redux store 中的`messages`
+The `Presentational` component should render the `messages` from the Redux store.
```js
async () => {
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/extract-state-logic-to-redux.md b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/extract-state-logic-to-redux.md
index 8a8a8b827c6..15ba65940aa 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/extract-state-logic-to-redux.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/extract-state-logic-to-redux.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036143
-title: 提取状态逻辑给 Redux
+title: Extract State Logic to Redux
challengeType: 6
forumTopicId: 301429
dashedName: extract-state-logic-to-redux
@@ -8,23 +8,23 @@ dashedName: extract-state-logic-to-redux
# --description--
-完成 React 组件后，我们需要把在本地`状态`执行的逻辑移到 Redux 中，这是为小规模 React 应用添加 Redux 的第一步。该应用的唯一功能是把用户的新消息添加到无序列表中。下面我们用简单的示例来演示 React 和 Redux 之间的配合。
+Now that you finished the React component, you need to move the logic it's performing locally in its `state` into Redux. This is the first step to connect the simple React app to Redux. The only functionality your app has is to add new messages from the user to an unordered list. The example is simple in order to demonstrate how React and Redux work together.
# --instructions--
-首先，定义 action 的类型 'ADD'，将其设置为常量`ADD`。接着，定义创建 action 的函数`addMessage()`，用该函数创建添加消息的 action，把`message`传给创建 action 的函数并返回包含该消息的`action`
+First, define an action type 'ADD' and set it to a const `ADD`. Next, define an action creator `addMessage()` which creates the action to add a message. You'll need to pass a `message` to this action creator and include the message in the returned `action`.
-接着，创建名为`messageReducer()`的 reducer 方法，为这些消息处理状态。初始状态应为空数组。reducer 向状态中的消息数组添加消息，或返回当前状态。最后，创建 Redux store 并传给 reducer。
+Then create a reducer called `messageReducer()` that handles the state for the messages. The initial state should equal an empty array. This reducer should add a message to the array of messages held in state, or return the current state. Finally, create your Redux store and pass it the reducer.
# --hints--
-应存在一个值为字符串`ADD`的常量`ADD`。
+The const `ADD` should exist and hold a value equal to the string `ADD`
```js
assert(ADD === 'ADD');
```
-创建 action 的函数`addMessage`应返回`type`等于`ADD`的对象，其返回的消息即被传入的消息。
+The action creator `addMessage` should return an object with `type` equal to `ADD` and `message` equal to the message that is passed in.
```js
assert(
@@ -35,13 +35,13 @@ assert(
);
```
-`messageReducer`应是一个函数。
+`messageReducer` should be a function.
```js
assert(typeof messageReducer === 'function');
```
-存在一个 store 且其初始状态为空数组。
+The store should exist and have an initial state set to an empty array.
```js
assert(
@@ -52,7 +52,7 @@ assert(
);
```
-分发`addMessage`到 store 应添加新消息到状态中消息数组。
+Dispatching `addMessage` against the store should immutably add a new message to the array of messages held in state.
```js
assert(
@@ -66,7 +66,7 @@ assert(
);
```
-`messageReducer`被其它任何 actions 调用时应返回当前状态。
+The `messageReducer` should return the current state if called with any other actions.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/getting-started-with-react-redux.md b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/getting-started-with-react-redux.md
index 48cef49f34c..8bf764191ff 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/getting-started-with-react-redux.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/getting-started-with-react-redux.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036141
-title: React 和 Redux 入门
+title: Getting Started with React Redux
challengeType: 6
forumTopicId: 301430
dashedName: getting-started-with-react-redux
@@ -8,19 +8,19 @@ dashedName: getting-started-with-react-redux
# --description--
-这一系列挑战介绍的是 Redux 和 React 的配合，我们先来回顾一下这两种技术的关键原则是什么。React 是提供数据的视图库，能以高效、可预测的方式渲染视图。Redux 是状态管理框架，可用于简化 APP 应用状态的管理。在 React Redux app 应用中，通常可创建单一的 Redux store 来管理整个应用的状态。React 组件仅订阅 store 中与其角色相关的数据，你可直接从 React 组件中分发 actions 以触发 store 对象的更新。
+This series of challenges introduces how to use Redux with React. First, here's a review of some of the key principles of each technology. React is a view library that you provide with data, then it renders the view in an efficient, predictable way. Redux is a state management framework that you can use to simplify the management of your application's state. Typically, in a React Redux app, you create a single Redux store that manages the state of your entire app. Your React components subscribe to only the pieces of data in the store that are relevant to their role. Then, you dispatch actions directly from React components, which then trigger store updates.
-React 组件可以在本地管理自己的状态，但是对于复杂的应用来说，它的状态最好是用 Redux 保存在单一位置，有特定本地状态的独立组件例外。最后一点是，Redux 没有内置的 React，需要安装`react-redux`包，通过这个方式把 Redux 的`state`和`dispatch`作为`props`传给组件。
+Although React components can manage their own state locally, when you have a complex app, it's generally better to keep the app state in a single location with Redux. There are exceptions when individual components may have local state specific only to them. Finally, because Redux is not designed to work with React out of the box, you need to use the `react-redux` package. It provides a way for you to pass Redux `state` and `dispatch` to your React components as `props`.
-在接下来的挑战中，先要创建一个可输入新文本消息的 React 组件，添加这些消息到数组里，在视图上显示数组。接着，创建 Redux store 和 actions 来管理消息数组的状态。最后，使用`react-redux`连接 Redux store 和组件，从而将本地状态提取到 Redux store 中。
+Over the next few challenges, first, you'll create a simple React component which allows you to input new text messages. These are added to an array that's displayed in the view. This should be a nice review of what you learned in the React lessons. Next, you'll create a Redux store and actions that manage the state of the messages array. Finally, you'll use `react-redux` to connect the Redux store with your component, thereby extracting the local state into the Redux store.
# --instructions--
-创建`DisplayMessages`组件，把构造函数添加到此组件中，使用含两个属性的状态初始化该组件，这两个属性为：input（设置为空字符串），`messages`（设置为空数组）。
+Start with a `DisplayMessages` component. Add a constructor to this component and initialize it with a state that has two properties: `input`, that's set to an empty string, and `messages`, that's set to an empty array.
# --hints--
-`DisplayMessages`组件应渲染空的`div`元素。
+The `DisplayMessages` component should render an empty `div` element.
```js
assert(
@@ -31,13 +31,13 @@ assert(
);
```
-`DisplayMessages`组件的构造函数应调用`super`，传入`props`。
+The `DisplayMessages` constructor should be called properly with `super`, passing in `props`.
```js
(getUserInput) =>
assert(
(function () {
- const noWhiteSpace = getUserInput('index').replace(/\s/g, '');
+ const noWhiteSpace = __helpers.removeWhiteSpace(getUserInput('index'));
return (
noWhiteSpace.includes('constructor(props)') &&
noWhiteSpace.includes('super(props')
@@ -46,7 +46,7 @@ assert(
);
```
-`DisplayMessages`组件的初始状态应是`{input: "", messages: []}`。
+The `DisplayMessages` component should have an initial state equal to `{input: "", messages: []}`.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/manage-state-locally-first.md b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/manage-state-locally-first.md
index 2022c9ac02c..ec7c618f3d3 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/manage-state-locally-first.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/manage-state-locally-first.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036142
-title: 首先在本地管理状态
+title: Manage State Locally First
challengeType: 6
forumTopicId: 301431
dashedName: manage-state-locally-first
@@ -8,19 +8,19 @@ dashedName: manage-state-locally-first
# --description--
-这一关的任务是完成`DisplayMessages`组件的创建。
+Here you'll finish creating the `DisplayMessages` component.
# --instructions--
-首先，在`render()`方法中，让组件渲染`input`、`button`、`ul`三个元素。`input`元素的改变会触发`handleChange()`方法。此外，`input`元素会渲染组件状态中`input`的值。点击按钮`button`需触发`submitMessage()`方法。
+First, in the `render()` method, have the component render an `input` element, `button` element, and `ul` element. When the `input` element changes, it should trigger a `handleChange()` method. Also, the `input` element should render the value of `input` that's in the component's state. The `button` element should trigger a `submitMessage()` method when it's clicked.
-接着，写出这两种方法。`handleChange()`方法会更新`input`为用户正在输入的内容。`submitMessage()`方法把当前存储在`input`的消息与本地状态的`messages`数组连接起来，并清除`input`的值。
+Second, write these two methods. The `handleChange()` method should update the `input` with what the user is typing. The `submitMessage()` method should concatenate the current message (stored in `input`) to the `messages` array in local state, and clear the value of the `input`.
-最后，在`ul`中展示`messages`数组，其中每个元素内容需放到`li`元素内。
+Finally, use the `ul` to map over the array of `messages` and render it to the screen as a list of `li` elements.
# --hints--
-`DisplayMessages`组件的初始状态应是`{ input: "", messages: [] }`。
+The `DisplayMessages` component should initialize with a state equal to `{ input: "", messages: [] }`.
```js
assert(
@@ -36,7 +36,7 @@ assert(
);
```
-`DisplayMessages`组件应渲染含`h2`、`button`、`ul`、`li`四个子元素的`div`。
+The `DisplayMessages` component should render a `div` containing an `h2` element, a `button` element, a `ul` element, and `li` elements as children.
```js
async () => {
@@ -58,13 +58,13 @@ async () => {
};
```
-`input`元素应渲染本地状态中的`input`值。
+`.map` should be used on the `messages` array.
```js
assert(code.match(/this\.state\.messages\.map/g));
```
-调用`handleChange`方法时应更新状态中的`input`值为当前输入。
+The `input` element should render the value of `input` in local state.
```js
async () => {
@@ -83,7 +83,7 @@ async () => {
};
```
-单击`Add message`按钮应调用`submitMessage`方法，添加当前`输入`到状态中的`消息`数组。
+Calling the method `handleChange` should update the `input` value in state to the current input.
```js
async () => {
@@ -106,7 +106,7 @@ async () => {
};
```
-`submitMessage`方法应清除当前输入。
+Clicking the `Add message` button should call the method `submitMessage` which should add the current `input` to the `messages` array in state.
```js
async () => {
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/map-dispatch-to-props.md b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/map-dispatch-to-props.md
index acef2a9a445..72dae912c22 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/map-dispatch-to-props.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/map-dispatch-to-props.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036146
-title: 映射 Dispatch 到 Props
+title: Map Dispatch to Props
challengeType: 6
forumTopicId: 301432
dashedName: map-dispatch-to-props
@@ -8,9 +8,9 @@ dashedName: map-dispatch-to-props
# --description--
-`mapDispatchToProps()`函数可为 React 组件提供特定的创建 action 的函数，以便组件可 dispatch actions，从而更改 Redux store 中的数据。该函数的结构跟上一挑战中的`mapStateToProps()`函数相似，它返回一个对象，把 dispatch actions 映射到属性名上，该属性名成为`props`。然而，每个属性都返回一个用 action creator 及与 action 相关的所有数据调用`dispatch`的函数，而不是返回`state`的一部分。你可以访问`dispatch`，因为在定义函数时，我们以参数形式把它传入`mapDispatchToProps()`了，这跟`state`传入`mapDispatchToProps()`是一样的。在幕后，React Redux 用 Redux 的`store.dispatch()`来管理这些含`mapDispatchToProps()`的dispatches，这跟它使用`store.subscribe()`来订阅映射到`state`的组件的方式类似。
+The `mapDispatchToProps()` function is used to provide specific action creators to your React components so they can dispatch actions against the Redux store. It's similar in structure to the `mapStateToProps()` function you wrote in the last challenge. It returns an object that maps dispatch actions to property names, which become component `props`. However, instead of returning a piece of `state`, each property returns a function that calls `dispatch` with an action creator and any relevant action data. You have access to this `dispatch` because it's passed in to `mapDispatchToProps()` as a parameter when you define the function, just like you passed `state` to `mapStateToProps()`. Behind the scenes, React Redux is using Redux's `store.dispatch()` to conduct these dispatches with `mapDispatchToProps()`. This is similar to how it uses `store.subscribe()` for components that are mapped to `state`.
-例如，创建 action 的函数`loginUser()`把`username`作为 action payload，`mapDispatchToProps()`返回给创建 action 的函数的对象如下：
+For example, you have a `loginUser()` action creator that takes a `username` as an action payload. The object returned from `mapDispatchToProps()` for this action creator would look something like:
```jsx
{
@@ -22,11 +22,11 @@ dashedName: map-dispatch-to-props
# --instructions--
-编辑器上提供的是创建 action 的函数`addMessage()`。写出接收`dispatch`为参数的函数`mapDispatchToProps()`，返回一个 dispatch 函数对象，其属性为`submitNewMessage`。该函数在 dispatch `addMessage()`时为新消息提供一个参数。
+The code editor provides an action creator called `addMessage()`. Write the function `mapDispatchToProps()` that takes `dispatch` as an argument, then returns an object. The object should have a property `submitNewMessage` set to the dispatch function, which takes a parameter for the new message to add when it dispatches `addMessage()`.
# --hints--
-`addMessage`应返回含`type`和`message`两个键的对象。
+`addMessage` should return an object with keys `type` and `message`.
```js
assert(
@@ -40,19 +40,19 @@ assert(
);
```
-`mapDispatchToProps`应为函数。
+`mapDispatchToProps` should be a function.
```js
assert(typeof mapDispatchToProps === 'function');
```
-`mapDispatchToProps`应返回一个对象。
+`mapDispatchToProps` should return an object.
```js
assert(typeof mapDispatchToProps() === 'object');
```
-从`mapDispatchToProps`通过`submitNewMessage`分发`addMessage`，应向 dispatch 函数返回一条消息。
+Dispatching `addMessage` with `submitNewMessage` from `mapDispatchToProps` should return a message to the dispatch function.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/map-state-to-props.md b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/map-state-to-props.md
index 445417e4c90..e75b1fb34cb 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/map-state-to-props.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/map-state-to-props.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036145
-title: 映射 State 到 Props
+title: Map State to Props
challengeType: 6
forumTopicId: 301433
dashedName: map-state-to-props
@@ -8,37 +8,37 @@ dashedName: map-state-to-props
# --description--
-`Provider`可向 React 组件提供`state`和`dispatch`，但你必须确切地指定所需要的 state 和 actions，以确保每个组件只能访问所需的 state。完成这个任务，你需要创建两个函数：`mapStateToProps()`、`mapDispatchToProps()`。
+The `Provider` component allows you to provide `state` and `dispatch` to your React components, but you must specify exactly what state and actions you want. This way, you make sure that each component only has access to the state it needs. You accomplish this by creating two functions: `mapStateToProps()` and `mapDispatchToProps()`.
-在这两个函数中，声明 state 中函数所要访问的部分及需要 dispatch 的创建 action 的函数。完成这些，我们就可以迎接下一个挑战，学习如何使用 React Redux 的`connect`方法来把函数连接到组件了。
+In these functions, you declare what pieces of state you want to have access to and which action creators you need to be able to dispatch. Once these functions are in place, you'll see how to use the React Redux `connect` method to connect them to your components in another challenge.
-**注意：** 在幕后，React Redux 用`store.subscribe()`方法来实现`mapStateToProps()`。
+**Note:** Behind the scenes, React Redux uses the `store.subscribe()` method to implement `mapStateToProps()`.
# --instructions--
-创建`mapStateToProps()`函数，以`state`为参数，然后返回一个对象，该对象把 state 映射到特定属性名上，这些属性能通过`props`访问组件。由于此示例把 app 应用的整个状态保存在单一数组中，你可把整个状态传给组件。在返回的对象中创建`messages`属性，并设置为`state`。
+Create a function `mapStateToProps()`. This function should take `state` as an argument, then return an object which maps that state to specific property names. These properties will become accessible to your component via `props`. Since this example keeps the entire state of the app in a single array, you can pass that entire state to your component. Create a property `messages` in the object that's being returned, and set it to `state`.
# --hints--
-常量`state`应为空数组。
+The const `state` should be an empty array.
```js
assert(Array.isArray(state) && state.length === 0);
```
-`mapStateToProps`应为函数。
+`mapStateToProps` should be a function.
```js
assert(typeof mapStateToProps === 'function');
```
-`mapStateToProps`应返还一个对象。
+`mapStateToProps` should return an object.
```js
assert(typeof mapStateToProps() === 'object');
```
-把 state 数组传入`mapStateToProps`后应返回赋值给`messages`键的数组。
+Passing an array as state to `mapStateToProps` should return this array assigned to a key of `messages`.
```js
assert(mapStateToProps(['messages']).messages.pop() === 'messages');
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/moving-forward-from-here.md b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/moving-forward-from-here.md
index 5efe9115bb9..77ef3860f5a 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/moving-forward-from-here.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/moving-forward-from-here.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403614a
-title: 从这里前进
+title: Moving Forward From Here
challengeType: 6
forumTopicId: 301434
dashedName: moving-forward-from-here
@@ -8,21 +8,21 @@ dashedName: moving-forward-from-here
# --description--
-恭喜你完成了 React 和 Redux 的所有课程！在继续前进前，还有一点值得我们注意。通常，我们不会在这样的编辑器中编写 React 应用代码。如果你在自己的计算机上使用 npm 和文件系统，这个挑战可让你一瞥 React 应用的语法之貌。除了使用`import`语句（这些语句引入了各挑战中提供的所有依赖关系），其代码看起来类似。“管理包（含 npm）”这一节更详细地介绍了 npm。
+Congratulations! You finished the lessons on React and Redux. There's one last item worth pointing out before you move on. Typically, you won't write React apps in a code editor like this. This challenge gives you a glimpse of what the syntax looks like if you're working with npm and a file system on your own machine. The code should look similar, except for the use of `import` statements (these pull in all of the dependencies that have been provided for you in the challenges). The "Managing Packages with npm" section covers npm in more detail.
-最后，写 React 和 Redux 的代码通常需要一些配置，且很快会变得复杂起来。如果你想在自己的机器上尝试写代码，点击链接
+Finally, writing React and Redux code generally requires some configuration. This can get complicated quickly. If you are interested in experimenting on your own machine, the
-[创建 React App](https://github.com/facebookincubator/create-react-app)可获取已配置好的现成代码。
+[Create React App](https://github.com/facebookincubator/create-react-app) comes configured and ready to go.
-另一种做法是在 CodePen 中启用 Babel 作为 JavaScript 预处理器，将 React 和 ReactDOM 添加为外部 JavaScript 资源，在那里编写应用。
+Alternatively, you can enable Babel as a JavaScript Preprocessor in CodePen, add React and ReactDOM as external JavaScript resources, and work there as well.
# --instructions--
-把`'Now I know React and Redux!'`这一消息输出到控制台。
+Log the message `'Now I know React and Redux!'` to the console.
# --hints--
-`Now I know React and Redux!`这一消息应输出到控制台。
+The message `Now I know React and Redux!` should be logged to the console.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/use-provider-to-connect-redux-to-react.md b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/use-provider-to-connect-redux-to-react.md
index 3f24706e468..f286f16b2ee 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/use-provider-to-connect-redux-to-react.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react-and-redux/use-provider-to-connect-redux-to-react.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036144
-title: 使用 Provider 连接 Redux 和 React
+title: Use Provider to Connect Redux to React
challengeType: 6
forumTopicId: 301435
dashedName: use-provider-to-connect-redux-to-react
@@ -8,9 +8,9 @@ dashedName: use-provider-to-connect-redux-to-react
# --description--
-在上一挑战中，你创建了 Redux store 和 action，分别用于处理消息数组和添加新消息。下一步要为 React 提供访问 Redux store 及发起更新所需的 actions。`react-redux`包可帮助我们完成这些任务。
+In the last challenge, you created a Redux store to handle the messages array and created an action for adding new messages. The next step is to provide React access to the Redux store and the actions it needs to dispatch updates. React Redux provides its `react-redux` package to help accomplish these tasks.
-React Redux 提供的 API 有两个关键的功能：`Provider`和`connect`。你会在另一个挑战中学`connect`。`Provider`是 React Redux 包装 React 应用的 wrapper 组件，它允许你访问整个组件树中的 Redux`store`及`dispatch（分发）`方法。`Provider`需要两个 props：Redux store 和 APP 应用的子组件。用于 APP 组件的`Provider`可这样定义：
+React Redux provides a small API with two key features: `Provider` and `connect`. Another challenge covers `connect`. The `Provider` is a wrapper component from React Redux that wraps your React app. This wrapper then allows you to access the Redux `store` and `dispatch` functions throughout your component tree. `Provider` takes two props, the Redux store and the child components of your app. Defining the `Provider` for an App component might look like this:
```jsx
<Provider store={store}>
@@ -20,13 +20,13 @@ React Redux 提供的 API 有两个关键的功能：`Provider`和`connect`。
# --instructions--
-此时，编辑器上显示的是过去几个挑战中所有代码，包括 Redux store、actions、`DisplayMessages`组件。新出现的代码是底部的`AppWrapper`组件，这个顶级组件可用于渲染`ReactRedux`的`Provider`，并把 Redux 的 store 作为 props 传入。接着，渲染`DisplayMessages`为子组件。完成这些任务后，你会看到 React 组件渲染到页面上。
+The code editor now shows all your Redux and React code from the past several challenges. It includes the Redux store, actions, and the `DisplayMessages` component. The only new piece is the `AppWrapper` component at the bottom. Use this top level component to render the `Provider` from `ReactRedux`, and pass the Redux store as a prop. Then render the `DisplayMessages` component as a child. Once you are finished, you should see your React component rendered to the page.
-**注意：** React Redux 在此可作全局变量，因此你可通过点号表示法访问 Provider。利用这一点，编辑器上的代码把`Provider`设置为常量，便于你在`AppWrapper`渲染方法中使用。
+**Note:** React Redux is available as a global variable here, so you can access the Provider with dot notation. The code in the editor takes advantage of this and sets it to a constant `Provider` for you to use in the `AppWrapper` render method.
# --hints--
-`AppWrapper`应渲染。
+The `AppWrapper` should render.
```js
assert(
@@ -37,21 +37,21 @@ assert(
);
```
-`Provider`组件应传入相当于 Redux store 的`store`参数。
+The `Provider` wrapper component should have a prop of `store` passed to it, equal to the Redux store.
```js
(getUserInput) =>
assert(
(function () {
const mockedComponent = Enzyme.mount(React.createElement(AppWrapper));
- return getUserInput('index')
- .replace(/\s/g, '')
+ return __helpers
+ .removeWhiteSpace(getUserInput('index'))
.includes('<Providerstore={store}>');
})()
);
```
-`DisplayMessages`应渲染为`AppWrapper`的子组件。
+`DisplayMessages` should render as a child of `AppWrapper`.
```js
assert(
@@ -64,7 +64,7 @@ assert(
);
```
-`DisplayMessages`组件应渲染 h2、input、button、`ul`四个元素。
+The `DisplayMessages` component should render an h2, input, button, and `ul` element.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/access-props-using-this.props.md b/curriculum/challenges/chinese/03-front-end-libraries/react/access-props-using-this.props.md
index 088238f2464..93d96e4f8c0 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/access-props-using-this.props.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/access-props-using-this.props.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403616e
-title: 使用 this.props 访问 Props
+title: Access Props Using this.props
challengeType: 6
forumTopicId: 301375
dashedName: access-props-using-this-props
@@ -8,17 +8,17 @@ dashedName: access-props-using-this-props
# --description--
-前几项挑战涵盖了将 props 传递给子组件的基本方法。但是，倘若接收 prop 的子组件不是无状态函数组件，而是一个 ES6 类组件又当如何呢？ES6 类组件访问 props 的方法略有不同。
+The last several challenges covered the basic ways to pass props to child components. But what if the child component that you're passing a prop to is an ES6 class component, rather than a stateless functional component? The ES6 class component uses a slightly different convention to access props.
-任何时候，只要引用类组件本身，就要使用`this`关键字。要访问类组件中的 props，你需要在在访问它的代码前面添加`this`。例如，如果 ES6 类组件有一个名为`data`的 prop，你可以在 JSX 中这样写：`{this.props.data}`。
+Anytime you refer to a class component within itself, you use the `this` keyword. To access props within a class component, you preface the code that you use to access it with `this`. For example, if an ES6 class component has a prop called `data`, you write `{this.props.data}` in JSX.
# --instructions--
-在父组件`ResetPassword`中渲染`ReturnTempPassword`组件的一个实例。在这里，为`ReturnTempPassword`提供一个`tempPassword`prop，并赋值给 prop 一个长度至少为 8 个字符的字符串。在子组件`ReturnTempPassword`中，访问`strong`标签中的`tempPassword`prop，以确保用户看到临时密码。
+Render an instance of the `ReturnTempPassword` component in the parent component `ResetPassword`. Here, give `ReturnTempPassword` a prop of `tempPassword` and assign it a value of a string that is at least 8 characters long. Within the child, `ReturnTempPassword`, access the `tempPassword` prop within the `strong` tags to make sure the user sees the temporary password.
# --hints--
-`ResetPassword`组件应该返回单个`div`元素。
+The `ResetPassword` component should return a single `div` element.
```js
assert(
@@ -29,7 +29,7 @@ assert(
);
```
-`ResetPassword`的第四个子组件应该是`ReturnTempPassword`组件。
+The fourth child of `ResetPassword` should be the `ReturnTempPassword` component.
```js
assert(
@@ -42,7 +42,7 @@ assert(
);
```
-`ReturnTempPassword`组件应该有一个名为`tempPassword`的属性。
+The `ReturnTempPassword` component should have a prop called `tempPassword`.
```js
assert(
@@ -53,7 +53,7 @@ assert(
);
```
-`ReturnTempPassword`组件的`tempPassword`prop 值应该是一个字符串，其长度至少为`8`。
+The `tempPassword` prop of `ReturnTempPassword` should be equal to a string of at least `8` characters.
```js
assert(
@@ -66,7 +66,7 @@ assert(
);
```
-`ReturnTempPassword`组件应该显示你作为`tempPassword`prop 创建的密码，并且密码显示在`strong`标签中。
+The `ReturnTempPassword` component should display the password you create as the `tempPassword` prop within `strong` tags.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/add-comments-in-jsx.md b/curriculum/challenges/chinese/03-front-end-libraries/react/add-comments-in-jsx.md
index 3b00f985877..94a2131b279 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/add-comments-in-jsx.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/add-comments-in-jsx.md
@@ -1,6 +1,6 @@
---
id: 5a24bbe0dba28a8d3cbd4c5e
-title: 在 JSX 中添加注释
+title: Add Comments in JSX
challengeType: 6
forumTopicId: 301376
dashedName: add-comments-in-jsx
@@ -8,35 +8,35 @@ dashedName: add-comments-in-jsx
# --description--
-JSX 是一种可以编译成有效 JavaScript 的语法。有时，为了便于阅读，你可能需要在代码中添加注释。像大多数编程语言一样，JSX 也有自己的方法来实现这一点。
+JSX is a syntax that gets compiled into valid JavaScript. Sometimes, for readability, you might need to add comments to your code. Like most programming languages, JSX has its own way to do this.
-要将注释放在 JSX 中，可以使用`{/* */}`语法来包裹注释文本。
+To put comments inside JSX, you use the syntax `{/* */}` to wrap around the comment text.
# --instructions--
-代码编辑器中的 JSX 元素与你在上一个挑战中创建的元素类似。在提供的`div`元素中的某处添加注释，而不修改现有的`h1`或`p`元素。
+The code editor has a JSX element similar to what you created in the last challenge. Add a comment somewhere within the provided `div` element, without modifying the existing `h1` or `p` elements.
# --hints--
-常量`JSX`应该返回一个`div`元素。
+The constant `JSX` should return a `div` element.
```js
assert(JSX.type === 'div');
```
-`div`应该包含一个`h1`标签作为第一个元素。
+The `div` should contain an `h1` tag as the first element.
```js
assert(JSX.props.children[0].type === 'h1');
```
-`div`应该包含一个`p`标签作为第二个元素。
+The `div` should contain a `p` tag as the second element.
```js
assert(JSX.props.children[1].type === 'p');
```
-当前的 `h1` 和 `p` 元素不能被修改。
+The existing `h1` and `p` elements should not be modified.
```js
assert(
@@ -45,7 +45,7 @@ assert(
);
```
-`JSX`应该包含一个注释。
+The `JSX` should use valid comment syntax.
```js
assert(/<div>[\s\S]*{\s*\/\*[\s\S]*\*\/\s*}[\s\S]*<\/div>/.test(code));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/add-event-listeners.md b/curriculum/challenges/chinese/03-front-end-libraries/react/add-event-listeners.md
index 2702c8dc88b..b1f8942c410 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/add-event-listeners.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/add-event-listeners.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403617e
-title: 添加事件侦听器
+title: Add Event Listeners
challengeType: 6
forumTopicId: 301377
dashedName: add-event-listeners
@@ -8,19 +8,19 @@ dashedName: add-event-listeners
# --description--
-`componentDidMount()`方法也是添加特定功能所需的任何事件监听器的最佳位置。React 提供了一个合成事件系统，它将本地事件系统封装在浏览器中。这意味着，不管用户的浏览器如何，合成事件系统的行为都完全相同--即使不同浏览器之间的本地事件的行为可能不同。
+The `componentDidMount()` method is also the best place to attach any event listeners you need to add for specific functionality. React provides a synthetic event system which wraps the native event system present in browsers. This means that the synthetic event system behaves exactly the same regardless of the user's browser - even if the native events may behave differently between different browsers.
-你已经使用了一些合成事件处理程序，如`onClick()`。React 的合成事件系统非常适合用于你在 DOM 元素上管理的大多数交互。但是，如果要将事件处理程序附加到 document 或 window 对象，则必须直接执行此操作。
+You've already been using some of these synthetic event handlers such as `onClick()`. React's synthetic event system is great to use for most interactions you'll manage on DOM elements. However, if you want to attach an event handler to the document or window objects, you have to do this directly.
# --instructions--
-在`componentDidMount()`方法中为`keydown`事件添加事件监听器，并让这些事件触发回调`handleKeyPress()`。你可以使用`document.addEventListener()`，它将事件（用引号括起来）作为第一个参数，将回调作为第二个参数。
+Attach an event listener in the `componentDidMount()` method for `keydown` events and have these events trigger the callback `handleKeyPress()`. You can use `document.addEventListener()` which takes the event (in quotes) as the first argument and the callback as the second argument.
-然后，在`componentWillUnmount()`中移除相同的事件监听器。你可以把相同的参数传递给`document.removeEventListener()`。在卸载和销毁 React 组件之前，最好使用这种生命周期方法对它们进行清理。移除事件监听器就是这样一个清理操作的例子。
+Then, in `componentWillUnmount()`, remove this same event listener. You can pass the same arguments to `document.removeEventListener()`. It's good practice to use this lifecycle method to do any clean up on React components before they are unmounted and destroyed. Removing event listeners is an example of one such clean up action.
# --hints--
-`MyComponent`应该渲染一个包含`h1`标签的`div`元素。
+`MyComponent` should render a `div` element which wraps an `h1` tag.
```js
assert(
@@ -31,7 +31,7 @@ assert(
);
```
-应该在`componentDidMount`中将 keydown 事件监听添加到到 document 上。
+A keydown listener should be attached to the document in `componentDidMount`.
```js
assert(
@@ -47,7 +47,7 @@ assert(
);
```
-应该在`componentWillUnmount`中将 document 上的 keydown 事件监听移除。
+The keydown listener should be removed from the document in `componentWillUnmount`.
```js
assert(
@@ -63,7 +63,7 @@ assert(
);
```
-一旦组件装载完毕，按`enter`应该会更新其 state 并渲染到`h1`标签。
+Once the component has mounted, pressing `enter` should update its state and the rendered `h1` tag.
```js
async () => {
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/add-inline-styles-in-react.md b/curriculum/challenges/chinese/03-front-end-libraries/react/add-inline-styles-in-react.md
index dc9035f13b3..90a51753101 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/add-inline-styles-in-react.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/add-inline-styles-in-react.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036182
-title: 在 React 中添加内联样式
+title: Add Inline Styles in React
challengeType: 6
forumTopicId: 301378
dashedName: add-inline-styles-in-react
@@ -8,41 +8,41 @@ dashedName: add-inline-styles-in-react
# --description--
-在上一次挑战中，你可能已经注意到，除了设置为 JavaScript 对象的`style`属性之外，与 HTML 内联样式相比，React 的内联样式还有其他几个语法差异。首先，某些 CSS 样式属性的名称使用驼峰式命名。例如，最后一个挑战用`fontSize`而不是`font-size`来设置字体的大小。对于 JavaScript 对象属性来说，像`font-size`这样的连字符命名是无效的语法，所以 React 使用驼峰式命名。通常，任何连字符的 style 属性在 JSX 中都是使用驼峰式命名的。
+You may have noticed in the last challenge that there were several other syntax differences from HTML inline styles in addition to the `style` attribute set to a JavaScript object. First, the names of certain CSS style properties use camel case. For example, the last challenge set the size of the font with `fontSize` instead of `font-size`. Hyphenated words like `font-size` are invalid syntax for JavaScript object properties, so React uses camel case. As a rule, any hyphenated style properties are written using camel case in JSX.
-除非另有规定，否则所有属性值是长度的（如`height`、`width`和`fontSize`）其单位都假定为`px`。例如，如果要使用`em`，可以用引号将值和单位括起来，例如`{fontSize: "4em"}`。除了默认为`px`的长度值之外，所有其他属性值都应该用引号括起来。
+All property value length units (like `height`, `width`, and `fontSize`) are assumed to be in `px` unless otherwise specified. If you want to use `em`, for example, you wrap the value and the units in quotes, like `{fontSize: "4em"}`. Other than the length values that default to `px`, all other property values should be wrapped in quotes.
# --instructions--
-如果你有大量样式，你可以将 style`对象`分配给一个常量，以保持代码的组织有序。取消对`styles`常量的注释，并声明具有三个样式属性及对应值的`对象`。使`div`的文字颜色为`"purple"`、字号为`40`、边框为`"2px solid purple"`。然后设置`style`属性，使其等于`styles`常量。
+If you have a large set of styles, you can assign a style `object` to a constant to keep your code organized. Initialize a `styles` constant and assign an `object` with three style properties and their values to it. Give the `div` a color of `"purple"`, a font-size of `40`, and a border of `"2px solid purple"`. Then set the `style` attribute equal to the `styles` constant.
# --hints--
-`styles`变量应该是具有三个属性的`对象`。
+The `styles` variable should be an `object` with three properties.
```js
assert(Object.keys(styles).length === 3);
```
-`styles`变量的`color`属性应该设置为`purple`。
+The `styles` variable should have a `color` property set to a value of `purple`.
```js
assert(styles.color === 'purple');
```
-`styles`变量应该将`fontSize`属性设置为`40`。
+The `styles` variable should have a `fontSize` property set to a value of `40`.
```js
assert(styles.fontSize === 40);
```
-`styles`变量的`border`属性应该设置为`2px solid purple`。
+The `styles` variable should have a `border` property set to a value of `2px solid purple`.
```js
assert(styles.border === '2px solid purple');
```
-组件应该渲染一个`div`元素。
+The component should render a `div` element.
```js
assert(
@@ -53,7 +53,7 @@ assert(
);
```
-`div`元素的样式应该由`styles`对象定义。
+The `div` element should have its styles defined by the `styles` object.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/bind-this-to-a-class-method.md b/curriculum/challenges/chinese/03-front-end-libraries/react/bind-this-to-a-class-method.md
index db5eb05a770..291e3712d7d 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/bind-this-to-a-class-method.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/bind-this-to-a-class-method.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036174
-title: 将 this 绑定到 Class 方法
+title: Bind 'this' to a Class Method
challengeType: 6
forumTopicId: 301379
dashedName: bind-this-to-a-class-method
@@ -8,23 +8,23 @@ dashedName: bind-this-to-a-class-method
# --description--
-除了设置和更新`state`之外，你还可以为组件类定义方法。类方法通常需要使用`this`关键字，以便它可以访问方法中类的属性（例如`state`和`props`）。有几种方法可以让你的类方法访问`this`。
+In addition to setting and updating `state`, you can also define methods for your component class. A class method typically needs to use the `this` keyword so it can access properties on the class (such as `state` and `props`) inside the scope of the method. There are a few ways to allow your class methods to access `this`.
-一种常见的方法是在构造函数中显式地绑定`this`，这样当组件初始化时，`this`就会绑定到类方法。你可能已经注意到上一个挑战使用了`this.handleClick = this.handleClick.bind(this)`用于其在构造函数中的`handleClick`方法。然后，当你在类方法中调用像`this.setState()`这样的函数时，`this`指的是这个类，而不是`undefined`。
+One common way is to explicitly bind `this` in the constructor so `this` becomes bound to the class methods when the component is initialized. You may have noticed the last challenge used `this.handleClick = this.handleClick.bind(this)` for its `handleClick` method in the constructor. Then, when you call a function like `this.setState()` within your class method, `this` refers to the class and will not be `undefined`.
-**注意：** `this`关键字是 JavaScript 中最令人困惑的方面之一，但它在 React 中扮演着重要的角色。虽然它的行为在这里是完全正常的，但是这些课程并不深入研究`this`，所以如果以上内容令你感到困惑，请参考其他课程！
+**Note:** The `this` keyword is one of the most confusing aspects of JavaScript but it plays an important role in React. Although its behavior here is totally normal, these lessons aren't the place for an in-depth review of `this` so please refer to other lessons if the above is confusing!
# --instructions--
-代码编辑器有一个带有`state`的组件，用于跟踪项目计数。它还有一个方法，允许你增加此项目计数。但是，该方法不起作用，因为它使用了未定义的`this`关键字。可以通过将`this`显式绑定到组件构造函数中的`addItem()`方法来修复它。
+The code editor has a component with a `state` that keeps track of the text. It also has a method which allows you to set the text to `"You clicked!"`. However, the method doesn't work because it's using the `this` keyword that is undefined. Fix it by explicitly binding `this` to the `handleClick()` method in the component's constructor.
-接下来，向 render 方法中的`button`元素添加一个单击处理程序。当按钮接收到单击事件时，它应该触发`addItem()`方法。记住，传递给`onClick`处理程序的方法需要使用花括号，因为它应该直接被解释为 JavaScript。
+Next, add a click handler to the `button` element in the render method. It should trigger the `handleClick()` method when the button receives a click event. Remember that the method you pass to the `onClick` handler needs curly braces because it should be interpreted directly as JavaScript.
-完成上述步骤后，你应该可以单击按钮并查看 HTML 中的项目计数增量。
+Once you complete the above steps you should be able to click the button and see `You clicked!`.
# --hints--
-`MyComponent`应返回`div`元素，该元素按顺序包含两个元素，一个按钮和一个`h1`元素。
+`MyComponent` should return a `div` element which wraps two elements, a button and an `h1` element, in that order.
```js
assert(
@@ -40,7 +40,7 @@ assert(
);
```
-`MyComponent`的 state 应该使用键值对`{ itemCount: 0 }`进行初始化。
+The state of `MyComponent` should initialize with the key value pair `{ text: "Hello" }`.
```js
assert(
@@ -48,7 +48,7 @@ assert(
);
```
-单击`button`元素应该运行`addItem`方法，并使 state`itemCount`的计数增加`1`。
+Clicking the `button` element should run the `handleClick` method and set the state `text` to `"You clicked!"`.
```js
async () => {
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/change-inline-css-conditionally-based-on-component-state.md b/curriculum/challenges/chinese/03-front-end-libraries/react/change-inline-css-conditionally-based-on-component-state.md
index 79bcf3fa2d8..9a088525940 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/change-inline-css-conditionally-based-on-component-state.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/change-inline-css-conditionally-based-on-component-state.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036189
-title: 根据组件状态有条件地更改内联 CSS
+title: Change Inline CSS Conditionally Based on Component State
challengeType: 6
forumTopicId: 301380
dashedName: change-inline-css-conditionally-based-on-component-state
@@ -8,17 +8,17 @@ dashedName: change-inline-css-conditionally-based-on-component-state
# --description--
-此时，你已经看到了一些条件渲染的应用程序和内联样式的使用。这里还有一个将这两个主题结合在一起的例子。你也可以根据 React 组件的 state 有条件地渲染 CSS。要执行此操作，请检查条件，如果满足该条件，则修改在 render 方法中分配给 JSX 元素的样式对象。
+At this point, you've seen several applications of conditional rendering and the use of inline styles. Here's one more example that combines both of these topics. You can also render CSS conditionally based on the state of a React component. To do this, you check for a condition, and if that condition is met, you modify the styles object that's assigned to the JSX elements in the render method.
-这个范例对于更加容易理解，因为相比传统的通过直接修改 DOM 元素来应用样式的方法（这在 jQuery 中非常常见），这种方法是一个戏剧性的转变。在传统方法中，你必须跟踪元素何时发生变化，并直接处理实际操作，这使得跟踪变化变得很困难，也可能会让你的用户界面变得不可预测。当你根据一个条件设置一个样式对象时，你描述了 UI 作为应用程序的状态函数应当如何展现。如此便有一个清晰的单向流动的信息流。这是使用 React 编写应用程序时的首选方法。
+This paradigm is important to understand because it is a dramatic shift from the more traditional approach of applying styles by modifying DOM elements directly (which is very common with jQuery, for example). In that approach, you must keep track of when elements change and also handle the actual manipulation directly. It can become difficult to keep track of changes, potentially making your UI unpredictable. When you set a style object based on a condition, you describe how the UI should look as a function of the application's state. There is a clear flow of information that only moves in one direction. This is the preferred method when writing applications with React.
# --instructions--
-代码编辑器有一个简单的带有边框样式的受控 input 组件。如果用户在输入框中键入超过 15 个字符的文本，你希望将此边框变成红色。添加一个条件来检查这一点，如果条件有效，则将 input 的边框样式设置为`3px solid red`。你可以通过在 input 中输入文本来尝试。
+The code editor has a simple controlled input component with a styled border. You want to style this border red if the user types more than 15 characters of text in the input box. Add a condition to check for this and, if the condition is valid, set the input border style to `3px solid red`. You can try it out by entering text in the input.
# --hints--
-`GateKeeper`组件应该渲染一个de>div</code>元素。
+The `GateKeeper` component should render a `div` element.
```js
assert(
@@ -29,7 +29,7 @@ assert(
);
```
-`GateKeeper`组件应使用设置为空字符串的 state `input`进行初始化。
+The `GateKeeper` component should be initialized with a state key `input` set to an empty string.
```js
assert(
@@ -40,7 +40,7 @@ assert(
);
```
-`GateKeeper`组件应该渲染一个`h3`标签和一个`input`标签。
+The `GateKeeper` component should render an `h3` tag and an `input` tag.
```js
assert(
@@ -54,7 +54,7 @@ assert(
);
```
-`input`标签`border`属性的样式应该初始化为`1px solid black`。
+The `input` tag should initially have a style of `1px solid black` for the `border` property.
```js
assert(
@@ -67,7 +67,7 @@ assert(
);
```
-如果 state 中 input 的值超过 15 个字符，则 `input` 标签的 border 样式应为`3px solid red`。
+The `input` tag should be styled with a border of `3px solid red` if the input value in state is longer than 15 characters.
```js
async () => {
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/compose-react-components.md b/curriculum/challenges/chinese/03-front-end-libraries/react/compose-react-components.md
index 06db11ae14a..f39e29008ac 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/compose-react-components.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/compose-react-components.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036166
-title: 组合 React 组件
+title: Compose React Components
challengeType: 6
forumTopicId: 301381
dashedName: compose-react-components
@@ -8,17 +8,17 @@ dashedName: compose-react-components
# --description--
-随着挑战继续，我们将组合使用更复杂的 React 组件和 JSX，有一点需要注意。在其他组件中渲染 ES6 风格的类组件和渲染你在过去几个挑战中使用的简单组件没有什么不同。你可以在其他组件中渲染 JSX 元素、无状态功能组件和 ES6 类组件。
+As the challenges continue to use more complex compositions with React components and JSX, there is one important point to note. Rendering ES6 style class components within other components is no different than rendering the simple components you used in the last few challenges. You can render JSX elements, stateless functional components, and ES6 class components within other components.
# --instructions--
-在代码编辑器中，`TypesOfFood`组件已经渲染了一个名为`Vegetables`的组件。此外，还有上次挑战中的`Fruits`组件。
+In the code editor, the `TypesOfFood` component is already rendering a component called `Vegetables`. Also, there is the `Fruits` component from the last challenge.
-在`Fruits`中嵌套两个组件，首先`NonCitrus`，然后是`Citrus`，这两个组件都是在后台为你提供的。接下来，将`Fruits`类组件嵌到`TypesOfFood`组件中，位于`h1`标题下方和`Vegetables`上方。结果应该是一系列嵌套的组件，它们使用两种不同的组件类型。
+Nest two components inside of `Fruits` — first `NonCitrus`, and then `Citrus`. Both of these components are provided for you behind the scenes. Next, nest the `Fruits` class component into the `TypesOfFood` component, below the `h1` header and above `Vegetables`. The result should be a series of nested components, which uses two different component types.
# --hints--
-`TypesOfFood`组件应该返回单个`div`元素。
+The `TypesOfFood` component should return a single `div` element.
```js
assert(
@@ -29,7 +29,7 @@ assert(
);
```
-`TypesOfFood`组件应该返回`Fruits`组件。
+The `TypesOfFood` component should return the `Fruits` component.
```js
assert(
@@ -40,7 +40,7 @@ assert(
);
```
-`Fruits`组件应该返回`NonCitrus`组件和`Citrus`组件。
+The `Fruits` component should return the `NonCitrus` component and the `Citrus` component.
```js
assert(
@@ -55,7 +55,7 @@ assert(
);
```
-`TypesOfFood`组件应该返回`Vegetables`组件，且其位于`Fruits`组件之下。
+The `TypesOfFood` component should return the `Vegetables` component below the `Fruits` component.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-complex-jsx-element.md b/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-complex-jsx-element.md
index 44ddb894384..9bc178406a0 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-complex-jsx-element.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-complex-jsx-element.md
@@ -1,6 +1,6 @@
---
id: 5a24bbe0dba28a8d3cbd4c5d
-title: 创建一个复杂的 JSX 元素
+title: Create a Complex JSX Element
challengeType: 6
forumTopicId: 301382
dashedName: create-a-complex-jsx-element
@@ -8,17 +8,17 @@ dashedName: create-a-complex-jsx-element
# --description--
-上一个挑战是 JSX 的一个简单示例，但 JSX 也可以表示更复杂的 HTML。
+The last challenge was a simple example of JSX, but JSX can represent more complex HTML as well.
-关于嵌套的 JSX，你需要知道的一件重要的事情，那就是它必须返回单个元素。
+One important thing to know about nested JSX is that it must return a single element.
-这个父元素将包裹所有其他级别的嵌套元素。
+This one parent element would wrap all of the other levels of nested elements.
-例如，几个作为兄弟元素而编写的JSX元素没有父元素包裹将不会被转换。
+For instance, several JSX elements written as siblings with no parent wrapper element will not transpile.
-这里是一个示例：
+Here's an example:
-**有效的 JSX：**
+**Valid JSX:**
```jsx
<div>
@@ -28,7 +28,7 @@ dashedName: create-a-complex-jsx-element
</div>
```
-**无效的 JSX：**
+**Invalid JSX:**
```jsx
<p>Paragraph One</p>
@@ -38,39 +38,39 @@ dashedName: create-a-complex-jsx-element
# --instructions--
-定义一个新的常量`JSX`，渲染一个`div`，其中依次包含以下元素：
+Define a new constant `JSX` that renders a `div` which contains the following elements in order:
-一个`h1`，一个`p`，一个包含三个`li`项的无序列表。你可以在每个元素中包含任何你想要的文本。
+An `h1`, a `p`, and an unordered list that contains three `li` items. You can include any text you want within each element.
-**注意：** 当像这样渲染多个元素时，你可以把它们都用圆括号括起来，但是这并不是必须的。还请注意，此挑战使用`div`标签把所有子元素包裹在里面。如果删除`div`，JSX 将不会编译这些元素。请记住这一点，因为当你在 React 组件中返回 JSX 元素时它也适用。
+**Note:** When rendering multiple elements like this, you can wrap them all in parentheses, but it's not strictly required. Also notice this challenge uses a `div` tag to wrap all the child elements within a single parent element. If you remove the `div`, the JSX will no longer transpile. Keep this in mind, since it will also apply when you return JSX elements in React components.
# --hints--
-常量`JSX`应该返回一个`div`元素。
+The constant `JSX` should return a `div` element.
```js
assert(JSX.type === 'div');
```
-`div`应该包含一个`p`标签作为第二个元素。
+The `div` should contain an `h1` tag as the first element.
```js
assert(JSX.props.children[0].type === 'h1');
```
-`div`应该包含一个`ul`标签作为第三个元素。
+The `div` should contain a `p` tag as the second element.
```js
assert(JSX.props.children[1].type === 'p');
```
-`div`应该包含一个`h1`标签作为第一个元素。
+The `div` should contain a `ul` tag as the third element.
```js
assert(JSX.props.children[2].type === 'ul');
```
-`ul`应该包含三个`li`元素。
+The `ul` should contain three `li` elements.
```js
assert(
@@ -89,7 +89,6 @@ ReactDOM.render(JSX, document.getElementById('root'))
```
## --seed-contents--
-
```jsx
```
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-component-with-composition.md b/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-component-with-composition.md
index e93aa05ab59..6abea21598d 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-component-with-composition.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-component-with-composition.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036164
-title: 用组合的方式创建一个 React 组件
+title: Create a Component with Composition
challengeType: 6
forumTopicId: 301383
dashedName: create-a-component-with-composition
@@ -8,9 +8,9 @@ dashedName: create-a-component-with-composition
# --description--
-现在我们来看看如何组合多个 React 组件。想象一下，你正在构建一个应用程序，并创建了三个组件：`Navbar`、`Dashboard`和`Footer`。
+Now we will look at how we can compose multiple React components together. Imagine you are building an App and have created three components, a `Navbar`, `Dashboard`, and `Footer`.
-要将这些组件组合在一起，你可以创建一个`App`*父组件*，将这三个组件分别渲染成为*子组件*。要在 React 组件中渲染一个子组件，你需要在 JSX 中包含作为自定义 HTML 标签编写的组件名称。例如，在`render`方法中，你可以这样编写：
+To compose these components together, you could create an `App` *parent* component which renders each of these three components as *children*. To render a component as a child in a React component, you include the component name written as a custom HTML tag in the JSX. For example, in the `render` method you could write:
```jsx
return (
@@ -22,17 +22,17 @@ return (
)
```
-当 React 遇到引用另一个组件的自定义 HTML 标签时（如本例所示，组件名称包含在`< />`中），它在标签的位置渲染该组件的标签。这可以说明`App`组件和`Navbar`、`Dashboard`以及`Footer`之间的父子关系。
+When React encounters a custom HTML tag that references another component (a component name wrapped in `< />` like in this example), it renders the markup for that component in the location of the tag. This should illustrate the parent/child relationship between the `App` component and the `Navbar`, `Dashboard`, and `Footer`.
# --instructions--
-在代码编辑器中，有一个名为`ChildComponent`的简单功能组件和一个名为`ParentComponent`的 React 组件。通过在`ParentComponent`中渲染`ChildComponent`来将两者组合在一起。确保使用正斜杠关闭`ChildComponent`标签。
+In the code editor, there is a simple functional component called `ChildComponent` and a class component called `ParentComponent`. Compose the two together by rendering the `ChildComponent` within the `ParentComponent`. Make sure to close the `ChildComponent` tag with a forward slash.
-**注意：** `ChildComponent`是使用 ES6 的箭头函数定义的，因为这是使用 React 时非常常见的做法。但是，要知道这只是一个函数。如果你不熟悉箭头函数语法，请参阅 JavaScript 部分。
+**Note:** `ChildComponent` is defined with an ES6 arrow function because this is a very common practice when using React. However, know that this is just a function. If you aren't familiar with the arrow function syntax, please refer to the JavaScript section.
# --hints--
-React 组件应该返回单个`div`元素。
+The React component should return a single `div` element.
```js
assert(
@@ -43,7 +43,7 @@ assert(
);
```
-组件应该返回两个嵌套的元素。
+The component should return two nested elements.
```js
assert(
@@ -54,7 +54,7 @@ assert(
);
```
-组件的第二个子元素应该是 ChildComponent。
+The component should return the ChildComponent as its second child.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-controlled-form.md b/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-controlled-form.md
index 4fd7c8fc96c..e488d2d26bc 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-controlled-form.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-controlled-form.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036179
-title: 创建一个可以控制的表单
+title: Create a Controlled Form
challengeType: 6
forumTopicId: 301384
dashedName: create-a-controlled-form
@@ -8,21 +8,21 @@ dashedName: create-a-controlled-form
# --description--
-上一个挑战展示了 React 能控制某些元素的内部 state，比如`input`和`textarea`，这使得这些元素成为受控组件。这也适用于其他表单元素，包括常规的 HTML 表单`form`元素。
+The last challenge showed that React can control the internal state for certain elements like `input` and `textarea`, which makes them controlled components. This applies to other form elements as well, including the regular HTML `form` element.
# --instructions--
-`MyForm`组件中是一个带有提交处理程序的空`form`元素，提交处理程序将在提交表单时被调用。
+The `MyForm` component is set up with an empty `form` with a submit handler. The submit handler will be called when the form is submitted.
-我们增加了一个提交表单的按钮。你可以看到它的`type`被设置为`submit`，表明它是控制表单的按钮。在表单中添加`input`元素，并像上次挑战一样设置其`value`和`onChange()`属性。然后，你应该完成`handleSubmit`方法，以便将组件 state 属性`submit`设置为本地`state`下的当前输入值。
+We've added a button which submits the form. You can see it has the `type` set to `submit` indicating it is the button controlling the form. Add the `input` element in the `form` and set its `value` and `onChange()` attributes like the last challenge. You should then complete the `handleSubmit` method so that it sets the component state property `submit` to the current input value in the local `state`.
-**注意：** 你还必须在提交处理程序中调用`event.preventDefault()`，以防止默认的表单提交行为刷新网页。
+**Note:** You also must call `event.preventDefault()` in the submit handler, to prevent the default form submit behavior which will refresh the web page.
-最后，在`form`元素之后创建一个`h1`标签，该标签从组件的`state`渲染`submit`的值。然后，你可以在表单中键入任何内容，然后单击按钮（或按 enter 键），你的输入会渲染到页面上。
+Finally, create an `h1` tag after the `form` which renders the `submit` value from the component's `state`. You can then type in the form and click the button (or press enter), and you should see your input rendered to the page.
# --hints--
-`MyForm`应该返回一个包含`form`和`h1`标签的`div`元素，其中，表单中应该包括一个`input`和一个`button`。
+`MyForm` should return a `div` element which contains a `form` and an `h1` tag. The form should include an `input` and a `button`.
```js
assert(
@@ -38,7 +38,7 @@ assert(
);
```
-`MyForm`的 state 应该用`input`和`submit`属性初始化，且两者都为空字符串。
+The state of `MyForm` should initialize with `input` and `submit` properties, both set to empty strings.
```js
assert(
@@ -47,42 +47,38 @@ assert(
);
```
-`input`元素中的输入应该会更新组件中 state 的`input`属性。
+Typing in the `input` element should update the `input` property of the component's state.
```js
-async () => {
- const waitForIt = (fn) =>
- new Promise((resolve, reject) => setTimeout(() => resolve(fn()), 250));
+(() => {
const mockedComponent = Enzyme.mount(React.createElement(MyForm));
const _1 = () => {
mockedComponent.setState({ input: '' });
- return waitForIt(() => mockedComponent.state('input'));
+ return mockedComponent.state('input');
};
const _2 = () => {
mockedComponent
.find('input')
.simulate('change', { target: { value: 'TestInput' } });
- return waitForIt(() => ({
+ return {
state: mockedComponent.state('input'),
inputVal: mockedComponent.find('input').props().value
- }));
+ };
};
- const before = await _1();
- const after = await _2();
+ const before = _1();
+ const after = _2();
assert(
before === '' &&
after.state === 'TestInput' &&
after.inputVal === 'TestInput'
);
-};
+})();
```
-提交表单应该运行`handleSubmit`，它应该将 state 中的`submit`属性设置为当前输入。
+Submitting the form should run `handleSubmit` which should set the `submit` property in state equal to the current input.
```js
-async () => {
- const waitForIt = (fn) =>
- new Promise((resolve, reject) => setTimeout(() => resolve(fn()), 250));
+(() => {
const mockedComponent = Enzyme.mount(React.createElement(MyForm));
const _1 = () => {
mockedComponent.setState({ input: '' });
@@ -90,24 +86,22 @@ async () => {
mockedComponent
.find('input')
.simulate('change', { target: { value: 'SubmitInput' } });
- return waitForIt(() => mockedComponent.state('submit'));
+ return mockedComponent.state('submit');
};
const _2 = () => {
mockedComponent.find('form').simulate('submit');
- return waitForIt(() => mockedComponent.state('submit'));
+ return mockedComponent.state('submit');
};
- const before = await _1();
- const after = await _2();
+ const before = _1();
+ const after = _2();
assert(before === '' && after === 'SubmitInput');
-};
+})();
```
-`h1`标题应该从组件的 state 渲染`submit`字段的值。
+The `h1` header should render the value of the `submit` field from the component's state.
```js
-async () => {
- const waitForIt = (fn) =>
- new Promise((resolve, reject) => setTimeout(() => resolve(fn()), 250));
+(() => {
const mockedComponent = Enzyme.mount(React.createElement(MyForm));
const _1 = () => {
mockedComponent.setState({ input: '' });
@@ -115,16 +109,16 @@ async () => {
mockedComponent
.find('input')
.simulate('change', { target: { value: 'TestInput' } });
- return waitForIt(() => mockedComponent.find('h1').text());
+ return mockedComponent.find('h1').text();
};
const _2 = () => {
mockedComponent.find('form').simulate('submit');
- return waitForIt(() => mockedComponent.find('h1').text());
+ return mockedComponent.find('h1').text();
};
- const before = await _1();
- const after = await _2();
+ const before = _1();
+ const after = _2();
assert(before === '' && after === 'TestInput');
-};
+})();
```
# --seed--
@@ -155,7 +149,7 @@ class MyForm extends React.Component {
}
handleSubmit(event) {
// Change code below this line
-
+
// Change code above this line
}
render() {
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-controlled-input.md b/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-controlled-input.md
index 7331a28f1bc..88c018d8079 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-controlled-input.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-controlled-input.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036178
-title: 创建一个可以控制的输入框
+title: Create a Controlled Input
challengeType: 6
forumTopicId: 301385
dashedName: create-a-controlled-input
@@ -8,23 +8,23 @@ dashedName: create-a-controlled-input
# --description--
-你的应用程序可能在`state`和渲染的 UI 之间有更复杂的交互。例如，用于文本输入的表单控件元素（如`input`和`textarea`）在用户键入时在 DOM 中维护自己的 state。通过 React，你可以将这种可变 state 转移到 React 组件的`state`中。用户的输入变成了应用程序`state`的一部分，因此 React 控制该输入字段的值。通常，如果你的 React 组件具有用户可以键入的输入字段，那么它将是一个受控的输入表单。
+Your application may have more complex interactions between `state` and the rendered UI. For example, form control elements for text input, such as `input` and `textarea`, maintain their own state in the DOM as the user types. With React, you can move this mutable state into a React component's `state`. The user's input becomes part of the application `state`, so React controls the value of that input field. Typically, if you have React components with input fields the user can type into, it will be a controlled input form.
# --instructions--
-代码编辑器具有一个名为`ControlledInput`的组件框架，用于创建受控的`input`元素。组件的`state`已经被包含空字符串的`input`属性初始化。此值表示用户在`input`字段中键入的文本。
+The code editor has the skeleton of a component called `ControlledInput` to create a controlled `input` element. The component's `state` is already initialized with an `input` property that holds an empty string. This value represents the text a user types into the `input` field.
-首先，创建一个名为`handleChange()`的方法，该方法具有一个名为`event`的参数。方法被调用时，它接收一个`event`对象，该对象包含一个来自`input`元素的字符串文本。你可以使用方法内的`event.target.value`来访问这个字符串。用这个新字符串更新组件的`state`的`input`属性。
+First, create a method called `handleChange()` that has a parameter called `event`. When the method is called, it receives an `event` object that contains a string of text from the `input` element. You can access this string with `event.target.value` inside the method. Update the `input` property of the component's `state` with this new string.
-在 render 方法中，在`h4`标签之上创建`input`元素。添加一个`value`属性，它等于组件的`state`的`input`属性。然后将`onChange()`事件处理程序设置到`handleChange()`方法。
+In the render method, create the `input` element above the `h4` tag. Add a `value` attribute which is equal to the `input` property of the component's `state`. Then add an `onChange()` event handler set to the `handleChange()` method.
-在输入框中键入时，该文本由`handleChange()`方法处理，该文本被设置为本地`state`中的`input`属性，并渲染在页面上的`input`框中。组件`state`是输入数据的唯一真实来源。
+When you type in the input box, that text is processed by the `handleChange()` method, set as the `input` property in the local `state`, and rendered as the value in the `input` box on the page. The component `state` is the single source of truth regarding the input data.
-最后也是最重要的，不要忘记在构造函数中添加必要的绑定。
+Last but not least, don't forget to add the necessary bindings in the constructor.
# --hints--
-`ControlledInput`应该返回包含一个`input`标签和`p`标签的`div`元素。
+`ControlledInput` should return a `div` element which contains an `input` and a `p` tag.
```js
assert(
@@ -39,7 +39,7 @@ assert(
);
```
-`ControlledInput`的 state 应该使用设置为空字符串的`input`属性初始化。
+The state of `ControlledInput` should initialize with an `input` property set to an empty string.
```js
assert.strictEqual(
@@ -48,7 +48,7 @@ assert.strictEqual(
);
```
-input 元素中的键入值应该更新 input 的 state 和值，并且`p`元素应该在输入时呈现 state。
+Typing in the input element should update the state and the value of the input, and the `p` element should render this state as you type.
```js
async () => {
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-react-component.md b/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-react-component.md
index 3cd1a40f3a2..6a92615c1db 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-react-component.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-react-component.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036163
-title: 创建一个 React 组件
+title: Create a React Component
challengeType: 6
forumTopicId: 301386
dashedName: create-a-react-component
@@ -8,7 +8,7 @@ dashedName: create-a-react-component
# --description--
-定义 React 组件的另一种方法是使用 ES6 的`class`语法。在以下示例中，`Kitten`扩展了`React.Component`：
+The other way to define a React component is with the ES6 `class` syntax. In the following example, `Kitten` extends `React.Component`:
```jsx
class Kitten extends React.Component {
@@ -24,21 +24,21 @@ class Kitten extends React.Component {
}
```
-这将创建一个 ES6 类`Kitten`，它扩展了`React.Component`类。因此，`Kitten`类现在可以访问许多有用的 React 功能，例如本地状态和生命周期钩子。如果你还不熟悉这些术语，请不要担心，在以后的挑战中我们将更详细地介绍它们。 另请注意，`Kitten`类中定义了一个调用`super()`方法的`constructor`。它使用`super()`调用父类的构造函数，即本例中的`React.Component`。构造函数是使用`class`关键字创建的特殊方法，它用在实例初始化之前。最佳做法是在组件的`constructor`里调用`super`，并将`props`传递给它们，这样可以保证组件能够正确地初始化。现在，你只需要知道这是标准的做法。很快你会看到构造函数的其他用途以及`props`。
+This creates an ES6 class `Kitten` which extends the `React.Component` class. So the `Kitten` class now has access to many useful React features, such as local state and lifecycle hooks. Don't worry if you aren't familiar with these terms yet, they will be covered in greater detail in later challenges. Also notice the `Kitten` class has a `constructor` defined within it that calls `super()`. It uses `super()` to call the constructor of the parent class, in this case `React.Component`. The constructor is a special method used during the initialization of objects that are created with the `class` keyword. It is best practice to call a component's `constructor` with `super`, and pass `props` to both. This makes sure the component is initialized properly. For now, know that it is standard for this code to be included. Soon you will see other uses for the constructor as well as `props`.
# --instructions--
-`MyComponent`是使用类语法在代码编辑器中定义的。完成`render`方法的编写，使其返回`div`元素，其中包含文本内容为`Hello React!`的`h1`元素。
+`MyComponent` is defined in the code editor using class syntax. Finish writing the `render` method so it returns a `div` element that contains an `h1` with the text `Hello React!`.
# --hints--
-该 React 组件应该返回一个`div`元素。
+The React component should return a `div` element.
```js
assert(Enzyme.shallow(React.createElement(MyComponent)).type() === 'div');
```
-返回的`div`中应该渲染一个`h1`标题。
+The returned `div` should render an `h1` header within it.
```js
assert(
@@ -48,7 +48,7 @@ assert(
);
```
-`h1`标题中应该包含字符串`Hello React!`。
+The `h1` header should contain the string `Hello React!`.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-simple-jsx-element.md b/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-simple-jsx-element.md
index 83abfc5fb61..ee4ee865a52 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-simple-jsx-element.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-simple-jsx-element.md
@@ -1,6 +1,6 @@
---
id: 587d7dbc367417b2b2512bb1
-title: 创建一个简单的 JSX 元素
+title: Create a Simple JSX Element
challengeType: 6
forumTopicId: 301390
dashedName: create-a-simple-jsx-element
@@ -8,29 +8,29 @@ dashedName: create-a-simple-jsx-element
# --description--
-**简介：** React 是由 Facebook 创建和维护的开源视图库。它是渲染当代 Web 应用程序用户界面（UI）的绝佳工具。
+**Intro:** React is an Open Source view library created and maintained by Facebook. It's a great tool to render the User Interface (UI) of modern web applications.
-React 使用名为 JSX 的 JavaScript 语法扩展，允许你直接在 JavaScript 中编写 HTML。这有几个好处。它允许你在 HTML 中使用 JavaScript 的完整程序功能，并有助于保持代码的可读性。在大多数情况下，JSX 类似于你已经学过的 HTML，但是在这些挑战中将会涉及一些关键差异。
+React uses a syntax extension of JavaScript called JSX that allows you to write HTML directly within JavaScript. This has several benefits. It lets you use the full programmatic power of JavaScript within HTML, and helps to keep your code readable. For the most part, JSX is similar to the HTML that you have already learned, however there are a few key differences that will be covered throughout these challenges.
-例如，因为 JSX 是 JavaScript 的语法扩展，所以你实际上可以直接在 JSX 中编写 JavaScript。要做到这一点，你只需在花括号中包含你希望被视为 JavaScript 的代码：`{“这被视为 JavaScript 代码”}`。请牢记这个写法，你将会在接下来的挑战中使用。
+For instance, because JSX is a syntactic extension of JavaScript, you can actually write JavaScript directly within JSX. To do this, you simply include the code you want to be treated as JavaScript within curly braces: `{ 'this is treated as JavaScript code' }`. Keep this in mind, since it's used in several future challenges.
-但是，由于浏览器不能解析 JSX，因此必须将 JSX 代码编译为 JavaScript。在这个过程中，转换器 Babel 是一个很受欢迎的工具。后续挑战已经在后台引入了 Babel，你可以直接写 JSX 代码。如果你的代码不符合 JSX 语法，那么挑战中的第一个测试就不会通过。
+However, because JSX is not valid JavaScript, JSX code must be compiled into JavaScript. The transpiler Babel is a popular tool for this process. For your convenience, it's already added behind the scenes for these challenges. If you happen to write syntactically invalid JSX, you will see the first test in these challenges fail.
-值得注意的是，这些挑战在底层调用`ReactDOM.render(JSX, document.getElementById('root'))`。这个函数调用是将你的 JSX 置于 React 自己的轻量级 DOM 中。然后，React 使用自己的 DOM 快照来优化更新实际 DOM 的特定部分。
+It's worth noting that under the hood the challenges are calling `ReactDOM.render(JSX, document.getElementById('root'))`. This function call is what places your JSX into React's own lightweight representation of the DOM. React then uses snapshots of its own DOM to optimize updating only specific parts of the actual DOM.
# --instructions--
-**说明：** 当前代码使用 JSX 将`div`元素赋值给常量`JSX`。将`div`替换为`h1`元素，并在其中添加文本`Hello JSX!`。
+**Instructions:** The current code uses JSX to assign a `div` element to the constant `JSX`. Replace the `div` with an `h1` element and add the text `Hello JSX!` inside it.
# --hints--
-常量`JSX`应该返回一个`h1`元素。
+The constant `JSX` should return an `h1` element.
```js
assert(JSX.type === 'h1');
```
-`h1`标签应该包含文本`Hello JSX!`。
+The `h1` tag should include the text `Hello JSX!`
```js
assert(Enzyme.shallow(JSX).contains('Hello JSX!'));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-stateful-component.md b/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-stateful-component.md
index a1c32fc3f20..86c86859cf1 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-stateful-component.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-stateful-component.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036170
-title: 创建一个有状态的组件
+title: Create a Stateful Component
challengeType: 6
forumTopicId: 301391
dashedName: create-a-stateful-component
@@ -8,9 +8,9 @@ dashedName: create-a-stateful-component
# --description--
-React中最重要的主题之一是`state`。 state 包含应用程序需要了解的任何数据，这些数据可能会随时间而变化。你希望应用程序能够响应 state 的变更，并在必要时显示更新后的 UI。React 为现代 Web 应用程序的状态管理提供了一个很好的解决方案。
+One of the most important topics in React is `state`. State consists of any data your application needs to know about, that can change over time. You want your apps to respond to state changes and present an updated UI when necessary. React offers a nice solution for the state management of modern web applications.
-你可以通过在`constructor`中的组件类上声明`state`属性来在 React 组件中创建 state，它在创建时使用`state`初始化组件。`state`属性必须设置为 JavaScript`对象`。声明如下：
+You create state in a React component by declaring a `state` property on the component class in its `constructor`. This initializes the component with `state` when it is created. The `state` property must be set to a JavaScript `object`. Declaring it looks like this:
```jsx
this.state = {
@@ -18,15 +18,15 @@ this.state = {
}
```
-你可以在组件的整个生命周期内访问`state`对象，你可以更新它、在 UI 中渲染它，也可以将其作为 props 传递给子组件。`state`对象的使用可以很简单，亦可以很复杂，就看你怎么用了。请注意，你必须通过扩展`React.Component`来创建类组件，以便像这样创建`state`。
+You have access to the `state` object throughout the life of your component. You can update it, render it in your UI, and pass it as props to child components. The `state` object can be as complex or as simple as you need it to be. Note that you must create a class component by extending `React.Component` in order to create `state` like this.
# --instructions--
-代码编辑器中有一个组件试图从其`state`中渲染一个`name`属性，但是`state`还没有定义。在`constructor`中使用`state`初始化组件，并将你的名字赋给`name`属性。
+There is a component in the code editor that is trying to render a `name` property from its `state`. However, there is no `state` defined. Initialize the component with `state` in the `constructor` and assign your name to a property of `name`.
# --hints--
-`StatefulComponent`应该存在并被渲染。
+`StatefulComponent` should exist and render.
```js
assert(
@@ -39,7 +39,7 @@ assert(
);
```
-`StatefulComponent`应该渲染一个`div`元素和一个`h1`元素。
+`StatefulComponent` should render a `div` and an `h1` element.
```js
assert(
@@ -55,7 +55,7 @@ assert(
);
```
-应使用被设置为字符串的`name`属性来初始化`StatefulComponent`的 state。
+The state of `StatefulComponent` should be initialized with a property `name` set to a string.
```js
assert(
@@ -71,7 +71,7 @@ assert(
);
```
-`StatefulComponent`中 state 的`name`属性应该渲染在`h1`元素里。
+The property `name` in the state of `StatefulComponent` should render in the `h1` element.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-stateless-functional-component.md b/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-stateless-functional-component.md
index 5f34a998fb4..b1b0925d7c4 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-stateless-functional-component.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/create-a-stateless-functional-component.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036162
-title: 创建一个无状态的函数组件
+title: Create a Stateless Functional Component
challengeType: 6
forumTopicId: 301392
dashedName: create-a-stateless-functional-component
@@ -8,11 +8,11 @@ dashedName: create-a-stateless-functional-component
# --description--
-组件是 React 的核心。React 中的所有内容都是一个组件，在这里你将学习如何创建一个组件。
+Components are the core of React. Everything in React is a component and here you will learn how to create one.
-有两种方法可以创建 React 组件。第一种方法是使用 JavaScript 函数。以这种方式定义组件会创建*无状态功能组件*。应用程序中的状态概念将在以后的挑战中介绍。目前，可以将无状态组件视为可以接收数据并对其进行渲染的组件，但是它不管理或跟踪对数据的更改，我们将在下一次挑战中介绍创建 React 组件的第二种方法。
+There are two ways to create a React component. The first way is to use a JavaScript function. Defining a component in this way creates a *stateless functional component*. The concept of state in an application will be covered in later challenges. For now, think of a stateless component as one that can receive data and render it, but does not manage or track changes to that data. (We'll cover the second way to create a React component in the next challenge.)
-要用函数创建组件，只需编写一个返回 JSX 或`null`的 JavaScript 函数。需要注意的一点是，React 要求你的函数名以大写字母开头。下面是一个无状态功能组件的示例，该组件在 JSX 中分配一个 HTML 的 class：
+To create a component with a function, you simply write a JavaScript function that returns either JSX or `null`. One important thing to note is that React requires your function name to begin with a capital letter. Here's an example of a stateless functional component that assigns an HTML class in JSX:
```jsx
// After being transpiled, the <div> will have a CSS class of 'customClass'
@@ -23,17 +23,17 @@ const DemoComponent = function() {
};
```
-因为 JSX 组件代表 HTML，所以你可以将几个组件放在一起以创建更复杂的 HTML 页面，这是 React 提供的组件架构的关键优势之一，它允许你用许多独立的组件组成 UI。这使得构建和维护复杂的用户界面变得更加容易。
+Because a JSX component represents HTML, you could put several components together to create a more complex HTML page. This is one of the key advantages of the component architecture React provides. It allows you to compose your UI from many separate, isolated components. This makes it easier to build and maintain complex user interfaces.
# --instructions--
-代码编辑器中有一个名为`MyComponent`的函数。完成此函数，使其返回包含一些文本字符串的单个`div`元素。
+The code editor has a function called `MyComponent`. Complete this function so it returns a single `div` element which contains some string of text.
-**注意：** 文本被视为是`div`的子元素，因此你将不能使用自闭合标签。
+**Note:** The text is considered a child of the `div` element, so you will not be able to use a self-closing tag.
# --hints--
-`MyComponent`应该返回 JSX。
+`MyComponent` should return JSX.
```js
assert(
@@ -44,7 +44,7 @@ assert(
);
```
-`MyComponent`应该返回一个`div`元素。
+`MyComponent` should return a `div` element.
```js
assert(
@@ -55,7 +55,7 @@ assert(
);
```
-`div`元素应该包含一个文本字符串。
+The `div` element should contain a string of text.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/define-an-html-class-in-jsx.md b/curriculum/challenges/chinese/03-front-end-libraries/react/define-an-html-class-in-jsx.md
index 9b0cfe1d649..2438f4b291f 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/define-an-html-class-in-jsx.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/define-an-html-class-in-jsx.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036160
-title: 在 JSX 中定义一个 HTML Class
+title: Define an HTML Class in JSX
challengeType: 6
forumTopicId: 301393
dashedName: define-an-html-class-in-jsx
@@ -8,27 +8,27 @@ dashedName: define-an-html-class-in-jsx
# --description--
-现在你已经习惯了编写 JSX，你可能想知道它与 HTML 有什么不同。
+Now that you're getting comfortable writing JSX, you may be wondering how it differs from HTML.
-到目前为止，HTML 和 JSX 似乎完全相同。
+So far, it may seem that HTML and JSX are exactly the same.
-JSX 的一个关键区别是你不能再使用`class`这个单词来定义 HTML 的 class 名。这是因为`class`是 JavaScript 中的关键字。JSX 使用`className`代替。
+One key difference in JSX is that you can no longer use the word `class` to define HTML classes. This is because `class` is a reserved word in JavaScript. Instead, JSX uses `className`.
-事实上，JSX 中所有 HTML 属性和事件引用的命名约定都变成了驼峰式。例如，JSX 中的单击事件是 `onClick`，而不是 `onclick`。同样，`onchange`变成了`onChange`。虽然这是一个微妙的差异，但请你一定要记住。
+In fact, the naming convention for all HTML attributes and event references in JSX become camelCase. For example, a click event in JSX is `onClick`, instead of `onclick`. Likewise, `onchange` becomes `onChange`. While this is a subtle difference, it is an important one to keep in mind moving forward.
# --instructions--
-将 class`myDiv` 应用于 JSX 提供的`div`上。
+Apply a class of `myDiv` to the `div` provided in the JSX code.
# --hints--
-常量`JSX`应该返回一个`div`元素。
+The constant `JSX` should return a `div` element.
```js
assert.strictEqual(JSX.type, 'div');
```
-`div`有一个`myDiv`class。
+The `div` should have a class of `myDiv`.
```js
assert.strictEqual(JSX.props.className, 'myDiv');
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/give-sibling-elements-a-unique-key-attribute.md b/curriculum/challenges/chinese/03-front-end-libraries/react/give-sibling-elements-a-unique-key-attribute.md
index ddf18670c65..49bf695b428 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/give-sibling-elements-a-unique-key-attribute.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/give-sibling-elements-a-unique-key-attribute.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403618b
-title: 给同级元素一个唯一的键属性
+title: Give Sibling Elements a Unique Key Attribute
challengeType: 6
forumTopicId: 301394
dashedName: give-sibling-elements-a-unique-key-attribute
@@ -8,17 +8,19 @@ dashedName: give-sibling-elements-a-unique-key-attribute
# --description--
-上一个挑战展示了如何使用`map`方法根据用户输入动态渲染多个元素。然而，这个例子中缺少一个重要的部分。创建元素数组时，每个元素都需要一个设置为唯一值的`key`属性。React 使用这些键来跟踪哪些项目被添加、更改或删除。这有助于在以任何方式修改列表时提高重新渲染过程的效率。请注意，键只需要在同级元素之间是唯一的，它们不需要在应用程序中是全局唯一的。
+The last challenge showed how the `map` method is used to dynamically render a number of elements based on user input. However, there was an important piece missing from that example. When you create an array of elements, each one needs a `key` attribute set to a unique value. React uses these keys to keep track of which items are added, changed, or removed. This helps make the re-rendering process more efficient when the list is modified in any way.
+
+**Note:** Keys only need to be unique between sibling elements, they don't need to be globally unique in your application.
# --instructions--
-代码编辑器有一个数组，它包含一些前端框架和一个名为`Frameworks()`的无状态函数组件。`Frameworks()`需要将数组映射到无序列表，就像上一个挑战一样。完成`map`回调，为`frontEndFrameworks`数组中的每个框架返回一个`li`元素。这次，确保给每个`li`的`key`属性设置一个唯一的值。
+The code editor has an array with some front end frameworks and a stateless functional component named `Frameworks()`. `Frameworks()` needs to map the array to an unordered list, much like in the last challenge. Finish writing the `map` callback to return an `li` element for each framework in the `frontEndFrameworks` array. This time, make sure to give each `li` a `key` attribute, set to a unique value. The `li` elements should also contain text from `frontEndFrameworks`.
-通常，你希望使 key 能唯一标识要渲染的元素。作为最后的手段，可以使用数组索引，但通常你应该尝试使用唯一标识。
+Normally, you want to make the key something that uniquely identifies the element being rendered. As a last resort the array index may be used, but typically you should try to use a unique identification.
# --hints--
-`Frameworks` 组件应该存在并渲染到页面。
+The `Frameworks` component should exist and render to the page.
```js
assert(
@@ -26,19 +28,19 @@ assert(
);
```
-`Frameworks`应该渲染一个`h1`元素。
+`Frameworks` should render an `h1` element.
```js
assert(Enzyme.mount(React.createElement(Frameworks)).find('h1').length === 1);
```
-`Frameworks`应该渲染一个`ul`元素。
+`Frameworks` should render a `ul` element.
```js
assert(Enzyme.mount(React.createElement(Frameworks)).find('ul').length === 1);
```
-`ul`标签应该渲染 6 个`li`子元素。
+The `ul` tag should render 6 child `li` elements.
```js
assert(
@@ -52,7 +54,7 @@ assert(
);
```
-每个列表项元素都应该有一个唯一的`key`属性。
+Each list item element should have a unique `key` attribute.
```js
assert(
@@ -71,7 +73,7 @@ assert(
);
```
-每个列表元素都应该包含 `frontEndFrameworks` 里的文本。
+Each list item element should contain text from `frontEndFrameworks`.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/introducing-inline-styles.md b/curriculum/challenges/chinese/03-front-end-libraries/react/introducing-inline-styles.md
index 2629675f972..70b4df7f929 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/introducing-inline-styles.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/introducing-inline-styles.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036181
-title: 介绍内联样式
+title: Introducing Inline Styles
challengeType: 6
forumTopicId: 301395
dashedName: introducing-inline-styles
@@ -8,29 +8,29 @@ dashedName: introducing-inline-styles
# --description--
-还有其他复杂的概念可以为你的 React 代码增加强大的功能。但是，你可能会想知道更简单的问题，比如：如何对在 React 中创建的 JSX 元素进行风格化。你可能知道，由于[将 class 应用于 JSX 元素的方式](/learn/front-end-libraries/react/define-an-html-class-in-jsx)与 HTML 中的使用并不完全相同。
+There are other complex concepts that add powerful capabilities to your React code. But you may be wondering about the more simple problem of how to style those JSX elements you create in React. You likely know that it won't be exactly the same as working with HTML because of [the way you apply classes to JSX elements](/learn/front-end-libraries/react/define-an-html-class-in-jsx).
-如果从样式表导入样式，它就没有太大的不同。使用`className`属性将 class 应用于 JSX 元素，并将样式应用于样式表中的 class。另一种选择是使用***内联***样式，这在 ReactJS 开发中非常常见。
+If you import styles from a stylesheet, it isn't much different at all. You apply a class to your JSX element using the `className` attribute, and apply styles to the class in your stylesheet. Another option is to apply inline styles, which are very common in ReactJS development.
-你将内联样式应用于 JSX 元素，类似于你在 HTML 中的操作方式，但有一些 JSX 差异。以下是 HTML 中内联样式的示例：
+You apply inline styles to JSX elements similar to how you do it in HTML, but with a few JSX differences. Here's an example of an inline style in HTML:
`<div style="color: yellow; font-size: 16px">Mellow Yellow</div>`
-JSX 元素使用`style`属性，但是由于 JSX 的传输方式，你不能将值设置为`字符串`。相反，你应将其设置为 JavaScript`对象`。这里有一个例子：
+JSX elements use the `style` attribute, but because of the way JSX is transpiled, you can't set the value to a `string`. Instead, you set it equal to a JavaScript `object`. Here's an example:
`<div style={{color: "yellow", fontSize: 16}}>Mellow Yellow</div>`
-注意我们使用驼峰式命名的 "fontSize" 属性，这是因为 React 不会接受样式对象中的连字符。React 将在 HTML 中为我们应用正确的属性名称。
+Notice how we camelCase the "fontSize" property? This is because React will not accept kebab-case keys in the style object. React will apply the correct property name for us in the HTML.
# --instructions--
-在代码编辑器的`div`中添加一个`style`属性，使文本颜色为红色，字体大小为 72px。
+Add a `style` attribute to the `div` in the code editor to give the text a color of red and font size of 72px.
-请注意，你可以选择将字体大小设置为数字，省略单位 "px"，或者将其写为 "72px"。
+Note that you can optionally set the font size to be a number, omitting the units "px", or write it as "72px".
# --hints--
-组件应该渲染一个`div`元素。
+The component should render a `div` element.
```js
assert(
@@ -41,7 +41,7 @@ assert(
);
```
-`div`元素应该是`红色`的。
+The `div` element should have a color of `red`.
```js
assert(
@@ -52,7 +52,7 @@ assert(
);
```
-`div`元素的字体大小应为`72px`。
+The `div` element should have a font size of `72px`.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/learn-about-self-closing-jsx-tags.md b/curriculum/challenges/chinese/03-front-end-libraries/react/learn-about-self-closing-jsx-tags.md
index fcbf1f6e4e9..31edb731835 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/learn-about-self-closing-jsx-tags.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/learn-about-self-closing-jsx-tags.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036161
-title: 了解自动闭合的 JSX 标记
+title: Learn About Self-Closing JSX Tags
challengeType: 6
forumTopicId: 301396
dashedName: learn-about-self-closing-jsx-tags
@@ -8,35 +8,35 @@ dashedName: learn-about-self-closing-jsx-tags
# --description--
-到目前为止，你已经看到 JSX 与 HTML 的不同之处在于使用`className`和使用`class`来定义 HTML 的 class。
+So far, you’ve seen how JSX differs from HTML in a key way with the use of `className` vs. `class` for defining HTML classes.
-JSX 不同于 HTML 的另一个重要方面是自闭合标签。
+Another important way in which JSX differs from HTML is in the idea of the self-closing tag.
-在HTML中，几乎所有的标签都有一个开始和结束标签：`<div></div>`，结束标签在你要关闭的标签名之前始终具有正斜杠。但是，HTML 中有一些称为“自闭合标签”的特殊实例，它们在另一个标签开始之前，不需要开始和结束标签都存在。
+In HTML, almost all tags have both an opening and closing tag: `<div></div>`; the closing tag always has a forward slash before the tag name that you are closing. However, there are special instances in HTML called “self-closing tags”, or tags that don’t require both an opening and closing tag before another tag can start.
-例如，换行标签可以写成`<br>`或者`<br />`，但是不应该写成`<br></br>`，因为它不包含任何内容。
+For example the line-break tag can be written as `<br>` or as `<br />`, but should never be written as `<br></br>`, since it doesn't contain any content.
-在 JSX 中，规则略有不同。任何 JSX 元素都可以使用自闭合标签编写，并且每个元素都必须关闭。例如，换行标签必须始终编写为`<br />`。另一方面`<div>`可以写成`<div />`或者`<div></div>`。不同之处在于，在第一个语法版本中，无法在`<div />`中包含任何内容。在后面的挑战中你会发现，这种语法在渲染 React 组件时非常有用。
+In JSX, the rules are a little different. Any JSX element can be written with a self-closing tag, and every element must be closed. The line-break tag, for example, must always be written as `<br />` in order to be valid JSX that can be transpiled. A `<div>`, on the other hand, can be written as `<div />` or `<div></div>`. The difference is that in the first syntax version there is no way to include anything in the `<div />`. You will see in later challenges that this syntax is useful when rendering React components.
# --instructions--
-修复代码编辑器中的错误，使其成为有效的 JSX 并成功转换。确保你不更改任何内容--你只需要在需要的地方关闭标签。
+Fix the errors in the code editor so that it is valid JSX and successfully transpiles. Make sure you don't change any of the content - you only need to close tags where they are needed.
# --hints--
-常量`JSX`应该返回一个`div`元素。
+The constant `JSX` should return a `div` element.
```js
assert.strictEqual(JSX.type, 'div');
```
-`div`应该包含一个`br`标签。
+The `div` should contain a `br` tag.
```js
assert(Enzyme.shallow(JSX).find('br').length === 1);
```
-`div`应该包含一个`hr`标签。
+The `div` should contain an `hr` tag.
```js
assert(Enzyme.shallow(JSX).find('hr').length === 1);
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/optimize-re-renders-with-shouldcomponentupdate.md b/curriculum/challenges/chinese/03-front-end-libraries/react/optimize-re-renders-with-shouldcomponentupdate.md
index 1b62f849964..661ea8e6ddd 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/optimize-re-renders-with-shouldcomponentupdate.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/optimize-re-renders-with-shouldcomponentupdate.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036180
-title: 使用 shouldComponentUpdate 优化重新渲染
+title: Optimize Re-Renders with shouldComponentUpdate
challengeType: 6
forumTopicId: 301398
dashedName: optimize-re-renders-with-shouldcomponentupdate
@@ -8,17 +8,17 @@ dashedName: optimize-re-renders-with-shouldcomponentupdate
# --description--
-到目前为止，如果任何组件接收到新的`state`或新的`props`，它会重新渲染自己及其所有子组件。这通常是好的。但是 React 提供了一种生命周期方法，当子组件接收到新的`state`或`props`时，你可以调用该方法，并特别声明组件是否应该更新。方法是`shouldComponentUpdate()`，它将`nextProps`和`nextState`作为参数。
+So far, if any component receives new `state` or new `props`, it re-renders itself and all its children. This is usually okay. But React provides a lifecycle method you can call when child components receive new `state` or `props`, and declare specifically if the components should update or not. The method is `shouldComponentUpdate()`, and it takes `nextProps` and `nextState` as parameters.
-这种方法是优化性能的有效方法。例如，默认行为是，当组件接收到新的`props`时，即使`props`没有改变，它也会重新渲染。你可以通过使用`shouldComponentUpdate()`比较`props`来防止这种情况。该方法必须返回一个布尔值，该值告诉 React 是否更新组件。你可以比较当前的 props（`this.props`）和下一个 props（`nextProps`），以确定你是否需要更新，并相应地返回`true`或`false`。
+This method is a useful way to optimize performance. For example, the default behavior is that your component re-renders when it receives new `props`, even if the `props` haven't changed. You can use `shouldComponentUpdate()` to prevent this by comparing the `props`. The method must return a `boolean` value that tells React whether or not to update the component. You can compare the current props (`this.props`) to the next props (`nextProps`) to determine if you need to update or not, and return `true` or `false` accordingly.
# --instructions--
-`shouldComponentUpdate()`方法添加到名为`OnlyEvens`的组件中。目前，该方法返回`true`，因此每次收到新的`props`时，`OnlyEvens`都会重新渲染。修改该方法，以便`OnlyEvens`仅在其新 props 的`value`为偶数时更新。单击`Add`按钮，在触发其他生命周期钩子时，在浏览器控制台中查看事件的顺序。
+The `shouldComponentUpdate()` method is added in a component called `OnlyEvens`. Currently, this method returns `true` so `OnlyEvens` re-renders every time it receives new `props`. Modify the method so `OnlyEvens` updates only if the `value` of its new props is even. Click the `Add` button and watch the order of events in your browser's console as the lifecycle hooks are triggered.
# --hints--
-`Controller`组件应该将`OnlyEvens`组件渲染为子组件。
+The `Controller` component should render the `OnlyEvens` component as a child.
```js
assert(
@@ -32,7 +32,7 @@ assert(
);
```
-应该在`OnlyEvens`组件上定义`shouldComponentUpdate`方法。
+The `shouldComponentUpdate` method should be defined on the `OnlyEvens` component.
```js
assert(
@@ -45,51 +45,47 @@ assert(
);
```
-`OnlyEvens`组件应该返回一个`h1`标签，该标签渲染`this.props.value`的值。
+The `OnlyEvens` component should return an `h1` tag which renders the value of `this.props.value`.
```js
-async () => {
- const waitForIt = (fn) =>
- new Promise((resolve, reject) => setTimeout(() => resolve(fn()), 250));
+(() => {
const mockedComponent = Enzyme.mount(React.createElement(Controller));
const first = () => {
mockedComponent.setState({ value: 1000 });
- return waitForIt(() => mockedComponent.find('h1').html());
+ return mockedComponent.find('h1').html();
};
const second = () => {
mockedComponent.setState({ value: 10 });
- return waitForIt(() => mockedComponent.find('h1').html());
+ return mockedComponent.find('h1').html();
};
- const firstValue = await first();
- const secondValue = await second();
+ const firstValue = first();
+ const secondValue = second();
assert(firstValue === '<h1>1000</h1>' && secondValue === '<h1>10</h1>');
-};
+})();
```
-只有在`nextProps.value`为偶数时，`OnlyEvens`才会重新渲染。
+`OnlyEvens` should re-render only when `nextProps.value` is even.
```js
-async () => {
- const waitForIt = (fn) =>
- new Promise((resolve, reject) => setTimeout(() => resolve(fn()), 250));
+(() => {
const mockedComponent = Enzyme.mount(React.createElement(Controller));
const first = () => {
mockedComponent.setState({ value: 8 });
- return waitForIt(() => mockedComponent.find('h1').text());
+ return mockedComponent.find('h1').text();
};
const second = () => {
mockedComponent.setState({ value: 7 });
- return waitForIt(() => mockedComponent.find('h1').text());
+ return mockedComponent.find('h1').text();
};
const third = () => {
mockedComponent.setState({ value: 42 });
- return waitForIt(() => mockedComponent.find('h1').text());
+ return mockedComponent.find('h1').text();
};
- const firstValue = await first();
- const secondValue = await second();
- const thirdValue = await third();
+ const firstValue = first();
+ const secondValue = second();
+ const thirdValue = third();
assert(firstValue === '8' && secondValue === '8' && thirdValue === '42');
-};
+})();
```
# --seed--
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/override-default-props.md b/curriculum/challenges/chinese/03-front-end-libraries/react/override-default-props.md
index 3f022294468..9a6afa97957 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/override-default-props.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/override-default-props.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403616c
-title: 覆盖默认的 Props
+title: Override Default Props
challengeType: 6
forumTopicId: 301399
dashedName: override-default-props
@@ -8,17 +8,17 @@ dashedName: override-default-props
# --description--
-在 React 中，设置默认的 props 是一个很有用的特性，显式设置组件的 prop 值即可覆盖默认 props。
+The ability to set default props is a useful feature in React. The way to override the default props is to explicitly set the prop values for a component.
# --instructions--
-`ShoppingCart`组件现在渲染了一个子组件`Items`。该`Items`组件有一个默认`quantity`prop，其值被设置为整数`0`。通过传入数值`10`来覆盖`quantity`的默认 prop。
+The `ShoppingCart` component now renders a child component `Items`. This `Items` component has a default prop `quantity` set to the integer `0`. Override the default prop by passing in a value of `10` for `quantity`.
-**注意：** 请记住，向组件添加 prop 的语法与添加 HTML 属性类似。但是，由于`quantity`的值是整数，所以它不会加引号，但应该用花括号括起来，例如`{100}`。这个语法告诉 JSX 直接将花括号中的值解释为 JavaScript。
+**Note:** Remember that the syntax to add a prop to a component looks similar to how you add HTML attributes. However, since the value for `quantity` is an integer, it won't go in quotes but it should be wrapped in curly braces. For example, `{100}`. This syntax tells JSX to interpret the value within the braces directly as JavaScript.
# --hints--
-应该渲染`ShoppingCart`组件。
+The component `ShoppingCart` should render.
```js
assert(
@@ -29,7 +29,7 @@ assert(
);
```
-应该渲染`Items`组件。
+The component `Items` should render.
```js
assert(
@@ -40,7 +40,7 @@ assert(
);
```
-`Items`组件应该有一个`{ quantity: 10 }`的prop，该 prop 是从`ShoppingCart`组件传递过去的。
+The `Items` component should have a prop of `{ quantity: 10 }` passed from the `ShoppingCart` component.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/pass-a-callback-as-props.md b/curriculum/challenges/chinese/03-front-end-libraries/react/pass-a-callback-as-props.md
index 349a98ab2d2..e542314e5f9 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/pass-a-callback-as-props.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/pass-a-callback-as-props.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403617b
-title: 传递回调作为 Props
+title: Pass a Callback as Props
challengeType: 6
forumTopicId: 301400
dashedName: pass-a-callback-as-props
@@ -8,17 +8,17 @@ dashedName: pass-a-callback-as-props
# --description--
-你可以将`state`作为 props 传递给子组件，但不仅限于传递数据。你也可以将处理函数或在 React 组件中定义的任何方法传递给子组件。这就是允许子组件与父组件交互的方式。你可以把方法像普通 prop 一样传递给子组件，它会被分配一个名字，你可以在子组件中的`this.props`下访问该方法的名字。
+You can pass `state` as props to child components, but you're not limited to passing data. You can also pass handler functions or any method that's defined on a React component to a child component. This is how you allow child components to interact with their parent components. You pass methods to a child just like a regular prop. It's assigned a name and you have access to that method name under `this.props` in the child component.
# --instructions--
-代码编辑器中列出了三个组件。`MyApp`是父组件，`GetInput`和`RenderInput`是它的子组件。将`GetInput`组件添加到`MyApp`的 render 方法，然后将`MyApp`的`state`中的`inputValue`传入名为`input`的 prop。还要创建一个名为`handleChange`的 prop，并将输入处理程序`handleChange`传递给它。
+There are three components outlined in the code editor. The `MyApp` component is the parent that will render the `GetInput` and `RenderInput` child components. Add the `GetInput` component to the render method in `MyApp`, then pass it a prop called `input` assigned to `inputValue` from `MyApp`'s `state`. Also create a prop called `handleChange` and pass the input handler `handleChange` to it.
-接下来，将`RenderInput`添加到`MyApp`中的 render 方法中，然后创建一个名为`input`的 prop，并将`state`中的`inputValue`传递给它。完成后，你将能够在`GetInput`组件中的`input`字段中键入内容，然后该组件通过 props 调用其父组件中的处理函数方法。这将更新处于父组件`state`中的 input，该 input 将作为 props 传递给两个子组件。观察数据如何在组件之间流动，以及单一数据源如何保持父组件`state`。诚然，这个示例有点做作，但是应该能用来说明数据和回调是如何在 React 组件之间传递的。
+Next, add `RenderInput` to the render method in `MyApp`, then create a prop called `input` and pass the `inputValue` from `state` to it. Once you are finished you will be able to type in the `input` field in the `GetInput` component, which then calls the handler method in its parent via props. This updates the input in the `state` of the parent, which is passed as props to both children. Observe how the data flows between the components and how the single source of truth remains the `state` of the parent component. Admittedly, this example is a bit contrived, but should serve to illustrate how data and callbacks can be passed between React components.
# --hints--
-应该渲染`MyApp`组件。
+The `MyApp` component should render.
```js
assert(
@@ -29,7 +29,7 @@ assert(
);
```
-应该渲染`GetInput`组件。
+The `GetInput` component should render.
```js
assert(
@@ -40,7 +40,7 @@ assert(
);
```
-应该渲染`RenderInput`组件。
+The `RenderInput` component should render.
```js
assert(
@@ -51,7 +51,7 @@ assert(
);
```
-`GetInput`组件应该接收`MyApp`的 state 属性`inputValue`作为 props，并包含一个修改`MyApp`state 的`input`元素。
+The `GetInput` component should receive the `MyApp` state property `inputValue` as props and contain an `input` element which modifies `MyApp` state.
```js
async () => {
@@ -74,7 +74,7 @@ async () => {
};
```
-`RenderInput`组件应该接收`MyApp`state 属性`inputValue`作为 props。
+The `RenderInput` component should receive the `MyApp` state property `inputValue` as props.
```js
async () => {
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/pass-an-array-as-props.md b/curriculum/challenges/chinese/03-front-end-libraries/react/pass-an-array-as-props.md
index c8589ef627f..a5b7170b309 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/pass-an-array-as-props.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/pass-an-array-as-props.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403616a
-title: 传递一个数组作为 Props
+title: Pass an Array as Props
challengeType: 6
forumTopicId: 301401
dashedName: pass-an-array-as-props
@@ -8,7 +8,7 @@ dashedName: pass-an-array-as-props
# --description--
-上一个挑战演示了如何将来自父组件的信息作为`props`传递给子组件。这个挑战着眼于如何将数组作为`props`传递。要将数组传递给 JSX 元素，必须将其视为 JavaScript 并用花括号括起来。
+The last challenge demonstrated how to pass information from a parent component to a child component as `props` or properties. This challenge looks at how arrays can be passed as `props`. To pass an array to a JSX element, it must be treated as JavaScript and wrapped in curly braces.
```jsx
<ParentComponent>
@@ -16,15 +16,15 @@ dashedName: pass-an-array-as-props
</ParentComponent>
```
-这样，子组件就可以访问数组属性`colors`。访问属性时可以使用`join()`等数组方法。 `const ChildComponent = (props) => <p>{props.colors.join(', ')}</p>` 这将把所有`colors`数组项连接成一个逗号分隔的字符串并生成： `<p>green, blue, red</p>` 稍后，我们将了解在 React 中渲染数组数据的其他常用方法。
+The child component then has access to the array property `colors`. Array methods such as `join()` can be used when accessing the property. `const ChildComponent = (props) => <p>{props.colors.join(', ')}</p>` This will join all `colors` array items into a comma separated string and produce: `<p>green, blue, red</p>` Later, we will learn about other common methods to render arrays of data in React.
# --instructions--
-代码编辑器中有`List`和`ToDo`组件。在`ToDo`组件中渲染每个`List`时，传入`tasks`属性并将其分配给待办任务数组，例如`["walk dog", "workout"]`。然后访问`List`组件中的`tasks`数组，在`p`元素中显示其值。使用`join(", ")`把`props.tasks`数组作为逗号分隔列表显示在`p`元素中。今天的列表应该至少有 2 个任务，明天应该至少有 3 个任务。
+There are `List` and `ToDo` components in the code editor. When rendering each `List` from the `ToDo` component, pass in a `tasks` property assigned to an array of to-do tasks, for example `["walk dog", "workout"]`. Then access this `tasks` array in the `List` component, showing its value within the `p` element. Use `join(", ")` to display the `props.tasks`array in the `p` element as a comma separated list. Today's list should have at least 2 tasks and tomorrow's should have at least 3 tasks.
# --hints--
-`ToDo`组件应该返回单个外部`div`。
+The `ToDo` component should return a single outer `div`.
```js
assert(
@@ -35,7 +35,7 @@ assert(
);
```
-`ToDo`组件的第三个子元素应该是`List`组件的一个实例。
+The third child of the `ToDo` component should be an instance of the `List` component.
```js
assert(
@@ -46,7 +46,7 @@ assert(
);
```
-`ToDo`组件的第五个子元素应该是`List`组件的一个实例。
+The fifth child of the `ToDo` component should be an instance of the `List` component.
```js
assert(
@@ -57,7 +57,7 @@ assert(
);
```
-`List`组件的两个实例都应该具有一个名为`tasks`的属性，并且`tasks`的类型应该是数组。
+Both instances of the `List` component should have a property called `tasks` and `tasks` should be of type array.
```js
assert(
@@ -71,7 +71,7 @@ assert(
);
```
-表示今天任务的第一个`List`组件应该有 2 个或更多项。
+The first `List` component representing the tasks for today should have 2 or more items.
```js
assert(
@@ -82,7 +82,7 @@ assert(
);
```
-表示明天任务的第二个`List`组件应该有 3 个或更多项。
+The second `List` component representing the tasks for tomorrow should have 3 or more items.
```js
assert(
@@ -93,17 +93,31 @@ assert(
);
```
-`List`组件应在`p`标签中渲染`tasks`属性的值。
+The `List` component should render the value from the `tasks` prop in the `p` tag.
```js
assert(
(function () {
const mockedComponent = Enzyme.mount(React.createElement(ToDo));
return (
- mockedComponent.find('p').get(0).props.children ===
- mockedComponent.find('List').get(0).props.tasks.join(', ') &&
- mockedComponent.find('p').get(1).props.children ===
- mockedComponent.find('List').get(1).props.tasks.join(', ')
+ mockedComponent
+ .find('p')
+ .get(0)
+ .props.children.replace(/\s*,\s*/g, ',') ===
+ mockedComponent
+ .find('List')
+ .get(0)
+ .props.tasks.join(',')
+ .replace(/\s*,\s*/g, ',') &&
+ mockedComponent
+ .find('p')
+ .get(1)
+ .props.children.replace(/\s*,\s*/g, ',') ===
+ mockedComponent
+ .find('List')
+ .get(1)
+ .props.tasks.join(',')
+ .replace(/\s*,\s*/g, ',')
);
})()
);
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/pass-props-to-a-stateless-functional-component.md b/curriculum/challenges/chinese/03-front-end-libraries/react/pass-props-to-a-stateless-functional-component.md
index ccd26594be0..b8f61177082 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/pass-props-to-a-stateless-functional-component.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/pass-props-to-a-stateless-functional-component.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036169
-title: 将 Props 传递给无状态函数组件
+title: Pass Props to a Stateless Functional Component
challengeType: 6
forumTopicId: 301402
dashedName: pass-props-to-a-stateless-functional-component
@@ -8,7 +8,7 @@ dashedName: pass-props-to-a-stateless-functional-component
# --description--
-之前的挑战涵盖了关于在 React 中创建和组合 JSX 元素、函数组件和 ES6 风格的类组件的很多内容。有了这个基础，现在是时候看看 React 中的另一个常见特性 **props** 了。在 React 中，你可以将属性传递给子组件。假设你有一个`App`组件，该组件渲染了一个名为`Welcome`的子组件，它是一个无状态函数组件。你可以通过以下方式给`Welcome`传递一个`user`属性：
+The previous challenges covered a lot about creating and composing JSX elements, functional components, and ES6 style class components in React. With this foundation, it's time to look at another feature very common in React: **props**. In React, you can pass props, or properties, to child components. Say you have an `App` component which renders a child component called `Welcome` which is a stateless functional component. You can pass `Welcome` a `user` property by writing:
```jsx
<App>
@@ -16,21 +16,21 @@ dashedName: pass-props-to-a-stateless-functional-component
</App>
```
-使用**自定义 HTML 属性**，React 支持将属性`user`传递给组件`Welcome`。由于`Welcome`是一个无状态函数组件，它可以像这样访问该值：
+You use **custom HTML attributes** created by you and supported by React to be passed to the component. In this case, the created property `user` is passed to the component `Welcome`. Since `Welcome` is a stateless functional component, it has access to this value like so:
```jsx
const Welcome = (props) => <h1>Hello, {props.user}!</h1>
```
-调用`props`这个值是常见做法，当处理无状态函数组件时，你基本上可以将其视为返回 JSX 的函数的参数。这样，你就可以在函数体中访问该值。但对于类组件，访问方式会略有不同。
+It is standard to call this value `props` and when dealing with stateless functional components, you basically consider it as an argument to a function which returns JSX. You can access the value of the argument in the function body. With class components, you will see this is a little different.
# --instructions--
-代码编辑器中有`Calendar`和`CurrentDate`组件。从`Calendar`组件渲染`CurrentDate`时，从 JavaScript 的`Date`对象分配当前日期，并将其作为`date`属性传入。然后访问`CurrentDate`组件的`prop`，并在`p`标签中显示其值。请注意，要将`prop`的值视为 JavaScript，必须将它们括在花括号中，例如`date={Date()}`。
+There are `Calendar` and `CurrentDate` components in the code editor. When rendering `CurrentDate` from the `Calendar` component, pass in a property of `date` assigned to the current date from JavaScript's `Date` object. Then access this `prop` in the `CurrentDate` component, showing its value within the `p` tags. Note that for `prop` values to be evaluated as JavaScript, they must be enclosed in curly brackets, for instance `date={Date()}`.
# --hints--
-`Calendar`组件应该返回单个`div`元素。
+The `Calendar` component should return a single `div` element.
```js
assert(
@@ -41,7 +41,7 @@ assert(
);
```
-`Calendar`组件的第二个子元素应该是`CurrentDate`组件。
+The second child of the `Calendar` component should be the `CurrentDate` component.
```js
assert(
@@ -52,7 +52,7 @@ assert(
);
```
-`CurrentDate`组件应该有一个名为`date`的属性。
+The `CurrentDate` component should have a prop called `date`.
```js
assert(
@@ -63,7 +63,7 @@ assert(
);
```
-`CurrentDate`的`date`属性应该包含一段文本字符串。
+The `date` prop of the `CurrentDate` should contain a string of text.
```js
assert(
@@ -75,13 +75,13 @@ assert(
);
```
-`CurrentDate`组件应该把`date`属性渲染在`p`标签内。
+The `date` prop should be generated by calling `Date()`
```js
-assert(/<CurrentDatedate={Date\(\)}\/>/.test(code.replace(/\s/g, '')));
+assert(/<CurrentDatedate={Date\(\)}\/>/.test(__helpers.removeWhiteSpace(code)));
```
-`CurrentDate` 组件应该把 `date` 属性渲染在 `p` 标签内。
+The `CurrentDate` component should render the value from the `date` prop in the `p` tag.
```js
let date = 'dummy date';
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/pass-state-as-props-to-child-components.md b/curriculum/challenges/chinese/03-front-end-libraries/react/pass-state-as-props-to-child-components.md
index c1463117bda..e0554a5a98e 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/pass-state-as-props-to-child-components.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/pass-state-as-props-to-child-components.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403617a
-title: 将 State 作为 Props 传递给子组件
+title: Pass State as Props to Child Components
challengeType: 6
forumTopicId: 301403
dashedName: pass-state-as-props-to-child-components
@@ -8,19 +8,19 @@ dashedName: pass-state-as-props-to-child-components
# --description--
-在之前的挑战中，你看到了很多将 props 传递给子 JSX 元素和子 React 组件的例子。你可能想知道那些 props 是从哪里来的。一个常见的模式是：有状态组件中包含对应用程序很重要的`state`，然后用它渲染子组件。你希望这些组件能够访问该`state`的某些部分，就把这些部分作为 props 传入。
+You saw a lot of examples that passed props to child JSX elements and child React components in previous challenges. You may be wondering where those props come from. A common pattern is to have a stateful component containing the `state` important to your app, that then renders child components. You want these components to have access to some pieces of that `state`, which are passed in as props.
-例如，也许你有一个`App`组件可以渲染`Navbar`以及其他组件。在你的`App`中，你的`state`中包含大量用户信息，但是`Navbar`只需要访问用户的用户名就可以显示出来，这时你将该`state`作为一个 prop 传递给`Navbar`组件即可。
+For example, maybe you have an `App` component that renders a `Navbar`, among other components. In your `App`, you have `state` that contains a lot of user information, but the `Navbar` only needs access to the user's username so it can display it. You pass that piece of `state` to the `Navbar` component as a prop.
-这个模式说明了 React 中的一些重要范例。第一个是*单向数据流*，state 沿着应用程序组件树的一个方向流动，从有状态父组件到子组件，子组件只接收它们需要的 state 数据。第二，复杂的有状态应用程序可以分解成几个，或者可能是一个单一的有状态组件。其余组件只是从父组件简单的接收 state 作为 props，并从该 state 渲染 UI。它开始创建一种分离，在这种分离中，state 管理在代码的一部分中处理，而 UI 渲染在另一部分中处理。将 state 逻辑与 UI 逻辑分离是 React 的关键原则之一。当它被正确使用时，它使得复杂的、有状态的应用程序的设计变得更容易管理。
+This pattern illustrates some important paradigms in React. The first is *unidirectional data flow*. State flows in one direction down the tree of your application's components, from the stateful parent component to child components. The child components only receive the state data they need. The second is that complex stateful apps can be broken down into just a few, or maybe a single, stateful component. The rest of your components simply receive state from the parent as props, and render a UI from that state. It begins to create a separation where state management is handled in one part of code and UI rendering in another. This principle of separating state logic from UI logic is one of React's key principles. When it's used correctly, it makes the design of complex, stateful applications much easier to manage.
# --instructions--
-`MyApp`组件是有状态的，它将`Navbar`组件渲染成它的为子组件。将`MyApp`组件`state`中的`name`属性向下传递给子组件，然后在`h1`标签中显示`name`，`name`是`Navbar`render 方法的一部分。`name` 应该显示在文字 `Hello, my name is:` 后面。
+The `MyApp` component is stateful and renders a `Navbar` component as a child. Pass the `name` property in its `state` down to the child component, then show the `name` in the `h1` tag that's part of the `Navbar` render method. `name` should appear after the text `Hello, my name is:`.
# --hints--
-`MyApp`组件应该在内部渲染一个`Navbar`组件。
+The `MyApp` component should render with a `Navbar` component inside.
```js
assert(
@@ -34,7 +34,7 @@ assert(
);
```
-`Navbar`组件应该接收`Navbar`的 state 中的`name`属性作为 props。
+The `Navbar` component should receive the `MyApp` state property `name` as props.
```js
async () => {
@@ -50,7 +50,7 @@ async () => {
};
```
-`Navbar`中的`h1`元素应该渲染 prop`name`。
+The `h1` element in `Navbar` should render the `name` prop.
```js
async () => {
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/render-a-class-component-to-the-dom.md b/curriculum/challenges/chinese/03-front-end-libraries/react/render-a-class-component-to-the-dom.md
index 2d43d5d1b7b..2e3a9a7a562 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/render-a-class-component-to-the-dom.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/render-a-class-component-to-the-dom.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036167
-title: 渲染 class 组件为 Dom 树
+title: Render a Class Component to the DOM
challengeType: 6
forumTopicId: 301404
dashedName: render-a-class-component-to-the-dom
@@ -8,19 +8,19 @@ dashedName: render-a-class-component-to-the-dom
# --description--
-你可能还记得在早期挑战中使用 ReactDOM API 将 JSX 元素渲染到 DOM，这与渲染 React 组件的过程十分相似。过去的几个挑战主要针对组件和组合，因此渲染是在幕后为你完成的。但是，如果不调用 ReactDOM API，你编写的任何 React 代码都不会渲染到 DOM。
+You may remember using the ReactDOM API in an earlier challenge to render JSX elements to the DOM. The process for rendering React components will look very similar. The past few challenges focused on components and composition, so the rendering was done for you behind the scenes. However, none of the React code you write will render to the DOM without making a call to the ReactDOM API.
-以下是语法的复习：`ReactDOM.render(componentToRender, targetNode)`。第一个参数是要渲染的 React 组件。第二个参数是要在其中渲染该组件的 DOM 节点。
+Here's a refresher on the syntax: `ReactDOM.render(componentToRender, targetNode)`. The first argument is the React component that you want to render. The second argument is the DOM node that you want to render that component within.
-React 组件传递到`ReactDOM.render()`与 JSX 元素略有不同。对于 JSX 元素，你传入的是要渲染的元素的名称。但是，对于 React 组件，你需要使用与渲染嵌套组件相同的语法，例如`ReactDOM.render(<ComponentToRender />, targetNode)`。你可以将此语法用于ES6类组件和函数组件。
+React components are passed into `ReactDOM.render()` a little differently than JSX elements. For JSX elements, you pass in the name of the element that you want to render. However, for React components, you need to use the same syntax as if you were rendering a nested component, for example `ReactDOM.render(<ComponentToRender />, targetNode)`. You use this syntax for both ES6 class components and functional components.
# --instructions--
-在后台为你定义了`Fruits`和`Vegetables`组件。将两个组件渲染为`TypesOfFood`组件的子组件，然后将`TypesOfFood`渲染到 DOM 节点，在这个挑战中，请渲染到 id 为`challenge-node`的`div`中。
+Both the `Fruits` and `Vegetables` components are defined for you behind the scenes. Render both components as children of the `TypesOfFood` component, then render `TypesOfFood` to the DOM. There is a `div` with `id='challenge-node'` available for you to use.
# --hints--
-`TypesOfFood`组件应该返回单个`div`元素。
+The `TypesOfFood` component should return a single `div` element.
```js
assert(
@@ -31,7 +31,7 @@ assert(
);
```
-`TypesOfFood`组件应该在`h1`元素之后渲染`Fruits`组件。
+The `TypesOfFood` component should render the `Fruits` component after the `h1` element.
```js
assert(
@@ -42,7 +42,7 @@ assert(
);
```
-`TypesOfFood`组件应该在`Fruits`组件之后渲染`Vegetables`组件。
+The `TypesOfFood` component should render the `Vegetables` component after `Fruits`.
```js
assert(
@@ -53,7 +53,7 @@ assert(
);
```
-`TypesOfFood`组件应该渲染到 id 为`challenge-node`的`div`中。
+The `TypesOfFood` component should render to the DOM within the `div` with the id `challenge-node`.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/render-conditionally-from-props.md b/curriculum/challenges/chinese/03-front-end-libraries/react/render-conditionally-from-props.md
index a4390e5c186..75979d54bc6 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/render-conditionally-from-props.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/render-conditionally-from-props.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036188
-title: 根据 Props 有条件地渲染
+title: Render Conditionally from Props
challengeType: 6
forumTopicId: 301405
dashedName: render-conditionally-from-props
@@ -8,21 +8,21 @@ dashedName: render-conditionally-from-props
# --description--
-到目前为止，你已经看到了如何使用`if/else`、`&&,`、`null`和三元运算符（`condition ? expressionIfTrue : expressionIfFalse`）对渲染什么和何时渲染做出有条件的判定。然而，还有一个重要的话题需要讨论，让你将这些概念中的任何一个或所有概念与另一个强大的 React 功能结合起来：props。使用 props 有条件地渲染代码在 React 开发人员中很常见--也就是说：他们使用给定 prop 的值来自动决定渲染什么。
+So far, you've seen how to use `if/else`, `&&,` `null` and the ternary operator (`condition ? expressionIfTrue : expressionIfFalse`) to make conditional decisions about what to render and when. However, there's one important topic left to discuss that lets you combine any or all of these concepts with another powerful React feature: props. Using props to conditionally render code is very common with React developers — that is, they use the value of a given prop to automatically make decisions about what to render.
-在这个挑战中，你将设置一个子组件来根据 props 做出渲染决定。你可以使用三元运算符，但是你可以看到过去几个挑战中涵盖的其他几个概念在这种情况下可能同样有用。
+In this challenge, you'll set up a child component to make rendering decisions based on props. You'll also use the ternary operator, but you can see how several of the other concepts that were covered in the last few challenges might be just as useful in this context.
# --instructions--
-代码编辑器有两个部分为你定义的组件：一个名为`GameOfChance`的父组件和一个名为`Results`的子组件。他们被用来创建一个简单的游戏，用户按下按钮来看他们是赢还是输。
+The code editor has two components that are partially defined for you: a parent called `GameOfChance`, and a child called `Results`. They are used to create a simple game where the user presses a button to see if they win or lose.
-首先，你需要一个简单的表达式，每次运行时都会随机返回一个不同的值。你可以使用`Math.random()`。每次调用此方法时，此方法返回`0`（包括）和`1`（不包括）之间的值。因此，对于50/50的几率，请在表达式中使用`Math.random() > .5`。从统计学上讲，这个表达式有 50％ 的几率返回`true`，另外 50％ 返回`false`。在第 render 方法里，用此表达式替换注释以完成变量声明。
+First, you'll need a simple expression that randomly returns a different value every time it is run. You can use `Math.random()`. This method returns a value between `0` (inclusive) and `1` (exclusive) each time it is called. So for 50/50 odds, use `Math.random() >= .5` in your expression. Statistically speaking, this expression will return `true` 50% of the time, and `false` the other 50%. In the render method, replace `null` with the above expression to complete the variable declaration.
-现在你有了一个表达式，可以用来在代码中做出随机决定，接下来你需要实现以下功能：将`Results`组件渲染为`GameOfChance`的子组件，并将`expression`作为 prop 传递出去，prop 的名字是`fiftyFifty`。在`Results`组件中，编写一个三元表达式基于从`GameOfChance`传来的 prop`fiftyFifty`来渲染文本`"You Win!"`或者`"You Lose!"`。最后，确保`handleClick()`方法正确计算每个回合，以便用户知道他们玩过多少次。这也可以让用户知道组件实际上已经更新，以防他们连续赢两次或输两次时自己不知道。
+Now you have an expression that you can use to make a randomized decision in the code. Next you need to implement this. Render the `Results` component as a child of `GameOfChance`, and pass in `expression` as a prop called `fiftyFifty`. In the `Results` component, write a ternary expression to render the `h1` element with the text `"You Win!"` or `"You Lose!"` based on the `fiftyFifty` prop that's being passed in from `GameOfChance`. Finally, make sure the `handleClick()` method is correctly counting each turn so the user knows how many times they've played. This also serves to let the user know the component has actually updated in case they win or lose twice in a row.
# --hints--
-`GameOfChance`组件应该存在并渲染到页面。
+The `GameOfChance` component should exist and render to the page.
```js
assert.strictEqual(
@@ -31,7 +31,7 @@ assert.strictEqual(
);
```
-`GameOfChance`应该返回单个`button`元素。
+`GameOfChance` should return a single `button` element.
```js
assert.strictEqual(
@@ -40,7 +40,7 @@ assert.strictEqual(
);
```
-`GameOfChance`应该返回`Results`组件的一个实例，它有一个名为`fiftyFifty`的 prop。
+`GameOfChance` should return a single instance of the `Results` component, which has a prop called `fiftyFifty`.
```js
assert(
@@ -53,7 +53,7 @@ assert(
);
```
-`GameOfChance`的 state 应该使用值为`1`的`counter`属性来初始化。
+`GameOfChance` state should be initialized with a property of `counter` set to a value of `1`.
```js
assert.strictEqual(
@@ -62,7 +62,7 @@ assert.strictEqual(
);
```
-当`GameOfChance`组件第一次渲染到 DOM 时，应该返回一个`p`元素，其内部文本为`Turn: 1`。
+When the `GameOfChance` component is first rendered to the DOM, a `p` element should be returned with the inner text of `Turn: 1`.
```js
assert.strictEqual(
@@ -71,12 +71,10 @@ assert.strictEqual(
);
```
-每次点击按钮，counter 应该增加 1，并且一个包含文本`"Turn: N"`的`p`元素应该渲染到DOM，其中`N`是 counter 的值。
+Each time the button is clicked, the counter state should be incremented by a value of 1, and a single `p` element should be rendered to the DOM that contains the text "Turn: N", where N is the value of the counter state.
```js
-async () => {
- const waitForIt = (fn) =>
- new Promise((resolve, reject) => setTimeout(() => resolve(fn()), 250));
+(() => {
const comp = Enzyme.mount(React.createElement(GameOfChance));
const simulate = () => {
comp.find('button').simulate('click');
@@ -87,29 +85,29 @@ async () => {
});
const _1 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _2 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _3 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _4 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _5 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
- const _1_val = await _1();
- const _2_val = await _2();
- const _3_val = await _3();
- const _4_val = await _4();
- const _5_val = await _5();
+ const _1_val = _1();
+ const _2_val = _2();
+ const _3_val = _3();
+ const _4_val = _4();
+ const _5_val = _5();
assert(
_1_val.count === 2 &&
_1_val.text === 'Turn: 2' &&
@@ -122,15 +120,13 @@ async () => {
_5_val.count === 6 &&
_5_val.text === 'Turn: 6'
);
-};
+})();
```
-当`GameOfChance`组件第一次挂载到 DOM 上时，每次按钮被点击，都应该返回一个`h1`元素，元素中随机渲染`You Win!`或者`You Lose!`。
+When the `GameOfChance` component is first mounted to the DOM and each time the button is clicked thereafter, a single `h1` element should be returned that randomly renders either `You Win!` or `You Lose!`.
```js
-async () => {
- const waitForIt = (fn) =>
- new Promise((resolve, reject) => setTimeout(() => resolve(fn()), 250));
+(() => {
const comp = Enzyme.mount(React.createElement(GameOfChance));
const simulate = () => {
comp.find('button').simulate('click');
@@ -142,49 +138,49 @@ async () => {
const _1 = result();
const _2 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _3 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _4 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _5 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _6 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _7 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _8 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _9 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _10 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
- const _2_val = await _2();
- const _3_val = await _3();
- const _4_val = await _4();
- const _5_val = await _5();
- const _6_val = await _6();
- const _7_val = await _7();
- const _8_val = await _8();
- const _9_val = await _9();
- const _10_val = await _10();
+ const _2_val = _2();
+ const _3_val = _3();
+ const _4_val = _4();
+ const _5_val = _5();
+ const _6_val = _6();
+ const _7_val = _7();
+ const _8_val = _8();
+ const _9_val = _9();
+ const _10_val = _10();
const __text = new Set([
_1.text,
_2_val.text,
@@ -210,7 +206,7 @@ async () => {
_10_val.h1
]);
assert(__text.size === 2 && __h1.size === 1);
-};
+})();
```
# --seed--
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/render-html-elements-to-the-dom.md b/curriculum/challenges/chinese/03-front-end-libraries/react/render-html-elements-to-the-dom.md
index 639d9e00696..d5167e4e138 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/render-html-elements-to-the-dom.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/render-html-elements-to-the-dom.md
@@ -1,6 +1,6 @@
---
id: 5a24bbe0dba28a8d3cbd4c5f
-title: 渲染 HTML 元素为 DOM 树
+title: Render HTML Elements to the DOM
challengeType: 6
forumTopicId: 301406
dashedName: render-html-elements-to-the-dom
@@ -8,37 +8,37 @@ dashedName: render-html-elements-to-the-dom
# --description--
-到目前为止，你已经了解到 JSX 是一种在 JavaScript 中编写可读 HTML 的便捷工具。在 React 中，我们可以使用它的的渲染 API（ReactDOM）将此 JSX 直接渲染到 HTML DOM。
+So far, you've learned that JSX is a convenient tool to write readable HTML within JavaScript. With React, we can render this JSX directly to the HTML DOM using React's rendering API known as ReactDOM.
-ReactDOM 提供了一个简单的方法来将 React 元素呈现给 DOM，如下所示：`ReactDOM.render(componentToRender, targetNode)`，其中第一个参数是要渲染的 React 元素或组件，第二个参数是要将组件渲染到的 DOM 节点。
+ReactDOM offers a simple method to render React elements to the DOM which looks like this: `ReactDOM.render(componentToRender, targetNode)`, where the first argument is the React element or component that you want to render, and the second argument is the DOM node that you want to render the component to.
-如你所料，必须在 JSX 元素声明之后调用`ReactDOM.render()`，就像你在使用变量之前必须声明它一样。
+As you would expect, `ReactDOM.render()` must be called after the JSX element declarations, just like how you must declare variables before using them.
# --instructions--
-代码编辑器有一个简单的 JSX 组件。使用`ReactDOM.render()`方法将该组件渲染到页面。可以将定义好的 JSX 元素直接作为第一个参数传入，并使用`document.getElementById()`来选择要渲染到的 DOM 节点，在这个挑战中，请渲染到 id 为`challenge-node`的`div`中。
+The code editor has a simple JSX component. Use the `ReactDOM.render()` method to render this component to the page. You can pass defined JSX elements directly in as the first argument and use `document.getElementById()` to select the DOM node to render them to. There is a `div` with `id='challenge-node'` available for you to use. Make sure you don't change the `JSX` constant.
# --hints--
-常量`JSX`应该返回一个`div`元素。
+The constant `JSX` should return a `div` element.
```js
assert(JSX.type === 'div');
```
-`div`应该包含一个`h1`标签作为第一个元素。
+The `div` should contain an `h1` tag as the first element.
```js
assert(JSX.props.children[0].type === 'h1');
```
-`div`应该包含一个`p`标签作为第二个元素。
+The `div` should contain a `p` tag as the second element.
```js
assert(JSX.props.children[1].type === 'p');
```
-提供的 JSX 元素应该渲染到 id 为`challenge-node`的 DOM 节点。
+The provided JSX element should render to the DOM node with id `challenge-node`.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/render-react-on-the-server-with-rendertostring.md b/curriculum/challenges/chinese/03-front-end-libraries/react/render-react-on-the-server-with-rendertostring.md
index dbc46d10fe2..bfced52d4b3 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/render-react-on-the-server-with-rendertostring.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/render-react-on-the-server-with-rendertostring.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403618d
-title: 用 renderToString 在服务器上渲染 React
+title: Render React on the Server with renderToString
challengeType: 6
forumTopicId: 301407
dashedName: render-react-on-the-server-with-rendertostring
@@ -8,17 +8,17 @@ dashedName: render-react-on-the-server-with-rendertostring
# --description--
-到目前为止，你已经能够在客户端上渲染 React 组件，一般来说我们都是这么做的。然而，在一些用例中，在服务器上渲染一个 React 组件是有意义的。由于 React 是一个 JavaScript 视图库，所以使用 Node 让 JavaScript 运行在服务器上是可行的。事实上，React 提供了一个可用于此目的的`renderToString()`方法。
+So far, you have been rendering React components on the client. Normally, this is what you will always do. However, there are some use cases where it makes sense to render a React component on the server. Since React is a JavaScript view library and you can run JavaScript on the server with Node, this is possible. In fact, React provides a `renderToString()` method you can use for this purpose.
-有两个关键原因可以解释为什么服务器上的渲染可能会在真实世界的应用程序中使用。首先，如果不这样做，你的 React 应用程序将包含一个代码量很少的 HTML 文件和一大堆 JavaScript，当它最初加载到浏览器时。这对于搜索引擎来说可能不太理想，因为它们试图为你的网页内容生成索引，以便人们可以找到你。如果在服务器上渲染初始 HTML 标记并将其发送到客户端，则初始页面加载的内容包含搜索引擎可以抓取的所有页面标记。其次，这创造了更快的初始页面加载体验，因为渲染的 HTML 代码量要比整个应用程序的 JavaScript 代码小。React 仍然能够识别你的应用并在初始加载后进行管理。
+There are two key reasons why rendering on the server may be used in a real world app. First, without doing this, your React apps would consist of a relatively empty HTML file and a large bundle of JavaScript when it's initially loaded to the browser. This may not be ideal for search engines that are trying to index the content of your pages so people can find you. If you render the initial HTML markup on the server and send this to the client, the initial page load contains all of the page's markup which can be crawled by search engines. Second, this creates a faster initial page load experience because the rendered HTML is smaller than the JavaScript code of the entire app. React will still be able to recognize your app and manage it after the initial load.
# --instructions--
-`renderToString()`方法由`ReactDOMServer`提供，在这里已为你定义成全局变量。这个方法接受一个 React 元素作为参数。用它来把`App`渲染成字符串。
+The `renderToString()` method is provided on `ReactDOMServer`, which is available here as a global object. The method takes one argument which is a React element. Use this to render `App` to a string.
# --hints--
-`App`组件应该使用`ReactDOMServer.renderToString`渲染一个字符串。
+The `App` component should render to a string using `ReactDOMServer.renderToString`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/render-state-in-the-user-interface-another-way.md b/curriculum/challenges/chinese/03-front-end-libraries/react/render-state-in-the-user-interface-another-way.md
index 41bcd7180e8..e1dd33ed044 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/render-state-in-the-user-interface-another-way.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/render-state-in-the-user-interface-another-way.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036172
-title: 以另一种方式在用户界面中渲染状态
+title: Render State in the User Interface Another Way
challengeType: 6
forumTopicId: 301408
dashedName: render-state-in-the-user-interface-another-way
@@ -8,17 +8,17 @@ dashedName: render-state-in-the-user-interface-another-way
# --description--
-还有另一种方法可以访问组件中的`state`。在`render()`方法中，在`return`语句之前，你可以直接编写 JavaScript。例如，你可以声明函数、从`state`或`props`访问数据、对此数据执行计算等。然后，你可以将任何数据赋值给你在`return`语句中可以访问的变量。
+There is another way to access `state` in a component. In the `render()` method, before the `return` statement, you can write JavaScript directly. For example, you could declare functions, access data from `state` or `props`, perform computations on this data, and so on. Then, you can assign any data to variables, which you have access to in the `return` statement.
# --instructions--
-在`MyComponent`的 render 方法中，定义一个名为`name`的`常量`，并将其设置为组件`state`中的 name 值。因为可以直接在代码部分编写 JavaScript，所以不需要用大括号括起来。
+In the `MyComponent` render method, define a `const` called `name` and set it equal to the name value in the component's `state`. Because you can write JavaScript directly in this part of the code, you don't have to enclose this reference in curly braces.
-接下来，在 return 语句中，在`h1`标签中渲染变量`name`的值。记住，在 return 语句中需要使用 JSX 语法（用到 JavaScript 的花括号）。
+Next, in the return statement, render this value in an `h1` tag using the variable `name`. Remember, you need to use the JSX syntax (curly braces for JavaScript) in the return statement.
# --hints--
-`MyComponent`应该有一个键`name`，其值`freeCodeCamp`存储在其 state 中。
+`MyComponent` should have a key `name` with value `freeCodeCamp` stored in its state.
```js
assert(
@@ -27,7 +27,7 @@ assert(
);
```
-`MyComponent`应该在`div`中渲染一个`h1`标题。
+`MyComponent` should render an `h1` header enclosed in a single `div`.
```js
assert(
@@ -37,14 +37,14 @@ assert(
);
```
-渲染的`h1`标签应该包含`{name}`的引用。
+The rendered `h1` tag should include a reference to `{name}`.
```js
(getUserInput) =>
assert(/<h1>\n*\s*\{\s*name\s*\}\s*\n*<\/h1>/.test(getUserInput('index')));
```
-渲染的`h1`标题中应该包含一段文本，这段文本是从组件的 state 中渲染出来的。
+The rendered `h1` header should contain text rendered from the component's state.
```js
async () => {
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/render-state-in-the-user-interface.md b/curriculum/challenges/chinese/03-front-end-libraries/react/render-state-in-the-user-interface.md
index 515ffc21a0d..e9a117d6564 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/render-state-in-the-user-interface.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/render-state-in-the-user-interface.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036171
-title: 在用户界面中渲染状态
+title: Render State in the User Interface
challengeType: 6
forumTopicId: 301409
dashedName: render-state-in-the-user-interface
@@ -8,23 +8,23 @@ dashedName: render-state-in-the-user-interface
# --description--
-一旦定义了组件的初始 state，你就可以在要渲染的 UI 中显示它的任何部分。如果组件是有状态的，它将始终可以访问`render()`方法中`state`的数据。你就可以使用`this.state`访问数据。
+Once you define a component's initial state, you can display any part of it in the UI that is rendered. If a component is stateful, it will always have access to the data in `state` in its `render()` method. You can access the data with `this.state`.
-如果你想在 render 方法的`return`中访问 state 值，你必须把这个值用花括号括起来。
+If you want to access a state value within the `return` of the render method, you have to enclose the value in curly braces.
-`state`是 React 组件中最强大的特性之一，它允许你跟踪应用程序中的重要数据，并根据数据的变化渲染 UI。如果你的数据发生变化，你的 UI 也会随之改变。React 使用所谓的虚拟 DOM 来跟踪幕后的变化。当 state 数据更新时，它会使用该数据触发组件的重新渲染--包括接收 prop 数据的子组件。React 只在必要的时候更新实际的DOM，这意味着你不必担心 DOM 的变更，只需声明 UI 的外观即可。
+`State` is one of the most powerful features of components in React. It allows you to track important data in your app and render a UI in response to changes in this data. If your data changes, your UI will change. React uses what is called a virtual DOM, to keep track of changes behind the scenes. When state data updates, it triggers a re-render of the components using that data - including child components that received the data as a prop. React updates the actual DOM, but only where necessary. This means you don't have to worry about changing the DOM. You simply declare what the UI should look like.
-注意，如果组件有状态，则没有其他组件知道它的`state`。它的`state`是完全封装的，或者是局限于组件本身的，除非你将 state 数据作为`props`传递给子组件。封装`state`的概念非常重要，因为它允许你编写特定的逻辑，然后将该逻辑包含并隔离在代码中的某个位置。
+Note that if you make a component stateful, no other components are aware of its `state`. Its `state` is completely encapsulated, or local to that component, unless you pass state data to a child component as `props`. This notion of encapsulated `state` is very important because it allows you to write certain logic, then have that logic contained and isolated in one place in your code.
# --instructions--
-在代码编辑器中，`MyComponent`是一个有状态组件，在组件的 render 方法中定义一个`h1`标签，该方法从组件的 state 渲染`name`的值。
+In the code editor, `MyComponent` is already stateful. Define an `h1` tag in the component's render method which renders the value of `name` from the component's state.
-**注意：** `h1`应该只渲染来自`state`的值。在 JSX 中，使用花括号`{ }`编写的任何代码都将被视为 JavaScript。因此，要访问`state`中的值，只需将引用括在花括号中即可。
+**Note:** The `h1` should only render the value from `state` and nothing else. In JSX, any code you write with curly braces `{ }` will be treated as JavaScript. So to access the value from `state` just enclose the reference in curly braces.
# --hints--
-`MyComponent`应该有一个键`name`，其值`freeCodeCamp`存储在其 state 中。
+`MyComponent` should have a key `name` with value `freeCodeCamp` stored in its state.
```js
assert(
@@ -33,7 +33,7 @@ assert(
);
```
-`MyComponent`应该在`div`中渲染一个`h1`标题。
+`MyComponent` should render an `h1` header enclosed in a single `div`.
```js
assert(
@@ -43,7 +43,7 @@ assert(
);
```
-渲染的`h1`标题中应该包含一段文本，这段文本是从组件的 state 中渲染出来的。
+The rendered `h1` header should contain text rendered from the component's state.
```js
async () => {
@@ -55,7 +55,8 @@ async () => {
return waitForIt(() => mockedComponent.html());
};
const firstValue = await first();
- assert(firstValue === '<div><h1>TestName</h1></div>');
+ const getValue = firstValue.replace(/\s/g, '');
+ assert(getValue === '<div><h1>TestName</h1></div>');
};
```
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/render-with-an-if-else-condition.md b/curriculum/challenges/chinese/03-front-end-libraries/react/render-with-an-if-else-condition.md
index 628e9b5b454..b56c814074b 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/render-with-an-if-else-condition.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/render-with-an-if-else-condition.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036184
-title: 使用 If-Else 条件进行渲染
+title: Render with an If-Else Condition
challengeType: 6
forumTopicId: 301410
dashedName: render-with-an-if-else-condition
@@ -8,17 +8,17 @@ dashedName: render-with-an-if-else-condition
# --description--
-使用 JavaScript 控制渲染视图的另一个应用是将渲染的元素绑定到一个条件。当条件为真时，将呈现一个视图，反之，则呈现另一种视图。你可以在 React 组件的`render()`方法中使用的标准`if/else`语句来实现这一点。
+Another application of using JavaScript to control your rendered view is to tie the elements that are rendered to a condition. When the condition is true, one view renders. When it's false, it's a different view. You can do this with a standard `if/else` statement in the `render()` method of a React component.
# --instructions--
-MyComponent 的 state 中包含一个`布尔值`，用于跟踪是否要在 UI 中显示某个元素。`按钮`切换此值的状态。目前，它每次都呈现相同的 UI。用`if/else`语句重写`render()`方法，如果`display`为`true`则返回当前标记。否则，返回不带`h1`元素的标记。
+MyComponent contains a `boolean` in its state which tracks whether you want to display some element in the UI or not. The `button` toggles the state of this value. Currently, it renders the same UI every time. Rewrite the `render()` method with an `if/else` statement so that if `display` is `true`, you return the current markup. Otherwise, return the markup without the `h1` element.
-**注意：** 写`if/else`语句才能通过测试，使用三元运算符是不会通过的。
+**Note:** You must write an `if/else` to pass the tests. Use of the ternary operator will not pass here.
# --hints--
-`MyComponent`应该存在并被渲染。
+`MyComponent` should exist and render.
```js
assert(
@@ -29,7 +29,7 @@ assert(
);
```
-当`display`被设置为`true`时，`div`、`button`和`h1`标签应该被渲染。
+When `display` is set to `true`, a `div`, `button`, and `h1` should render.
```js
async () => {
@@ -50,7 +50,7 @@ async () => {
};
```
-当`display`被设置为`false`时，只有`div`和`button`应该被渲染。
+When `display` is set to `false`, only a `div` and `button` should render.
```js
async () => {
@@ -71,7 +71,7 @@ async () => {
};
```
-render 方法中应该使用`if/else`语句来检查`this.state.display`的条件。
+The render method should use an `if/else` statement to check the condition of `this.state.display`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/review-using-props-with-stateless-functional-components.md b/curriculum/challenges/chinese/03-front-end-libraries/react/review-using-props-with-stateless-functional-components.md
index e986c0843ee..c9d751ce8d2 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/review-using-props-with-stateless-functional-components.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/review-using-props-with-stateless-functional-components.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403616f
-title: 复习如何使用 Props 和无状态函数组件
+title: Review Using Props with Stateless Functional Components
challengeType: 6
forumTopicId: 301411
dashedName: review-using-props-with-stateless-functional-components
@@ -8,19 +8,19 @@ dashedName: review-using-props-with-stateless-functional-components
# --description--
-除了上一个挑战，你一直在将 props 传递给无状态的函数组件。这些组件就像纯函数，它们接收 props 作为输入，并在每次传递相同 props 时返回相同的视图。你可能会想知道什么是状态，下一个挑战将会更详细地讲述它。在此之前，我们先来回顾一下组件的术语。
+Except for the last challenge, you've been passing props to stateless functional components. These components act like pure functions. They accept props as input and return the same view every time they are passed the same props. You may be wondering what state is, and the next challenge will cover it in more detail. Before that, here's a review of the terminology for components.
-*无状态函数组件*是一个函数，它接收 props 作为输入并返回 JSX。另一方面，*无状态组件*是一个类，它扩展了`React.Component`，但是不使用内部状态（下一个挑战中讨论）。最后，*状态组件*是指维护其自身内部状态的组件，它简称组件或 React 组件。
+A *stateless functional component* is any function you write which accepts props and returns JSX. A *stateless component*, on the other hand, is a class that extends `React.Component`, but does not use internal state (covered in the next challenge). Finally, a *stateful component* is a class component that does maintain its own internal state. You may see stateful components referred to simply as components or React components.
-一种常见的应用模式是尽可能减少状态组件并创建无状态的函数组件。这有助于将状态管理包含到应用程序的特定区域。反过来，通过更容易地跟踪状态变化如何影响其行为，可以改进应用程序的开发和维护。
+A common pattern is to try to minimize statefulness and to create stateless functional components wherever possible. This helps contain your state management to a specific area of your application. In turn, this improves development and maintenance of your app by making it easier to follow how changes to state affect its behavior.
# --instructions--
-在代码编辑器中有一个`CampSite`组件，它把`Camper`组件渲染为自己的子组件。定义`Camper`组件，并为其分配默认 props`{ name: 'CamperBot' }`。你可以在`Camper`组件内部渲染任何你想要的代码，但是要确保有一个`p`元素，它只包含作为`prop`传递的`name`值。最后，在`Camper`组件上定义`propTypes`，要求提供`name`作为 prop，并验证它是`string`类型。
+The code editor has a `CampSite` component that renders a `Camper` component as a child. Define the `Camper` component and assign it default props of `{ name: 'CamperBot' }`. Inside the `Camper` component, render any code that you want, but make sure to have one `p` element that includes only the `name` value that is passed in as a `prop`. Finally, define `propTypes` on the `Camper` component to require `name` to be provided as a prop and verify that it is of type `string`.
# --hints--
-应该渲染`CampSite`组件。
+The `CampSite` component should render.
```js
assert(
@@ -31,7 +31,7 @@ assert(
);
```
-应该渲染`Camper`组件。
+The `Camper` component should render.
```js
assert(
@@ -42,27 +42,27 @@ assert(
);
```
-`Camper`组件应该包含默认 props，它将字符串`CamperBot`赋值给关键字`name`。
+The `Camper` component should include default props which assign the string `CamperBot` to the key `name`.
```js
assert(
/Camper.defaultProps={name:(['"`])CamperBot\1,?}/.test(
- code.replace(/\s/g, '')
+ __helpers.removeWhiteSpace(code)
)
);
```
-`Camper`组件应该包含`string`类型的`name`prop。
+The `Camper` component should include prop types which require the `name` prop to be of type `string`.
```js
assert(
/Camper.propTypes={name:PropTypes.string.isRequired,?}/.test(
- code.replace(/\s/g, '')
+ __helpers.removeWhiteSpace(code)
)
);
```
-`Camper`组件应该包含一个`p`元素，元素内是来自prop`name`的唯一文本。
+The `Camper` component should contain a `p` element with only the text from the `name` prop.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/set-state-with-this.setstate.md b/curriculum/challenges/chinese/03-front-end-libraries/react/set-state-with-this.setstate.md
index 985b41ded63..93b387efc95 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/set-state-with-this.setstate.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/set-state-with-this.setstate.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036173
-title: 用 this.setState 设置状态
+title: Set State with this.setState
challengeType: 6
forumTopicId: 301412
dashedName: set-state-with-this-setstate
@@ -8,7 +8,7 @@ dashedName: set-state-with-this-setstate
# --description--
-前面的挑战涵盖了组件的`state`以及如何在`constructor`中初始化 state。还有一种方法可以更改组件的`state`，React 提供了`setState`方法来更新组件的`state`。在组件类中调用`setState`方法如下所示：`this.setState()`，传入键值对的对象，其中键是 state 属性，值是更新后的 state 数据。例如，如果我们在 state 中存储`username`，并想要更新它，代码如下所示：
+The previous challenges covered component `state` and how to initialize state in the `constructor`. There is also a way to change the component's `state`. React provides a method for updating component `state` called `setState`. You call the `setState` method within your component class like so: `this.setState()`, passing in an object with key-value pairs. The keys are your state properties and the values are the updated state data. For instance, if we were storing a `username` in state and wanted to update it, it would look like this:
```jsx
this.setState({
@@ -16,17 +16,17 @@ this.setState({
});
```
-React 希望你永远不要直接修改`state`，而是在 state 发生改变时始终使用`this.setState()`。此外，你应该注意，React 可以批量处理多个 state 更新以提高性能。这意味着通过`setState`方法进行的 state 更新可以是异步的。`setState`方法有一种替代语法可以解决异步问题，虽然这很少用到，但是最好还是记住它！有关详细信息，请参阅[React 文档](https://facebook.github.io/react/docs/state-and-lifecycle.html)。
+React expects you to never modify `state` directly, instead always use `this.setState()` when state changes occur. Also, you should note that React may batch multiple state updates in order to improve performance. What this means is that state updates through the `setState` method can be asynchronous. There is an alternative syntax for the `setState` method which provides a way around this problem. This is rarely needed but it's good to keep it in mind! Please consult the [React documentation](https://facebook.github.io/react/docs/state-and-lifecycle.html) for further details.
# --instructions--
-代码编辑器中有一个`button`元素，它有一个`onClick()`处理程序。当`button`在浏览器中接收到单击事件时触发此处理程序，并运行`MyComponent`中定义的`handleClick`方法。在`handleClick`方法中，使用`this.setState()`更新组件的`state`。设置`state`中的`name`属性为字符串`React Rocks!`。
+There is a `button` element in the code editor which has an `onClick()` handler. This handler is triggered when the `button` receives a click event in the browser, and runs the `handleClick` method defined on `MyComponent`. Within the `handleClick` method, update the component `state` using `this.setState()`. Set the `name` property in `state` to equal the string `React Rocks!`.
-单击按钮查看渲染的 state 的更新。如果你不完全理解单击处理程序代码在此时的工作方式，请不要担心。在接下来的挑战中会有讲述。
+Click the button and watch the rendered state update. Don't worry if you don't fully understand how the click handler code works at this point. It's covered in upcoming challenges.
# --hints--
-`MyComponent`的 state 应该使用键值对 `{ name: Initial State }` 来初始化。
+The state of `MyComponent` should initialize with the key value pair `{ name: Initial State }`.
```js
assert(
@@ -35,13 +35,13 @@ assert(
);
```
-`MyComponent`应该渲染一个`h1`标题。
+`MyComponent` should render an `h1` header.
```js
assert(Enzyme.mount(React.createElement(MyComponent)).find('h1').length === 1);
```
-渲染的`h1`标题中应该包含一段文本，这段文本是从组件的 state 中渲染出来的。
+The rendered `h1` header should contain text rendered from the component's state.
```js
async () => {
@@ -57,7 +57,7 @@ async () => {
};
```
-调用`MyComponent`的`handleClick`方法应该将 state 的 name 属性设置为`React Rocks!`。
+Calling the `handleClick` method on `MyComponent` should set the name property in state to equal `React Rocks!`.
```js
async () => {
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/use--for-a-more-concise-conditional.md b/curriculum/challenges/chinese/03-front-end-libraries/react/use--for-a-more-concise-conditional.md
index 5784dccb399..a103a216256 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/use--for-a-more-concise-conditional.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/use--for-a-more-concise-conditional.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036185
-title: 使用 && 获得更简洁的条件
+title: Use && for a More Concise Conditional
challengeType: 6
forumTopicId: 301413
dashedName: use--for-a-more-concise-conditional
@@ -8,19 +8,19 @@ dashedName: use--for-a-more-concise-conditional
# --description--
-if/else 语句在上一次挑战中是有效的，但是有一种更简洁的方法可以达到同样的结果。假设你正在跟踪组件中的几个条件，并且希望根据这些条件中的每一个来渲染不同的元素。如果你写了很多`else if`语句来返回稍微不同的 UI，你可能会写很多重复代码，这就留下了出错的空间。相反，你可以使用`&&`逻辑运算符以更简洁的方式执行条件逻辑。这是完全可行的，因为你希望检查条件是否为真，如果为真，则返回一些标记。这里有一个例子：
+The if/else statements worked in the last challenge, but there's a more concise way to achieve the same result. Imagine that you are tracking several conditions in a component and you want different elements to render depending on each of these conditions. If you write a lot of `else if` statements to return slightly different UIs, you may repeat code which leaves room for error. Instead, you can use the `&&` logical operator to perform conditional logic in a more concise way. This is possible because you want to check if a condition is `true`, and if it is, return some markup. Here's an example:
`{condition && <p>markup</p>}`
-如果`condition`为 true，则返回标记。如果 condition 为 false，操作将在判断`condition`后立即返回`false`，并且不返回任何内容。你可以将这些语句直接包含在 JSX 中，并通过在每个条件后面写`&&`来将多个条件串在一起。这允许你在`render()`方法中处理更复杂的条件逻辑，而无需重复大量代码。
+If the `condition` is `true`, the markup will be returned. If the condition is `false`, the operation will immediately return `false` after evaluating the `condition` and return nothing. You can include these statements directly in your JSX and string multiple conditions together by writing `&&` after each one. This allows you to handle more complex conditional logic in your `render()` method without repeating a lot of code.
# --instructions--
-再来看看前面的示例，`h1`还是在`display`为`true`时被渲染，但使用`&&`逻辑运算符代替`if/else`语句。
+Solve the previous example again, so the `h1` only renders if `display` is `true`, but use the `&&` logical operator instead of an `if/else` statement.
# --hints--
-`MyComponent`应该存在并被渲染。
+`MyComponent` should exist and render.
```js
assert(
@@ -31,7 +31,7 @@ assert(
);
```
-当`display`被设置为`true`时，`div`、`button`和`h1`标签应该被渲染。
+When `display` is set to `true`, a `div`, `button`, and `h1` should render.
```js
async () => {
@@ -52,7 +52,7 @@ async () => {
};
```
-当`display`被设置为`false`时，只有`div`和`button`应该被渲染。
+When `display` is set to `false`, only a `div` and `button` should render.
```js
async () => {
@@ -73,7 +73,7 @@ async () => {
};
```
-render 方法应该使用`&&`逻辑运算符来检查`this.state.display`的条件。
+The render method should use the && logical operator to check the condition of this.state.display.
```js
(getUserInput) => assert(getUserInput('index').includes('&&'));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/use-a-ternary-expression-for-conditional-rendering.md b/curriculum/challenges/chinese/03-front-end-libraries/react/use-a-ternary-expression-for-conditional-rendering.md
index 9078dd1a248..33292905237 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/use-a-ternary-expression-for-conditional-rendering.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/use-a-ternary-expression-for-conditional-rendering.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036187
-title: 使用三元表达式进行条件渲染
+title: Use a Ternary Expression for Conditional Rendering
challengeType: 6
forumTopicId: 301414
dashedName: use-a-ternary-expression-for-conditional-rendering
@@ -8,21 +8,21 @@ dashedName: use-a-ternary-expression-for-conditional-rendering
# --description--
-在继续使用动态渲染技术之前，还有最后一种方法可以渲染你想要的东西，它使用内置的 JavaScript 条件：***三元运算符***。三元运算符经常被用作 JavaScript 中`if/else`语句的缩写。它们不像传统的`if/else`语句那样健壮，但是在 React 开发人员中非常流行，原因之一就是 JSX 的编译原理，`if/else`语句不能直接插入到 JSX 代码中。你可能在前几个挑战就注意到了这一点--当需要`if/else`语句时，它总是在`return`语句*外面*。如果你想在 JSX 中实现条件逻辑，三元表达式是一个很好的选择。回想一下，三元运算符有三个部分，但是你可以将多个三元表达式组合在一起。以下是基本语法：
+Before moving on to dynamic rendering techniques, there's one last way to use built-in JavaScript conditionals to render what you want: the <dfn>ternary operator</dfn>. The ternary operator is often utilized as a shortcut for `if/else` statements in JavaScript. They're not quite as robust as traditional `if/else` statements, but they are very popular among React developers. One reason for this is because of how JSX is compiled, `if/else` statements can't be inserted directly into JSX code. You might have noticed this a couple challenges ago — when an `if/else` statement was required, it was always *outside* the `return` statement. Ternary expressions can be an excellent alternative if you want to implement conditional logic within your JSX. Recall that a ternary operator has three parts, but you can combine several ternary expressions together. Here's the basic syntax:
-```js
-condition ? expressionIfTrue : expressionIfFalse
+```jsx
+condition ? expressionIfTrue : expressionIfFalse;
```
# --instructions--
-代码编辑器在`CheckUserAge`组件的`render()`方法中定义了三个常量，它们分别是`buttonOne`、`buttonTwo`和`buttonThree`。每个都分配了一个表示按钮元素的简单 JSX 表达式。首先，使用`input`和`userAge`初始化`CheckUserAge`的 state，并将其值设置为空字符串。
+The code editor has three constants defined within the `CheckUserAge` component's `render()` method. They are called `buttonOne`, `buttonTwo`, and `buttonThree`. Each of these is assigned a simple JSX expression representing a button element. First, initialize the state of `CheckUserAge` with `input` and `userAge` both set to values of an empty string.
-一旦组件将信息渲染给页面，用户应该有一种方法与之交互。在组件的`return`语句中，设置一个实现以下逻辑的三元表达式：当页面首次加载时，将提交按钮`buttonOne`渲染到页面。然后，当用户输入年龄并点击该按钮时，根据年龄渲染不同的按钮。如果用户输入的数字小于`18`，则渲染`buttonThree`。如果用户输入的数字大于或等于`18`，则渲染`buttonTwo`。
+Once the component is rendering information to the page, users should have a way to interact with it. Within the component's `return` statement, set up a ternary expression that implements the following logic: when the page first loads, render the submit button, `buttonOne`, to the page. Then, when a user enters their age and clicks the button, render a different button based on the age. If a user enters a number less than `18`, render `buttonThree`. If a user enters a number greater than or equal to `18`, render `buttonTwo`.
# --hints--
-`CheckUserAge`组件应该渲染出单个`input`元素和单个`button`元素。
+The `CheckUserAge` component should render with a single `input` element and a single `button` element.
```js
assert(
@@ -33,7 +33,7 @@ assert(
);
```
-`CheckUserAge`组件的 state 应该用`userAge`属性和`input`属性初始化，并且这两个属性的值都被设置为空字符串。
+The `CheckUserAge` component's state should be initialized with a property of `userAge` and a property of `input`, both set to a value of an empty string.
```js
assert(
@@ -42,7 +42,7 @@ assert(
);
```
-当`CheckUserAge`组件首次渲染到 DOM 时，`按钮`内部的文本应为 Submit。
+When the `CheckUserAge` component is first rendered to the DOM, the `button`'s inner text should be Submit.
```js
assert(
@@ -51,12 +51,10 @@ assert(
);
```
-当小于 18 的数字输入到`input`元素中并点击该`按钮`时，该`按钮`的内部文本应该是`You Shall Not Pass`。
+When a number of less than 18 is entered into the `input` element and the `button` is clicked, the `button`'s inner text should read `You Shall Not Pass`.
```js
-async () => {
- const waitForIt = (fn) =>
- new Promise((resolve, reject) => setTimeout(() => resolve(fn()), 250));
+(() => {
const mockedComponent = Enzyme.mount(React.createElement(CheckUserAge));
const initialButton = mockedComponent.find('button').text();
const enter3AndClickButton = () => {
@@ -64,37 +62,31 @@ async () => {
.find('input')
.simulate('change', { target: { value: '3' } });
mockedComponent.find('button').simulate('click');
- return waitForIt(() => {
- mockedComponent.update();
- return mockedComponent.find('button').text();
- });
+ mockedComponent.update();
+ return mockedComponent.find('button').text();
};
const enter17AndClickButton = () => {
mockedComponent
.find('input')
.simulate('change', { target: { value: '17' } });
mockedComponent.find('button').simulate('click');
- return waitForIt(() => {
- mockedComponent.update();
- return mockedComponent.find('button').text();
- });
+ mockedComponent.update();
+ return mockedComponent.find('button').text();
};
- const userAge3 = await enter3AndClickButton();
- const userAge17 = await enter17AndClickButton();
+ const userAge3 = enter3AndClickButton();
+ const userAge17 = enter17AndClickButton();
assert(
initialButton === 'Submit' &&
userAge3 === 'You Shall Not Pass' &&
userAge17 === 'You Shall Not Pass'
);
-};
+})();
```
-当大于或等于 18 的数字输入到`input`元素中并点击该`按钮`时，该`按钮`的内部文本应该是`You May Enter`。
+When a number greater than or equal to 18 is entered into the `input` element and the `button` is clicked, the `button`'s inner text should read `You May Enter`.
```js
-async () => {
- const waitForIt = (fn) =>
- new Promise((resolve, reject) => setTimeout(() => resolve(fn()), 250));
+(() => {
const mockedComponent = Enzyme.mount(React.createElement(CheckUserAge));
const initialButton = mockedComponent.find('button').text();
const enter18AndClickButton = () => {
@@ -102,81 +94,69 @@ async () => {
.find('input')
.simulate('change', { target: { value: '18' } });
mockedComponent.find('button').simulate('click');
- return waitForIt(() => {
- mockedComponent.update();
- return mockedComponent.find('button').text();
- });
+ mockedComponent.update();
+ return mockedComponent.find('button').text();
};
const enter35AndClickButton = () => {
mockedComponent
.find('input')
.simulate('change', { target: { value: '35' } });
mockedComponent.find('button').simulate('click');
- return waitForIt(() => {
- mockedComponent.update();
- return mockedComponent.find('button').text();
- });
+ mockedComponent.update();
+ return mockedComponent.find('button').text();
};
- const userAge18 = await enter18AndClickButton();
- const userAge35 = await enter35AndClickButton();
+ const userAge18 = enter18AndClickButton();
+ const userAge35 = enter35AndClickButton();
assert(
initialButton === 'Submit' &&
userAge18 === 'You May Enter' &&
userAge35 === 'You May Enter'
);
-};
+})();
```
-一旦提交了一个数字，并再次更改了`input`的值，该`按钮`内部文本应该变回`Submit`。
+Once a number has been submitted, and the value of the `input` is once again changed, the `button` should return to reading `Submit`.
```js
-async () => {
- const waitForIt = (fn) =>
- new Promise((resolve, reject) => setTimeout(() => resolve(fn()), 250));
+(() => {
const mockedComponent = Enzyme.mount(React.createElement(CheckUserAge));
const enter18AndClickButton = () => {
mockedComponent
.find('input')
.simulate('change', { target: { value: '18' } });
mockedComponent.find('button').simulate('click');
- return waitForIt(() => {
- mockedComponent.update();
- return mockedComponent.find('button').text();
- });
+ mockedComponent.update();
+ return mockedComponent.find('button').text();
};
const changeInputDontClickButton = () => {
mockedComponent
.find('input')
.simulate('change', { target: { value: '5' } });
- return waitForIt(() => {
- mockedComponent.update();
- return mockedComponent.find('button').text();
- });
+ mockedComponent.update();
+ return mockedComponent.find('button').text();
};
const enter10AndClickButton = () => {
mockedComponent
.find('input')
.simulate('change', { target: { value: '10' } });
mockedComponent.find('button').simulate('click');
- return waitForIt(() => {
- mockedComponent.update();
- return mockedComponent.find('button').text();
- });
+ mockedComponent.update();
+ return mockedComponent.find('button').text();
};
- const userAge18 = await enter18AndClickButton();
- const changeInput1 = await changeInputDontClickButton();
- const userAge10 = await enter10AndClickButton();
- const changeInput2 = await changeInputDontClickButton();
+ const userAge18 = enter18AndClickButton();
+ const changeInput1 = changeInputDontClickButton();
+ const userAge10 = enter10AndClickButton();
+ const changeInput2 = changeInputDontClickButton();
assert(
userAge18 === 'You May Enter' &&
changeInput1 === 'Submit' &&
userAge10 === 'You Shall Not Pass' &&
changeInput2 === 'Submit'
);
-};
+})();
```
-你的代码不应该包含任何`if/else`语句。
+Your code should not contain any `if/else` statements.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/use-advanced-javascript-in-react-render-method.md b/curriculum/challenges/chinese/03-front-end-libraries/react/use-advanced-javascript-in-react-render-method.md
index bff06b7f3f5..6efebc2ddb0 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/use-advanced-javascript-in-react-render-method.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/use-advanced-javascript-in-react-render-method.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036183
-title: 在 React Render 方法中使用 JavaScript
+title: Use Advanced JavaScript in React Render Method
challengeType: 6
forumTopicId: 301415
dashedName: use-advanced-javascript-in-react-render-method
@@ -8,17 +8,17 @@ dashedName: use-advanced-javascript-in-react-render-method
# --description--
-在之前的挑战中，你学习了如何使用大括号`{ }`将 JavaScript 代码插入到 JSX 代码中，用于访问 props、传递 props、访问 state、在代码中插入注释以及最近学习的定制组件样式等任务。这些都是将 JavaScript 放在 JSX 中的常见用例，但是在 React 组件中使用 JavaScript 代码还有其他方式。
+In previous challenges, you learned how to inject JavaScript code into JSX code using curly braces, `{ }`, for tasks like accessing props, passing props, accessing state, inserting comments into your code, and most recently, styling your components. These are all common use cases to put JavaScript in JSX, but they aren't the only way that you can utilize JavaScript code in your React components.
-在`render`方法中编写 JavaScript，你可以把 JavaScript 直接放在`return`语句之前，而***不必***将其插入大括号中。这是因为它还不在 JSX 代码中。当你稍后想在`return`语句中的 JSX 代码中使用变量时，可以将变量名放在大括号中。
+You can also write JavaScript directly in your `render` methods, before the `return` statement, ***without*** inserting it inside of curly braces. This is because it is not yet within the JSX code. When you want to use a variable later in the JSX code *inside* the `return` statement, you place the variable name inside curly braces.
# --instructions--
-在提供的代码中，`render`方法中有一个包含 20 个短语的数组，用于表示 20 世纪 80 年代经典魔术八球玩具中的答案。绑定`ask`方法到按钮的单击事件，每次单击该按钮时，将生成随机数并将其存储为 state 中的`randomIndex`。在第 52 行，删除字符串`"change me!"`并重新分配`answer`常量，以便每次组件更新时，你的代码随机访问`possibleAnswers`数组的不同索引。最后，在`p`标签内插入`answer`常量。
+In the code provided, the `render` method has an array that contains 20 phrases to represent the answers found in the classic 1980's Magic Eight Ball toy. The button click event is bound to the `ask` method, so each time the button is clicked a random number will be generated and stored as the `randomIndex` in state. On line 52, delete the string `"change me!"` and reassign the `answer` const so your code randomly accesses a different index of the `possibleAnswers` array each time the component updates. Finally, insert the `answer` const inside the `p` tags.
# --hints--
-`MagicEightBall`组件应该存在并被渲染到页面。
+The `MagicEightBall` component should exist and should render to the page.
```js
assert.strictEqual(
@@ -28,7 +28,7 @@ assert.strictEqual(
);
```
-`MagicEightBall`的第一个子元素应该是`input`元素。
+`MagicEightBall`'s first child should be an `input` element.
```js
assert.strictEqual(
@@ -40,7 +40,7 @@ assert.strictEqual(
);
```
-`MagicEightBall`的第三个子元素应该是`button`元素。
+`MagicEightBall`'s third child should be a `button` element.
```js
assert.strictEqual(
@@ -52,7 +52,7 @@ assert.strictEqual(
);
```
-`MagicEightBall`的 state 应该用`userInput`属性和`randomIndex`属性初始化，并且这两个属性的值都应该是空字符串。
+`MagicEightBall`'s state should be initialized with a property of `userInput` and a property of `randomIndex` both set to a value of an empty string.
```js
assert(
@@ -62,7 +62,7 @@ assert(
);
```
-当`MagicEightBall`第一次加载到 DOM 中时，它应该返回一个空的`p`元素。
+When `MagicEightBall` is first mounted to the DOM, it should return an empty `p` element.
```js
assert(
@@ -71,12 +71,10 @@ assert(
);
```
-当文本被输入到`input`元素中并点击按钮时，`MagicEightBall`组件应该返回一个`p`元素，该元素包含数组`possibleAnswers`中的随机一个元素。
+When text is entered into the `input` element and the button is clicked, the `MagicEightBall` component should return a `p` element that contains a random element from the `possibleAnswers` array.
```js
-async () => {
- const waitForIt = (fn) =>
- new Promise((resolve, reject) => setTimeout(() => resolve(fn()), 250));
+(() => {
const comp = Enzyme.mount(React.createElement(MagicEightBall));
const simulate = () => {
comp.find('input').simulate('change', { target: { value: 'test?' } });
@@ -85,54 +83,54 @@ async () => {
const result = () => comp.find('p').text();
const _1 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _2 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _3 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _4 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _5 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _6 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _7 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _8 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _9 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
const _10 = () => {
simulate();
- return waitForIt(() => result());
+ return result();
};
- const _1_val = await _1();
- const _2_val = await _2();
- const _3_val = await _3();
- const _4_val = await _4();
- const _5_val = await _5();
- const _6_val = await _6();
- const _7_val = await _7();
- const _8_val = await _8();
- const _9_val = await _9();
- const _10_val = await _10();
+ const _1_val = _1();
+ const _2_val = _2();
+ const _3_val = _3();
+ const _4_val = _4();
+ const _5_val = _5();
+ const _6_val = _6();
+ const _7_val = _7();
+ const _8_val = _8();
+ const _9_val = _9();
+ const _10_val = _10();
const actualAnswers = [
_1_val,
_2_val,
@@ -150,7 +148,7 @@ async () => {
);
const notAllEqual = new Set(actualAnswers);
assert(notAllEqual.size > 1 && hasIndex.length === 10);
-};
+})();
```
# --seed--
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/use-array.filter-to-dynamically-filter-an-array.md b/curriculum/challenges/chinese/03-front-end-libraries/react/use-array.filter-to-dynamically-filter-an-array.md
index 65f7518f72f..2d8c9a7a2cc 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/use-array.filter-to-dynamically-filter-an-array.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/use-array.filter-to-dynamically-filter-an-array.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403618c
-title: 使用 Array.Filter() 动态过滤数组
+title: Use Array.filter() to Dynamically Filter an Array
challengeType: 6
forumTopicId: 301416
dashedName: use-array-filter-to-dynamically-filter-an-array
@@ -8,17 +8,17 @@ dashedName: use-array-filter-to-dynamically-filter-an-array
# --description--
-`map`数组方法是一个强大的工具，在使用 React 时经常使用。与`map`相关的另一种方法是`filter`，它根据条件过滤数组的内容，然后返回一个新数组。例如，如果你有一个 users 数组，每个数组元素都有一个可以设置为`true`或`false`的`online`属性，你可以只过滤那些在线的用户：
+The `map` array method is a powerful tool that you will use often when working with React. Another method related to `map` is `filter`, which filters the contents of an array based on a condition, then returns a new array. For example, if you have an array of users that all have a property `online` which can be set to `true` or `false`, you can filter only those users that are online by writing:
`let onlineUsers = users.filter(user => user.online);`
# --instructions--
-在代码编辑器中，`MyComponent`的`state`由一个 users 数组初始化。有些用户在线，有些则不在线。过滤数组，以便你只看到在线用户。为此，首先使用`filter`返回一个新数组，该数组只包含`online`属性为`true`的用户。然后，在`renderOnline`变量中，映射经过过滤的数组，并为每个用户返回一个包含它们`username`文本的`li`元素。确保像上一个挑战一样包含一个独特的`key`。
+In the code editor, `MyComponent`'s `state` is initialized with an array of users. Some users are online and some aren't. Filter the array so you see only the users who are online. To do this, first use `filter` to return a new array containing only the users whose `online` property is `true`. Then, in the `renderOnline` variable, map over the filtered array, and return a `li` element for each user that contains the text of their `username`. Be sure to include a unique `key` as well, like in the last challenges.
# --hints--
-`MyComponent`应该存在并被渲染到页面。
+`MyComponent` should exist and render to the page.
```js
assert.strictEqual(
@@ -27,7 +27,7 @@ assert.strictEqual(
);
```
-`MyComponent`的 state 应该初始化为包含 6 个用户的数组。
+`MyComponent`'s state should be initialized to an array of six users.")
```js
assert(
@@ -38,12 +38,10 @@ assert(
);
```
-`MyComponent`应该返回一个`div`、一个`h1`和一个包含`li`元素的无序列表，该列表用于展示在线状态为`true`的每个用户。
+`MyComponent` should return a `div`, an `h1`, and then an unordered list containing `li` elements for every user whose online status is set to `true`.
```js
-async () => {
- const waitForIt = (fn) =>
- new Promise((resolve, reject) => setTimeout(() => resolve(fn()), 250));
+(() => {
const comp = Enzyme.mount(React.createElement(MyComponent));
const users = (bool) => ({
users: [
@@ -58,19 +56,19 @@ async () => {
const _1 = result();
const _2 = () => {
comp.setState(users(true));
- return waitForIt(() => result());
+ return result();
};
const _3 = () => {
comp.setState(users(false));
- return waitForIt(() => result());
+ return result();
};
const _4 = () => {
comp.setState({ users: [] });
- return waitForIt(() => result());
+ return result();
};
- const _2_val = await _2();
- const _3_val = await _3();
- const _4_val = await _4();
+ const _2_val = _2();
+ const _3_val = _3();
+ const _4_val = _4();
assert(
comp.find('div').length === 1 &&
comp.find('h1').length === 1 &&
@@ -80,15 +78,13 @@ async () => {
_3_val === 0 &&
_4_val === 0
);
-};
+})();
```
-`MyComponent`应该渲染包含每个在线用户用户名的`li`元素。
+`MyComponent` should render `li` elements that contain the username of each online user.
```js
-async () => {
- const waitForIt = (fn) =>
- new Promise((resolve, reject) => setTimeout(() => resolve(fn()), 250));
+(() => {
const comp = Enzyme.mount(React.createElement(MyComponent));
const users = (bool) => ({
users: [
@@ -101,17 +97,17 @@ async () => {
});
const ul = () => {
comp.setState(users(true));
- return waitForIt(() => comp.find('ul').html());
+ return comp.find('ul').html();
};
- const html = await ul();
+ const html = ul();
assert(
html ===
'<ul><li>Jeff</li><li>Alan</li><li>Mary</li><li>Jim</li><li>Laura</li></ul>'
);
-};
+})();
```
-每个列表项元素都应该有一个唯一的`key`属性。
+Each list item element should have a unique `key` attribute.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/use-array.map-to-dynamically-render-elements.md b/curriculum/challenges/chinese/03-front-end-libraries/react/use-array.map-to-dynamically-render-elements.md
index fdac107e0ba..0d6cca62f61 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/use-array.map-to-dynamically-render-elements.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/use-array.map-to-dynamically-render-elements.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403618a
-title: 使用 Array.map() 动态渲染元素
+title: Use Array.map() to Dynamically Render Elements
challengeType: 6
forumTopicId: 301417
dashedName: use-array-map-to-dynamically-render-elements
@@ -8,21 +8,21 @@ dashedName: use-array-map-to-dynamically-render-elements
# --description--
-条件渲染很有用，但是你可能需要组件来渲染未知数量的元素。通常在响应式编程中，程序员在应用程序运行时之前无法知道其 state，因为这在很大程度上取决于用户与该程序的交互。程序员需要提前编写代码来正确处理未知状态。在 React 中使用`Array.map()`阐明了这个概念。
+Conditional rendering is useful, but you may need your components to render an unknown number of elements. Often in reactive programming, a programmer has no way to know what the state of an application is until runtime, because so much depends on a user's interaction with that program. Programmers need to write their code to correctly handle that unknown state ahead of time. Using `Array.map()` in React illustrates this concept.
-例如，你创建一个简单的“To Do List”应用程序。作为程序员，你无法知道用户可能在其列表中有多少项。你需要设置组件，以便在使用该程序的人决定今天今日待办事项之前***动态渲染***正确数量的列表元素。
+For example, you create a simple "To Do List" app. As the programmer, you have no way of knowing how many items a user might have on their list. You need to set up your component to dynamically render the correct number of list elements long before someone using the program decides that today is laundry day.
# --instructions--
-代码编辑器完成了`MyToDoList`组件的大部分设置。如果你完成了受控表单挑战，这些代码中的一些应该看起来很熟悉。你会注意到一个`textarea`和一个`button`，以及一些跟踪它们状态的方法，但是页面当前还没有任何东西被渲染。
+The code editor has most of the `MyToDoList` component set up. Some of this code should look familiar if you completed the controlled form challenge. You'll notice a `textarea` and a `button`, along with a couple of methods that track their states, but nothing is rendered to the page yet.
-在`constructor`中，创建一个`this.state`对象并定义两个 state：`userInput`应该初始化为空字符串，`toDoList`应该初始化为空数组。接下来，删除`items`变量旁边`render()`方法中的注释。取而代之的是，将存储在组件内部 state 中的`toDoList`数组一一映射并相应的动态呈现`li`元素。尝试在`textarea`中输入`eat, code, sleep, repeat`，然后点击按钮，看看会发生什么。
+Inside the `constructor`, create a `this.state` object and define two states: `userInput` should be initialized as an empty string, and `toDoList` should be initialized as an empty array. Next, delete the comment in the `render()` method next to the `items` variable. In its place, map over the `toDoList` array stored in the component's internal state and dynamically render a `li` for each item. Try entering the string `eat, code, sleep, repeat` into the `textarea`, then click the button and see what happens.
-**注意：** 你可能知道，像这样的映射操作创建的所有兄弟子元素都需要提供唯一的`key`属性。别担心，这是下一个挑战的主题。
+**Note:** You may know that all sibling child elements created by a mapping operation like this do need to be supplied with a unique `key` attribute. Don't worry, this is the topic of the next challenge.
# --hints--
-`MyToDoList`组件应该存在并渲染到页面。
+The MyToDoList component should exist and render to the page.
```js
assert(
@@ -33,7 +33,7 @@ assert(
);
```
-`MyToDoList`组件的第一个子元素应该是`textarea`元素。
+The first child of `MyToDoList` should be a `textarea` element.
```js
assert(
@@ -47,7 +47,7 @@ assert(
);
```
-`MyToDoList`组件的第二个子元素应该是`br`元素。
+The second child of `MyToDoList` should be a `br` element.
```js
assert(
@@ -60,7 +60,7 @@ assert(
);
```
-`MyToDoList`组件的第三个子元素应该是`button`元素。
+The third child of `MyToDoList` should be a `button` element.
```js
assert(
@@ -74,7 +74,7 @@ assert(
);
```
-`MyToDoList`的 state 应该使用被设置为空数组的`toDoList`进行初始化。
+The state of `MyToDoList` should be initialized with `toDoList` as an empty array.
```js
assert(
@@ -89,7 +89,7 @@ assert(
);
```
-`MyToDoList`的 state 应该使用被设置为空字符串的`userInput`进行初始化。
+The state of `MyToDoList` should be initialized with `userInput` as an empty string.
```js
assert(
@@ -104,61 +104,57 @@ assert(
);
```
-单击`Create List`按钮时，`MyToDoList`组件应该动态返回一个无序列表，该列表包含输入到`textarea`元素中的逗号分隔列表的每个项目的列表项目元素。
+When the `Create List` button is clicked, the `MyToDoList` component should dynamically return an unordered list that contains a list item element for every item of a comma-separated list entered into the `textarea` element.
```js
-async () => {
- const waitForIt = (fn) =>
- new Promise((resolve, reject) => setTimeout(() => resolve(fn()), 100));
+(() => {
const mockedComponent = Enzyme.mount(React.createElement(MyToDoList));
const simulateChange = (el, value) =>
el.simulate('change', { target: { value } });
const state_1 = () => {
- return waitForIt(() => mockedComponent.find('ul').find('li'));
+ return mockedComponent.find('ul').find('li');
};
const setInput = () => {
- return waitForIt(() =>
- simulateChange(mockedComponent.find('textarea'), 'testA, testB, testC')
+ return simulateChange(
+ mockedComponent.find('textarea'),
+ 'testA, testB, testC'
);
};
const click = () => {
- return waitForIt(() => mockedComponent.find('button').simulate('click'));
+ return mockedComponent.find('button').simulate('click');
};
const state_2 = () => {
- return waitForIt(() => {
- const nodes = mockedComponent.find('ul').find('li');
- return { nodes, text: nodes.reduce((t, n) => t + n.text(), '') };
- });
+ const nodes = mockedComponent.find('ul').find('li');
+ return { nodes, text: nodes.reduce((t, n) => t + n.text().trim(), '') };
};
const setInput_2 = () => {
- return waitForIt(() =>
- simulateChange(mockedComponent.find('textarea'), 't1, t2, t3, t4, t5, t6')
+ return simulateChange(
+ mockedComponent.find('textarea'),
+ 't1, t2, t3, t4, t5, t6'
);
};
const click_1 = () => {
- return waitForIt(() => mockedComponent.find('button').simulate('click'));
+ return mockedComponent.find('button').simulate('click');
};
const state_3 = () => {
- return waitForIt(() => {
- const nodes = mockedComponent.find('ul').find('li');
- return { nodes, text: nodes.reduce((t, n) => t + n.text(), '') };
- });
+ const nodes = mockedComponent.find('ul').find('li');
+ return { nodes, text: nodes.reduce((t, n) => t + n.text().trim(), '') };
};
- const awaited_state_1 = await state_1();
- const awaited_setInput = await setInput();
- const awaited_click = await click();
- const awaited_state_2 = await state_2();
- const awaited_setInput_2 = await setInput_2();
- const awaited_click_1 = await click_1();
- const awaited_state_3 = await state_3();
+ const awaited_state_1 = state_1();
+ const awaited_setInput = setInput();
+ const awaited_click = click();
+ const awaited_state_2 = state_2();
+ const awaited_setInput_2 = setInput_2();
+ const awaited_click_1 = click_1();
+ const awaited_state_3 = state_3();
assert(
awaited_state_1.length === 0 &&
awaited_state_2.nodes.length === 3 &&
awaited_state_3.nodes.length === 6 &&
- awaited_state_2.text === 'testA testB testC' &&
- awaited_state_3.text === 't1 t2 t3 t4 t5 t6'
+ awaited_state_2.text === 'testAtestBtestC' &&
+ awaited_state_3.text === 't1t2t3t4t5t6'
);
-};
+})();
```
# --seed--
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/use-default-props.md b/curriculum/challenges/chinese/03-front-end-libraries/react/use-default-props.md
index 35e145b367d..1f048ada95f 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/use-default-props.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/use-default-props.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403616b
-title: 使用默认的 Props
+title: Use Default Props
challengeType: 6
forumTopicId: 301418
dashedName: use-default-props
@@ -8,15 +8,15 @@ dashedName: use-default-props
# --description--
-React 还有一个设置默认 props 的选项。你可以将默认 props 作为组件本身的属性分配给组件，React 会在必要时分配默认 prop。如果没有显式的提供任何值，这允许你指定 prop 值应该是什么。例如，如果你声明`MyComponent.defaultProps = { location: 'San Francisco' }`，即定义一个 location 属性，并且其值在没有另行制定的情况下被设置为字符串`San Francisco`。如果 props 未定义，则 React 会分配默认 props，但如果你将`null`作为 prop 的值，它将保持`null`。
+React also has an option to set default props. You can assign default props to a component as a property on the component itself and React assigns the default prop if necessary. This allows you to specify what a prop value should be if no value is explicitly provided. For example, if you declare `MyComponent.defaultProps = { location: 'San Francisco' }`, you have defined a location prop that's set to the string `San Francisco`, unless you specify otherwise. React assigns default props if props are undefined, but if you pass `null` as the value for a prop, it will remain `null`.
# --instructions--
-代码编辑器中有一个`ShoppingCart`组件。在这个组件上定义默认 props，它指定一个`items`prop，其值为`0`。
+The code editor shows a `ShoppingCart` component. Define default props on this component which specify a prop `items` with a value of `0`.
# --hints--
-应该渲染`ShoppingCart`组件。
+The `ShoppingCart` component should render.
```js
assert(
@@ -27,7 +27,7 @@ assert(
);
```
-`ShoppingCart`组件应该有一个`{ items: 0 }`的默认 prop。
+The `ShoppingCart` component should have a default prop of `{ items: 0 }`.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/use-proptypes-to-define-the-props-you-expect.md b/curriculum/challenges/chinese/03-front-end-libraries/react/use-proptypes-to-define-the-props-you-expect.md
index 1cc3fc2579b..49abf8b2c9b 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/use-proptypes-to-define-the-props-you-expect.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/use-proptypes-to-define-the-props-you-expect.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403616d
-title: 使用 PropTypes 来定义你期望的 Props
+title: Use PropTypes to Define the Props You Expect
challengeType: 6
forumTopicId: 301419
dashedName: use-proptypes-to-define-the-props-you-expect
@@ -8,23 +8,23 @@ dashedName: use-proptypes-to-define-the-props-you-expect
# --description--
-React 提供了有用的类型检查特性，以验证组件是否接收了正确类型的 props。例如，你的应用程序调用 API 来检索你希望在数组中的数据，然后将数据作为 prop 传递给组件。你可以在组件上设置`propTypes`，以要求数据的类型为`array`。当数据是任何其他类型时，都会抛出警告。
+React provides useful type-checking features to verify that components receive props of the correct type. For example, your application makes an API call to retrieve data that you expect to be in an array, which is then passed to a component as a prop. You can set `propTypes` on your component to require the data to be of type `array`. This will throw a useful warning when the data is of any other type.
-当你提前知道 prop 的类型时，最好的做法是设置`propTypes`。可以为组件定义`propTypes`属性，方法与定义`defaultProps`相同。这样做将检查给定键的 prop 是否是给定类型。这里有一个示例，名为`handleClick`的 prop 应为`function`类型：
+It's considered a best practice to set `propTypes` when you know the type of a prop ahead of time. You can define a `propTypes` property for a component in the same way you defined `defaultProps`. Doing this will check that props of a given key are present with a given type. Here's an example to require the type `function` for a prop called `handleClick`:
`MyComponent.propTypes = { handleClick: PropTypes.func.isRequired }`
-在上面的示例中，`PropTypes.func`部分检查`handleClick`是否为函数。添加`isRequired`是为了告诉 React`handleClick`是该组件的必需属性。如果未提供该 prop，你将看到警告信息。另请注意，`func`表示`function`。在 7 种 JavaScript 基本类型中，`function`和`boolean`（写为`bool`）是仅有的使用异常拼写的两种类型。除了基本类型，还有其他类型可用。例如，你可以检查 prop 是否为 React 组件，请参阅文档以获取所有选项。
+In the example above, the `PropTypes.func` part checks that `handleClick` is a function. Adding `isRequired` tells React that `handleClick` is a required property for that component. You will see a warning if that prop isn't provided. Also notice that `func` represents `function`. Among the seven JavaScript primitive types, `function` and `boolean` (written as `bool`) are the only two that use unusual spelling. In addition to the primitive types, there are other types available. For example, you can check that a prop is a React element. Please refer to the [documentation](https://reactjs.org/docs/jsx-in-depth.html#specifying-the-react-element-type) for all of the options.
-**注意：** 在 React v15.5.0 版本中, `PropTypes`可以从 React 中单独引入，如下所示： `import React, { PropTypes } from 'react';`
+**Note:** As of React v15.5.0, `PropTypes` is imported independently from React, like this: `import PropTypes from 'prop-types';`
# --instructions--
-为`Items`组件定义`propTypes`，要求`quantity`作为 prop，并验证它是`number`类型。
+Define `propTypes` for the `Items` component to require `quantity` as a prop and verify that it is of type `number`.
# --hints--
-应该渲染`ShoppingCart`组件。
+The `ShoppingCart` component should render.
```js
assert(
@@ -35,7 +35,7 @@ assert(
);
```
-应该渲染`Items`组件。
+The `Items` component should render.
```js
assert(
@@ -46,7 +46,7 @@ assert(
);
```
-`Items`组件应该包含一个`propTypes`，以检查`quantity`是`number`类型。
+The `Items` component should include a `propTypes` check to require a value for `quantity` and ensure that its value is a number.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/use-react-to-render-nested-components.md b/curriculum/challenges/chinese/03-front-end-libraries/react/use-react-to-render-nested-components.md
index efc9dbad31a..3f98507054f 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/use-react-to-render-nested-components.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/use-react-to-render-nested-components.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036165
-title: 使用 React 渲染嵌套组件
+title: Use React to Render Nested Components
challengeType: 6
forumTopicId: 301420
dashedName: use-react-to-render-nested-components
@@ -8,23 +8,23 @@ dashedName: use-react-to-render-nested-components
# --description--
-上一个挑战显示了组合两个组件的简单方法，但是有许多不同的方法可以把 React 组件组合在一起。
+The last challenge showed a simple way to compose two components, but there are many different ways you can compose components with React.
-组件组合是 React 的强大功能之一。当你使用 React 时，应当先用组件的思路考虑清楚用户界面的结构（如上一个挑战中的 App 示例）。可以将 UI 分解为基本的构建块，这些构建块就是组件。这样做有助于将负责 UI 的代码与负责处理应用程序逻辑的代码分开，并可以大大简化复杂项目的开发和维护。
+Component composition is one of React's powerful features. When you work with React, it is important to start thinking about your user interface in terms of components like the App example in the last challenge. You break down your UI into its basic building blocks, and those pieces become the components. This helps to separate the code responsible for the UI from the code responsible for handling your application logic. It can greatly simplify the development and maintenance of complex projects.
# --instructions--
-代码编辑器中定义了两个功能组件，分别是`TypesOfFruit`和`Fruits`。请把`TypesOfFruit`组件放到`Fruits`组件中，然后把`Fruits`组件放到`TypesOfFood`组件中。结果应该是子组件嵌套在父组件中，父组件嵌套在它本身的父组件中！
+There are two functional components defined in the code editor, called `TypesOfFruit` and `Fruits`. Take the `TypesOfFruit` component and compose it, or *nest* it, within the `Fruits` component. Then take the `Fruits` component and nest it within the `TypesOfFood` component. The result should be a child component, nested within a parent component, which is nested within a parent component of its own!
# --hints--
-`TypesOfFood`组件应该返回单个`div`元素。
+The `TypesOfFood` component should return a single `div` element.
```js
assert(Enzyme.shallow(React.createElement(TypesOfFood)).type() === 'div');
```
-`TypesOfFood`组件应该返回`Fruits`组件。
+The `TypesOfFood` component should return the `Fruits` component.
```js
assert(
@@ -33,7 +33,7 @@ assert(
);
```
-`Fruits`组件应该返回`TypesOfFruit`组件。
+The `Fruits` component should return the `TypesOfFruit` component.
```js
assert(
@@ -42,7 +42,7 @@ assert(
);
```
-`TypesOfFruit`组件应该返回`h2`和`ul`元素。
+The `TypesOfFruit` component should return the `h2` and `ul` elements.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/use-state-to-toggle-an-element.md b/curriculum/challenges/chinese/03-front-end-libraries/react/use-state-to-toggle-an-element.md
index 360eeeb73a5..534aa278e4e 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/use-state-to-toggle-an-element.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/use-state-to-toggle-an-element.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036176
-title: 使用 State 切换元素
+title: Use State to Toggle an Element
challengeType: 6
forumTopicId: 301421
dashedName: use-state-to-toggle-an-element
@@ -8,7 +8,7 @@ dashedName: use-state-to-toggle-an-element
# --description--
-有时可能在更新状态的时候想知道上一个状态是什么。但是状态更新是异步的，这意味着 React 可能会把多个 `setState()` 集中在一起批量更新。所以设置 `this.state` 或者 `this.props` 后值后可能没有立即更新。所以最好不要写如下的代码：
+Sometimes you might need to know the previous state when updating the state. However, state updates may be asynchronous - this means React may batch multiple `setState()` calls into a single update. This means you can't rely on the previous value of `this.state` or `this.props` when calculating the next value. So, you should not use code like this:
```jsx
this.setState({
@@ -16,7 +16,7 @@ this.setState({
});
```
-正确的做法是，给 `setState` 传入一个函数，这个函数可以访问 state 和 props。给 `setState` 传入函数可以返回赋值后的 state 和 props。代码可以重写为这样：
+Instead, you should pass `setState` a function that allows you to access state and props. Using a function with `setState` guarantees you are working with the most current values of state and props. This means that the above should be rewritten as:
```jsx
this.setState((state, props) => ({
@@ -24,7 +24,7 @@ this.setState((state, props) => ({
}));
```
-如果只需要 `state`，那么用下面的格式也是可以的：
+You can also use a form without `props` if you need only the `state`:
```jsx
this.setState(state => ({
@@ -32,21 +32,21 @@ this.setState(state => ({
}));
```
-注意一定要把 object 放在括号里，否则 JavaScript 会认为这只是代码片段。
+Note that you have to wrap the object literal in parentheses, otherwise JavaScript thinks it's a block of code.
# --instructions--
-`MyComponent`有一个初始值为`false`的`visibility`属性。如果`visibility`的值为 true，render 方法返回一个视图，如果为 false，返回另一个视图。
+`MyComponent` has a `visibility` property which is initialized to `false`. The render method returns one view if the value of `visibility` is true, and a different view if it is false.
-目前，无法更新组件的`state`中的`visibility`属性，该值应在 true 和 false 之间来回切换。按钮上有一个单击处理程序，它触发一个名为`toggleVisibility()`的类方法。定义此方法，以便`visibility`的`state`在调用方法时切换到相反的值。如果`visibility`是`false`，则该方法将其设置为`true`，反之亦然。
+Currently, there is no way of updating the `visibility` property in the component's `state`. The value should toggle back and forth between true and false. There is a click handler on the button which triggers a class method called `toggleVisibility()`. Pass a function to `setState` to define this method so that the `state` of `visibility` toggles to the opposite value when the method is called. If `visibility` is `false`, the method sets it to `true`, and vice versa.
-最后，单击按钮以查看基于其`state`的组件的条件渲染。
+Finally, click the button to see the conditional rendering of the component based on its `state`.
-**提示：** 不要忘记将`this`关键字绑定到构造函数中的方法上！
+**Hint:** Don't forget to bind the `this` keyword to the method in the constructor!
# --hints--
-`MyComponent`应该返回一个`div`元素，其中包含一个`button`元素。
+`MyComponent` should return a `div` element which contains a `button`.
```js
assert.strictEqual(
@@ -56,7 +56,7 @@ assert.strictEqual(
);
```
-`MyComponent`应该使用设置为`false`的`visibility`属性来初始化其 state。
+The state of `MyComponent` should initialize with a `visibility` property set to `false`.
```js
assert.strictEqual(
@@ -65,37 +65,34 @@ assert.strictEqual(
);
```
-单击按钮元素应在`true`和`false`之间切换`visibility`属性的状态。
+Clicking the button element should toggle the `visibility` property in state between `true` and `false`.
```js
-async () => {
- const waitForIt = (fn) =>
- new Promise((resolve, reject) => setTimeout(() => resolve(fn()), 250));
+(() => {
const mockedComponent = Enzyme.mount(React.createElement(MyComponent));
const first = () => {
mockedComponent.setState({ visibility: false });
- return waitForIt(() => mockedComponent.state('visibility'));
+ return mockedComponent.state('visibility');
};
const second = () => {
mockedComponent.find('button').simulate('click');
- return waitForIt(() => mockedComponent.state('visibility'));
+ return mockedComponent.state('visibility');
};
const third = () => {
mockedComponent.find('button').simulate('click');
- return waitForIt(() => mockedComponent.state('visibility'));
+ return mockedComponent.state('visibility');
};
- const firstValue = await first();
- const secondValue = await second();
- const thirdValue = await third();
+ const firstValue = first();
+ const secondValue = second();
+ const thirdValue = third();
assert(!firstValue && secondValue && !thirdValue);
-};
+})();
```
-应该传入`setState` 一个匿名函数。
+An anonymous function should be passed to `setState`.
```js
const paramRegex = '[a-zA-Z$_]\\w*(,[a-zA-Z$_]\\w*)?';
-const noSpaces = code.replace(/\s/g, '');
assert(
new RegExp(
'this\\.setState\\((function\\(' +
@@ -103,11 +100,11 @@ assert(
'\\){|([a-zA-Z$_]\\w*|\\(' +
paramRegex +
'\\))=>)'
- ).test(noSpaces)
+ ).test(__helpers.removeWhiteSpace(code))
);
```
-不要在 `setState` 里面使用 `this`。
+`this` should not be used inside `setState`
```js
assert(!/this\.setState\([^}]*this/.test(code));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/use-the-lifecycle-method-componentdidmount.md b/curriculum/challenges/chinese/03-front-end-libraries/react/use-the-lifecycle-method-componentdidmount.md
index 3c1536f7c15..f20b192ce80 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/use-the-lifecycle-method-componentdidmount.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/use-the-lifecycle-method-componentdidmount.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403617d
-title: 使用生命周期方法：componentDidMount
+title: Use the Lifecycle Method componentDidMount
challengeType: 6
forumTopicId: 301422
dashedName: use-the-lifecycle-method-componentdidmount
@@ -8,17 +8,17 @@ dashedName: use-the-lifecycle-method-componentdidmount
# --description--
-某些时候，大多数 web 开发人员需要调用 API 端点来检索数据。如果你正在使用 React，知道在哪里执行这个动作是很重要的。
+Most web developers, at some point, need to call an API endpoint to retrieve data. If you're working with React, it's important to know where to perform this action.
-React 的最佳实践是在生命周期方法`componentDidMount()`中对服务器进行 API 调用或任何其他调用。将组件装载到 DOM 后会调用此方法。此处对`setState()`的任何调用都将触发组件的重新渲染。在此方法中调用 API 并使用 API 返回的数据设置 state 时，一旦收到数据，它将自动触发更新。
+The best practice with React is to place API calls or any calls to your server in the lifecycle method `componentDidMount()`. This method is called after a component is mounted to the DOM. Any calls to `setState()` here will trigger a re-rendering of your component. When you call an API in this method, and set your state with the data that the API returns, it will automatically trigger an update once you receive the data.
# --instructions--
-`componentDidMount()` 中有一个模拟 API 调用。它在 2.5 秒后设置 state，以模拟调用服务器检索数据。本示例请求站点的当前活动用户总数。在 render 方法中，把 `activeUsers` 渲染到文字 `Active Users:` 后的 `h1` 标签中。观看预览中发生的事情，随意更改超时时间以查看不同的效果。
+There is a mock API call in `componentDidMount()`. It sets state after 2.5 seconds to simulate calling a server to retrieve data. This example requests the current total active users for a site. In the render method, render the value of `activeUsers` in the `h1` after the text `Active Users:`. Watch what happens in the preview, and feel free to change the timeout to see the different effects.
# --hints--
-`MyComponent`应该渲染一个包含`h1`标签的`div`元素。
+`MyComponent` should render a `div` element which wraps an `h1` tag.
```js
assert(
@@ -32,7 +32,7 @@ assert(
);
```
-组件 state 应该用`componentDidMount`中的延时函数来更新。
+Component state should be updated with a timeout function in `componentDidMount`.
```js
assert(
@@ -45,10 +45,10 @@ assert(
);
```
-`h1`标签应该从`MyComponent`的 state 渲染`activeUsers`值。
+The `h1` tag should render the `activeUsers` value from `MyComponent`'s state.
```js
-async () => {
+(() => {
const mockedComponent = Enzyme.mount(React.createElement(MyComponent));
const first = () => {
mockedComponent.setState({ activeUsers: 1237 });
@@ -59,7 +59,7 @@ async () => {
return mockedComponent.find('h1').text();
};
assert(new RegExp('1237').test(first()) && new RegExp('1000').test(second()));
-};
+})();
```
# --seed--
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/use-the-lifecycle-method-componentwillmount.md b/curriculum/challenges/chinese/03-front-end-libraries/react/use-the-lifecycle-method-componentwillmount.md
index 66cb18b2b1a..6095a48eed3 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/use-the-lifecycle-method-componentwillmount.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/use-the-lifecycle-method-componentwillmount.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403617c
-title: 使用生命周期方法：componentWillMount
+title: Use the Lifecycle Method componentWillMount
challengeType: 6
forumTopicId: 301423
dashedName: use-the-lifecycle-method-componentwillmount
@@ -8,17 +8,17 @@ dashedName: use-the-lifecycle-method-componentwillmount
# --description--
-React 组件有几种特殊方法，可以在组件生命周期的特定点执行操作。这些称为生命周期方法或生命周期钩子，允许你在特定时间点捕获组件。这可以在渲染之前、更新之前、接收 props 之前、卸载之前等等。以下是一些主要生命周期方法的列表： `componentWillMount()` `componentDidMount()` `shouldComponentUpdate()` `componentDidUpdate()` `componentWillUnmount()` 接下来的几节课将讲述这些生命周期方法的一些基本用例。
+React components have several special methods that provide opportunities to perform actions at specific points in the lifecycle of a component. These are called lifecycle methods, or lifecycle hooks, and allow you to catch components at certain points in time. This can be before they are rendered, before they update, before they receive props, before they unmount, and so on. Here is a list of some of the main lifecycle methods: `componentWillMount()` `componentDidMount()` `shouldComponentUpdate()` `componentDidUpdate()` `componentWillUnmount()` The next several lessons will cover some of the basic use cases for these lifecycle methods.
-**注意：** `componentWillMount` 生命周期方法会在版本 16.X 废弃在版本 17 移除 [(Source)](https://reactjs.org/blog/2018/03/27/update-on-async-rendering.html)
+**Note:** The `componentWillMount` Lifecycle method will be deprecated in a future version of 16.X and removed in version 17. [(Source)](https://reactjs.org/blog/2018/03/27/update-on-async-rendering.html)
# --instructions--
-当组件被挂载到 DOM 时，`componentWillMount()`方法在`render()`方法之前被调用。在`componentWillMount()`中将一些内容记录到控制台--你需要打开浏览器控制台以查看输出。
+The `componentWillMount()` method is called before the `render()` method when a component is being mounted to the DOM. Log something to the console within `componentWillMount()` - you may want to have your browser console open to see the output.
# --hints--
-`MyComponent`应该渲染一个`div`元素。
+`MyComponent` should render a `div` element.
```js
assert(
@@ -29,7 +29,7 @@ assert(
);
```
-应该在`componentWillMount`中调用`console.log`。
+`console.log` should be called in `componentWillMount`.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/write-a-react-component-from-scratch.md b/curriculum/challenges/chinese/03-front-end-libraries/react/write-a-react-component-from-scratch.md
index ee0cea9ebfc..83cf8603ab8 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/write-a-react-component-from-scratch.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/write-a-react-component-from-scratch.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036168
-title: 从零开始写一个 React 组件
+title: Write a React Component from Scratch
challengeType: 6
forumTopicId: 301424
dashedName: write-a-react-component-from-scratch
@@ -8,28 +8,28 @@ dashedName: write-a-react-component-from-scratch
# --description--
-现在你已经了解了 JSX 和 React 组件的基础知识，现在是时候自己编写一个组件了。React 组件是 React 应用程序的核心组成部分，因此熟练编写它们是非常重要的。记住，典型的 React 组件是 ES6`class`，它扩展了`React.Component`。它有一个返回 HTML（从 JSX 返回）或`null`的渲染方法，这是 React 组件的基本形式。一旦你深刻地理解了这一点，你就可以开始构建更复杂的 React 项目了。
+Now that you've learned the basics of JSX and React components, it's time to write a component on your own. React components are the core building blocks of React applications so it's important to become very familiar with writing them. Remember, a typical React component is an ES6 `class` which extends `React.Component`. It has a render method that returns HTML (from JSX) or `null`. This is the basic form of a React component. Once you understand this well, you will be prepared to start building more complex React projects.
# --instructions--
-定义一个`MyComponent`类，它是`React.Component`的扩展。它的渲染方法应该返回一个`div`，其中包含一个`h1`标签，标签文本为：`My First React Component!`。请确保文本内容、大小写和标点符号正确，以及调用了组件的构造函数。
+Define a class `MyComponent` that extends `React.Component`. Its render method should return a `div` that contains an `h1` tag with the text: `My First React Component!` in it. Use this text exactly, the case and punctuation matter. Make sure to call the constructor for your component, too.
-使用`ReactDOM.render()`把该组件渲染到 DOM 中。有一个`id='challenge-node'`的`div`可供你使用。
+Render this component to the DOM using `ReactDOM.render()`. There is a `div` with `id='challenge-node'` available for you to use.
# --hints--
-应该有一个名为`MyComponent`的React组件。
+There should be a React component called `MyComponent`.
```js
(getUserInput) =>
assert(
- getUserInput('index')
- .replace(/\s/g, '')
+ __helpers
+ .removeWhiteSpace(getUserInput('index'))
.includes('classMyComponentextendsReact.Component{')
);
```
-`MyComponent`应该包含一个`h1`标签，标签的文本为`My First React Component!`，区分大小写并有标点符号。
+`MyComponent` should contain an `h1` tag with text `My First React Component!` Case and punctuation matter.
```js
assert(
@@ -40,13 +40,13 @@ assert(
);
```
-`MyComponent`应该渲染到 DOM。
+`MyComponent` should render to the DOM.
```js
assert(document.getElementById('challenge-node').childNodes.length === 1);
```
-`MyComponent` 应该有一个构造器，里面调用了传参 `props` 的 `super` 函数。
+`MyComponent` should have a constructor calling `super` with `props`.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/react/write-a-simple-counter.md b/curriculum/challenges/chinese/03-front-end-libraries/react/write-a-simple-counter.md
index 6dec25d1450..cfc063e0efc 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/react/write-a-simple-counter.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/react/write-a-simple-counter.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036177
-title: 写一个简单的计数器
+title: Write a Simple Counter
challengeType: 6
forumTopicId: 301425
dashedName: write-a-simple-counter
@@ -8,17 +8,17 @@ dashedName: write-a-simple-counter
# --description--
-你可以结合目前为止所涵盖的概念来设计更复杂的有状态组件。这包括初始化`state`，编写设置`state`的方法，以及指定单击处理程序来触发这些方法。
+You can design a more complex stateful component by combining the concepts covered so far. These include initializing `state`, writing methods that set `state`, and assigning click handlers to trigger these methods.
# --instructions--
-`Counter`组件跟踪`state`中的`count`值。有两个按钮分别调用`increment()`和`decrement()`方法。编写这些方法，使计数器值在单击相应按钮时增加或减少 1。另外，创建一个`reset()`方法，当单击 reset 按钮时，把计数设置为 0。
+The `Counter` component keeps track of a `count` value in `state`. There are two buttons which call methods `increment()` and `decrement()`. Write these methods so the counter value is incremented or decremented by 1 when the appropriate button is clicked. Also, create a `reset()` method so when the reset button is clicked, the count is set to 0.
-**注意：** 确保你没有修改按钮的`classNames`。另外，请记住在构造函数中为新创建的方法添加必要的绑定。
+**Note:** Make sure you don't modify the `classNames` of the buttons. Also, remember to add the necessary bindings for the newly-created methods in the constructor.
# --hints--
-`Counter`应该返回一个`div`元素，它包含三个按钮，按钮内容依次是`Increment!`、`Decrement!`、`Reset`。
+`Counter` should return a `div` element which contains three buttons with text content in this order `Increment!`, `Decrement!`, `Reset`.
```js
assert(
@@ -33,14 +33,14 @@ assert(
);
```
-`Counter`应该使用设置为`0`的`count`属性初始化 state。
+The state of `Counter` should initialize with a `count` property set to `0`.
```js
const mockedComponent = Enzyme.mount(React.createElement(Counter));
assert(mockedComponent.find('h1').text() === 'Current Count: 0');
```
-单击 increment 按钮应将计数增加`1`。
+Clicking the increment button should increment the count by `1`.
```js
const mockedComponent = Enzyme.mount(React.createElement(Counter));
@@ -48,7 +48,7 @@ mockedComponent.find('.inc').simulate('click');
assert(mockedComponent.find('h1').text() === 'Current Count: 1');
```
-单击 decrement 按钮应将计数减少`1`。
+Clicking the decrement button should decrement the count by `1`.
```js
const mockedComponent = Enzyme.mount(React.createElement(Counter));
@@ -56,7 +56,7 @@ mockedComponent.find('.dec').simulate('click');
assert(mockedComponent.find('h1').text() === 'Current Count: -1');
```
-单击 reset 按钮应将计数重置为`0`。
+Clicking the reset button should reset the count to `0`.
```js
const mockedComponent = Enzyme.mount(React.createElement(Counter));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/redux/combine-multiple-reducers.md b/curriculum/challenges/chinese/03-front-end-libraries/redux/combine-multiple-reducers.md
index 82ec12af80a..741bfc9c6af 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/redux/combine-multiple-reducers.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/redux/combine-multiple-reducers.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036154
-title: 组合多个 Reduces
+title: Combine Multiple Reducers
challengeType: 6
forumTopicId: 301436
dashedName: combine-multiple-reducers
@@ -8,11 +8,11 @@ dashedName: combine-multiple-reducers
# --description--
-当你应用程序的状态开始变得越来越复杂时，将状态划分为多个部分可能是个更好的选择。相反，请记住 Redux 的第一个原则：所有应用程序状态都保存在 store 中的一个简单的 state 对象中。因此，Redux 提供 reducer 组合作为复杂状态模型的解决方案。定义多个 reducer 来处理应用程序状态的不同部分，然后将这些 reducer 组合成一个 root reducer。然后将 root reducer 传递给 Redux `createStore()`方法。
+When the state of your app begins to grow more complex, it may be tempting to divide state into multiple pieces. Instead, remember the first principle of Redux: all app state is held in a single state object in the store. Therefore, Redux provides reducer composition as a solution for a complex state model. You define multiple reducers to handle different pieces of your application's state, then compose these reducers together into one root reducer. The root reducer is then passed into the Redux `createStore()` method.
-为了让我们将可以将多个 reducer 组合在一起，Redux 提供了`combineReducers()`方法。该方法接受一个对象作为参数，在该参数中定义一个将键与特定 reducer 函数关联的属性。Redux 将使用你给的键值作为关联状态的名称。
+In order to let us combine multiple reducers together, Redux provides the `combineReducers()` method. This method accepts an object as an argument in which you define properties which associate keys to specific reducer functions. The name you give to the keys will be used by Redux as the name for the associated piece of state.
-通常情况下，当它们在某种程度上是独一无二的，为每个应用程序的 state 创建一个减少器是一个很好的做法。例如，在一个带有用户身份验证的记笔记应用程序中，一个 reducer 可以处理身份验证而另一个处理用户提交的文本和注释。对于这样的应用程序，我们可能会编写`combineReducers()`方法，如下所示：
+Typically, it is a good practice to create a reducer for each piece of application state when they are distinct or unique in some way. For example, in a note-taking app with user authentication, one reducer could handle authentication while another handles the text and notes that the user is submitting. For such an application, we might write the `combineReducers()` method like this:
```js
const rootReducer = Redux.combineReducers({
@@ -21,15 +21,15 @@ const rootReducer = Redux.combineReducers({
});
```
-现在，`notes`键将包含与我们的注释相关联的所有状态，并由我们的`notesReducer`处理。这就是如何组合多个 reducer 来管理更复杂的应用程序状态，在此示例中，Redux store 中保存的状态将是一个包含`auth`和`notes`属性的简单对象。
+Now, the key `notes` will contain all of the state associated with our notes and handled by our `notesReducer`. This is how multiple reducers can be composed to manage more complex application state. In this example, the state held in the Redux store would then be a single object containing `auth` and `notes` properties.
# --instructions--
-代码编辑器中提供了`counterReducer()`和`authReducer()`函数以及 Redux store。使用`Redux.combineReducers()`方法编写完成`rootReducer()`函数。将`counterReducer`分配给一个叫做`count`的键，将`authReducer`分配给一个叫做`auth`的键。
+There are `counterReducer()` and `authReducer()` functions provided in the code editor, along with a Redux store. Finish writing the `rootReducer()` function using the `Redux.combineReducers()` method. Assign `counterReducer` to a key called `count` and `authReducer` to a key called `auth`.
# --hints--
-`counterReducer`应该递增和递减`state`。
+The `counterReducer` should increment and decrement the `state`.
```js
assert(
@@ -45,7 +45,7 @@ assert(
);
```
-`authenticated`的`state`值应该在`true`和`false`之间切换。
+The `authReducer` should toggle the `state` of `authenticated` between `true` and `false`.
```js
assert(
@@ -59,7 +59,7 @@ assert(
);
```
-store `state`应该有两个 key：一个是`count`，它包含一个数字。另一个`auth`，它包含一个对象。`auth`对象应该具有`authenticated`的属性，该属性的值应该为布尔值。
+The store `state` should have two keys: `count`, which holds a number, and `auth`, which holds an object. The `auth` object should have a property of `authenticated`, which holds a boolean.
```js
assert(
@@ -74,13 +74,13 @@ assert(
);
```
-`rootReducer`应该是一个合并了`counterReducer`和`authReducer`的函数。
+The `rootReducer` should be a function that combines the `counterReducer` and the `authReducer`.
```js
(getUserInput) =>
assert(
(function () {
- const noWhiteSpace = getUserInput('index').replace(/\s/g, '');
+ const noWhiteSpace = __helpers.removeWhiteSpace(getUserInput('index'));
return (
typeof rootReducer === 'function' &&
noWhiteSpace.includes('Redux.combineReducers')
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/redux/copy-an-object-with-object.assign.md b/curriculum/challenges/chinese/03-front-end-libraries/redux/copy-an-object-with-object.assign.md
index b361aa25873..95f447d9089 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/redux/copy-an-object-with-object.assign.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/redux/copy-an-object-with-object.assign.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403615b
-title: 使用 Object.assign 拷贝对象
+title: Copy an Object with Object.assign
challengeType: 6
forumTopicId: 301437
dashedName: copy-an-object-with-object-assign
@@ -8,19 +8,19 @@ dashedName: copy-an-object-with-object-assign
# --description--
-最后几个挑战适用于数组，但是当状态是`object`时，有一些方法可以帮助强制执行状态不变性。处理对象的一个方法是`Object.assign()`。`Object.assign()`获取目标对象和源对象，并将源对象中的属性映射到目标对象。任何匹配的属性都会被源对象中的属性覆盖。通常用于通过传递一个空对象作为第一个参数，然后是要用复制的对象来制作对象的浅表副本。这是一个例子：
+The last several challenges worked with arrays, but there are ways to help enforce state immutability when state is an `object`, too. A useful tool for handling objects is the `Object.assign()` utility. `Object.assign()` takes a target object and source objects and maps properties from the source objects to the target object. Any matching properties are overwritten by properties in the source objects. This behavior is commonly used to make shallow copies of objects by passing an empty object as the first argument followed by the object(s) you want to copy. Here's an example:
`const newObject = Object.assign({}, obj1, obj2);`
-这会创建`newObject`作为新的`object`，其中包含`obj1`和`obj2`中当前存在的属性。
+This creates `newObject` as a new `object`, which contains the properties that currently exist in `obj1` and `obj2`.
# --instructions--
-Redux 状态和 action 被修改为处理`state`的`对象`。编辑代码以返回一个新的`state`对象，用于类型的 action`ONLINE`，它将`status`属性设置为字符串`online`。尝试使用`Object.assign()`来完成挑战。
+The Redux state and actions were modified to handle an `object` for the `state`. Edit the code to return a new `state` object for actions with type `ONLINE`, which set the `status` property to the string `online`. Try to use `Object.assign()` to complete the challenge.
# --hints--
-Redux store 应该存在并使用与第 1 行声明的`defaultState`对象相同的状态进行初始化。
+The Redux store should exist and initialize with a state that is equivalent to the `defaultState` object declared on line 1.
```js
assert(
@@ -37,13 +37,13 @@ assert(
);
```
-`wakeUp`和`immutableReducer`都应该是函数。
+`wakeUp` and `immutableReducer` both should be functions.
```js
assert(typeof wakeUp === 'function' && typeof immutableReducer === 'function');
```
-dispatch 一个类型为`ONLINE`的 action 应该将状态`status`更新为`online`，并且不应该改变状态。
+Dispatching an action of type `ONLINE` should update the property `status` in state to `online` and should NOT mutate state.
```js
assert(
@@ -63,7 +63,7 @@ assert(
);
```
-`Object.assign`应该被用于返回一个新状态。
+`Object.assign` should be used to return new state.
```js
(getUserInput) => assert(getUserInput('index').includes('Object.assign'));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/redux/create-a-redux-store.md b/curriculum/challenges/chinese/03-front-end-libraries/redux/create-a-redux-store.md
index 791f93091c3..7881ad686d3 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/redux/create-a-redux-store.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/redux/create-a-redux-store.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403614b
-title: 创建一个 Redux Store
+title: Create a Redux Store
challengeType: 6
forumTopicId: 301439
dashedName: create-a-redux-store
@@ -8,29 +8,29 @@ dashedName: create-a-redux-store
# --description--
-Redux 是一个状态管理框架，可以与包括 React 在内的许多不同的 Web 技术一起使用。
+Redux is a state management framework that can be used with a number of different web technologies, including React.
-在 Redux 中，有一个状态对象负责应用程序的整个状态，这意味着如果你有一个包含十个组件且每个组件都有自己的本地状态的 React 项目，那么这个项目的整个状态将通过 Redux`store`被定义为单个状态对象，这是学习 Redux 时要理解的第一个重要原则：Redux store 是应用程序状态的唯一真实来源。
+In Redux, there is a single state object that's responsible for the entire state of your application. This means if you had a React app with ten components, and each component had its own local state, the entire state of your app would be defined by a single state object housed in the Redux `store`. This is the first important principle to understand when learning Redux: the Redux store is the single source of truth when it comes to application state.
-这也意味着，如果你的应用程序想要更新状态，只能通过 Redux store 执行，单向数据流可以更轻松地对应用程序中的状态进行监测管理。
+This also means that any time any piece of your app wants to update state, it **must** do so through the Redux store. The unidirectional data flow makes it easier to track state management in your app.
# --instructions--
-Redux `store`是一个保存和管理应用程序状态的`state`，你可以使用 Redux 对象中的`createStore()`来创建一个 redux`store`，此方法将`reducer`函数作为必需参数，`reducer`函数将在后面的挑战中介绍。该函数已在代码编辑器中为你定义，它只需将`state`作为参数并返回一个`state`即可。
+The Redux `store` is an object which holds and manages application `state`. There is a method called `createStore()` on the Redux object, which you use to create the Redux `store`. This method takes a `reducer` function as a required argument. The `reducer` function is covered in a later challenge, and is already defined for you in the code editor. It simply takes `state` as an argument and returns `state`.
-声明一个`store`变量并把它分配给`createStore()`方法，然后把`reducer`作为一个参数传入即可。
+Declare a `store` variable and assign it to the `createStore()` method, passing in the `reducer` as an argument.
-注意: 编辑器中的代码使用 ES6 默认参数语法将 state 的值初始化为`5`， 如果你不熟悉默认参数，你可以参考[ES6 全部课程](http://beta.freecodecamp.com/en/challenges/es6/set-default-parameters-for-your-functions)，它里面涵盖了这个内容。
+**Note:** The code in the editor uses ES6 default argument syntax to initialize this state to hold a value of `5`. If you're not familiar with default arguments, you can refer to the [ES6 section in the Curriculum](https://learn.freecodecamp.org/javascript-algorithms-and-data-structures/es6/set-default-parameters-for-your-functions) which covers this topic.
# --hints--
-redux store 应当存在。
+The redux store should exist.
```js
assert(typeof store.getState === 'function');
```
-redux store 的 state 的值为 5。
+The redux store should have a value of 5 for the state.
```js
assert(store.getState() === 5);
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/redux/define-a-redux-action.md b/curriculum/challenges/chinese/03-front-end-libraries/redux/define-a-redux-action.md
index 05418e11d7a..0c33ade1a00 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/redux/define-a-redux-action.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/redux/define-a-redux-action.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403614d
-title: 定义一个 Redux Action
+title: Define a Redux Action
challengeType: 6
forumTopicId: 301440
dashedName: define-a-redux-action
@@ -8,17 +8,17 @@ dashedName: define-a-redux-action
# --description--
-由于 Redux 是一个状态管理框架，因此更新状态是其核心任务之一。在 Redux 中，所有状态更新都由 dispatch action 触发，action 只是一个 JavaScript 对象，其中包含有关已发生的 action 事件的信息。Redux store 接收这些 action 对象，然后更新相应的状态。有时，Redux action 也会携带一些数据。例如，在用户登录后携带用户名，虽然数据是可选的，但 action 必须带有`type`属性，该属性表示此 action 的类型。
+Since Redux is a state management framework, updating state is one of its core tasks. In Redux, all state updates are triggered by dispatching actions. An action is simply a JavaScript object that contains information about an action event that has occurred. The Redux store receives these action objects, then updates its state accordingly. Sometimes a Redux action also carries some data. For example, the action carries a username after a user logs in. While the data is optional, actions must carry a `type` property that specifies the 'type' of action that occurred.
-我们可以将 Redux action 视为信使，将有关应用程序中发生的事件信息提供给 Redux store，然后 store 根据发生的 action 进行状态的更新。
+Think of Redux actions as messengers that deliver information about events happening in your app to the Redux store. The store then conducts the business of updating state based on the action that occurred.
# --instructions--
-编写 Redux action 就像声明具有 type 属性的对象一样简单，声明一个对象`action`并为它设置一个属性`type`，并将他的值设置成字符串`'LOGIN'`。
+Writing a Redux action is as simple as declaring an object with a type property. Declare an object `action` and give it a property `type` set to the string `'LOGIN'`.
# --hints--
-应该有一个 action 对象。
+An action object should exist.
```js
assert(
@@ -28,7 +28,7 @@ assert(
);
```
-该 action 的值应该为`LOGIN`。
+The action should have a key property type with value `LOGIN`.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/redux/define-an-action-creator.md b/curriculum/challenges/chinese/03-front-end-libraries/redux/define-an-action-creator.md
index 87fbbe1c91e..13f0f279ece 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/redux/define-an-action-creator.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/redux/define-an-action-creator.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403614e
-title: 定义一个 Action Creator
+title: Define an Action Creator
challengeType: 6
forumTopicId: 301441
dashedName: define-an-action-creator
@@ -8,27 +8,27 @@ dashedName: define-an-action-creator
# --description--
-创建 action 后要将 action 发送到 Redux store，以便它可以更新其状态。在 Redux 中，你可以定义动作创建器来完成此任务，action creator 只是一个返回动作的 JavaScript 函数，换句话说，action creator 创建表示动作事件的对象。
+After creating an action, the next step is sending the action to the Redux store so it can update its state. In Redux, you define action creators to accomplish this. An action creator is simply a JavaScript function that returns an action. In other words, action creators create objects that represent action events.
# --instructions--
-定义名为`actionCreator()`的函数，该函数在调用时返回`action`对象。
+Define a function named `actionCreator()` that returns the `action` object when called.
# --hints--
-函数`actionCreator`应该存在。
+The function `actionCreator` should exist.
```js
assert(typeof actionCreator === 'function');
```
-运行`actionCreator`函数应返回 action 对象。
+Running the `actionCreator` function should return the action object.
```js
assert(typeof action === 'object');
```
-返回的 action 应具有值为`LOGIN`的键值类型。
+The returned action should have a key property type with value `LOGIN`.
```js
assert(action.type === 'LOGIN');
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/redux/dispatch-an-action-event.md b/curriculum/challenges/chinese/03-front-end-libraries/redux/dispatch-an-action-event.md
index 128b3cb98a9..46bfbbb12ea 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/redux/dispatch-an-action-event.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/redux/dispatch-an-action-event.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403614f
-title: 分发 Action Event
+title: Dispatch an Action Event
challengeType: 6
forumTopicId: 301442
dashedName: dispatch-an-action-event
@@ -8,9 +8,9 @@ dashedName: dispatch-an-action-event
# --description--
-`dispatch`方法用于将 action 分派给 Redux store，调用`store.dispatch()`将从 action creator 返回的值发送回 store。
+`dispatch` method is what you use to dispatch actions to the Redux store. Calling `store.dispatch()` and passing the value returned from an action creator sends an action back to the store.
-action creator 返回一个具有 type 属性的对象，该属性指定已发生的 action，然后，该方法将 action 对象 dispatch 到 Redux store，根据之前的挑战示例，以下内容是等效的，并且都 dispatch 类型为`LOGIN`的 action：
+Recall that action creators return an object with a type property that specifies the action that has occurred. Then the method dispatches an action object to the Redux store. Based on the previous challenge's example, the following lines are equivalent, and both dispatch the action of type `LOGIN`:
```js
store.dispatch(actionCreator());
@@ -19,23 +19,23 @@ store.dispatch({ type: 'LOGIN' });
# --instructions--
-代码编辑器中的 Redux store 具有初始化状态对象{login:'false'}，还有一个名为`loginAction()`的 action creator，它返回类型为`LOGIN`的 action，然后通过调用`dispatch`方法将`LOGIN`的 action dispatch 给 Redux store，并传递`loginAction()`创建的 action。
+The Redux store in the code editor has an initialized state that's an object containing a `login` property currently set to `false`. There's also an action creator called `loginAction()` which returns an action of type `LOGIN`. Dispatch the `LOGIN` action to the Redux store by calling the `dispatch` method, and pass in the action created by `loginAction()`.
# --hints--
-调用函数`loginAction`应该返回一个对象{type:"LOGIN"}。
+Calling the function `loginAction` should return an object with `type` property set to the string `LOGIN`.
```js
assert(loginAction().type === 'LOGIN');
```
-store 应该初始化一个对象 {login:false}。
+The store should be initialized with an object with property `login` set to `false`.
```js
assert(store.getState().login === false);
```
-`store.dispatch()`方法应该被用于 dispatch 一个类型为`LOGIN`的 action。
+The `store.dispatch()` method should be used to dispatch an action of type `LOGIN`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/redux/get-state-from-the-redux-store.md b/curriculum/challenges/chinese/03-front-end-libraries/redux/get-state-from-the-redux-store.md
index 8f9738d4f01..c7a02905de3 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/redux/get-state-from-the-redux-store.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/redux/get-state-from-the-redux-store.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403614c
-title: 从 Redux Store 获取状态
+title: Get State from the Redux Store
challengeType: 6
forumTopicId: 301443
dashedName: get-state-from-the-redux-store
@@ -8,21 +8,21 @@ dashedName: get-state-from-the-redux-store
# --description--
-Redux store 对象提供了几种允许你与之交互的方法，你可以使用`getState()`方法检索 Redux store 对象中保存的当前的`state`。
+The Redux store object provides several methods that allow you to interact with it. For example, you can retrieve the current `state` held in the Redux store object with the `getState()` method.
# --instructions--
-在代码编辑器中可以将上一个挑战中的代码更简洁地重写，在`store`中使用`store.getState()`检索`state`，并将其分配给新变量`currentState`。
+The code from the previous challenge is re-written more concisely in the code editor. Use `store.getState()` to retrieve the `state` from the `store`, and assign this to a new variable `currentState`.
# --hints--
-redux store 的 state 应该有一个初始值 5。
+The redux store should have a value of 5 for the initial state.
```js
assert(store.getState() === 5);
```
-应该存在一个变量`currentState`，并为其分配 Redux store 的当前状态。
+A variable `currentState` should exist and should be assigned the current state of the Redux store.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/redux/handle-an-action-in-the-store.md b/curriculum/challenges/chinese/03-front-end-libraries/redux/handle-an-action-in-the-store.md
index b5a7a4b1e59..d33750b9c35 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/redux/handle-an-action-in-the-store.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/redux/handle-an-action-in-the-store.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036150
-title: 在 Store 里处理 Action
+title: Handle an Action in the Store
challengeType: 6
forumTopicId: 301444
dashedName: handle-an-action-in-the-store
@@ -8,29 +8,29 @@ dashedName: handle-an-action-in-the-store
# --description--
-在一个 action 被创建并 dispatch 之后，Redux store 需要知道如何响应该操作。这就是`reducer`函数存在的意义。Redux 中的 Reducers 负责响应 action 然后进行状态的修改。`reducer`将`state`和`action`作为参数，并且它总是返回一个新的`state`。我们要知道这是 reducer 的**唯一**的作用。它不应有任何其他的作用：比如它不应调用 API 接口，也不应存在任何潜在的副作用。reducer 只是一个接受状态和动作，然后返回新状态的纯函数。
+After an action is created and dispatched, the Redux store needs to know how to respond to that action. This is the job of a `reducer` function. Reducers in Redux are responsible for the state modifications that take place in response to actions. A `reducer` takes `state` and `action` as arguments, and it always returns a new `state`. It is important to see that this is the **only** role of the reducer. It has no side effects — it never calls an API endpoint and it never has any hidden surprises. The reducer is simply a pure function that takes state and action, then returns new state.
-Redux 的另一个关键原则是`state`是只读的。换句话说，`reducer`函数必须**始终**返回一个新的`state`，并且永远不会直接修改状态。Redux 不强制改变状态，但是你需要在你的 reducer 函数的代码中强制执行它，你将在以后的挑战中练习这一点。
+Another key principle in Redux is that `state` is read-only. In other words, the `reducer` function must **always** return a new copy of `state` and never modify state directly. Redux does not enforce state immutability, however, you are responsible for enforcing it in the code of your reducer functions. You'll practice this in later challenges.
# --instructions--
-代码编辑器中具有前面的示例以及一个`reducer`函数。你需要完善`reducer`函数的内容，使得它如果收到类型为`'LOGIN'`的action，它将返回一个将`login`设置为`true`的 state 对象。否则，它就返回当前的`state`。请注意，当前`state`和dispatch的`action`将被传递给reducer，因此你可以使用`action.type`直接获取 action 的类型。
+The code editor has the previous example as well as the start of a `reducer` function for you. Fill in the body of the `reducer` function so that if it receives an action of type `'LOGIN'` it returns a state object with `login` set to `true`. Otherwise, it returns the current `state`. Note that the current `state` and the dispatched `action` are passed to the reducer, so you can access the action's type directly with `action.type`.
# --hints--
-调用`loginAction`函数应该返回一个对象 {type:"LOGIN"}。
+Calling the function `loginAction` should return an object with type property set to the string `LOGIN`.
```js
assert(loginAction().type === 'LOGIN');
```
-store 应该初始化一个对象 {login:false}。
+The store should be initialized with an object with property `login` set to `false`.
```js
assert(store.getState().login === false);
```
-dispatch `loginAction`后应将 store 中 state 的`login`值更新为`true`。
+Dispatching `loginAction` should update the `login` property in the store state to `true`.
```js
assert(
@@ -43,7 +43,7 @@ assert(
);
```
-如果 action 的类型不是`LOGIN`，那么 store 应返回当前的 state。
+If the action is not of type `LOGIN`, the store should return the current state.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/redux/never-mutate-state.md b/curriculum/challenges/chinese/03-front-end-libraries/redux/never-mutate-state.md
index 4b23a9a9856..002a7c4da28 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/redux/never-mutate-state.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/redux/never-mutate-state.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036158
-title: 永不改变状态
+title: Never Mutate State
challengeType: 6
forumTopicId: 301445
dashedName: never-mutate-state
@@ -8,19 +8,19 @@ dashedName: never-mutate-state
# --description--
-这些最后的挑战描述了在 Redux 中强制执行状态不变性关键原则的几种方法。不可变状态意味着你永远不会直接修改状态，而是返回一个新的状态副本。
+These final challenges describe several methods of enforcing the key principle of state immutability in Redux. Immutable state means that you never modify state directly, instead, you return a new copy of state.
-如果你拍摄 Redux 应用程序状态的快照，你会看到类似`state 1`，`state 2`，`state 3`，`state 4`，`...`等等，每个状态可能与最后一个状态相似，但每个状态都是一个独特的数据。事实上，这种不变性是什么提供了你可能听说过的时间旅行调试等功能。
+If you took a snapshot of the state of a Redux app over time, you would see something like `state 1`, `state 2`, `state 3`,`state 4`, `...` and so on where each state may be similar to the last, but each is a distinct piece of data. This immutability, in fact, is what provides such features as time-travel debugging that you may have heard about.
-Redux 并没有积极地在其 store 或者 reducer 中强制执行状态不变性，责任落在程序员身上。幸运的是，JavaScript（尤其是 ES6）提供了一些有用的工具，可以用来强制执行状态的不变性，无论是`string`，`number`，`array`或`object`。请注意，字符串和数字是原始值，并且本质上是不可变的。换句话说，3 总是 3，你不能改变数字 3 的值。然而，`array`或`object`是可变的。实际上，你的状态可能包括`array`或`object`，因为它们在表示许多类型信息的数据结构时非常有用。
+Redux does not actively enforce state immutability in its store or reducers, that responsibility falls on the programmer. Fortunately, JavaScript (especially ES6) provides several useful tools you can use to enforce the immutability of your state, whether it is a `string`, `number`, `array`, or `object`. Note that strings and numbers are primitive values and are immutable by nature. In other words, 3 is always 3. You cannot change the value of the number 3. An `array` or `object`, however, is mutable. In practice, your state will probably consist of an `array` or `object`, as these are useful data structures for representing many types of information.
# --instructions--
-代码编辑器中有一个`store`和`reducer`，用于管理待办事项。完成在 reducer 中编写`ADD_TO_DO`的情况，使用标准 JavaScript 或 ES6 可以通过几种方法来实现这一目标。看看是否可以找到一种方法来返回一个新数组，其中来自`action.todo`的项目添加到数组的末尾。
+There is a `store` and `reducer` in the code editor for managing to-do items. Finish writing the `ADD_TO_DO` case in the reducer to append a new to-do to the state. There are a few ways to accomplish this with standard JavaScript or ES6. See if you can find a way to return a new array with the item from `action.todo` appended to the end.
# --hints--
-Redux store 应该在代码编辑器中存在并使用名字为`todos`的数组进行状态初始化。
+The Redux store should exist and initialize with a state equal to the `todos` array in the code editor.
```js
assert(
@@ -39,13 +39,13 @@ assert(
);
```
-`addToDo`和`immutableReducer`都应该是函数。
+`addToDo` and `immutableReducer` both should be functions.
```js
assert(typeof addToDo === 'function' && typeof immutableReducer === 'function');
```
-在 Redux store 上 dispatch 一个类型为`ADD_TO_DO`的aciton应该添加一个`todo`项，并且不应该改变 state。
+Dispatching an action of type `ADD_TO_DO` on the Redux store should add a `todo` item and should NOT mutate state.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/redux/register-a-store-listener.md b/curriculum/challenges/chinese/03-front-end-libraries/redux/register-a-store-listener.md
index 0986390658d..56d06178d0d 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/redux/register-a-store-listener.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/redux/register-a-store-listener.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036153
-title: 注册 Store 监听器
+title: Register a Store Listener
challengeType: 6
forumTopicId: 301446
dashedName: register-a-store-listener
@@ -8,15 +8,15 @@ dashedName: register-a-store-listener
# --description--
-在 Redux `store`对象上访问数据的另一种方法是`store.subscribe()`。这允许你将监听器函数订阅到 store，只要一个 action 被 dispatch 就会调用它们。这个方法的一个简单用途是为你的 store 订阅一个函数，它只是在每次收到一个 action 并且更新 store 时记录一条消息。
+Another method you have access to on the Redux `store` object is `store.subscribe()`. This allows you to subscribe listener functions to the store, which are called whenever an action is dispatched against the store. One simple use for this method is to subscribe a function to your store that simply logs a message every time an action is received and the store is updated.
# --instructions--
-编写一个回调函数，每次 store 收到一个 action 时，它会递增全局变量`count`，并将此函数传递给`store.subscribe()`方法。你将会看到`store.dispatch()`连续三次被调用，每次都直接传入一个 action 对象。观察 dispatch action 之间的控制台输出，看看是否发生了更新。
+Write a callback function that increments the global variable `count` every time the store receives an action, and pass this function in to the `store.subscribe()` method. You'll see that `store.dispatch()` is called three times in a row, each time directly passing in an action object. Watch the console output between the action dispatches to see the updates take place.
# --hints--
-在 store 上 dispatch `ADD` action 应该使计数器增加`1`。
+Dispatching the `ADD` action on the store should increment the state by `1`.
```js
assert(
@@ -29,13 +29,13 @@ assert(
);
```
-应该有一个监听函数`store.subscribe`订阅 store。
+There should be a listener function subscribed to the store using `store.subscribe`.
```js
(getUserInput) => assert(getUserInput('index').includes('store.subscribe('));
```
-在更新 store 时，`store.subscribe`应该在回调中使全局变量`count`增加。
+The callback to `store.subscribe` should also increment the global `count` variable as the store is updated.
```js
assert(store.getState() === count);
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/redux/remove-an-item-from-an-array.md b/curriculum/challenges/chinese/03-front-end-libraries/redux/remove-an-item-from-an-array.md
index 24e6ec2e519..16bd461669b 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/redux/remove-an-item-from-an-array.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/redux/remove-an-item-from-an-array.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d403615a
-title: 从数组中删除项目
+title: Remove an Item from an Array
challengeType: 6
forumTopicId: 301447
dashedName: remove-an-item-from-an-array
@@ -8,15 +8,15 @@ dashedName: remove-an-item-from-an-array
# --description--
-是时候练习从数组中删除项目了。扩展运算符也可以在这里使用。其他有用的JavaScript方法包括`slice()`和`concat()`。
+Time to practice removing items from an array. The spread operator can be used here as well. Other useful JavaScript methods include `slice()` and `concat()`.
# --instructions--
-reducer 和 action creator 被修改为根据项目的索引从数组中删除一个项目。完成编写 reducer 以便返回一个新的状态数组，并删除特定索引处的项目。
+The reducer and action creator were modified to remove an item from an array based on the index of the item. Finish writing the reducer so a new state array is returned with the item at the specific index removed.
# --hints--
-Redux store 应该存在并初始化一个`[0,1,2,3,4,5]`的状态。
+The Redux store should exist and initialize with a state equal to `[0,1,2,3,4,5]`
```js
assert(
@@ -30,7 +30,7 @@ assert(
);
```
-`removeItem`和`immutableReducer`都应该是一个函数。
+`removeItem` and `immutableReducer` both should be functions.
```js
assert(
@@ -38,7 +38,7 @@ assert(
);
```
-dispatch `removeItem` action creator 应该从 state 中删除项目，不应该改变 state。
+Dispatching the `removeItem` action creator should remove items from the state and should NOT mutate state.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/redux/send-action-data-to-the-store.md b/curriculum/challenges/chinese/03-front-end-libraries/redux/send-action-data-to-the-store.md
index 1bb2aeaacd3..ccfda74b7af 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/redux/send-action-data-to-the-store.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/redux/send-action-data-to-the-store.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036155
-title: 发送 Action Data 给 Store
+title: Send Action Data to the Store
challengeType: 6
forumTopicId: 301448
dashedName: send-action-data-to-the-store
@@ -8,19 +8,19 @@ dashedName: send-action-data-to-the-store
# --description--
-到目前为止，你已经学会了如何将 action dispatch 给 Redux store，但到目前为止，这些 action 并未包含除 `type`之外的任何信息。你还可以发送特定数据和 action 一起。事实上，这是非常常见的，因为 action 通常源于一些用户交互，并且往往会携带一些数据，Redux store 经常需要知道这些数据。
+By now you've learned how to dispatch actions to the Redux store, but so far these actions have not contained any information other than a `type`. You can also send specific data along with your actions. In fact, this is very common because actions usually originate from some user interaction and tend to carry some data with them. The Redux store often needs to know about this data.
# --instructions--
-在代码编辑器中定义了一个基础的`notesReducer()`和`addNoteText()` action creator。完成`addNoteText()`函数的主体，这样它就会返回一个`action`对象。该对象应该包含一个`type`属性，其值为`ADD_NOTE`，还有一个`text`属性通过 action creator 将值设置为`note`。当你调用 action creator 时，你需要传入可以访问该对象的特定注释信息。
+There's a basic `notesReducer()` and an `addNoteText()` action creator defined in the code editor. Finish the body of the `addNoteText()` function so that it returns an `action` object. The object should include a `type` property with a value of `ADD_NOTE`, and also a `text` property set to the `note` data that's passed into the action creator. When you call the action creator, you'll pass in specific note information that you can access for the object.
-接下来，完成在`notesReducer()`中编写的`switch`语句。你需要添加一个处理`addNoteText()`操作的选项。只要存在`ADD_NOTE`类型的 action，就应该触发 case，并且它应该在传入的`action`上返回`text`属性作为新的`state`
+Next, finish writing the `switch` statement in the `notesReducer()`. You need to add a case that handles the `addNoteText()` actions. This case should be triggered whenever there is an action of type `ADD_NOTE` and it should return the `text` property on the incoming `action` as the new `state`.
-这个 action 将在代码底部发送。一旦完成后，运行代码并观察控制台。这就是将特定于 action 的数据发送到 store 并在更新 store `state`时使用它所需的全部内容。
+The action is dispatched at the bottom of the code. Once you're finished, run the code and watch the console. That's all it takes to send action-specific data to the store and use it when you update store `state`.
# --hints--
-action creator `addNoteText`应该返回一个包含`type`和`text`的对象。
+The action creator `addNoteText` should return an object with keys `type` and `text`.
```js
assert(
@@ -31,7 +31,7 @@ assert(
);
```
-dispatch 一个 action creator 是`addNoteText`的action`ADD_NOTE`，应将`state`更新为 action creator 传递的字符串。
+Dispatching an action of type `ADD_NOTE` with the `addNoteText` action creator should update the `state` to the string passed to the action creator.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/redux/use-a-switch-statement-to-handle-multiple-actions.md b/curriculum/challenges/chinese/03-front-end-libraries/redux/use-a-switch-statement-to-handle-multiple-actions.md
index 01a20e2c528..604bc416f70 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/redux/use-a-switch-statement-to-handle-multiple-actions.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/redux/use-a-switch-statement-to-handle-multiple-actions.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036151
-title: 使用 Switch 语句处理多个 Actions
+title: Use a Switch Statement to Handle Multiple Actions
challengeType: 6
forumTopicId: 301449
dashedName: use-a-switch-statement-to-handle-multiple-actions
@@ -8,35 +8,35 @@ dashedName: use-a-switch-statement-to-handle-multiple-actions
# --description--
-你可以定义 Redux store 如何处理多种 action 类型。比如你正在 Redux store 中进行用户身份验证，如果你希望用户在登录和注销时具有状态的响应，你可以使用具有`authenticated`属性的单个的 state 对象。你还需要使用 action creators 创建与用户登录和用户注销相对应的 action，以及 action 对象本身。
+You can tell the Redux store how to handle multiple action types. Say you are managing user authentication in your Redux store. You want to have a state representation for when users are logged in and when they are logged out. You represent this with a single state object with the property `authenticated`. You also need action creators that create actions corresponding to user login and user logout, along with the action objects themselves.
# --instructions--
-代码编辑器为你创建了 store、actions、action creators。通过编写`reducer`函数来处理多个身份验证操作。可以在`reducer`通过使用 JavaScript 的`switch`来响应不同的 action 事件。这是编写 Redux reducer 时的标准模式，switch 语句选择`action.type`中的一个值并返回相应的身份验证状态。
+The code editor has a store, actions, and action creators set up for you. Fill in the `reducer` function to handle multiple authentication actions. Use a JavaScript `switch` statement in the `reducer` to respond to different action events. This is a standard pattern in writing Redux reducers. The switch statement should switch over `action.type` and return the appropriate authentication state.
-**注意：** 此时，不要担心 state 的不变性，因为在这个示例中它很小而且很简单。所以对于每个操作你都可以返回一个新对象，比如`{authenticated:true}`。另外，不要忘记在 switch 语句中写一个`default` case，返回当前的`state`。这是很重要的，因为一旦你的程序有多个 reducer，当一个 action 被 dispatch 时它们都会运行，即使 action 与该 reducer 无关。在这种情况下，你要确保返回当前的`state`
+**Note:** At this point, don't worry about state immutability, since it is small and simple in this example. For each action, you can return a new object — for example, `{authenticated: true}`. Also, don't forget to write a `default` case in your switch statement that returns the current `state`. This is important because once your app has multiple reducers, they are all run any time an action dispatch is made, even when the action isn't related to that reducer. In such a case, you want to make sure that you return the current `state`.
# --hints--
-调用`loginUser`函数应该返回一个 {type:"LOGIN"} 对象。
+Calling the function `loginUser` should return an object with type property set to the string `LOGIN`.
```js
assert(loginUser().type === 'LOGIN');
```
-调用`logoutUser`函数应该返回一个 {type:"LOGOUT"} 对象。
+Calling the function `logoutUser` should return an object with type property set to the string `LOGOUT`.
```js
assert(logoutUser().type === 'LOGOUT');
```
-store 应该设置一个初始化对象 {authenticated:false}。
+The store should be initialized with an object with an `authenticated` property set to `false`.
```js
assert(store.getState().authenticated === false);
```
-dispatch `loginUser`应该将 store 中的 state 的`authenticated`值更新为`true`。
+Dispatching `loginUser` should update the `authenticated` property in the store state to `true`.
```js
assert(
@@ -51,7 +51,7 @@ assert(
);
```
-dispatch `logoutUser`应该将 store 中的 state 的`authenticated`值更新为`false`。
+Dispatching `logoutUser` should update the `authenticated` property in the store state to `false`.
```js
assert(
@@ -67,7 +67,7 @@ assert(
);
```
-`authReducer`函数应该使用`switch`语句处理多个 action 类型。
+The `authReducer` function should handle multiple action types with a `switch` statement.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/redux/use-const-for-action-types.md b/curriculum/challenges/chinese/03-front-end-libraries/redux/use-const-for-action-types.md
index 36f385a16bf..5df85e5c20f 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/redux/use-const-for-action-types.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/redux/use-const-for-action-types.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036152
-title: 使用 const 声明 Action Types
+title: Use const for Action Types
challengeType: 6
forumTopicId: 301450
dashedName: use-const-for-action-types
@@ -8,35 +8,35 @@ dashedName: use-const-for-action-types
# --description--
-在使用 Redux 时的一个常见做法是将操作类型指定为只读，然后在任何使用它们的地方引用这些常量。你可以通过将 action types 使用`const`声明重构你正在使用的代码。
+A common practice when working with Redux is to assign action types as read-only constants, then reference these constants wherever they are used. You can refactor the code you're working with to write the action types as `const` declarations.
# --instructions--
-将`LOGIN`和`LOGOUT`声明为`const`类型的值，并为它们分别分配字符串`'LOGIN'`和`'LOGOUT'`。然后，编辑`authReducer()`和 action creators 来引用这些常量而不是字符串值。
+Declare `LOGIN` and `LOGOUT` as `const` values and assign them to the strings `'LOGIN'` and `'LOGOUT'`, respectively. Then, edit the `authReducer()` and the action creators to reference these constants instead of string values.
-**注意：** 通常以全部大写形式写出常量，这也是 Redux 的标准做法。
+**Note:** It's generally a convention to write constants in all uppercase, and this is standard practice in Redux as well.
# --hints--
-调用`loginUser`函数应该返回一个 {type:"LOGIN"} 对象。
+Calling the function `loginUser` should return an object with `type` property set to the string `LOGIN`.
```js
assert(loginUser().type === 'LOGIN');
```
-调用`logoutUser`函数应该返回一个 {type:\\"LOGOUT\\"} 对象。
+Calling the function `logoutUser` should return an object with `type` property set to the string `LOGOUT`.
```js
assert(logoutUser().type === 'LOGOUT');
```
-store 应该初始化一个对象 {login：false}。
+The store should be initialized with an object with property `login` set to `false`.
```js
assert(store.getState().authenticated === false);
```
-dispatch `loginUser`以后应将 store 中的 state 的`login`值更新为`true`。
+Dispatching `loginUser` should update the `login` property in the store state to `true`.
```js
assert(
@@ -51,7 +51,7 @@ assert(
);
```
-dispatch `logoutUser`应将 store 中的 state 的`login`值更新为`false`。
+Dispatching `logoutUser` should update the `login` property in the store state to `false`.
```js
assert(
@@ -67,7 +67,7 @@ assert(
);
```
-`authReducer`函数应该使用 switch 语句处理多个 action 类型。
+The `authReducer` function should handle multiple action types with a switch statement.
```js
(getUserInput) =>
@@ -83,23 +83,25 @@ assert(
);
```
-应该使用`const LOGIN="LOGIN"`和`const LOGOUT="LOGOUT"`的方式声明`LOGIN`和`LOGOUT`。
+`LOGIN` and `LOGOUT` should be declared as `const` values and should be assigned strings of `LOGIN`and `LOGOUT`.
```js
-const noWhiteSpace = code.replace(/\s/g, '');
+const noWhiteSpace = __helpers.removeWhiteSpace(code);
assert(
/constLOGIN=(['"`])LOGIN\1/.test(noWhiteSpace) &&
/constLOGOUT=(['"`])LOGOUT\1/.test(noWhiteSpace)
);
```
-action creator 和 reducer 中应该引用`LOGIN`和`LOGOUT`常量。
+The action creators and the reducer should reference the `LOGIN` and `LOGOUT` constants.
```js
(getUserInput) =>
assert(
(function () {
- const noWhiteSpace = getUserInput('index').toString().replace(/\s/g, '');
+ const noWhiteSpace = __helpers.removeWhiteSpace(
+ getUserInput('index').toString()
+ );
return (
noWhiteSpace.includes('caseLOGIN:') &&
noWhiteSpace.includes('caseLOGOUT:') &&
@@ -126,13 +128,11 @@ const defaultState = {
const authReducer = (state = defaultState, action) => {
switch (action.type) {
-
- case 'LOGIN':
+ case 'LOGIN': // Change this line
return {
authenticated: true
}
-
- case 'LOGOUT':
+ case 'LOGOUT': // Change this line
return {
authenticated: false
}
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/redux/use-middleware-to-handle-asynchronous-actions.md b/curriculum/challenges/chinese/03-front-end-libraries/redux/use-middleware-to-handle-asynchronous-actions.md
index 7c8b7d60e68..12c66b6acb6 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/redux/use-middleware-to-handle-asynchronous-actions.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/redux/use-middleware-to-handle-asynchronous-actions.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036156
-title: 使用中间件处理异步操作
+title: Use Middleware to Handle Asynchronous Actions
challengeType: 6
forumTopicId: 301451
dashedName: use-middleware-to-handle-asynchronous-actions
@@ -8,39 +8,39 @@ dashedName: use-middleware-to-handle-asynchronous-actions
# --description--
-目前为止的挑战都在避免讨论异步操作，但它们是 Web 开发中不可避免的一部分。在某些时候，你需要在 Redux 应用程序中使用异步请求，那么如何处理这些类型的请求？Redux 中间件专为此目的而设计，称为 Redux Thunk 中间件。这里简要介绍如何在 Redux 中使用它。
+So far these challenges have avoided discussing asynchronous actions, but they are an unavoidable part of web development. At some point you'll need to call asynchronous endpoints in your Redux app, so how do you handle these types of requests? Redux provides middleware designed specifically for this purpose, called Redux Thunk middleware. Here's a brief description how to use this with Redux.
-如果要使用 Redux Thunk 中间件，请将其作为参数传递给`Redux.applyMiddleware()`。然后将此函数作为第二个可选参数提供给`createStore()`函数，看一下编辑器底部的代码，然后，要创建一个异步的 action，你需要在 action creator 中返回一个以`dispatch`为参数的函数。在这个函数中，你可以 dispatch action 并执行异步请求。
+To include Redux Thunk middleware, you pass it as an argument to `Redux.applyMiddleware()`. This statement is then provided as a second optional parameter to the `createStore()` function. Take a look at the code at the bottom of the editor to see this. Then, to create an asynchronous action, you return a function in the action creator that takes `dispatch` as an argument. Within this function, you can dispatch actions and perform asynchronous requests.
-在此示例中，使用`setTimeout()`调用模拟异步请求。通常在执行异步行为之前 dispatch action，以便应用程序状态知道正在请求某些数据（例如，这个状态可以显示加载图标）。然后，一旦收到数据，就会发送另一个 action，这个 action 完成的时间将作为数据的一个有效值。
+In this example, an asynchronous request is simulated with a `setTimeout()` call. It's common to dispatch an action before initiating any asynchronous behavior so that your application state knows that some data is being requested (this state could display a loading icon, for instance). Then, once you receive the data, you dispatch another action which carries the data as a payload along with information that the action is completed.
-请记住，你正在通过将`dispatch`作为参数传递给这个特殊的 action creator。你用于 dispatch 你的 action 时只需将 action 直接传递给 dispatch，中间件就可以处理其余的部分。
+Remember that you're passing `dispatch` as a parameter to this special action creator. This is what you'll use to dispatch your actions, you simply pass the action directly to dispatch and the middleware takes care of the rest.
# --instructions--
-在`handleAsync()`的 action creator 中编写两个 dispatch 事件。在`setTimeout()`（模拟 API 调用）之前 dispatch`requestingData()`。然后在收到（模拟）数据后，dispatch`receivedData()`action，传入这些数据。现在你知道如何处理 Redux 中的异步操作，其他所有操作都继续像以前一样。
+Write both dispatches in the `handleAsync()` action creator. Dispatch `requestingData()` before the `setTimeout()` (the simulated API call). Then, after you receive the (pretend) data, dispatch the `receivedData()` action, passing in this data. Now you know how to handle asynchronous actions in Redux. Everything else continues to behave as before.
# --hints--
-`requesData` action creator 应返回类型和值都为`REQUESTING_DATA`的对象。
+The `requestingData` action creator should return an object of type equal to the value of `REQUESTING_DATA`.
```js
assert(requestingData().type === REQUESTING_DATA);
```
-`receivedData` action creator 应返回类型和值都等于`RECEIVED_DATA`的对象。
+The `receivedData` action creator should return an object of type equal to the value of `RECEIVED_DATA`.
```js
assert(receivedData('data').type === RECEIVED_DATA);
```
-`asyncDataReducer`必须是个函数。
+`asyncDataReducer` should be a function.
```js
assert(typeof asyncDataReducer === 'function');
```
-分发 requestedData 后 action creator 应该将 store 中的`state`的 fetching 的值更新为 `true`。
+Dispatching the requestingData action creator should update the store `state` property of fetching to `true`.
```js
assert(
@@ -53,12 +53,12 @@ assert(
);
```
-如果要 dispatch `handleAsync` 应先 dispatch 数据请求的 action，然后在收到请求结果后再 dispatch 接收的数据 action。
+Dispatching `handleAsync` should dispatch the data request action and then dispatch the received data action after a delay.
```js
assert(
(function () {
- const noWhiteSpace = handleAsync.toString().replace(/\s/g, '');
+ const noWhiteSpace = __helpers.removeWhiteSpace(handleAsync.toString());
return (
noWhiteSpace.includes('dispatch(requestingData())') === true &&
noWhiteSpace.includes('dispatch(receivedData(data))') === true
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/redux/use-the-spread-operator-on-arrays.md b/curriculum/challenges/chinese/03-front-end-libraries/redux/use-the-spread-operator-on-arrays.md
index ad11a289efc..a03547ccb61 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/redux/use-the-spread-operator-on-arrays.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/redux/use-the-spread-operator-on-arrays.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036159
-title: 在数组中使用扩展运算符
+title: Use the Spread Operator on Arrays
challengeType: 6
forumTopicId: 301452
dashedName: use-the-spread-operator-on-arrays
@@ -8,19 +8,19 @@ dashedName: use-the-spread-operator-on-arrays
# --description--
-ES6 中有助于在 Redux 中强制执行状态不变性的一个解决方案是扩展运算符：`...`。扩展运算符具有很多的应用，其中一种非常适合通过一个已有的数组生成一个新数组。这是相对较新的，但常用的语法。例如，如果你有一个数组`myArray`并写：
+One solution from ES6 to help enforce state immutability in Redux is the spread operator: `...`. The spread operator has a variety of applications, one of which is well-suited to the previous challenge of producing a new array from an existing array. This is relatively new, but commonly used syntax. For example, if you have an array `myArray` and write:
`let newArray = [...myArray];`
-`newArray`现在是`myArray`的克隆。两个数组仍然在内存中单独存在。如果你执行像`newArray.push(5)`这样的突变，`myArray`不会改变。`...`有效*将`myArray`中的值*传播到新数组中。要克隆数组但在新数组中添加其他值，可以编写`[... myArray，'new value']`。这将返回一个由`中的值组成的新数组。`myArray和字符串`'new value'`作为最后一个值。扩展语法可以像这样在数组组合中多次使用，但重要的是要注意它只做一个浅拷贝这就是说，它只为一维数组提供了不可变的数组操作。
+`newArray` is now a clone of `myArray`. Both arrays still exist separately in memory. If you perform a mutation like `newArray.push(5)`, `myArray` doesn't change. The `...` effectively *spreads* out the values in `myArray` into a new array. To clone an array but add additional values in the new array, you could write `[...myArray, 'new value']`. This would return a new array composed of the values in `myArray` and the string `'new value'` as the last value. The spread syntax can be used multiple times in array composition like this, but it's important to note that it only makes a shallow copy of the array. That is to say, it only provides immutable array operations for one-dimensional arrays.
# --instructions--
-添加待办事项时，使用 spread 运算符返回新的状态副本。
+Use the spread operator to return a new copy of state when a to-do is added.
# --hints--
-Redux store 应该存在并初始化一个`[Do not mutate state！]`的状态。
+The Redux store should exist and initialize with a state equal to `[Do not mutate state!]`.
```js
assert(
@@ -34,13 +34,13 @@ assert(
);
```
-`addToDo`和`immutableReducer`都应该是一个函数。
+`addToDo` and `immutableReducer` both should be functions.
```js
assert(typeof addToDo === 'function' && typeof immutableReducer === 'function');
```
-在 Redux store 上 dispatch 一个类型为`ADD_TO_DO` aciton 应该添加一个`todo`项，并且不应该改变 state。
+Dispatching an action of type `ADD_TO_DO` on the Redux store should add a `todo` item and should NOT mutate state.
```js
assert(
@@ -55,7 +55,7 @@ assert(
);
```
-应使用扩展运算符返回新的 state。
+The spread operator should be used to return new state.
```js
(getUserInput) => assert(getUserInput('index').includes('...state'));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/redux/write-a-counter-with-redux.md b/curriculum/challenges/chinese/03-front-end-libraries/redux/write-a-counter-with-redux.md
index 97549adba2a..9bf25454c3c 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/redux/write-a-counter-with-redux.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/redux/write-a-counter-with-redux.md
@@ -1,6 +1,6 @@
---
id: 5a24c314108439a4d4036157
-title: 用 Redux 写一个计数器
+title: Write a Counter with Redux
challengeType: 6
forumTopicId: 301453
dashedName: write-a-counter-with-redux
@@ -8,33 +8,33 @@ dashedName: write-a-counter-with-redux
# --description--
-现在你已经了解了 Redux 的所有核心原则！你已经了解了如何创建 action 和 action creator，创建 Redux store，通过 store dispatch action，以及使用纯粹的 reducer 设计状态更新。你甚至已经看到过如何使用 reducer 组合管理复杂状态并处理异步操作。这些例子很简单，但这些概念是 Redux 的核心原则。如果你理解它们，你就可以开始构建自己的 Redux 应用了。接下来的挑战包括关于`state`不变性的一些细节，但是，这里是对你到目前为止学到的所有内容的回顾。
+Now you've learned all the core principles of Redux! You've seen how to create actions and action creators, create a Redux store, dispatch your actions against the store, and design state updates with pure reducers. You've even seen how to manage complex state with reducer composition and handle asynchronous actions. These examples are simplistic, but these concepts are the core principles of Redux. If you understand them well, you're ready to start building your own Redux app. The next challenges cover some of the details regarding `state` immutability, but first, here's a review of everything you've learned so far.
# --instructions--
-在本课程中，你将从头开始使用 Redux 实现一个简单的计数器。基本知识在代码编辑器中提供，但你必须完成详细的内容！使用提供给你的名称并定义`incAction`和`decActio` action creator `counterReducer()`，`INCREMENT`和`DECREMENT` action 类型，最后是 Redux `store`。一旦完成，你应该能够 dispatch `INCREMENT`或`DECREMENT`动作来增加或减少`store`中保存的状态。开始构建你的第一个 Redux 应用程序吧，祝你好运！
+In this lesson, you'll implement a simple counter with Redux from scratch. The basics are provided in the code editor, but you'll have to fill in the details! Use the names that are provided and define `incAction` and `decAction` action creators, the `counterReducer()`, `INCREMENT` and `DECREMENT` action types, and finally the Redux `store`. Once you're finished you should be able to dispatch `INCREMENT` or `DECREMENT` actions to increment or decrement the state held in the `store`. Good luck building your first Redux app!
# --hints--
-action creator `incAction`应返回一个 action 对象 {type:"INCREMENT"}。
+The action creator `incAction` should return an action object with `type` equal to the value of `INCREMENT`
```js
assert(incAction().type === INCREMENT);
```
-action creator `decAction`应返回一个 action 对象 {type:"DECREMENT"}。
+The action creator `decAction` should return an action object with `type` equal to the value of `DECREMENT`
```js
assert(decAction().type === DECREMENT);
```
-Redux store 应该将`state`初始化为 0。
+The Redux store should initialize with a `state` of 0.
```js
assert(store.getState() === 0);
```
-在 Redux store 上 dispatch `incAction`应该将`state`增加 1。
+Dispatching `incAction` on the Redux store should increment the `state` by 1.
```js
assert(
@@ -47,7 +47,7 @@ assert(
);
```
-在 Redux store 上 dispatch `decAction`应该将`state`减少 1。
+Dispatching `decAction` on the Redux store should decrement the `state` by 1.
```js
assert(
@@ -60,7 +60,7 @@ assert(
);
```
-`counterReducer`必须是一个函数。
+`counterReducer` should be a function
```js
assert(typeof counterReducer === 'function');
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/sass/apply-a-style-until-a-condition-is-met-with-while.md b/curriculum/challenges/chinese/03-front-end-libraries/sass/apply-a-style-until-a-condition-is-met-with-while.md
index 5444bb08532..88f02b1441f 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/sass/apply-a-style-until-a-condition-is-met-with-while.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/sass/apply-a-style-until-a-condition-is-met-with-while.md
@@ -1,6 +1,6 @@
---
id: 587d7dbf367417b2b2512bbb
-title: 使用 @while 当条件满足时样式生效
+title: Apply a Style Until a Condition is Met with @while
challengeType: 0
forumTopicId: 301454
dashedName: apply-a-style-until-a-condition-is-met-with-while
@@ -8,9 +8,9 @@ dashedName: apply-a-style-until-a-condition-is-met-with-while
# --description--
-Sass 中的`@while`指令与 JavaScript 中的`while`类似。只要满足条件，它就会一直创建 CSS 代码。
+The `@while` directive is an option with similar functionality to the JavaScript `while` loop. It creates CSS rules until a condition is met.
-`@for`挑战提供了一个创建简单网格系统的示例。这也适用于`@while`。
+The `@for` challenge gave an example to create a simple grid system. This can also work with `@while`.
```scss
$x: 1;
@@ -20,59 +20,59 @@ $x: 1;
}
```
-首先，定义变量`$x`并将其设置为 1。接下来，使用`@while`指令，当`$x`小于 13 时创建网格系统 。 在设置`width`的 CSS 规则后，`$x`增加 1 以避免无限循环。
+First, define a variable `$x` and set it to 1. Next, use the `@while` directive to create the grid system *while* `$x` is less than 13. After setting the CSS rule for `width`, `$x` is incremented by 1 to avoid an infinite loop.
# --instructions--
-使用`@while`创建一系列具有不同`font-sizes`的 class。
+Use `@while` to create a series of classes with different `font-sizes`.
-从`text-1`到`text-10`应该有 10 个不同的 class。然后将`font-size`设置为 15px 乘以当前索引号。注意不要写成死循环！
+There should be 5 different classes from `text-1` to `text-5`. Then set `font-size` to `15px` multiplied by the current index number. Make sure to avoid an infinite loop!
# --hints--
-你的代码应使用`@while`指令。
+Your code should use the `@while` directive.
```js
assert(code.match(/@while /g));
```
-你的代码应将索引变量设置为 1 才能启动。
+Your code should use an index variable which starts at an index of 1.
```js
assert(code.match(/\$.*:\s*?1;/gi));
```
-你的代码应该递增计数器变量。
+Your code should increment the counter variable.
```js
assert(code.match(/\$(.*)\s*?:\s*\$\1\s*\+\s*1\s*;/gi));
```
-`.text-1`class 的`font-size`应为 15px。
+Your `.text-1` class should have a `font-size` of 15px.
```js
assert($('.text-1').css('font-size') == '15px');
```
-`.text-2`class 的`font-size`应为 30px。
+Your `.text-2` class should have a `font-size` of 30px.
```js
assert($('.text-2').css('font-size') == '30px');
```
-`.text-3`class 的`font-size`应为 45px。
+Your `.text-3` class should have a `font-size` of 45px.
```js
assert($('.text-3').css('font-size') == '45px');
```
-`.text-4`class 的`font-size`应为 60px。
+Your `.text-4` class should have a `font-size` of 60px.
```js
assert($('.text-4').css('font-size') == '60px');
```
-`.text-5`class 的`font-size`应为 75px。
+Your `.text-5` class should have a `font-size` of 75px.
```js
assert($('.text-5').css('font-size') == '75px');
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/sass/create-reusable-css-with-mixins.md b/curriculum/challenges/chinese/03-front-end-libraries/sass/create-reusable-css-with-mixins.md
index 38cf20c2a89..d84a69bbc69 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/sass/create-reusable-css-with-mixins.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/sass/create-reusable-css-with-mixins.md
@@ -1,6 +1,6 @@
---
id: 587d7dbd367417b2b2512bb6
-title: 用 Mixins 创建可重用 CSS
+title: Create Reusable CSS with Mixins
challengeType: 0
forumTopicId: 301455
dashedName: create-reusable-css-with-mixins
@@ -8,9 +8,9 @@ dashedName: create-reusable-css-with-mixins
# --description--
-在 Sass 中，`mixin`是一组 CSS 声明，可以在整个样式表中重复使用。
+In Sass, a <dfn>mixin</dfn> is a group of CSS declarations that can be reused throughout the style sheet.
-CSS 的新功能需要一段时间适配后，所有浏览器后才能完全使用。随着浏览器的不断升级，使用这些 CSS 规则时可能需要添加浏览器前缀。比如 "box-shadow":
+Newer CSS features take time before they are fully adopted and ready to use in all browsers. As features are added to browsers, CSS rules using them may need vendor prefixes. Consider "box-shadow":
```scss
div {
@@ -21,7 +21,7 @@ div {
}
```
-对于所有具有`box-shadow`属性的元素重写此规则，或者更改每个值以测试不同的效果，需要花费大量的精力。 `Mixins`就像 CSS 的函数。以下是一个例子：
+It's a lot of typing to re-write this rule for all the elements that have a `box-shadow`, or to change each value to test different effects. Mixins are like functions for CSS. Here is how to write one:
```scss
@mixin box-shadow($x, $y, $blur, $c){
@@ -32,7 +32,7 @@ div {
}
```
-定义以`@mixin`开头，后跟自定义名称。参数（`$x`，`$y`，`$blur`，以及上例中的`$c`是可选的。 现在，只要在需要 `box-shadow` 的地方使用`@include`调用上面定义的`mixin`，就可以自动添加好所有的浏览器前缀：`mixin`使用`@include`指令调用：
+The definition starts with `@mixin` followed by a custom name. The parameters (the `$x`, `$y`, `$blur`, and `$c` in the example above) are optional. Now any time a `box-shadow` rule is needed, only a single line calling the mixin replaces having to type all the vendor prefixes. A mixin is called with the `@include` directive:
```scss
div {
@@ -42,17 +42,17 @@ div {
# --instructions--
-为`border-radius`写一个`mixin`，并给它一个`$radius`参数。它应该使用示例中的所有浏览器前缀，然后使用`border-radius mixin`为`#awesome`元素提供 15px 的边框半径。
+Write a mixin for `border-radius` and give it a `$radius` parameter. It should use all the vendor prefixes from the example. Then use the `border-radius` mixin to give the `#awesome` element a border radius of 15px.
# --hints--
-你应声明名为`border-radius`的`mixin`，其中包含名为`$radius`的参数。
+Your code should declare a mixin named `border-radius` which has a parameter named `$radius`.
```js
assert(code.match(/@mixin\s+?border-radius\s*?\(\s*?\$radius\s*?\)\s*?{/gi));
```
-你应该给`$radius`添加`-webkit-border-radius`浏览器前缀。
+Your code should include the `-webkit-border-radius` vendor prefix that uses the `$radius` parameter.
```js
assert(
@@ -60,7 +60,7 @@ assert(
);
```
-你应该给`$radius`添加`-moz-border-radius`浏览器前缀。
+Your code should include the `-moz-border-radius` vendor prefix that uses the `$radius` parameter.
```js
assert(
@@ -68,13 +68,13 @@ assert(
);
```
-你应该给`$radius`添加`-ms-border-radius`浏览器前缀。
+Your code should include the `-ms-border-radius` vendor prefix that uses the `$radius` parameter.
```js
assert(__helpers.removeWhiteSpace(code).match(/-ms-border-radius:\$radius;/gi));
```
-你应该给`$radius`添加`border-radius`。
+Your code should include the general `border-radius` rule that uses the `$radius` parameter.
```js
assert(
@@ -83,7 +83,7 @@ assert(
);
```
-你应使用`@include`关键字调用`border-radius mixin`，并将其设置为 15px。
+Your code should call the `border-radius mixin` using the `@include` keyword, setting it to 15px.
```js
assert(code.match(/@include\s+?border-radius\(\s*?15px\s*?\)\s*;/gi));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/sass/extend-one-set-of-css-styles-to-another-element.md b/curriculum/challenges/chinese/03-front-end-libraries/sass/extend-one-set-of-css-styles-to-another-element.md
index 8a51f480fd9..c5723d66d13 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/sass/extend-one-set-of-css-styles-to-another-element.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/sass/extend-one-set-of-css-styles-to-another-element.md
@@ -1,6 +1,6 @@
---
id: 587d7fa5367417b2b2512bbd
-title: 将一组CSS样式扩展到另一个元素
+title: Extend One Set of CSS Styles to Another Element
challengeType: 0
forumTopicId: 301456
dashedName: extend-one-set-of-css-styles-to-another-element
@@ -8,9 +8,9 @@ dashedName: extend-one-set-of-css-styles-to-another-element
# --description--
-Sass 有一个名为`extend`的功能，可以很容易地从一个元素中借用 CSS 规则并在另一个元素上重用它们。
+Sass has a feature called `extend` that makes it easy to borrow the CSS rules from one element and build upon them in another.
-例如，下面的 CSS 规则块设置了`.panel`class 的样式。它有`background-color`，`height`和`border`。
+For example, the below block of CSS rules style a `.panel` class. It has a `background-color`, `height` and `border`.
```scss
.panel{
@@ -20,7 +20,7 @@ Sass 有一个名为`extend`的功能，可以很容易地从一个元素中借
}
```
-现在你需要另一个名为`.big-panel`的面板。它具有与`.panel`相同的基本属性，但还需要`width`和`font-size`。 可以从`.panel`复制并粘贴初始 CSS 规则，但是当你添加更多类型的面板时，代码会变得重复。 `extend`指令是一种重用为一个元素编写的规则的简单方法，然后为另一个元素添加更多：
+Now you want another panel called `.big-panel`. It has the same base properties as `.panel`, but also needs a `width` and `font-size`. It's possible to copy and paste the initial CSS rules from `.panel`, but the code becomes repetitive as you add more types of panels. The `extend` directive is a simple way to reuse the rules written for one element, then add more for another:
```scss
.big-panel{
@@ -30,15 +30,15 @@ Sass 有一个名为`extend`的功能，可以很容易地从一个元素中借
}
```
-除了新样式之外，`.big-panel`将具有与`.panel`相同的属性。
+The `.big-panel` will have the same properties as `.panel` in addition to the new styles.
# --instructions--
-创建一个扩展`.info`的 class`.info-important`，并将`background-color`设置为洋红色。
+Make a class `.info-important` that extends `.info` and also has a `background-color` set to magenta.
# --hints--
-`info-important`class 应该将`background-color`设置为`magenta`。
+Your `info-important` class should have a `background-color` set to `magenta`.
```js
assert(
@@ -48,7 +48,7 @@ assert(
);
```
-`info-important`class 应使用`@extend`继承`info`class 的样式。
+Your `info-important` class should use `@extend` to inherit the styling from the `info` class.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/sass/nest-css-with-sass.md b/curriculum/challenges/chinese/03-front-end-libraries/sass/nest-css-with-sass.md
index c3c51d790e2..1e7b8493c6b 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/sass/nest-css-with-sass.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/sass/nest-css-with-sass.md
@@ -1,6 +1,6 @@
---
id: 587d7dbd367417b2b2512bb5
-title: 用 Sass 嵌套 CSS
+title: Nest CSS with Sass
challengeType: 0
forumTopicId: 301457
dashedName: nest-css-with-sass
@@ -8,9 +8,9 @@ dashedName: nest-css-with-sass
# --description--
-Sass 允许 CSS 规则的`嵌套`，这在写样式表的时候会很有用。
+Sass allows nesting of CSS rules, which is a useful way of organizing a style sheet.
-在 CSS 里，每个元素的样式都需要写在独立的代码块中，如下所示：
+Normally, each element is targeted on a different line to style it, like so:
```scss
nav {
@@ -26,7 +26,7 @@ nav ul li {
}
```
-对于一个大型项目，CSS 规则会很复杂。这时，引入`嵌套`功能（即在对应的父元素中写子元素的样式）可以有效地简化代码：
+For a large project, the CSS file will have many lines and rules. This is where nesting can help organize your code by placing child style rules within the respective parent elements:
```scss
nav {
@@ -45,11 +45,11 @@ nav {
# --instructions--
-根据上面关于嵌套的例子，来简化<code>.blog-post<code>中两个子元素的 CSS 代码。出于测试目的，<code>h1</code>应该出现在<code>p</code>元素之前。</code></code>
+Use the nesting technique shown above to re-organize the CSS rules for both children of `.blog-post` element. For testing purposes, the `h1` should come before the `p` element.
# --hints--
-你应重新组织 CSS 规则，以便`h1`和`p`嵌套在`.blog-post`父元素中。
+Your code should re-organize the CSS rules so the `h1` and `p` are nested in the `.blog-post` parent element.
```js
assert(
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/sass/split-your-styles-into-smaller-chunks-with-partials.md b/curriculum/challenges/chinese/03-front-end-libraries/sass/split-your-styles-into-smaller-chunks-with-partials.md
index f6bd6d2725b..fa421a0cb02 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/sass/split-your-styles-into-smaller-chunks-with-partials.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/sass/split-your-styles-into-smaller-chunks-with-partials.md
@@ -1,6 +1,6 @@
---
id: 587d7dbf367417b2b2512bbc
-title: 用 Partials 将你的样式分成小块
+title: Split Your Styles into Smaller Chunks with Partials
challengeType: 0
forumTopicId: 301459
dashedName: split-your-styles-into-smaller-chunks-with-partials
@@ -8,11 +8,11 @@ dashedName: split-your-styles-into-smaller-chunks-with-partials
# --description--
-Sass 中的`Partials`是包含 CSS 代码段的单独文件。这些是在其他 Sass 文件中导入和使用的。我们可以把类似代码放到模块中，以保持代码结构规整且易于管理。
+<dfn>Partials</dfn> in Sass are separate files that hold segments of CSS code. These are imported and used in other Sass files. This is a great way to group similar code into a module to keep it organized.
-`partials`的名称以下划线（`_`）字符开头，告诉 Sass 它是 CSS 的一小部分，而不是将其转换为 CSS 文件。此外，Sass 文件以`.scss`文件扩展名结尾。要将`partial`中的代码放入另一个 Sass 文件中，请使用`@import`指令。
+Names for partials start with the underscore (`_`) character, which tells Sass it is a small segment of CSS and not to convert it into a CSS file. Also, Sass files end with the `.scss` file extension. To bring the code in the partial into another Sass file, use the `@import` directive.
-例如，如果所有`mixins`都保存在名为 "\_mixins.scss " 的`partial`中，并且在"main.scss "文件中需要它们，这是如何在主文件中使用它们：
+For example, if all your mixins are saved in a partial named "\_mixins.scss", and they are needed in the "main.scss" file, this is how to use them in the main file:
```scss
// In the main.scss file
@@ -20,15 +20,15 @@ Sass 中的`Partials`是包含 CSS 代码段的单独文件。这些是在其他
@import 'mixins'
```
-请注意，`import`语句中不需要下划线——Sass 知道它是`partial`。将`partial`导入文件后，可以使用所有变量`mixins`和其他代码。
+Note that the underscore and file extension are not needed in the `import` statement - Sass understands it is a partial. Once a partial is imported into a file, all variables, mixins, and other code are available to use.
# --instructions--
-编写`@import`语句，将名为`_variables.scss`的`partial`导入 main.scss 文件。
+Write an `@import` statement to import a partial named `_variables.scss` into the main.scss file.
# --hints--
-你的代码应使用`@import`指令，并且不应在文件名中包含下划线。
+Your code should use the `@import` directive, and should not include the underscore in the file name.
```js
assert(code.match(/@import\s+?('|")variables\1/gi));
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/sass/store-data-with-sass-variables.md b/curriculum/challenges/chinese/03-front-end-libraries/sass/store-data-with-sass-variables.md
index 07344ac0d56..6829bc9477f 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/sass/store-data-with-sass-variables.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/sass/store-data-with-sass-variables.md
@@ -1,6 +1,6 @@
---
id: 587d7dbd367417b2b2512bb4
-title: 用 Sass 变量存储数据
+title: Store Data with Sass Variables
challengeType: 0
forumTopicId: 301460
dashedName: store-data-with-sass-variables
@@ -8,11 +8,11 @@ dashedName: store-data-with-sass-variables
# --description--
-Sass 不同于 CSS 的一个特点是它允许使用变量。我们可以在 Sass 中声明变量，并为它赋值，就像我们在 JavaScript 中一样。
+One feature of Sass that's different than CSS is it uses variables. They are declared and set to store data, similar to JavaScript.
-在 JavaScript 中，变量是使用`let`和`const`关键字定义的。在 Sass 中，变量以`$`开头的，后跟变量名。
+In JavaScript, variables are defined using the `let` and `const` keywords. In Sass, variables start with a `$` followed by the variable name.
-这里有几个例子：
+Here are a couple examples:
```scss
$main-fonts: Arial, sans-serif;
@@ -25,33 +25,33 @@ h1 {
}
```
-当需要把多个元素设置成相同颜色时，变量就会很有用。一旦我们需要更改颜色，只需要改变这个变量的值就好。
+One example where variables are useful is when a number of elements need to be the same color. If that color is changed, the only place to edit the code is the variable value.
# --instructions--
-创建一个变量`$text-color`并将其设置为红色。然后更改`.blog-post`和`h2`的`color`属性的值为`$text-color`变量。
+Create a variable `$text-color` and set it to red. Then change the value of the `color` property for the `.blog-post` and `h2` to the `$text-color` variable.
# --hints--
-你应该为`$text-color`声明一个值为红色的 Sass 变量。
+Your code should have a Sass variable declared for `$text-color` with a value of red.
```js
assert(code.match(/\$text-color:\s*?red;/g));
```
-你应使用`$text-color`变量来更改`.blog-post`和`h2`的`颜色`。
+Your code should use the `$text-color` variable to change the `color` for the `.blog-post` and `h2` items.
```js
assert(code.match(/color:\s*?\$text-color;/g));
```
-`.blog-post`元素应为红色。
+Your `.blog-post` element should have a `color` of red.
```js
assert($('.blog-post').css('color') == 'rgb(255, 0, 0)');
```
-`h2`元素应为红色。
+Your `h2` elements should have a `color` of red.
```js
assert($('h2').css('color') == 'rgb(255, 0, 0)');
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/sass/use-each-to-map-over-items-in-a-list.md b/curriculum/challenges/chinese/03-front-end-libraries/sass/use-each-to-map-over-items-in-a-list.md
index 1b171b5c7a9..963960aa092 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/sass/use-each-to-map-over-items-in-a-list.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/sass/use-each-to-map-over-items-in-a-list.md
@@ -1,6 +1,6 @@
---
id: 587d7dbf367417b2b2512bba
-title: 使用 @each 遍历列表中的项目
+title: Use @each to Map Over Items in a List
challengeType: 0
forumTopicId: 301461
dashedName: use-each-to-map-over-items-in-a-list
@@ -8,7 +8,7 @@ dashedName: use-each-to-map-over-items-in-a-list
# --description--
-最后一个挑战显示了`@for`指令如何使用起始值和结束值循环一定次数。Sass 还提供`@each`指令，该指令循环遍历列表或映射中的每个项目。 在每次迭代时，变量将从列表映射赋值给当前值。
+The last challenge showed how the `@for` directive uses a starting and ending value to loop a certain number of times. Sass also offers the `@each` directive which loops over each item in a list or map. On each iteration, the variable gets assigned to the current value from the list or map.
```scss
@each $color in blue, red, green {
@@ -16,7 +16,7 @@ dashedName: use-each-to-map-over-items-in-a-list
}
```
-map 的语法略有不同。这是一个例子：
+A map has slightly different syntax. Here's an example:
```scss
$colors: (color1: blue, color2: red, color3: green);
@@ -26,7 +26,7 @@ $colors: (color1: blue, color2: red, color3: green);
}
```
-请注意，需要`$key`变量来引用 map 中的键。否则，编译后的 CSS 将包含`color1`，`color2`...... 以上两个代码示例都转换为以下 CSS：
+Note that the `$key` variable is needed to reference the keys in the map. Otherwise, the compiled CSS would have `color1`, `color2`... in it. Both of the above code examples are converted into the following CSS:
```scss
.blue-text {
@@ -44,29 +44,29 @@ $colors: (color1: blue, color2: red, color3: green);
# --instructions--
-编写一个`@each`指令，通过一个列表：`blue,black,red`并将每个变量分配给`.color-bg`class, 其中每个项目的“颜色”部分都会发生变化。 每个 class 都应该将`background-color`设置为相应的颜色。
+Write an `@each` directive that goes through a list: `blue, black, red` and assigns each variable to a `.color-bg` class, where the "color" part changes for each item. Each class should set the `background-color` the respective color.
# --hints--
-你的代码应使用`@each`指令。
+Your code should use the `@each` directive.
```js
assert(code.match(/@each /g));
```
-`.blue-bg`class 背景色应为蓝色。
+Your `.blue-bg` class should have a `background-color` of blue.
```js
assert($('.blue-bg').css('background-color') == 'rgb(0, 0, 255)');
```
-`.black-bg`class 背景色应为黑色。
+Your `.black-bg` class should have a `background-color` of black.
```js
assert($('.black-bg').css('background-color') == 'rgb(0, 0, 0)');
```
-`.red-bg`class 背景色应为红色。
+Your `.red-bg` class should have a `background-color` of red.
```js
assert($('.red-bg').css('background-color') == 'rgb(255, 0, 0)');
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/sass/use-for-to-create-a-sass-loop.md b/curriculum/challenges/chinese/03-front-end-libraries/sass/use-for-to-create-a-sass-loop.md
index dd92b5e55de..f2d8bd73b8a 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/sass/use-for-to-create-a-sass-loop.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/sass/use-for-to-create-a-sass-loop.md
@@ -1,6 +1,6 @@
---
id: 587d7dbe367417b2b2512bb9
-title: 使用 @for 创建一个 Sass 循环
+title: Use @for to Create a Sass Loop
challengeType: 0
forumTopicId: 301462
dashedName: use-for-to-create-a-sass-loop
@@ -8,11 +8,11 @@ dashedName: use-for-to-create-a-sass-loop
# --description--
-你可以在 Sass 中使用`@for`循环，它的表现类似与 JavaScript 中的`for`循环。
+The `@for` directive adds styles in a loop, very similar to a `for` loop in JavaScript.
-`@for`以两种方式使用："开始 through 到" 或 "开始 to 结束"。主要区别在于“开始结束”*排除* 结束号码，而“开始结束”*包括* 结束号码。
+`@for` is used in two ways: "start through end" or "start to end". The main difference is that the "start **to** end" *excludes* the end number as part of the count, and "start **through** end" *includes* the end number as part of the count.
-这是一个开始 **到** 结束示例：
+Here's a start **through** end example:
```scss
@for $i from 1 through 12 {
@@ -20,7 +20,7 @@ dashedName: use-for-to-create-a-sass-loop
}
```
-`#{$i}`部分是将变量（`i`）与文本组合成字符串的语法。当 Sass 文件转换为 CSS 时，它看起来像这样：
+The `#{$i}` part is the syntax to combine a variable (`i`) with text to make a string. When the Sass file is converted to CSS, it looks like this:
```scss
.col-1 {
@@ -38,47 +38,47 @@ dashedName: use-for-to-create-a-sass-loop
}
```
-这是创建网格布局的有效方法。现在，你有 12 个可用作 CSS classes 的列宽选项。
+This is a powerful way to create a grid layout. Now you have twelve options for column widths available as CSS classes.
# --instructions--
-编写`@for`指令，使`$j`的值为从 1（包含）到 6（不包含）。
+Write a `@for` directive that takes a variable `$j` that goes from 1 **to** 6.
-它应该创建 5 个名为`.text-1`的 classes 到`.text-5`，其中每个 class 的`font-size`设置为 15px 乘以索引。
+It should create 5 classes called `.text-1` to `.text-5` where each has a `font-size` set to 15px multiplied by the index.
# --hints--
-你应使用`@for`指令。
+Your code should use the `@for` directive.
```js
assert(code.match(/@for /g));
```
-`.text-1`class 的`font-size`应为 15px。
+Your `.text-1` class should have a `font-size` of 15px.
```js
assert($('.text-1').css('font-size') == '15px');
```
-`.text-2`class 的`font-size`应为 30px。
+Your `.text-2` class should have a `font-size` of 30px.
```js
assert($('.text-2').css('font-size') == '30px');
```
-`.text-3`class 的`font-size`应为 45px。
+Your `.text-3` class should have a `font-size` of 45px.
```js
assert($('.text-3').css('font-size') == '45px');
```
-`.text-4`class 的`font-size`应为 60px。
+Your `.text-4` class should have a `font-size` of 60px.
```js
assert($('.text-4').css('font-size') == '60px');
```
-`.text-5`class 的`font-size`应为 75px。
+Your `.text-5` class should have a `font-size` of 75px.
```js
assert($('.text-5').css('font-size') == '75px');
diff --git a/curriculum/challenges/chinese/03-front-end-libraries/sass/use-if-and-else-to-add-logic-to-your-styles.md b/curriculum/challenges/chinese/03-front-end-libraries/sass/use-if-and-else-to-add-logic-to-your-styles.md
index 775a9ab0c86..5192bbacbae 100644
--- a/curriculum/challenges/chinese/03-front-end-libraries/sass/use-if-and-else-to-add-logic-to-your-styles.md
+++ b/curriculum/challenges/chinese/03-front-end-libraries/sass/use-if-and-else-to-add-logic-to-your-styles.md
@@ -1,6 +1,6 @@
---
id: 587d7dbe367417b2b2512bb8
-title: 使用 @if 和 @else 为你的样式添加逻辑
+title: Use @if and @else to Add Logic To Your Styles
challengeType: 0
forumTopicId: 301463
dashedName: use-if-and-else-to-add-logic-to-your-styles
@@ -8,7 +8,7 @@ dashedName: use-if-and-else-to-add-logic-to-your-styles
# --description--
-Sass 中的`@if`指令对于测试特定情况非常有用--它的工作方式与 JavaScript中的`if`语句类似。
+The `@if` directive in Sass is useful to test for a specific case - it works just like the `if` statement in JavaScript.
```scss
@mixin make-bold($bool) {
@@ -18,7 +18,7 @@ Sass 中的`@if`指令对于测试特定情况非常有用--它的工作方式
}
```
-类似 JavaScript，你可以在 Sass 中使用`@else if`和`@else`添加更多条件：
+And just like in JavaScript, `@else if` and `@else` test for more conditions:
```scss
@mixin text-effect($val) {
@@ -39,7 +39,7 @@ Sass 中的`@if`指令对于测试特定情况非常有用--它的工作方式
# --instructions--
-创建一个名为`border-stroke`的`mixin`，它接受一个参数`$val`。`mixin`应使用`@if`，`@else if`和`@else`检查以下条件：
+Create a mixin called `border-stroke` that takes a parameter `$val`. The mixin should check for the following conditions using `@if`, `@else if`, and `@else`:
```scss
light - 1px solid black
@@ -47,17 +47,17 @@ medium - 3px solid black
heavy - 6px solid black
```
-如果 `$val` 不是 `light`、`medium` 或者 `heavy`，border 应该设置为 `none`。
+If `$val` is not `light`, `medium`, or `heavy`, the border should be set to `none`.
# --hints--
-你应该声明一个名为`border-stroke`的`mixin`，它有一个名为`$val`的参数。
+Your code should declare a mixin named `border-stroke` which has a parameter named `$val`.
```js
assert(code.match(/@mixin\s+?border-stroke\s*?\(\s*?\$val\s*?\)\s*?{/gi));
```
-`mixin`应该有一个`@if`语句来检查`$val`是否很轻，并将`border`设置为 1px 纯黑色。
+Your mixin should have an `@if` statement to check if `$val` is light, and to set the `border` to 1px solid black.
```js
assert(
@@ -67,7 +67,7 @@ assert(
);
```
-`mixin`应该有一个`@else if`语句来检查`$val`是否中等，并设置`border`为3px 纯黑色。
+Your mixin should have an `@else if` statement to check if `$val` is medium, and to set the `border` to 3px solid black.
```js
assert(
@@ -77,7 +77,7 @@ assert(
);
```
-`mixin`应该有一个`@else if`语句来检查`$val`是否很重，并设置`border`为6px 纯黑色。
+Your mixin should have an `@else if` statement to check if `$val` is heavy, and to set the `border` to 6px solid black.
```js
assert(
@@ -87,7 +87,7 @@ assert(
);
```
-`mixin`应该有一个`@else`语句来将`border`设置为 none。
+Your mixin should have an `@else` statement to set the `border` to none.
```js
assert(code.match(/@else\s*?{\s*?border\s*?:\s*?none\s*?;\s*?}/gi));
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-bar-chart.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-bar-chart.md
index 21d207a4153..4add1bf6b46 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-bar-chart.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-bar-chart.md
@@ -1,6 +1,6 @@
---
id: bd7168d8c242eddfaeb5bd13
-title: 用条形图可视化数据
+title: Visualize Data with a Bar Chart
challengeType: 3
forumTopicId: 301464
dashedName: visualize-data-with-a-bar-chart
@@ -8,43 +8,43 @@ dashedName: visualize-data-with-a-bar-chart
# --description--
-**目标：** 在 [CodePen.io](https://codepen.io) 上实现一个功能类似 <https://codepen.io/freeCodeCamp/full/GrZVaM> 的 App。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/GrZVaM>.
-在满足以下[需求](https://en.wikipedia.org/wiki/User_story)并能通过所有测试的前提下，你可以根据自己的喜好来美化你的 app。
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story) and get all of the tests to pass. Give it your own personal style.
-你可以使用 HTML、JavaScript、CSS、以及基于 svg 的 D3 可视化库来完成这个挑战。该任务需要使用 D3 的坐标轴属性生成坐标轴，这个属性会自动生成沿轴的刻度。这些刻度是通过 D3 测试所必需的，因为它们的位置是用来确定图表元素的对齐方式。你可以在这里 <https://github.com/d3/d3/blob/master/API.md#axes-d3-axis> 获取关于生成坐标轴的信息。每次测试查询的元素都必须是非虚拟 DOM。 如果你使用了前端框架（例如 Vue），那么对于动态的内容测试结果可能不准确。我们希望最终能够兼容这些框架，但 D3 项目目前还不支持它们。
+You can use HTML, JavaScript, CSS, and the D3 svg-based visualization library. The tests require axes to be generated using the D3 axis property, which automatically generates ticks along the axis. These ticks are required for passing the D3 tests because their positions are used to determine alignment of graphed elements. You will find information about generating axes at <https://github.com/d3/d3/blob/master/API.md#axes-d3-axis>. Required (non-virtual) DOM elements are queried on the moment of each test. If you use a frontend framework (like Vue for example), the test results may be inaccurate for dynamic content. We hope to accommodate them eventually, but these frameworks are not currently supported for D3 projects.
-**需求 #1：** 我的图表应该有一个具有`id="title"`属性的标题。
+**User Story #1:** My chart should have a title with a corresponding `id="title"`.
-**需求 #2：** 我的图表应该有一个`g`元素作为 x 轴，并具有`id="x-axis"`属性。
+**User Story #2:** My chart should have a `g` element x-axis with a corresponding `id="x-axis"`.
-**需求 #3：** 我的图表应该有一个`g`元素作为 y 轴，并具有`id="y-axis"`属性。
+**User Story #3:** My chart should have a `g` element y-axis with a corresponding `id="y-axis"`.
-**需求 #4：** 两个轴都应包含多个刻度标签，每个标签具有`class="tick"`属性
+**User Story #4:** Both axes should contain multiple tick labels, each with the corresponding `class="tick"`.
-**需求 #5：** 在我的图表里，每个数据点都应该有一个具有`class="bar"`属性的`rect`元素来展示数据。
+**User Story #5:** My chart should have a `rect` element for each data point with a corresponding `class="bar"` displaying the data.
-**需求 #6：** 每一栏应该具有包含日期的`data-date`属性以及包含 GDP 值的`data-gdp`属性。
+**User Story #6:** Each bar should have the properties `data-date` and `data-gdp` containing date and GDP values.
-**需求 #7：** 条形元素的`data-date`属性应与提供的数据的顺序相匹配。
+**User Story #7:** The bar elements' `data-date` properties should match the order of the provided data.
-**需求 #8：** 条形元素的`data-gdp`属性应与提供的数据的顺序相匹配。
+**User Story #8:** The bar elements' `data-gdp` properties should match the order of the provided data.
-**需求 #9：** 每个条形元素的高度应准确地表示其数据所对应的 GDP 值。
+**User Story #9:** Each bar element's height should accurately represent the data's corresponding GDP.
-**需求 #10：** `data-date`属性和它对应的条形元素应与 x 轴上的相应的值对齐。
+**User Story #10:** The `data-date` attribute and its corresponding bar element should align with the corresponding value on the x-axis.
-**需求 #11：** `data-gdp`属性和它对应的条形元素应与 y 轴上的相应的值对齐。
+**User Story #11:** The `data-gdp` attribute and its corresponding bar element should align with the corresponding value on the y-axis.
-**需求 #12：** 我可以将鼠标悬停在某个区域上，并查看具有`id="tooltip"`属性的提示框，它会显示有关该区域的更多信息。
+**User Story #12:** I can mouse over an area and see a tooltip with a corresponding `id="tooltip"` which displays more information about the area.
-**需求 #13：** 我的提示框应该有一个`data-date`属性，它对应了当前激活区域的`data-date`属性。
+**User Story #13:** My tooltip should have a `data-date` property that corresponds to the `data-date` of the active area.
-以下是完成此项目所需的数据：`https://raw.githubusercontent.com/freeCodeCamp/ProjectReferenceData/master/GDP-data.json`
+Here is the dataset you will need to complete this project: `https://raw.githubusercontent.com/freeCodeCamp/ProjectReferenceData/master/GDP-data.json`
-你可以 fork [这个 CodePen pen 项目](https://codepen.io/freeCodeCamp/pen/MJjpwO)来构建你的项目。或者你可以在任何你喜欢的环境中使用以下 CDN 链接来运行测试：`https://gitcdn.link/repo/freeCodeCamp/testable-projects-fcc/master/build/bundle.js`.
+You can build your project by forking [this CodePen pen](https://codepen.io/freeCodeCamp/pen/MJjpwO). Or you can use this CDN link to run the tests in any environment you like: `https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`.
-一旦你完成了本项目并且该项目所有测试运行通过，请提交项目的 URL。
+Once you're done, submit the URL to your working project with all its tests passing.
# --solutions--
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-choropleth-map.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-choropleth-map.md
index 022abac3fed..5d4c6cf7717 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-choropleth-map.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-choropleth-map.md
@@ -1,6 +1,6 @@
---
id: 587d7fa6367417b2b2512bbf
-title: 用等值区域图可视化数据
+title: Visualize Data with a Choropleth Map
challengeType: 3
forumTopicId: 301465
dashedName: visualize-data-with-a-choropleth-map
@@ -8,42 +8,42 @@ dashedName: visualize-data-with-a-choropleth-map
# --description--
-**目标：** 在 [CodePen.io](https://codepen.io) 上实现一个功能类似 <https://codepen.io/freeCodeCamp/full/EZKqza> 的 App。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/EZKqza>.
-在满足以下[需求](https://en.wikipedia.org/wiki/User_story)并能通过所有测试的前提下，你可以根据自己的喜好来美化你的 app。
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story) and get all of the tests to pass. Give it your own personal style.
-你可以使用 HTML、JavaScript、CSS、以及基于 svg 的 D3 可视化库来完成这个挑战。每次测试查询的元素都必须是非虚拟 DOM。 如果你使用了前端框架（例如 Vue），那么对于动态的内容测试结果可能不准确。我们希望最终能够兼容这些框架，但 D3 项目目前还不支持它们。
+You can use HTML, JavaScript, CSS, and the D3 svg-based visualization library. Required (non-virtual) DOM elements are queried on the moment of each test. If you use a frontend framework (like Vue for example), the test results may be inaccurate for dynamic content. We hope to accommodate them eventually, but these frameworks are not currently supported for D3 projects.
-**需求 #1：** 等值区域图包含一个具有`id="title"`属性的标题。
+**User Story #1:** My choropleth should have a title with a corresponding `id="title"`.
-**需求 #2：** 等值区域图包含一个具有`id="description"`属性的描述内容。
+**User Story #2:** My choropleth should have a description element with a corresponding `id="description"`.
-**需求 #3：** 等值区域图包含一些州县来展示数据，这些州县应该具有`class="county"`属性。
+**User Story #3:** My choropleth should have counties with a corresponding `class="county"` that represent the data.
-**需求 #4：** 这些州县至少应该有 4 种不同的填充颜色。
+**User Story #4:** There should be at least 4 different fill colors used for the counties.
-**需求 #5：** 每个州县都应该具有`data-fips`和`data-education`属性，分别包含他们相应的 fips 值和教育值。
+**User Story #5:** My counties should each have `data-fips` and `data-education` properties containing their corresponding fips and education values.
-**需求 #6：** 在等值区域图中，每一个提供的数据点都应该有一个对应的州县。
+**User Story #6:** My choropleth should have a county for each provided data point.
-**需求 #7：** 各个州县应该具有与样本数据匹配的 fips 值和教育值。
+**User Story #7:** The counties should have data-fips and data-education values that match the sample data.
-**需求 #8：** 等值区域图包含一个具有`id="legend"`属性的图例。
+**User Story #8:** My choropleth should have a legend with a corresponding `id="legend"`.
-**需求 #9：** 图例至少应该使用 4 种不同的填充颜色。
+**User Story #9:** There should be at least 4 different fill colors used for the legend.
-**需求 #10：** 将鼠标悬停在某个区域上时，可以看到具有`id="tooltip"`属性的提示框，它会显示有关该区域的更多信息。
+**User Story #10:** I can mouse over an area and see a tooltip with a corresponding `id="tooltip"` which displays more information about the area.
-**需求 #11：** 提示框应该有`data-education`属性，它对应了当前激活区域的`data-education`属性。
+**User Story #11:** My tooltip should have a `data-education` property that corresponds to the `data-education` of the active area.
-以下是完成此项目所需的数据集：
+Here are the datasets you will need to complete this project:
-- **US Education Data:** `https://raw.githubusercontent.com/no-stack-dub-sack/testable-projects-fcc/master/src/data/choropleth_map/for_user_education.json`
-- **US County Data:** `https://raw.githubusercontent.com/no-stack-dub-sack/testable-projects-fcc/master/src/data/choropleth_map/counties.json`
+- **US Education Data:**`https://cdn.freecodecamp.org/testable-projects-fcc/data/choropleth_map/for_user_education.json`
+- **US County Data:**`https://cdn.freecodecamp.org/testable-projects-fcc/data/choropleth_map/counties.json`
-你可以 fork [这个 CodePen 项目](https://codepen.io/freeCodeCamp/pen/MJjpwO)来构建你的项目。或者你可以在任何你喜欢的环境中使用以下 CDN 链接来运行测试：`https://gitcdn.link/repo/freeCodeCamp/testable-projects-fcc/master/build/bundle.js`.
+You can build your project by forking [this CodePen pen](https://codepen.io/freeCodeCamp/pen/MJjpwO). Or you can use this CDN link to run the tests in any environment you like: `https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`
-一旦你完成了本项目并且该项目所有测试运行通过，请提交项目的 URL。
+Once you're done, submit the URL to your working project with all its tests passing.
# --solutions--
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-heat-map.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-heat-map.md
index 078021b6951..4ca060d07f7 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-heat-map.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-heat-map.md
@@ -1,6 +1,6 @@
---
id: bd7188d8c242eddfaeb5bd13
-title: 用热图可视化数据
+title: Visualize Data with a Heat Map
challengeType: 3
forumTopicId: 301466
dashedName: visualize-data-with-a-heat-map
@@ -8,51 +8,51 @@ dashedName: visualize-data-with-a-heat-map
# --description--
-**目标：** 在 [CodePen.io](https://codepen.io) 上实现一个功能类似 <https://codepen.io/freeCodeCamp/full/JEXgeY> 的 App。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/JEXgeY>.
-在满足以下[需求](https://en.wikipedia.org/wiki/User_story)并能通过所有测试的前提下，你可以根据自己的喜好来美化你的 app。
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story) and get all of the tests to pass. Give it your own personal style.
-你可以使用 HTML、JavaScript、CSS、以及基于 svg 的 D3 可视化库来完成这个挑战。每次测试查询的元素都必须是非虚拟 DOM。 如果你使用了前端框架（例如 Vue），那么对于动态的内容测试结果可能不准确。我们希望最终能够兼容这些框架，但 D3 项目目前还不支持它们。
+You can use HTML, JavaScript, CSS, and the D3 svg-based visualization library. Required (non-virtual) DOM elements are queried on the moment of each test. If you use a frontend framework (like Vue for example), the test results may be inaccurate for dynamic content. We hope to accommodate them eventually, but these frameworks are not currently supported for D3 projects.
-**需求 #1：** 热度图包含一个具有`id="title"`属性的标题。
+**User Story #1:** My heat map should have a title with a corresponding `id="title"`.
-**需求 #2：** 热度图包含一个具有`id="description"`属性的描述内容。
+**User Story #2:** My heat map should have a description with a corresponding `id="description"`.
-**需求 #3：** 热度图包含一个具有`id="x-axis"`属性的 x 轴。
+**User Story #3:** My heat map should have an x-axis with a corresponding `id="x-axis"`.
-**需求 #4：** 热度图包含一个具有`id="y-axis"`属性的 y 轴。
+**User Story #4:** My heat map should have a y-axis with a corresponding `id="y-axis"`.
-**需求 #5：** 热度图包含一些`rect`元素来展示数据，他们具有`class="cell"`属性。
+**User Story #5:** My heat map should have `rect` elements with a `class="cell"` that represent the data.
-**需求 #6：** 这些单元格元素至少应该有 4 种不同的填充颜色。
+**User Story #6:** There should be at least 4 different fill colors used for the cells.
-**需求 #7：** 每个单元格元素都有这些属性`data-month`，`data-year`，`data-temp`，包含了它们相应的月份、年份和温度值。
+**User Story #7:** Each cell will have the properties `data-month`, `data-year`, `data-temp` containing their corresponding month, year, and temperature values.
-**需求 #8：** 每个元素的`data-month`，`data-year`属性应该在数据范围内。
+**User Story #8:** The `data-month`, `data-year` of each cell should be within the range of the data.
-**需求 #9：** 热度图包含与 y 轴上的相应月份对齐的单元格。
+**User Story #9:** My heat map should have cells that align with the corresponding month on the y-axis.
-**需求 #10：** 热度图包含与 x 轴上相应年份对齐的单元格。
+**User Story #10:** My heat map should have cells that align with the corresponding year on the x-axis.
-**需求 #11：** 热度图在 y 轴上有多个刻度标签，并带有完整的月份名称。
+**User Story #11:** My heat map should have multiple tick labels on the y-axis with the full month name.
-**需求 #12：** 热度图在 x 轴上有多个刻度标签，年份在 1754 到 2015 之间。
+**User Story #12:** My heat map should have multiple tick labels on the x-axis with the years between 1754 and 2015.
-**需求 #13：** 热度图包含一个具有`id="legend"`属性的图例。
+**User Story #13:** My heat map should have a legend with a corresponding `id="legend"`.
-**需求 #14：** 图例包含一些`rect`元素。
+**User Story #14:** My legend should contain `rect` elements.
-**需求 #15：** 图例中的这些`rect`元素应该至少使用 4 种不同的填充颜色。
+**User Story #15:** The `rect` elements in the legend should use at least 4 different fill colors.
-**需求 #16：** 将鼠标悬停在某个区域上时，可以看到具有`id="tooltip"`属性的提示框，它会显示有关该区域的更多信息。
+**User Story #16:** I can mouse over an area and see a tooltip with a corresponding `id="tooltip"` which displays more information about the area.
-**需求 #17：** 提示框应该有`data-year`属性，它对应了当前激活区域的`data-year`属性。
+**User Story #17:** My tooltip should have a `data-year` property that corresponds to the `data-year` of the active area.
-以下是完成此项目所需的数据：`https://raw.githubusercontent.com/freeCodeCamp/ProjectReferenceData/master/global-temperature.json`
+Here is the dataset you will need to complete this project: `https://raw.githubusercontent.com/freeCodeCamp/ProjectReferenceData/master/global-temperature.json`
-你可以 fork [这个 CodePen 项目](https://codepen.io/freeCodeCamp/pen/MJjpwO)来构建你的项目。或者你可以在任何你喜欢的环境中使用以下 CDN 链接来运行测试：`https://gitcdn.link/repo/freeCodeCamp/testable-projects-fcc/master/build/bundle.js`.
+You can build your project by forking [this CodePen pen](https://codepen.io/freeCodeCamp/pen/MJjpwO). Or you can use this CDN link to run the tests in any environment you like: `https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`
-一旦你完成了本项目并且该项目所有测试运行通过，请提交项目的 URL。
+Once you're done, submit the URL to your working project with all its tests passing.
# --solutions--
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-scatterplot-graph.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-scatterplot-graph.md
index 6da7a69d06d..2c07a9b5aa5 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-scatterplot-graph.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-scatterplot-graph.md
@@ -1,6 +1,6 @@
---
id: bd7178d8c242eddfaeb5bd13
-title: 用散点图可视化数据
+title: Visualize Data with a Scatterplot Graph
challengeType: 3
forumTopicId: 301467
dashedName: visualize-data-with-a-scatterplot-graph
@@ -8,47 +8,47 @@ dashedName: visualize-data-with-a-scatterplot-graph
# --description--
-**目标：** 在 [CodePen.io](https://codepen.io) 上实现一个功能类似 <https://codepen.io/freeCodeCamp/full/bgpXyK> 的 App。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/bgpXyK>.
-在满足以下[需求](https://en.wikipedia.org/wiki/User_story)并能通过所有测试的前提下，你可以根据自己的喜好来美化你的 app。
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story) and get all of the tests to pass. Give it your own personal style.
-你可以使用 HTML、JavaScript、CSS、以及基于 svg 的 D3 可视化库来完成这个挑战。该任务需要使用 D3 的坐标轴属性生成坐标轴，这个属性会自动生成沿轴的刻度。这些刻度是通过 D3 测试所必需的，因为它们的位置是用来确定图表元素的对齐方式。你可以在这里 <https://github.com/d3/d3/blob/master/API.md#axes-d3-axis> 获取关于生成坐标轴的信息。每次测试查询的元素都必须是非虚拟 DOM。 如果你使用了前端框架（例如 Vue），那么对于动态的内容测试结果可能不准确。我们希望最终能够兼容这些框架，但 D3 项目目前还不支持它们。
+You can use HTML, JavaScript, CSS, and the D3 svg-based visualization library. The tests require axes to be generated using the D3 axis property, which automatically generates ticks along the axis. These ticks are required for passing the D3 tests because their positions are used to determine alignment of graphed elements. You will find information about generating axes at <https://github.com/d3/d3/blob/master/API.md#axes-d3-axis>. Required (non-virtual) DOM elements are queried on the moment of each test. If you use a frontend framework (like Vue for example), the test results may be inaccurate for dynamic content. We hope to accommodate them eventually, but these frameworks are not currently supported for D3 projects.
-**需求 #1：** 散点图包含一个具有`id="title"`属性的标题元素。
+**User Story #1:** I can see a title element that has a corresponding `id="title"`.
-**需求 #2：** 散点图包含一个具有`id="x-axis"`属性的 x 轴。
+**User Story #2:** I can see an x-axis that has a corresponding `id="x-axis"`.
-**需求 #3：** 散点图包含一个具有`id="y-axis"`属性的 y 轴。
+**User Story #3:** I can see a y-axis that has a corresponding `id="y-axis"`.
-**需求 #4：** 散点图包含一些点，每个点都有一个值为`dot`的 class 属性，它代表了被绘制的数据。
+**User Story #4:** I can see dots, that each have a class of `dot`, which represent the data being plotted.
-**需求 #5：** 每个点都应具有`data-xvalue`属性和`data-yvalue`属性，对应这个点的 x 和 y 值。
+**User Story #5:** Each dot should have the properties `data-xvalue` and `data-yvalue` containing their corresponding x and y values.
-**需求 #6：** 每个点的`data-xvalue`属性和`data-yvalue`属性应该在实际数据的范围内，并且数据格式应该正确无误。对于`data-xvalue`属性，可以接受整数（全年）或 Date 对象进行测试评估。对于`data-yvalue`属性（分钟），请使用 Date 对象。
+**User Story #6:** The `data-xvalue` and `data-yvalue` of each dot should be within the range of the actual data and in the correct data format. For `data-xvalue`, integers (full years) or Date objects are acceptable for test evaluation. For `data-yvalue` (minutes), use Date objects.
-**需求 #7：** `data-xvalue`属性和它对应的点应该和 x 轴上的点或值对齐。
+**User Story #7:** The `data-xvalue` and its corresponding dot should align with the corresponding point/value on the x-axis.
-**需求 #8：** `data-yvalue`属性和它对应的点应该和 y 轴上的点或值对齐。
+**User Story #8:** The `data-yvalue` and its corresponding dot should align with the corresponding point/value on the y-axis.
-**需求 #9：** 散点图的 y 轴上有多个时间格式为`%M:%S`的刻度标签。
+**User Story #9:** I can see multiple tick labels on the y-axis with `%M:%S` time format.
-**需求 #10：** 散点图的 x 轴上有多个显示年份的刻度标签。
+**User Story #10:** I can see multiple tick labels on the x-axis that show the year.
-**需求 #11：** 散点图的 x 轴标签的范围在实际 x 轴数据的范围内。
+**User Story #11:** I can see that the range of the x-axis labels are within the range of the actual x-axis data.
-**需求 #12：** 散点图的 y 轴标签的范围在实际 y 轴数据的范围内。
+**User Story #12:** I can see that the range of the y-axis labels are within the range of the actual y-axis data.
-**需求 #13：** 散点图包含一个包含描述性文字的图例，它具有`id="legend"`属性。
+**User Story #13:** I can see a legend containing descriptive text that has `id="legend"`.
-**需求 #14：** 将鼠标悬停在某个区域上时，可以看到具有`id="tooltip"`属性的提示框，它会显示有关该区域的更多信息。
+**User Story #14:** I can mouse over an area and see a tooltip with a corresponding `id="tooltip"` which displays more information about the area.
-**需求 #15：** 提示框应该有`data-year`属性，它对应了当前激活区域的`data-xvalue`属性。
+**User Story #15:** My tooltip should have a `data-year` property that corresponds to the `data-xvalue` of the active area.
-以下是完成此项目所需的数据：`https://raw.githubusercontent.com/freeCodeCamp/ProjectReferenceData/master/cyclist-data.json`
+Here is the dataset you will need to complete this project: `https://raw.githubusercontent.com/freeCodeCamp/ProjectReferenceData/master/cyclist-data.json`
-你可以 fork [这个 CodePen 项目](https://codepen.io/freeCodeCamp/pen/MJjpwO)来构建你的项目。或者你可以在任何你喜欢的环境中使用以下 CDN 链接来运行测试：`https://gitcdn.link/repo/freeCodeCamp/testable-projects-fcc/master/build/bundle.js`.
+You can build your project by forking [this CodePen pen](https://codepen.io/freeCodeCamp/pen/MJjpwO). Or you can use this CDN link to run the tests in any environment you like: `https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`
-一旦你完成了本项目并且通过所有测试，请提交项目的 URL。
+Once you're done, submit the URL to your working project with all its tests passing.
# --solutions--
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-treemap-diagram.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-treemap-diagram.md
index 56500dc6e1f..85a5139f97f 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-treemap-diagram.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-projects/visualize-data-with-a-treemap-diagram.md
@@ -1,6 +1,6 @@
---
id: 587d7fa6367417b2b2512bc0
-title: 用树形图可视化数据
+title: Visualize Data with a Treemap Diagram
challengeType: 3
forumTopicId: 301468
dashedName: visualize-data-with-a-treemap-diagram
@@ -8,43 +8,43 @@ dashedName: visualize-data-with-a-treemap-diagram
# --description--
-**目标：** 在 [CodePen.io](https://codepen.io) 上实现一个功能类似 <https://codepen.io/freeCodeCamp/full/KaNGNR> 的 App。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/KaNGNR>.
-在满足以下[需求](https://en.wikipedia.org/wiki/User_story)并能通过所有测试的前提下，你可以根据自己的喜好来美化你的 app。
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story) and get all of the tests to pass. Give it your own personal style.
-你可以使用 HTML、JavaScript、CSS、以及基于 svg 的 D3 可视化库来完成这个挑战。该任务需要使用 D3 的坐标轴属性生成坐标轴，这个属性会自动生成沿轴的刻度。这些刻度是通过 D3 测试所必需的，因为它们的位置是用来确定图表元素的对齐方式。你可以在这里 <https://github.com/d3/d3/blob/master/API.md#axes-d3-axis> 获取关于生成坐标轴的信息。每次测试查询的元素都必须是非虚拟 DOM。 如果你使用了前端框架（例如 Vue），那么对于动态的内容测试结果可能不准确。我们希望最终能够兼容这些框架，但 D3 项目目前还不支持它们。
+You can use HTML, JavaScript, CSS, and the D3 svg-based visualization library. The tests require axes to be generated using the D3 axis property, which automatically generates ticks along the axis. These ticks are required for passing the D3 tests because their positions are used to determine alignment of graphed elements. You will find information about generating axes at <https://github.com/d3/d3/blob/master/API.md#axes-d3-axis>. Required (non-virtual) DOM elements are queried on the moment of each test. If you use a frontend framework (like Vue for example), the test results may be inaccurate for dynamic content. We hope to accommodate them eventually, but these frameworks are not currently supported for D3 projects.
-**需求 #1：** 矩阵树图包含一个具有`id="title"`属性的标题。
+**User Story #1:** My tree map should have a title with a corresponding `id="title"`.
-**需求 #2：** 矩阵树图包含一个具有`id="description"`属性的描述内容。
+**User Story #2:** My tree map should have a description with a corresponding `id="description"`.
-**需求 #3：** 矩阵树图包含一些具有`class="tile"`属性的`rect`元素来展示数据。
+**User Story #3:** My tree map should have `rect` elements with a corresponding `class="tile"` that represent the data.
-**需求 #4：** 这些矩形块元素至少应该有 2 种不同的填充颜色。
+**User Story #4:** There should be at least 2 different fill colors used for the tiles.
-**需求 #5：** 每一块矩形元素应该具有`data-name`，`data-category`，以及`data-value`这些属性，包含了它们相应的名称，分类，以及数值。
+**User Story #5:** Each tile should have the properties `data-name`, `data-category`, and `data-value` containing their corresponding name, category, and value.
-**需求 #6：** 每个矩形块的面积和它的 data-value 属性值相对应：data-value 值越大的矩形块面积越大。
+**User Story #6:** The area of each tile should correspond to the data-value amount: tiles with a larger data-value should have a bigger area.
-**需求 #7：** 矩阵树图包含一个具有`id="legend"`属性的图例。
+**User Story #7:** My tree map should have a legend with corresponding `id="legend"`.
-**需求 #8：** 图例包含一些具有`class="legend-item"`属性的`rect`元素。
+**User Story #8:** My legend should have `rect` elements with a corresponding `class="legend-item"`.
-**需求 #9：** 图例中的这些`rect`元素至少应该使用 2 种不同的填充颜色。
+**User Story #9:** The `rect` elements in the legend should use at least 2 different fill colors.
-**需求 #10：** 将鼠标悬停在某个区域上时，可以看到具有`id="tooltip"`属性的提示框，它会显示有关该区域的更多信息。
+**User Story #10:** I can mouse over an area and see a tooltip with a corresponding `id="tooltip"` which displays more information about the area.
-**需求 #11：** 提示框应该有`data-value`属性，它对应了当前激活区域的`data-value`属性。
+**User Story #11:** My tooltip should have a `data-value` property that corresponds to the `data-value` of the active area.
-对于此项目，您可以使用以下任何数据集：
+For this project you can use any of the following datasets:
-- **Kickstarter Pledges：** `https://cdn.rawgit.com/freeCodeCamp/testable-projects-fcc/a80ce8f9/src/data/tree_map/kickstarter-funding-data.json`
-- **Movie Sales：** `https://cdn.rawgit.com/freeCodeCamp/testable-projects-fcc/a80ce8f9/src/data/tree_map/movie-data.json`
-- **Video Game Sales：** `https://cdn.rawgit.com/freeCodeCamp/testable-projects-fcc/a80ce8f9/src/data/tree_map/video-game-sales-data.json`
+- **Kickstarter Pledges:** `https://cdn.freecodecamp.org/testable-projects-fcc/data/tree_map/kickstarter-funding-data.json`
+- **Movie Sales:** `https://cdn.freecodecamp.org/testable-projects-fcc/data/tree_map/movie-data.json`
+- **Video Game Sales:** `https://cdn.freecodecamp.org/testable-projects-fcc/data/tree_map/video-game-sales-data.json`
-你可以 fork [这个 CodePen 项目](https://codepen.io/freeCodeCamp/pen/MJjpwO)来构建你的项目。或者你可以在任何你喜欢的环境中使用以下 CDN 链接来运行测试：`https://gitcdn.link/repo/freeCodeCamp/testable-projects-fcc/master/build/bundle.js`.
+You can build your project by forking [this CodePen pen](https://codepen.io/freeCodeCamp/pen/MJjpwO). Or you can use this CDN link to run the tests in any environment you like: `https://cdn.freecodecamp.org/testable-projects-fcc/v1/bundle.js`
-一旦你完成了本项目并且该项目所有测试运行通过，请提交项目的 URL。
+Once you're done, submit the URL to your working project with all its tests passing.
# --solutions--
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-a-hover-effect-to-a-d3-element.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-a-hover-effect-to-a-d3-element.md
index 7a6c3919bda..85a34c34a19 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-a-hover-effect-to-a-d3-element.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-a-hover-effect-to-a-d3-element.md
@@ -1,6 +1,6 @@
---
id: 587d7faa367417b2b2512bd4
-title: 给 D3 元素添加悬停效果
+title: Add a Hover Effect to a D3 Element
challengeType: 6
forumTopicId: 301469
dashedName: add-a-hover-effect-to-a-d3-element
@@ -8,17 +8,17 @@ dashedName: add-a-hover-effect-to-a-d3-element
# --description--
-我们可以为用户的鼠标悬停行为添加高亮显示的效果。到目前为止，矩形的样式应用了内置的 D3 和 SVG 方法，但是你也可以使用 CSS 来实现。
+It's possible to add effects that highlight a bar when the user hovers over it with the mouse. So far, the styling for the rectangles is applied with the built-in D3 and SVG methods, but you can use CSS as well.
-你可以使用 `attr()` 方法在 SVG 元素上设置 CSS 类。然后用 `:hover` 伪类为你新添加的 CSS 类设置鼠标悬停的效果。
+You set the CSS class on the SVG elements with the `attr()` method. Then the `:hover` pseudo-class for your new class holds the style rules for any hover effects.
# --instructions--
-用 `attr()` 方法给所有的 `rect` 元素都添加 `bar` 类。当鼠标悬停在元素上时，它的 `fill` 将变为棕色。
+Use the `attr()` method to add a class of `bar` to all the `rect` elements. This changes the `fill` color of the bar to brown when you mouse over it.
# --hints--
-`rect` 元素应该有 `bar` 类。
+Your `rect` elements should have a class of `bar`.
```js
assert($('rect').attr('class') == 'bar');
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-a-tooltip-to-a-d3-element.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-a-tooltip-to-a-d3-element.md
index 9ce5fa9da44..5cbd990ad55 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-a-tooltip-to-a-d3-element.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-a-tooltip-to-a-d3-element.md
@@ -1,6 +1,6 @@
---
id: 587d7faa367417b2b2512bd6
-title: 给 D3 元素添加工具提示
+title: Add a Tooltip to a D3 Element
challengeType: 6
forumTopicId: 301470
dashedName: add-a-tooltip-to-a-d3-element
@@ -8,71 +8,71 @@ dashedName: add-a-tooltip-to-a-d3-element
# --description--
-当用户在一个对象上悬停时，提示框可以显示关于这个对象更多的信息。在可视化中有多种方法添加提示框，这个挑战将使用 SVG 的 `title` 元素。
+A tooltip shows more information about an item on a page when the user hovers over that item. There are several ways to add a tooltip to a visualization, this challenge uses the SVG `title` element.
-请使用 `tile` 和 `text()` 方法一起给每组动态添加数据。
+`title` pairs with the `text()` method to dynamically add data to the bars.
# --instructions--
-在每个 `rect` 节点下面添加一个 `title` 元素，然后用回调函数调用 `text()` 方法使它的文本显示数据值。
+Append a `title` element under each `rect` node. Then call the `text()` method with a callback function so the text displays the data value.
# --hints--
-你应该有 9 个 `title` 元素。
+Your code should have 9 `title` elements.
```js
assert($('title').length == 9);
```
-第一个 `title` 元素的提示框文本应为 12。
+The first `title` element should have tooltip text of 12.
```js
assert($('title').eq(0).text() == '12');
```
-第二个 `title` 元素的提示框文本应为 31。
+The second `title` element should have tooltip text of 31.
```js
assert($('title').eq(1).text() == '31');
```
-第三个 `title` 元素的提示框文本应为 22。
+The third `title` element should have tooltip text of 22.
```js
assert($('title').eq(2).text() == '22');
```
-第四个 `title` 元素的提示框文本应为 17。
+The fourth `title` element should have tooltip text of 17.
```js
assert($('title').eq(3).text() == '17');
```
-第五个 `title` 元素的提示框文本应为 25。
+The fifth `title` element should have tooltip text of 25.
```js
assert($('title').eq(4).text() == '25');
```
-第六个 `title` 元素的提示框文本应为 18。
+The sixth `title` element should have tooltip text of 18.
```js
assert($('title').eq(5).text() == '18');
```
-第七个 `title` 元素的提示框文本应为 29。
+The seventh `title` element should have tooltip text of 29.
```js
assert($('title').eq(6).text() == '29');
```
-第八个 `title` 元素的提示框文本应为 14。
+The eighth `title` element should have tooltip text of 14.
```js
assert($('title').eq(7).text() == '14');
```
-第九个 `title` 元素的提示框文本应为 9。
+The ninth `title` element should have tooltip text of 9.
```js
assert($('title').eq(8).text() == '9');
@@ -160,7 +160,7 @@ assert($('title').eq(8).text() == '9');
.attr("class", "bar")
.append("title")
.text((d) => d)
-
+
svg.selectAll("text")
.data(dataset)
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-attributes-to-the-circle-elements.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-attributes-to-the-circle-elements.md
index bae88e5e63c..7ad073a32a3 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-attributes-to-the-circle-elements.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-attributes-to-the-circle-elements.md
@@ -1,6 +1,6 @@
---
id: 587d7fab367417b2b2512bd8
-title: 给 Circle 元素添加属性
+title: Add Attributes to the Circle Elements
challengeType: 6
forumTopicId: 301471
dashedName: add-attributes-to-the-circle-elements
@@ -8,27 +8,27 @@ dashedName: add-attributes-to-the-circle-elements
# --description--
-上个挑战为 `dataset` 中的每个点都创建了 `circle` 元素，并将它们添加到 SVG 画布上。但是 D3 需要更多关于位置和 `circle` 大小的信息来正确的显示它们。
+The last challenge created the `circle` elements for each point in the `dataset`, and appended them to the SVG canvas. But D3 needs more information about the position and size of each `circle` to display them correctly.
-在 SVG 中 `circle` 有三个主要的属性。`cx` 和 `cy` 属性是坐标，它们告诉 D3 将图形的*中心*放在 SVG 画布的何处。半径（ `r` 属性）给出 `circle` 的大小。
+A `circle` in SVG has three main attributes. The `cx` and `cy` attributes are the coordinates. They tell D3 where to position the *center* of the shape on the SVG canvas. The radius (`r` attribute) gives the size of the `circle`.
-就像 `rect` 的 `y` 坐标，`circle` 的 `cy` 属性是从 SVG 画布的顶端开始测量的，而不是从底端。
+Just like the `rect` `y` coordinate, the `cy` attribute for a `circle` is measured from the top of the SVG canvas, not from the bottom.
-所有的属性都可以用回调函数来动态设值。记住，所有串联在 `data(dataset)` 后面的方法为 `dataset` 中的每个对象都运行一次。回调函数中的 `d` 参数指在 `dataset` 中的当前对象，对每个点来说都是一个数组。就像 `d[0]`，你可以使用方括号的方式来访问数组中的值。
+All three attributes can use a callback function to set their values dynamically. Remember that all methods chained after `data(dataset)` run once per item in `dataset`. The `d` parameter in the callback function refers to the current item in `dataset`, which is an array for each point. You use bracket notation, like `d[0]`, to access the values in that array.
# --instructions--
-为 `circle` 元素添加 `cx`、`cy`、`r` 属性。`cx` 的值应该是 `dataset` 中每个对象的数组的第一个数，`cy` 的值应该根据数据中的第二个数，但是要确保正向显示图表而不是倒转。所有圆圈的 `r` 的值应该为 5。
+Add `cx`, `cy`, and `r` attributes to the `circle` elements. The `cx` value should be the first number in the array for each item in `dataset`. The `cy` value should be based off the second number in the array, but make sure to show the chart right-side-up and not inverted. The `r` value should be 5 for all circles.
# --hints--
-你应该有 10 个 `circle` 元素。
+Your code should have 10 `circle` elements.
```js
assert($('circle').length == 10);
```
-第一个 `circle` 元素的 `cx` 值应该为 34，`cy` 值应该为 422，`r` 值应该为 5。
+The first `circle` element should have a `cx` value of 34, a `cy` value of 422, and an `r` value of 5.
```js
assert(
@@ -38,7 +38,7 @@ assert(
);
```
-第二个 `circle` 元素的 `cx` 值应该为 109，`cy` 值应该为 220，`r` 值应该为 5。
+The second `circle` element should have a `cx` value of 109, a `cy` value of 220, and an `r` value of 5.
```js
assert(
@@ -48,7 +48,7 @@ assert(
);
```
-第三个 `circle` 元素的 `cx` 值应该为 310，`cy` 值应该为 380，`r` 值应该为 5。
+The third `circle` element should have a `cx` value of 310, a `cy` value of 380, and an `r` value of 5.
```js
assert(
@@ -58,7 +58,7 @@ assert(
);
```
-第四个 `circle` 元素的 `cx` 值应该为 79，`cy` 值应该为 89，`r` 值应该为 5。
+The fourth `circle` element should have a `cx` value of 79, a `cy` value of 89, and an `r` value of 5.
```js
assert(
@@ -68,7 +68,7 @@ assert(
);
```
-第五个 `circle` 元素的 `cx` 值应该为 420，`cy` 值应该为 280，`r` 值应该为 5。
+The fifth `circle` element should have a `cx` value of 420, a `cy` value of 280, and an `r` value of 5.
```js
assert(
@@ -78,7 +78,7 @@ assert(
);
```
-第六个 `circle` 元素的 `cx` 值应该为 233，`cy` 值应该为 355，`r` 值应该为 5。
+The sixth `circle` element should have a `cx` value of 233, a `cy` value of 355, and an `r` value of 5.
```js
assert(
@@ -88,7 +88,7 @@ assert(
);
```
-第七个 `circle` 元素的 `cx` 值应该为 333，`cy` 值应该为 404，`r` 值应该为 5。
+The seventh `circle` element should have a `cx` value of 333, a `cy` value of 404, and an `r` value of 5.
```js
assert(
@@ -98,7 +98,7 @@ assert(
);
```
-第八个 `circle` 元素的 `cx` 值应该为 222，`cy` 值应该为 167，`r` 值应该为 5。
+The eighth `circle` element should have a `cx` value of 222, a `cy` value of 167, and an `r` value of 5.
```js
assert(
@@ -108,7 +108,7 @@ assert(
);
```
-第九个 `circle` 元素的 `cx` 值应该为 78，`cy` 值应该为 180，`r` 值应该为 5。
+The ninth `circle` element should have a `cx` value of 78, a `cy` value of 180, and an `r` value of 5.
```js
assert(
@@ -118,7 +118,7 @@ assert(
);
```
-第十个 `circle` 元素的 `cx` 值应该为 21，`cy` 值应该为 377，`r` 值应该为 5。
+The tenth `circle` element should have a `cx` value of 21, a `cy` value of 377, and an `r` value of 5.
```js
assert(
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-axes-to-a-visualization.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-axes-to-a-visualization.md
index 3ecfe749c98..f4d9f3fc4b7 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-axes-to-a-visualization.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-axes-to-a-visualization.md
@@ -1,6 +1,6 @@
---
id: 587d7fad367417b2b2512bdf
-title: 添加坐标轴到视图中
+title: Add Axes to a Visualization
challengeType: 6
forumTopicId: 301472
dashedName: add-axes-to-a-visualization
@@ -8,21 +8,15 @@ dashedName: add-axes-to-a-visualization
# --description--
-另一种改进散点图的方法是添加 x 轴和 y 轴。
+Another way to improve the scatter plot is to add an x-axis and a y-axis.
-D3 有两种方法来渲染 y 轴和 x 轴，分别是 `axisLeft()` 和 `axisBottom()`。下面是一个基于上个挑战中的 `xScale` 创建 x 轴的例子：
+D3 has two methods, `axisLeft()` and `axisBottom()`, to render the y-axis and x-axis, respectively. Here's an example to create the x-axis based on the `xScale` in the previous challenges:
```js
const xAxis = d3.axisBottom(xScale);
```
-下一步是在 SVG 画布上渲染 x 轴。为此，你可以使用一个 SVG 组件， `g` 元素，`g` 是英文中组(group)的缩写。
-
-不同于 `rect`、`circle`、`text`，在渲染时，轴只是一条直线。因为它是一个简单的图形，所以可以用 `g` 。
-
-最后一步是使用 `transform` 属性将轴放置在 SVG 画布的正确位置上。否则，轴将会沿着 SVG 画布的边缘渲染，从而不可见。
-
-SVG 支持多种 `transform`，但是定位轴需要使用 `translate` 属性。当它应用在 `g` 元素上时，它根据给出的总量移动整组。下面是一个例子：
+The next step is to render the axis on the SVG canvas. To do so, you can use a general SVG component, the `g` element. The `g` stands for group. Unlike `rect`, `circle`, and `text`, an axis is just a straight line when it's rendered. Because it is a simple shape, using `g` works. The last step is to apply a `transform` attribute to position the axis on the SVG canvas in the right place. Otherwise, the line would render along the border of SVG canvas and wouldn't be visible. SVG supports different types of `transforms`, but positioning an axis needs `translate`. When it's applied to the `g` element, it moves the whole group over and down by the given amounts. Here's an example:
```js
const xAxis = d3.axisBottom(xScale);
@@ -32,21 +26,21 @@ svg.append("g")
.call(xAxis);
```
-上部分代码将 x 轴放置在 SVG 画布的底端。然后 x 轴作为参数被传递给 `call` 方法。 除了 `translate` 的参数变成 (x,0) ，y 轴的定位也是一样的。因为 `translate` 是 `attr` 方法中的一个字符串，你可以在参数中使用字符串的连接将变量值包括进去。
+The above code places the x-axis at the bottom of the SVG canvas. Then it's passed as an argument to the `call()` method. The y-axis works in the same way, except the `translate` argument is in the form (x, 0). Because `translate` is a string in the `attr()` method above, you can use concatenation to include variable values for its arguments.
# --instructions--
-现在散点图有 x 轴了。用 `axisLeft` 方法创建 y 轴并赋值给 `yAxis` 变量，然后通过 `g` 元素渲染 y 轴。确保用 `transform` 属性将 y 轴向右平移 padding 个单位，向下平移 0 个单位。记得 `call()` y 轴。
+The scatter plot now has an x-axis. Create a y-axis in a variable named `yAxis` using the `axisLeft()` method. Then render the axis using a `g` element. Make sure to use a `transform` attribute to translate the axis by the amount of padding units right, and 0 units down. Remember to `call()` the axis.
# --hints--
-你应该使用参数为 `yScale` 的 `axisLeft()` 方法。
+Your code should use the `axisLeft()` method with `yScale` passed as the argument.
```js
assert(code.match(/\.axisLeft\(yScale\)/g));
```
-y 轴的 `g`元素应该有一个`transform` 属性来将 y 轴平移（60，0）。
+The y-axis `g` element should have a `transform` attribute to translate the axis by (60, 0).
```js
assert(
@@ -57,7 +51,7 @@ assert(
);
```
-你应该调用(call) `yAxis` 。
+Your code should call the `yAxis`.
```js
assert(code.match(/\.call\(\s*yAxis\s*\)/g));
@@ -187,9 +181,9 @@ assert(code.match(/\.call\(\s*yAxis\s*\)/g));
.attr("y", (d) => yScale(d[1]))
const xAxis = d3.axisBottom(xScale);
-
+
const yAxis = d3.axisLeft(yScale);
-
+
svg.append("g")
.attr("transform", "translate(0," + (h - padding) + ")")
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-classes-with-d3.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-classes-with-d3.md
index d591e58e4b4..a097263ca4e 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-classes-with-d3.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-classes-with-d3.md
@@ -1,6 +1,6 @@
---
id: 587d7fa7367417b2b2512bc8
-title: 用 D3 添加 Class
+title: Add Classes with D3
challengeType: 6
forumTopicId: 301473
dashedName: add-classes-with-d3
@@ -8,25 +8,27 @@ dashedName: add-classes-with-d3
# --description--
-即使对小型 app 来说在 HTML 元素中大量使用内联样式表也十分难以管理。更方便的是给元素添加遵守 CSS 规则的类。D3 中的 `attr()` 方法可以给元素添加任何 HTML 属性，包括类名称。
+Using a lot of inline styles on HTML elements gets hard to manage, even for smaller apps. It's easier to add a class to elements and style that class one time using CSS rules. D3 has the `attr()` method to add any HTML attribute to an element, including a class name.
-`attr()` 方法和 `style()` 的使用方法一样。它以逗号分隔的键值对为参数使用回调函数。这里是一个给选中元素添加类名为 "container" 的例子：`selection.attr("class", "container");`
+The `attr()` method works the same way that `style()` does. It takes comma-separated values, and can use a callback function. Here's an example to add a class of "container" to a selection:
-注意当 "container" 元素改变或者添加一个 class 时，"class" 参数会保持不变。
+`selection.attr("class", "container");`
+
+Note that the "class" parameter will remain the same whenever you need to add a class and only the "container" parameter will change.
# --instructions--
-在编辑器中添加 `attr()` 方法，给 `div` 元素添加类名 `bar`。
+Add the `attr()` method to the code in the editor and put a class of `bar` on the `div` elements.
# --hints--
-你的 `div`元素应该有一个`bar` 类。
+Your `div` elements should have a class of `bar`.
```js
assert($('div').attr('class') == 'bar');
```
-你应该使用 `attr()` 方法。
+Your code should use the `attr()` method.
```js
assert(code.match(/\.attr/g));
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-document-elements-with-d3.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-document-elements-with-d3.md
index 3e1dae6d7b6..a1107dbf11d 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-document-elements-with-d3.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-document-elements-with-d3.md
@@ -1,6 +1,6 @@
---
id: 587d7fa6367417b2b2512bc2
-title: 用 D3 给文档添加元素
+title: Add Document Elements with D3
challengeType: 6
forumTopicId: 301474
dashedName: add-document-elements-with-d3
@@ -8,21 +8,21 @@ dashedName: add-document-elements-with-d3
# --description--
-D3 有多种方法可以用来在文档中增加元素、修改元素。
+D3 has several methods that let you add and change elements in your document.
-`select()` 方法用来从文档中选择元素，它以你查询的元素名称作为参数，返回第一个符合条件的 HTML 节点。以下是一个例子：
+The `select()` method selects one element from the document. It takes an argument for the name of the element you want and returns an HTML node for the first element in the document that matches the name. Here's an example:
`const anchor = d3.select("a");`
-上面这个例子找到页面上的第一个 a 标签（锚标签），将它作为一个 HTML 节点保存在变量 `anchor` 中。你也可以用其他的方法选择页面上的元素。例子中的 "d3" 是对 D3 对象的引用，可以通过它来访问 D3 的方法。
+The above example finds the first anchor tag on the page and saves an HTML node for it in the variable `anchor`. You can use the selection with other methods. The "d3" part of the example is a reference to the D3 object, which is how you access D3 methods.
-还可以用 `append()` 和 `text()` 方法。
+Two other useful methods are `append()` and `text()`.
-`append()` 方法以你想添加到文档中的元素作为参数，给选中的元素添加一个 HTML 节点，返回那个节点的句柄。
+The `append()` method takes an argument for the element you want to add to the document. It appends an HTML node to a selected item, and returns a handle to that node.
-`text()` 方法既可以给节点设置新的文本，也可以获取节点的当前文本。 如果要设置文字内容，需要在圆括号中传入一个 string（字符串）类型的参数。
+The `text()` method either sets the text of the selected node, or gets the current text. To set the value, you pass a string as an argument inside the parentheses of the method.
-下面是一个选择无序列表、添加列表项和文字的例子：
+Here's an example that selects an unordered list, appends a list item, and adds text:
```js
d3.select("ul")
@@ -30,45 +30,45 @@ d3.select("ul")
.text("Very important item");
```
-在 D3 中可以使用英文句号将多个方法串联在一起执行多个操作。
+D3 allows you to chain several methods together with periods to perform a number of actions in a row.
# --instructions--
-使用 `select` 方法选择文本中的 `body` 标签。然后用 `append` 方法为它添加一个 `h1` 标签，同时在 `h1` 中添加文本 "Learning D3"。
+Use the `select` method to select the `body` tag in the document. Then `append` an `h1` tag to it, and add the text "Learning D3" into the `h1` element.
# --hints--
-你的 `body` 元素应该包含元素 `h1`。
+The `body` should have one `h1` element.
```js
assert($('body').children('h1').length == 1);
```
-你的 `h1` 元素应该包含文本 'Learning D3'。
+The `h1` element should have the text "Learning D3" in it.
```js
assert($('h1').text() == 'Learning D3');
```
-你应该能访问 `d3` 的对象。
+Your code should access the `d3` object.
```js
assert(code.match(/d3/g));
```
-你应该使用 `select` 方法。
+Your code should use the `select` method.
```js
assert(code.match(/\.select/g));
```
-你应该使用 `append` 方法。
+Your code should use the `append` method.
```js
assert(code.match(/\.append/g));
```
-你应该使用 `text` 方法。
+Your code should use the `text` method.
```js
assert(code.match(/\.text/g));
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-inline-styling-to-elements.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-inline-styling-to-elements.md
index 28771901cd4..88c1b74f176 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-inline-styling-to-elements.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-inline-styling-to-elements.md
@@ -1,6 +1,6 @@
---
id: 587d7fa7367417b2b2512bc6
-title: 给元素添加内联样式
+title: Add Inline Styling to Elements
challengeType: 6
forumTopicId: 301475
dashedName: add-inline-styling-to-elements
@@ -8,25 +8,25 @@ dashedName: add-inline-styling-to-elements
# --description--
-D3 可以使用 `style()` 方法为动态元素添加内联 CSS 样式表。
+D3 lets you add inline CSS styles on dynamic elements with the `style()` method.
-`style()` 方法以用逗号分隔的键值对作为参数。这里是一个将选中文本的颜色设为蓝色的例子：
+The `style()` method takes a comma-separated key-value pair as an argument. Here's an example to set the selection's text color to blue:
`selection.style("color","blue");`
# --instructions--
-在编辑器中添加 `style()` 方法，使所有显示的文本的 `font-family` 为 `verdana`。
+Add the `style()` method to the code in the editor to make all the displayed text have a `font-family` of `verdana`.
# --hints--
-你的 `h2` 元素的 `font-family` 应该为 verdana。
+Your `h2` elements should have a `font-family` of verdana.
```js
assert($('h2').css('font-family') == 'verdana');
```
-你应该使用 `style()` 方法。
+Your code should use the `style()` method.
```js
assert(code.match(/\.style/g));
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-labels-to-d3-elements.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-labels-to-d3-elements.md
index 719f414bcae..060f4e59c1a 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-labels-to-d3-elements.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-labels-to-d3-elements.md
@@ -1,6 +1,6 @@
---
id: 587d7faa367417b2b2512bd2
-title: 给 D3 元素添加标签
+title: Add Labels to D3 Elements
challengeType: 6
forumTopicId: 301476
dashedName: add-labels-to-d3-elements
@@ -8,70 +8,70 @@ dashedName: add-labels-to-d3-elements
# --description--
-D3 允许使用 SVG 的 `text` 元素给图形元素贴标签，例如给条形图中的各组都贴上标签。
+D3 lets you label a graph element, such as a bar, using the SVG `text` element.
-和 `rect` 元素类似，`text` 元素也需要 `x` 和 `y` 属性来指定其放置在 SVG 画布上的位置，它也需要能够获取数据来显示数据值。
+Like the `rect` element, a `text` element needs to have `x` and `y` attributes, to place it on the SVG canvas. It also needs to access the data to display those values.
-关于如何给组贴标签，D3 给了你很高的控制权。
+D3 gives you a high level of control over how you label your bars.
# --instructions--
-编辑器中的代码已经将数据绑定到每个新的 `text` 元素。首先，在 `svg` 中添加 `text` 节点。然后，添加 `x` 和 `y` 坐标属性，除了 `text` 的 `y` 值应该使标签比组的高 `y` 3 个单位，其余值的计算方法应该和 `rect` 中计算方法一样。最后，用 D3 的 `text()` 方法将标签的文本设置为和数据点相等的值。
+The code in the editor already binds the data to each new `text` element. First, append `text` nodes to the `svg`. Next, add attributes for the `x` and `y` coordinates. They should be calculated the same way as the `rect` ones, except the `y` value for the `text` should make the label sit 3 units higher than the bar. Finally, use the D3 `text()` method to set the label equal to the data point value.
-**提示**
-关于标签比组高，想一想 `text` 的 `y` 值应该比组的 `y` 值大 3 还是小 3。
+**Note**
+For the label to sit higher than the bar, decide if the `y` value for the `text` should be 3 greater or 3 less than the `y` value for the bar.
# --hints--
-第一个 `text` 元素应该有一个值为 12 的标签并且 `y` 值为 61。
+The first `text` element should have a label of 12 and a `y` value of 61.
```js
assert($('text').eq(0).text() == '12' && $('text').eq(0).attr('y') == '61');
```
-第二个 `text` 元素应该有一个值为 31 的标签并且 `y` 值为 4。
+The second `text` element should have a label of 31 and a `y` value of 4.
```js
assert($('text').eq(1).text() == '31' && $('text').eq(1).attr('y') == '4');
```
-第三个 `text` 元素应该有一个值为 22 的标签并且 `y` 值为 31。
+The third `text` element should have a label of 22 and a `y` value of 31.
```js
assert($('text').eq(2).text() == '22' && $('text').eq(2).attr('y') == '31');
```
-第四个 `text` 元素应该有一个值为 17 的标签并且 `y` 值为 46。
+The fourth `text` element should have a label of 17 and a `y` value of 46.
```js
assert($('text').eq(3).text() == '17' && $('text').eq(3).attr('y') == '46');
```
-第五个 `text` 元素应该有一个值为 25 的标签并且 `y` 值为 22。
+The fifth `text` element should have a label of 25 and a `y` value of 22.
```js
assert($('text').eq(4).text() == '25' && $('text').eq(4).attr('y') == '22');
```
-第六个 `text` 元素应该有一个值为 18 的标签并且 `y` 值为 43。
+The sixth `text` element should have a label of 18 and a `y` value of 43.
```js
assert($('text').eq(5).text() == '18' && $('text').eq(5).attr('y') == '43');
```
-第七个 `text` 元素应该有一个值为 29 的标签并且 `y` 值为 10。
+The seventh `text` element should have a label of 29 and a `y` value of 10.
```js
assert($('text').eq(6).text() == '29' && $('text').eq(6).attr('y') == '10');
```
-第八个 `text` 元素应该有一个值为 14 的标签并且 `y` 值为 55。
+The eighth `text` element should have a label of 14 and a `y` value of 55.
```js
assert($('text').eq(7).text() == '14' && $('text').eq(7).attr('y') == '55');
```
-第九个 `text` 元素应该有一个值为 9 的标签并且 `y` 值为 70。
+The ninth `text` element should have a label of 9 and a `y` value of 70.
```js
assert($('text').eq(8).text() == '9' && $('text').eq(8).attr('y') == '70');
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-labels-to-scatter-plot-circles.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-labels-to-scatter-plot-circles.md
index df32454db1a..8697e5117e0 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-labels-to-scatter-plot-circles.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/add-labels-to-scatter-plot-circles.md
@@ -1,6 +1,6 @@
---
id: 587d7fab367417b2b2512bd9
-title: 向散点图的 Circles 添加标签
+title: Add Labels to Scatter Plot Circles
challengeType: 6
forumTopicId: 301477
dashedName: add-labels-to-scatter-plot-circles
@@ -8,25 +8,25 @@ dashedName: add-labels-to-scatter-plot-circles
# --description--
-你可以添加文本为散点图中的点创建标签。
+You can add text to create labels for the points in a scatter plot.
-目标是显示以逗号分隔的值，分别为 `dataset` 中每个对象的第一个(`x`)和第二个(`y`)字段
+The goal is to display the comma-separated values for the first (`x`) and second (`y`) fields of each item in `dataset`.
-`text` 节点需要 `x` 和 `y` 属性来指定放置在 SVG 画布中的位置。在这个挑战中，`y` 值（决定高度）可以用和 `circle` 的 `cy` 属性相同的值，`x` 值可以比 `circle` 的 `cx` 值稍微大一些，这样标签才是可见的并且被放置在散点的右边。
+The `text` nodes need `x` and `y` attributes to position it on the SVG canvas. In this challenge, the `y` value (which determines height) can use the same value that the `circle` uses for its `cy` attribute. The `x` value can be slightly larger than the `cx` value of the `circle`, so the label is visible. This will push the label to the right of the plotted point.
# --instructions--
-用 `text` 元素为散点图中的每个点创建标签。标签的文本应该为用一个逗号和一个空格分隔开的两个值，例如，第一个点的标签为 "34, 78"。设置 `x` 属性比 `circle` 的 `cx` 属性大 5 个单位，设置 `y` 属性和 `circle` 的 `cy` 属性相同。
+Label each point on the scatter plot using the `text` elements. The text of the label should be the two values separated by a comma and a space. For example, the label for the first point is "34, 78". Set the `x` attribute so it's 5 units more than the value you used for the `cx` attribute on the `circle`. Set the `y` attribute the same way that's used for the `cy` value on the `circle`.
# --hints--
-你应该有 10 个 `text` 元素。
+Your code should have 10 `text` elements.
```js
assert($('text').length == 10);
```
-第一个标签的文本应该为 '34, 78'， `x` 值应该为 39， `y` 应该为 422。
+The first label should have text of "34, 78", an `x` value of 39, and a `y` value of 422.
```js
assert(
@@ -36,7 +36,7 @@ assert(
);
```
-第二个标签的文本应该为 '109, 280'， `x` 值应该为 114， `y` 应该为 220。
+The second label should have text of "109, 280", an `x` value of 114, and a `y` value of 220.
```js
assert(
@@ -46,7 +46,7 @@ assert(
);
```
-第三个标签的文本应该为 '310, 120'， `x` 值应该为 315， `y` 应该为 380。
+The third label should have text of "310, 120", an `x` value of 315, and a `y` value of 380.
```js
assert(
@@ -56,7 +56,7 @@ assert(
);
```
-第四个标签的文本应该为 '79, 411'， `x` 值应该为 84， `y` 应该为 89。
+The fourth label should have text of "79, 411", an `x` value of 84, and a `y` value of 89.
```js
assert(
@@ -66,7 +66,7 @@ assert(
);
```
-第五个标签的文本应该为 '420, 220'， `x` 值应该为 425， `y` 应该为 280。
+The fifth label should have text of "420, 220", an `x` value of 425, and a `y` value of 280.
```js
assert(
@@ -76,7 +76,7 @@ assert(
);
```
-第六个标签的文本应该为 '233, 145'， `x` 值应该为 238， `y` 应该为 355。
+The sixth label should have text of "233, 145", an `x` value of 238, and a `y` value of 355.
```js
assert(
@@ -86,7 +86,7 @@ assert(
);
```
-第七个标签的文本应该为 '333, 96'， `x` 值应该为 338， `y` 应该为 404。
+The seventh label should have text of "333, 96", an `x` value of 338, and a `y` value of 404.
```js
assert(
@@ -96,7 +96,7 @@ assert(
);
```
-第八个标签的文本应该为 '222, 333'， `x` 值应该为 227， `y` 应该为 167。
+The eighth label should have text of "222, 333", an `x` value of 227, and a `y` value of 167.
```js
assert(
@@ -106,7 +106,7 @@ assert(
);
```
-第九个标签的文本应该为 '78, 320'， `x` 值应该为 83， `y` 应该为 180。
+The ninth label should have text of "78, 320", an `x` value of 83, and a `y` value of 180.
```js
assert(
@@ -116,7 +116,7 @@ assert(
);
```
-第十个标签的文本应该为 '21, 123'， `x` 值应该为 26， `y` 应该为 377。
+The tenth label should have text of "21, 123", an `x` value of 26, and a `y` value of 377.
```js
assert(
@@ -218,7 +218,7 @@ assert(
.attr("x", (d) => d[0] + 5)
.attr("y", (d) => h - d[1])
.text((d) => (d[0] + ", " + d[1]))
-
+
</script>
</body>
```
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/change-styles-based-on-data.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/change-styles-based-on-data.md
index ea8f397c5ac..348828f163a 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/change-styles-based-on-data.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/change-styles-based-on-data.md
@@ -1,6 +1,6 @@
---
id: 587d7fa7367417b2b2512bc7
-title: 根据数据更改样式
+title: Change Styles Based on Data
challengeType: 6
forumTopicId: 301479
dashedName: change-styles-based-on-data
@@ -8,9 +8,9 @@ dashedName: change-styles-based-on-data
# --description--
-D3 是关于可视化和展示数据的。如果你期望基于数据来改变元素的样式，你可以在 `style()` 方法中使用回调函数为不同元素改变样式。
+D3 is about visualization and presentation of data. It's likely you'll want to change the styling of elements based on the data. You can use a callback function in the `style()` method to change the styling for different elements.
-例如，你想将值小于 20 的数据点设置为蓝色，其余设置为红色。你可以在 `style()` 方法中使用包含条件逻辑的回调函数。回调函数以 `d` 作为参数来表示一个数据点：
+For example, you may want to color a data point blue if it has a value less than 20, and red otherwise. You can use a callback function in the `style()` method and include the conditional logic. The callback function uses the `d` parameter to represent the data point:
```js
selection.style("color", (d) => {
@@ -18,66 +18,66 @@ selection.style("color", (d) => {
});
```
-`style()` 方法不仅仅可以设置 `color`——它也适用于其他 CSS 属性。
+The `style()` method is not limited to setting the `color` - it can be used with other CSS properties as well.
# --instructions--
-在编辑器中添加 `style()` 方法，根据条件设置 `h2` 元素的 `color` 属性。写一个回调函数，如果值小于 20 返回 "red"，否则返回 "green"。
+Add the `style()` method to the code in the editor to set the `color` of the `h2` elements conditionally. Write the callback function so if the data value is less than 20, it returns "red", otherwise it returns "green".
-**提示**
-你可以使用 if-else 语句或者三目操作符。
+**Note**
+You can use if-else logic, or the ternary operator.
# --hints--
-第一个 `h2` 的 `color` 应该为 red。
+The first `h2` should have a `color` of red.
```js
assert($('h2').eq(0).css('color') == 'rgb(255, 0, 0)');
```
-第二个 `h2` 的 `color` 应该为 green。
+The second `h2` should have a `color` of green.
```js
assert($('h2').eq(1).css('color') == 'rgb(0, 128, 0)');
```
-第三个 `h2` 的 `color` 应该为 green。
+The third `h2` should have a `color` of green.
```js
assert($('h2').eq(2).css('color') == 'rgb(0, 128, 0)');
```
-第四个 `h2` 的 `color` 应该为 red。
+The fourth `h2` should have a `color` of red.
```js
assert($('h2').eq(3).css('color') == 'rgb(255, 0, 0)');
```
-第五个 `h2` 的 `color` 应该为 green。
+The fifth `h2` should have a `color` of green.
```js
assert($('h2').eq(4).css('color') == 'rgb(0, 128, 0)');
```
-第六个 `h2` 的 `color` 应该为 red。
+The sixth `h2` should have a `color` of red.
```js
assert($('h2').eq(5).css('color') == 'rgb(255, 0, 0)');
```
-第七个 `h2` 的 `color` 应该为 green。
+The seventh `h2` should have a `color` of green.
```js
assert($('h2').eq(6).css('color') == 'rgb(0, 128, 0)');
```
-第八个 `h2` 的 `color` 应该为 red。
+The eighth `h2` should have a `color` of red.
```js
assert($('h2').eq(7).css('color') == 'rgb(255, 0, 0)');
```
-第九个 `h2` 的 `color` 应该为 red。
+The ninth `h2` should have a `color` of red.
```js
assert($('h2').eq(8).css('color') == 'rgb(255, 0, 0)');
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/change-the-color-of-an-svg-element.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/change-the-color-of-an-svg-element.md
index 90e32316c9f..d49e0a571f3 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/change-the-color-of-an-svg-element.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/change-the-color-of-an-svg-element.md
@@ -1,6 +1,6 @@
---
id: 587d7fa9367417b2b2512bd1
-title: 更改 SVG 元素的颜色
+title: Change the Color of an SVG Element
challengeType: 6
forumTopicId: 301480
dashedName: change-the-color-of-an-svg-element
@@ -8,17 +8,17 @@ dashedName: change-the-color-of-an-svg-element
# --description--
-所有组都在正确的位置上了，但是它们都是一样的黑色。SVG 可以改变组的颜色。
+The bars are in the right position, but they are all the same black color. SVG has a way to change the color of the bars.
-在 SVG 中， `rect` 图形用 `fill` 属性着色，它支持十六进制代码、颜色名称、rgb 值以及更复杂的选项，比如渐变和透明。
+In SVG, a `rect` shape is colored with the `fill` attribute. It supports hex codes, color names, and rgb values, as well as more complex options like gradients and transparency.
# --instructions--
-添加 `attr()` 方法，将所有组的 "fill" 设置为 "navy"。
+Add an `attr()` method to set the "fill" of all the bars to the color "navy".
# --hints--
-所有组的 `fill` 颜色都应该是 navy。
+The bars should all have a `fill` color of navy.
```js
assert($('rect').css('fill') == 'rgb(0, 0, 128)');
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/change-the-presentation-of-a-bar-chart.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/change-the-presentation-of-a-bar-chart.md
index f4e946d8aeb..00ce9fa19f4 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/change-the-presentation-of-a-bar-chart.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/change-the-presentation-of-a-bar-chart.md
@@ -1,6 +1,6 @@
---
id: 587d7fa8367417b2b2512bca
-title: 更改条形图的显示方式
+title: Change the Presentation of a Bar Chart
challengeType: 6
forumTopicId: 301481
dashedName: change-the-presentation-of-a-bar-chart
@@ -8,22 +8,22 @@ dashedName: change-the-presentation-of-a-bar-chart
# --description--
-这里有一些格式的改变可以美化上个挑战中创建的条形图：
+The last challenge created a bar chart, but there are a couple of formatting changes that could improve it:
-1) 通过在 CSS 中为 `bar` 的类添加 margin 属性，为每一组之间添加空格以直观的将它们分开
+1) Add space between each bar to visually separate them, which is done by adding a margin to the CSS for the `bar` class
-2) 通过给每个值乘以一个数来缩放高度，增加高度以更好地显示值的差异
+2) Increase the height of the bars to better show the difference in values, which is done by multiplying the value by a number to scale the height
# --instructions--
-首先，在 `style` 标签中为 `bar` 类添加 2px 的 `margin` 属性。然后，在 `style()` 方法中修改回调函数，使它返回 10 倍原数值的值（在后面加上 "px"）。
+First, add a `margin` of 2px to the `bar` class in the `style` tag. Next, change the callback function in the `style()` method so it returns a value 10 times the original data value (plus the "px").
-**提示**
-通过给每一个数值点乘以*相同的*常量值仅仅改变比例。这就像放大，它不会改变底层数据的含义。
+**Note**
+Multiplying each data point by the *same* constant only alters the scale. It's like zooming in, and it doesn't change the meaning of the underlying data.
# --hints--
-第一个 `div` 的 `height` 应该为 120 个像素，`margin` 应该为 2 个像素。
+The first `div` should have a `height` of 120 pixels and a `margin` of 2 pixels.
```js
assert(
@@ -32,7 +32,7 @@ assert(
);
```
-第二个 `div` 的 `height` 应该为 310 个像素，`margin` 应该为 2 个像素。
+The second `div` should have a `height` of 310 pixels and a `margin` of 2 pixels.
```js
assert(
@@ -41,7 +41,7 @@ assert(
);
```
-第三个 `div` 的 `height` 应该为 220 个像素，`margin` 应该为 2 个像素。
+The third `div` should have a `height` of 220 pixels and a `margin` of 2 pixels.
```js
assert(
@@ -50,7 +50,7 @@ assert(
);
```
-第四个 `div` 的 `height` 应该为 170 个像素，`margin` 应该为 2 个像素。
+The fourth `div` should have a `height` of 170 pixels and a `margin` of 2 pixels.
```js
assert(
@@ -59,7 +59,7 @@ assert(
);
```
-第五个 `div` 的 `height` 应该为 250 个像素，`margin` 应该为 2 个像素。
+The fifth `div` should have a `height` of 250 pixels and a `margin` of 2 pixels.
```js
assert(
@@ -68,7 +68,7 @@ assert(
);
```
-第六个 `div` 的 `height` 应该为 180 个像素，`margin` 应该为 2 个像素。
+The sixth `div` should have a `height` of 180 pixels and a `margin` of 2 pixels.
```js
assert(
@@ -77,7 +77,7 @@ assert(
);
```
-第七个 `div` 的 `height` 应该为 290 个像素，`margin` 应该为 2 个像素。
+The seventh `div` should have a `height` of 290 pixels and a `margin` of 2 pixels.
```js
assert(
@@ -86,7 +86,7 @@ assert(
);
```
-第八个 `div` 的 `height` 应该为 140 个像素，`margin` 应该为 2 个像素。
+The eighth `div` should have a `height` of 140 pixels and a `margin` of 2 pixels.
```js
assert(
@@ -95,7 +95,7 @@ assert(
);
```
-第九个 `div` 的 `height` 应该为 90 个像素，`margin` 应该为 2 个像素。
+The ninth `div` should have a `height` of 90 pixels and a `margin` of 2 pixels.
```js
assert(
@@ -115,7 +115,7 @@ assert(
height: 100px;
/* Only change code below this line */
-
+
/* Only change code above this line */
display: inline-block;
background-color: blue;
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/create-a-bar-for-each-data-point-in-the-set.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/create-a-bar-for-each-data-point-in-the-set.md
index d31862c8738..3712daf3f60 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/create-a-bar-for-each-data-point-in-the-set.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/create-a-bar-for-each-data-point-in-the-set.md
@@ -1,6 +1,6 @@
---
id: 587d7fa8367417b2b2512bcd
-title: 为集合中的每个数据点创建一个数据条
+title: Create a Bar for Each Data Point in the Set
challengeType: 6
forumTopicId: 301482
dashedName: create-a-bar-for-each-data-point-in-the-set
@@ -8,9 +8,9 @@ dashedName: create-a-bar-for-each-data-point-in-the-set
# --description--
-上个挑战在 `svg` 中只添加了一个矩形来表示数据条。接下来你将结合到目前为止所学的关于 `data()`、`enter()` 和 SVG 图形的知识，为 `dataset` 中的每一个数据点创建并且添加一个矩形。
+The last challenge added only one rectangle to the `svg` element to represent a bar. Here, you'll combine what you've learned so far about `data()`, `enter()`, and SVG shapes to create and append a rectangle for each data point in `dataset`.
-之前的挑战展示了如何为 `dataset` 中的每个对象创建并添加一个 `div`：
+A previous challenge showed the format for how to create and append a `div` for each item in `dataset`:
```js
d3.select("body").selectAll("div")
@@ -19,33 +19,33 @@ d3.select("body").selectAll("div")
.append("div")
```
-操作 `rect` 元素和 `div` 有一些不同，`rect` 必须添加在 `svg` 元素内，而不能直接添加在 `body` 内。同时，你需要告诉 D3 将 `rect` 放在 `svg` 区域的哪个位置。条形的放置会在下一个挑战中讲到。
+There are a few differences working with `rect` elements instead of `divs`. The `rects` must be appended to an `svg` element, not directly to the `body`. Also, you need to tell D3 where to place each `rect` within the `svg` area. The bar placement will be covered in the next challenge.
# --instructions--
-用 `data()`、`enter()`、`append()` 方法为 `dataset` 中的每一个对象创建并添加一个 `rect` 。每个数据条都将直接显示在上一个数据条的上面，将上一个数据条覆盖，这将会在下一个挑战中得到修改。
+Use the `data()`, `enter()`, and `append()` methods to create and append a `rect` for each item in `dataset`. The bars should display all on top of each other; this will be fixed in the next challenge.
# --hints--
-你的代码应该包含 9 个 `rect` 元素。
+Your document should have 9 `rect` elements.
```js
assert($('rect').length == 9);
```
-你应该使用 `data()` 方法。
+Your code should use the `data()` method.
```js
assert(code.match(/\.data/g));
```
-你应该使用 `enter()` 方法。
+Your code should use the `enter()` method.
```js
assert(code.match(/\.enter/g));
```
-你应该使用 `append()` 方法。
+Your code should use the `append()` method.
```js
assert(code.match(/\.append/g));
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/create-a-linear-scale-with-d3.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/create-a-linear-scale-with-d3.md
index f1c85e983bb..3f75bf3e285 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/create-a-linear-scale-with-d3.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/create-a-linear-scale-with-d3.md
@@ -1,6 +1,6 @@
---
id: 587d7fab367417b2b2512bda
-title: 用 D3 创建线性比例
+title: Create a Linear Scale with D3
challengeType: 6
forumTopicId: 301483
dashedName: create-a-linear-scale-with-d3
@@ -8,39 +8,39 @@ dashedName: create-a-linear-scale-with-d3
# --description--
-条形图和散点图都直接在 SVG 画布上绘制数据。但是，如果一组的高或者其中一个数据点比 SVG 的高或宽更大，它将跑到 SVG 区域外。
+The bar and scatter plot charts both plotted data directly onto the SVG canvas. However, if the height of a bar or one of the data points were larger than the SVG height or width values, it would go outside the SVG area.
-D3 中，比例尺可帮助布局数据。 `Scales` 是告诉程序如何将一组原始数据点映射到 SVG 画布上像素的函数。
+In D3, there are scales to help plot data. `Scales` are functions that tell the program how to map a set of raw data points onto the pixels of the SVG canvas.
-例如，假设你有一个 100x500 大小的 SVG 画布，你想为许多国家绘制国内生产总值(GDP)的图表。这组数据将在十亿美元或万亿美元的范围内。你给 D3 提供一种缩放方法告诉它如何将大的 GDP 值放置在 100x500 大小的区域。
+For example, say you have a 100x500-sized SVG canvas and you want to plot Gross Domestic Product (GDP) for a number of countries. The set of numbers would be in the billion or trillion-dollar range. You provide D3 a type of scale to tell it how to place the large GDP values into that 100x500-sized area.
-你不太可能按原样绘制原始数据，在绘制之前，将整个数据集缩放，这样 `x` 和 `y` 值才适合你画布的宽高。
+It's unlikely you would plot raw data as-is. Before plotting it, you set the scale for your entire data set, so that the `x` and `y` values fit your canvas width and height.
-D3 有几种缩放类型。对于线性缩放（通常使用于定量数据），使用 D3 的 `scaleLinear()` 方法：
+D3 has several scale types. For a linear scale (usually used with quantitative data), there is the D3 method `scaleLinear()`:
`const scale = d3.scaleLinear()`
-默认情况下，比例尺使用一比一的比例，输出的值和输入的值相同。在后面的章节中将涉及如何改变默认比例。
+By default, a scale uses the identity relationship. The value of the input is the same as the value of the output. A separate challenge covers how to change this.
# --instructions--
-改变 `scale` 变量的值创建线性缩放，然后将 `output` 变量设置为 `scale` 函数（参数为 50）。
+Change the `scale` variable to create a linear scale. Then set the `output` variable to the scale called with an input argument of 50.
# --hints--
-`h2` 的文本应该为 50。
+The text in the `h2` should be 50.
```js
assert($('h2').text() == '50');
```
-你应该使用 `scaleLinear` 方法。
+Your code should use the `scaleLinear()` method.
```js
assert(code.match(/\.scaleLinear/g));
```
-`output` 变量应该调用输入参数为 50 的 scale 的值。
+The `output` variable should call `scale` with an argument of 50.
```js
assert(output == 50 && code.match(/scale\(\s*?50\s*?\)/g));
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/create-a-scatterplot-with-svg-circles.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/create-a-scatterplot-with-svg-circles.md
index d13345f67fe..dea96419de2 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/create-a-scatterplot-with-svg-circles.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/create-a-scatterplot-with-svg-circles.md
@@ -1,6 +1,6 @@
---
id: 587d7fab367417b2b2512bd7
-title: 使用 SVG Circles 创建散点图
+title: Create a Scatterplot with SVG Circles
challengeType: 6
forumTopicId: 301484
dashedName: create-a-scatterplot-with-svg-circles
@@ -8,20 +8,20 @@ dashedName: create-a-scatterplot-with-svg-circles
# --description--
-散点图是另一种形式的可视化。它用圆圈来映射数据点，每个数据点有两个值，这两个值和 `x` 和 `y` 轴相关联，在可视化中用来给圆圈定位。
+A scatter plot is another type of visualization. It usually uses circles to map data points, which have two values each. These values tie to the `x` and `y` axes, and are used to position the circle in the visualization.
-SVG 用 `circle` 标签来创建圆形，它和之前用来构建条形图的 `rect` 非常相像。
+SVG has a `circle` tag to create the circle shape. It works a lot like the `rect` elements you used for the bar chart.
# --instructions--
-使用 `data()`、`enter()`、`append()` 方法将 `dataset` 和新添加到 SVG 画布上的 `circle` 元素绑定起来。
+Use the `data()`, `enter()`, and `append()` methods to bind `dataset` to new `circle` elements that are appended to the SVG canvas.
-**注意**
-circles 并不可见，因为我们还没有设置它们的属性。我们会在下一个挑战来设置它。
+**Note**
+The circles won't be visible because we haven't set their attributes yet. We'll do that in the next challenge.
# --hints--
-你应该有 10 个 `circle` 元素。
+Your code should have 10 `circle` elements.
```js
assert($('circle').length == 10);
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/display-shapes-with-svg.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/display-shapes-with-svg.md
index 83b3d133cc0..5e24ba82634 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/display-shapes-with-svg.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/display-shapes-with-svg.md
@@ -1,6 +1,6 @@
---
id: 587d7fa8367417b2b2512bcc
-title: 用 SVG 显示形状
+title: Display Shapes with SVG
challengeType: 6
forumTopicId: 301485
dashedName: display-shapes-with-svg
@@ -8,47 +8,47 @@ dashedName: display-shapes-with-svg
# --description--
-上个挑战用给定的宽和高创建了一个 `svg` 元素，因为在它的 `style` 标签中有 `background-color`，所以它是可见的。这一段代码为给定的宽和高腾出空间。
+The last challenge created an `svg` element with a given width and height, which was visible because it had a `background-color` applied to it in the `style` tag. The code made space for the given width and height.
-下一步是在 `svg` 区域中创建图形。SVG 支持多种图形，比如矩形和圆形，并用它们来显示数据。例如，在条形图中一个矩形（`<rect>`）SVG 图形可以创建一个组。
+The next step is to create a shape to put in the `svg` area. There are a number of supported shapes in SVG, such as rectangles and circles. They are used to display data. For example, a rectangle (`<rect>`) SVG shape could create a bar in a bar chart.
-当把图形放入 `svg` 区域中时，你可以用 `x` 和 `y` 坐标来指定它的位置。起始点 (0,0) 是在左上角。`x` 正值将图形右移，`y` 正值将图形从原点下移
+When you place a shape into the `svg` area, you can specify where it goes with `x` and `y` coordinates. The origin point of (0, 0) is in the upper-left corner. Positive values for `x` push the shape to the right, and positive values for `y` push the shape down from the origin point.
-若要把一个图形放在上个挑战的 500（宽）x 100（高）的 `svg` 中心，可将 `x` 坐标设置为 250，`y` 坐标设置为 50。
+To place a shape in the middle of the 500 (width) x 100 (height) `svg` from last challenge, the `x` coordinate would be 250 and the `y` coordinate would be 50.
-SVG 的 `rect` 有四个属性。`x` 和 `y` 坐标指定图形放在 `svg` 区域的位置，`height` 和 `width` 指定图形大小。
+An SVG `rect` has four attributes. There are the `x` and `y` coordinates for where it is placed in the `svg` area. It also has a `height` and `width` to specify the size.
# --instructions--
-用 `append()` 方法给 `svg` 添加一个 `rect` 图形。将它的 `width` 属性设置为 25，`height` 属性设置为 100，`x` 和 `y` 属性都设置为 0。
+Add a `rect` shape to the `svg` using `append()`, and give it a `width` attribute of 25 and `height` attribute of 100. Also, give the `rect` `x` and `y` attributes each set to 0.
# --hints--
-你的文档应该有 1 个 `rect` 元素。
+Your document should have 1 `rect` element.
```js
assert($('rect').length == 1);
```
-`rect` 元素的 `width` 属性应该为 25。
+The `rect` element should have a `width` attribute set to 25.
```js
assert($('rect').attr('width') == '25');
```
-`rect` 元素的 `height` 属性应该为 100。
+The `rect` element should have a `height` attribute set to 100.
```js
assert($('rect').attr('height') == '100');
```
-`rect` 元素的 `x` 属性应该为 0。
+The `rect` element should have an `x` attribute set to 0.
```js
assert($('rect').attr('x') == '0');
```
-`rect` 元素的 `y` 属性应该为 0。
+The `rect` element should have a `y` attribute set to 0.
```js
assert($('rect').attr('y') == '0');
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/dynamically-change-the-height-of-each-bar.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/dynamically-change-the-height-of-each-bar.md
index cbfdfbc750d..dcf94bee445 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/dynamically-change-the-height-of-each-bar.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/dynamically-change-the-height-of-each-bar.md
@@ -1,6 +1,6 @@
---
id: 587d7fa9367417b2b2512bcf
-title: 动态更改每个条的高度
+title: Dynamically Change the Height of Each Bar
challengeType: 6
forumTopicId: 301486
dashedName: dynamically-change-the-height-of-each-bar
@@ -8,7 +8,7 @@ dashedName: dynamically-change-the-height-of-each-bar
# --description--
-和动态设置 `x` 值一样，每组的高也可以被设置成数组中数据点的值。
+The height of each bar can be set to the value of the data point in the array, similar to how the `x` value was set dynamically.
```js
selection.attr("property", (d, i) => {
@@ -21,62 +21,62 @@ selection.attr("property", (d, i) => {
# --instructions--
-改变 `height` 属性的回调函数，让它返回数据值乘以 3 的值。
+Change the callback function for the `height` attribute to return the data value times 3.
-**提示**
-记住，把所有数据点乘以相同的常数来对数据进行缩放（就像放大）。这有利于看清例子中每组之间的差异。
+**Note**
+Remember that multiplying all data points by the same constant scales the data (like zooming in). It helps to see the differences between bar values in this example.
# --hints--
-第一个 `rect` 的 `height` 应该为 36。
+The first `rect` should have a `height` of 36.
```js
assert($('rect').eq(0).attr('height') == '36');
```
-第二个 `rect` 的 `height` 应该为 93。
+The second `rect` should have a `height` of 93.
```js
assert($('rect').eq(1).attr('height') == '93');
```
-第三个 `rect` 的 `height` 应该为 66。
+The third `rect` should have a `height` of 66.
```js
assert($('rect').eq(2).attr('height') == '66');
```
-第四个 `rect` 的 `height` 应该为 51。
+The fourth `rect` should have a `height` of 51.
```js
assert($('rect').eq(3).attr('height') == '51');
```
-第五个 `rect` 的 `height` 应该为 75。
+The fifth `rect` should have a `height` of 75.
```js
assert($('rect').eq(4).attr('height') == '75');
```
-第六个 `rect` 的 `height` 应该为 54。
+The sixth `rect` should have a `height` of 54.
```js
assert($('rect').eq(5).attr('height') == '54');
```
-第七个 `rect` 的 `height` 应该为 87。
+The seventh `rect` should have a `height` of 87.
```js
assert($('rect').eq(6).attr('height') == '87');
```
-第八个 `rect` 的 `height` 应该为 42。
+The eighth `rect` should have a `height` of 42.
```js
assert($('rect').eq(7).attr('height') == '42');
```
-第九个 `rect` 的 `height` 应该为 27。
+The ninth `rect` should have a `height` of 27.
```js
assert($('rect').eq(8).attr('height') == '27');
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/dynamically-set-the-coordinates-for-each-bar.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/dynamically-set-the-coordinates-for-each-bar.md
index 5ad06855184..eb0cff20518 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/dynamically-set-the-coordinates-for-each-bar.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/dynamically-set-the-coordinates-for-each-bar.md
@@ -1,6 +1,6 @@
---
id: 587d7fa9367417b2b2512bce
-title: 动态设置每个 Bar 的坐标
+title: Dynamically Set the Coordinates for Each Bar
challengeType: 6
forumTopicId: 301487
dashedName: dynamically-set-the-coordinates-for-each-bar
@@ -8,83 +8,83 @@ dashedName: dynamically-set-the-coordinates-for-each-bar
# --description--
-上个挑战在 `svg` 元素中为 `dataset` 的每一个数据点创建并且添加了一个矩形。其中一个矩形表示一组，但是它们相互重叠。
+The last challenge created and appended a rectangle to the `svg` element for each point in `dataset` to represent a bar. Unfortunately, they were all stacked on top of each other.
-矩形的摆放是由 `x` 和 `y` 属性处理的，它们告诉 D3 在 `svg` 区域的哪个位置开始绘制图形。上个挑战将它们都设置为 0，所以所有组都摆放在左上角。
+The placement of a rectangle is handled by the `x` and `y` attributes. They tell D3 where to start drawing the shape in the `svg` area. The last challenge set them each to 0, so every bar was placed in the upper-left corner.
-对于条形图，所有组应该处于相同的垂直水平上，也就是说所有组的 `y` 值相同（为 0），但是 `x` 值需要随着增添新的组而变化。注意 `x` 值越大图形就越靠近右边，所以当遍历 `dataset` 中的元素时，`x` 的值应该递增。
+For a bar chart, all of the bars should sit on the same vertical level, which means the `y` value stays the same (at 0) for all bars. The `x` value, however, needs to change as you add new bars. Remember that larger `x` values push items farther to the right. As you go through the array elements in `dataset`, the x value should increase.
-D3 的 `attr()` 方法可接收一个回调函数来动态设置属性。这个回调函数有两个参数，一个是数据点本身（通常是 `d`），另一个是该数据点在数组中的下标，这个参数是可选的。下面是其格式：
+The `attr()` method in D3 accepts a callback function to dynamically set that attribute. The callback function takes two arguments, one for the data point itself (usually `d`) and one for the index of the data point in the array. The second argument for the index is optional. Here's the format:
```js
selection.attr("property", (d, i) => {
/*
- * d 是数据点的值
- * i 是数据点在数组中的下标
+ * d is the data point value
+ * i is the index of the data point in the array
*/
})
```
-值得注意的是，你不需要写 `for` 循环或者用 `forEach` 迭代数据集中的对象。`data()` 方法会解析数据集，任何链接在它后面的方法都会为数据集中的每个对象运行一次。
+It's important to note that you do NOT need to write a `for` loop or use `forEach()` to iterate over the items in the data set. Recall that the `data()` method parses the data set, and any method that's chained after `data()` is run once for each item in the data set.
# --instructions--
-改变 `x` 属性的回调函数，让它返回下标乘以 30 的值。
+Change the `x` attribute callback function so it returns the index times 30.
-**提示**
-每组的宽为 25，所以每次 `x` 增加 30 可在每组之间留出一些空隙。在这个例子中任何比 25 大的数也同样适用。
+**Note**
+Each bar has a width of 25, so increasing each `x` value by 30 adds some space between the bars. Any value greater than 25 would work in this example.
# --hints--
-第一个 `rect` 的 `x` 应该为 0。
+The first `rect` should have an `x` value of 0.
```js
assert($('rect').eq(0).attr('x') == '0');
```
-第二个 `rect` 的 `x` 应该为 30。
+The second `rect` should have an `x` value of 30.
```js
assert($('rect').eq(1).attr('x') == '30');
```
-第三个 `rect` 的 `x` 应该为 60。
+The third `rect` should have an `x` value of 60.
```js
assert($('rect').eq(2).attr('x') == '60');
```
-第四个 `rect` 的 `x` 应该为 90。
+The fourth `rect` should have an `x` value of 90.
```js
assert($('rect').eq(3).attr('x') == '90');
```
-第五个 `rect` 的 `x` 应该为 120。
+The fifth `rect` should have an `x` value of 120.
```js
assert($('rect').eq(4).attr('x') == '120');
```
-第六个 `rect` 的 `x` 应该为 150。
+The sixth `rect` should have an `x` value of 150.
```js
assert($('rect').eq(5).attr('x') == '150');
```
-第七个 `rect` 的 `x` 应该为 180。
+The seventh `rect` should have an `x` value of 180.
```js
assert($('rect').eq(6).attr('x') == '180');
```
-第八个 `rect` 的 `x` 应该为 210。
+The eighth `rect` should have an `x` value of 210.
```js
assert($('rect').eq(7).attr('x') == '210');
```
-第九个 `rect` 的 `x` 应该为 240。
+The ninth `rect` should have an `x` value of 240.
```js
assert($('rect').eq(8).attr('x') == '240');
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/invert-svg-elements.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/invert-svg-elements.md
index 9224f23ab74..794e45b8680 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/invert-svg-elements.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/invert-svg-elements.md
@@ -1,6 +1,6 @@
---
id: 587d7fa9367417b2b2512bd0
-title: 反转 SVG 元素
+title: Invert SVG Elements
challengeType: 6
forumTopicId: 301488
dashedName: invert-svg-elements
@@ -8,74 +8,74 @@ dashedName: invert-svg-elements
# --description--
-你可能已经注意到了常见的条形图像是把这个翻转或者颠倒过来。这是因为 SVG 的 (x, y) 坐标有些特别。
+You may have noticed the bar chart looked like it's upside-down, or inverted. This is because of how SVG uses (x, y) coordinates.
-在 SVG 中，坐标轴的原点在左上角。`x` 坐标为 0 将图形放在 SVG 区域的左边缘，`y` 坐标为 0 将图形放在 SVG 区域的上边缘。`x` 值增大矩形将向右移动，`y` 值增大矩形将向下移动。
+In SVG, the origin point for the coordinates is in the upper-left corner. An `x` coordinate of 0 places a shape on the left edge of the SVG area. A `y` coordinate of 0 places a shape on the top edge of the SVG area. Higher `x` values push the rectangle to the right. Higher `y` values push the rectangle down.
-为了使条形图向上，需要改变 `y` 坐标计算的方式。这需要计算组的高度和 SVG 区域的总高度。
+To make the bars right-side-up, you need to change the way the `y` coordinate is calculated. It needs to account for both the height of the bar and the total height of the SVG area.
-SVG 区域的高度为 100。如果在集合中一个数据点的值为 0，这组将从 SVG 区域的最底端开始（而不是顶端）。为此，`y` 坐标的值应为 100。如果数据点的值为 1，你将从 `y` 坐标为 100 开始来将这组设置在底端，然后需要考虑该组的高度为 1，所以最终的 `y` 坐标将是 99。
+The height of the SVG area is 100. If you have a data point of 0 in the set, you would want the bar to start at the bottom of the SVG area (not the top). To do this, the `y` coordinate needs a value of 100. If the data point value were 1, you would start with a `y` coordinate of 100 to set the bar at the bottom. Then you need to account for the height of the bar of 1, so the final `y` coordinate would be 99.
-`y` 坐标为 `y = heightOfSVG - heightOfBar` 会将条形图向上放置。
+The `y` coordinate that is `y = heightOfSVG - heightOfBar` would place the bars right-side-up.
# --instructions--
-改变 `y` 属性的回调函数，让条形图向上放置。`height` 的值是 3 倍 `d` 的值。
+Change the callback function for the `y` attribute to set the bars right-side-up. Remember that the `height` of the bar is 3 times the data value `d`.
-**提示**
-通常，高度关系为 `y = h - m * d`，其中 `m` 是对数据点进行缩放的比例。
+**Note**
+In general, the relationship is `y = h - m * d`, where `m` is the constant that scales the data points.
# --hints--
-第一个 `rect` 的 `y` 值应该为 64。
+The first `rect` should have a `y` value of 64.
```js
assert($('rect').eq(0).attr('y') == h - dataset[0] * 3);
```
-第二个 `rect` 的 `y` 值应该为 7。
+The second `rect` should have a `y` value of 7.
```js
assert($('rect').eq(1).attr('y') == h - dataset[1] * 3);
```
-第三个 `rect` 的 `y` 值应该为 34。
+The third `rect` should have a `y` value of 34.
```js
assert($('rect').eq(2).attr('y') == h - dataset[2] * 3);
```
-第四个 `rect` 的 `y` 值应该为 49。
+The fourth `rect` should have a `y` value of 49.
```js
assert($('rect').eq(3).attr('y') == h - dataset[3] * 3);
```
-第五个 `rect` 的 `y` 值应该为 25。
+The fifth `rect` should have a `y` value of 25.
```js
assert($('rect').eq(4).attr('y') == h - dataset[4] * 3);
```
-第六个 `rect` 的 `y` 值应该为 46。
+The sixth `rect` should have a `y` value of 46.
```js
assert($('rect').eq(5).attr('y') == h - dataset[5] * 3);
```
-第七个 `rect` 的 `y` 值应该为 13。
+The seventh `rect` should have a `y` value of 13.
```js
assert($('rect').eq(6).attr('y') == h - dataset[6] * 3);
```
-第八个 `rect` 的 `y` 值应该为 58。
+The eighth `rect` should have a `y` value of 58.
```js
assert($('rect').eq(7).attr('y') == h - dataset[7] * 3);
```
-第九个 `rect` 的 `y` 值应该为 73。
+The ninth `rect` should have a `y` value of 73.
```js
assert($('rect').eq(8).attr('y') == h - dataset[8] * 3);
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/learn-about-svg-in-d3.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/learn-about-svg-in-d3.md
index b5f28030a52..a9894790692 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/learn-about-svg-in-d3.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/learn-about-svg-in-d3.md
@@ -1,6 +1,6 @@
---
id: 587d7fa8367417b2b2512bcb
-title: 了解 D3 中的 SVG
+title: Learn About SVG in D3
challengeType: 6
forumTopicId: 301489
dashedName: learn-about-svg-in-d3
@@ -8,36 +8,36 @@ dashedName: learn-about-svg-in-d3
# --description--
-SVG 是 `Scalable Vector Graphics` 的缩写。
+<dfn>SVG</dfn> stands for <dfn>Scalable Vector Graphics</dfn>.
-"scalable" 的意思是如果放大或缩小一个对象，它不会像素化。不管是在小的移动手机屏幕上还是在大的电视显示器上它都会随着显示系统缩放。
+Here "scalable" means that, if you zoom in or out on an object, it would not appear pixelated. It scales with the display system, whether it's on a small mobile screen or a large TV monitor.
-SVG 用于制作常见的几何图形。由于 D3 将数据映射成可视化表达，它用 SVG 来创建可视化的图形。网页上的 SVG 图形必须在 HTML 的 `svg` 标签中。
+SVG is used to make common geometric shapes. Since D3 maps data into a visual representation, it uses SVG to create the shapes for the visualization. SVG shapes for a web page must go within an HTML `svg` tag.
-当使用相对单位（例如 `vh`、`vw` 或者百分比）时，CSS 是可伸缩的。但是在实现数据可视化的时候 SVG 更加的灵活。
+CSS can be scalable when styles use relative units (such as `vh`, `vw`, or percentages), but using SVG is more flexible to build data visualizations.
# --instructions--
-用 `append()` 给 `body` 加一个 `svg` 节点。分别使用 `attr()` 给这个 `svg` 一个 `width` 属性和一个 `height` 属性，并分别将它们设置为给定的常量 `w` 和给定的常量 `h`。你可以在输出中看见它，因为在 `style` 标签中它的 `background-color` 设置为 pink。
+Add an `svg` node to the `body` using `append()`. Give it a `width` attribute set to the provided `w` constant and a `height` attribute set to the provided `h` constant using the `attr()` or `style()` methods for each. You'll see it in the output because there's a `background-color` of pink applied to it in the `style` tag.
-**提示**
-`width` 和 `height` 属性没有单位，它们是来定义缩放的。但无论怎么缩放，这个 `svg` 元素的宽高比永远是 5:1 。
+**Note**
+When using `attr()` width and height attributes do not have units. This is the building block of scaling - the element will always have a 5:1 width to height ratio, no matter what the zoom level is.
# --hints--
-你的文档应该有 1 个 `svg` 元素。
+Your document should have 1 `svg` element.
```js
assert($('svg').length == 1);
```
-`svg` 元素的 `width` 属性应该为 500。
+The `svg` element should have a `width` attribute set to 500 or styled to have a width of 500px.
```js
assert($('svg').attr('width') == '500' || $('svg').css('width') == '500px');
```
-`svg` 元素的 `height` 属性应该为 100。
+The `svg` element should have a `height` attribute set to 100 or styled to have a height of 100px.
```js
assert($('svg').attr('height') == '100' || $('svg').css('height') == '100px');
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/select-a-group-of-elements-with-d3.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/select-a-group-of-elements-with-d3.md
index 9bb3edeff9f..d6c9240e9da 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/select-a-group-of-elements-with-d3.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/select-a-group-of-elements-with-d3.md
@@ -1,6 +1,6 @@
---
id: 587d7fa6367417b2b2512bc3
-title: 用 D3 选择一组元素
+title: Select a Group of Elements with D3
challengeType: 6
forumTopicId: 301490
dashedName: select-a-group-of-elements-with-d3
@@ -8,19 +8,19 @@ dashedName: select-a-group-of-elements-with-d3
# --description--
-`selectAll()` 方法选择一组元素。它以 HTML 节点数组的形式返回该文本中所有匹配所输入字符串的对象。以下是一个选择文本中所有锚标签的例子：
+D3 also has the `selectAll()` method to select a group of elements. It returns an array of HTML nodes for all the items in the document that match the input string. Here's an example to select all the anchor tags in a document:
`const anchors = d3.selectAll("a");`
-像 `select()` 方法一样，`selectAll()` 也支持链式调用，你可以在它之后调用其他方法。
+Like the `select()` method, `selectAll()` supports method chaining, and you can use it with other methods.
# --instructions--
-选择所有的 `li` 标签，通过 `.text()` 方法将它的文本改为 "list item" 。
+Select all of the `li` tags in the document, and change their text to "list item" by chaining the `.text()` method.
# --hints--
-页面上应该有 3 个 `li` 元素，每个元素的文本内容应为 'list item'。大小写和空格必须一致。
+There should be 3 `li` elements on the page, and the text in each one should say "list item". Capitalization and spacing should match exactly.
```js
assert(
@@ -30,13 +30,13 @@ assert(
);
```
-你应该能访问 `d3` 的对象。
+Your code should access the `d3` object.
```js
assert(code.match(/d3/g));
```
-你应该使用 `selectAll` 方法。
+Your code should use the `selectAll` method.
```js
assert(code.match(/\.selectAll/g));
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/set-a-domain-and-a-range-on-a-scale.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/set-a-domain-and-a-range-on-a-scale.md
index 5617da9b219..27b93e18b02 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/set-a-domain-and-a-range-on-a-scale.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/set-a-domain-and-a-range-on-a-scale.md
@@ -1,6 +1,6 @@
---
id: 587d7fac367417b2b2512bdb
-title: 按比例设置域和范围
+title: Set a Domain and a Range on a Scale
challengeType: 6
forumTopicId: 301491
dashedName: set-a-domain-and-a-range-on-a-scale
@@ -8,64 +8,64 @@ dashedName: set-a-domain-and-a-range-on-a-scale
# --description--
-默认情况下，比例尺使用同一关系(identity relationship)，即输入值直接映射为输出值。但是比例尺可以更灵活更有趣。
+By default, scales use the identity relationship. This means the input value maps to the output value. However, scales can be much more flexible and interesting.
-假设有一个数据集范围为 50 到 480，这是缩放比例尺的输入信息，也被称为域(domain)。
+Say a dataset has values ranging from 50 to 480. This is the input information for a scale, also known as the <dfn>domain</dfn>.
-你想沿着 `x` 轴将这些点映射到 SVG 画布上，位置介于 10 个单位到 500 个单位之间。这是输出信息，也被称为范围(range)。
+You want to map those points along the `x` axis on the SVG canvas, between 10 units and 500 units. This is the output information, also known as the <dfn>range</dfn>.
-`domain()` 和 `range()` 方法设置比例尺的值，它们都接受一个至少有两个元素的数组作为参数。下面是一个例子：
+The `domain()` and `range()` methods set these values for the scale. Both methods take an array of at least two elements as an argument. Here's an example:
```js
-// 设置域
-// 域覆盖了一组输入值
+// Set a domain
+// The domain covers the set of input values
scale.domain([50, 480]);
-// 设置范围
-// 范围覆盖了一组输出值
+// Set a range
+// The range covers the set of output values
scale.range([10, 500]);
-scale(50) // 返回 10
-scale(480) // 返回 500
-scale(325) // 返回 323.37
-scale(750) // 返回 807.67
+scale(50) // Returns 10
+scale(480) // Returns 500
+scale(325) // Returns 323.37
+scale(750) // Returns 807.67
d3.scaleLinear()
```
-注意，比例尺使用了域和范围之间的线性关系来找出给定数字的输出值。域中的最小值(50)映射为范围中的最小值(10)。
+Notice that the scale uses the linear relationship between the domain and range values to figure out what the output should be for a given number. The minimum value in the domain (50) maps to the minimum value (10) in the range.
# --instructions--
-创建一个比例尺，将它的域设置为 `[250, 500]`，范围设置为 `[10, 150]`。
+Create a scale and set its domain to `[250, 500]` and range to `[10, 150]`.
-**提示**
-你可以将 `domain()` 和 `range()` 方法链在 `scale` 变量后。
+**Note**
+You can chain the `domain()` and `range()` methods onto the `scale` variable.
# --hints--
-你应该使用 `domain()` 方法。
+Your code should use the `domain()` method.
```js
assert(code.match(/\.domain/g));
```
-比例尺的 `domain()` 应该设置为 `[250, 500]`。
+The `domain()` of the scale should be set to `[250, 500]`.
```js
assert(JSON.stringify(scale.domain()) == JSON.stringify([250, 500]));
```
-你应该使用 `range()` 方法。
+Your code should use the `range()` method.
```js
assert(code.match(/\.range/g));
```
-比例尺的 `range()` 应该设置为 `[10, 150]`。
+The `range()` of the scale should be set to `[10, 150]`.
```js
assert(JSON.stringify(scale.range()) == JSON.stringify([10, 150]));
```
-`h2` 的文本应该为 -102。
+The text in the `h2` should be -102.
```js
assert($('h2').text() == '-102');
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/style-d3-labels.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/style-d3-labels.md
index 93961d614c4..31dc5162fd3 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/style-d3-labels.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/style-d3-labels.md
@@ -1,6 +1,6 @@
---
id: 587d7faa367417b2b2512bd3
-title: 给 D3 标签添加样式
+title: Style D3 Labels
challengeType: 6
forumTopicId: 301492
dashedName: style-d3-labels
@@ -8,21 +8,21 @@ dashedName: style-d3-labels
# --description--
-D3 可以将样式添加到组标签中。`fill` 属性为 `text` 节点设置文本颜色，`style()` 方法设置其它样式的 CSS 规则，例如 "font-family"、"font-size"。
+D3 methods can add styles to the bar labels. The `fill` attribute sets the color of the text for a `text` node. The `style()` method sets CSS rules for other styles, such as "font-family" or "font-size".
# --instructions--
-将 `text` 元素的 `font-size` 设置为 25px，文本颜色设置为 red。
+Set the `font-size` of the `text` elements to 25px, and the color of the text to red.
# --hints--
-所有标签的 `fill` 颜色应该是 red。
+The labels should all have a `fill` color of red.
```js
assert($('text').css('fill') == 'rgb(255, 0, 0)');
```
-所有标签的 `font-size` 应该为 25 个像素。
+The labels should all have a `font-size` of 25 pixels.
```js
assert($('text').css('font-size') == '25px');
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/update-the-height-of-an-element-dynamically.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/update-the-height-of-an-element-dynamically.md
index c73a4ae1a4d..5e7f8c3ff94 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/update-the-height-of-an-element-dynamically.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/update-the-height-of-an-element-dynamically.md
@@ -1,6 +1,6 @@
---
id: 587d7fa8367417b2b2512bc9
-title: 动态更新元素的高度
+title: Update the Height of an Element Dynamically
challengeType: 6
forumTopicId: 301493
dashedName: update-the-height-of-an-element-dynamically
@@ -8,71 +8,71 @@ dashedName: update-the-height-of-an-element-dynamically
# --description--
-之前的挑战包括如何从数组中显示数据和如何添加 CSS 类。将这些课程的内容结合起来只需两步你就能创建出一个简单的条形图：
+The previous challenges covered how to display data from an array and how to add CSS classes. You can combine these lessons to create a simple bar chart. There are two steps to this:
-1) 为每一个数组中的数据点都创建一个 `div`
+1) Create a `div` for each data point in the array
-2) 为每个 `div` 动态分配高度值，在 `style()` 方法中使用回调函数将高度值设置为数据大小
+2) Give each `div` a dynamic height, using a callback function in the `style()` method that sets height equal to the data value
-回想使用回调函数设置样式的格式：
+Recall the format to set a style using a callback function:
`selection.style("cssProperty", (d) => d)`
# --instructions--
-在编辑器中添加 `style()` 方法给每个元素设置 `height` 属性。使用回调函数返回数据点的值加上字符串 "px"。
+Add the `style()` method to the code in the editor to set the `height` property for each element. Use a callback function to return the value of the data point with the string "px" added to it.
# --hints--
-第一个 `div` 的 `height` 应该为 12 个像素。
+The first `div` should have a `height` of 12 pixels.
```js
assert($('div').eq(0)[0].style.height === '12px');
```
-第二个 `div` 的 `height` 应该为 31 个像素。
+The second `div` should have a `height` of 31 pixels.
```js
assert($('div').eq(1)[0].style.height === '31px');
```
-第三个 `div` 的 `height` 应该为 22 个像素。
+The third `div` should have a `height` of 22 pixels.
```js
assert($('div').eq(2)[0].style.height === '22px');
```
-第四个 `div` 的 `height` 应该为 17 个像素。
+The fourth `div` should have a `height` of 17 pixels.
```js
assert($('div').eq(3)[0].style.height === '17px');
```
-第五个 `div` 的 `height` 应该为 25 个像素。
+The fifth `div` should have a `height` of 25 pixels.
```js
assert($('div').eq(4)[0].style.height === '25px');
```
-第六个 `div` 的 `height` 应该为 18 个像素。
+The sixth `div` should have a `height` of 18 pixels.
```js
assert($('div').eq(5)[0].style.height === '18px');
```
-第七个 `div` 的 `height` 应该为 29 个像素。
+The seventh `div` should have a `height` of 29 pixels.
```js
assert($('div').eq(6)[0].style.height === '29px');
```
-第八个 `div` 的 `height` 应该为 14 个像素。
+The eighth `div` should have a `height` of 14 pixels.
```js
assert($('div').eq(7)[0].style.height === '14px');
```
-第九个 `div` 的 `height` 应该为 9 个像素。
+The ninth `div` should have a `height` of 9 pixels.
```js
assert($('div').eq(8)[0].style.height === '9px');
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/use-a-pre-defined-scale-to-place-elements.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/use-a-pre-defined-scale-to-place-elements.md
index 7a9ee794410..61ffe6a29e7 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/use-a-pre-defined-scale-to-place-elements.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/use-a-pre-defined-scale-to-place-elements.md
@@ -1,6 +1,6 @@
---
id: 587d7fac367417b2b2512bde
-title: 使用预定义的比例放置元素
+title: Use a Pre-Defined Scale to Place Elements
challengeType: 6
forumTopicId: 301494
dashedName: use-a-pre-defined-scale-to-place-elements
@@ -8,32 +8,32 @@ dashedName: use-a-pre-defined-scale-to-place-elements
# --description--
-当比例尺建立好后，是时候重新映射散点图了。比例尺就像操作函数一样，将 x 和 y 的原数据值变为适合并可在 SVG 画布上正确渲染的值。它们使数据在屏幕的布局区域内部。
+With the scales set up, it's time to map the scatter plot again. The scales are like processing functions that turn the x and y raw data into values that fit and render correctly on the SVG canvas. They keep the data within the screen's plotting area.
-用比例尺函数为 SVG 图形设置坐标属性值。这包括 `rect` 或者 `text` 元素的 `x` 和 `y` 属性，或者 `circles` 的 `cx` 和 `cy`。以下是一个例子：
+You set the coordinate attribute values for an SVG shape with the scaling function. This includes `x` and `y` attributes for `rect` or `text` elements, or `cx` and `cy` for `circles`. Here's an example:
```js
shape
.attr("x", (d) => xScale(d[0]))
```
-比例尺设置图形坐标属性来将数据点放置在 SVG 画布上。当你显示实际数据值时，不用使用比例尺，例如，在提示框或标签中的 `text()` 方法。
+Scales set shape coordinate attributes to place the data points onto the SVG canvas. You don't need to apply scales when you display the actual data value, for example, in the `text()` method for a tooltip or label.
# --instructions--
-使用 `xScale` 和 `yScale` 将 `circle` 和 `text` 图形放置在 SVG 画布上。对于 `circles`，使用比例尺设置 `cx` 和 `cy` 属性，半径为 5 个单位。
+Use `xScale` and `yScale` to position both the `circle` and `text` shapes onto the SVG canvas. For the `circles`, apply the scales to set the `cx` and `cy` attributes. Give them a radius of 5 units, too.
-对于 `text` 元素，使用比例尺设置 `x` 和 `y` 属性。标签应该标注在点的右边，为此，在将 x 数据值传递给 `xScale` 之前要将它加上 10 个单位。
+For the `text` elements, apply the scales to set the `x` and `y` attributes. The labels should be offset to the right of the dots. To do this, add 10 units to the x data value before passing it to the `xScale`.
# --hints--
-你应该有 10 个 `circle` 元素。
+Your code should have 10 `circle` elements.
```js
assert($('circle').length == 10);
```
-在使用比例尺后第一个 `circle` 元素的 `cx` 值应该大约为 91，`cy` 值应该大约为 368。它的 `r` 值应该为 5。
+The first `circle` element should have a `cx` value of approximately 91 and a `cy` value of approximately 368 after applying the scales. It should also have an `r` value of 5.
```js
assert(
@@ -43,7 +43,7 @@ assert(
);
```
-在使用比例尺后第二个 `circle` 元素的 `cx` 值应该大约为 159，`cy` 值应该大约为 181。它的 `r` 值也应该为 5。
+The second `circle` element should have a `cx` value of approximately 159 and a `cy` value of approximately 181 after applying the scales. It should also have an `r` value of 5.
```js
assert(
@@ -53,7 +53,7 @@ assert(
);
```
-在使用比例尺后第三个 `circle` 元素的 `cx` 值应该大约为 340，`cy` 值应该大约为 329。它的 `r` 值也应该为 5。
+The third `circle` element should have a `cx` value of approximately 340 and a `cy` value of approximately 329 after applying the scales. It should also have an `r` value of 5.
```js
assert(
@@ -63,7 +63,7 @@ assert(
);
```
-在使用比例尺后第四个 `circle` 元素的 `cx` 值应该大约为 131，`cy` 值应该大约为 60。它的 `r` 值也应该为 5。
+The fourth `circle` element should have a `cx` value of approximately 131 and a `cy` value of approximately 60 after applying the scales. It should also have an `r` value of 5.
```js
assert(
@@ -73,7 +73,7 @@ assert(
);
```
-在使用比例尺后第五个 `circle` 元素的 `cx` 值应该大约为 440，`cy` 值应该大约为 237。它的 `r` 值也应该为 5。
+The fifth `circle` element should have a `cx` value of approximately 440 and a `cy` value of approximately 237 after applying the scales. It should also have an `r` value of 5.
```js
assert(
@@ -83,7 +83,7 @@ assert(
);
```
-在使用比例尺后第六个 `circle` 元素的 `cx` 值应该大约为 271，`cy` 值应该大约为 306。它的 `r` 值也应该为 5。
+The sixth `circle` element should have a `cx` value of approximately 271 and a `cy` value of approximately 306 after applying the scales. It should also have an `r` value of 5.
```js
assert(
@@ -93,7 +93,7 @@ assert(
);
```
-在使用比例尺后第七个 `circle` 元素的 `cx` 值应该大约为 361，`cy` 值应该大约为 351。它的 `r` 值也应该为 5。
+The seventh `circle` element should have a `cx` value of approximately 361 and a `cy` value of approximately 351 after applying the scales. It should also have an `r` value of 5.
```js
assert(
@@ -103,7 +103,7 @@ assert(
);
```
-在使用比例尺后第八个 `circle` 元素的 `cx` 值应该大约为 261，`cy` 值应该大约为 132。它的 `r` 值也应该为 5。
+The eighth `circle` element should have a `cx` value of approximately 261 and a `cy` value of approximately 132 after applying the scales. It should also have an `r` value of 5.
```js
assert(
@@ -113,7 +113,7 @@ assert(
);
```
-在使用比例尺后第九个 `circle` 元素的 `cx` 值应该大约为 131，`cy` 值应该大约为 144。它的 `r` 值也应该为 5。
+The ninth `circle` element should have a `cx` value of approximately 131 and a `cy` value of approximately 144 after applying the scales. It should also have an `r` value of 5.
```js
assert(
@@ -123,7 +123,7 @@ assert(
);
```
-在使用比例尺后第十个 `circle` 元素的 `cx` 值应该大约为 79，`cy` 值应该大约为 326。它的 `r` 值也应该为 5。
+The tenth `circle` element should have a `cx` value of approximately 79 and a `cy` value of approximately 326 after applying the scales. It should also have an `r` value of 5.
```js
assert(
@@ -133,13 +133,13 @@ assert(
);
```
-你应该有 10 个 `text` 元素。
+Your code should have 10 `text` elements.
```js
assert($('text').length == 10);
```
-在使用比例尺后第一个标签的 `x` 值应该大约为 100，`y` 值应该大约为 368。
+The first label should have an `x` value of approximately 100 and a `y` value of approximately 368 after applying the scales.
```js
assert(
@@ -148,7 +148,7 @@ assert(
);
```
-在使用比例尺后第二个标签的 `x` 值应该大约为 168，`y` 值应该大约为 181。
+The second label should have an `x` value of approximately 168 and a `y` value of approximately 181 after applying the scales.
```js
assert(
@@ -157,7 +157,7 @@ assert(
);
```
-在使用比例尺后第三个标签的 `x` 值应该大约为 350，`y` 值应该大约为 329。
+The third label should have an `x` value of approximately 350 and a `y` value of approximately 329 after applying the scales.
```js
assert(
@@ -166,7 +166,7 @@ assert(
);
```
-在使用比例尺后第四个标签的 `x` 值应该大约为 141，`y` 值应该大约为 60。
+The fourth label should have an `x` value of approximately 141 and a `y` value of approximately 60 after applying the scales.
```js
assert(
@@ -175,7 +175,7 @@ assert(
);
```
-在使用比例尺后第五个标签的 `x` 值应该大约为 449，`y` 值应该大约为 237。
+The fifth label should have an `x` value of approximately 449 and a `y` value of approximately 237 after applying the scales.
```js
assert(
@@ -184,7 +184,7 @@ assert(
);
```
-在使用比例尺后第六个标签的 `x` 值应该大约为 280，`y` 值应该大约为 306。
+The sixth label should have an `x` value of approximately 280 and a `y` value of approximately 306 after applying the scales.
```js
assert(
@@ -193,7 +193,7 @@ assert(
);
```
-在使用比例尺后第七个标签的 `x` 值应该大约为 370，`y` 值应该大约为 351。
+The seventh label should have an `x` value of approximately 370 and a `y` value of approximately 351 after applying the scales.
```js
assert(
@@ -202,7 +202,7 @@ assert(
);
```
-在使用比例尺后第八个标签的 `x` 值应该大约为 270，`y` 值应该大约为 132。
+The eighth label should have an `x` value of approximately 270 and a `y` value of approximately 132 after applying the scales.
```js
assert(
@@ -211,7 +211,7 @@ assert(
);
```
-在使用比例尺后第九个标签的 `x` 值应该大约为 140，`y` 值应该大约为 144。
+The ninth label should have an `x` value of approximately 140 and a `y` value of approximately 144 after applying the scales.
```js
assert(
@@ -220,7 +220,7 @@ assert(
);
```
-在使用比例尺后第十个标签的 `x` 值应该大约为 88，`y` 值应该大约为 326。
+The tenth label should have an `x` value of approximately 88 and a `y` value of approximately 326 after applying the scales.
```js
assert(
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/use-dynamic-scales.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/use-dynamic-scales.md
index b6aa10826d2..3ec73e8a6d6 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/use-dynamic-scales.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/use-dynamic-scales.md
@@ -1,6 +1,6 @@
---
id: 587d7fac367417b2b2512bdd
-title: 使用动态比例
+title: Use Dynamic Scales
challengeType: 6
forumTopicId: 301495
dashedName: use-dynamic-scales
@@ -8,13 +8,13 @@ dashedName: use-dynamic-scales
# --description--
-D3 的 `min()` 和 `max()` 方法在设置比例尺时十分有用。
+The D3 `min()` and `max()` methods are useful to help set the scale.
-对于一个复杂的数据集，首要是设置比例尺，这样可视化才能适合 SVG 容器的宽和高。所有数据都应布局在 SVG 画布内部，这样它们在页面上才是可见的。
+Given a complex data set, one priority is to set the scale so the visualization fits the SVG container's width and height. You want all the data plotted inside the SVG canvas so it's visible on the web page.
-下面这个例子为散点图设置了 x 轴的比例尺。`domain()` 方法给比例尺传递关于散点图原数据值的信息，`range()` 方法给出在页面上进行可视化的实际空间信息。
+The example below sets the x-axis scale for scatter plot data. The `domain()` method passes information to the scale about the raw data values for the plot. The `range()` method gives it information about the actual space on the web page for the visualization.
-在这个例子中，domain 是从 0 到数据集中的最大值，它使用 `max()` 方法和基于数组中 x 值的回调函数。range 使用 SVG 画布的宽(`w`)并包含 padding，这将在散点图和 SVG 画布边缘之间添加空隙。
+In the example, the domain goes from 0 to the maximum in the set. It uses the `max()` method with a callback function based on the x values in the arrays. The range uses the SVG canvas' width (`w`), but it includes some padding, too. This puts space between the scatter plot dots and the edge of the SVG canvas.
```js
const dataset = [
@@ -32,37 +32,37 @@ const dataset = [
const w = 500;
const h = 500;
-// SVG 画布边缘和散点图之间的 padding
+// Padding between the SVG canvas boundary and the plot
const padding = 30;
const xScale = d3.scaleLinear()
.domain([0, d3.max(dataset, (d) => d[0])])
.range([padding, w - padding]);
```
-在一开始可能很难理解 padding。想象 x 轴是一条从 0 到 500 （SVG 画布宽的值）的水平直线。在 `range()` 方法中包含 padding 使散点图沿着这条直线从 30 （而不是 0）开始，在 470 （而不是 500）结束。
+The padding may be confusing at first. Picture the x-axis as a horizontal line from 0 to 500 (the width value for the SVG canvas). Including the padding in the `range()` method forces the plot to start at 30 along that line (instead of 0), and end at 470 (instead of 500).
# --instructions--
-使用 `yScale` 变量创建一个线性的 y 轴比例尺。`domain` 应该从 0 开始到数据集中 y 的最大值，range 应该使用 SVG 的高(`h`)并包含 padding。
+Use the `yScale` variable to create a linear y-axis scale. The domain should start at zero and go to the maximum y value in the set. The range should use the SVG height (`h`) and include padding.
-**提示**
-记得正向布局。当你为 y 坐标设置 range 时，大的值（height 减去 padding）是第一个参数，小的值是第二个参数。
+**Note**
+Remember to keep the plot right-side-up. When you set the range for the y coordinates, the higher value (height minus padding) is the first argument, and the lower value is the second argument.
# --hints--
-`h2` 中的文本应该是 30。
+The text in the `h2` should be 30.
```js
assert(output == 30 && $('h2').text() == '30');
```
-yScale 的 `domain()` 应该等于 `[0, 411]`。
+The `domain()` of yScale should be equivalent to `[0, 411]`.
```js
assert(JSON.stringify(yScale.domain()) == JSON.stringify([0, 411]));
```
-yScale 的 `range()` 应该等于 `[470, 30]`。
+The `range()` of yScale should be equivalent to `[470, 30]`.
```js
assert(JSON.stringify(yScale.range()) == JSON.stringify([470, 30]));
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/use-the-d3.max-and-d3.min-functions-to-find-minimum-and-maximum-values-in-a-dataset.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/use-the-d3.max-and-d3.min-functions-to-find-minimum-and-maximum-values-in-a-dataset.md
index 464ff7f12ba..f30962512e6 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/use-the-d3.max-and-d3.min-functions-to-find-minimum-and-maximum-values-in-a-dataset.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/use-the-d3.max-and-d3.min-functions-to-find-minimum-and-maximum-values-in-a-dataset.md
@@ -1,6 +1,7 @@
---
id: 587d7fac367417b2b2512bdc
-title: 使用 d3.max 和 d3.min 函数在数据集中查找最小值和最大值
+title: >-
+ Use the d3.max and d3.min Functions to Find Minimum and Maximum Values in a Dataset
challengeType: 6
forumTopicId: 301496
dashedName: >-
@@ -9,45 +10,45 @@ dashedName: >-
# --description--
-D3 的 `domain()` 和 `range()` 方法根据数据设置比例尺的信息。下面有几种更简单的方法。
+The D3 methods `domain()` and `range()` set that information for your scale based on the data. There are a couple methods to make that easier.
-通常当你设置域的时候，你会想用数据集中的最小值和最大值。试图手动的找到这些值，尤其是在很大的数据集中，可能会出错。
+Often when you set the domain, you'll want to use the minimum and maximum values within the data set. Trying to find these values manually, especially in a large data set, may cause errors.
-D3 有两个方法——`min()` 和 `max()` 来返回这些值。下面是一个例子：
+D3 has two methods - `min()` and `max()` to return this information. Here's an example:
```js
const exampleData = [34, 234, 73, 90, 6, 52];
-d3.min(exampleData) // 返回 6
-d3.max(exampleData) // 返回 234
+d3.min(exampleData) // Returns 6
+d3.max(exampleData) // Returns 234
```
-像在散点图的例子中的 \[x, y] 坐标对一样，数据集有可能嵌套数组。在这种情况下，你需要告诉 D3 怎么计算最大值和最小值。幸运的是，`min()` 和 `max()` 都可以使用回调函数。 在下面这个例子中，回调函数的参数 `d` 是当前的内部数组。回调函数需要从内部数组中返回你想计算最大/小值的元素（x 还是 y 值）。下面是一个如何找到二维数组的最大值和最小值的例子：
+A dataset may have nested arrays, like the \[x, y] coordinate pairs that were in the scatter plot example. In that case, you need to tell D3 how to calculate the maximum and minimum. Fortunately, both the `min()` and `max()` methods take a callback function. In this example, the callback function's argument `d` is for the current inner array. The callback needs to return the element from the inner array (the x or y value) over which you want to compute the maximum or minimum. Here's an example for how to find the min and max values with an array of arrays:
```js
const locationData = [[1, 7],[6, 3],[8, 3]];
-// 返回第一个元素中的最小值s
+// Returns the smallest number out of the first elements
const minX = d3.min(locationData, (d) => d[0]);
-// 在 1，6，8 中 minX 为 1
+// minX compared 1, 6, and 8 and is set to 1
```
# --instructions--
-`positionData` 数组中有包含 x、y、z 坐标的子数组。用 D3 的方法去找到数组中 z 坐标（第三个值）的最大值并将它保存在 `output` 变量中。
+The `positionData` array holds sub arrays of x, y, and z coordinates. Use a D3 method to find the maximum value of the z coordinate (the third value) from the arrays and save it in the `output` variable.
# --hints--
-`h2` 的文本应该为 8。
+The text in the `h2` should be 8.
```js
assert(output == 8 && $('h2').text() == '8');
```
-你应该使用 `max()` 方法。
+Your code should use the `max()` method.
```js
assert(
code.match(/\.max/g),
- 'Your code should use the <code>max()</code> method.'
+ 'Your code should use the <code>max()</code> method.'
);
```
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/work-with-data-in-d3.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/work-with-data-in-d3.md
index 6989932fcff..de8a225c842 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/work-with-data-in-d3.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/work-with-data-in-d3.md
@@ -1,6 +1,6 @@
---
id: 587d7fa7367417b2b2512bc4
-title: 使用 D3 中的数据
+title: Work with Data in D3
challengeType: 6
forumTopicId: 301497
dashedName: work-with-data-in-d3
@@ -8,15 +8,15 @@ dashedName: work-with-data-in-d3
# --description--
-D3 是数据驱动的库,可以使用 D3 的方法将数组形式的数据显示在页面上。
+The D3 library focuses on a data-driven approach. When you have a set of data, you can apply D3 methods to display it on the page. Data comes in many formats, but this challenge uses a simple array of numbers.
-第一步是让 D3 知道数据。`data` 方法选择连接着数据的 DOM 元素,数据集作为参数传递给该方法。
+The first step is to make D3 aware of the data. The `data()` method is used on a selection of DOM elements to attach the data to those elements. The data set is passed as an argument to the method.
-常见的方法是在文档中为数据集中的每一个数据创建一个元素，为此，你可以使用 D3 的 `enter()` 方法
+A common workflow pattern is to create a new element in the document for each piece of data in the set. D3 has the `enter()` method for this purpose.
-当 `enter()` 和 `data()` 方法一起使用时，它把从页面中选择的元素和数据集中的元素作比较。如果页面中选择的元素较少则创建缺少的元素。
+When `enter()` is combined with the `data()` method, it looks at the selected elements from the page and compares them to the number of data items in the set. If there are fewer elements than data items, it creates the missing elements.
-以下是一个选择 `ul` 元素并根据添加的数组创建新的列表项的例子。
+Here is an example that selects a `ul` element and creates a new list item based on the number of entries in the array:
```html
<body>
@@ -32,21 +32,21 @@ D3 是数据驱动的库,可以使用 D3 的方法将数组形式的数据显示
</body>
```
-选择不存在的 li 元素似乎有些难以理解。事实上，这段代码先选择页面上的 ul 元素，再选择所有的列表项——li，它将返回空。然后data()方法接收数组作为参数，并运行三次后面的代码，每次对应数组中的一个对象。enter()方法发现页面中没有 li 元素，但是需要 3 个（每个对应dataset中的一个对象）。它将在 ul 中添加带有文本 "New item" 的 li 元素。
+It may seem confusing to select elements that don't exist yet. This code is telling D3 to first select the `ul` on the page. Next, select all list items, which returns an empty selection. Then the `data()` method reviews the dataset and runs the following code three times, once for each item in the array. The `enter()` method sees there are no `li` elements on the page, but it needs 3 (one for each piece of data in `dataset`). New `li` elements are appended to the `ul` and have the text "New item".
# --instructions--
-选择 `body` 节点，然后选择所有的 `h2` 元素。让 D3 为 `dataset` 数组中的每一个对象创建并添加文本为 "New Title" 的 `h2` 标签。你应该使用 `data()` 和 `enter()` 方法。
+Select the `body` node, then select all `h2` elements. Have D3 create and append an `h2` tag for each item in the `dataset` array. The text in the `h2` should say "New Title". Your code should use the `data()` and `enter()` methods.
# --hints--
-你的文档应该有 9 个 `h2` 元素。
+Your document should have 9 `h2` elements.
```js
assert($('h2').length == 9);
```
-`h2` 元素中的文本应为 'New Title'。大小写和空格必须一致。
+The text in the `h2` elements should say "New Title". The capitalization and spacing should match exactly.
```js
assert(
@@ -56,13 +56,13 @@ assert(
);
```
-你应该使用 `data()` 方法。
+Your code should use the `data()` method.
```js
assert(code.match(/\.data/g));
```
-你应该使用 `enter()` 方法。
+Your code should use the `enter()` method.
```js
assert(code.match(/\.enter/g));
diff --git a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/work-with-dynamic-data-in-d3.md b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/work-with-dynamic-data-in-d3.md
index c437f671f10..e8c80ac4386 100644
--- a/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/work-with-dynamic-data-in-d3.md
+++ b/curriculum/challenges/chinese/04-data-visualization/data-visualization-with-d3/work-with-dynamic-data-in-d3.md
@@ -1,6 +1,6 @@
---
id: 587d7fa7367417b2b2512bc5
-title: 使用 D3 中的动态数据
+title: Work with Dynamic Data in D3
challengeType: 6
forumTopicId: 301498
dashedName: work-with-dynamic-data-in-d3
@@ -8,73 +8,73 @@ dashedName: work-with-dynamic-data-in-d3
# --description--
-最后两个挑战涉及到使用 D3 的 `data()` 和 `enter()` 方法来动态展示数据。它们以数据集为参数，和 `append()` 方法一起使用，为数据集中的每一个元素对象创建一个 DOM 元素。
+The last two challenges cover the basics of displaying data dynamically with D3 using the `data()` and `enter()` methods. These methods take a data set and, together with the `append()` method, create a new DOM element for each entry in the data set.
-在之前的挑战中，你为 `dataset` 数组中的每一个对象创建了一个新的 `h2` 元素，但是它们的文本都是相同的 "New Title"。 这是因为你还没有使用和每个 `h2` 元素关联的数据。
+In the previous challenge, you created a new `h2` element for each item in the `dataset` array, but they all contained the same text, "New Title". This is because you have not made use of the data that is bound to each of the `h2` elements.
-`text()` 方法以字符串或者回调函数作为参数：
+The D3 `text()` method can take a string or a callback function as an argument:
`selection.text((d) => d)`
-上面这个例子中的参数 `d` 指关联数据集的一个对象。
+In the example above, the parameter `d` refers to a single entry in the dataset that a selection is bound to.
-以当前例子为例，第一个 `h2` 元素和 12 关联，第二个 `h2` 元素和 31 关联，第三个 `h2` 元素和 22 关联，以此类推。
+Using the current example as context, the first `h2` element is bound to 12, the second `h2` element is bound to 31, the third `h2` element is bound to 22, and so on.
# --instructions--
-改变 `text()` 方法使得每个 `h2` 元素显示 `dataset` 数组中的对应值加上一个空格和 "USD"。例如，第一个标题应该为 "12 USD"。
+Change the `text()` method so that each `h2` element displays the corresponding value from the `dataset` array with a single space and "USD". For example, the first heading should be "12 USD".
# --hints--
-第一个 <code>h2<code> 的文本应该为 '12 USD'。
+The first `h2` should have the text "12 USD".
```js
assert($('h2').eq(0).text() == '12 USD');
```
-第二个 <code>h2<code> 的文本应该为 '31 USD'。
+The second `h2` should have the text "31 USD".
```js
assert($('h2').eq(1).text() == '31 USD');
```
-第三个 <code>h2<code> 的文本应该为 '22 USD'。
+The third `h2` should have the text "22 USD".
```js
assert($('h2').eq(2).text() == '22 USD');
```
-第四个 <code>h2<code> 的文本应该为 '17 USD'。
+The fourth `h2` should have the text "17 USD".
```js
assert($('h2').eq(3).text() == '17 USD');
```
-第五个 <code>h2<code> 的文本应该为 '25 USD'。
+The fifth `h2` should have the text "25 USD".
```js
assert($('h2').eq(4).text() == '25 USD');
```
-第六个 <code>h2<code> 的文本应该为 '18 USD'。
+The sixth `h2` should have the text "18 USD".
```js
assert($('h2').eq(5).text() == '18 USD');
```
-第七个 <code>h2<code> 的文本应该为 '29 USD'。
+The seventh `h2` should have the text "29 USD".
```js
assert($('h2').eq(6).text() == '29 USD');
```
-第八个 <code>h2<code> 的文本应该为 '14 USD'。
+The eighth `h2` should have the text "14 USD".
```js
assert($('h2').eq(7).text() == '14 USD');
```
-第九个 <code>h2<code> 的文本应该为 '9 USD'。
+The ninth `h2` should have the text "9 USD".
```js
assert($('h2').eq(8).text() == '9 USD');
diff --git a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/access-the-json-data-from-an-api.md b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/access-the-json-data-from-an-api.md
index 310e74d84c2..99a331e3ff5 100644
--- a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/access-the-json-data-from-an-api.md
+++ b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/access-the-json-data-from-an-api.md
@@ -1,6 +1,6 @@
---
id: 587d7fae367417b2b2512be4
-title: 访问来自 API 的 JSON 数据
+title: Access the JSON Data from an API
challengeType: 6
forumTopicId: 301499
dashedName: access-the-json-data-from-an-api
@@ -8,23 +8,23 @@ dashedName: access-the-json-data-from-an-api
# --description--
-在前面的挑战中，你了解了如何从 freeCodeCamp Cat Photo API 获取 JSON 数据。
+In the previous challenge, you saw how to get JSON data from the freeCodeCamp Cat Photo API.
-现在，你将进一步观察返回的数据来更好的了解 JSON 格式，回忆一下 JavaScript 中的一些符号：
+Now you'll take a closer look at the returned data to better understand the JSON format. Recall some notation in JavaScript:
-<blockquote>[ ] -> 方括号表示数组<br>{ } -> 大括号表示对象<br>" " -> 双引号表示字符串，它们还用于表示 JSON 中的键名</blockquote>
+<blockquote>[ ] -> Square brackets represent an array<br>{ } -> Curly brackets represent an object<br>" " -> Double quotes represent a string. They are also used for key names in JSON</blockquote>
-理解 API 返回数据的结构是必需的，它将影响你如何获取你所需的值。
+Understanding the structure of the data that an API returns is important because it influences how you retrieve the values you need.
-在右侧，点击 "Get Message" 按钮，将 freeCodeCamp Cat Photo API JSON 加载到页面中。
+On the right, click the "Get Message" button to load the freeCodeCamp Cat Photo API JSON into the HTML.
-你看到在 JSON 数据中的第一个和最后一个字符是中括号`[ ]`，这意味着返回的数据是一个数组。JSON 数据中的第二个符号是一个大括号`{`，这意味着是一个对象。再仔细看，你会发现有三个独立的对象，这个 JSON 数据是一个包含三个对象的数组，它们各自都包含了 cat photo 的信息。
+The first and last character you see in the JSON data are square brackets `[ ]`. This means that the returned data is an array. The second character in the JSON data is a curly `{` bracket, which starts an object. Looking closely, you can see that there are three separate objects. The JSON data is an array of three objects, where each object contains information about a cat photo.
-你之前了解了对象包含了用逗号分隔的 "键值对"。在 Cat Photo 示例中，第一个对象的`"id":0`"id" 是键，0 是其对应的值，类似的，"imageLink", "altText", 和 "codeNames" 都是对应的键。每个 cat photo 对象具有相同的键，但具有不同的值
+You learned earlier that objects contain "key-value pairs" that are separated by commas. In the Cat Photo example, the first object has `"id":0` where "id" is a key and 0 is its corresponding value. Similarly, there are keys for "imageLink", "altText", and "codeNames". Each cat photo object has these same keys, but with different values.
-在第一个对象中有一个有趣的 "键值对" 它是`"codeNames":["Juggernaut","Mrs. Wallace","ButterCup"]`。 "codeNames" 是键，它的值是包含三个字符串的数组。对象数组以及数组作为键可以作为值
+Another interesting "key-value pair" in the first object is `"codeNames":["Juggernaut","Mrs. Wallace","ButterCup"]`. Here "codeNames" is the key and its value is an array of three strings. It's possible to have arrays of objects as well as a key with an array as a value.
-记住如何访问数组和对象中的数据，数组使用括号表示法来访问项目的特定索引，对象使用括号或点表示法来访问给定属性的值。这是打印第一张 cat photo 的“altText”的示例 - 请注意，编辑器中解析的 JSON 数据保存在名为`json`的变量中：
+Remember how to access data in arrays and objects. Arrays use bracket notation to access a specific index of an item. Objects use either bracket or dot notation to access the value of a given property. Here's an example that prints the "altText" of the first cat photo - note that the parsed JSON data in the editor is saved in a variable called `json`:
```js
console.log(json[0].altText);
@@ -33,11 +33,11 @@ console.log(json[0].altText);
# --instructions--
-对于 "id" 是 2 的猫， 打印`codeNames`数组的第二个值到控制台，你应该在对象上使用括号或者点表示法（保存在变量`json`中）来访问该值。
+For the cat with the "id" of 2, print to the console the second value in the `codeNames` array. You should use bracket and dot notation on the object (which is saved in the variable `json`) to access the value.
# --hints--
-你的代码应该使用括号和点表示法来读取正确的代码名称，并将 'Loki' 打印到控制台。
+Your code should use bracket and dot notation to access the proper code name, and print "Loki" to the console.
```js
assert(
diff --git a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/change-text-with-click-events.md b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/change-text-with-click-events.md
index 9562e31d3d6..3ae5a6fe527 100644
--- a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/change-text-with-click-events.md
+++ b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/change-text-with-click-events.md
@@ -1,6 +1,6 @@
---
id: 587d7fad367417b2b2512be2
-title: 通过单击事件更改文本
+title: Change Text with click Events
challengeType: 6
forumTopicId: 301500
dashedName: change-text-with-click-events
@@ -8,21 +8,21 @@ dashedName: change-text-with-click-events
# --description--
-当点击事件发生时，你可以使用 JavaScript 更新 HTML 元素。
+When the click event happens, you can use JavaScript to update an HTML element.
-例如，当用户点击 "Get Message" 按钮时，它将改变类名`message`元素的文本为 "Here is the message"。
+For example, when a user clicks the "Get Message" button, it changes the text of the element with the class `message` to say "Here is the message".
-通过在点击事件内添加以下代码实现：
+This works by adding the following code within the click event:
`document.getElementsByClassName('message')[0].textContent="Here is the message";`
# --instructions--
-在`onclick`事件处理器中添加代码，改变`message`元素内的文字为 "Here is the message"。
+Add code inside the `onclick` event handler to change the text inside the `message` element to say "Here is the message".
# --hints--
-你的代码应该使用`document.getElementsByClassName`方法来选择类名为`message`的元素，然后将其`innerHTML`改为给定文字。
+Your code should use the `document.getElementsByClassName` method to select the element with class `message` and set its `textContent` to the given string.
```js
assert(
diff --git a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/convert-json-data-to-html.md b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/convert-json-data-to-html.md
index 0b5f37d1bf7..ac56debface 100644
--- a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/convert-json-data-to-html.md
+++ b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/convert-json-data-to-html.md
@@ -1,6 +1,6 @@
---
id: 587d7fae367417b2b2512be5
-title: 将 JSON 数据转换为 HTML
+title: Convert JSON Data to HTML
challengeType: 6
forumTopicId: 16807
dashedName: convert-json-data-to-html
@@ -8,15 +8,15 @@ dashedName: convert-json-data-to-html
# --description--
-现在你从 JSON API 获取了数据，可以在 HTML 中显示它们了。
+Now that you're getting data from a JSON API, you can display it in the HTML.
-既然 cat photo 对象都保存在数组里，你可以使用 `forEach` 方法来遍历它们。当你拿到每个对象时，你就可以修改 HTML 元素了。
+You can use a `forEach` method to loop through the data since the cat photo objects are held in an array. As you get to each item, you can modify the HTML elements.
-首先，通过 `let html = "";` 声明一个 html 变量。
+First, declare an html variable with `let html = "";`.
-接着，遍历 JSON，将用 `strong` 标签包裹的键名和后面跟着值的 html 元素添加给变量。当循环结束后渲染它。
+Then, loop through the JSON, adding HTML to the variable that wraps the key names in `strong` tags, followed by the value. When the loop is finished, you render it.
-这是执行此操作的代码：
+Here's the code that does this:
```js
let html = "";
@@ -30,13 +30,13 @@ json.forEach(function(val) {
});
```
-**注意：** 在本挑战中，你需要给页面添加新的 HTML 元素，不能依赖 `textContent` 方法，这个方法容易遭受跨站脚本攻击，可以用 `innerHTML` 来完成挑战。
+**Note:** For this challenge, you need to add new HTML elements to the page, so you cannot rely on `textContent`. Instead, you need to use `innerHTML`, which can make a site vulnerable to Cross-site scripting attacks.
# --instructions--
-添加一个 `forEach` 循环来遍历 JSON 数据，并创建 HTML 元素以显示它。
+Add a `forEach` method to loop over the JSON data and create the HTML elements to display it.
-以下是一个 JSON 示例：
+Here is some example JSON
```json
[
@@ -52,19 +52,19 @@ json.forEach(function(val) {
# --hints--
-你的代码应该将数据保存在 `html` 变量中。
+Your code should store the data in the `html` variable
```js
assert(code.match(/html\s+?(\+=|=\shtml\s\+)/g));
```
-你的代码应该使用 `forEach` 方法来遍历 API 中的 JSON 数据。
+Your code should use a `forEach` method to loop over the JSON data from the API.
```js
assert(code.match(/json\.forEach/g));
```
-你的代码应该用 `strong` 标签包裹键名。
+Your code should wrap the key names in `strong` tags.
```js
assert(code.match(/<strong>.+<\/strong>/g));
diff --git a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/get-geolocation-data-to-find-a-users-gps-coordinates.md b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/get-geolocation-data-to-find-a-users-gps-coordinates.md
index 7d12a673e39..45b0abe4283 100644
--- a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/get-geolocation-data-to-find-a-users-gps-coordinates.md
+++ b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/get-geolocation-data-to-find-a-users-gps-coordinates.md
@@ -1,6 +1,6 @@
---
id: 587d7faf367417b2b2512be8
-title: 根据地理位置数据找到用户的 GPS 坐标
+title: Get Geolocation Data to Find A User's GPS Coordinates
challengeType: 6
forumTopicId: 18188
dashedName: get-geolocation-data-to-find-a-users-gps-coordinates
@@ -8,15 +8,15 @@ dashedName: get-geolocation-data-to-find-a-users-gps-coordinates
# --description--
-你还能做一件很酷的事就是访问你用户当前的地理位置，每个浏览器都有内置的导航器，可以为你提供这些信息。
+Another cool thing you can do is access your user's current location. Every browser has a built in navigator that can give you this information.
-导航器会获取用户当前的经度和纬度。
+The navigator will get the user's current longitude and latitude.
-您将看到允许或阻止此站点了解您当前位置的提示。只要代码正确，挑战就可以以任何一种方式完成。
+You will see a prompt to allow or block this site from knowing your current location. The challenge can be completed either way, as long as the code is correct.
-通过选择允许，你将看到输出手机上的文本更改为你的纬度和经度
+By selecting allow, you will see the text on the output phone change to your latitude and longitude.
-这是执行此操作的代码：
+Here's code that does this:
```js
if (navigator.geolocation){
@@ -26,33 +26,33 @@ if (navigator.geolocation){
}
```
-首先，它检查`navigator.geolocation`对象是否存在。如果是，`getCurrentPosition`则调用该对象上的方法，该方法启动对用户位置的异步请求。如果请求成功，则运行方法中的回调函数。此函数`position`使用点表示法访问对象的纬度和经度值，并更新页面。
+First, it checks if the `navigator.geolocation` object exists. If it does, the `getCurrentPosition` method on that object is called, which initiates an asynchronous request for the user's position. If the request is successful, the callback function in the method runs. This function accesses the `position` object's values for latitude and longitude using dot notation and updates the HTML.
# --instructions--
-在`script`标记内添加示例代码以检查用户的当前位置并将其插入 HTML
+Add the example code inside the `script` tags to check a user's current location and insert it into the HTML.
# --hints--
-你的代码应该用于`navigator.geolocation`访问用户的当前位置。
+Your code should use `navigator.geolocation` to access the user's current location.
```js
assert(code.match(/navigator\.geolocation\.getCurrentPosition/g));
```
-你的代码应该用于`position.coords.latitude`显示用户的纬度位置。
+Your code should use `position.coords.latitude` to display the user's latitudinal location.
```js
assert(code.match(/position\.coords\.latitude/g));
```
-你的代码应该用于`position.coords.longitude`显示用户的经度位置。
+Your code should use `position.coords.longitude` to display the user's longitudinal location.
```js
assert(code.match(/position\.coords\.longitude/g));
```
-你应该在`data`div 元素中显示用户的位置。
+You should display the user's position within the `data` div element.
```js
assert(
diff --git a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/get-json-with-the-javascript-fetch-method.md b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/get-json-with-the-javascript-fetch-method.md
index f706874a071..776f077ac83 100644
--- a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/get-json-with-the-javascript-fetch-method.md
+++ b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/get-json-with-the-javascript-fetch-method.md
@@ -1,6 +1,6 @@
---
id: 5ccfad82bb2dc6c965a848e5
-title: 使用 JavaScript 的 fetch 方法获取 JSON
+title: Get JSON with the JavaScript fetch method
challengeType: 6
forumTopicId: 301501
dashedName: get-json-with-the-javascript-fetch-method
@@ -8,59 +8,59 @@ dashedName: get-json-with-the-javascript-fetch-method
# --description--
-请求外部数据的另一个方法是使用 JavaScript 的 `fetch()` 方法。它的作用和 XMLHttpRequest 一致，但是语法更易理解。
+Another way to request external data is to use the JavaScript `fetch()` method. It is equivalent to XMLHttpRequest, but the syntax is considered easier to understand.
-下面是使用 GET 请求 `/json/cats.json` 数据的例子。
+Here is the code for making a GET request to `/json/cats.json`
```js
fetch('/json/cats.json')
- .then(response => response.json())
- .then(data => {
- document.getElementById('message').innerHTML = JSON.stringify(data);
- })
+ .then(response => response.json())
+ .then(data => {
+ document.getElementById('message').innerHTML = JSON.stringify(data);
+ })
```
-逐行解释一下代码。
+Take a look at each piece of this code.
-第一行是发起请求。也就是说，`fetch(URL)` 向指定的 URL 发起 GET 请求。方法返回一个 Promise。
+The first line is the one that makes the request. So, `fetch(URL)` makes a GET request to the URL specified. The method returns a Promise.
-当 Promise 返回后，如果请求成功，会执行 `then` 方法，该方法把响应转换为 JSON 格式。
+After a Promise is returned, if the request was successful, the `then` method is executed, which takes the response and converts it to JSON format.
-`then` 方法返回的也是 Promise，会被下一个 `then` 方法捕获。第二个 `then` 方法传入的参数就是最终的 JSON 对象。
+The `then` method also returns a Promise, which is handled by the next `then` method. The argument in the second `then` is the JSON object you are looking for!
-接着，使用 `document.getElementById()` 选择将要接收数据的元素。然后插入请求返回的 JSON 对象创建的字符串修改元素的 HTML 代码。
+Now, it selects the element that will receive the data by using `document.getElementById()`. Then it modifies the HTML code of the element by inserting a string created from the JSON object returned from the request.
# --instructions--
-更新代码，创建一个 "GET" 请求向 freeCodeCamp Cat Photo API 请求数据。这次使用 `fetch` 方法而不是 `XMLHttpRequest`。
+Update the code to create and send a "GET" request to the freeCodeCamp Cat Photo API. But this time, using the `fetch` method instead of `XMLHttpRequest`.
# --hints--
-应该使用 `fetch` 来发起 GET 请求。
+Your code should make a GET request with `fetch`.
```js
assert(code.match(/fetch\s*\(\s*('|")\/json\/cats\.json\1\s*\)/g));
```
-应该在 `then` 里面将响应转换为 JSON。
+Your code should use `then` to convert the response to JSON.
```js
assert(
code.match(
- /\.then\s*\(\s*(response|\(\s*response\s*\))\s*=>\s*response\s*\.json\s*\(\s*\)\s*\)/g
+ /\.then\s*\(\s*\(?(?<var>\w+)\)?\s*=>\s*\k<var>\s*\.json\s*\(\s*\)\s*\)/g
)
);
```
-应该使用另一个 `then` 接收 `then` 转换的 JSON。
+Your code should use `then` to handle the data converted to JSON by the other `then`.
```js
assert(code.match(/\.then\s*\(\s*(data|\(\s*data\s*\))\s*=>\s*{[^}]*}\s*\)/g));
```
-代码应该选择 id 为 `message` 的元素然后把它的内部 HTML 改成 JSON data 的字符串。
+Your code should get the element with id `message` and change its inner HTML to the string of JSON data.
```js
assert(
diff --git a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/get-json-with-the-javascript-xmlhttprequest-method.md b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/get-json-with-the-javascript-xmlhttprequest-method.md
index 6800e174fbd..634902febc8 100644
--- a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/get-json-with-the-javascript-xmlhttprequest-method.md
+++ b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/get-json-with-the-javascript-xmlhttprequest-method.md
@@ -1,6 +1,6 @@
---
id: 587d7fae367417b2b2512be3
-title: 使用 XMLHttpRequest 方法获取 JSON
+title: Get JSON with the JavaScript XMLHttpRequest Method
challengeType: 6
forumTopicId: 301502
dashedName: get-json-with-the-javascript-xmlhttprequest-method
@@ -8,19 +8,19 @@ dashedName: get-json-with-the-javascript-xmlhttprequest-method
# --description--
-你还可以从外部链接请求数据，这就是 API 发挥作用的地方。
+You can also request data from an external source. This is where APIs come into play.
-请记住，API（或叫应用程序编程接口）是计算机用来互相通信的工具。你将学习如何通过 AJAX技术 从 API 获得的数据来更新 HTML。
+Remember that APIs - or Application Programming Interfaces - are tools that computers use to communicate with one another. You'll learn how to update HTML with the data we get from APIs using a technology called AJAX.
-大部分 web APIs 以 JSON 格式传输数据。JSON 是 JavaScript Object Notation 的简写。
+Most web APIs transfer data in a format called JSON. JSON stands for JavaScript Object Notation.
-JSON 语法与 JavaScript 对象字面符号非常相似，JSON 具有对象属性以及其当前值，夹在`{` 和 `}`之间。
+JSON syntax looks very similar to JavaScript object literal notation. JSON has object properties and their current values, sandwiched between a `{` and a `}`.
-这些属性及其值通常称为 "键值对"。
+These properties and their values are often referred to as "key-value pairs".
-但是，JSON 是由 API 以`bytes` 形式传输的，你的程序以`string`接受它。它们能转换成为 JavaScript 对象，但默认情况下它们不是 JavaScript 对象。 `JSON.parse`方法解析字符串并构造它描述的 JavaScript 对象。
+However, JSON transmitted by APIs are sent as `bytes`, and your application receives it as a `string`. These can be converted into JavaScript objects, but they are not JavaScript objects by default. The `JSON.parse` method parses the string and constructs the JavaScript object described by it.
-你可以从 freeCodeCamp 的 Cat Photo API 请求 JSON，以下是你可以在点击事件中添加的代码：
+You can request the JSON from freeCodeCamp's Cat Photo API. Here's the code you can put in your click event to do this:
```js
const req = new XMLHttpRequest();
@@ -32,21 +32,21 @@ req.onload = function(){
};
```
-这里介绍每行代码的作用，JavaScript `XMLHttpRequest` 对象具有许多用于传输数据的属性和方法。首先，创建一个`XMLHttpRequest`对象实例，并保存在`req`变量里 。 接着, `open` 方法初始化请求 - 此示例从 API 请求数据，因此是个 "GET" 请求。第二个参数 `open` 是你要从中请求数据的 API 的 URL。第三个参数是一个布尔值， `true`使其成为异步请求。 该`send`方法发送请求，最后，`onload`事件处理程序解析返回的数据并应用该`JSON.stringify`方法将JavaScript对象转换为字符串，然后将此字符串作为消息文本插入。
+Here's a review of what each piece is doing. The JavaScript `XMLHttpRequest` object has a number of properties and methods that are used to transfer data. First, an instance of the `XMLHttpRequest` object is created and saved in the `req` variable. Next, the `open` method initializes a request - this example is requesting data from an API, therefore is a "GET" request. The second argument for `open` is the URL of the API you are requesting data from. The third argument is a Boolean value where `true` makes it an asynchronous request. The `send` method sends the request. Finally, the `onload` event handler parses the returned data and applies the `JSON.stringify` method to convert the JavaScript object into a string. This string is then inserted as the message text.
# --instructions--
-更新代码以创建并向 freeCodeCamp Cat Photo API 发送 "GET" 请求。然后点击 "Get Message" 按钮，你的 AJAX 函数将使用 API 的原生 JSON 替换 "The message will go here" 的文本。
+Update the code to create and send a "GET" request to the freeCodeCamp Cat Photo API. Then click the "Get Message" button. Your AJAX function will replace the "The message will go here" text with the raw JSON output from the API.
# --hints--
-你的代码应该创建一个新的`XMLHttpRequest`。
+Your code should create a new `XMLHttpRequest`.
```js
assert(code.match(/new\s+?XMLHttpRequest\(\s*?\)/g));
```
-你的代码应使用该`open`方法初始化对 freeCodeCamp Cat Photo API 的 'GET' 请求。
+Your code should use the `open` method to initialize a "GET" request to the freeCodeCamp Cat Photo API.
```js
assert(
@@ -56,13 +56,13 @@ assert(
);
```
-你的代码应使用该`send`方法发送请求。
+Your code should use the `send` method to send the request.
```js
assert(code.match(/\.send\(\s*\)/g));
```
-你的代码应该有一个`onload`设置为函数的事件处理程序。
+Your code should have an `onload` event handler set to a function.
```js
assert(
@@ -70,13 +70,13 @@ assert(
);
```
-你的代码应该使用该`JSON.parse`方法来解析`responseText`。
+Your code should use the `JSON.parse` method to parse the `responseText`.
```js
assert(code.match(/JSON\s*\.parse\(\s*.*\.responseText\s*\)/g));
```
-你的代码应该用`message`获取元素，并将其内部 HTML 转换为 JSON 数据字符串。
+Your code should get the element with class `message` and change its inner HTML to the string of JSON data.
```js
assert(
diff --git a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/handle-click-events-with-javascript-using-the-onclick-property.md b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/handle-click-events-with-javascript-using-the-onclick-property.md
index 75984788321..5d57a07236d 100644
--- a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/handle-click-events-with-javascript-using-the-onclick-property.md
+++ b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/handle-click-events-with-javascript-using-the-onclick-property.md
@@ -1,6 +1,6 @@
---
id: 587d7fad367417b2b2512be1
-title: 使用 onclick 属性处理点击事件
+title: Handle Click Events with JavaScript using the onclick property
challengeType: 6
forumTopicId: 301503
dashedName: handle-click-events-with-javascript-using-the-onclick-property
@@ -8,7 +8,7 @@ dashedName: handle-click-events-with-javascript-using-the-onclick-property
# --description--
-你希望代码仅在页面完成加载后执行。为此，你可将名为`DOMContentLoaded`的 JavaScript 事件附加到文档中。以下是实现的代码：
+You want your code to execute only once your page has finished loading. For that purpose, you can attach a JavaScript event to the document called `DOMContentLoaded`. Here's the code that does this:
```js
document.addEventListener('DOMContentLoaded', function() {
@@ -16,7 +16,7 @@ document.addEventListener('DOMContentLoaded', function() {
});
```
-你可以在`DOMContentLoaded`函数内部添加事件处理方法。你可以添加`onclick`事件处理器，当用户点击 id 为`getMessage`的元素时会触发事件。添加以下代码：
+You can implement event handlers that go inside of the `DOMContentLoaded` function. You can implement an `onclick` event handler which triggers when the user clicks on the element with id `getMessage`, by adding the following code:
```js
document.getElementById('getMessage').onclick = function(){};
@@ -24,17 +24,17 @@ document.getElementById('getMessage').onclick = function(){};
# --instructions--
-在`DOMContentLoaded`函数内为 id 为`getMessage`的元素添加一个 click 事件处理器。
+Add a click event handler inside of the `DOMContentLoaded` function for the element with id of `getMessage`.
# --hints--
-你的代码应该用`document.getElementById`方法来选择`getMessage`元素。
+Your code should use the `document.getElementById` method to select the `getMessage` element.
```js
assert(code.match(/document\s*\.getElementById\(\s*?('|")getMessage\1\s*?\)/g));
```
-你的代码应该添加`onclick`事件处理器。
+Your code should add an `onclick` event handler.
```js
assert(typeof document.getElementById('getMessage').onclick === 'function');
diff --git a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/post-data-with-the-javascript-xmlhttprequest-method.md b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/post-data-with-the-javascript-xmlhttprequest-method.md
index ee277baad74..ef071a15ab3 100644
--- a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/post-data-with-the-javascript-xmlhttprequest-method.md
+++ b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/post-data-with-the-javascript-xmlhttprequest-method.md
@@ -1,6 +1,6 @@
---
id: 587d7faf367417b2b2512be9
-title: 使用 XMLHttpRequest 方法发送数据
+title: Post Data with the JavaScript XMLHttpRequest Method
challengeType: 6
forumTopicId: 301504
dashedName: post-data-with-the-javascript-xmlhttprequest-method
@@ -8,9 +8,9 @@ dashedName: post-data-with-the-javascript-xmlhttprequest-method
# --description--
-在前面的示例中，你通过外部资源获取数据。此外，你也可以将数据发送到外部资源，只要该资源支持 AJAX 请求并且你知道 URL。
+In the previous examples, you received data from an external resource. You can also send data to an external resource, as long as that resource supports AJAX requests and you know the URL.
-JavaScript 的`XMLHttpRequest`方法也用于将数据发布到服务器，这是个例子：
+JavaScript's `XMLHttpRequest` method is also used to post data to a server. Here's an example:
```js
const xhr = new XMLHttpRequest();
@@ -26,27 +26,27 @@ const body = JSON.stringify({ userName: userName, suffix: ' loves cats!' });
xhr.send(body);
```
-你之前已经见过这些方法。`open` 方法根据给定的外部资源的 URL 初始化一个 POST 请求，参数 `true` 表示请求是异步执行的。 `setRequestHeader` 方法设置了 HTTP 请求标头的值，该标头包含有关发送人和请求的信息。它必须在 `open` 方法之后、`send` 方法之前调用。它的两个参数表示标头的内容类型和标头数据将被设置成什么值。 接下来，`onreadystatechange` 事件监听器监听请求状态的更改。`readyState` 为 4 表示操作完成，`status` 为 201 表示请求成功，此时文档的 HTML 可以更新了。 最后，`send` 方法发送带有 `body` 值的请求，其中 `userName` 的值由用户在 `input` 字段中输入。
+You've seen several of these methods before. Here the `open` method initializes the request as a "POST" to the given URL of the external resource, and uses the `true` Boolean to make it asynchronous. The `setRequestHeader` method sets the value of an HTTP request header, which contains information about the sender and the request. It must be called after the `open` method, but before the `send` method. The two parameters are the name of the header and the value to set as the body of that header. Next, the `onreadystatechange` event listener handles a change in the state of the request. A `readyState` of 4 means the operation is complete, and a `status` of 201 means it was a successful request. The document's HTML can be updated. Finally, the `send` method sends the request with the `body` value, which the `userName` key was given by the user in the `input` field.
# --instructions--
-更新代码以创建并发送 "POST" 请求。然后在输入框中输入你的姓名，并点击 "Send Message"。你的 AJAX 函数会用服务器返回的数据替换 "Reply from Server will be here"。修改返回的请求结果，在你的名字前加上 " loves cats"。
+Update the code so it makes a "POST" request to the API endpoint. Then type your name in the input field and click "Send Message". Your AJAX function should replace "Reply from Server will be here." with data from the server. Format the response to display your name appended with " loves cats".
# --hints--
-你的代码应该创建一个新的 `XMLHttpRequest`。
+Your code should create a new `XMLHttpRequest`.
```js
assert(code.match(/new\s+?XMLHttpRequest\(\s*?\)/g));
```
-你的代码应该使用 `open` 方法初始化一个发送给服务器的 'POST' 请求。
+Your code should use the `open` method to initialize a "POST" request to the server.
```js
assert(code.match(/\.open\(\s*?('|")POST\1\s*?,\s*?url\s*?,\s*?true\s*?\)/g));
```
-你的代码应该使用 `setRequestHeader` 方法。
+Your code should use the `setRequestHeader` method.
```js
assert(
@@ -56,13 +56,13 @@ assert(
);
```
-你的代码应该有一个 `onreadystatechange` 的事件监听器。
+Your code should have an `onreadystatechange` event handler set to a function.
```js
assert(code.match(/\.onreadystatechange\s*?=/g));
```
-你的代码应该获取 class 为 `message` 的元素，并将它的 `textContent` 更改为 "`userName` loves cats"。
+Your code should get the element with class `message` and change its `textContent` to "`userName` loves cats"
```js
assert(
@@ -72,7 +72,7 @@ assert(
);
```
-你的代码应该使用 `send` 方法。
+Your code should use the `send` method.
```js
assert(code.match(/\.send\(\s*?body\s*?\)/g));
diff --git a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/pre-filter-json-to-get-the-data-you-need.md b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/pre-filter-json-to-get-the-data-you-need.md
index c320932b70d..c8a31451f28 100644
--- a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/pre-filter-json-to-get-the-data-you-need.md
+++ b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/pre-filter-json-to-get-the-data-you-need.md
@@ -1,6 +1,6 @@
---
id: 587d7fae367417b2b2512be7
-title: 预先过滤 JSON 以获得所需的数据
+title: Pre-filter JSON to Get the Data You Need
challengeType: 6
forumTopicId: 18257
dashedName: pre-filter-json-to-get-the-data-you-need
@@ -8,11 +8,11 @@ dashedName: pre-filter-json-to-get-the-data-you-need
# --description--
-如果你不希望渲染每张从 freeCodeCamp Cat Photo API 取回的猫照片，你可以在循环先预先过滤 JSON 数据。
+If you don't want to render every cat photo you get from the freeCodeCamp Cat Photo API, you can pre-filter the JSON before looping through it.
-鉴于 JSON 数据都存储在数组中，你可以使用`filter`方法过滤掉 "id" 键值为 1 的猫。
+Given that the JSON data is stored in an array, you can use the `filter` method to filter out the cat whose "id" key has a value of 1.
-这是执行此操作的代码：
+Here's the code to do this:
```js
json = json.filter(function(val) {
@@ -22,11 +22,11 @@ json = json.filter(function(val) {
# --instructions--
-将`filter`代码添加到 JSON 数据中来移除 "id" 值为 1 的猫。
+Add code to `filter` the json data to remove the cat with the "id" value of 1.
# --hints--
-你的代码应该使用`filter`方法。
+Your code should use the `filter` method.
```js
assert(code.match(/json\.filter/g));
diff --git a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/render-images-from-data-sources.md b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/render-images-from-data-sources.md
index 1c67301e5ce..e4a9372d586 100644
--- a/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/render-images-from-data-sources.md
+++ b/curriculum/challenges/chinese/04-data-visualization/json-apis-and-ajax/render-images-from-data-sources.md
@@ -1,6 +1,6 @@
---
id: 587d7fae367417b2b2512be6
-title: 渲染数据源的图像
+title: Render Images from Data Sources
challengeType: 6
forumTopicId: 18265
dashedName: render-images-from-data-sources
@@ -8,27 +8,27 @@ dashedName: render-images-from-data-sources
# --description--
-前几个挑战中表明，JSON 数组中的每个对象都包含一个`imageLink`键，其值为猫图像的 URL。
+The last few challenges showed that each object in the JSON array contains an `imageLink` key with a value that is the URL of a cat's image.
-当你遍历这些对象的时候，你可以使用`imageLink`属性在`img`元素中显示此图像。
+When you're looping through these objects, you can use this `imageLink` property to display this image in an `img` element.
-这是执行此操作的代码：
+Here's the code that does this:
`html += "<img src = '" + val.imageLink + "' " + "alt='" + val.altText + "'>";`
# --instructions--
-添加代码以在`img`标签中使用`imageLink`和`altText`属性。
+Add code to use the `imageLink` and `altText` properties in an `img` tag.
# --hints--
-你应该使用`imageLink`属性来显示图像。
+You should use the `imageLink` property to display the images.
```js
assert(code.match(/val\.imageLink/g));
```
-应该用 `altText` 做为图片的 alt 属性。
+You should use the `altText` for the alt attribute values of the images.
```js
assert(code.match(/val\.altText/g));
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/exercise-tracker.md b/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/exercise-tracker.md
index 02187267a45..32398aa57f4 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/exercise-tracker.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/exercise-tracker.md
@@ -1,6 +1,6 @@
---
id: 5a8b073d06fa14fcfde687aa
-title: 运动跟踪器
+title: Exercise Tracker
challengeType: 4
forumTopicId: 301505
dashedName: exercise-tracker
@@ -8,26 +8,28 @@ dashedName: exercise-tracker
# --description--
-构建一个功能类似于 <https://fuschia-custard.glitch.me/> 的 JavaScript 全栈应用。
+Build a full stack JavaScript app that is functionally similar to this: <https://exercise-tracker.freecodecamp.rocks/>. Working on this project will involve you writing your code using one of the following methods:
-在开发这个项目时，我们推荐你在 [Glitch](https://glitch.com/) 上编码。编码完成之后，你可以把应用主页的链接复制到屏幕的输入框中，测试你的代码是否能通过项目需求。当然你也可以基于其他的平台来完成自己的项目，只要提供一个公开的主页便于我们测试就行。
+- Clone [this GitHub repo](https://github.com/freeCodeCamp/boilerplate-project-exercisetracker/) and complete your project locally.
+- Use [our repl.it starter project](https://repl.it/github/freeCodeCamp/boilerplate-project-exercisetracker) to complete your project.
+- Use a site builder of your choice to complete the project. Be sure to incorporate all the files from our GitHub repo.
-参考示例：你可以通过 [这个链接](https://glitch.com/#!/import/github/freeCodeCamp/boilerplate-project-exercisetracker/) 访问在 Glitch 上的项目，或者从 GitHub 上 clone [这个仓库的代码](https://github.com/freeCodeCamp/boilerplate-project-exercisetracker/)。如果你使用 Glitch，请记住将项目链接保存到妥当的地方。
+When you are done, make sure a working demo of your project is hosted somewhere public. Then submit the URL to it in the `Solution Link` field. Optionally, also submit a link to your project's source code in the `GitHub Link` field.
# --hints--
-提供独立的项目, 而不是例程 url.
+You should provide your own project, not the example URL.
```js
(getUserInput) => {
const url = getUserInput('url');
assert(
- !new RegExp('.*/fuschia-custard.glitch.me.*').test(getUserInput('url'))
+ !/.*\/exercise-tracker\.freecodecamp\.rocks/.test(getUserInput('url'))
);
};
```
-通过指定 `/api/exercise/new-user` 的 username 参数来创建用户, 并且返回一个具有 username 和 \_id 的对象。
+You can `POST` to `/api/exercise/new-user` with form data `username` to create a new user. The returned response will be an object with `username` and `_id` properties.
```js
async (getUserInput) => {
@@ -47,7 +49,7 @@ async (getUserInput) => {
};
```
-通过指定 `api/exercise/users` 的参数为创建用户时相同的信息，来获取到一个用户的数组。
+You can make a `GET` request to `api/exercise/users` to get an array of all users. Each element in the array is an object containing a user's `username` and `_id`.
```js
async (getUserInput) => {
@@ -56,13 +58,15 @@ async (getUserInput) => {
if (res.ok) {
const data = await res.json();
assert.isArray(data);
+ assert.isString(data[0].username);
+ assert.isString(data[0]._id);
} else {
throw new Error(`${res.status} ${res.statusText}`);
}
};
```
-通过指定 `/api/exercise/add` 的 userId、description、duration 和 date（可选）参数, 去添加一条练习记录。 当没有传入 date 的时候，默认采用当前日期。 应用程序将返回添加了练习字段的用户对象。
+You can `POST` to `/api/exercise/add` with form data `userId=_id`, `description`, `duration`, and optionally `date`. If no date is supplied, the current date will be used. The response returned will be the user object with the exercise fields added.
```js
async (getUserInput) => {
@@ -98,7 +102,7 @@ async (getUserInput) => {
};
```
-通过指定 `/api/exercise/log` 的 userId 参数，可以获取任意用户的练习日志和总的练习次数。
+You can make a `GET` request to `/api/exercise/log` with a parameter of `userId=_id` to retrieve a full exercise log of any user. The returned response will be the user object with a `log` array of all the exercises added. Each log item has the `description`, `duration`, and `date` properties.
```js
async (getUserInput) => {
@@ -140,7 +144,48 @@ async (getUserInput) => {
};
```
-我还可以通过传递可选参数 from & to 或 limit 来检索任何用户的部分日志。(日期格式如： yyyy-mm-dd, limit = int)
+A request to a user's log (`/api/exercise/log`) returns an object with a `count` property representing the number of exercises returned.
+
+```js
+async (getUserInput) => {
+ const url = getUserInput('url');
+ const res = await fetch(url + '/api/exercise/new-user', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/x-www-form-urlencoded' },
+ body: `username=fcc_test_${Date.now()}`.substr(0, 29)
+ });
+ if (res.ok) {
+ const { _id, username } = await res.json();
+ const expected = {
+ username,
+ description: 'test',
+ duration: 60,
+ _id,
+ date: new Date().toDateString()
+ };
+ const addRes = await fetch(url + '/api/exercise/add', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/x-www-form-urlencoded' },
+ body: `userId=${_id}&description=${expected.description}&duration=${expected.duration}`
+ });
+ if (addRes.ok) {
+ const logRes = await fetch(url + `/api/exercise/log?userId=${_id}`);
+ if (logRes.ok) {
+ const { count } = await logRes.json();
+ assert(count);
+ } else {
+ throw new Error(`${logRes.status} ${logRes.statusText}`);
+ }
+ } else {
+ throw new Error(`${addRes.status} ${addRes.statusText}`);
+ }
+ } else {
+ throw new Error(`${res.status} ${res.statusText}`);
+ }
+};
+```
+
+You can add `from`, `to` and `limit` parameters to a `/api/exercise/log` request to retrieve part of the log of any user. `from` and `to` are dates in `yyyy-mm-dd` format. `limit` is an integer of how many logs to send back.
```js
async (getUserInput) => {
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/file-metadata-microservice.md b/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/file-metadata-microservice.md
index db45610171f..3c4c93f13fa 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/file-metadata-microservice.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/file-metadata-microservice.md
@@ -1,6 +1,6 @@
---
id: bd7158d8c443edefaeb5bd0f
-title: 文件元数据
+title: File Metadata Microservice
challengeType: 4
forumTopicId: 301506
dashedName: file-metadata-microservice
@@ -8,24 +8,73 @@ dashedName: file-metadata-microservice
# --description--
-构建一个功能类似于 <https://purple-paladin.glitch.me/> 的 JavaScript 全栈应用。
+Build a full stack JavaScript app that is functionally similar to this: <https://file-metadata-microservice.freecodecamp.rocks/>. Working on this project will involve you writing your code using one of the following methods:
-在开发这个项目时，我们推荐你在 [Glitch](https://glitch.com/) 上编码。编码完成之后，你可以把应用主页的链接复制到屏幕的输入框中，测试你的代码是否能通过项目需求。当然你也可以基于其他的平台来完成自己的项目，只要提供一个公开的主页便于我们测试就行。
+- Clone [this GitHub repo](https://github.com/freeCodeCamp/boilerplate-project-filemetadata/) and complete your project locally.
+- Use [our repl.it starter project](https://repl.it/github/freeCodeCamp/boilerplate-project-filemetadata) to complete your project.
+- Use a site builder of your choice to complete the project. Be sure to incorporate all the files from our GitHub repo.
-参考示例：你可以通过 [这个链接](https://glitch.com/#!/import/github/freeCodeCamp/boilerplate-project-filemetadata/) 访问在 Glitch 上的项目，或者从 GitHub 上 clone [这个仓库的代码](https://github.com/freeCodeCamp/boilerplate-project-filemetadata/)。如果你使用 Glitch，请记住将项目链接保存到妥当的地方。
+When you are done, make sure a working demo of your project is hosted somewhere public. Then submit the URL to it in the `Solution Link` field. Optionally, also submit a link to your projects source code in the `GitHub Link` field.
+
+# --instructions--
+
+**HINT:** You can use the `multer` npm package to handle file uploading.
# --hints--
-我可以提交包含文件上传的 FormData 对象。
+You should provide your own project, not the example URL.
```js
-
+(getUserInput) => {
+ assert(
+ !/.*\/file-metadata-microservice\.freecodecamp\.rocks/.test(
+ getUserInput('url')
+ )
+ );
+};
```
-当我提交某些内容时，我将在 JSON 响应中收到以 bytes（字节）为单位的文件大小。
+You can submit a form that includes a file upload.
```js
+async (getUserInput) => {
+ const site = await fetch(getUserInput('url'));
+ const data = await site.text();
+ const doc = new DOMParser().parseFromString(data, 'text/html');
+ assert(doc.querySelector('input[type="file"]'));
+};
+```
+The form file input field has the `name` attribute set to `upfile`.
+
+```js
+async (getUserInput) => {
+ const site = await fetch(getUserInput('url'));
+ const data = await site.text();
+ const doc = new DOMParser().parseFromString(data, 'text/html');
+ assert(doc.querySelector('input[name="upfile"]'));
+};
+```
+
+When you submit a file, you receive the file `name`, `type`, and `size` in bytes within the JSON response.
+
+```js
+async (getUserInput) => {
+ const formData = new FormData();
+ const fileData = await fetch(
+ 'https://cdn.freecodecamp.org/weather-icons/01d.png'
+ );
+ const file = await fileData.blob();
+ formData.append('upfile', file, 'icon');
+ const data = await fetch(getUserInput('url') + '/api/fileanalyse', {
+ method: 'POST',
+ body: formData
+ });
+ const parsed = await data.json();
+ assert.property(parsed, 'size');
+ assert.equal(parsed.name, 'icon');
+ assert.equal(parsed.type, 'image/png');
+};
```
# --solutions--
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/request-header-parser-microservice.md b/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/request-header-parser-microservice.md
index e82d679f74d..530d0283005 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/request-header-parser-microservice.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/request-header-parser-microservice.md
@@ -1,6 +1,6 @@
---
id: bd7158d8c443edefaeb5bdff
-title: 请求头解析器
+title: Request Header Parser Microservice
challengeType: 4
forumTopicId: 301507
dashedName: request-header-parser-microservice
@@ -8,15 +8,29 @@ dashedName: request-header-parser-microservice
# --description--
-构建一个功能类似于 <https://dandelion-roar.glitch.me/> 的 JavaScript 全栈应用。
+Build a full stack JavaScript app that is functionally similar to this: <https://request-header-parser-microservice.freecodecamp.rocks/>. Working on this project will involve you writing your code using one of the following methods:
-在开发这个项目时，我们推荐你在 [Glitch](https://glitch.com/) 上编码。编码完成之后，你可以把应用主页的链接复制到屏幕的输入框中，测试你的代码是否能通过项目需求。当然你也可以基于其他的平台来完成自己的项目，只要提供一个公开的主页便于我们测试就行。
+- Clone [this GitHub repo](https://github.com/freeCodeCamp/boilerplate-project-headerparser/) and complete your project locally.
+- Use [our repl.it starter project](https://repl.it/github/freeCodeCamp/boilerplate-project-headerparser) to complete your project.
+- Use a site builder of your choice to complete the project. Be sure to incorporate all the files from our GitHub repo.
-参考示例：你可以通过 [这个链接](https://glitch.com/#!/import/github/freeCodeCamp/boilerplate-project-headerparser/) 访问在 Glitch 上的项目，或者从 GitHub 上 clone [这个仓库的代码](https://github.com/freeCodeCamp/boilerplate-project-headerparser/)。如果你使用 Glitch，请记住将项目链接保存到妥当的地方。
+When you are done, make sure a working demo of your project is hosted somewhere public. Then submit the URL to it in the `Solution Link` field. Optionally, also submit a link to your project's source code in the `GitHub Link` field.
# --hints--
-我可以获得浏览器的IP地址、语言和操作系统信息。
+You should provide your own project, not the example URL.
+
+```js
+(getUserInput) => {
+ assert(
+ !/.*\/request-header-parser-microservice\.freecodecamp\.rocks/.test(
+ getUserInput('url')
+ )
+ );
+};
+```
+
+A request to `/api/whoami` should return a JSON object with your IP address in the `ipaddress` key.
```js
(getUserInput) =>
@@ -28,7 +42,7 @@ dashedName: request-header-parser-microservice
);
```
-首选语言应该在 `language` 键里返回。
+A request to `/api/whoami` should return a JSON object with your preferred language in the `language` key.
```js
(getUserInput) =>
@@ -40,7 +54,7 @@ dashedName: request-header-parser-microservice
);
```
-software 应该在 `software` 键里返回。
+A request to `/api/whoami` should return a JSON object with your software in the `software` key.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/timestamp-microservice.md b/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/timestamp-microservice.md
index 1acab03c390..df5332d7c67 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/timestamp-microservice.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/timestamp-microservice.md
@@ -1,6 +1,6 @@
---
id: bd7158d8c443edefaeb5bdef
-title: 时间戳
+title: Timestamp Microservice
challengeType: 4
forumTopicId: 301508
dashedName: timestamp-microservice
@@ -8,15 +8,27 @@ dashedName: timestamp-microservice
# --description--
-构建一个功能类似于 <https://curse-arrow.glitch.me/> 的 JavaScript 全栈应用。
+Build a full stack JavaScript app that is functionally similar to this: <https://timestamp-microservice.freecodecamp.rocks/>. Working on this project will involve you writing your code using one of the following methods:
-在开发这个项目时，我们推荐你在 [Glitch](https://glitch.com/) 上编码。编码完成之后，你可以把应用主页的链接复制到屏幕的输入框中，测试你的代码是否能通过项目需求。当然你也可以基于其他的平台来完成自己的项目，只要提供一个公开的主页便于我们测试就行。
+- Clone [this GitHub repo](https://github.com/freeCodeCamp/boilerplate-project-timestamp/) and complete your project locally.
+- Use [our repl.it starter project](https://repl.it/github/freeCodeCamp/boilerplate-project-timestamp) to complete your project.
+- Use a site builder of your choice to complete the project. Be sure to incorporate all the files from our GitHub repo.
-参考示例：你可以通过 [这个链接](https://glitch.com/#!/import/github/freeCodeCamp/boilerplate-project-timestamp/) 访问在 Glitch 上的项目，或者从 GitHub 上 clone [这个仓库的代码](https://github.com/freeCodeCamp/boilerplate-project-timestamp/)。如果你使用 Glitch，请记住将项目链接保存到妥当的地方。
+When you are done, make sure a working demo of your project is hosted somewhere public. Then submit the URL to it in the `Solution Link` field. Optionally, also submit a link to your projects source code in the `GitHub Link` field.
# --hints--
-当处理一个有效的日期，返回正确的 unix 时间戳。
+You should provide your own project, not the example URL.
+
+```js
+(getUserInput) => {
+ assert(
+ !/.*\/timestamp-microservice\.freecodecamp\.rocks/.test(getUserInput('url'))
+ );
+};
+```
+
+A request to `/api/timestamp/:date?` with a valid date should return a JSON object with a `unix` key that is a Unix timestamp of the input date in milliseconds
```js
(getUserInput) =>
@@ -34,7 +46,7 @@ dashedName: timestamp-microservice
);
```
-当处理一个有效的日期， 返回正确的 UTC 字符串。
+A request to `/api/timestamp/:date?` with a valid date should return a JSON object with a `utc` key that is a string of the input date in the format: `Thu, 01 Jan 1970 00:00:00 GMT`
```js
(getUserInput) =>
@@ -52,13 +64,16 @@ dashedName: timestamp-microservice
);
```
-当处理一个有效的 unix 格式的日期, 返回正确的 unix 时间戳。
+A request to `/api/timestamp/1451001600000` should return `{ unix: 1451001600000, utc: "Fri, 25 Dec 2015 00:00:00 GMT" }`
```js
(getUserInput) =>
- $.get(getUserInput('url') + '/api/timestamp/1482624000000').then(
+ $.get(getUserInput('url') + '/api/timestamp/1451001600000').then(
(data) => {
- assert.equal(data.unix, 1482624000000);
+ assert(
+ data.unix === 1451001600000 &&
+ data.utc === 'Fri, 25 Dec 2015 00:00:00 GMT'
+ );
},
(xhr) => {
throw new Error(xhr.responseText);
@@ -66,7 +81,24 @@ dashedName: timestamp-microservice
);
```
-这个程序应该返回无效日期的预期错误消息
+Your project can handle dates that can be successfully parsed by `new Date(date_string)`
+
+```js
+(getUserInput) =>
+ $.get(getUserInput('url') + '/api/timestamp/05 October 2011').then(
+ (data) => {
+ assert(
+ data.unix === 1317772800000 &&
+ data.utc === 'Wed, 05 Oct 2011 00:00:00 GMT'
+ );
+ },
+ (xhr) => {
+ throw new Error(xhr.responseText);
+ }
+ );
+```
+
+If the input date string is invalid, the api returns an object having the structure `{ error : "Invalid Date" }`
```js
(getUserInput) =>
@@ -80,7 +112,7 @@ dashedName: timestamp-microservice
);
```
-当处理一个空的日期参数的时候，返回当前时间的 unix 格式。
+An empty date parameter should return the current time in a JSON object with a `unix` key
```js
(getUserInput) =>
@@ -95,7 +127,7 @@ dashedName: timestamp-microservice
);
```
-当处理一个空的日期参数的时候，返回当前时间的 UTC 格式。
+An empty date parameter should return the current time in a JSON object with a `utc` key
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/url-shortener-microservice.md b/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/url-shortener-microservice.md
index 1f59d4b9aed..c5786a7a8c4 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/url-shortener-microservice.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/apis-and-microservices-projects/url-shortener-microservice.md
@@ -1,6 +1,6 @@
---
id: bd7158d8c443edefaeb5bd0e
-title: 短网址
+title: URL Shortener Microservice
challengeType: 4
forumTopicId: 301509
dashedName: url-shortener-microservice
@@ -8,30 +8,110 @@ dashedName: url-shortener-microservice
# --description--
-构建一个功能类似于 <https://thread-paper.glitch.me/> 的 JavaScript 全栈应用。
+Build a full stack JavaScript app that is functionally similar to this: <https://url-shortener-microservice.freecodecamp.rocks/>. Working on this project will involve you writing your code using one of the following methods:
-在开发这个项目时，我们推荐你在 [Glitch](https://glitch.com/) 上编码。编码完成之后，你可以把应用主页的链接复制到屏幕的输入框中，测试你的代码是否能通过项目需求。当然你也可以基于其他的平台来完成自己的项目，只要提供一个公开的主页便于我们测试就行。
+- Clone [this GitHub repo](https://github.com/freeCodeCamp/boilerplate-project-urlshortener/) and complete your project locally.
+- Use [our repl.it starter project](https://repl.it/github/freeCodeCamp/boilerplate-project-urlshortener) to complete your project.
+- Use a site builder of your choice to complete the project. Be sure to incorporate all the files from our GitHub repo.
-参考示例：你可以通过 [这个链接](https://glitch.com/#!/import/github/freeCodeCamp/boilerplate-project-urlshortener/) 访问在 Glitch 上的项目，或者从 GitHub 上 clone [这个仓库的代码](https://github.com/freeCodeCamp/boilerplate-project-urlshortener/)。如果你使用 Glitch，请记住将项目链接保存到妥当的地方。
+When you are done, make sure a working demo of your project is hosted somewhere public. Then submit the URL to it in the `Solution Link` field. Optionally, also submit a link to your projects source code in the `GitHub Link` field.
+
+# --instructions--
+
+**HINT:** Do not forget to use a body parsing middleware to handle the POST requests. Also, you can use the function `dns.lookup(host, cb)` from the `dns` core module to verify a submitted URL.
# --hints--
-当我传入一个 url 作为参数时，我将在JSON响应中收到缩短的URL。
+You should provide your own project, not the example URL.
```js
-
+(getUserInput) => {
+ assert(
+ !/.*\/url-shortener-microservice\.freecodecamp\.rocks/.test(
+ getUserInput('url')
+ )
+ );
+};
```
-如果我传入一个无效的链接，则会返回一个包含 “没有遵循如 `http://www.example.com` 的有效格式” 的错误信息的 JSON 响应。
+You can POST a URL to `/api/shorturl/new` and get a JSON response with `original_url` and `short_url` properties. Here's an example: `{ original_url : 'https://freeCodeCamp.org', short_url : 1}`
```js
-
+async (getUserInput) => {
+ const url = getUserInput('url');
+ const urlVariable = Date.now();
+ const res = await fetch(url + '/api/shorturl/new/', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/x-www-form-urlencoded' },
+ body: `url=https://timestamp-microservice.freecodecamp.rocks/api/timestamp/${urlVariable}`
+ });
+ if (res.ok) {
+ const { short_url, original_url } = await res.json();
+ assert.isNotNull(short_url);
+ assert.match(
+ original_url,
+ new RegExp(
+ `https://timestamp-microservice.freecodecamp.rocks/api/timestamp/${urlVariable}`
+ )
+ );
+ } else {
+ throw new Error(`${res.status} ${res.statusText}`);
+ }
+};
```
-当我访问这个短 URL 时, 将重定向到我原来的链接。
+When you visit `/api/shorturl/<short_url>`, you will be redirected to the original URL.
```js
+async (getUserInput) => {
+ const url = getUserInput('url');
+ const urlVariable = Date.now();
+ let shortenedUrlVariable;
+ const postResponse = await fetch(url + '/api/shorturl/new/', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/x-www-form-urlencoded' },
+ body: `url=https://timestamp-microservice.freecodecamp.rocks/api/timestamp/${urlVariable}`
+ });
+ if (postResponse.ok) {
+ const { short_url } = await postResponse.json();
+ shortenedUrlVariable = short_url;
+ } else {
+ throw new Error(`${postResponse.status} ${postResponse.statusText}`);
+ }
+ const getResponse = await fetch(
+ url + '/api/shorturl/' + shortenedUrlVariable
+ );
+ if (getResponse) {
+ const { redirected, url } = getResponse;
+ assert.isTrue(redirected);
+ assert.strictEqual(
+ url,
+ `https://timestamp-microservice.freecodecamp.rocks/api/timestamp/${urlVariable}`
+ );
+ } else {
+ throw new Error(`${getResponse.status} ${getResponse.statusText}`);
+ }
+};
+```
+If you pass an invalid URL that doesn't follow the valid `http://www.example.com` format, the JSON response will contain `{ error: 'invalid url' }`
+
+```js
+async (getUserInput) => {
+ const url = getUserInput('url');
+ const res = await fetch(url + '/api/shorturl/new/', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/x-www-form-urlencoded' },
+ body: `url=ftp:/john-doe.org`
+ });
+ if (res.ok) {
+ const { error } = await res.json();
+ assert.isNotNull(error);
+ assert.strictEqual(error.toLowerCase(), 'invalid url');
+ } else {
+ throw new Error(`${res.status} ${res.statusText}`);
+ }
+};
```
# --solutions--
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/chain-middleware-to-create-a-time-server.md b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/chain-middleware-to-create-a-time-server.md
index 8419a638f24..d68ab003f34 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/chain-middleware-to-create-a-time-server.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/chain-middleware-to-create-a-time-server.md
@@ -1,6 +1,6 @@
---
id: 587d7fb1367417b2b2512bf4
-title: 使用链中间件来创建时间服务器
+title: Chain Middleware to Create a Time Server
challengeType: 2
forumTopicId: 301510
dashedName: chain-middleware-to-create-a-time-server
@@ -8,9 +8,9 @@ dashedName: chain-middleware-to-create-a-time-server
# --description--
-使用`app.METHOD(path, middlewareFunction)`可以将中间件挂载到指定的路由。中间件也可以在路由定义中链接。
+Middleware can be mounted at a specific route using `app.METHOD(path, middlewareFunction)`. Middleware can also be chained inside route definition.
-请看以下示例：
+Look at the following example:
```js
app.get('/user', function(req, res, next) {
@@ -21,17 +21,17 @@ app.get('/user', function(req, res, next) {
});
```
-此方法可用于将服务操作拆分为较小的单元。这样可以让应用拥有更好的结构，以便于在不同的位置上复用代码。此方法还可用于对数据执行某些验证。在每一个中间件堆栈中，你都可以阻止当前链的执行，并将控制权传递给专门设计用于处理错误的函数。或者你可以将控制权传递给下一个匹配的路径，以处理特殊情况。我们将在高级 Express 章节中看到。
+This approach is useful to split the server operations into smaller units. That leads to a better app structure, and the possibility to reuse code in different places. This approach can also be used to perform some validation on the data. At each point of the middleware stack you can block the execution of the current chain and pass control to functions specifically designed to handle errors. Or you can pass control to the next matching route, to handle special cases. We will see how in the advanced Express section.
# --instructions--
-在路由`app.get('/now', ...)`链中，在中间件函数中，你应该在`req.time`里将当前时间添加到请求对象中。你可以使用`new Date().toString()`。在处理函数中，使用`{time: req.time}`结构的 JSON 对象来响应。
+In the route `app.get('/now', ...)` chain a middleware function and the final handler. In the middleware function you should add the current time to the request object in the `req.time` key. You can use `new Date().toString()`. In the handler, respond with a JSON object, taking the structure `{time: req.time}`.
-提示: 如果不链接中间件，测试将不能通过。如果将中间件函数挂载在其他地方，即使输出结果正确，测试也会失败。
+**Note:** The test will not pass if you don’t chain the middleware. If you mount the function somewhere else, the test will fail, even if the output result is correct.
# --hints--
-路由 /now 应该已经挂载了中间件
+The /now endpoint should have mounted middleware
```js
(getUserInput) =>
@@ -49,7 +49,7 @@ app.get('/user', function(req, res, next) {
);
```
-路由 /now 应该返回一个从现在开始 +/-20 秒的时间
+The /now endpoint should return a time that is +/- 20 secs from now
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/get-data-from-post-requests.md b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/get-data-from-post-requests.md
index 4581285918e..8398c13973f 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/get-data-from-post-requests.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/get-data-from-post-requests.md
@@ -1,6 +1,6 @@
---
id: 587d7fb2367417b2b2512bf8
-title: 从 POST 请求中获取数据
+title: Get Data from POST Requests
challengeType: 2
forumTopicId: 301511
dashedName: get-data-from-post-requests
@@ -8,27 +8,27 @@ dashedName: get-data-from-post-requests
# --description--
-在路径`/name`处挂载一个 POST 处理方法。和前面一样。我们已经在 html 首页准备了一份表格。它将提交与练习 10 相同的数据（查询字符串）。如果 body-parser 正确配置好了，你就可以在`req.body`对象中找到请求的参数。来看看一个常规的请求 /library 例子：
+Mount a POST handler at the path `/name`. It’s the same path as before. We have prepared a form in the html frontpage. It will submit the same data of exercise 10 (Query string). If the body-parser is configured correctly, you should find the parameters in the object `req.body`. Have a look at the usual library example:
<blockquote>route: POST '/library'<br>urlencoded_body: userId=546&bookId=6754 <br>req.body: {userId: '546', bookId: '6754'}</blockquote>
-和前面一样响应一个 JSON 对象`{name: 'firstname lastname'}`。你可以使用首页应用提供的 html 表单，来测试你的 API 是否正常工作。
+Respond with the same JSON object as before: `{name: 'firstname lastname'}`. Test if your endpoint works using the html form we provided in the app frontpage.
-提示: 除了 GET 和 POST，还有其他几种 http 方法。按照惯例，http 动词之间有对应关系，它们分别对应你在服务端执行的某种操作，传统的对应关系：
+Tip: There are several other http methods other than GET and POST. And by convention there is a correspondence between the http verb, and the operation you are going to execute on the server. The conventional mapping is:
-POST (有时候是 PUT) - 使用请求发送信息，以创建新资源，
+POST (sometimes PUT) - Create a new resource using the information sent with the request,
-GET - 读取已存在的资源，不用修改它，
+GET - Read an existing resource without modifying it,
-PUT 或者 PATCH (有时候是 POST) - 发送数据，以更新资源，
+PUT or PATCH (sometimes POST) - Update a resource using the data sent,
-DELETE => 删除一个资源。
+DELETE => Delete a resource.
-还有一些其他方法，常用于与服务进行交互。除了 GET 之外，上面列出的所有方法都可以负载数据（换言之，数据都能在请求体中找到）。也可以使用 body-parser 来正常工作。
+There are also a couple of other methods which are used to negotiate a connection with the server. Except from GET, all the other methods listed above can have a payload (i.e. the data into the request body). The body-parser middleware works with these methods as well.
# --hints--
-测试 1：你的 API 应该使用正确的名称来响应
+Test 1 : Your API endpoint should respond with the correct name
```js
(getUserInput) =>
@@ -46,7 +46,7 @@ DELETE => 删除一个资源。
);
```
-测试 2：你的 API 应该使用正确的名称来响应
+Test 2 : Your API endpoint should respond with the correct name
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/get-query-parameter-input-from-the-client.md b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/get-query-parameter-input-from-the-client.md
index 1121ce9a8b1..0d301a9f5e3 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/get-query-parameter-input-from-the-client.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/get-query-parameter-input-from-the-client.md
@@ -1,6 +1,6 @@
---
id: 587d7fb2367417b2b2512bf6
-title: 从客户端获取查询参数输入
+title: Get Query Parameter Input from the Client
challengeType: 2
forumTopicId: 301512
dashedName: get-query-parameter-input-from-the-client
@@ -8,19 +8,19 @@ dashedName: get-query-parameter-input-from-the-client
# --description--
-从客户端获取输入的另一种常见方式是使用查询字符串对路由路径中的数据进行编码。查询字符串使用标记 (?) 分隔，并且包含一对`field=value`。每一对键值使用符号 (&) 分隔。Express 能够从查询字符串中分析这些数据，并且把它放到`req.query`对象中。有些字符不能在出现在 URL 中，它们在发送前必须以[不同的格式](https://en.wikipedia.org/wiki/Percent-encoding)进行编码。如果你使用来自 JavaScript 的 API，你可以使用特定的方法来编码/解码这些字符。
+Another common way to get input from the client is by encoding the data after the route path, using a query string. The query string is delimited by a question mark (?), and includes field=value couples. Each couple is separated by an ampersand (&). Express can parse the data from the query string, and populate the object `req.query`. Some characters, like the percent (%), cannot be in URLs and have to be encoded in a different format before you can send them. If you use the API from JavaScript, you can use specific methods to encode/decode these characters.
<blockquote>route_path: '/library'<br>actual_request_URL: '/library?userId=546&bookId=6754' <br>req.query: {userId: '546', bookId: '6754'}</blockquote>
# --instructions--
-构建一个 API ，使用路由挂载`GET /name`。使用一个 JSON 文件来响应，它的结构是这样的：`{ name: 'firstname lastname'}`。名字和姓氏参数应该编码在查询参数中，举个例子：`?first=firstname&last=lastname`.
+Build an API endpoint, mounted at `GET /name`. Respond with a JSON document, taking the structure `{ name: 'firstname lastname'}`. The first and last name parameters should be encoded in a query string e.g. `?first=firstname&last=lastname`.
-提示: 在下面的练习中，我们将从相同的`/name`路由路径中 POST 请求接收数据。如果你愿意，你可以使用`app.route(path).get(handler).post(handler)`这中写法，此语法允许你在同一路径路由上链接不同的 HTTP 动词处理函数。可以节省一点打字时间，并且可以让代码看起来更清晰。
+**Note:** In the following exercise you are going to receive data from a POST request, at the same `/name` route path. If you want, you can use the method `app.route(path).get(handler).post(handler)`. This syntax allows you to chain different verb handlers on the same path route. You can save a bit of typing, and have cleaner code.
# --hints--
-测试 1：你的 API 应该使用正确的名称来响应
+Test 1 : Your API endpoint should respond with the correct name
```js
(getUserInput) =>
@@ -38,7 +38,7 @@ dashedName: get-query-parameter-input-from-the-client
);
```
-测试 2：你的 API 应该使用正确的名称来响应
+Test 2 : Your API endpoint should respond with the correct name
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/get-route-parameter-input-from-the-client.md b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/get-route-parameter-input-from-the-client.md
index 1ba66bf4d62..b0055f569ea 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/get-route-parameter-input-from-the-client.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/get-route-parameter-input-from-the-client.md
@@ -1,6 +1,6 @@
---
id: 587d7fb2367417b2b2512bf5
-title: 从客户端获取路由参数输入
+title: Get Route Parameter Input from the Client
challengeType: 2
forumTopicId: 301513
dashedName: get-route-parameter-input-from-the-client
@@ -8,17 +8,17 @@ dashedName: get-route-parameter-input-from-the-client
# --description--
-在构建 API 时，我们要让用户告诉我们他们想从服务中获取什么。举个例子，如果客户请求数据库中存储的用户信息，他们需要一种方法让我们知道他们对哪个用户感兴趣。实现这个需求的的方式就是使用路由参数。路由参数是由斜杠 (/) 分隔的 URL 命名段。每一小段能捕获与其位置匹配的 URL 部分的值。捕获的值能够在`req.params`对象中找到。
+When building an API, we have to allow users to communicate to us what they want to get from our service. For example, if the client is requesting information about a user stored in the database, they need a way to let us know which user they're interested in. One possible way to achieve this result is by using route parameters. Route parameters are named segments of the URL, delimited by slashes (/). Each segment captures the value of the part of the URL which matches its position. The captured values can be found in the `req.params` object.
<blockquote>route_path: '/user/:userId/book/:bookId'<br>actual_request_URL: '/user/546/book/6754' <br>req.params: {userId: '546', bookId: '6754'}</blockquote>
# --instructions--
-在路由中`GET /:word/echo`构建一个 echo 服务，响应一个采用`{echo: word}`结构的 JSON 对象。你可以在`req.params.word`中找到要重复的单词。你可以在浏览器的地址栏测试你的路由，访问一些匹配的路由，比如：your-app-rootpath/freecodecamp/echo
+Build an echo server, mounted at the route `GET /:word/echo`. Respond with a JSON object, taking the structure `{echo: word}`. You can find the word to be repeated at `req.params.word`. You can test your route from your browser's address bar, visiting some matching routes, e.g. `your-app-rootpath/freecodecamp/echo`.
# --hints--
-测试 1：你的 echo 服务应该正确地重复单词
+Test 1 : Your echo server should repeat words correctly
```js
(getUserInput) =>
@@ -36,7 +36,7 @@ dashedName: get-route-parameter-input-from-the-client
);
```
-测试 2：你的 echo 服务应该正确地重复单词
+Test 2 : Your echo server should repeat words correctly
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/implement-a-root-level-request-logger-middleware.md b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/implement-a-root-level-request-logger-middleware.md
index cd90a310ece..2a2faa35c39 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/implement-a-root-level-request-logger-middleware.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/implement-a-root-level-request-logger-middleware.md
@@ -1,6 +1,6 @@
---
id: 587d7fb1367417b2b2512bf3
-title: 实现一个根层的请求记录器中间件
+title: Implement a Root-Level Request Logger Middleware
challengeType: 2
forumTopicId: 301514
dashedName: implement-a-root-level-request-logger-middleware
@@ -8,9 +8,9 @@ dashedName: implement-a-root-level-request-logger-middleware
# --description--
-前面我们介绍了`express.static()`中间件函数。现在是时候更详细地了解什么是中间件了。中间件是一个接收 3 个参数的函数：请求对象、响应对象和在应用请求响应循环中的下一个函数。这些函数执行一些可能对应用程序产生副作用的代码，通常还会在请求对象或者响应对象里添加一些信息。当满足某些条件时，它们也可以结束发送响应的循环。如果它们没有发送响应，那么当它们完成时就会开始执行堆栈中的下一个函数。这将触发调用第 3 个参数`next()`。更多信息请查看 [express 文档](http://expressjs.com/en/guide/using-middleware.html)。
+Earlier, you were introduced to the `express.static()` middleware function. Now it’s time to see what middleware is, in more detail. Middleware functions are functions that take 3 arguments: the request object, the response object, and the next function in the application’s request-response cycle. These functions execute some code that can have side effects on the app, and usually add information to the request or response objects. They can also end the cycle by sending a response when some condition is met. If they don’t send the response when they are done, they start the execution of the next function in the stack. This triggers calling the 3rd argument, `next()`.
-看看下面的例子：
+Look at the following example:
```js
function(req, res, next) {
@@ -19,17 +19,17 @@ function(req, res, next) {
}
```
-假设我们在某个路由上安装了这个中间件函数。当一个请求与路由匹配时，它会显示字符串"我是中间件..."。然后它执行堆栈中的下一个函数。 在这个练习中，我们将构建根级中间件。正如我们在挑战 4 中看到的，要在根层级安装中间件函数，我们可以使用`app.use(<mware-function>)`方法。在这种情况下，该函数将对所有请求执行，但是你还是可以设置成更具体的条件来执行。举个例子，如果你希望某个函数只针对 POST 请求执行，可以使用`app.post(<mware-function>)`方法。所有 http 动作都有类似的方法，比如 GET、DELETE、PUT 等等。
+Let’s suppose you mounted this function on a route. When a request matches the route, it displays the string “I’m a middleware…”, then it executes the next function in the stack. In this exercise, you are going to build root-level middleware. As you have seen in challenge 4, to mount a middleware function at root level, you can use the `app.use(<mware-function>)` method. In this case, the function will be executed for all the requests, but you can also set more specific conditions. For example, if you want a function to be executed only for POST requests, you could use `app.post(<mware-function>)`. Analogous methods exist for all the HTTP verbs (GET, DELETE, PUT, …).
# --instructions--
-构建一个简单的日志记录器。对于每个请求，它应该在控制台中记录一个采用以下格式的字符串：`method path - ip`。一个简单的日志看起来就像这样：`GET /json - ::ffff:127.0.0.1`。注意`method`和`path`有一个空格，并且`path`和`ip`中间的破折号两边都有空格。在请求对象中，可以使用`req.method`、`req.path`和`req.ip`获取请求方法（http 动词）、路由相对路径和请求者的 IP 信息。记住，当你完成时，要调用`next()`方法，否则你的服务器将一直处于挂起状态。请确保 "Logs" 是打开的，观察一下当一些请求被响应时会发生什么……
+Build a simple logger. For every request, it should log to the console a string taking the following format: `method path - ip`. An example would look like this: `GET /json - ::ffff:127.0.0.1`. Note that there is a space between `method` and `path` and that the dash separating `path` and `ip` is surrounded by a space on both sides. You can get the request method (http verb), the relative route path, and the caller’s ip from the request object using `req.method`, `req.path` and `req.ip`. Remember to call `next()` when you are done, or your server will be stuck forever. Be sure to have the ‘Logs’ opened, and see what happens when some request arrives.
-提示: Express 按照函数在代码中出现的顺序来评估函数。中间件也是如此。如果你想让中间件函数适用于所有路由，那么应该在路由之前配置好中间件。
+**Note:** Express evaluates functions in the order they appear in the code. This is true for middleware too. If you want it to work for all the routes, it should be mounted before them.
# --hints--
-根级记录器中间件应该被激活
+Root level logger middleware should be active
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/meet-the-node-console.md b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/meet-the-node-console.md
index 21401bf4b0a..e85e90bda71 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/meet-the-node-console.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/meet-the-node-console.md
@@ -1,6 +1,6 @@
---
id: 587d7fb0367417b2b2512bed
-title: 认识 Node 的控制台
+title: Meet the Node console
challengeType: 2
forumTopicId: 301515
dashedName: meet-the-node-console
@@ -8,19 +8,21 @@ dashedName: meet-the-node-console
# --description--
-在开发过程中，能够随时看到代码的运行结果是非常重要的。Node 只是一个 JavaScript 环境。与客户端 JavaScript 一样，你可以使用控制台输出有用的调试信息。在本地计算机上，你可以在终端中输出调试信息。在 Glitch 上，你可以打开屏幕下方的日志。使用 "Logs" 按钮切换日志面板（在左上角，应用名称的下面）。
+During the development process, it is important to be able to check what’s going on in your code.
-我们建议在做这些挑战题时，保持日志面板处于打开状态。通过这些错误日志，你可能会发现这些错误的本质原因。
+Node is just a JavaScript environment. Like client side JavaScript, you can use the console to display useful debug information. On your local machine, you would see console output in a terminal. On Repl.it, a terminal is open in the right pane by default.
+
+We recommend to keep the terminal open while working at these challenges. By reading the output in the terminal, you can see any errors that may occur.
# --instructions--
-如果还没准备好，请阅读[介绍](/learn/apis-and-microservices/basic-node-and-express/)里面的指引，在 Glitch 里通过这个[链接](https://glitch.com/edit/#!/remix/clone-from-repo?REPO_URL=https://github.com/freeCodeCamp/boilerplate-express/)创建一个新项目。
+If you have not already done so, please read the instructions in [the introduction](/learn/apis-and-microservices/basic-node-and-express/) and start a new project on Repl.it using [this link](https://repl.it/github/freeCodeCamp/boilerplate-express).
-修改 `myApp.js` 文件，在控制台打印出经典的 "Hello World" 即可。
+Modify the `myApp.js` file to log "Hello World" to the console.
# --hints--
-控制台应该输出：`'Hello World'`
+`"Hello World"` should be in the console
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/serve-an-html-file.md b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/serve-an-html-file.md
index 7b746e1821f..1d618673227 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/serve-an-html-file.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/serve-an-html-file.md
@@ -1,6 +1,6 @@
---
id: 587d7fb0367417b2b2512bef
-title: 服务 HTML 文件
+title: Serve an HTML File
challengeType: 2
forumTopicId: 301516
dashedName: serve-an-html-file
@@ -8,7 +8,7 @@ dashedName: serve-an-html-file
# --description--
-我们可以使用`res.sendFile(path)`方法来响应一个文件。你可以把响应一个文件的方法放到路由处理程序中：`app.get('/', ...)`。在后台，这个方法会根据你想发送的文件的类型，设置适当的 headers 头信息来告诉浏览器如何处理它。然后它会读取并发送文件。此方法需要文件的绝对路径。我们建议你使用 Node.js 的全局变量`__dirname`来计算出这个文件的绝对路径。
+You can respond to requests with a file using the `res.sendFile(path)` method. You can put it inside the `app.get('/', ...)` route handler. Behind the scenes, this method will set the appropriate headers to instruct your browser on how to handle the file you want to send, according to its type. Then it will read and send the file. This method needs an absolute file path. We recommend you to use the Node global variable `__dirname` to calculate the path like this:
```js
absolutePath = __dirname + relativePath/file.ext
@@ -16,13 +16,13 @@ absolutePath = __dirname + relativePath/file.ext
# --instructions--
-要发送的文件是`/views/index.html`。在 app 中点击 "Show Live" 按钮，你会看到一个大的 HTML 标题（以及我们稍后将使用的表单…），目前它们还没有任何样式。
+Send the `/views/index.html` file as a response to GET requests to the `/` path. If you view your live app, you should see a big HTML heading (and a form that we will use later…), with no style applied.
-**注意:**你可以编辑上一个挑战的解题代码，或者创建一个新的挑战。如果你重写了之前的代码，请注意 Express 会从上到下重新解析对应的路由方法。它执行第一个匹配的路由处理方法。你必须注释掉前面的代码，否则服务器还是响应之前的字符串。
+**Note:** You can edit the solution of the previous challenge or create a new one. If you create a new solution, keep in mind that Express evaluates routes from top to bottom, and executes the handler for the first match. You have to comment out the preceding solution, or the server will keep responding with a string.
# --hints--
-app 应该响应 views/index.html 文件
+Your app should serve the file views/index.html
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/serve-json-on-a-specific-route.md b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/serve-json-on-a-specific-route.md
index ea1971bbf13..1e977061c7c 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/serve-json-on-a-specific-route.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/serve-json-on-a-specific-route.md
@@ -1,6 +1,6 @@
---
id: 587d7fb1367417b2b2512bf1
-title: 服务指定路由上的 JSON
+title: Serve JSON on a Specific Route
challengeType: 2
forumTopicId: 301517
dashedName: serve-json-on-a-specific-route
@@ -8,17 +8,17 @@ dashedName: serve-json-on-a-specific-route
# --description--
-HTML 服务器提供 HTML，API 服务器提供数据。<dfn>REST</dfn>（表现层状态转换）API 允许使用更简易的方式交换数据，从而不需要客户端知道服务器的实现细节。客户端只要知道需请求资源对应的 URL 是什么，以及需要对这个 URL 进行何种操作就够了。比如 GET 这个动作，就是从服务器上获取某些信息，它不会修改任何数据。如今，在 web 上传输数据的首选数据格式是 JSON。简言之，JSON 是一种将 JavaScript 对象表示为字符串的简易方式，因此可以很容易地传输这些数据。
+While an HTML server serves HTML, an API serves data. A <dfn>REST</dfn> (REpresentational State Transfer) API allows data exchange in a simple way, without the need for clients to know any detail about the server. The client only needs to know where the resource is (the URL), and the action it wants to perform on it (the verb). The GET verb is used when you are fetching some information, without modifying anything. These days, the preferred data format for moving information around the web is JSON. Simply put, JSON is a convenient way to represent a JavaScript object as a string, so it can be easily transmitted.
-我们来创建一个简单的 API，一个路径为`/json`且返回数据是 JSON 格式的路由，你可以像之前那样通过`app.get()`方法来做，然后在路由处理部分使用`res.json()`方法返回 JSON 格式的数据，这个方法可以接收一个配置对象。这个方法会结束请求响应循环（request-response loop），然后返回数据。`res.json()`将一个有效的 JavaScript 对象转化为字符串，然后会设置适当的头信息（headers）来告诉浏览器，这是一个 JSON 数据，最后返回给客户端进行处理。一个有效的对象通常是这种结构：`{key: data}`。数据可以是数字、字符串、嵌套对象或数组。也可以是变量或者函数返回值，在这种情况下，会以它们的执行结果为基准，再转成字符串。
+Let's create a simple API by creating a route that responds with JSON at the path `/json`. You can do it as usual, with the `app.get()` method. Inside the route handler, use the method `res.json()`, passing in an object as an argument. This method closes the request-response loop, returning the data. Behind the scenes, it converts a valid JavaScript object into a string, then sets the appropriate headers to tell your browser that you are serving JSON, and sends the data back. A valid object has the usual structure `{key: data}`. `data` can be a number, a string, a nested object or an array. `data` can also be a variable or the result of a function call, in which case it will be evaluated before being converted into a string.
# --instructions--
-当 GET 请求路由`/json`时，将对象`{"message": "Hello json"}`作为 JSON 格式返回给客户端。然后在浏览器里输入完整的 URL，比如`your-app-url/json`，就可以在屏幕上看到这个消息了。
+Serve the object `{"message": "Hello json"}` as a response, in JSON format, to GET requests to the `/json` route. Then point your browser to `your-app-url/json`, you should see the message on the screen.
# --hints--
-服务端`/json`应该返回一个 JSON 对象`{"message": "Hello json"}`
+The endpoint `/json` should serve the json object `{"message": "Hello json"}`
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/serve-static-assets.md b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/serve-static-assets.md
index bf01f6489eb..44bf643496c 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/serve-static-assets.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/serve-static-assets.md
@@ -1,6 +1,6 @@
---
id: 587d7fb0367417b2b2512bf0
-title: 服务静态资源
+title: Serve Static Assets
challengeType: 2
forumTopicId: 301518
dashedName: serve-static-assets
@@ -8,17 +8,17 @@ dashedName: serve-static-assets
# --description--
-HTML 服务器通常有一个或多个用户可以访问的目录。你可以将应用程序所需的静态资源 (样式表、脚本、图片) 放在那里。在 Express 中你可以使用中间件`express.static(path)`来设置此功能，它的参数就是静态资源文件的绝对路径。如果你不知道什么是中间件，也不用担心。我们稍后将详细讨论此事。一个最基本的中间件可以看做是一个函数，它拦截路由处理方法，并在里面添加了一点别的信息。使用`app.use(path, middlewareFunction)`方法来加载一个中间件。它的第一个参数是可选的，如果没设置第一个参数，那么应用的所有请求都会经过这个中间件处理。
+An HTML server usually has one or more directories that are accessible by the user. You can place there the static assets needed by your application (stylesheets, scripts, images). In Express, you can put in place this functionality using the middleware `express.static(path)`, where the `path` parameter is the absolute path of the folder containing the assets. If you don’t know what middleware is... don’t worry, we will discuss in detail later. Basically, middleware are functions that intercept route handlers, adding some kind of information. A middleware needs to be mounted using the method `app.use(path, middlewareFunction)`. The first `path` argument is optional. If you don’t pass it, the middleware will be executed for all requests.
# --instructions--
-使用`app.use()`来加载`express.static()`中间件，让所有的请求都能访问我们的静态资源目录。静态资源的绝对路径是`__dirname + /public`。
+Mount the `express.static()` middleware for all requests with `app.use()`. The absolute path to the assets folder is `__dirname + /public`.
-现在 app 就能正常返回一个样式文件了，访问 app 根路径时，静态资源目录的文件就会被加载进来，现在你的首页看起来应该好些了！
+Now your app should be able to serve a CSS stylesheet. From outside, the public folder will appear mounted to the root directory. Your front-page should look a little better now!
# --hints--
-应用的静态资源文件应该来自`/public`目录
+Your app should serve asset files from the `/public` directory
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/start-a-working-express-server.md b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/start-a-working-express-server.md
index b820868d206..9a587bfb403 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/start-a-working-express-server.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/start-a-working-express-server.md
@@ -1,6 +1,6 @@
---
id: 587d7fb0367417b2b2512bee
-title: 启动一个 Express 服务器
+title: Start a Working Express Server
challengeType: 2
forumTopicId: 301519
dashedName: start-a-working-express-server
@@ -8,11 +8,9 @@ dashedName: start-a-working-express-server
# --description--
-在 myApp.js 文件的前两行中，你可以看到创建一个 Express 应用对象是很简单的。这个 Express 对象有几个方法，在这些挑战中，我们将会学习它们更多的方法。一个基础的方法是`app.listen(port)`。它告诉服务器监听指定的端口，并且让这个服务处于运行状态。你可以在文件的底部看到它们。它在注释里面，因为出于测试原因，我们需要知道这个应用正在后台运行。你想要添加的所有代码，都放在这两部分之间。Glitch 将端口号存储在环境变量`process.env.PORT`中。它的值是`3000`。
+In the first two lines of the file `myApp.js`, you can see how easy it is to create an Express app object. This object has several methods, and you will learn many of them in these challenges. One fundamental method is `app.listen(port)`. It tells your server to listen on a given port, putting it in running state. For testing reasons, we need the app to be running in the background so we added this method in the `server.js` file for you.
-让我们在服务端输出第一个字符串！在 Express 中，路由采用这种结构：`app.METHOD(PATH, HANDLER)`。`METHOD`是小写的 http 方法。`PATH`是服务器上的相对路径（它是一个字符串，甚至是正则表达式）`HANDLER`是 Express 匹配路由时调用的处理函数。
-
-处理函数采用这种形式：`function(req, res) {...}`，在这个处理函数的参数中`req`是请求对象。`res`是响应对象。举个例子，处理函数：
+Let’s serve our first string! In Express, routes takes the following structure: `app.METHOD(PATH, HANDLER)`. METHOD is an http method in lowercase. PATH is a relative path on the server (it can be a string, or even a regular expression). HANDLER is a function that Express calls when the route is matched. Handlers take the form `function(req, res) {...}`, where req is the request object, and res is the response object. For example, the handler
```js
function(req, res) {
@@ -20,17 +18,17 @@ function(req, res) {
}
```
-将会响应一个字符串 'Response String'。
+will serve the string 'Response String'.
# --instructions--
-当 GET 请求根路由（ "/" ）时，使用`app.get()`方法端响应一个 "Hello Express" 字符串。
+Use the `app.get()` method to serve the string "Hello Express" to GET requests matching the `/` (root) path. Be sure that your code works by looking at the logs, then see the results in the preview if you are using Repl.it.
-**注意：** 在 Glitch 中点击 "Show Live" 按钮，通过查看日志确保你的代码正常运行，然后在浏览器中查看结果。
+**Note:** All the code for these lessons should be added in between the few lines of code we have started you off with.
# --hints--
-app 应该输出 "Hello Express" 字符串
+Your app should serve the string 'Hello Express'
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/use-body-parser-to-parse-post-requests.md b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/use-body-parser-to-parse-post-requests.md
index aa58e4225df..9e0c3a879bf 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/use-body-parser-to-parse-post-requests.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/use-body-parser-to-parse-post-requests.md
@@ -1,6 +1,6 @@
---
id: 587d7fb2367417b2b2512bf7
-title: 使用 body-parser 来解析POST请求
+title: Use body-parser to Parse POST Requests
challengeType: 2
forumTopicId: 301520
dashedName: use-body-parser-to-parse-post-requests
@@ -8,9 +8,9 @@ dashedName: use-body-parser-to-parse-post-requests
# --description--
-除了 GET 还有另一个常见的 http 动词，它是 POST。POST 是使用 HTML 表单发送客户端数据的默认方法。在 REST 规范中，POST 常用于发送数据，以便在数据库中创建新项目（新用户或新博客文章）。我们在这个项目中没有使用数据库，我们将学习如何处理 POST 请求。
+Besides GET, there is another common HTTP verb, it is POST. POST is the default method used to send client data with HTML forms. In REST convention, POST is used to send data to create new items in the database (a new user, or a new blog post). You don’t have a database in this project, but you are going to learn how to handle POST requests anyway.
-在这些类型的请求中，数据不会出现在 URL 中，而是隐藏在请求正文中。这也是 HTML 请求的一部分，被称为负载。因为 HTML 是基于文本的，你看不到数据，这并不意味着它们是加密的。HTTP POST 请求的原始内容如下所示：
+In these kind of requests, the data doesn’t appear in the URL, it is hidden in the request body. The body is a part of the HTTP request, also called the payload. Even though the data is not visible in the URL, this does not mean that it is private. To see why, look at the raw content of an HTTP POST request:
```http
POST /path/subpath HTTP/1.0
@@ -18,20 +18,21 @@ From: john@example.com
User-Agent: someBrowser/1.0
Content-Type: application/x-www-form-urlencoded
Content-Length: 20
+
name=John+Doe&age=25
```
-正如你所看到的，正文被编码成了查询字符串。这是 HTML 表单使用的默认格式。使用 Ajax，我们还可以使用 JSON 来处理具有更复杂结构的数据。还有另一种类型的编码：multipart/form-data。它用来上传二进制文件。 在本练习中，我们将使用网址编码 body。要解析来自 POST 请求的数据，你必须安装一个包：body-parser。这个包允许你使用一套可以解码不同格式数据的中间件，在[这里](https://github.com/expressjs/body-parser)查看文档。
+As you can see, the body is encoded like the query string. This is the default format used by HTML forms. With Ajax, you can also use JSON to handle data having a more complex structure. There is also another type of encoding: multipart/form-data. This one is used to upload binary files. In this exercise, you will use a urlencoded body. To parse the data coming from POST requests, you have to install the `body-parser` package. This package allows you to use a series of middleware, which can decode data in different formats.
# --instructions--
-在 package.json 中安装 body-parser 模块，然后在文件顶部 require 进来，用变量 bodyParser 保存它。
+Install the `body-parser` module in your `package.json`. Then, `require` it at the top of the file. Store it in a variable named `bodyParser`. The middleware to handle urlencoded data is returned by `bodyParser.urlencoded({extended: false})`. Pass to `app.use()` the function returned by the previous method call. As usual, the middleware must be mounted before all the routes which need it.
-处理 URL 编码数据通过中间件的`bodyParser.urlencoded({extended: false})`方法。`extended=false`是一个配置选项，告诉解析器使用经典编码。当你使用它时，值只能是字符串或者数组。继承版使用起来数据更加灵活，它比 JSON 更好。传递给`app.use()`上一次方法调用返回的函数。通常中间件必须挂载在所有需要它的路由之前。
+**Note:** `extended=false` is a configuration option that tells the parser to use the classic encoding. When using it, values can be only strings or arrays. The extended version allows more data flexibility, but it is outmatched by JSON.
# --hints--
-"body-parser" 中间件应该被挂载
+The 'body-parser' middleware should be mounted
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/use-the-.env-file.md b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/use-the-.env-file.md
index 586c2ba8957..ad83674c7ae 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/use-the-.env-file.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/basic-node-and-express/use-the-.env-file.md
@@ -1,6 +1,6 @@
---
id: 587d7fb1367417b2b2512bf2
-title: 使用 .env 文件
+title: Use the .env File
challengeType: 2
forumTopicId: 301521
dashedName: use-the--env-file
@@ -8,19 +8,19 @@ dashedName: use-the--env-file
# --description--
-`.env`文件是一个隐藏文件，用于将环境变量传给应用程序。这是一个私密文件，除了你之外没人可以访问它，它可以用来存储你想保持私有或者隐藏的数据。举个例子，可以存储第三方服务 API 密钥或者数据库 URI。你也可以使用它来存储配置选项。通过设置配置选项，你可以改变应用程序的行为，而无需重写一些代码。
+The `.env` file is a hidden file that is used to pass environment variables to your application. This file is secret, no one but you can access it, and it can be used to store data that you want to keep private or hidden. For example, you can store API keys from external services or your database URI. You can also use it to store configuration options. By setting configuration options, you can change the behavior of your application, without the need to rewrite some code.
-在应用程序中可以通过`process.env.VAR_NAME`访问到环境变量。`process.env`是 Node 程序中的一个全局对象，可以给这个变量传字符串。按照惯例，变量名都是大写的，单词之间用下划线隔开。`.env`是一个 shell 文件，因此不需要用给变量名和值加引号。还有一点需要注意，当你给变量赋值时，等号周围不能有空格，举个例子：`VAR_NAME=value`。通常来讲，每一个变量会单独定义在新的一行。
+The environment variables are accessible from the app as `process.env.VAR_NAME`. The `process.env` object is a global Node object, and variables are passed as strings. By convention, the variable names are all uppercase, with words separated by an underscore. The `.env` is a shell file, so you don’t need to wrap names or values in quotes. It is also important to note that there cannot be space around the equals sign when you are assigning values to your variables, e.g. `VAR_NAME=value`. Usually, you will put each variable definition on a separate line.
# --instructions--
-让我们添加一个环境变量作为配置选项。
+Let's add an environment variable as a configuration option.
-在`.env`文件中保存变量`MESSAGE_STYLE=uppercase`。它的作用是，告诉上一次挑战中的路由处理程序，当我们 GET 方法请求 /JSON 时，如果`process.env.MESSAGE_STYLE`的值为`uppercase`，那么返回的对象则应该是`{"message": "HELLO JSON"}`.
+Store the variable `MESSAGE_STYLE=uppercase` in the `.env` file. Then tell the GET `/json` route handler that you created in the last challenge to transform the response object’s message to uppercase if `process.env.MESSAGE_STYLE` equals `uppercase`. The response object should become `{"message": "HELLO JSON"}`.
# --hints--
-客户端响应`/json`的值，应该随着环境变量`MESSAGE_STYLE`的变化而改变
+The response of the endpoint `/json` should change according to the environment variable `MESSAGE_STYLE`
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/add-a-description-to-your-package.json.md b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/add-a-description-to-your-package.json.md
index 139e380c62e..d4f4da4d681 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/add-a-description-to-your-package.json.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/add-a-description-to-your-package.json.md
@@ -1,6 +1,6 @@
---
id: 587d7fb3367417b2b2512bfc
-title: 给 package.json 添加一个描述
+title: Add a Description to Your package.json
challengeType: 2
forumTopicId: 301522
dashedName: add-a-description-to-your-package-json
@@ -8,11 +8,11 @@ dashedName: add-a-description-to-your-package-json
# --description--
-一个好的 package.json 文件的下一部分就是 description 字段，通过简洁的文字来描述你的项目。
+The next part of a good package.json file is the `description` field; where a short, but informative description about your project belongs.
-如果你计划将来把这个包发布到 npm，请注意 description 字段的作用是告知用户这个包的用途，这样用户就可以决定是否要安装你发布的包。然而，这并不是描述信息的唯一使用场景：它也是一种很好的总结项目的方式，对于一个普通的 Node.js 项目来说，它可以帮助其它开发者、未来的维护者以及你自己快速地了解项目，这同样非常重要。
+If you some day plan to publish a package to npm, this is the string that should sell your idea to the user when they decide whether to install your package or not. However, that’s not the only use case for the description, it’s a great way to summarize what a project does. It’s just as important in any Node.js project to help other developers, future maintainers or even your future self understand the project quickly.
-无论你如何计划项目，都建议你使用描述信息。我们来添加类似这样的信息：
+Regardless of what you plan for your project, a description is definitely recommended. Here's an example:
```json
"description": "A project that does something awesome",
@@ -20,13 +20,13 @@ dashedName: add-a-description-to-your-package-json
# --instructions--
-在 Glitch 项目的 package.json 中添加描述。
+Add a `description` to the package.json file of your project.
-请记住使用（"）包裹字段名并且使用（,）分隔字段。
+**Note:** Remember to use double-quotes for field-names (") and commas (,) to separate fields.
# --hints--
-package.json 应该有一个有效的 'description' 键。
+package.json should have a valid "description" key
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/add-a-license-to-your-package.json.md b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/add-a-license-to-your-package.json.md
index a1907717dc9..d97b0134019 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/add-a-license-to-your-package.json.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/add-a-license-to-your-package.json.md
@@ -1,6 +1,6 @@
---
id: 587d7fb4367417b2b2512bfe
-title: 给 package.json 添加许可证
+title: Add a License to Your package.json
challengeType: 2
forumTopicId: 301523
dashedName: add-a-license-to-your-package-json
@@ -8,27 +8,21 @@ dashedName: add-a-license-to-your-package-json
# --description--
-license 字段是你告知用户允许他们拿这个项目干什么的地方。
+The `license` field is where you inform users of what they are allowed to do with your project.
-常见的开源协议是 MIT 和 BSD。如果你想了解更多适合你项目的许可证的信息，那么 `http://choosealicense.com` 是一个不错的网站。
-
-许可证信息并不是必须的。大多数国家的版权法会默认让你拥有自己创作的作品的所有权。但是，明确说明用户可以做什么和不能做什么会是一个很好的做法。
-
-以下是一个 license 字段的示例：
-
-\`\`\`json
+Some common licenses for open source projects include MIT and BSD. License information is not required, and copyright laws in most countries will give you ownership of what you create by default. However, it’s always a good practice to explicitly state what users can and can’t do. Here's an example of the license field:
+```json
"license": "MIT",
-
-\`\`\`
+```
# --instructions--
-在 Glitch 项目的 package.json 中填写合适的 license 字段。
+Fill the `license` field in the package.json file of your project as you find suitable.
# --hints--
-package.json 应该有一个有效的 'license' 键。
+package.json should have a valid "license" key
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/add-a-version-to-your-package.json.md b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/add-a-version-to-your-package.json.md
index 432db506b85..a5b3b8b2453 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/add-a-version-to-your-package.json.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/add-a-version-to-your-package.json.md
@@ -1,6 +1,6 @@
---
id: 587d7fb4367417b2b2512bff
-title: 给 package.json 添加版本号
+title: Add a Version to Your package.json
challengeType: 2
forumTopicId: 301525
dashedName: add-a-version-to-your-package-json
@@ -8,7 +8,7 @@ dashedName: add-a-version-to-your-package-json
# --description--
-在 package.json 中 version 和 name 是所必填的字段之一。version 字段描述了当前项目的版本。
+A `version` is one of the required fields of your package.json file. This field describes the current version of your project. Here's an example:
```json
"version": "1.2.0",
@@ -16,11 +16,11 @@ dashedName: add-a-version-to-your-package-json
# --instructions--
-在 Glitch 项目中的 package.json 中添加一个版本号。
+Add a `version` to the package.json file of your project.
# --hints--
-package.json 应该包含一个有效的 'version' 键。
+package.json should have a valid "version" key
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/add-keywords-to-your-package.json.md b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/add-keywords-to-your-package.json.md
index 6a6f3f2e48e..33c9742e8bc 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/add-keywords-to-your-package.json.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/add-keywords-to-your-package.json.md
@@ -1,6 +1,6 @@
---
id: 587d7fb4367417b2b2512bfd
-title: 给 package.json 添加关键词
+title: Add Keywords to Your package.json
challengeType: 2
forumTopicId: 301526
dashedName: add-keywords-to-your-package-json
@@ -8,23 +8,23 @@ dashedName: add-keywords-to-your-package-json
# --description--
-你可以在 keywords 字段中使用相关的关键字描述项目。
+The `keywords` field is where you can describe your project using related keywords. Here's an example:
```json
"keywords": [ "descriptive", "related", "words" ],
```
-正如你所见，这个字段的结构是一个由双引号字符串组成的数组。
+As you can see, this field is structured as an array of double-quoted strings.
# --instructions--
-在 Glitch 项目的 package.json 中，给 keywords 添加适当的字符串数组。
+Add an array of suitable strings to the `keywords` field in the package.json file of your project.
-关键词之一应该是 freecodecamp。
+One of the keywords should be "freecodecamp".
# --hints--
-package.json 应该有一个有效的 'keywords' 键。
+package.json should have a valid "keywords" key
```js
(getUserInput) =>
@@ -39,7 +39,7 @@ package.json 应该有一个有效的 'keywords' 键。
);
```
-'keywords' 应该是一个数组。
+"keywords" field should be an Array
```js
(getUserInput) =>
@@ -54,7 +54,7 @@ package.json 应该有一个有效的 'keywords' 键。
);
```
-'keywords' 中应该包含关键词 'freecodecamp'。
+"keywords" should include "freecodecamp"
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/expand-your-project-with-external-packages-from-npm.md b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/expand-your-project-with-external-packages-from-npm.md
index faf77155be5..18523e17aad 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/expand-your-project-with-external-packages-from-npm.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/expand-your-project-with-external-packages-from-npm.md
@@ -1,6 +1,6 @@
---
id: 587d7fb4367417b2b2512c00
-title: 使用 npm 的外部软件包扩展您的项目
+title: Expand Your Project with External Packages from npm
challengeType: 2
forumTopicId: 301527
dashedName: expand-your-project-with-external-packages-from-npm
@@ -8,9 +8,9 @@ dashedName: expand-your-project-with-external-packages-from-npm
# --description--
-使用包管理器的最大原因之一是它们强大的依赖管理特性。在新的计算机上开始一个项目时，不用手动确认你已安装所有的依赖，npm 会自动为你安装它们。但是 npm 如何准确的知道你项目需要哪些依赖呢？我们来看看 package.json 中 dependencies 这一部分。
+One of the biggest reasons to use a package manager, is their powerful dependency management. Instead of manually having to make sure that you get all dependencies whenever you set up a project on a new computer, npm automatically installs everything for you. But how can npm know exactly what your project needs? Meet the `dependencies` section of your package.json file.
-在 dependencies 这一部分，你的项目需要按照下面这种格式来存储这些依赖包：
+In this section, packages your project requires are stored using the following format:
```json
"dependencies": {
@@ -22,13 +22,13 @@ dashedName: expand-your-project-with-external-packages-from-npm
# --instructions--
-在 package.json 的 dependencies 字段中添加一个版本号为 2.14.0 的 moment 包。
+Add version "2.14.0" of the "moment" package to the `dependencies` field of your package.json file.
-Moment 是一个非常方便的库，它用来处理时间和日期。
+**Note:** Moment is a handy library for working with time and dates.
# --hints--
-'dependencies' 应该包含 'moment'。
+"dependencies" should include "moment"
```js
(getUserInput) =>
@@ -47,7 +47,7 @@ Moment 是一个非常方便的库，它用来处理时间和日期。
);
```
-'moment' 的版本应该是 '2.14.0'。
+"moment" version should be "2.14.0"
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/how-to-use-package.json-the-core-of-any-node.js-project-or-npm-package.md b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/how-to-use-package.json-the-core-of-any-node.js-project-or-npm-package.md
index 938bcd2b89c..6222d50f0f9 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/how-to-use-package.json-the-core-of-any-node.js-project-or-npm-package.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/how-to-use-package.json-the-core-of-any-node.js-project-or-npm-package.md
@@ -1,6 +1,6 @@
---
id: 587d7fb3367417b2b2512bfb
-title: 如何使用 package.json —— 任何 Node.js 项目或 npm 包的核心
+title: 'How to Use package.json, the Core of Any Node.js Project or npm Package'
challengeType: 2
forumTopicId: 301528
dashedName: how-to-use-package-json-the-core-of-any-node-js-project-or-npm-package
@@ -8,13 +8,11 @@ dashedName: how-to-use-package-json-the-core-of-any-node-js-project-or-npm-packa
# --description--
-package.json 是任何 Node.js 项目或 npm 包的中心。它存储项目的相关信息，就像 HTML 文档中的 <head> 区域是用来描述网页的通用信息（元信息）一样。package.json 由单个 JSON 对象组成，它以键值对的形式存储项目的信息。一个最小的 package.json 文件至少包含两个必须字段：name 和 version——但是提供有关项目的附加信息是一个更好的做法，在以后它可能对你的用户或者维护者有所帮助。
+The `package.json` file is the center of any Node.js project or npm package. It stores information about your project, similar to how the <head> section of an HTML document describes the content of a webpage. It consists of a single JSON object where information is stored in key-value pairs. There are only two required fields; "name" and "version", but it’s good practice to provide additional information about your project that could be useful to future users or maintainers.
-author 字段：
+If you look at the file tree of your project, you will find the package.json file on the top level of the tree. This is the file that you will be improving in the next couple of challenges.
-如果你转到之前设置的 Glitch 项目并查看屏幕左侧，你会看到一个文件树，你可以在其中查看项目中各种文件的概述。在文件树的末尾部分，你会看到 package.json——我们将在接下来的几个挑战中改进这个文件。
-
-这个文件中最常见的信息之一是 author 字段，它指定了谁是项目的创建者。它可以是字符串，也可以是带有联系人详细信息的对象。对于较大的项目，建议使用对象，但是在我们的例子中，一个简单的字符串就可以了，比如下面的例子。
+One of the most common pieces of information in this file is the `author` field. It specifies who created the project, and can consist of a string or an object with contact or other details. An object is recommended for bigger projects, but a simple string like the following example will do for this project.
```json
"author": "Jane Doe",
@@ -22,13 +20,13 @@ author 字段：
# --instructions--
-将你的名字添加到 Glitch 项目中，package.json 的 author 字段。
+Add your name as the `author` of the project in the package.json file.
-**注意：** 请记住，你正在编写 JSON 文件。所有的字段名必须使用双引号（"）包裹, 比如："author"，所有字段必须用逗号（,）分隔。
+**Note:** Remember that you’re writing JSON, so all field names must use double-quotes (") and be separated with a comma (,).
# --hints--
-package.json 应该有一个有效的 'author' 键。
+package.json should have a valid "author" key
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/manage-npm-dependencies-by-understanding-semantic-versioning.md b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/manage-npm-dependencies-by-understanding-semantic-versioning.md
index 7e6e61b1403..c1b6abac8a0 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/manage-npm-dependencies-by-understanding-semantic-versioning.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/manage-npm-dependencies-by-understanding-semantic-versioning.md
@@ -1,6 +1,6 @@
---
id: 587d7fb5367417b2b2512c01
-title: 通过理解语义版本化来管理 npm 依赖关系
+title: Manage npm Dependencies By Understanding Semantic Versioning
challengeType: 2
forumTopicId: 301529
dashedName: manage-npm-dependencies-by-understanding-semantic-versioning
@@ -8,23 +8,23 @@ dashedName: manage-npm-dependencies-by-understanding-semantic-versioning
# --description--
-在 package.json 中，dependencies 这部分的 npm 包的版本号遵循所谓的语义化版本（SemVer），它是一种软件版本控制的行业标准，旨在使管理依赖项更加容易。在 npm 上发布的库、框架或其它工具都应该使用语义化版本（SemVer），以便让使用该依赖包的用户能够在依赖包需要升级时，提前规划自己的项目需要为之做出的改动。
+`Versions` of the npm packages in the dependencies section of your package.json file follow what’s called Semantic Versioning (SemVer), an industry standard for software versioning aiming to make it easier to manage dependencies. Libraries, frameworks or other tools published on npm should use SemVer in order to clearly communicate what kind of changes projects can expect if they update.
-在开发使用外部依赖项的软件（大多数情况都是这样）时，了解语义化版本（SemVer）会很有用。有一天你会明白这些数字的含义，在项目中它可以避免你意外地引入一些非向下兼容的更改，同时也能避免“昨天还能好好的运行，今天就不行了”这种情况发生。
+Knowing SemVer can be useful when you develop software that uses external dependencies (which you almost always do). One day, your understanding of these numbers will save you from accidentally introducing breaking changes to your project without understanding why things that worked yesterday suddenly don’t work today. This is how Semantic Versioning works according to the official website:
```json
"package": "MAJOR.MINOR.PATCH"
```
-主版本号 MAJOR：当你做了不向下兼容的公共 API 修改， 次版本号 MINOR：当你添加了向下兼容的新功能， 修订号 PATCH：当你做了向下兼容的问题修正。 这意味着修订号是用来修复错误的，次版本号则是添加了新功能，但它们都没有破坏之前的功能。最后，主版本号的变更则是添加了对早期版本不兼容的更改。
+The MAJOR version should increment when you make incompatible API changes. The MINOR version should increment when you add functionality in a backwards-compatible manner. The PATCH version should increment when you make backwards-compatible bug fixes. This means that PATCHes are bug fixes and MINORs add new features but neither of them break what worked before. Finally, MAJORs add changes that won’t work with earlier versions.
# --instructions--
-在 package.json 中，修改 dependencies 里的 moment 的版本号，让它的主版本是 2，次版本号是 10，修订号是 2。
+In the dependencies section of your package.json file, change the `version` of moment to match MAJOR version 2, MINOR version 10 and PATCH version 2
# --hints--
-'dependencies' 应该包含 'moment'。
+"dependencies" should include "moment"
```js
(getUserInput) =>
@@ -43,7 +43,7 @@ dashedName: manage-npm-dependencies-by-understanding-semantic-versioning
);
```
-'moment' 的版本号应该是 '2.10.2'。
+"moment" version should be "2.10.2"
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/remove-a-package-from-your-dependencies.md b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/remove-a-package-from-your-dependencies.md
index 435dfe9ae08..b1b7b32731b 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/remove-a-package-from-your-dependencies.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/remove-a-package-from-your-dependencies.md
@@ -1,6 +1,6 @@
---
id: 587d7fb5367417b2b2512c04
-title: 从依赖项中删除包
+title: Remove a Package from Your Dependencies
challengeType: 2
forumTopicId: 301530
dashedName: remove-a-package-from-your-dependencies
@@ -8,21 +8,21 @@ dashedName: remove-a-package-from-your-dependencies
# --description--
-在 package.json 中 dependencies 这一部分，目前尝试了一些管理依赖的方式。你已经添加了一些外部的依赖包到项目中，甚至通过一些特殊的字符比如波浪符号（~）或者插入符号（^）来告诉 npm 你想要的版本类型。
+You have now tested a few ways you can manage dependencies of your project by using the package.json's dependencies section. You have also included external packages by adding them to the file and even told npm what types of versions you want, by using special characters such as the tilde or the caret.
-但是，如果想要删除不需要的依赖包，该怎么办呢？你可能已经猜到了——只需要删除 dependencies 中的键值对就行了。
+But what if you want to remove an external package that you no longer need? You might already have guessed it, just remove the corresponding key-value pair for that package from your dependencies.
-同样的方法也适用于删除 package.json 中的其它字段
+This same method applies to removing other fields in your package.json as well
# --instructions--
-删除 moment 依赖包。
+Remove the moment package from your dependencies.
-删除依赖包后，确保没有多余的逗号。
+**Note:** Make sure you have the right amount of commas after removing it.
# --hints--
-'dependencies' 不包含 'moment'。
+"dependencies" should not include "moment"
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/use-the-caret-character-to-use-the-latest-minor-version-of-a-dependency.md b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/use-the-caret-character-to-use-the-latest-minor-version-of-a-dependency.md
index 7a1b7627823..a39adcd533d 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/use-the-caret-character-to-use-the-latest-minor-version-of-a-dependency.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/use-the-caret-character-to-use-the-latest-minor-version-of-a-dependency.md
@@ -1,6 +1,6 @@
---
id: 587d7fb5367417b2b2512c03
-title: 用插入字符可以使用依赖项的最新次要版本
+title: Use the Caret-Character to Use the Latest Minor Version of a Dependency
challengeType: 2
forumTopicId: 301531
dashedName: use-the-caret-character-to-use-the-latest-minor-version-of-a-dependency
@@ -8,25 +8,25 @@ dashedName: use-the-caret-character-to-use-the-latest-minor-version-of-a-depende
# --description--
-和上一个挑战中我们学到的波浪符号（~）来安装最新的修订版依赖一样，使用插入符号（^）允许 npm 来安装功能更新。它们的不同之处在于：插入符号（^）允许次版本和修订版更新。
+Similar to how the tilde we learned about in the last challenge allows npm to install the latest PATCH for a dependency, the caret (`^`) allows npm to install future updates as well. The difference is that the caret will allow both MINOR updates and PATCHes.
-此刻，你项目中的 moment 依赖包的版本应该是 ~2.10.2，这意味着 npm 可以安装 2.10.x 版的 moment，如果使用插入符号（^）来替换版本号的前缀，那么 npm 可以安装的版本则是 2.x.x。
+Your current version of moment should be "~2.10.2" which allows npm to install to the latest 2.10.x version. If you were to use the caret (^) as a version prefix instead, npm would be allowed to update to any 2.x.x version.
```json
"package": "^1.3.8"
```
-这会更新 package 到任意的 1.x.x 版本。
+This would allow updates to any 1.x.x version of the package.
# --instructions--
-使用插入符号（^）为依赖项中的 moment 版本添加前缀，允许 npm 更新依赖包到任意向下兼容的新功能版。
+Use the caret (`^`) to prefix the version of moment in your dependencies and allow npm to update it to any new MINOR release.
-请注意，原本的版本号不用更改。
+**Note:** The version numbers themselves should not be changed.
# --hints--
-'dependencies' 应包含 'moment'。
+"dependencies" should include "moment"
```js
(getUserInput) =>
@@ -45,7 +45,7 @@ dashedName: use-the-caret-character-to-use-the-latest-minor-version-of-a-depende
);
```
-'moment' 的版本应匹配 '^2.x.x'。
+"moment" version should match "^2.x.x"
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/use-the-tilde-character-to-always-use-the-latest-patch-version-of-a-dependency.md b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/use-the-tilde-character-to-always-use-the-latest-patch-version-of-a-dependency.md
index 630bd475c74..564df74a9d4 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/use-the-tilde-character-to-always-use-the-latest-patch-version-of-a-dependency.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/managing-packages-with-npm/use-the-tilde-character-to-always-use-the-latest-patch-version-of-a-dependency.md
@@ -1,6 +1,6 @@
---
id: 587d7fb5367417b2b2512c02
-title: 用波浪字符始终使用依赖项的最新补丁版本
+title: Use the Tilde-Character to Always Use the Latest Patch Version of a Dependency
challengeType: 2
forumTopicId: 301532
dashedName: use-the-tilde-character-to-always-use-the-latest-patch-version-of-a-dependency
@@ -8,9 +8,9 @@ dashedName: use-the-tilde-character-to-always-use-the-latest-patch-version-of-a-
# --description--
-在最后一个挑战中，我们告诉 npm 只包含特定版本的依赖包。如果想让项目各个部分保持相互兼容，锁定依赖包版本是一个有效的办法。但是大多数情况下，我们并不希望错过依赖项的问题修复，因为它们通常包含重要的安全补丁，而且它们理论上也会兼容我们既有的代码。
+In the last challenge, you told npm to only include a specific version of a package. That’s a useful way to freeze your dependencies if you need to make sure that different parts of your project stay compatible with each other. But in most use cases, you don’t want to miss bug fixes since they often include important security patches and (hopefully) don’t break things in doing so.
-为了让 npm 依赖项更新到最新的修订版，你可以在依赖包的版本号前加一个波浪符号（~）。在 package.json 中，我们当前的 moment 依赖包更新规则是：仅使用特定版本（2.10.2），但我们想用它最新的 2.10.x 版本。
+To allow an npm dependency to update to the latest PATCH version, you can prefix the dependency’s version with the tilde (`~`) character. Here's an example of how to allow updates to any 1.3.x version.
```json
"package": "~1.3.8"
@@ -18,15 +18,15 @@ dashedName: use-the-tilde-character-to-always-use-the-latest-patch-version-of-a-
# --instructions--
-在 package.json 文件中，当前有关 npm 如何升级的规则是使用特定版本（2.10.2）。 但是现在，要允许使用最新的2.10.x版本。
+In the package.json file, your current rule for how npm may upgrade moment is to use a specific version (2.10.2). But now, you want to allow the latest 2.10.x version.
-在 dependencies 中，给 moment 的版本号添加波浪符号（~）前缀，允许 npm 将其更新为最新的修订版。
+Use the tilde (`~`) character to prefix the version of moment in your dependencies, and allow npm to update it to any new PATCH release.
-请注意，原本的版本号不用更改。
+**Note:** The version numbers themselves should not be changed.
# --hints--
-'dependencies' 应该包含 'moment'。
+"dependencies" should include "moment"
```js
(getUserInput) =>
@@ -45,7 +45,7 @@ dashedName: use-the-tilde-character-to-always-use-the-latest-patch-version-of-a-
);
```
-'moment' 的版本号应该是 '~2.10.2'。
+"moment" version should match "~2.10.2"
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/chain-search-query-helpers-to-narrow-search-results.md b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/chain-search-query-helpers-to-narrow-search-results.md
index 78473e2771b..12be099f8ab 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/chain-search-query-helpers-to-narrow-search-results.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/chain-search-query-helpers-to-narrow-search-results.md
@@ -1,6 +1,6 @@
---
id: 587d7fb9367417b2b2512c12
-title: 通过链式调用搜索查询辅助函数来缩小搜索结果
+title: Chain Search Query Helpers to Narrow Search Results
challengeType: 2
forumTopicId: 301533
dashedName: chain-search-query-helpers-to-narrow-search-results
@@ -8,24 +8,15 @@ dashedName: chain-search-query-helpers-to-narrow-search-results
# --description--
-如果不给 `Model.find()`（或者别的搜索方法）的最后一个参数传入回调函数, 查询将不会执行。你可以将查询条件存储在变量中供以后使用，我们也可以通过链式调用这类变量来构建新的查询字段。实际的数据库操作会在最终调用 `.exec()` 方法时执行。需要注意的是，你必须给 `exec()` 传一个回调方法。并且你必须把回调函数传给最后一个方法。Mongoose 为我们提供了许多查询辅助函数, 这里我们使用最著名的一种。
+If you don’t pass the callback as the last argument to `Model.find()` (or to the other search methods), the query is not executed. You can store the query in a variable for later use. This kind of object enables you to build up a query using chaining syntax. The actual db search is executed when you finally chain the method `.exec()`. You always need to pass your callback to this last method. There are many query helpers, here we'll use the most commonly used.
# --instructions--
-修改 `queryChain` 函数来用变量 `foodToSearch` 来查询喜欢指定食物的人。找出喜欢 `burrito` 的人并按 `name` 进行排序。同时，我们需要隐藏他们的 `age` 属性。结果应限制在两个 document 内。请链式调用 `.find()`、`.sort()`、`.limit()` 和 `.select()`，在最后调用 `.exec()` 并给它传入 `done(err, data)` 回调函数。
-
-## 拓展阅读
-
-如果你想进一步了解，你可以参看：
-
-- 索引( Indexes， 对查询效率非常重要 ),
-- Pre/Post hooks,
-- Validation,
-- Schema 虚拟、模型、静态和实例方法
+Modify the `queryChain` function to find people who like the food specified by the variable named `foodToSearch`. Sort them by `name`, limit the results to two documents, and hide their age. Chain `.find()`, `.sort()`, `.limit()`, `.select()`, and then `.exec()`. Pass the `done(err, data)` callback to `exec()`.
# --hints--
-链接查询辅助函数应该成功
+Chaining query helpers should succeed
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/create-a-model.md b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/create-a-model.md
index 2327b64226d..10158433697 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/create-a-model.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/create-a-model.md
@@ -1,6 +1,6 @@
---
id: 587d7fb6367417b2b2512c07
-title: 创建一个模型
+title: Create a Model
challengeType: 2
forumTopicId: 301535
dashedName: create-a-model
@@ -8,43 +8,43 @@ dashedName: create-a-model
# --description--
-**C**RUD 第一小节——CREATE
+**C**RUD Part I - CREATE
-首先，我们需要一个 Schema（架构、模式或纲要，详情请参阅[维基百科](https://en.wikipedia.org/wiki/Database_schema)），每一个 Schema 都对应一个 MongoDB 的 [collection](https://docs.mongodb.com/manual/reference/glossary/#term-collection)，并且在相应的 collection 里定义 [documents](https://docs.mongodb.com/manual/reference/glossary/#term-document) 的“样子”。 Schema 用于组成 Model（模型），我们甚至可以通过嵌套 Schema 来创建复杂的模型。但是这里我们只学习 Schema 的基础用法。 我们可以根据模型创建实例，模型实例化后的对象称为 documents。
+First of all we need a Schema. Each schema maps to a MongoDB collection. It defines the shape of the documents within that collection. Schemas are building block for Models. They can be nested to create complex models, but in this case we'll keep things simple. A model allows you to create instances of your objects, called documents.
-Repl.it 是一个真实的服务器。习惯上，我们会编写函数来处理外界与服务器中数据库的交互，这些函数会在特定事件发生（比如有人调用了我们的服务器 API）时执行。接下来的挑战题目即是以此为基础。`done()` 是一个回调函数，它的作用是在一个异步操作（比如对数据库进行插入、查询、更新或删除）执行完成时，通知我们可以继续执行后续的其它代码。这与 Node.js 中的处理方式十分类似，在 Node.js 中，我们会在（异步操作）成功时调用 `done(null, data)`，在失败时调用 `done(err)`。
+Repl.it is a real server, and in real servers the interactions with the database happen in handler functions. These functions are executed when some event happens (e.g. someone hits an endpoint on your API). We’ll follow the same approach in these exercises. The `done()` function is a callback that tells us that we can proceed after completing an asynchronous operation such as inserting, searching, updating, or deleting. It's following the Node convention, and should be called as `done(null, data)` on success, or `done(err)` on error.
-注意：与远程服务器进行交互时，我们需要考虑到发生错误的可能
+Warning - When interacting with remote services, errors may occur!
```js
-/* 示例 */
+/* Example */
-var someFunc = function(done) {
+const someFunc = function(done) {
//... do something (risky) ...
- if(error) return done(error);
+ if (error) return done(error);
done(null, result);
};
```
# --instructions--
-按下面的原型信息创建一个名为 `personSchema` 的 schema：
+Create a person schema called `personSchema` having this prototype:
```markup
-- Person 的原型 -
+- Person Prototype -
--------------------
name : string [required]
-age : number
+age : number
favoriteFoods : array of strings (*)
```
-采用 Mongoose 基础 schema 类型。你如果还想添加更多的键，就请使用 required 或 unique 等简单的验证器（validators），并设置默认值。详情请参考 [Mongoose 文档](http://mongoosejs.com/docs/guide.html)。
+Use the Mongoose basic schema types. If you want you can also add more fields, use simple validators like required or unique, and set default values. See the [Mongoose docs](http://mongoosejs.com/docs/guide.html).
-请从 `personSchema` 创建一个名为 `Person` 的 model。
+Now, create a model called `Person` from the `personSchema`.
# --hints--
-应当成功地通过 Mongoose schema 创建实例
+Creating an instance from a mongoose schema should succeed
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/create-and-save-a-record-of-a-model.md b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/create-and-save-a-record-of-a-model.md
index 8b10fd4724d..ae385b6b21d 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/create-and-save-a-record-of-a-model.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/create-and-save-a-record-of-a-model.md
@@ -1,6 +1,6 @@
---
id: 587d7fb6367417b2b2512c09
-title: 创建并保存模型记录
+title: Create and Save a Record of a Model
challengeType: 2
forumTopicId: 301536
dashedName: create-and-save-a-record-of-a-model
@@ -8,14 +8,14 @@ dashedName: create-and-save-a-record-of-a-model
# --description--
-在这个挑战中，你需要给之前创建的模型添加一条数据。
+In this challenge you will have to create and save a record of a model.
# --instructions--
-用我们在上一个挑战中写好的 Person 构造函数创建 document 实例：将包含 `name`、`age` 和 `favoriteFoods` 的对象传给构造函数。它们的数据类型必须符合我们在 Person 原型中定义的类型。然后在返回的 document 实例上调用方法 `document.save()`。同时，我们需要像之前提到过的那样，为它传一个回调函数。在 Mongoose 中，所有的 CRUD 方法接收的最后一个参数都是回调函数。
+Within the `createAndSavePerson` function, create a document instance using the `Person` model constructor you built before. Pass to the constructor an object having the fields `name`, `age`, and `favoriteFoods`. Their types must conform to the ones in the `personSchema`. Then, call the method `document.save()` on the returned document instance. Pass to it a callback using the Node convention. This is a common pattern; all the following CRUD methods take a callback function like this as the last argument.
```js
-/* 示例 */
+/* Example */
// ...
person.save(function(err, data) {
@@ -25,7 +25,7 @@ person.save(function(err, data) {
# --hints--
-应成功地创建数据，并保存到数据库
+Creating and saving a db item should succeed
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/create-many-records-with-model.create.md b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/create-many-records-with-model.create.md
index 7e129b3a0ab..08812efbd1b 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/create-many-records-with-model.create.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/create-many-records-with-model.create.md
@@ -1,6 +1,6 @@
---
id: 587d7fb7367417b2b2512c0a
-title: 使用model.create（）创建许多记录
+title: Create Many Records with model.create()
challengeType: 2
forumTopicId: 301537
dashedName: create-many-records-with-model-create
@@ -8,17 +8,17 @@ dashedName: create-many-records-with-model-create
# --description--
-在一些情况下，比如进行数据库初始化，你会需要创建很多 model 实例来用作初始数据。`Model.create()` 接受一组像 `[{name:'John', ...}, {...}, ...]` 的数组作为第一个参数，并将其保存到数据库。
+Sometimes you need to create many instances of your models, e.g. when seeding a database with initial data. `Model.create()` takes an array of objects like `[{name: 'John', ...}, {...}, ...]` as the first argument, and saves them all in the db.
# --instructions--
-修改 `createManyPeople` 方法，使用 `arrayOfPeople` 作为 `Model.create()` 的参数来创建多个 people 实例。
+Modify the `createManyPeople` function to create many people using `Model.create()` with the argument `arrayOfPeople`.
-**注意：** 你可以使用在上一个挑战中创建的 model 来完成当前挑战。
+**Note:** You can reuse the model you instantiated in the previous exercise.
# --hints--
-应当成功地一次性创建多个数据
+Creating many db items at once should succeed
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/delete-many-documents-with-model.remove.md b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/delete-many-documents-with-model.remove.md
index 9cc9360ca98..7709b5fd5d3 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/delete-many-documents-with-model.remove.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/delete-many-documents-with-model.remove.md
@@ -1,6 +1,6 @@
---
id: 587d7fb8367417b2b2512c11
-title: 使用 model.remove() 删除多个 document
+title: Delete Many Documents with model.remove()
challengeType: 2
forumTopicId: 301538
dashedName: delete-many-documents-with-model-remove
@@ -8,24 +8,18 @@ dashedName: delete-many-documents-with-model-remove
# --description--
-`Model.remove()` 可以用于删除符合给定匹配条件的所有 document。
+`Model.remove()` is useful to delete all the documents matching given criteria.
# --instructions--
-如果想要删除所有叫 "Mary" 的人，我们可以使用 `Model.remove()`，并给它传入一个包含 `name` 字段的对象作为查询条件。当然，我们还需要给它传入一个回调函数。
+Modify the `removeManyPeople` function to delete all the people whose name is within the variable `nameToRemove`, using `Model.remove()`. Pass it to a query document with the `name` field set, and a callback.
-**注意：** `Model.remove()` 不会返回被删除的 document，而是会返回一个包含操作结果以及受影响的数据数量的 JSON 对象。不要忘记将它传入 `done()` 回调，因为我们需要在挑战的测试中调用它。
+**Note:** The `Model.remove()` doesn’t return the deleted document, but a JSON object containing the outcome of the operation, and the number of items affected. Don’t forget to pass it to the `done()` callback, since we use it in tests.
# --hints--
Deleting many items at once should succeed
-```js
-
-```
-
-应成功地删除多条数据
-
```js
(getUserInput) =>
$.ajax({
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/delete-one-document-using-model.findbyidandremove.md b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/delete-one-document-using-model.findbyidandremove.md
index 94df108e16a..f010b51c40e 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/delete-one-document-using-model.findbyidandremove.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/delete-one-document-using-model.findbyidandremove.md
@@ -1,6 +1,6 @@
---
id: 587d7fb8367417b2b2512c10
-title: 使用 model.findByIdAndRemove 删除一个 document
+title: Delete One Document Using model.findByIdAndRemove
challengeType: 2
forumTopicId: 301539
dashedName: delete-one-document-using-model-findbyidandremove
@@ -8,11 +8,15 @@ dashedName: delete-one-document-using-model-findbyidandremove
# --description--
-如果想在我们的数据库中通过 `_id` 删除一个人的数据，我们可以使用 `findByIdAndRemove()` 或 `findOneAndRemove()` 方法，它的使用方式和上面的更新方法很像。在这个挑战中，请使用 `personId` 找到对应的数据并把它删除。
+`findByIdAndRemove` and `findOneAndRemove` are like the previous update methods. They pass the removed document to the db. As usual, use the function argument `personId` as the search key.
+
+# --instructions--
+
+Modify the `removeById` function to delete one person by the person's `_id`. You should use one of the methods `findByIdAndRemove()` or `findOneAndRemove()`.
# --hints--
-应当成功地删除一条数据
+Deleting an item should succeed
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/install-and-set-up-mongoose.md b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/install-and-set-up-mongoose.md
index 58b04088241..2067e13d909 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/install-and-set-up-mongoose.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/install-and-set-up-mongoose.md
@@ -1,6 +1,6 @@
---
id: 587d7fb6367417b2b2512c06
-title: 安装和设置 Mongoose
+title: Install and Set Up Mongoose
challengeType: 2
forumTopicId: 301540
dashedName: install-and-set-up-mongoose
@@ -8,15 +8,19 @@ dashedName: install-and-set-up-mongoose
# --description--
-在 package.json 文件中添加 mongodb 和 mongoose 作为项目依赖，然后引入 Mongoose。之后，将 MongoDB Atlas 的 URI 作为 MONGO_URI 字段存储在私有的 `.env` 文件中。然后使用单引号或双引号包裹 URI，最后通过以下的代码片段来连接数据库：
+In this challenge, you will import the required projects, and connect to your Atlas database.
+
+# --instructions--
+
+Add `mongodb` and `mongoose` to the project’s `package.json`. Then, require mongoose as `mongoose` in `myApp.js`. Store your MongoDB Atlas database URI in a private `.env` file as `MONGO_URI`. Surround the the URI with single or double quotes, and make sure no space exists between both the variable and the `=`, and the value and `=`. Connect to the database using the following syntax:
```js
-mongoose.connect(<Your URI>, { useNewUrlParser: true, useUnifiedTopology: true });
+mongoose.connect(<Your URI>, { useNewUrlParser: true, useUnifiedTopology: true });
```
# --hints--
-"mongodb" 应在 package.json 中作为依赖项定义
+"mongodb" dependency should be in package.json
```js
(getUserInput) =>
@@ -31,7 +35,7 @@ mongoose.connect(<Your URI>, { useNewUrlParser: true, useUnifiedTopology: true }
);
```
-"mongoose" 应在 package.json 中作为依赖项定义
+"mongoose" dependency should be in package.json
```js
(getUserInput) =>
@@ -46,7 +50,7 @@ mongoose.connect(<Your URI>, { useNewUrlParser: true, useUnifiedTopology: true }
);
```
-应使用 "mongoose" 连接数据库
+"mongoose" should be connected to a database
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/perform-classic-updates-by-running-find-edit-then-save.md b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/perform-classic-updates-by-running-find-edit-then-save.md
index 322b785d470..ecbb514dac6 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/perform-classic-updates-by-running-find-edit-then-save.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/perform-classic-updates-by-running-find-edit-then-save.md
@@ -1,6 +1,6 @@
---
id: 587d7fb8367417b2b2512c0e
-title: 通过执行查找、编辑、保存来执行更新
+title: 'Perform Classic Updates by Running Find, Edit, then Save'
challengeType: 2
forumTopicId: 301541
dashedName: perform-classic-updates-by-running-find-edit-then-save
@@ -8,17 +8,17 @@ dashedName: perform-classic-updates-by-running-find-edit-then-save
# --description--
-在传统的应用中，如果想要编辑 document 并在其它地方使用它（比如放到服务器的返回数据中），就必须执行查找、编辑和保存。Mongoose 有一个专用的更新方法 `Model.update()`，它与底层的 mongo 驱动绑定。通过这个方法，我们可以批量编辑符合特定条件的多个 document。但问题在于，这个方法不会返回更新后的 document，而是返回状态信息。此外，由于它是直接调用 mongo 的底层驱动，因此它让处理 model 的验证变得更加棘手。
+In the good old days, this was what you needed to do if you wanted to edit a document, and be able to use it somehow (e.g. sending it back in a server response). Mongoose has a dedicated updating method: `Model.update()`. It is bound to the low-level mongo driver. It can bulk-edit many documents matching certain criteria, but it doesn’t send back the updated document, only a 'status' message. Furthermore, it makes model validations difficult, because it just directly calls the mongo driver.
# --instructions--
-在这个挑战中，请使用参数 `personId` 作为字段，在数据库中通过 `_id` 找到相应的 person（你可以使用之前挑战中的任何一种方法）。然后，将 "hamburger" 添加到它的 `favoriteFoods` 中（你可以使用 `Array.push()`）。之后，在 `.find()` 的回调里通过 `.save()` 方法更新数据库。
+Modify the `findEditThenSave` function to find a person by `_id` (use any of the above methods) with the parameter `personId` as search key. Add `"hamburger"` to the list of the person's `favoriteFoods` (you can use `Array.push()`). Then - inside the find callback - `save()` the updated `Person`.
-**提示：** 如果你在 Schema 中将 `favoriteFoods` 声明为一个 Array（数组）并且没有指定数组的类型(如 `[String]`)，那么此时，`favoriteFoods` 就会是默认的 Mixed 类型。如果想编辑它，就必须执行 `document.markModified('edited-field')`。请参阅 [Mongoose 文档](https://mongoosejs.com/docs/schematypes.html#Mixed)
+**Note:** This may be tricky, if in your Schema, you declared `favoriteFoods` as an Array, without specifying the type (i.e. `[String]`). In that case, `favoriteFoods` defaults to Mixed type, and you have to manually mark it as edited using `document.markModified('edited-field')`. See [Mongoose documentation](https://mongoosejs.com/docs/schematypes.html#Mixed)
# --hints--
-应成功地对一条数据进行查找、编辑和更新
+Find-edit-update an item should succeed
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/perform-new-updates-on-a-document-using-model.findoneandupdate.md b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/perform-new-updates-on-a-document-using-model.findoneandupdate.md
index 445aa03caa9..ddd4ecd80ad 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/perform-new-updates-on-a-document-using-model.findoneandupdate.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/perform-new-updates-on-a-document-using-model.findoneandupdate.md
@@ -1,6 +1,6 @@
---
id: 587d7fb8367417b2b2512c0f
-title: 使用 model.findOneAndUpdate() 对更新 document
+title: Perform New Updates on a Document Using model.findOneAndUpdate()
challengeType: 2
forumTopicId: 301542
dashedName: perform-new-updates-on-a-document-using-model-findoneandupdate
@@ -8,17 +8,17 @@ dashedName: perform-new-updates-on-a-document-using-model-findoneandupdate
# --description--
-最近发布的 mongoose 版本简化了 document 的更新方式，但同时，一些高级功能（如 pre/post hook, 验证）的使用方式也变得和以前不同。因此，在很多情景下，上一个挑战中提到的老方法其实更常用。新方法的加入，可以让我们使用 `findByIdAndUpdate()` 来进行基于 Id 的搜索。
+Recent versions of Mongoose have methods to simplify documents updating. Some more advanced features (i.e. pre/post hooks, validation) behave differently with this approach, so the classic method is still useful in many situations. `findByIdAndUpdate()` can be used when searching by id.
# --instructions--
-用 `personName` 作为 `Name` 的查找条件，并将查到的 person 年龄设为 20 岁。
+Modify the `findAndUpdate` function to find a person by `Name` and set the person's age to `20`. Use the function parameter `personName` as the search key.
-**提示：** 你需要返回更新后的 document。你可以把 `findOneAndUpdate()` 的第三个参数设置为 `{new: true}`。默认情况下，这个方法会返回修改前的数据。
+**Note:** You should return the updated document. To do that, you need to pass the options document `{ new: true }` as the 3rd argument to `findOneAndUpdate()`. By default, these methods return the unmodified object.
# --hints--
-应成功地使用 findOneAndUpdate 更新数据
+findOneAndUpdate an item should succeed
```js
(getUserInput) =>
@@ -38,7 +38,7 @@ dashedName: perform-new-updates-on-a-document-using-model-findoneandupdate
assert.equal(
data.__v,
0,
- 'findOneAndUpdate does not increment version by design !!!'
+ 'findOneAndUpdate does not increment version by design!'
);
},
(xhr) => {
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/use-model.find-to-search-your-database.md b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/use-model.find-to-search-your-database.md
index 01aaf5255bf..a3e22a5b1eb 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/use-model.find-to-search-your-database.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/use-model.find-to-search-your-database.md
@@ -1,6 +1,6 @@
---
id: 587d7fb7367417b2b2512c0b
-title: 使用 model.find() 查询数据库
+title: Use model.find() to Search Your Database
challengeType: 2
forumTopicId: 301543
dashedName: use-model-find-to-search-your-database
@@ -8,17 +8,17 @@ dashedName: use-model-find-to-search-your-database
# --description--
-我们尝试一种最简单的用法，`Model.find()` 接收一个查询 document（一个 JSON 对象）作为第一个参数，一个回调函数作为第二个参数，它会返回由匹配到的数据组成的数组。这个方法支持很多搜索选项，详情请参阅文档。
+In its simplest usage, `Model.find()` accepts a query document (a JSON object) as the first argument, then a callback. It returns an array of matches. It supports an extremely wide range of search options. Read more in the docs.
# --instructions--
-请使用 `Model.find() -> [Person]` 修改 `findPeopleByName` 函数，以查询全部拥有指定姓名的人。
+Modify the `findPeopleByName` function to find all the people having a given name, using <code>Model.find() -\> [Person]</code>
-请使用函数参数中的 `personName` 作为搜索条件。
+Use the function argument `personName` as the search key.
# --hints--
-应成功地找到所有符合条件的数据
+Find all items corresponding to a criteria should succeed
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/use-model.findbyid-to-search-your-database-by-id.md b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/use-model.findbyid-to-search-your-database-by-id.md
index ddcb9900339..36c2b551e1d 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/use-model.findbyid-to-search-your-database-by-id.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/use-model.findbyid-to-search-your-database-by-id.md
@@ -1,6 +1,6 @@
---
id: 587d7fb7367417b2b2512c0d
-title: 使用 model.findById() 方法，根据 _id 来搜索数据
+title: Use model.findById() to Search Your Database By _id
challengeType: 2
forumTopicId: 301544
dashedName: use-model-findbyid-to-search-your-database-by-id
@@ -8,15 +8,15 @@ dashedName: use-model-findbyid-to-search-your-database-by-id
# --description--
-在保存 document 的时候，MongoDB 会自动为它添加 `_id` 字段，并给该字段设置一个唯一的仅包含数字和字母的值。通过 `_id` 搜索是一个十分常见的操作，为此，Mongoose 提供了一个专门的方法。
+When saving a document, MongoDB automatically adds the field `_id`, and set it to a unique alphanumeric key. Searching by `_id` is an extremely frequent operation, so Mongoose provides a dedicated method for it.
# --instructions--
-执行`Model.findById() -> Person`，使用 `personId` 作为搜索参数，根据 `_id` 找到对应的（唯一的）person。
+Modify the `findPersonById` to find the only person having a given `_id`, using `Model.findById() -> Person`. Use the function argument `personId` as the search key.
# --hints--
-应成功地根据 Id 找到对应的数据
+Find an item by Id should succeed
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/use-model.findone-to-return-a-single-matching-document-from-your-database.md b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/use-model.findone-to-return-a-single-matching-document-from-your-database.md
index 9b5b589fc0e..e7f9c78278b 100644
--- a/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/use-model.findone-to-return-a-single-matching-document-from-your-database.md
+++ b/curriculum/challenges/chinese/05-apis-and-microservices/mongodb-and-mongoose/use-model.findone-to-return-a-single-matching-document-from-your-database.md
@@ -1,6 +1,6 @@
---
id: 587d7fb7367417b2b2512c0c
-title: 使用 model.findOne() 从数据库中返回一个单一匹配的 Document
+title: Use model.findOne() to Return a Single Matching Document from Your Database
challengeType: 2
forumTopicId: 301545
dashedName: use-model-findone-to-return-a-single-matching-document-from-your-database
@@ -8,15 +8,15 @@ dashedName: use-model-findone-to-return-a-single-matching-document-from-your-dat
# --description--
-`Model.findOne()` 与 `Model.find()` 十分类似。但就算数据库中有很多条数据可以匹配查询条件，`Model.findOne()` 也只返回一个 document，而不会返回一个数组。如果查询条件是声明为唯一值的属性，`Model.findOne()` 会更加适用。
+`Model.findOne()` behaves like `Model.find()`, but it returns only one document (not an array), even if there are multiple items. It is especially useful when searching by properties that you have declared as unique.
# --instructions--
-修改 `findOneByFood` 函数，用 `Model.findOne() -> Person` 来查询在收藏夹中有某种食物的一个人。将函数参数中的 `food` 作为检索条件。
+Modify the `findOneByFood` function to find just one person which has a certain food in the person's favorites, using `Model.findOne() -> Person`. Use the function argument `food` as search key.
# --hints--
-应成功地找到一个数据
+Find one item should succeed
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/announce-new-users.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/announce-new-users.md
index 09da91097fc..fca0105f3da 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/announce-new-users.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/announce-new-users.md
@@ -1,6 +1,6 @@
---
id: 589fc832f9fc0f352b528e78
-title: 宣布新用户
+title: Announce New Users
challengeType: 2
forumTopicId: 301546
dashedName: announce-new-users
@@ -8,7 +8,9 @@ dashedName: announce-new-users
# --description--
-许多聊天室都有这个功能：所有已连接到服务器的在线用户都会看到有人加入或退出的提醒。我们已经写好了处理连接和断开事件的代码，只要对这个方法稍作修改就可以实现这个功能。在事件中，我们需要发送这三条信息：连接或断开的用户名、当前用户数量、事件类型（即需要知道用户究竟是连接还是断开）。 请将事件名称更改为 `'user'`，其中应包含如下字段：'name'、'currentUsers'、'connected'（布尔值，连接上即为 `true`，断开则是 `false`）。记得要修改之前我们写好的处理 'user count' 的那部分代码，现在我们应在用户连接时传入布尔值 `true`；在用户断开连接是传入布尔值 `false`。
+Many chat rooms are able to announce when a user connects or disconnects and then display that to all of the connected users in the chat. Seeing as though you already are emitting an event on connect and disconnect, you will just have to modify this event to support such a feature. The most logical way of doing so is sending 3 pieces of data with the event: the name of the user who connected/disconnected, the current user count, and if that name connected or disconnected.
+
+Change the event name to `'user'`, and pass an object along containing the fields 'name', 'currentUsers', and 'connected' (to be `true` in case of connection, or `false` for disconnection of the user sent). Be sure to change both 'user count' events and set the disconnect one to send `false` for the field 'connected' instead of `true` like the event emitted on connect.
```js
io.emit('user', {
@@ -18,7 +20,9 @@ io.emit('user', {
});
```
-现在，我们的客户端已经有足够的信息显示现有用户数量和发送用户上下线通知。接下来我们需要在客户端监听 'user' 事件，然后使用 jQuery 把 `#num-users` 节点的文本内容更新为 '{NUMBER} users online'。同时，我们需要为 `<ul>` 添加一个 id 为 'messages' 且带有 '{NAME} has {joined/left} the chat.' 文本的`<li>`。 一种实现方式如下：
+Now your client will have all the necessary information to correctly display the current user count and announce when a user connects or disconnects! To handle this event on the client side we should listen for `'user'`, then update the current user count by using jQuery to change the text of `#num-users` to `'{NUMBER} users online'`, as well as append a `<li>` to the unordered list with id `messages` with `'{NAME} has {joined/left} the chat.'`.
+
+An implementation of this could look like the following:
```js
socket.on('user', data => {
@@ -30,11 +34,11 @@ socket.on('user', data => {
});
```
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/bf95a0f74b756cf0771cd62c087b8286) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/bf95a0f74b756cf0771cd62c087b8286).
# --hints--
-`'user'` 事件应发送包含 name、currentUsers、connected 字段的对象。
+Event `'user'` should be emitted with name, currentUsers, and connected.
```js
(getUserInput) =>
@@ -52,7 +56,7 @@ socket.on('user', data => {
);
```
-客户端应处理和显示 `'user'` 对象中的信息
+Client should properly handle and display the new data from event `'user'`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/authentication-strategies.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/authentication-strategies.md
index 398a9afd32a..08f791012b4 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/authentication-strategies.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/authentication-strategies.md
@@ -1,6 +1,6 @@
---
id: 5895f70df9fc0f352b528e68
-title: 身份验证策略
+title: Authentication Strategies
challengeType: 2
forumTopicId: 301547
dashedName: authentication-strategies
@@ -8,11 +8,11 @@ dashedName: authentication-strategies
# --description--
-现在，我们需要构建验证用户的策略，策略的选择有很多。比如，如果我们已经让用户在注册时填写了用户信息，那我们就可以基于这些信息验证；或者也可以引入第三方登录，如 Google 或者 Github。为此，你可以参考 [Passports 中提供的策略插件](http://passportjs.org/)。对于这个项目的验证策略，我们会采用自己搭建的方式完成。
+A strategy is a way of authenticating a user. You can use a strategy for allowing users to authenticate based on locally saved information (if you have them register first) or from a variety of providers such as Google or GitHub. For this project, we will set up a local strategy. To see a list of the hundreds of strategies, visit Passport's site [here](http://passportjs.org/).
-首先，我们需要引入 *passport-local* 作为依赖，然后将它添加到服务器，就像这样：`const LocalStrategy = require('passport-local');`
+Add `passport-local` as a dependency and add it to your server as follows: `const LocalStrategy = require('passport-local');`
-然后，我们需要让 passport 使用实例化的本地策略对象。请注意，接下来的所有代码都应写在连接数据库的回调中，因为它们的执行都依赖数据库。
+Now you will have to tell passport to **use** an instantiated LocalStrategy object with a few settings defined. Make sure this (as well as everything from this point on) is encapsulated in the database connection since it relies on it!
```js
passport.use(new LocalStrategy(
@@ -28,11 +28,17 @@ passport.use(new LocalStrategy(
));
```
-这就是我们的用户验证逻辑：首先根据用户输入的用户名在数据库中寻找用户；然后检查密码是否匹配；最后如果没有发生错误，那么就会返回用户对象并通过验证。 我们也可以采用上面链接中提供的验证策略，一般来说，根据该策略仓库中的 README 来进行配置就足够了。一个很好的例子是 Github 策略，在该策略中，我们不需要写用户名或密码的相关验证逻辑，因为用户将被引导到 Github 页面进行验证。只要他们登录并同意，Github 就会返回他们的个人信息供我们使用。 以上就是本次挑战的内容。在下一个挑战中，我们会基于表单数据调用上面写好的验证策略。 完成上述要求后，你可以在下方提交你的页面链接。
+This is defining the process to use when we try to authenticate someone locally. First, it tries to find a user in our database with the username entered, then it checks for the password to match, then finally, if no errors have popped up that we checked for, like an incorrect password, the `user`'s object is returned and they are authenticated.
+
+Many strategies are set up using different settings, but generally it is easy to set it up based on the README in that strategy's repository. A good example of this is the GitHub strategy where we don't need to worry about a username or password because the user will be sent to GitHub's auth page to authenticate. As long as they are logged in and agree then GitHub returns their profile for us to use.
+
+In the next step, we will set up how to actually call the authentication strategy to validate a user based on form data!
+
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/53b495c02b92adeee0aa1bd3f3be8a4b).
# --hints--
-你的项目需要使用 `passport-local` 作为依赖。
+Passport-local should be a dependency.
```js
(getUserInput) =>
@@ -51,7 +57,7 @@ passport.use(new LocalStrategy(
);
```
-应正确地引入和设置 `passport-local`。
+Passport-local should be correctly required and setup.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/authentication-with-socket.io.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/authentication-with-socket.io.md
index 98d34707806..ca8cbd7cd95 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/authentication-with-socket.io.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/authentication-with-socket.io.md
@@ -1,6 +1,6 @@
---
id: 589fc831f9fc0f352b528e77
-title: 使用Socket.IO进行身份验证
+title: Authentication with Socket.IO
challengeType: 2
forumTopicId: 301548
dashedName: authentication-with-socket-io
@@ -8,9 +8,9 @@ dashedName: authentication-with-socket-io
# --description--
-目前，我们还无法确定连接到服务器的用户身份。虽然 `req.user` 包含用户信息，但这个只在用户直接与服务器交互（即不通过 web socket 访问服务器资源）时产生。当我们的用户通过 web socket 与服务器连接时，由于不存在 `req` 对象，因此我们无法获取用户数据。解决这个问题的方法之一是通过读取和解析请求中包含 passport session 的 cookie，然后反序列化，进而获取用户信息对象。幸运的是，npm 上有可以让这个复杂的流程简单化的库。
+Currently, you cannot determine who is connected to your web socket. While `req.user` contains the user object, that's only when your user interacts with the web server, and with web sockets you have no `req` (request) and therefore no user data. One way to solve the problem of knowing who is connected to your web socket is by parsing and decoding the cookie that contains the passport session then deserializing it to obtain the user object. Luckily, there is a package on NPM just for this that turns a once complex task into something simple!
-添加 `passport.socketio`、`connect-mongo`、`cookie-parser` 作为依赖，把它们分别赋值给 `passportSocketIo`、`MongoStore`、`cookieParser`。同时，我们需要从之前引入的 `express-session` 中开辟新的内存空间，就像接下来这样：
+Add `passport.socketio`, `connect-mongo`, and `cookie-parser` as dependencies and require them as `passportSocketIo`, `MongoStore`, and `cookieParser` respectively. Also, we need to initialize a new memory store, from `express-session` which we previously required. It should look like this:
```js
const MongoStore = require('connect-mongo')(session);
@@ -18,7 +18,7 @@ const URI = process.env.MONGO_URI;
const store = new MongoStore({ url: URI });
```
-现在我们只需要让 Socket.IO 调用它并进行相应的配置即可。请注意，以上这些都必须放在现有的 socket 相关代码之前，而且不能放到连接的监听回调函数里。你的服务器代码应类似于这样：
+Now we just have to tell Socket.IO to use it and set the options. Be sure this is added before the existing socket code and not in the existing connection listener. For your server, it should look like this:
```js
io.use(
@@ -33,11 +33,11 @@ io.use(
);
```
-记得要把 `key` 和 `store` 加到 app 的 `session` 中间件。这样，SocketIO 才知道该对哪个 session 执行此配置。
+Be sure to add the `key` and `store` to the `session` middleware mounted on the app. This is necessary to tell *SocketIO* which session to relate to.
-<hr>
+<hr />
-接下来，我们可以定义 `success` 与 `fail` 的回调函数：
+Now, define the `success`, and `fail` callback functions:
```js
function onAuthorizeSuccess(data, accept) {
@@ -53,19 +53,19 @@ function onAuthorizeFail(data, message, error, accept) {
}
```
-现在，我们可以通过 socket 对象通过 `socket.request.user` 访问 `user` 对象。为此，你可以这样做：
+The user object is now accessible on your socket object as `socket.request.user`. For example, now you can add the following:
```js
console.log('user ' + socket.request.user.name + ' connected');
```
-这样，我们可以在 console 里看到谁连接到了我们的服务器。
+It will log to the server console who has connected!
-完成上述要求后，你可以在下方提交你的页面链接。 完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/1414cc9433044e306dd7fd0caa1c6254) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project up to this point [here](https://gist.github.com/camperbot/1414cc9433044e306dd7fd0caa1c6254).
# --hints--
-应添加 `passport.socketio` 作为依赖。
+`passport.socketio` should be a dependency.
```js
(getUserInput) =>
@@ -84,7 +84,7 @@ console.log('user ' + socket.request.user.name + ' connected');
);
```
-应添加 `cookie-parser` 作为依赖。
+`cookie-parser` should be a dependency.
```js
(getUserInput) =>
@@ -103,7 +103,7 @@ console.log('user ' + socket.request.user.name + ' connected');
);
```
-应正确引入 passportSocketIo。
+passportSocketIo should be properly required.
```js
(getUserInput) =>
@@ -121,7 +121,7 @@ console.log('user ' + socket.request.user.name + ' connected');
);
```
-应正确配置 passportSocketIo。
+passportSocketIo should be properly setup.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/clean-up-your-project-with-modules.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/clean-up-your-project-with-modules.md
index 777bf5001b4..13321446e96 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/clean-up-your-project-with-modules.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/clean-up-your-project-with-modules.md
@@ -1,6 +1,6 @@
---
id: 589690e6f9fc0f352b528e6e
-title: 使用模块清理项目
+title: Clean Up Your Project with Modules
challengeType: 2
forumTopicId: 301549
dashedName: clean-up-your-project-with-modules
@@ -8,7 +8,9 @@ dashedName: clean-up-your-project-with-modules
# --description--
-目前，我们把所有的代码都放到了 server.js 文件里，这会导致代码难以维护，且扩展性差。 现在让我们来创建两个新文件：`routes.js` 和 `auth.js` 在每个文件的开头，我们都需要写上这段代码：
+Right now, everything you have is in your `server.js` file. This can lead to hard to manage code that isn't very expandable. Create 2 new files: `routes.js` and `auth.js`
+
+Both should start with the following code:
```js
module.exports = function (app, myDataBase) {
@@ -16,19 +18,19 @@ module.exports = function (app, myDataBase) {
}
```
-然后，在 server.js 文件的开头，我们需要像这样引入文件：`const routes = require('./routes.js');` 在成功连接数据库之后，我们需要像这样进行实例化：`routes(app, myDataBase)`。 最后，我们需要把所有路由相关的代码从 server.js 移动到新文件 routes.js。不要忘了，`ensureAuthenticated` 方法的定义也要移动到新文件中，这个是我们在之前的挑战中，为在路由中判断用户是否已登录创建的函数。然后，我们还需要在 routes.js 文件开头添加所需要的依赖，如：`const passport = require('passport');`。
+Now, in the top of your server file, require these files like so: `const routes = require('./routes.js');` Right after you establish a successful connection with the database, instantiate each of them like so: `routes(app, myDataBase)`
-如果在这些步骤后没有报错，那么恭喜你，你已成功地从 server.js 文件中分离出了路由文件（**尽管 catch 那部分还是包含了路由的逻辑**）。
+Finally, take all of the routes in your server and paste them into your new files, and remove them from your server file. Also take the `ensureAuthenticated` function, since it was specifically created for routing. Now, you will have to correctly add the dependencies in which are used, such as `const passport = require('passport');`, at the very top, above the export line in your `routes.js` file.
-现在，我们来把 server.js 中与验证相关的代码分离到 auth.js 中，例如序列化，设置验证策略等。请正确添加依赖，并在 server.js 中调用`auth(app,myDataBase)`。另外，由于我们的注册路由依赖 passport，所以我们需要先调用`auth(app, myDataBase)`，再调用`routes(app, myDataBase)`。
+Keep adding them until no more errors exist, and your server file no longer has any routing (**except for the route in the catch block**)!
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/2d06ac5c7d850d8cf073d2c2c794cc92) 的答案。
+Now do the same thing in your auth.js file with all of the things related to authentication such as the serialization and the setting up of the local strategy and erase them from your server file. Be sure to add the dependencies in and call `auth(app, myDataBase)` in the server in the same spot.
Submit your page when you think you've got it right. If you're running into errors, you can check out an example of the completed project [here](https://gist.github.com/camperbot/2d06ac5c7d850d8cf073d2c2c794cc92).
# --hints--
-应正确引入新文件。
+Modules should be present.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/communicate-by-emitting.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/communicate-by-emitting.md
index 05a6dbdbcae..1b9e03a3b50 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/communicate-by-emitting.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/communicate-by-emitting.md
@@ -1,6 +1,6 @@
---
id: 589fc831f9fc0f352b528e75
-title: 通过 Emitting 通信
+title: Communicate by Emitting
challengeType: 2
forumTopicId: 301550
dashedName: communicate-by-emitting
@@ -8,27 +8,27 @@ dashedName: communicate-by-emitting
# --description--
-<dfn>Emit</dfn> 是你会用到的，最常见的通信方式。如果我们从服务器发送信息给 'io'，就相当于把事件的名称和数据发送给了所有处于连接状态的 socket。我们可以利用这个特性实现这样的功能：只要有新用户连接到服务器，我们就可以把目前连接的总用户数发给所有已连接的用户，这样所有用户随时都可以看到实时的在线人数。
+<dfn>Emit</dfn> is the most common way of communicating you will use. When you emit something from the server to 'io', you send an event's name and data to all the connected sockets. A good example of this concept would be emitting the current count of connected users each time a new user connects!
-首先，我们需要在监听连接的地方之前添加一个用于追踪用户数的变量：
+Start by adding a variable to keep track of the users, just before where you are currently listening for connections.
```js
let currentUsers = 0;
```
-然后，只要有人连接到服务器，我们需要在发出用户数量之前先给这个变量加 1：
+Now, when someone connects, you should increment the count before emitting the count. So, you will want to add the incrementer within the connection listener.
```js
++currentUsers;
```
-最后，在监听连接的地方发出（emit）该事件即可。这个事件应命名为 'user count'，且数据应该为 `currentUsers`：
+Finally, after incrementing the count, you should emit the event (still within the connection listener). The event should be named 'user count', and the data should just be the `currentUsers`.
```js
io.emit('user count', currentUsers);
```
-接下来，我们还需要让客户端监听从服务端发出的事件。为此，我们还是需要用到 *on* 这个方法：
+Now, you can implement a way for your client to listen for this event! Similar to listening for a connection on the server, you will use the `on` keyword.
```js
socket.on('user count', function(data) {
@@ -36,13 +36,13 @@ socket.on('user count', function(data) {
});
```
-现在你可以尝试启动你的 app 并登录，你会看到在客户端的控制台打印出了 1，这就表示目前连接到服务器的用户数为 1。你可以试着通过打开多个 app 来验证数量是否会增加。
+Now, try loading up your app, authenticate, and you should see in your client console '1' representing the current user count! Try loading more clients up, and authenticating to see the number go up.
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/28ef7f1078f56eb48c7b1aeea35ba1f5) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/28ef7f1078f56eb48c7b1aeea35ba1f5).
# --hints--
-应定义 currentUsers。
+currentUsers should be defined.
```js
(getUserInput) =>
@@ -60,7 +60,7 @@ socket.on('user count', function(data) {
);
```
-服务器应在有新的连接时 emit 当前用户数量。
+Server should emit the current user count at each new connection.
```js
(getUserInput) =>
@@ -78,7 +78,7 @@ socket.on('user count', function(data) {
);
```
-客户端应监听 'user count' 事件。
+Your client should be listening for 'user count' event.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/create-new-middleware.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/create-new-middleware.md
index 306c15ae26c..e34d3a458a7 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/create-new-middleware.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/create-new-middleware.md
@@ -1,6 +1,6 @@
---
id: 5895f70df9fc0f352b528e6a
-title: 创建新的中间件
+title: Create New Middleware
challengeType: 2
forumTopicId: 301551
dashedName: create-new-middleware
@@ -8,9 +8,9 @@ dashedName: create-new-middleware
# --description--
-无论是否登录，或者哪怕用户试图访问其他页面，目前都会跳转到 `/profile`。为了解决这个问题，我们需要在 profile 页面渲染之前进行用户验证，创建中间件就可以实现这个功能。
+As is, any user can just go to `/profile` whether they have authenticated or not, by typing in the url. We want to prevent this, by checking if the user is authenticated first before rendering the profile page. This is the perfect example of when to create a middleware.
-这个挑战的目标是创建`ensureAuthenticated(req, res, next)`中间件方法，通过在 *request* 上调用 passports 的`isAuthenticated` 方法，我们可以检查 *req.user* 是否定义，从而确定用户是否通过认证。如果用户已通过验证，就会调用 *next()*，否则我们应重定向到主页并让用户登录。该中间件的实现如下：
+The challenge here is creating the middleware function `ensureAuthenticated(req, res, next)`, which will check if a user is authenticated by calling passport's `isAuthenticated` method on the `request` which, in turn, checks if `req.user` is defined. If it is, then `next()` should be called, otherwise, we can just respond to the request with a redirect to our homepage to login. An implementation of this middleware is:
```js
function ensureAuthenticated(req, res, next) {
@@ -21,7 +21,7 @@ function ensureAuthenticated(req, res, next) {
};
```
-然后，我们需要把 *ensureAuthenticated* 中间件添加到处理请求的回调之前：
+Now add *ensureAuthenticated* as a middleware to the request for the profile page before the argument to the get request containing the function that renders the page.
```js
app
@@ -31,11 +31,11 @@ app
});
```
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/ae49b8778cab87e93284a91343da0959) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/ae49b8778cab87e93284a91343da0959).
# --hints--
-`ensureAuthenticated` 中间件应添加到 `/profile`路由中。
+Middleware ensureAuthenticated should be implemented and on our /profile route.
```js
(getUserInput) =>
@@ -58,7 +58,7 @@ app
);
```
-如果没有通过验证，对 /profile 的 GET 请求应重定向到 /
+A Get request to /profile should correctly redirect to / since we are not authenticated.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/handle-a-disconnect.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/handle-a-disconnect.md
index 22e0711508a..2873c18ff27 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/handle-a-disconnect.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/handle-a-disconnect.md
@@ -1,6 +1,6 @@
---
id: 589fc831f9fc0f352b528e76
-title: 处理连接断开
+title: Handle a Disconnect
challengeType: 2
forumTopicId: 301552
dashedName: handle-a-disconnect
@@ -8,9 +8,9 @@ dashedName: handle-a-disconnect
# --description--
-你也许注意到，目前为止我们只处理用户数量的增加，没有处理减少。事实上，处理用户断开连接也很简单。区别在于，新连接的监听是发生在整个服务器上，但连接断开的监听是发生在每个 socket 上。
+You may notice that up to now you have only been increasing the user count. Handling a user disconnecting is just as easy as handling the initial connect, except you have to listen for it on each socket instead of on the whole server.
-为此，我们需要在目前的监听回调里面监听 socket 断开连接的事件。在断开连接的回调函数中，我们可以不传任何参数，但你可以在这里添加连接断开的测试代码：
+To do this, add another listener inside the existing `'connect'` listener that listens for `'disconnect'` on the socket with no data passed through. You can test this functionality by just logging that a user has disconnected to the console.
```js
socket.on('disconnect', () => {
@@ -18,15 +18,15 @@ socket.on('disconnect', () => {
});
```
-为确保客户端可以看到实时的用户数量，显然，我们应该在用户断开时让 currentUsers 减 1，然后发送 'user count' 事件，并使用修改后的用户数量。
+To make sure clients continuously have the updated count of current users, you should decrease the currentUsers by 1 when the disconnect happens then emit the 'user count' event with the updated count!
-**注意：** 和 `'disconnect'` 类似，所有 socket 可以发送到服务器的事件，我们都应该在有 'socket' 定义的连接监听器里处理。
+**Note:** Just like `'disconnect'`, all other events that a socket can emit to the server should be handled within the connecting listener where we have 'socket' defined.
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/ab1007b76069884fb45b215d3c4496fa) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/ab1007b76069884fb45b215d3c4496fa).
# --hints--
-服务器应处理断开 socket 连接的事件。
+Server should handle the event disconnect from a socket.
```js
(getUserInput) =>
@@ -40,7 +40,7 @@ socket.on('disconnect', () => {
);
```
-客户端应监听 'user count' 事件。
+Your client should be listening for 'user count' event.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/hashing-your-passwords.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/hashing-your-passwords.md
index e1963eda78f..ee33328d673 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/hashing-your-passwords.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/hashing-your-passwords.md
@@ -1,6 +1,6 @@
---
id: 58a25c98f9fc0f352b528e7f
-title: 哈希密码
+title: Hashing Your Passwords
challengeType: 2
forumTopicId: 301553
dashedName: hashing-your-passwords
@@ -8,13 +8,13 @@ dashedName: hashing-your-passwords
# --description--
-回过头来看信息安全，你也许记得在数据库中存储明文密码是*绝对*禁止的。现在，我们需要引入 BCrypt 来解决这个问题。
+Going back to the information security section, you may remember that storing plaintext passwords is *never* okay. Now it is time to implement BCrypt to solve this issue.
-添加 BCrypt 作为依赖，并通过`require`添加到服务器代码中。你需要在两个步骤中使用哈希运算：注册和保存新账户，以及登录时检查密码是否正确。
+Add BCrypt as a dependency, and require it in your server. You will need to handle hashing in 2 key areas: where you handle registering/saving a new account, and when you check to see that a password is correct on login.
-目前处理注册的路由中，我们是这样把密码添加到数据库的：`password: req.body.password`。我们可以通过这段代码创建哈希值：`var hash = bcrypt.hashSync(req.body.password, 12);`，然后就可以把`passsword: req.body.password`替换为`password: hash`。
+Currently on our registration route, you insert a user's password into the database like so: `password: req.body.password`. An easy way to implement saving a hash instead is to add the following before your database logic `const hash = bcrypt.hashSync(req.body.password, 12);`, and replacing the `req.body.password` in the database saving with just `password: hash`.
-最后，在验证逻辑中，我们已经有这样一段代码执行检查：`if (password !== user.password) { return done(null, false); }`。但我们现在存储的密码`user.password`已经是哈希值了。由于目前的检测机制是密码不匹配时就返回未认证，因此修改后，用于比对用户密码哈希值的代码应该是这样：
+Finally, on our authentication strategy, we check for the following in our code before completing the process: `if (password !== user.password) { return done(null, false); }`. After making the previous changes, now `user.password` is a hash. Before making a change to the existing code, notice how the statement is checking if the password is **not** equal then return non-authenticated. With this in mind, your code could look as follows to properly check the password entered against the hash:
```js
if (!bcrypt.compareSync(password, user.password)) {
@@ -22,13 +22,13 @@ if (!bcrypt.compareSync(password, user.password)) {
}
```
-当你需要存储密码时，这样做可以有效地提升网站的安全性。
+That is all it takes to implement one of the most important security features when you have to store passwords!
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/dc16cca09daea4d4151a9c36a1fab564) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/dc16cca09daea4d4151a9c36a1fab564).
# --hints--
-应存在 BCrypt 依赖。
+BCrypt should be a dependency.
```js
(getUserInput) =>
@@ -47,7 +47,7 @@ if (!bcrypt.compareSync(password, user.password)) {
);
```
-BCrypt 应正确地引入和调用。
+BCrypt should be correctly required and implemented.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/how-to-put-a-profile-together.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/how-to-put-a-profile-together.md
index 4c4c0c09474..bd17bbb9a66 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/how-to-put-a-profile-together.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/how-to-put-a-profile-together.md
@@ -1,6 +1,6 @@
---
id: 5895f70ef9fc0f352b528e6b
-title: 如何将 Profile 放在一起
+title: How to Put a Profile Together
challengeType: 2
forumTopicId: 301554
dashedName: how-to-put-a-profile-together
@@ -8,17 +8,27 @@ dashedName: how-to-put-a-profile-together
# --description--
-现在，只有通过验证的用户才能进入 */profile* 页面，这样我们就可以在页面上使用 'req.user' 里的信息了。
+Now that we can ensure the user accessing the `/profile` is authenticated, we can use the information contained in `req.user` on our page!
-请在变量中包含 *username* 键，值为 'req.user.username'，并通过 render 方法传给 profile 页面。然后在 'profile.pug' 页面，添加这行 `h2.center#welcome Welcome, #{username}!` 代码来创建 class 为 `center`、id 为 `welcome` 且文本内容为 'Welcome, ' 后加用户名的 h2 元素。
+Pass an object containing the property `username` and value of `req.user.username` as the second argument for the render method of the profile view. Then, go to your `profile.pug` view, and add the following line below the existing `h1` element, and at the same level of indentation:
-以及，请在 profile 里添加 */logout* 链接，后续会用于处理用户退出登录的逻辑：`a(href='/logout') Logout`
+```pug
+h2.center#welcome Welcome, #{username}!
+```
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/136b3ad611cc80b41cab6f74bb460f6a) 的答案。
+This creates an `h2` element with the class '`center`' and id '`welcome`' containing the text '`Welcome,`' followed by the username.
+
+Also, in `profile.pug`, add a link referring to the `/logout` route, which will host the logic to unauthenticate a user.
+
+```pug
+a(href='/logout') Logout
+```
+
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/136b3ad611cc80b41cab6f74bb460f6a).
# --hints--
-应在 Pug render 中给 /profile 传一个变量。
+You should correctly add a Pug render variable to /profile.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/how-to-use-passport-strategies.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/how-to-use-passport-strategies.md
index db93316dbb8..ba78302a8ca 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/how-to-use-passport-strategies.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/how-to-use-passport-strategies.md
@@ -1,6 +1,6 @@
---
id: 5895f70df9fc0f352b528e69
-title: 如何使用 Passport 策略
+title: How to Use Passport Strategies
challengeType: 2
forumTopicId: 301555
dashedName: how-to-use-passport-strategies
@@ -8,21 +8,21 @@ dashedName: how-to-use-passport-strategies
# --description--
-在提供的 index.pug 文件里有一个登录表单。因为这个表单中存在行内 JavaScript 代码 `if showLogin`，因此它是隐藏的。因为 showLogin 未定义，所以表单不会渲染。如果在该页面的 `res.render()` 里添加一个包含 `showLogin: true` 的对象，你就可以在刷新页面后看到表单。当你点击 login 时，表单会向服务器的 */login 发送 POST 请求，此时服务器端就可以接受 POST 请求信息并进行用户验证。*
+In the `index.pug` file supplied, there is actually a login form. It has previously been hidden because of the inline JavaScript `if showLogin` with the form indented after it. Before `showLogin` as a variable was never defined, so it never rendered the code block containing the form. Go ahead and on the `res.render` for that page add a new variable to the object `showLogin: true`. When you refresh your page, you should then see the form! This form is set up to **POST** on `/login`, so this is where we should set up to accept the POST and authenticate the user.
-在这个挑战中，你需要为 POST 请求添加路由`/login`。为了用这个路由进行验证，你需要添加一个中间件，中间件应作为参数添加到用于处理请求的回调函数 `function(req,res)` 之前。对于 passport 的验证中间件，应这样调用：`passport.authenticate('local')`。
+For this challenge you should add the route `/login` to accept a POST request. To authenticate on this route, you need to add a middleware to do so before then sending a response. This is done by just passing another argument with the middleware before your `function(req,res)` with your response! The middleware to use is `passport.authenticate('local')`.
-*passport.authenticate* 也接收选项作为参数，这些选项用于设置验证，例如`{ failureRedirect: '/' }`就很有用，请记得添加到你的代码中。如果中间件验证通过，我们就应该提供相应的后续处理。在这个挑战中，我们需要让用户重定到 */profile*，这样 `profile.pug` 页面就会渲染。
+`passport.authenticate` can also take some options as an argument such as: `{ failureRedirect: '/' }` which is incredibly useful, so be sure to add that in as well. The response after using the middleware (which will only be called if the authentication middleware passes) should be to redirect the user to `/profile` and that route should render the view `profile.pug`.
-如果验证通过，用户对象将会储存到 *req.user* 中。
+If the authentication was successful, the user object will be saved in `req.user`.
-这时，如果你在表单里输入了用户名和密码，路由将会重定向到主页 */*，在服务端将会打印 'User {USERNAME} attempted to log in.'，由于现在我们还没有实现注册功能，因此所有登录尝试都会失败。
+At this point, if you enter a username and password in the form, it should redirect to the home page `/`, and the console of your server should display `'User {USERNAME} attempted to log in.'`, since we currently cannot login a user who isn't registered.
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/7ad011ac54612ad53188b500c5e99cb9) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/7ad011ac54612ad53188b500c5e99cb9).
# --hints--
-server.js 中应正确执行所有步骤。
+All steps should be correctly implemented in the server.js.
```js
(getUserInput) =>
@@ -50,7 +50,7 @@ server.js 中应正确执行所有步骤。
);
```
-到 /login 的 POST 请求应重定向到 /
+A POST request to /login should correctly redirect to /.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/implement-the-serialization-of-a-passport-user.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/implement-the-serialization-of-a-passport-user.md
index 4316c544588..03d737f4aaa 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/implement-the-serialization-of-a-passport-user.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/implement-the-serialization-of-a-passport-user.md
@@ -1,6 +1,6 @@
---
id: 5895f70cf9fc0f352b528e67
-title: 实现 Passport 用户的序列化
+title: Implement the Serialization of a Passport User
challengeType: 2
forumTopicId: 301556
dashedName: implement-the-serialization-of-a-passport-user
@@ -8,13 +8,11 @@ dashedName: implement-the-serialization-of-a-passport-user
# --description--
-截至目前，我们还没有配置完数据库，因此还无法加载用户数据。实现这个的方式很多，但对于我们的项目，一旦服务器启动，那么只要有 app 实例在运行，数据库就应一直处于连接状态。
+Right now, we're not loading an actual user object since we haven't set up our database. This can be done many different ways, but for our project we will connect to the database once when we start the server and keep a persistent connection for the full life-cycle of the app. To do this, add your database's connection string (for example: `mongodb+srv://:@cluster0-jvwxi.mongodb.net/?retryWrites=true&w=majority`) to the environment variable `MONGO_URI`. This is used in the `connection.js` file.
-为此，你需要在环境变量 `MONGO_URI` 中添加你的数据库地址（比如：`mongodb+srv://:@cluster0-jvwxi.mongodb.net/?retryWrites=true&w=majority`），我们会在 *connection.js* 中调用它。
+*You can set up a free database on [MongoDB Atlas](https://www.mongodb.com/cloud/atlas).*
-*你可以在 [MongoDB Atlas](https://www.mongodb.com/cloud/atlas) 创建一个免费的数据库。*
-
-在连接数据库之后，我们才能让服务器开始监听请求，这样做可以保证服务器在数据库连接前或数据库发生错误时不接受任何请求。为此，我们需要这样写：
+Now we want to connect to our database then start listening for requests. The purpose of this is to not allow requests before our database is connected or if there is a database error. To accomplish this, you will want to encompass your serialization and your app routes in the following code:
```js
myDB(async client => {
@@ -40,13 +38,13 @@ myDB(async client => {
// app.listen out here...
```
-记得要取消 deserializeUser 中 `myDataBase` 的注释，并把 `doc` 添加到 `done(null, null)`。
+Be sure to uncomment the `myDataBase` code in `deserializeUser`, and edit your `done(null, null)` to include the `doc`.
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/175f2f585a2d8034044c7e8857d5add7) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/175f2f585a2d8034044c7e8857d5add7).
# --hints--
-应存在数据库连接。
+Database connection should be present.
```js
(getUserInput) =>
@@ -64,7 +62,7 @@ myDB(async client => {
);
```
-反序列化应正确使用，且应正确调用 `done(null, null)`。
+Deserialization should now be correctly using the DB and `done(null, null)` should be called with the `doc`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/implementation-of-social-authentication-ii.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/implementation-of-social-authentication-ii.md
index e8700409e8e..8b9c8e8d954 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/implementation-of-social-authentication-ii.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/implementation-of-social-authentication-ii.md
@@ -1,6 +1,6 @@
---
id: 589a69f5f9fc0f352b528e71
-title: 实现社交账号登陆 (2)
+title: Implementation of Social Authentication II
challengeType: 2
forumTopicId: 301557
dashedName: implementation-of-social-authentication-ii
@@ -8,11 +8,11 @@ dashedName: implementation-of-social-authentication-ii
# --description--
-设置 GitHub 验证的最后一步是创建策略本身。为此，你需要在 `auth.js` 中`require`'passport-github'，且实例化为 GithubStrategy：`const GithubStrategy = require('passport-github').Strategy;`。别忘了在 `dotenv` 中修改环境变量。
+The last part of setting up your GitHub authentication is to create the strategy itself. For this, you will need to add the dependency of 'passport-github' to your project and require it in your `auth.js` as `GithubStrategy` like this: `const GitHubStrategy = require('passport-github').Strategy;`. Do not forget to require and configure `dotenv` to use your environment variables.
-为了设置 GitHub 策略，我们需要在 **passport** 中使用实例化的 **GithubStrategy**，它可以接收两个参数：一个对象（包括 *clientID*, *clientSecret* 和 *callbackURL*），以及一个回调函数。在这个回调函数中，我们要处理验证成功时，判断用户是否已经在数据库中存在的逻辑，还有如果数据库中不存在，把用户数据添加到数据库的代码。这种处理方式适用于绝大部分第三方验证策略，但有些策略会需要我们提供更多的信息，详情请参考相关策略的 README。例如，Google 的验证策略会要求你提供一个 *scope*，用于标示用户成功登录后，你需要从返回的对象中获取那些信息。以及，这也需要经过用户同意，你才可以获取到。你可以在 [这里](https://github.com/jaredhanson/passport-github/) 了解当前我们使用的验证策略的用法，不过我们也会在这里进行详细讲解。
+To set up the GitHub strategy, you have to tell Passport to use an instantiated `GitHubStrategy`, which accepts 2 arguments: an object (containing `clientID`, `clientSecret`, and `callbackURL`) and a function to be called when a user is successfully authenticated, which will determine if the user is new and what fields to save initially in the user's database object. This is common across many strategies, but some may require more information as outlined in that specific strategy's GitHub README. For example, Google requires a *scope* as well which determines what kind of information your request is asking to be returned and asks the user to approve such access. The current strategy we are implementing has its usage outlined [here](https://github.com/jaredhanson/passport-github/), but we're going through it all right here on freeCodeCamp!
-你的新策略应该这样去实现：
+Here's how your new strategy should look at this point:
```js
passport.use(new GitHubStrategy({
@@ -27,13 +27,13 @@ passport.use(new GitHubStrategy({
));
```
-目前，你的验证部分不会成功。由于没有数据库的逻辑和回调函数，你的代码目前还会报错。但如果你试一试，就可以在右边的控制台里看到输出了你的 GitHub 的个人信息。
+Your authentication won't be successful yet, and it will actually throw an error without the database logic and callback, but it should log your GitHub profile to your console if you try it!
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/ff3a1166684c1b184709ac0bee30dee6) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/ff3a1166684c1b184709ac0bee30dee6).
# --hints--
-应正确添加依赖 passport-github。
+passport-github dependency should be added.
```js
(getUserInput) =>
@@ -52,7 +52,7 @@ passport.use(new GitHubStrategy({
);
```
-应正确引入依赖 passport-github。
+passport-github should be required.
```js
(getUserInput) =>
@@ -70,7 +70,7 @@ passport.use(new GitHubStrategy({
);
```
-到目前为止，Github 策略应正确设置。
+GitHub strategy should be setup correctly thus far.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/implementation-of-social-authentication-iii.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/implementation-of-social-authentication-iii.md
index 1ccca394a43..6e6b65cf56e 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/implementation-of-social-authentication-iii.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/implementation-of-social-authentication-iii.md
@@ -1,6 +1,6 @@
---
id: 589a8eb3f9fc0f352b528e72
-title: 实现社交账号登陆 (3)
+title: Implementation of Social Authentication III
challengeType: 2
forumTopicId: 301558
dashedName: implementation-of-social-authentication-iii
@@ -8,40 +8,45 @@ dashedName: implementation-of-social-authentication-iii
# --description--
-验证策略的最后一部分是处理从 GitHub 返回的个人信息。如果用户存在，我们就需要从数据库中读取用户数据并在 profile 页面加载；否则，我们需要把用户信息添加到数据库。GitHub 在用户信息中为我们提供了独一无二的 *id*，我们可以通过序列化的 id 在数据库中搜索用户（已实现）。以下是这个逻辑的实现示例，我们应该把它传到新策略的第二个参数，就是目前 `console.log(profile);` 的下方：
+The final part of the strategy is handling the profile returned from GitHub. We need to load the user's database object if it exists, or create one if it doesn't, and populate the fields from the profile, then return the user's object. GitHub supplies us a unique *id* within each profile which we can use to search with to serialize the user with (already implemented). Below is an example implementation you can use in your project--it goes within the function that is the second argument for the new strategy, right below where `console.log(profile);` currently is:
```js
-myDataBase.findAndModify(
- {id: profile.id},
- {},
- {$setOnInsert:{
- id: profile.id,
- name: profile.displayName || 'John Doe',
- photo: profile.photos[0].value || '',
- email: Array.isArray(profile.emails) ? profile.emails[0].value : 'No public email',
- created_on: new Date(),
- provider: profile.provider || ''
- },$set:{
- last_login: new Date()
- },$inc:{
- login_count: 1
- }},
- {upsert:true, new: true},
+myDataBase.findOneAndUpdate(
+ { id: profile.id },
+ {
+ $setOnInsert: {
+ id: profile.id,
+ name: profile.displayName || 'John Doe',
+ photo: profile.photos[0].value || '',
+ email: Array.isArray(profile.emails)
+ ? profile.emails[0].value
+ : 'No public email',
+ created_on: new Date(),
+ provider: profile.provider || ''
+ },
+ $set: {
+ last_login: new Date()
+ },
+ $inc: {
+ login_count: 1
+ }
+ },
+ { upsert: true, new: true },
(err, doc) => {
return cb(null, doc.value);
}
);
```
-`findAndModify` 的作用是在数据库中查询对象并更新，如果对象不存在，我们也可以 `upsert`（注，upsert 可以理解为 update + insert）它，然后我们可以在回调方法里获取到插入数据后的新对象。在这个例子中，我们会把 last_login 设置成为 now，而且总会为 login_count 加 1。只有在插入一个新对象（新用户）时，我们才会初始化这些字段。另外，还需要注意默认值的使用。有时返回的用户信息可能不全，可能是因为用户没有填写，也可能是因为用户选择不公开一部分信息。在这种情况下，我们需要进行相应的处理，以防我们的 app 报错。
+`findOneAndUpdate` allows you to search for an object and update it. If the object doesn't exist, it will be inserted and made available to the callback function. In this example, we always set `last_login`, increment the `login_count` by `1`, and only populate the majority of the fields when a new object (new user) is inserted. Notice the use of default values. Sometimes a profile returned won't have all the information filled out or the user will keep it private. In this case, you handle it to prevent an error.
-你现在应该可以登录你的应用了，试试吧。
+You should be able to login to your app now--try it!
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/183e968f0e01d81dde015d45ba9d2745) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/183e968f0e01d81dde015d45ba9d2745).
# --hints--
-GitHub 策略应配置完成。
+GitHub strategy setup should be complete.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/implementation-of-social-authentication.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/implementation-of-social-authentication.md
index 4acae8da91f..66ab3a79dbf 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/implementation-of-social-authentication.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/implementation-of-social-authentication.md
@@ -1,6 +1,6 @@
---
id: 589a69f5f9fc0f352b528e70
-title: 实现社交账号登陆 (1)
+title: Implementation of Social Authentication
challengeType: 2
forumTopicId: 301559
dashedName: implementation-of-social-authentication
@@ -8,21 +8,21 @@ dashedName: implementation-of-social-authentication
# --description--
-第三方用户验证的实现逻辑如下：
+The basic path this kind of authentication will follow in your app is:
-1. 在用户点击按钮或者链接后，进入验证页面，通过第三方平台（如 GitHub）来进行用户验证。
-2. 需要在路由中调用`passport.authenticate('github')`，跳转至 GitHub 验证页面。
-3. 页面跳转到 GitHub 上，如果用户未登录 GitHub，就需要在这里进行登录。登录成功后，会出现向用户申请访问权限的确认页。
-4. 如果用户同意访问，则用户会回到我们提供的回调地址，带着 GitHub 那边提供的用户信息回到我们的 app 中。
-5. 验证已完成。在我们的应用中，我们需要查询这个用户是否已经存在。如果是新用户，那我们需要把他的用户信息存储到数据库。
+1. User clicks a button or link sending them to our route to authenticate using a specific strategy (e.g. GitHub).
+2. Your route calls `passport.authenticate('github')` which redirects them to GitHub.
+3. The page the user lands on, on GitHub, allows them to login if they aren't already. It then asks them to approve access to their profile from our app.
+4. The user is then returned to our app at a specific callback url with their profile if they are approved.
+5. They are now authenticated, and your app should check if it is a returning profile, or save it in your database if it is not.
-在 OAuth 验证策略中，我们至少需要提供 *Client ID* 和 *Client Secret*，这样第三方平台就会获悉验证请求的来源，以及这个来源是否有效。为此，需要去我们使用的第三方验证平台（比如 GitHub）获取这两个字段的值。注意，我们获取到的这个值是唯一的，且仅对我们的当前 app 有效——**因此，千万不要分享给别人**，更不要上传到公共仓库或者直接写在代码里。通常，我们会在 *.env* 文件里配置，并在 Node.js 里通过：`process.env.GITHUB_CLIENT_ID`获取。对于这次挑战，我们将会使用 GitHub 作为验证平台。
+Strategies with OAuth require you to have at least a *Client ID* and a *Client Secret* which is a way for the service to verify who the authentication request is coming from and if it is valid. These are obtained from the site you are trying to implement authentication with, such as GitHub, and are unique to your app--**THEY ARE NOT TO BE SHARED** and should never be uploaded to a public repository or written directly in your code. A common practice is to put them in your `.env` file and reference them like so: `process.env.GITHUB_CLIENT_ID`. For this challenge we're going to use the GitHub strategy.
-首先，你需要进入账户设置里的 [Developer settings](https://github.com/settings/developers)板块，然后在 OAuth Apps 获取 *Client ID* 和 *Client Secret*。点击 'Register a new application'，设置你的应用名称，然后把你的 glitch 主页地址（**注意，不是项目代码的地址**）粘贴到 Homepage URL。然后，回调 url 需要设置成上面 Homepage URL 里你粘贴的地址，但后面要加上 '/auth/github/callback'。这样在用户通过 Github 验证后才能跳转到我们指定的页面。别忘了，我们还需要在 .env 文件里配置好 'GITHUB_CLIENT_ID' 和 'GITHUB_CLIENT_SECRET'。
+Obtaining your *Client ID and Secret* from GitHub is done in your account profile settings under 'developer settings', then '[OAuth applications](https://github.com/settings/developers)'. Click 'Register a new application', name your app, paste in the url to your Repl.it homepage (**Not the project code's url**), and lastly, for the callback url, paste in the same url as the homepage but with `/auth/github/callback` added on. This is where users will be redirected for us to handle after authenticating on GitHub. Save the returned information as `'GITHUB_CLIENT_ID'` and `'GITHUB_CLIENT_SECRET'` in your `.env` file.
-然后，请在你现在的项目里，为 /auth/github 和 /auth/github/callback 创建两个接收 GET 请求的路由。第一个只需要通过 passport 来调用 'github' 验证，第二个需要调用 passport 来验证 'github'，但需要在失败时跳转回主页 '/'，成功是跳转到用户页面 '/profile'。跳转的逻辑与上一个项目中的逻辑一样。
+In your `routes.js` file, add `showSocialAuth: true` to the homepage route, after `showRegistration: true`. Now, create 2 routes accepting GET requests: `/auth/github` and `/auth/github/callback`. The first should only call passport to authenticate `'github'`. The second should call passport to authenticate `'github'` with a failure redirect to `/`, and then if that is successful redirect to `/profile` (similar to our last project).
-例如 '/auth/github/callback' 应该像我们处理在上一个项目中一般的登录一样：
+An example of how `/auth/github/callback` should look is similar to how we handled a normal login:
```js
app.route('/login')
@@ -31,19 +31,19 @@ app.route('/login')
});
```
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/1f7f6f76adb178680246989612bea21e) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project up to this point [here](https://gist.github.com/camperbot/1f7f6f76adb178680246989612bea21e).
# --hints--
-路由 /auth/github 应正确配置。
+Route /auth/github should be correct.
```js
(getUserInput) =>
$.get(getUserInput('url') + '/_api/routes.js').then(
(data) => {
assert.match(
- data,
- /('|")\/auth\/github('|")[^]*get.*passport.authenticate.*github/gi,
+ data.replace(/\s/g, ''),
+ /('|")\/auth\/github\/?\1[^]*?get.*?passport.authenticate.*?github/gi,
'Route auth/github should only call passport.authenticate with github'
);
},
@@ -53,15 +53,15 @@ app.route('/login')
);
```
-路由 /auth/github/callback 应正确配置。
+Route /auth/github/callback should be correct.
```js
(getUserInput) =>
$.get(getUserInput('url') + '/_api/routes.js').then(
(data) => {
assert.match(
- data,
- /('|")\/auth\/github\/callback('|")[^]*get.*passport.authenticate.*github.*failureRedirect:( |)("|')\/\2/gi,
+ data.replace(/\s/g, ''),
+ /('|")\/auth\/github\/callback\/?\1[^]*?get.*?passport.authenticate.*?github.*?failureRedirect:("|')\/\2/gi,
'Route auth/github/callback should accept a get request and call passport.authenticate for github with a failure redirect to home'
);
},
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/logging-a-user-out.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/logging-a-user-out.md
index dbad5387c6a..ba023ef1b89 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/logging-a-user-out.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/logging-a-user-out.md
@@ -1,6 +1,6 @@
---
id: 58965611f9fc0f352b528e6c
-title: 用户退出登录
+title: Logging a User Out
challengeType: 2
forumTopicId: 301560
dashedName: logging-a-user-out
@@ -8,9 +8,9 @@ dashedName: logging-a-user-out
# --description--
-创建退出登录的逻辑是比较简单的。只要用户尝试退出登录，路由就应重定向到主页，而不应该显示任何其他页面。
+Creating the logout logic is easy. The route should just unauthenticate the user and redirect to the home page instead of rendering any view.
-在 passport 里，只需要在重定向前调用 `req.logout();` 即可完成用户的退出登录。
+In passport, unauthenticating a user is as easy as just calling `req.logout();` before redirecting.
```js
app.route('/logout')
@@ -20,7 +20,7 @@ app.route('/logout')
});
```
-你可能注意到我们还没有处理 404 错误，这个错误码代表页面无法找到。在 Node.js 中我们通常会用如下的中间件来处理，请在所有路由之后添加这段代码：
+You may have noticed that we're not handling missing pages (404). The common way to handle this in Node is with the following middleware. Go ahead and add this in after all your other routes:
```js
app.use((req, res, next) => {
@@ -30,11 +30,11 @@ app.use((req, res, next) => {
});
```
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/c3eeb8a3ebf855e021fd0c044095a23b) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/c3eeb8a3ebf855e021fd0c044095a23b).
# --hints--
-`req.Logout` 应在 `/logout` 中调用。
+`req.Logout` should be called in your `/logout` route.
```js
(getUserInput) =>
@@ -52,7 +52,7 @@ app.use((req, res, next) => {
);
```
-退出登录后应重定向到主页 /
+Logout should redirect to the home page.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/registration-of-new-users.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/registration-of-new-users.md
index 52ff5eea5e4..a18599d868e 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/registration-of-new-users.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/registration-of-new-users.md
@@ -1,6 +1,6 @@
---
id: 58966a17f9fc0f352b528e6d
-title: 注册新用户
+title: Registration of New Users
challengeType: 2
forumTopicId: 301561
dashedName: registration-of-new-users
@@ -8,11 +8,11 @@ dashedName: registration-of-new-users
# --description--
-现在我们需要为新用户添加注册帐号的功能，首先我们需要在主页的 res.render 接收的变量对象中添加 `showRegistration: true`。此时刷新页面，你会看到页面上已经显示了我们在 index.pug 文件中定义的注册表单。这个表单设置了请求路径 */register*，并将请求方法设置成 **POST**，所以我们需要在服务器接受 **POST** 请求，且在数据库中创建用户对象。
+Now we need to allow a new user on our site to register an account. On the `res.render` for the home page add a new variable to the object passed along--`showRegistration: true`. When you refresh your page, you should then see the registration form that was already created in your `index.pug` file! This form is set up to **POST** on `/register`, so this is where we should set up to accept the **POST** and create the user object in the database.
-用户注册的逻辑如下：注册新用户 > 认证新用户 > 重定向到 /profile。
+The logic of the registration route should be as follows: Register the new user > Authenticate the new user > Redirect to /profile
-对于步骤一的注册新用户，详细逻辑如下：用 findOne 命令查询数据库 > 如果返回了用户对象，则表示用户存在，然后返回主页；如果用户未定义且没有报错，则会将包含用户名和密码的用户对象通过 `insertOne` 添加到数据库，只要没有报错则会继续下一步：认证新用户——我们已经在 /login 路由的 POST 请求中写好了这部分逻辑。
+The logic of step 1, registering the new user, should be as follows: Query database with a findOne command > if user is returned then it exists and redirect back to home *OR* if user is undefined and no error occurs then 'insertOne' into the database with the username and password, and, as long as no errors occur, call *next* to go to step 2, authenticating the new user, which we've already written the logic for in our POST */login* route.
```js
app.route('/register')
@@ -47,13 +47,13 @@ app.route('/register')
);
```
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/b230a5b3bbc89b1fa0ce32a2aa7b083e) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/b230a5b3bbc89b1fa0ce32a2aa7b083e).
-**注意：** 接下来的挑战可能会在运行 *picture-in-picture*（画中画）模式的浏览器中出现问题。如果你使用的线上 IDE 提供在 IDE 内预览 app 的功能，请考虑打开新的标签页预览。
+**NOTE:** From this point onwards, issues can arise relating to the use of the *picture-in-picture* browser. If you are using an online IDE which offers a preview of the app within the editor, it is recommended to open this preview in a new tab.
# --hints--
-注册路由和显示主页。
+You should register route and display on home.
```js
(getUserInput) =>
@@ -76,7 +76,7 @@ app.route('/register')
);
```
-注册功能应可以正常运行。
+Registering should work.
```js
async (getUserInput) => {
@@ -104,7 +104,7 @@ async (getUserInput) => {
};
```
-登录功能应可以正常运行。
+Login should work.
```js
async (getUserInput) => {
@@ -153,7 +153,7 @@ async (getUserInput) => {
};
```
-退出登录功能应可以正常运行。
+Logout should work.
```js
(getUserInput) =>
@@ -171,7 +171,7 @@ async (getUserInput) => {
);
```
-退出登录后，profile 页面应无法访问。
+Profile should no longer work after logout.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/send-and-display-chat-messages.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/send-and-display-chat-messages.md
index ad5d6185522..e74551d1f99 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/send-and-display-chat-messages.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/send-and-display-chat-messages.md
@@ -1,6 +1,6 @@
---
id: 589fc832f9fc0f352b528e79
-title: 发送和显示聊天消息
+title: Send and Display Chat Messages
challengeType: 2
forumTopicId: 301562
dashedName: send-and-display-chat-messages
@@ -8,7 +8,7 @@ dashedName: send-and-display-chat-messages
# --description--
-现在，我们可以开始实现聊天室功能了。整体逻辑很简单，只需要获取用户发给服务端的消息，再通过服务端给所有客户端发送信息就可以了。在 client.js 文件里，你应该已经注意到了这段提交表单的代码：
+It's time you start allowing clients to send a chat message to the server to emit to all the clients! In your `client.js` file, you should see there is already a block of code handling when the message form is submitted.
```js
$('form').submit(function() {
@@ -16,23 +16,23 @@ $('form').submit(function() {
});
```
-现在我们需要处理事件的 emit，它应该发生在定义 `messageToSend` 之后，以及清除 `#m` 中的文本之前。我们称这个事件叫 `'chat message'`，需发送的数据叫 `messageToSend`：
+Within the form submit code, you should emit an event after you define `messageToSend` but before you clear the text box `#m`. The event should be named `'chat message'` and the data should just be `messageToSend`.
```js
socket.emit('chat message', messageToSend);
```
-在服务端，我们需要监听包含 `message` 的 `'chat message'` 事件。一旦事件发生，我们就通过`io.emit` 把包含 `name` 和 `message` 的 `'chat message'` 事件发送给所有已连接的 socket。
+Now, on your server, you should be listening to the socket for the event `'chat message'` with the data being named `message`. Once the event is received, it should emit the event `'chat message'` to all sockets `io.emit` with the data being an object containing `name` and `message`.
-回到客户端，我们需要监听 `'chat message'` 事件。只要接收到这个事件，就把包含名字和消息的内容（注意：需要在名字后添加冒号）通过 `<li>` 添加到 `#messages`。
+In `client.js`, you should now listen for event `'chat message'` and, when received, append a list item to `#messages` with the name, a colon, and the message!
-至此，我们已经完成发送信息到所有客户端的功能。
+At this point, the chat should be fully functional and sending messages across all clients!
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/d7af9864375207e254f73262976d2016) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/d7af9864375207e254f73262976d2016).
# --hints--
-服务端应监听 `'chat message'`，且应在监听到后 emit。
+Server should listen for `'chat message'` and then emit it properly.
```js
(getUserInput) =>
@@ -50,7 +50,7 @@ socket.emit('chat message', messageToSend);
);
```
-客户端应正确处理和展示从 `'chat message'` 事件发来的新数据。
+Client should properly handle and display the new data from event `'chat message'`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/serialization-of-a-user-object.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/serialization-of-a-user-object.md
index 6b1d5c37360..f3967b928a2 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/serialization-of-a-user-object.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/serialization-of-a-user-object.md
@@ -1,6 +1,6 @@
---
id: 5895f70cf9fc0f352b528e66
-title: 用户对象的序列化
+title: Serialization of a User Object
challengeType: 2
forumTopicId: 301563
dashedName: serialization-of-a-user-object
@@ -8,11 +8,11 @@ dashedName: serialization-of-a-user-object
# --description--
-序列化和反序列化在身份认证中是很重要的概念。序列化一个对象就是将其内容转换成一个体积很小的 *key*，后续可以通过这个 *key* 反序列化为原始对象。这样，服务器就可以在用户未登录时识别用户，或者说给这个用户一个唯一标识，用户也不需要在每次访问不同页面时都给服务器发送用户名和密码。
+Serialization and deserialization are important concepts in regards to authentication. To serialize an object means to convert its contents into a small *key* that can then be deserialized into the original object. This is what allows us to know who has communicated with the server without having to send the authentication data, like the username and password, at each request for a new page.
-我们需要用到序列化和反序列化的方法来进行配置。passport 为我们提供了 `passport.serializeUser( OURFUNCTION )` 和 `passport.deserializeUser( OURFUNCTION )` 两个方法。
+To set this up properly, we need to have a serialize function and a deserialize function. In Passport, we create these with `passport.serializeUser( OURFUNCTION )` and `passport.deserializeUser( OURFUNCTION )`
-code>serializeUser方法接收两个参数，分别是表示用户的对象和一个回调函数。其中，回调函数的返回值应为这个用户的唯一标识符：最简单的写法就是让它返回用户的`_id`，这个`_id`属性是 MongoDB 为用户创建的唯一字段。类似地，反序列化也接收两个参数，分别是在序列化时生成的标识符以及一个回调函数。在回调函数里，我们需要根据根据传入的标识符（比如 \_id）返回表示用户的对象。为了在 MongoDB 中通过 query（查询语句）获取 `_id` 字段，首先我们需要引入 MongoDB 的`ObjectID`方法：`const ObjectID = require('mongodb').ObjectID;`；然后调用它：`new ObjectID(THE_ID)`。当然，这一切的前提都是先引入 MongoDB 作为依赖。你可以在下面的例子中看到：
+The `serializeUser` is called with 2 arguments, the full user object and a callback used by passport. A unique key to identify that user should be returned in the callback, the easiest one to use being the user's `_id` in the object. It should be unique as it generated by MongoDB. Similarly, `deserializeUser` is called with that key and a callback function for passport as well, but, this time, we have to take that key and return the full user object to the callback. To make a query search for a Mongo `_id`, you will have to create `const ObjectID = require('mongodb').ObjectID;`, and then to use it you call `new ObjectID(THE_ID)`. Be sure to add MongoDB as a dependency. You can see this in the examples below:
```js
passport.serializeUser((user, done) => {
@@ -26,13 +26,13 @@ passport.deserializeUser((id, done) => {
});
```
-注意：在完全配置好 MongoDB 前，`deserializeUser` 会抛出错误。因此，现在请先注释掉上面的代码，在 `deserializeUser` 中仅调用 `done(null, null)` 即可。
+NOTE: This `deserializeUser` will throw an error until we set up the DB in the next step, so for now comment out the whole block and just call `done(null, null)` in the function `deserializeUser`.
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/7068a0d09e61ec7424572b366751f048) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/7068a0d09e61ec7424572b366751f048).
# --hints--
-应存在正确的 `serializeUser` 方法。
+You should serialize user function correctly.
```js
(getUserInput) =>
@@ -55,7 +55,7 @@ passport.deserializeUser((id, done) => {
);
```
-应存在正确的 `deserializeUser` 方法。
+You should deserialize user function correctly.
```js
(getUserInput) =>
@@ -78,7 +78,7 @@ passport.deserializeUser((id, done) => {
);
```
-MongoDB 应作为项目的依赖。
+MongoDB should be a dependency.
```js
(getUserInput) =>
@@ -97,7 +97,7 @@ MongoDB 应作为项目的依赖。
);
```
-注释掉的代码中应包含 `ObjectId`。
+Mongodb should be properly required including the ObjectId.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/set-up-a-template-engine.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/set-up-a-template-engine.md
index 6dbc95645ec..eb90238236c 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/set-up-a-template-engine.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/set-up-a-template-engine.md
@@ -1,6 +1,6 @@
---
id: 5895f700f9fc0f352b528e63
-title: 设置模板引擎
+title: Set up a Template Engine
challengeType: 2
forumTopicId: 301564
dashedName: set-up-a-template-engine
@@ -8,25 +8,25 @@ dashedName: set-up-a-template-engine
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-advancednode), or clone from [GitHub](https://github.com/freeCodeCamp/boilerplate-advancednode/).
-你可以在应用的模版引擎中使用静态模板文件（如那些写在 *Pug* 里的）。在运行时，模版引擎会用服务端的真实数据替换掉模版文件中的变量，然后将模版转译成发送给客户端的 HTML 静态文件。这样可以轻松地构造 HTML 页面，允许在页面直接显示变量内容而不需要发送 API 请求。
+A template engine enables you to use static template files (such as those written in *Pug*) in your app. At runtime, the template engine replaces variables in a template file with actual values which can be supplied by your server. Then it transforms the template into a static HTML file that is sent to the client. This approach makes it easier to design an HTML page and allows for displaying variables on the page without needing to make an API call from the client.
-为了在项目中使用 *Pug*，你需要在 package.json 中添加依赖 `"pug": "^0.1.0"`。注意，依赖的名称和版本号都要添加。
+Add `pug@~3.0.0` as a dependency in your `package.json` file.
-为了在 Express 中使用 pug 作为模版引擎，你需要在 express 中将 **app** 的 "view-engine" 设置为 "pug"，就像这样：`app.set('view engine', 'pug')`。
+Express needs to know which template engine you are using. We will use the `set` method to assign `pug` as the `view engine` property's value: `app.set('view engine', 'pug')`
-如果没有正确的 *render* *'views/pug'* 路径下的 index 文件，页面将不会被加载。
+Your page will not load until you correctly render the index file in the `views/pug` directory.
-最后, 你需要修改 `res.render()` 方法，设置 `/` 的响应。`res.render()` 方法接收一个文件路径作为参数，这个路径既可以是相对路径（相对于 views），也可以是绝对路径。而且，我们不需要给它添加文件扩展名（文件后缀名）。
+Change the argument of the `res.render()` declaration in the `/` route to be the file path to the `views/pug` directory. The path can be a relative path (relative to views), or an absolute path, and does not require a file extension.
-如果一切顺利，刷新一下应用的主页就不会看到 "Pug template is not defined." 的报错了，而是会看到 Pug 成功加载的提示。
+If all went as planned, your app home page will stop showing the message "`Pug template is not defined.`" and will now display a message indicating you've successfully rendered the Pug template!
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/3515cd676ea4dfceab4e322f59a37791) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/3515cd676ea4dfceab4e322f59a37791).
# --hints--
-项目中应使用 Pug 作为依赖。
+Pug should be a dependency.
```js
(getUserInput) =>
@@ -45,7 +45,7 @@ dashedName: set-up-a-template-engine
);
```
-项目中应使用 Pug 作为模版引擎。
+View engine should be Pug.
```js
(getUserInput) =>
@@ -63,7 +63,7 @@ dashedName: set-up-a-template-engine
);
```
-在 Response 里使用正确的 ExpressJS 方法渲染主页面。
+Use the correct ExpressJS method to render the index page from the response.
```js
(getUserInput) =>
@@ -81,7 +81,7 @@ dashedName: set-up-a-template-engine
);
```
-Pug 应该生效。
+Pug should be working.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/set-up-passport.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/set-up-passport.md
index fb639227a97..852d12c28cc 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/set-up-passport.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/set-up-passport.md
@@ -1,6 +1,6 @@
---
id: 5895f70cf9fc0f352b528e65
-title: 设置 Passport
+title: Set up Passport
challengeType: 2
forumTopicId: 301565
dashedName: set-up-passport
@@ -8,15 +8,15 @@ dashedName: set-up-passport
# --description--
-现在我们来创建 *Passport*，最终我们需要用它来实现用户注册和登录。除了 Passport，我们还会用 express-session 来处理 session（会话）。在客户端，我们可以用这个中间件把 session id 储存到 cookie。同时，我们可以在服务器上通过这个 id 访问 session 数据。通过这种方式，我们无需把用户的个人信息存到 cookie 来只完成用户的验证，只需要用这个 id 作为 *key* 来访问服务器上用户的数据即可。
+It's time to set up *Passport* so we can finally start allowing a user to register or login to an account! In addition to Passport, we will use Express-session to handle sessions. Using this middleware saves the session id as a cookie in the client and allows us to access the session data using that id on the server. This way we keep personal account information out of the cookie used by the client to verify to our server they are authenticated and just keep the *key* to access the data stored on the server.
-为了在你的项目中使用 Passport，首先你需要在 package.json 文件中添加依赖：`"passport": "^0.3.2"`
+To set up Passport for use in your project, you will need to add it as a dependency first in your package.json. `"passport": "^0.3.2"`
-此外，还需要添加 express-session 作为依赖，就像这样：`"express-session": "^1.15.0"`。express-session 有许多高级特性，但我们暂时只需要了解其基础功能。
+In addition, add Express-session as a dependency now as well. Express-session has a ton of advanced features you can use but for now we're just going to use the basics! `"express-session": "^1.15.0"`
-现在，我们需要配置 session 并初始化 Passport。请先创建变量 `session` 和 `passport` 来引入 express-session 和 passport。
+You will need to set up the session settings now and initialize Passport. Be sure to first create the variables 'session' and 'passport' to require 'express-session' and 'passport' respectively.
-为了让 express 应用可以使用 session，我们需要添加一些基础选项。请在 .env 文件中添加字段 'SESSION_SECRET'，并给它赋一个随机值，便于加密 cookie、计算哈希。
+To set up your express app to use the session we'll define just a few basic options. Be sure to add 'SESSION_SECRET' to your .env file and give it a random value. This is used to compute the hash used to encrypt your cookie!
```js
app.use(session({
@@ -27,13 +27,13 @@ app.use(session({
}));
```
-还有，我们需要让 express 使用 `passport.initialize()` 和 `passport.session()`。为此，你可以这样写：`app.use(passport.initialize());`。
+As well you can go ahead and tell your express app to **use** 'passport.initialize()' and 'passport.session()'. (For example, `app.use(passport.initialize());`)
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/4068a7662a2f9f5d5011074397d6788c) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/4068a7662a2f9f5d5011074397d6788c).
# --hints--
-应添加 Passort 和 express-session 作为依赖。
+Passport and Express-session should be dependencies.
```js
(getUserInput) =>
@@ -57,7 +57,7 @@ app.use(session({
);
```
-依赖应正确引入。
+Dependencies should be correctly required.
```js
(getUserInput) =>
@@ -80,7 +80,7 @@ app.use(session({
);
```
-express 应调用 passport 的方法。
+Express app should use new dependencies.
```js
(getUserInput) =>
@@ -103,7 +103,7 @@ express 应调用 passport 的方法。
);
```
-应正确设置 session 和 session secret。
+Session and session secret should be correctly set up.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/set-up-the-environment.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/set-up-the-environment.md
index 487ee878d1d..12fc60fb52e 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/set-up-the-environment.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/set-up-the-environment.md
@@ -1,6 +1,6 @@
---
id: 589fc830f9fc0f352b528e74
-title: 设置环境
+title: Set up the Environment
challengeType: 2
forumTopicId: 301566
dashedName: set-up-the-environment
@@ -8,20 +8,20 @@ dashedName: set-up-the-environment
# --description--
-在接下来的挑战中，我们将会用到 `chat.pug` 文件。首先，你需要在你的 `routes.js` 文件中为 `/chat` 添加一个处理 GET 请求的路由，并给它传入 `ensureAuthenticated`。在回调函数中，我们需要让它 render `chat.pug` 文件，并在响应中包含 `{ user: req.user }` 信息。现在，请修改 `/auth/github/callback` 路由，让它可以像这样设置 user_id：`req.session.user_id = req.user.id`，并在设置完成后重定向至 `/chat`。
+The following challenges will make use of the `chat.pug` file. So, in your `routes.js` file, add a GET route pointing to `/chat` which makes use of `ensureAuthenticated`, and renders `chat.pug`, with `{ user: req.user }` passed as an argument to the response. Now, alter your existing `/auth/github/callback` route to set the `req.session.user_id = req.user.id`, and redirect to `/chat`.
-我们还需要添加 `http` 和 `socket.io` 两个依赖项，并且像这样引入：
+Add `http` and `socket.io` as a dependency and require/instantiate them in your server defined as follows:
```javascript
const http = require('http').createServer(app);
const io = require('socket.io')(http);
```
-现在我们的 *express* 应用已经包含了 *http* 服务，接下来我们需要监听 *http* 服务的事件。为此，我们需要把 `app.listen` 更新为 `http.listen`。
+Now that the *http* server is mounted on the *express app*, you need to listen from the *http* server. Change the line with `app.listen` to `http.listen`.
-我们需要处理的第一件事是监听从客户端发出的连接事件，我们可以调用 <dfn>on</dfn> 方法来监听具体的事件。它接收两个参数：一个是发出的事件的标题字符串，另一个是后续用来传递数据的回调函数。在这个回调函数中，我们用 *socket* 来代表它所包含的数据。简单来说，socket 就是指已连接到服务器的客户端。
+The first thing needing to be handled is listening for a new connection from the client. The <dfn>on</dfn> keyword does just that- listen for a specific event. It requires 2 arguments: a string containing the title of the event thats emitted, and a function with which the data is passed though. In the case of our connection listener, we use *socket* to define the data in the second argument. A socket is an individual client who is connected.
-为了可以监听服务器的连接事件，我们在数据库连接的部分加入如下代码：
+To listen for connections to your server, add the following within your database connection:
```javascript
io.on('connection', socket => {
@@ -29,24 +29,24 @@ io.on('connection', socket => {
});
```
-对于发出连接事件的客户端，只需要在 `client.js` 中添加以下内容：
+Now for the client to connect, you just need to add the following to your `client.js` which is loaded by the page after you've authenticated:
```js
/*global io*/
let socket = io();
```
-注意，这个 `client.js` 文件是在用户通过验证后才加载到客户端的。在这个文件中，我们没有定义 io 变量，但第一行的注释会阻止运行时产生的报错。不过，我们在 chat.pug 的页面上已经为你添加好了 Socket.IO 库的 CDN。
+The comment suppresses the error you would normally see since 'io' is not defined in the file. We've already added a reliable CDN to the Socket.IO library on the page in chat.pug.
-现在你可以重启一下你的 app，尝试一下验证用户，然后你应该会看到服务器的 console 里输出了 'A user has connected'。
+Now try loading up your app and authenticate and you should see in your server console 'A user has connected'!
-**注意：** 只有在连接到处于同一个 url/server 上的 socket 时，`io()`才可以正常执行。如果需要连接到外部的 socket，就需要这样调用：`io.connect('URL');`。
+**Note:**`io()` works only when connecting to a socket hosted on the same url/server. For connecting to an external socket hosted elsewhere, you would use `io.connect('URL');`.
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/aae41cf59debc1a4755c9a00ee3859d1) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/aae41cf59debc1a4755c9a00ee3859d1).
# --hints--
-应添加 Socket.IO 作为依赖。
+`socket.io` should be a dependency.
```js
(getUserInput) =>
@@ -65,7 +65,7 @@ let socket = io();
);
```
-应正确引入 `http`，并示例化为 `http`。
+You should correctly require and instantiate `http` as `http`.
```js
(getUserInput) =>
@@ -83,7 +83,7 @@ let socket = io();
);
```
-应正确引入 `socket.io`，并示例化为 `io`。
+You should correctly require and instantiate `socket.io` as `io`.
```js
(getUserInput) =>
@@ -101,7 +101,7 @@ let socket = io();
);
```
-Socket.IO 应监听连接。
+Socket.IO should be listening for connections.
```js
(getUserInput) =>
@@ -119,7 +119,7 @@ Socket.IO 应监听连接。
);
```
-客户端应连接到服务器。
+Your client should connect to your server.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/use-a-template-engines-powers.md b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/use-a-template-engines-powers.md
index e6ab8b6e633..3f605fa3abf 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/use-a-template-engines-powers.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/advanced-node-and-express/use-a-template-engines-powers.md
@@ -1,6 +1,6 @@
---
id: 5895f70bf9fc0f352b528e64
-title: 使用模板引擎
+title: Use a Template Engine's Powers
challengeType: 2
forumTopicId: 301567
dashedName: use-a-template-engines-powers
@@ -8,25 +8,23 @@ dashedName: use-a-template-engines-powers
# --description--
-模版引擎最大的特点之一就是在 HTML 页面展示之前，可以从服务端传变量到模版文件。
+One of the greatest features of using a template engine is being able to pass variables from the server to the template file before rendering it to HTML.
-在 Pug 文件中，你可以用变量名来调用变量，比如写成 `#{variable_name}` 来实现行内调用，或像 `p= variable_name` 把元素与变量直接写在一起，这表示 p 元素的内容等价于这个变量。
+In your Pug file, you're able to use a variable by referencing the variable name as `#{variable_name}` inline with other text on an element or by using an equal sign on the element without a space such as `p=variable_name` which assigns the variable's value to the p element's text.
-建议大家在 [Pug 的 README](https://github.com/pugjs/pug) 里看看它的语法和用法，这样你写出的代码会相对简练。另外要注意，Pug 使用缩进来表示嵌套的代码块。
+We strongly recommend looking at the syntax and structure of Pug [here](https://github.com/pugjs/pug) on GitHub's README. Pug is all about using whitespace and tabs to show nested elements and cutting down on the amount of code needed to make a beautiful site.
-在 pug 的 'index.pug' 文件中，我们使用了 *title* 和 *message* 两个变量。
+Looking at our pug file 'index.pug' included in your project, we used the variables *title* and *message*.
-为了从服务器传递这些信息，你需要给 *res.render* 的第二个参数传入一个对象，其中包含变量对应的值。比如，如果你想传递对象 `{title: 'Hello', message: 'Please login'` 到你的主页，那么应该这样写：
+To pass those along from our server, you will need to add an object as a second argument to your *res.render* with the variables and their values. For example, pass this object along setting the variables for your index view: `{title: 'Hello', message: 'Please login'}`
-`res.render(process.cwd() + '/views/pug/index', {title: 'Hello', message: 'Please login'});`
+It should look like: `res.render(process.cwd() + '/views/pug/index', {title: 'Hello', message: 'Please login'});` Now refresh your page and you should see those values rendered in your view in the correct spot as laid out in your index.pug file!
-刷新页面，如果页面中数据显示正确，你就可以提交你的页面了。
-
-完成上述要求后，你可以在下方提交你的页面链接。如果你遇到了问题，可以参考 [这里](https://gist.github.com/camperbot/4af125119ed36e6e6a8bb920db0c0871) 的答案。
+Submit your page when you think you've got it right. If you're running into errors, you can check out the project completed up to this point [here](https://gist.github.com/camperbot/4af125119ed36e6e6a8bb920db0c0871).
# --hints--
-Pug 应正确地展示变量。
+Pug should correctly render variables.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/assert-deep-equality-with-.deepequal-and-.notdeepequal.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/assert-deep-equality-with-.deepequal-and-.notdeepequal.md
index 143450be342..2a49a6e67b3 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/assert-deep-equality-with-.deepequal-and-.notdeepequal.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/assert-deep-equality-with-.deepequal-and-.notdeepequal.md
@@ -1,6 +1,6 @@
---
id: 587d824c367417b2b2512c4c
-title: 用 Assert.deepEqual() 和 Assert.notDeepEqual() 断言深度相等
+title: Assert Deep Equality with .deepEqual and .notDeepEqual
challengeType: 2
forumTopicId: 301587
dashedName: assert-deep-equality-with--deepequal-and--notdeepequal
@@ -8,17 +8,17 @@ dashedName: assert-deep-equality-with--deepequal-and--notdeepequal
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
-`deepEqual()` 断言两个对象是否 deep equal
+`deepEqual()` asserts that two objects are deep equal.
# --instructions--
-使用 `assert.deepEqual()` 或者 `assert.notDeepEqual()` 来让测试通过。
+Within `tests/1_unit-tests.js` under the test labelled `#7` in the `Equality` suite, change each `assert` to either `assert.deepEqual` or `assert.notDeepEqual` to make the test pass (should evaluate to `true`). Do not alter the arguments passed to the asserts.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -32,7 +32,7 @@ dashedName: assert-deep-equality-with--deepequal-and--notdeepequal
);
```
-请选择正确的断言—deepEqual 或 notDeepEqual
+You should choose the correct method for the first assertion - `deepEqual` vs. `notDeepEqual`.
```js
(getUserInput) =>
@@ -50,7 +50,7 @@ dashedName: assert-deep-equality-with--deepequal-and--notdeepequal
);
```
-请选择正确的断言—deepEqual 或 notDeepEqual
+You should choose the correct method for the second assertion - `deepEqual` vs. `notDeepEqual`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/compare-the-properties-of-two-elements.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/compare-the-properties-of-two-elements.md
index 78281718a37..49e53e01b3c 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/compare-the-properties-of-two-elements.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/compare-the-properties-of-two-elements.md
@@ -1,6 +1,6 @@
---
id: 587d824c367417b2b2512c4d
-title: 比较两个元素的属性
+title: Compare the Properties of Two Elements
challengeType: 2
forumTopicId: 301588
dashedName: compare-the-properties-of-two-elements
@@ -8,15 +8,15 @@ dashedName: compare-the-properties-of-two-elements
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
# --instructions--
-使用 `assert.isAbove()`（大于）或者 `assert.isAtMost()`（小于或等于）来让测试通过。
+Within `tests/1_unit-tests.js` under the test labelled `#8` in the `Comparisons` suite, change each `assert` to either `assert.isAbove` or `assert.isAtMost` to make the test pass (should evaluate to `true`). Do not alter the arguments passed to the asserts.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -30,7 +30,7 @@ dashedName: compare-the-properties-of-two-elements
);
```
-请选择正确的断言—isAbove 或 isAtMost
+You should choose the correct method for the first assertion - `isAbove` vs. `isAtMost`.
```js
(getUserInput) =>
@@ -48,7 +48,7 @@ dashedName: compare-the-properties-of-two-elements
);
```
-请选择正确的断言—isAbove 或 isAtMost
+You should choose the correct method for the second assertion - `isAbove` vs. `isAtMost`.
```js
(getUserInput) =>
@@ -62,7 +62,7 @@ dashedName: compare-the-properties-of-two-elements
);
```
-请选择正确的断言—isAbove 或 isAtMost
+You should choose the correct method for the third assertion - `isAbove` vs. `isAtMost`.
```js
(getUserInput) =>
@@ -80,7 +80,7 @@ dashedName: compare-the-properties-of-two-elements
);
```
-请选择正确的断言—isAbove 或 isAtMost
+You should choose the correct method for the fourth assertion - `isAbove` vs. `isAtMost`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/learn-how-javascript-assertions-work.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/learn-how-javascript-assertions-work.md
index 6b6236e4842..411c4032d56 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/learn-how-javascript-assertions-work.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/learn-how-javascript-assertions-work.md
@@ -1,6 +1,6 @@
---
id: 587d824a367417b2b2512c46
-title: 了解 JavaScript 断言是如何使用的
+title: Learn How JavaScript Assertions Work
challengeType: 2
forumTopicId: 301589
dashedName: learn-how-javascript-assertions-work
@@ -8,15 +8,15 @@ dashedName: learn-how-javascript-assertions-work
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
# --instructions--
-使用 assert.isNull() 或 assert.isNotNull() 来让测试通过。
+Within `tests/1_unit-tests.js` under the test labelled `#1` in the `Basic Assertions` suite, change each `assert` to either `assert.isNull` or `assert.isNotNull` to make the test pass (should evaluate to `true`). Do not alter the arguments passed to the asserts.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -30,7 +30,7 @@ dashedName: learn-how-javascript-assertions-work
);
```
-请选择正确的断言—isNull 或 isNotNull
+You should choose the correct method for the first assertion - `isNull` vs. `isNotNull`.
```js
(getUserInput) =>
@@ -44,7 +44,7 @@ dashedName: learn-how-javascript-assertions-work
);
```
-You should choose the right assertion - isNull vs. isNotNull.
+You should choose the correct method for the second assertion - `isNull` vs. `isNotNull`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-on-an-api-response-using-chai-http-iii---put-method.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-on-an-api-response-using-chai-http-iii---put-method.md
index 30e6a91a5b2..fef7dae3002 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-on-an-api-response-using-chai-http-iii---put-method.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-on-an-api-response-using-chai-http-iii---put-method.md
@@ -1,6 +1,6 @@
---
id: 587d824f367417b2b2512c5a
-title: 使用 Chai-HTTP 测试 API 响应结果 (3)—PUT 方法
+title: Run Functional Tests on an API Response using Chai-HTTP III - PUT method
challengeType: 2
forumTopicId: 301590
dashedName: run-functional-tests-on-an-api-response-using-chai-http-iii---put-method
@@ -8,11 +8,9 @@ dashedName: run-functional-tests-on-an-api-response-using-chai-http-iii---put-me
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
-接下来，我们将了解如何使用请求的 payload（body）发送数据。
-
-我们需要测试一个 PUT 请求，`'/travellers'` 接收如下的 JSON 对象：
+In the next example we'll see how to send data in a request payload (body). We are going to test a PUT request. The `'/travellers'` endpoint accepts a JSON object taking the structure:
```json
{
@@ -20,7 +18,7 @@ dashedName: run-functional-tests-on-an-api-response-using-chai-http-iii---put-me
}
```
-响应数据如下：
+The route responds with :
```json
{
@@ -28,11 +26,13 @@ dashedName: run-functional-tests-on-an-api-response-using-chai-http-iii---put-me
}
```
-更多细节请查看服务器代码。
+See the server code for more details.
# --instructions--
-发送
+Within `tests/2_functional-tests.js`, alter the `'send {surname: "Colombo"}'` test (`// #3`):
+
+Send the following JSON response as a payload:
```json
{
@@ -40,11 +40,18 @@ dashedName: run-functional-tests-on-an-api-response-using-chai-http-iii---put-me
}
```
-替换 `assert.fail()` 使测试通过。 分别测试 1) `status`，2) `type`，3) `body.name`，4) `body.surname` 请按照以上顺序书写断言，顺序错误会影响此挑战的判定。
+Check for the following, within the `request.end` callback:
+
+1. `status`
+2. `type`
+3. `body.name`
+4. `body.surname`
+
+Follow the assertion order above - we rely on it. Be sure to remove `assert.fail()`, once complete.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -58,7 +65,7 @@ dashedName: run-functional-tests-on-an-api-response-using-chai-http-iii---put-me
);
```
-你需要测试 'res.status' 是否为 200
+You should test for 'res.status' to be 200.
```js
(getUserInput) =>
@@ -74,7 +81,7 @@ dashedName: run-functional-tests-on-an-api-response-using-chai-http-iii---put-me
);
```
-你需要测试 'res.type' 是否为 'application/json'
+You should test for 'res.type' to be 'application/json'.
```js
(getUserInput) =>
@@ -90,7 +97,7 @@ dashedName: run-functional-tests-on-an-api-response-using-chai-http-iii---put-me
);
```
-你需要测试 'res.body.name' 是否为 'Cristoforo'
+You should test for 'res.body.name' to be 'Cristoforo'.
```js
(getUserInput) =>
@@ -106,7 +113,7 @@ dashedName: run-functional-tests-on-an-api-response-using-chai-http-iii---put-me
);
```
-你需要测试 'res.body.surname' 是否为 'Colombo'
+You should test for 'res.body.surname' to be 'Colombo'.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-on-an-api-response-using-chai-http-iv---put-method.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-on-an-api-response-using-chai-http-iv---put-method.md
index 5a8dc95fde0..e52c9af7fbc 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-on-an-api-response-using-chai-http-iv---put-method.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-on-an-api-response-using-chai-http-iv---put-method.md
@@ -1,6 +1,6 @@
---
id: 587d824f367417b2b2512c5b
-title: 使用 Chai-HTTP 测试 API 响应结果 (4)—PUT 方法
+title: Run Functional Tests on an API Response using Chai-HTTP IV - PUT method
challengeType: 2
forumTopicId: 301591
dashedName: run-functional-tests-on-an-api-response-using-chai-http-iv---put-method
@@ -8,13 +8,15 @@ dashedName: run-functional-tests-on-an-api-response-using-chai-http-iv---put-met
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/). This exercise is similar to the preceding one. Look at it for the details.
-这个练习与上一个类似，我们详细看看。
+Now that you have seen how it is done, it is your turn to do it from scratch.
# --instructions--
-发送
+Within `tests/2_functional-tests.js`, alter the `'send {surname: "da Verrazzano"}'` test (`// #4`):
+
+Send the following JSON response as a payload to the `/travellers` route:
```json
{
@@ -22,11 +24,18 @@ dashedName: run-functional-tests-on-an-api-response-using-chai-http-iv---put-met
}
```
-替换 assert.fail()，使测试通过。 分别测试 1) `status`，2) `type`，3) `body.name`，4) `body.surname` 请按照以上顺序书写断言，顺序错误会影响此挑战的判定。
+Check for the following, within a `request.end` callback:
+
+1. `status`
+2. `type`
+3. `body.name`
+4. `body.surname`
+
+Follow the assertion order above - we rely on it. Be sure to remove `assert.fail()`, once complete.
# --hints--
-不应有未通过的测试
+All tests should pass
```js
(getUserInput) =>
@@ -40,7 +49,7 @@ dashedName: run-functional-tests-on-an-api-response-using-chai-http-iv---put-met
);
```
-你需要测试 'res.status' 是否为 200
+You should test for 'res.status' to be 200
```js
(getUserInput) =>
@@ -56,7 +65,7 @@ dashedName: run-functional-tests-on-an-api-response-using-chai-http-iv---put-met
);
```
-你需要测试 'res.type' 是否为 'application/json'
+You should test for 'res.type' to be 'application/json'
```js
(getUserInput) =>
@@ -72,7 +81,7 @@ dashedName: run-functional-tests-on-an-api-response-using-chai-http-iv---put-met
);
```
-你需要测试 'res.body.name' 是否为 'Giovanni'
+You should test for 'res.body.name' to be 'Giovanni'
```js
(getUserInput) =>
@@ -88,7 +97,7 @@ dashedName: run-functional-tests-on-an-api-response-using-chai-http-iv---put-met
);
```
-你需要测试 'res.body.surname' 是否为 'da Verrazzano'
+You should test for 'res.body.surname' to be 'da Verrazzano'
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-on-api-endpoints-using-chai-http-ii.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-on-api-endpoints-using-chai-http-ii.md
index 657f38c08d9..cf47e5ae87b 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-on-api-endpoints-using-chai-http-ii.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-on-api-endpoints-using-chai-http-ii.md
@@ -1,6 +1,6 @@
---
id: 587d824f367417b2b2512c59
-title: 使用 Chai-HTTP 进行 API 功能测试 (2)
+title: Run Functional Tests on API Endpoints using Chai-HTTP II
challengeType: 2
forumTopicId: 301592
dashedName: run-functional-tests-on-api-endpoints-using-chai-http-ii
@@ -8,17 +8,17 @@ dashedName: run-functional-tests-on-api-endpoints-using-chai-http-ii
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
# --instructions--
-测试 status 和 `text.response`。替换 `assert.fail()` 使测试通过。
+Within `tests/2_functional-tests.js`, alter the `'Test GET /hello with your name'` test (`// #2`) to assert the `status` and the `text` response to make the test pass.
-请把 name 的值添加到 query 结尾：`?name=<your_name>`，后端将会返回 `'hello <your_name>'`。
+Send your name in the query, appending `?name=<your_name>` to the route. The endpoint responds with `'hello <your_name>'`.
# --hints--
-不应有未通过的测试
+All tests should pass
```js
(getUserInput) =>
@@ -32,7 +32,7 @@ dashedName: run-functional-tests-on-api-endpoints-using-chai-http-ii
);
```
-你需要测试 'res.status' 是否为 200
+You should test for 'res.status' == 200
```js
(getUserInput) =>
@@ -48,7 +48,7 @@ dashedName: run-functional-tests-on-api-endpoints-using-chai-http-ii
);
```
-你需要测试 'res.text' 是否为 'hello Guest'
+You should test for 'res.text' == 'hello Guest'
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-on-api-endpoints-using-chai-http.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-on-api-endpoints-using-chai-http.md
index 92e198976f0..e48d66852ee 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-on-api-endpoints-using-chai-http.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-on-api-endpoints-using-chai-http.md
@@ -1,6 +1,6 @@
---
id: 587d824e367417b2b2512c58
-title: 使用 Chai-HTTP 进行 API 功能测试 (1)
+title: Run Functional Tests on API Endpoints using Chai-HTTP
challengeType: 2
forumTopicId: 301593
dashedName: run-functional-tests-on-api-endpoints-using-chai-http
@@ -8,17 +8,43 @@ dashedName: run-functional-tests-on-api-endpoints-using-chai-http
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
+
+Mocha allows testing asyncronous operations. There is a small (BIG) difference. Can you spot it?
+
+We can test our API endpoints using a plugin, called `chai-http`. Let's see how it works. And remember, API calls are asynchronous.
+
+The following is an example of a test using `chai-http` for the `'GET /hello?name=[name] => "hello [name]"'` suite. The test sends a name string in a url query string (`?name=John`) using a `GET`request to the `server`. In the `end` method's callback function, the response object (`res`) is received and contains the `status` property. The first `assert.equal` checks if the status is equal to `200`. The second `assert.equal` checks that the response string (`res.text`) is equal to `"hello John"`.
+
+```js
+suite('GET /hello?name=[name] => "hello [name]"', function () {
+ test("?name=John", function (done) {
+ chai
+ .request(server)
+ .get("/hello?name=John")
+ .end(function (err, res) {
+ assert.equal(res.status, 200, "response status should be 200");
+ assert.equal(
+ res.text,
+ "hello John",
+ 'response should be "hello John"'
+ );
+ done();
+ });
+ });
+```
+
+Notice the `done` parameter in the test's callback function. Calling it at the end without an argument is necessary to signal successful asynchronous completion.
# --instructions--
-测试 status 和 `text.response`。替换 `assert.fail()` 使测试通过。
+Within `tests/2_functional-tests.js`, alter the `'Test GET /hello with no name'` test (`// #1`) to assert the `status` and the `text` response to make the test pass. Do not alter the arguments passed to the asserts.
-不要在 query 中传入 name，后端将会返回 "hello Guest"。
+There should be no name in the query; the endpoint responds with `hello Guest`.
# --hints--
-不应有未通过的测试
+All tests should pass
```js
(getUserInput) =>
@@ -32,7 +58,7 @@ dashedName: run-functional-tests-on-api-endpoints-using-chai-http
);
```
-你需要测试 'res.status' 是否为 200
+You should test for 'res.status' == 200
```js
(getUserInput) =>
@@ -48,7 +74,7 @@ dashedName: run-functional-tests-on-api-endpoints-using-chai-http
);
```
-你需要测试 'res.text' 是否为 'hello Guest'
+You should test for 'res.text' == 'hello Guest'
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-using-a-headless-browser-ii.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-using-a-headless-browser-ii.md
index 28b20257918..815edc8c247 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-using-a-headless-browser-ii.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-using-a-headless-browser-ii.md
@@ -1,6 +1,6 @@
---
-id: 587d8250367417b2b2512c5d
-title: 使用 headless 浏览器运行功能测试 (2)
+id: 5f8884f4c46685731aabfc41
+title: Run Functional Tests Using a Headless Browser II
challengeType: 2
forumTopicId: 301594
dashedName: run-functional-tests-using-a-headless-browser-ii
@@ -8,17 +8,27 @@ dashedName: run-functional-tests-using-a-headless-browser-ii
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
# --instructions--
-此挑战与之前的挑战十分类似。
+Within `tests/2_functional-tests.js`, in the `'submit "surname" : "Vespucci" - write your e2e test...'` test (`// #6`), automate filling-in and submitting the form from scratch:
-请阅读此挑战给出的代码指引，按顺序书写断言。顺序错误会影响此挑战的判定。
+1. Fill in the form with the `surname` of `Vespucci`
+2. Submit it pressing `'submit'` button
+
+Within the callback:
+
+1. assert that status is `200`
+2. assert that the text inside the element `span#name` is `'Amerigo'`
+3. assert that the text inside the element `span#surname` is `'Vespucci'`
+4. assert that the element(s) `span#dates` exist and their count is `1`
+
+Do not forget to to remove the `assert.fail()` call.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -32,7 +42,7 @@ dashedName: run-functional-tests-using-a-headless-browser-ii
);
```
-Headless browser 应成功执行请求
+You should assert that the headless browser request succeeded.
```js
(getUserInput) =>
@@ -46,7 +56,7 @@ Headless browser 应成功执行请求
);
```
-元素 'span#name' 中的文字应为 'Amerigo'.
+You should assert that the text inside the element 'span#name' is 'Amerigo'.
```js
(getUserInput) =>
@@ -62,7 +72,7 @@ Headless browser 应成功执行请求
);
```
-元素 'span#surname' 中的文字应为 'Vespucci'.
+You should assert that the text inside the element 'span#surname' is 'Vespucci'.
```js
(getUserInput) =>
@@ -78,7 +88,7 @@ Headless browser 应成功执行请求
);
```
-元素 'span#dates' 应存在，并且只有一个
+You should assert that the element 'span#dates' exist and its count is 1.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-using-a-headless-browser.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-using-a-headless-browser.md
index cb76388d7f1..2c15f1d1dc3 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-using-a-headless-browser.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/run-functional-tests-using-a-headless-browser.md
@@ -1,6 +1,6 @@
---
-id: 587d824f367417b2b2512c5c
-title: 使用 headless 浏览器运行功能测试 (1)
+id: 587d8250367417b2b2512c5d
+title: Run Functional Tests using a Headless Browser
challengeType: 2
forumTopicId: 301595
dashedName: run-functional-tests-using-a-headless-browser
@@ -8,21 +8,52 @@ dashedName: run-functional-tests-using-a-headless-browser
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
-在这一项挑战中，我们将会使用 Headless Browser（无头浏览器）模拟人机交互。
+In the HTML main view we provided a input form. It sends data to the `PUT /travellers` endpoint that we used above with an Ajax request. When the request successfully completes, the client code appends a `<div>` containing the info returned by the call to the DOM. Here is an example of how to interact with this form:
-[无头浏览器](https://en.wikipedia.org/wiki/Headless_browser) 是一种没有图形用户界面的浏览器。这类工具对网页调试特别有效，因为它们可以跟普通浏览器一样理解和渲染 HTML，CSS 和 JavaScript。
+```js
+test('#test - submit the input "surname" : "Polo"', function (done) {
+ browser.fill('surname', 'Polo').pressButton('submit', function () {
+ browser.assert.success();
+ browser.assert.text('span#name', 'Marco');
+ browser.assert.text('span#surname', 'Polo');
+ browser.assert.element('span#dates', 1);
+ done();
+ });
+}
+```
+
+First, the `fill` method of the `browser` object fills the `surname` field of the form with the value `'Polo'`. Immediately after, the `pressButton` method invokes the `submit` event listener of the form. The `pressButton` method is asynchronous.
+
+Then, once a response is received from the AJAX request, a few assertions are made confirming:
+
+1. The status of the response is `200`
+2. The text within the `<span id='name'></span>` element matches `'Marco'`
+3. The text within the `<span id='surname'></span>` element matches `'Polo'`
+4. The there is `1` `<span id='dates'></span>` element.
+
+Finally, the `done` callback is invoked, which is needed due to the asynchronous test.
# --instructions--
-在这个挑战中，我们将使用 Zombie.JS。它是一款完全基于 JS 的轻量级浏览器，不需要安装其他二进制文件。这个特性让它轻松地在如 Repl.it 的环境下使用。除此之外，还有很多（更强大的）Headless Browser 选项。
+Within `tests/2_functional-tests.js`, in the `'submit "surname" : "Colombo" - write your e2e test...'` test (`// #5`), automate filling-in and submitting the form:
-请阅读此挑战给出的代码指引，按顺序书写断言。顺序错误会影响此挑战的判定。
+1. Fill in the form
+2. Submit it pressing `'submit'` button.
+
+Within the callback:
+
+1. assert that status is OK `200`
+2. assert that the text inside the element `span#name` is `'Cristoforo'`
+3. assert that the text inside the element `span#surname` is `'Colombo'`
+4. assert that the element(s) `span#dates` exist and their count is `1`
+
+Do not forget to remove the `assert.fail()` call.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -36,7 +67,7 @@ dashedName: run-functional-tests-using-a-headless-browser
);
```
-Headless browser 应成功执行请求
+You should assert that the headless browser request succeeded.
```js
(getUserInput) =>
@@ -50,7 +81,7 @@ Headless browser 应成功执行请求
);
```
-元素 'span#name' 中的文字应为 'Cristoforo'
+You should assert that the text inside the element 'span#name' is 'Cristoforo'.
```js
(getUserInput) =>
@@ -66,7 +97,7 @@ Headless browser 应成功执行请求
);
```
-元素 'span#surname' 中的文字应为 'Colombo'.
+You should assert that the text inside the element 'span#surname' is 'Colombo'.
```js
(getUserInput) =>
@@ -82,7 +113,7 @@ Headless browser 应成功执行请求
);
```
-元素 'span#dates' 应存在，并且只有一个
+You should assert that the element 'span#dates' exist and its count is 1.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/simulate-actions-using-a-headless-browser.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/simulate-actions-using-a-headless-browser.md
index aaa0425a0c5..c5a21227db1 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/simulate-actions-using-a-headless-browser.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/simulate-actions-using-a-headless-browser.md
@@ -1,5 +1,5 @@
---
-id: 5f8884f4c46685731aabfc41
+id: 587d824f367417b2b2512c5c
title: Simulate Actions Using a Headless Browser
challengeType: 2
dashedName: simulate-actions-using-a-headless-browser
@@ -9,21 +9,43 @@ dashedName: simulate-actions-using-a-headless-browser
As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
+In the next challenges we are going to simulate the human interaction with a page using a device called 'Headless Browser'.
+
+A headless browser is a web browser without a graphical user interface. This kind of tool is particularly useful for testing web pages, as it is able to render and understand HTML, CSS, and JavaScript the same way a browser would.
+
+For these challenges we are using Zombie.JS. It's a lightweight browser which is totally based on JS, without relying on additional binaries to be installed. This feature makes it usable in an environment such as Repl.it. There are many other (more powerful) options.
+
+Mocha allows you to prepare the ground running some code before the actual tests. This can be useful for example to create items in the database, which will be used in the successive tests.
+
+With a headless browser, before the actual testing, we need to **visit** the page we are going to inspect. The `suiteSetup` 'hook' is executed only once at the suite startup. Other different hook types can be executed before each test, after each test, or at the end of a suite. See the Mocha docs for more information.
+
# --instructions--
-Within `tests/2_functional-tests.js`, in the `'submit "surname" : "Vespucci" - write your e2e test...'` test (`// #6`), automate filling-in and submitting the form from scratch:
+Within `tests/2_functional-tests.js`, immediately after the `Browser` declaration, add your project URL to the `site` property of the variable:
-1. Fill in the form with the `surname` of `Vespucci`
-2. Submit it pressing `'submit'` button
+```js
+Browser.site = 'https://sincere-cone.gomix.me'; // Your URL here
+```
-Within the callback:
+If you are testing on a local environment replace the line above with
-1. assert that status is `200`
-2. assert that the text inside the element `span#name` is `'Amerigo'`
-3. assert that the text inside the element `span#surname` is `'Vespucci'`
-4. assert that the element(s) `span#dates` exist and their count is `1`
+```js
+Browser.localhost('example.com', process.env.PORT || 3000);
+```
-Do not forget to to remove the `assert.fail()` call.
+Within `tests/2_functional-tests.js`, at the root level of the `'Functional Tests with Zombie.js'` suite, instantiate a new instance of the `Browser` object with the following code:
+
+```js
+const browser = new Browser();
+```
+
+Then, use the `suiteSetup` hook to direct the `browser` to the `/` route with the following code:
+
+```js
+suiteSetup(function(done) {
+ return browser.visit('/', done);
+});
+```
# --hints--
@@ -31,71 +53,11 @@ All tests should pass.
```js
(getUserInput) =>
- $.get(getUserInput('url') + '/_api/get-tests?type=functional&n=6').then(
+ $.get(getUserInput('url') + '/_api/get-tests?type=functional').then(
(data) => {
- assert.equal(data.state, 'passed');
- },
- (xhr) => {
- throw new Error(xhr.responseText);
- }
- );
-```
-
-You should assert that the headless browser request succeeded.
-
-```js
-(getUserInput) =>
- $.get(getUserInput('url') + '/_api/get-tests?type=functional&n=6').then(
- (data) => {
- assert.equal(data.assertions[0].method, 'browser.success');
- },
- (xhr) => {
- throw new Error(xhr.responseText);
- }
- );
-```
-
-You should assert that the text inside the element 'span#name' is 'Amerigo'.
-
-```js
-(getUserInput) =>
- $.get(getUserInput('url') + '/_api/get-tests?type=functional&n=6').then(
- (data) => {
- assert.equal(data.assertions[1].method, 'browser.text');
- assert.match(data.assertions[1].args[0], /('|")span#name\1/);
- assert.match(data.assertions[1].args[1], /('|")Amerigo\1/);
- },
- (xhr) => {
- throw new Error(xhr.responseText);
- }
- );
-```
-
-You should assert that the text inside the element 'span#surname' is 'Vespucci'.
-
-```js
-(getUserInput) =>
- $.get(getUserInput('url') + '/_api/get-tests?type=functional&n=6').then(
- (data) => {
- assert.equal(data.assertions[2].method, 'browser.text');
- assert.match(data.assertions[2].args[0], /('|")span#surname\1/);
- assert.match(data.assertions[2].args[1], /('|")Vespucci\1/);
- },
- (xhr) => {
- throw new Error(xhr.responseText);
- }
- );
-```
-
-You should assert that the element 'span#dates' exist and its count is 1.
-
-```js
-(getUserInput) =>
- $.get(getUserInput('url') + '/_api/get-tests?type=functional&n=6').then(
- (data) => {
- assert.equal(data.assertions[3].method, 'browser.element');
- assert.match(data.assertions[3].args[0], /('|")span#dates\1/);
- assert.equal(data.assertions[3].args[1], 1);
+ data.slice(0, 4).forEach((test) => {
+ assert.equal(test.state, 'passed');
+ })
},
(xhr) => {
throw new Error(xhr.responseText);
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-for-truthiness.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-for-truthiness.md
index 87f0fcd84e7..d06c838a5a2 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-for-truthiness.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-for-truthiness.md
@@ -1,32 +1,32 @@
---
id: 587d824b367417b2b2512c49
-title: 测试真实性
+title: Test for Truthiness
challengeType: 2
-videoUrl: ''
+forumTopicId: 301596
dashedName: test-for-truthiness
---
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
-`isTrue()` 仅当给出的值为 Boolean 的 `true` 时可以通过测试；`isNotTrue()` 则会在给出除 `true` 以外的值时通过测试。
+`isTrue()` will test for the boolean value `true` and `isNotTrue()` will pass when given anything but the boolean value of `true`.
```js
-assert.isTrue(true, '可以通过，因为值的类型是布尔且值为 true');
-assert.isTrue('true', '不能通过，因为值的类型是字符串);
-assert.isTrue(1, '不过能通过，因为值是数字 1');
+assert.isTrue(true, 'this will pass with the boolean value true');
+assert.isTrue('true', 'this will NOT pass with the string value "true"');
+assert.isTrue(1, 'this will NOT pass with the number value 1');
```
-`isFalse()` 和 `isNotFalse()` 同样存在，表现与上面提到的两个方法类似。
+`isFalse()` and `isNotFalse()` also exist, and behave similarly to their true counterparts except they look for the boolean value of `false`.
# --instructions--
-使用 `assert.isTrue()` 或 `assert.isNotTrue()` 让测试通过。
+Within `tests/1_unit-tests.js` under the test labelled `#4` in the `Basic Assertions` suite, change each `assert` to either `assert.isTrue` or `assert.isNotTrue` to make the test pass (should evaluate to `true`). Do not alter the arguments passed to the asserts.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -40,7 +40,7 @@ assert.isTrue(1, '不过能通过，因为值是数字 1');
);
```
-请选择正确的断言—isTrue 或 isNotTrue
+You should choose the correct method for the first assertion - `isTrue` vs. `isNotTrue`.
```js
(getUserInput) =>
@@ -54,7 +54,7 @@ assert.isTrue(1, '不过能通过，因为值是数字 1');
);
```
-请选择正确的断言—isTrue 或 isNotTrue
+You should choose the correct method for the second assertion - `isTrue` vs. `isNotTrue`.
```js
(getUserInput) =>
@@ -72,7 +72,7 @@ assert.isTrue(1, '不过能通过，因为值是数字 1');
);
```
-请选择正确的断言—isTrue 或 isNotTrue
+You should choose the correct method for the third assertion - `isTrue` vs. `isNotTrue`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-string-contains-a-substring.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-string-contains-a-substring.md
index a60cd86f22d..3db5ce9dabc 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-string-contains-a-substring.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-string-contains-a-substring.md
@@ -1,6 +1,6 @@
---
id: 587d824d367417b2b2512c53
-title: 测试字符串是否包含子字符串
+title: Test if a String Contains a Substring
challengeType: 2
forumTopicId: 301597
dashedName: test-if-a-string-contains-a-substring
@@ -8,17 +8,17 @@ dashedName: test-if-a-string-contains-a-substring
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
-`include()` 和 `notInclude()` 同样可以用于字符串。 `include()` 用于断言字符串中包含某个子字符串。
+`include()` and `notInclude()` work for strings too! `include()` asserts that the actual string contains the expected substring.
# --instructions--
-使用 `assert.include()` 或 `assert.notInclude()` 让测试通过。
+Within `tests/1_unit-tests.js` under the test labelled `#14` in the `Strings` suite, change each `assert` to either `assert.include` or `assert.notInclude` to make the test pass (should evaluate to `true`). Do not alter the arguments passed to the asserts.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -32,7 +32,7 @@ dashedName: test-if-a-string-contains-a-substring
);
```
-请选择正确的断言—include 或 notInclude
+You should choose the correct method for the first assertion - `include` vs. `notInclude`.
```js
(getUserInput) =>
@@ -50,7 +50,7 @@ dashedName: test-if-a-string-contains-a-substring
);
```
-请选择正确的断言—include 或 notInclude
+You should choose the correct method for the second assertion - `include` vs. `notInclude`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-value-falls-within-a-specific-range.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-value-falls-within-a-specific-range.md
index f080e11990a..7ee2db2e831 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-value-falls-within-a-specific-range.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-value-falls-within-a-specific-range.md
@@ -1,6 +1,6 @@
---
id: 587d824c367417b2b2512c4f
-title: 测试某个值是否在特定范围内
+title: Test if a Value Falls within a Specific Range
challengeType: 2
forumTopicId: 301598
dashedName: test-if-a-value-falls-within-a-specific-range
@@ -8,19 +8,23 @@ dashedName: test-if-a-value-falls-within-a-specific-range
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
-`.approximately(actual, expected, delta, [message])` 断言待测试的值处于 `expected` +/- `delta` 的范围内。
+```javascript
+.approximately(actual, expected, delta, [message])
+```
+
+Asserts that the `actual` is equal to `expected`, to within a +/- `delta` range.
# --instructions--
-使用 `assert.approximately()` 让测试通过。
+Within `tests/1_unit-tests.js` under the test labelled `#10` in the `Comparisons` suite, change each `assert` to `assert.approximately` to make the test pass (should evaluate to `true`).
-选择最小范围（第三个参数）来通过所有测试。它应该小于 1。
+Choose the minimum range (3rd parameter) to make the test always pass. It should be less than 1.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -34,7 +38,7 @@ dashedName: test-if-a-value-falls-within-a-specific-range
);
```
-你应使用 approximately(actual, expected, range) 并选择正确的范围
+You should choose the correct range for the first assertion - `approximately(actual, expected, range)`.
```js
(getUserInput) =>
@@ -53,7 +57,7 @@ dashedName: test-if-a-value-falls-within-a-specific-range
);
```
-你应使用 approximately(actual, expected, range) 并选择正确的范围
+You should choose the correct range for the second assertion - `approximately(actual, expected, range)`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-value-is-a-string.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-value-is-a-string.md
index 2a51d767f29..d51fae0ff74 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-value-is-a-string.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-value-is-a-string.md
@@ -1,6 +1,6 @@
---
id: 587d824d367417b2b2512c52
-title: 测试某个值是否为字符串
+title: Test if a Value is a String
challengeType: 2
forumTopicId: 301599
dashedName: test-if-a-value-is-a-string
@@ -8,17 +8,17 @@ dashedName: test-if-a-value-is-a-string
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
-`isString` 或 `isNotString` 断言一个值是否为字符串。
+`isString` or `isNotString` asserts that the actual value is a string.
# --instructions--
-使用 `assert.isString()` 或 `assert.isNotString()` 让所有测试通过。
+Within `tests/1_unit-tests.js` under the test labelled `#13` in the `Strings` suite, change each `assert` to either `assert.isString` or `assert.isNotString` to make the test pass (should evaluate to `true`). Do not alter the arguments passed to the asserts.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -32,7 +32,7 @@ dashedName: test-if-a-value-is-a-string
);
```
-请选择正确的断言—isString 或 isNotString
+You should choose the correct method for the first assertion - `isString` vs. `isNotString`.
```js
(getUserInput) =>
@@ -50,7 +50,7 @@ dashedName: test-if-a-value-is-a-string
);
```
-请选择正确的断言—isString 或 isNotString
+You should choose the correct method for the second assertion - `isString` vs. `isNotString`.
```js
(getUserInput) =>
@@ -68,7 +68,7 @@ dashedName: test-if-a-value-is-a-string
);
```
-请选择正确的断言—isString 或 isNotString
+You should choose the correct method for the third assertion - `isString` vs. `isNotString`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-value-is-an-array.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-value-is-an-array.md
index 7770b924751..aa9ba1e31a7 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-value-is-an-array.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-value-is-an-array.md
@@ -1,6 +1,6 @@
---
id: 587d824d367417b2b2512c50
-title: 测试某个值是否为数组
+title: Test if a Value is an Array
challengeType: 2
forumTopicId: 301600
dashedName: test-if-a-value-is-an-array
@@ -8,15 +8,15 @@ dashedName: test-if-a-value-is-an-array
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
# --instructions--
-使用 `assert.isArray()` 或 `assert.isNotArray()` 让所有测试通过。
+Within `tests/1_unit-tests.js` under the test labelled `#11` in the `Arrays` suite, change each `assert` to either `assert.isArray` or `assert.isNotArray` to make the test pass (should evaluate to `true`). Do not alter the arguments passed to the asserts.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -30,7 +30,7 @@ dashedName: test-if-a-value-is-an-array
);
```
-请选择正确的断言—isArray 或 isNotArray
+You should choose the correct method for the first assertion - `isArray` vs. `isNotArray`.
```js
(getUserInput) =>
@@ -48,7 +48,7 @@ dashedName: test-if-a-value-is-an-array
);
```
-请选择正确的断言—isArray 或 isNotArray
+You should choose the correct method for the second assertion - `isArray` vs. `isNotArray`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-value-is-of-a-specific-data-structure-type.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-value-is-of-a-specific-data-structure-type.md
index 5467d714622..d40e7da38aa 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-value-is-of-a-specific-data-structure-type.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-value-is-of-a-specific-data-structure-type.md
@@ -1,6 +1,6 @@
---
id: 587d824e367417b2b2512c56
-title: 测试值是否为特定数据结构类型
+title: Test if a Value is of a Specific Data Structure Type
challengeType: 2
forumTopicId: 301601
dashedName: test-if-a-value-is-of-a-specific-data-structure-type
@@ -8,17 +8,17 @@ dashedName: test-if-a-value-is-of-a-specific-data-structure-type
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
-`#typeOf` 断言一个值的类型符合给定的类型，这个类型与 `Object.prototype.toString` 一致。
+`#typeOf` asserts that value's type is the given string, as determined by `Object.prototype.toString`.
# --instructions--
-使用 `assert.typeOf()` 或 `assert.notTypeOf()` 让所有测试通过。
+Within `tests/1_unit-tests.js` under the test labelled `#17` in the `Objects` suite, change each `assert` to either `assert.typeOf` or `assert.notTypeOf` to make the test pass (should evaluate to `true`). Do not alter the arguments passed to the asserts.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -32,7 +32,7 @@ dashedName: test-if-a-value-is-of-a-specific-data-structure-type
);
```
-请选择正确的断言—typeOf 或 notTypeOf
+You should choose the correct method for the first assertion - `typeOf` vs. `notTypeOf`.
```js
(getUserInput) =>
@@ -50,7 +50,7 @@ dashedName: test-if-a-value-is-of-a-specific-data-structure-type
);
```
-请选择正确的断言—typeOf 或 notTypeOf
+You should choose the correct method for the second assertion - `typeOf` vs. `notTypeOf`.
```js
(getUserInput) =>
@@ -68,7 +68,7 @@ dashedName: test-if-a-value-is-of-a-specific-data-structure-type
);
```
-请选择正确的断言—typeOf 或 notTypeOf
+You should choose the correct method for the third assertion - `typeOf` vs. `notTypeOf`.
```js
(getUserInput) =>
@@ -86,7 +86,7 @@ dashedName: test-if-a-value-is-of-a-specific-data-structure-type
);
```
-请选择正确的断言—typeOf 或 notTypeOf
+You should choose the correct method for the fourth assertion - `typeOf` vs. `notTypeOf`.
```js
(getUserInput) =>
@@ -104,7 +104,7 @@ dashedName: test-if-a-value-is-of-a-specific-data-structure-type
);
```
-请选择正确的断言—typeOf 或 notTypeOf
+You should choose the correct method for the fifth assertion - `typeOf` vs. `notTypeOf`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-variable-or-function-is-defined.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-variable-or-function-is-defined.md
index 61b7d739918..e90101f3357 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-variable-or-function-is-defined.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-a-variable-or-function-is-defined.md
@@ -1,6 +1,6 @@
---
id: 587d824b367417b2b2512c47
-title: 测试变量或函数是否已定义
+title: Test if a Variable or Function is Defined
challengeType: 2
forumTopicId: 301602
dashedName: test-if-a-variable-or-function-is-defined
@@ -8,15 +8,15 @@ dashedName: test-if-a-variable-or-function-is-defined
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
# --instructions--
-使用 `assert.isDefined()` 或 `assert.isUndefined()` 让测试通过。
+Within `tests/1_unit-tests.js` under the test labelled `#2` in the `Basic Assertions` suite, change each `assert` to either `assert.isDefined()` or `assert.isUndefined()` to make the test pass (should evaluate to `true`). Do not alter the arguments passed to the asserts.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -30,7 +30,7 @@ dashedName: test-if-a-variable-or-function-is-defined
);
```
-请选择正确的断言—idDefined 或 idUndefined
+You should choose the correct method for the first assertion - `isDefined` vs. `isUndefined`.
```js
(getUserInput) =>
@@ -48,7 +48,7 @@ dashedName: test-if-a-variable-or-function-is-defined
);
```
-请选择正确的断言—idDefined 或 idUndefined
+You should choose the correct method for the second assertion - `isDefined` vs. `isUndefined`.
```js
(getUserInput) =>
@@ -66,7 +66,7 @@ dashedName: test-if-a-variable-or-function-is-defined
);
```
-请选择正确的断言—idDefined 或 idUndefined
+You should choose the correct method for the third assertion - `isDefined` vs. `isUndefined`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-an-array-contains-an-item.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-an-array-contains-an-item.md
index 0797063602e..320f7b6edbf 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-an-array-contains-an-item.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-an-array-contains-an-item.md
@@ -1,6 +1,6 @@
---
id: 587d824d367417b2b2512c51
-title: 测试数组是否包含项目
+title: Test if an Array Contains an Item
challengeType: 2
forumTopicId: 301603
dashedName: test-if-an-array-contains-an-item
@@ -8,15 +8,15 @@ dashedName: test-if-an-array-contains-an-item
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
# --instructions--
-使用 `assert.include()` 或 `assert.notInclude()` 让所有测试通过。
+Within `tests/1_unit-tests.js` under the test labelled `#12` in the `Arrays` suite, change each `assert` to either `assert.include` or `assert.notInclude` to make the test pass (should evaluate to `true`). Do not alter the arguments passed to the asserts.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -30,7 +30,7 @@ dashedName: test-if-an-array-contains-an-item
);
```
-请选择正确的断言—include 或 notInclude
+You should choose the correct method for the first assertion - `include` vs. `notInclude`.
```js
(getUserInput) =>
@@ -48,7 +48,7 @@ dashedName: test-if-an-array-contains-an-item
);
```
-请选择正确的断言—include 或 notInclude
+You should choose the correct method for the second assertion - `include` vs. `notInclude`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-an-object-has-a-property.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-an-object-has-a-property.md
index 01004565eee..34d88a9ff4f 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-an-object-has-a-property.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-an-object-has-a-property.md
@@ -1,6 +1,6 @@
---
id: 587d824e367417b2b2512c55
-title: 测试对象是否具有某个属性
+title: Test if an Object has a Property
challengeType: 2
forumTopicId: 301604
dashedName: test-if-an-object-has-a-property
@@ -8,17 +8,17 @@ dashedName: test-if-an-object-has-a-property
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
-`property` 断言一个对象含有给定属性。
+`property` asserts that the actual object has a given property.
# --instructions--
-使用 `assert.property()` 或者 `assert.notProperty()` 让所有测试通过。
+Within `tests/1_unit-tests.js` under the test labelled `#16` in the `Objects` suite, change each `assert` to either `assert.property` or `assert.notProperty` to make the test pass (should evaluate to `true`). Do not alter the arguments passed to the asserts.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -32,7 +32,7 @@ dashedName: test-if-an-object-has-a-property
);
```
-请选择正确的断言—property 或 notProperty
+You should choose the correct method for the first assertion - `property` vs. `notProperty`.
```js
(getUserInput) =>
@@ -50,7 +50,7 @@ dashedName: test-if-an-object-has-a-property
);
```
-请选择正确的断言—property 或 notProperty
+You should choose the correct method for the second assertion - `property` vs. `notProperty`.
```js
(getUserInput) =>
@@ -68,7 +68,7 @@ dashedName: test-if-an-object-has-a-property
);
```
-请选择正确的断言—property 或 notProperty
+You should choose the correct method for the third assertion - `property` vs. `notProperty`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-an-object-is-an-instance-of-a-constructor.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-an-object-is-an-instance-of-a-constructor.md
index fc342377544..cccba2ce79a 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-an-object-is-an-instance-of-a-constructor.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-an-object-is-an-instance-of-a-constructor.md
@@ -1,6 +1,6 @@
---
id: 587d824e367417b2b2512c57
-title: 测试对象是否是构造函数的实例
+title: Test if an Object is an Instance of a Constructor
challengeType: 2
forumTopicId: 301605
dashedName: test-if-an-object-is-an-instance-of-a-constructor
@@ -8,17 +8,17 @@ dashedName: test-if-an-object-is-an-instance-of-a-constructor
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
-`#instanceOf` 断言一个对象是一个构造器的实例。
+`#instanceOf` asserts that an object is an instance of a constructor.
# --instructions--
-使用 `assert.instanceOf()` 或 `assert.notInstanceOf()` 让测试通过。
+Within `tests/1_unit-tests.js` under the test labelled `#18` in the `Objects` suite, change each `assert` to either `assert.instanceOf` or `assert.notInstanceOf` to make the test pass (should evaluate to `true`). Do not alter the arguments passed to the asserts.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -32,7 +32,7 @@ dashedName: test-if-an-object-is-an-instance-of-a-constructor
);
```
-请选择正确的断言—instanceOf 或 notInstanceOf
+You should choose the correct method for the first assertion - `instanceOf` vs. `notInstanceOf`.
```js
(getUserInput) =>
@@ -50,7 +50,7 @@ dashedName: test-if-an-object-is-an-instance-of-a-constructor
);
```
-请选择正确的断言—instanceOf 或 notInstanceOf
+You should choose the correct method for the second assertion - `instanceOf` vs. `notInstanceOf`.
```js
(getUserInput) =>
@@ -68,7 +68,7 @@ dashedName: test-if-an-object-is-an-instance-of-a-constructor
);
```
-请选择正确的断言—instanceOf 或 notInstanceOf
+You should choose the correct method for the third assertion - `instanceOf` vs. `notInstanceOf`.
```js
(getUserInput) =>
@@ -86,7 +86,7 @@ dashedName: test-if-an-object-is-an-instance-of-a-constructor
);
```
-请选择正确的断言—instanceOf 或 notInstanceOf
+You should choose the correct method for the fourth assertion - `instanceOf` vs. `notInstanceOf`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-one-value-is-below-or-at-least-as-large-as-another.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-one-value-is-below-or-at-least-as-large-as-another.md
index 400adc92ed7..be4dcadd852 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-one-value-is-below-or-at-least-as-large-as-another.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/test-if-one-value-is-below-or-at-least-as-large-as-another.md
@@ -1,6 +1,6 @@
---
id: 587d824c367417b2b2512c4e
-title: 测试一个值是否小于或等于另一个值
+title: Test if One Value is Below or At Least as Large as Another
challengeType: 2
forumTopicId: 301606
dashedName: test-if-one-value-is-below-or-at-least-as-large-as-another
@@ -8,17 +8,15 @@ dashedName: test-if-one-value-is-below-or-at-least-as-large-as-another
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
# --instructions--
-Use `assert.isBelow()` (i.e. less than) or `assert.isAtLeast()` (i.e. greater than or equal) to make the tests pass.
-
-使用 `assert.isBelow()`（小于）或 `assert.isAtLeast()`（大于等于）让所有测试通过。
+Within `tests/1_unit-tests.js` under the test labelled `#9` in the `Comparisons` suite, change each `assert` to either `assert.isBelow` or `assert.isAtLeast` to make the test pass (should evaluate to `true`). Do not alter the arguments passed to the asserts.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -32,7 +30,7 @@ Use `assert.isBelow()` (i.e. less than) or `assert.isAtLeast()` (i.e. greater th
);
```
-请选择正确的断言—isBelow 或 isAtLeast
+You should choose the correct method for the first assertion - `isBelow` vs. `isAtLeast`.
```js
(getUserInput) =>
@@ -50,7 +48,7 @@ Use `assert.isBelow()` (i.e. less than) or `assert.isAtLeast()` (i.e. greater th
);
```
-请选择正确的断言—isBelow 或 isAtLeast
+You should choose the correct method for the second assertion - `isBelow` vs. `isAtLeast`.
```js
(getUserInput) =>
@@ -68,7 +66,7 @@ Use `assert.isBelow()` (i.e. less than) or `assert.isAtLeast()` (i.e. greater th
);
```
-请选择正确的断言—isBelow 或 isAtLeast
+You should choose the correct method for the third assertion - `isBelow` vs. `isAtLeast`.
```js
(getUserInput) =>
@@ -82,7 +80,7 @@ Use `assert.isBelow()` (i.e. less than) or `assert.isAtLeast()` (i.e. greater th
);
```
-请选择正确的断言—isBelow 或 isAtLeast
+You should choose the correct method for the fourth assertion - `isBelow` vs. `isAtLeast`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/use-assert.isok-and-assert.isnotok.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/use-assert.isok-and-assert.isnotok.md
index 4cf765cbaa8..d7cd8030532 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/use-assert.isok-and-assert.isnotok.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/use-assert.isok-and-assert.isnotok.md
@@ -1,6 +1,6 @@
---
id: 587d824b367417b2b2512c48
-title: 使用 Assert.isOk() 和 Assert.isNotOK()
+title: Use Assert.isOK and Assert.isNotOK
challengeType: 2
forumTopicId: 301607
dashedName: use-assert-isok-and-assert-isnotok
@@ -8,17 +8,19 @@ dashedName: use-assert-isok-and-assert-isnotok
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
-`isOk()` 用来测试值是否为 `truthy`；`isNotOk()` 用来测试值是否为 `falsy`。 如果你想了解更多关于 truthy 和 falsy 值的内容，请参阅 [Falsy Bouncer](https://www.freecodecamp.org/learn/javascript-algorithms-and-data-structures/basic-algorithm-scripting/falsy-bouncer) 这道挑战题目。
+`isOk()` will test for a truthy value, and `isNotOk()` will test for a falsy value.
+
+To learn more about truthy and falsy values, try our [Falsy Bouncer](https://www.freecodecamp.org/learn/javascript-algorithms-and-data-structures/basic-algorithm-scripting/falsy-bouncer) challenge.
# --instructions--
-使用 `assert.isOk()` 或 `assert.isNotOk()` 让测试通过。
+Within `tests/1_unit-tests.js` under the test labelled `#3` in the `Basic Assertions` suite, change each `assert` to either `assert.isOk()` or `assert.isNotOk()` to make the test pass (should evaluate to `true`). Do not alter the arguments passed to the asserts.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -32,7 +34,7 @@ dashedName: use-assert-isok-and-assert-isnotok
);
```
-请选择正确的断言—isOk 或 isNotOk
+You should choose the correct method for the first assertion - `isOk` vs. `isNotOk`.
```js
(getUserInput) =>
@@ -46,7 +48,7 @@ dashedName: use-assert-isok-and-assert-isnotok
);
```
-请选择正确的断言—isOk 或 isNotOk
+You should choose the correct method for the second assertion - `isOk` vs. `isNotOk`.
```js
(getUserInput) =>
@@ -60,7 +62,7 @@ dashedName: use-assert-isok-and-assert-isnotok
);
```
-请选择正确的断言—isOk 或 isNotOk
+You should choose the correct method for the third assertion - `isOk` vs. `isNotOk`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/use-regular-expressions-to-test-a-string.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/use-regular-expressions-to-test-a-string.md
index f32b4e827c0..bfd745c7ee4 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/use-regular-expressions-to-test-a-string.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/use-regular-expressions-to-test-a-string.md
@@ -1,6 +1,6 @@
---
id: 587d824d367417b2b2512c54
-title: 使用正则表达式测试字符串
+title: Use Regular Expressions to Test a String
challengeType: 2
forumTopicId: 301608
dashedName: use-regular-expressions-to-test-a-string
@@ -8,17 +8,17 @@ dashedName: use-regular-expressions-to-test-a-string
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
-`match()` 断言一个值匹配一个正则表达式（第二个参数）。
+`match()` asserts that the actual value matches the second argument regular expression.
# --instructions--
-使用 `assert.match()` 让所有测试通过。
+Within `tests/1_unit-tests.js` under the test labelled `#15` in the `Strings` suite, change each `assert` to either `assert.match` or `assert.notMatch` to make the test pass (should evaluate to `true`). Do not alter the arguments passed to the asserts.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -32,7 +32,7 @@ dashedName: use-regular-expressions-to-test-a-string
);
```
-请选择正确的断言—match 或 notMatch
+You should choose the correct method for the first assertion - `match` vs. `notMatch`.
```js
(getUserInput) =>
@@ -41,7 +41,7 @@ dashedName: use-regular-expressions-to-test-a-string
assert.equal(
data.assertions[0].method,
'match',
- "'# name: John Doe, age: 35' matches the regex"
+ "'# name:John Doe, age:35' matches the regex"
);
},
(xhr) => {
@@ -50,7 +50,7 @@ dashedName: use-regular-expressions-to-test-a-string
);
```
-请选择正确的断言—match 或 notMatch
+You should choose the correct method for the second assertion - `match` vs. `notMatch`.
```js
(getUserInput) =>
@@ -59,7 +59,7 @@ dashedName: use-regular-expressions-to-test-a-string
assert.equal(
data.assertions[1].method,
'notMatch',
- "'# name: Paul Smith III, age: twenty-four' does not match the regex (the age must be numeric)"
+ "'# name:Paul Smith III, age:twenty-four' does not match the regex (the age must be numeric)"
);
},
(xhr) => {
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/use-the-double-equals-to-assert-equality.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/use-the-double-equals-to-assert-equality.md
index cc9d298df8f..eab031cce78 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/use-the-double-equals-to-assert-equality.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/use-the-double-equals-to-assert-equality.md
@@ -1,6 +1,6 @@
---
id: 587d824b367417b2b2512c4a
-title: 用两个等号断言相等
+title: Use the Double Equals to Assert Equality
challengeType: 2
forumTopicId: 301609
dashedName: use-the-double-equals-to-assert-equality
@@ -8,17 +8,17 @@ dashedName: use-the-double-equals-to-assert-equality
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
-`equal()` 使用 `==` 比较对象。
+`equal()` compares objects using `==`.
# --instructions--
-使用 `assert.equal()` 或 `assert.notEqual()` 让所有测试通过。
+Within `tests/1_unit-tests.js` under the test labelled `#5` in the `Equality` suite, change each `assert` to either `assert.equal` or `assert.notEqual` to make the test pass (should evaluate to `true`). Do not alter the arguments passed to the asserts.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -32,7 +32,7 @@ dashedName: use-the-double-equals-to-assert-equality
);
```
-请选择正确的断言—equal 或 notEqual
+You should choose the correct method for the first assertion - `equal` vs. `notEqual`.
```js
(getUserInput) =>
@@ -50,7 +50,7 @@ dashedName: use-the-double-equals-to-assert-equality
);
```
-请选择正确的断言—equal 或 notEqual
+You should choose the correct method for the second assertion - `equal` vs. `notEqual`.
```js
(getUserInput) =>
@@ -68,7 +68,7 @@ dashedName: use-the-double-equals-to-assert-equality
);
```
-请选择正确的断言—equal 或 notEqual
+You should choose the correct method for the third assertion - `equal` vs. `notEqual`.
```js
(getUserInput) =>
@@ -86,7 +86,7 @@ dashedName: use-the-double-equals-to-assert-equality
);
```
-请选择正确的断言—equal 或 notEqual
+You should choose the correct method for the fourth assertion - `equal` vs. `notEqual`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/use-the-triple-equals-to-assert-strict-equality.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/use-the-triple-equals-to-assert-strict-equality.md
index b14ff4a101f..de0259762c1 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/use-the-triple-equals-to-assert-strict-equality.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-and-testing-with-chai/use-the-triple-equals-to-assert-strict-equality.md
@@ -1,6 +1,6 @@
---
id: 587d824b367417b2b2512c4b
-title: 用三个等号断言严格相等
+title: Use the Triple Equals to Assert Strict Equality
challengeType: 2
forumTopicId: 301610
dashedName: use-the-triple-equals-to-assert-strict-equality
@@ -8,17 +8,17 @@ dashedName: use-the-triple-equals-to-assert-strict-equality
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 的基础上进行开发。你也可以从 [GitHub](https://repl.it/github/freeCodeCamp/boilerplate-mochachai) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mochachai), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-mochachai/).
-`strictEqual()` 使用 `===` 比较对象。
+`strictEqual()` compares objects using `===`.
# --instructions--
-使用 `assert.strictEqual()` 或 `assert.notStrictEqual()` 让所有测试通过。
+Within `tests/1_unit-tests.js` under the test labelled `#6` in the `Equality` suite, change each `assert` to either `assert.strictEqual` or `assert.notStrictEqual` to make the test pass (should evaluate to `true`). Do not alter the arguments passed to the asserts.
# --hints--
-不应有未通过的测试
+All tests should pass.
```js
(getUserInput) =>
@@ -32,7 +32,7 @@ dashedName: use-the-triple-equals-to-assert-strict-equality
);
```
-请选择正确的断言—strictEqual 或 notStrictEqual
+You should choose the correct method for the first assertion - `strictEqual` vs. `notStrictEqual`.
```js
(getUserInput) =>
@@ -50,7 +50,7 @@ dashedName: use-the-triple-equals-to-assert-strict-equality
);
```
-请选择正确的断言—strictEqual 或 notStrictEqual
+You should choose the correct method for the second assertion - `strictEqual` vs. `notStrictEqual`.
```js
(getUserInput) =>
@@ -64,7 +64,7 @@ dashedName: use-the-triple-equals-to-assert-strict-equality
);
```
-请选择正确的断言—strictEqual 或 notStrictEqual
+You should choose the correct method for the third assertion - `strictEqual` vs. `notStrictEqual`.
```js
(getUserInput) =>
@@ -82,7 +82,7 @@ dashedName: use-the-triple-equals-to-assert-strict-equality
);
```
-请选择正确的断言—strictEqual 或 notStrictEqual
+You should choose the correct method for the fourth assertion - `strictEqual` vs. `notStrictEqual`.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/american-british-translator.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/american-british-translator.md
index 0779a948ee1..c2e5448f12b 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/american-british-translator.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/american-british-translator.md
@@ -1,77 +1,328 @@
---
id: 5e601c0d5ac9d0ecd8b94afe
+title: American British Translator
challengeType: 4
dashedName: american-british-translator
---
# --description--
-Build a full stack JavaScript app that is functionally similar to this: <https://youthful-grave-scabiosa.glitch.me/>.
+Build a full stack JavaScript app that is functionally similar to this: <https://american-british-translator.freecodecamp.rocks/>. Working on this project will involve you writing your code using one of the following methods:
-Working on this project will involve you writing your code on Glitch on our starter project. After completing this project you can copy your public glitch url (to the homepage of your app) into this screen to test it! Optionally you may choose to write your project on another platform but it must be publicly visible for our testing.
+- Clone [this GitHub repo](https://github.com/freeCodeCamp/boilerplate-project-american-british-english-translator/) and complete your project locally.
+- Use [our repl.it starter project](https://repl.it/github/freeCodeCamp/boilerplate-project-american-british-english-translator) to complete your project.
+- Use a site builder of your choice to complete the project. Be sure to incorporate all the files from our GitHub repo.
-Start this project on Glitch using [this link](https://glitch.com/edit/#!/remix/clone-from-repo?REPO_URL=https://github.com/freeCodeCamp/boilerplate-project-american-british-english-translator/) or clone [this repository](https://github.com/freeCodeCamp/boilerplate-project-american-british-english-translator/) on GitHub! If you use Glitch, remember to save the link to your project somewhere safe!
+When you are done, make sure a working demo of your project is hosted somewhere public. Then submit the URL to it in the `Solution Link` field. Optionally, also submit a link to your project's source code in the `GitHub Link` field.
+
+# --instructions--
+
+- All logic can go into `/components/translator.js`
+- Complete the `/api/translate` route in `/routes/api.js`
+- Create all of the unit/functional tests in `tests/1_unit-tests.js` and `tests/2_functional-tests.js`
+- See the JavaScript files in `/components` for the different spelling and terms your application should translate
+- To run the tests on Repl.it, set `NODE_ENV` to `test` without quotes in the `.env` file
+- To run the tests in the console, use the command `npm run test`. To open the Repl.it console, press Ctrl+Shift+P (Cmd if on a Mac) and type "open shell"
+
+Write the following tests in `tests/1_unit-tests.js`:
+
+- Translate `Mangoes are my favorite fruit.` to British English
+- Translate `I ate yogurt for breakfast.` to British English
+- Translate `We had a party at my friend's condo.` to British English
+- Translate `Can you toss this in the trashcan for me?` to British English
+- Translate `The parking lot was full.` to British English
+- Translate `Like a high tech Rube Goldberg machine.` to British English
+- Translate `To play hooky means to skip class or work.` to British English
+- Translate `No Mr. Bond, I expect you to die.` to British English
+- Translate `Dr. Grosh will see you now.` to British English
+- Translate `Lunch is at 12:15 today.` to British English
+- Translate `We watched the footie match for a while.` to American English
+- Translate `Paracetamol takes up to an hour to work.` to American English
+- Translate `First, caramelise the onions.` to American English
+- Translate `I spent the bank holiday at the funfair.` to American English
+- Translate `I had a bicky then went to the chippy.` to American English
+- Translate `I've just got bits and bobs in my bum bag.` to American English
+- Translate `The car boot sale at Boxted Airfield was called off.` to American English
+- Translate `Have you met Mrs Kalyani?` to American English
+- Translate `Prof Joyner of King's College, London.` to American English
+- Translate `Tea time is usually around 4 or 4.30.` to American English
+- Highlight translation in `Mangoes are my favorite fruit.`
+- Highlight translation in `I ate yogurt for breakfast.`
+- Highlight translation in `We watched the footie match for a while.`
+- Highlight translation in `Paracetamol takes up to an hour to work.`
+
+Write the following tests in `tests/2_functional-tests.js`:
+
+- Translation with text and locale fields: POST request to `/api/translate`
+- Translation with text and invalid locale field: POST request to `/api/translate`
+- Translation with missing text field: POST request to `/api/translate`
+- Translation with missing locale field: POST request to `/api/translate`
+- Translation with empty text: POST request to `/api/translate`
+- Translation with text that needs no translation: POST request to `/api/translate`
# --hints--
-I can enter a simple sentence into the text area and select whether to translate to British or American English from the dropdown menu.
+I can provide my own project, not the example URL.
```js
-
+(getUserInput) => {
+ assert(
+ !/.*\/american-british-translator\.freecodecamp\.rocks/.test(
+ getUserInput('url')
+ )
+ );
+};
```
-When the "Translate" button is pressed, append the translated sentence to the `translated-sentence` `div`. See the JavaScript files in `/public` for the different spelling and terms your application should translate.
+You can `POST` to `/api/translate` with a body containing `text` with the text to translate and `locale` with either `american-to-british` or `british-to-american`. The returned object should contain the submitted `text` and `translation` with the translated text.
```js
-
+async (getUserInput) => {
+ try {
+ const text = 'Mangoes are my favorite fruit.';
+ const locale = 'american-to-british';
+ const output = {
+ text: 'Mangoes are my favorite fruit.',
+ translation:
+ 'Mangoes are my <span class="highlight">favourite</span> fruit.'
+ };
+ let data = await fetch(getUserInput('url') + '/api/translate', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ text, locale })
+ });
+ let parsed = await data.json();
+ assert.isObject(parsed);
+ assert.property(parsed, 'text');
+ assert.property(parsed, 'translation');
+ assert.deepEqual(parsed, output);
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-Your application should handle the way time is written in American and British English. For example, ten thirty is written as "10.30" in British English and "10:30" in American English.
+The `/api/translate` route should handle the way time is written in American and British English. For example, ten thirty is written as "10.30" in British English and "10:30" in American English. The `span` element should wrap the entire time string, i.e. `<span class="highlight">10:30</span>`.
```js
-
+async (getUserInput) => {
+ try {
+ const text = 'Lunch is at 12:15 today.';
+ const locale = 'american-to-british';
+ const output = {
+ text: text,
+ translation: 'Lunch is at <span class="highlight">12.15</span> today.'
+ };
+ let data = await fetch(getUserInput('url') + '/api/translate', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ text, locale })
+ });
+ let parsed = await data.json();
+ assert.isObject(parsed);
+ assert.property(parsed, 'text');
+ assert.property(parsed, 'translation');
+ assert.deepEqual(parsed, output);
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-Your application should also handle the way titles/honorifics are abbreviated in American and British English. For example, Doctor Wright is abbreviated as "Dr Wright" in British English and "Dr. Wright" in American English. See `/public/american-to-british-titles.js` for the different titles your application should handle.
+The `/api/translate` route should also handle the way titles/honorifics are abbreviated in American and British English. For example, Doctor Wright is abbreviated as "Dr Wright" in British English and "Dr. Wright" in American English. See `/public/american-to-british-titles.js` for the different titles your application should handle.
```js
-
+async (getUserInput) => {
+ try {
+ const text = 'Dr. Grosh will see you now.';
+ const locale = 'american-to-british';
+ const output = {
+ text: text,
+ translation: '<span class="highlight">Dr</span> Grosh will see you now.'
+ };
+ let data = await fetch(getUserInput('url') + '/api/translate', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ text, locale })
+ });
+ let parsed = await data.json();
+ assert.isObject(parsed);
+ assert.property(parsed, 'text');
+ assert.property(parsed, 'translation');
+ assert.deepEqual(parsed, output);
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
Wrap any translated spelling or terms with `<span class="highlight">...</span>` tags so they appear in green.
```js
-
+async (getUserInput) => {
+ try {
+ const text = 'Mangoes are my favorite fruit.';
+ const locale = 'american-to-british';
+ const output = {
+ text: 'Mangoes are my favorite fruit.',
+ translation:
+ 'Mangoes are my <span class="highlight">favourite</span> fruit.'
+ };
+ let data = await fetch(getUserInput('url') + '/api/translate', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ text, locale })
+ });
+ let parsed = await data.json();
+ assert.isObject(parsed);
+ assert.property(parsed, 'text');
+ assert.property(parsed, 'translation');
+ assert.deepEqual(parsed, output);
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-If the sentence in the text area has no spelling or terms that should be translated, append the message "Everything looks good to me!" to the `translated-sentence` `div`.
+If one or more of the required fields is missing, return `{ error: 'Required field(s) missing' }`.
```js
-
+async (getUserInput) => {
+ try {
+ const locale = 'american-to-british';
+ let data = await fetch(getUserInput('url') + '/api/translate', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ locale })
+ });
+ let parsed = await data.json();
+ assert.isObject(parsed);
+ assert.property(parsed, 'error');
+ assert.equal(parsed.error, 'Required field(s) missing');
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-If there is no text in the text area, append the message "Error: No text to translate." to the `error-msg` `div` so the text appears in red.
+If `text` is empty, return `{ error: 'No text to translate' }`
```js
-
+async (getUserInput) => {
+ try {
+ const locale = 'american-to-british';
+ let data = await fetch(getUserInput('url') + '/api/translate', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ text: '', locale })
+ });
+ let parsed = await data.json();
+ assert.isObject(parsed);
+ assert.property(parsed, 'error');
+ assert.equal(parsed.error, 'No text to translate');
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-I can press the "Clear Input" button to remove all text from the text area and the `translated-sentence` `div`.
+If `locale` does not match one of the two specified locales, return `{ error: 'Invalid value for locale field' }`.
```js
-
+async (getUserInput) => {
+ try {
+ const text = "Ceci n'est pas une pipe";
+ const locale = 'french-to-american';
+ let data = await fetch(getUserInput('url') + '/api/translate', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ text, locale })
+ });
+ let parsed = await data.json();
+ assert.isObject(parsed);
+ assert.property(parsed, 'error');
+ assert.equal(parsed.error, 'Invalid value for locale field');
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-All 20 unit tests are complete and passing. See `/tests/1_unit-tests.js` for the sentences you should write tests for.
+If `text` requires no translation, return `"Everything looks good to me!"` for the `translation` value.
```js
-
+async (getUserInput) => {
+ try {
+ const locale = 'british-to-american';
+ const output = {
+ text: 'SaintPeter and nhcarrigan give their regards!',
+ translation: 'Everything looks good to me!'
+ };
+ let data = await fetch(getUserInput('url') + '/api/translate', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ text: output.text, locale })
+ });
+ let parsed = await data.json();
+ assert.isObject(parsed);
+ assert.isObject(parsed);
+ assert.property(parsed, 'text');
+ assert.property(parsed, 'translation');
+ assert.deepEqual(parsed, output);
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-All 4 functional tests are complete and passing. See `/tests/2_functional-tests.js` for the functionality you should write tests for.
+All 24 unit tests are complete and passing. See `/tests/1_unit-tests.js` for the expected behavior you should write tests for.
```js
+async (getUserInput) => {
+ try {
+ const getTests = await $.get(getUserInput('url') + '/_api/get-tests');
+ assert.isArray(getTests);
+ const unitTests = getTests.filter((test) => {
+ return !!test.context.match(/Unit Tests/gi);
+ });
+ assert.isAtLeast(unitTests.length, 24, 'At least 24 tests passed');
+ unitTests.forEach((test) => {
+ assert.equal(test.state, 'passed', 'Tests in Passed State');
+ assert.isAtLeast(
+ test.assertions.length,
+ 1,
+ 'At least one assertion per test'
+ );
+ });
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
+```
+All 6 functional tests are complete and passing. See `/tests/2_functional-tests.js` for the functionality you should write tests for.
+
+```js
+async (getUserInput) => {
+ try {
+ const getTests = await $.get(getUserInput('url') + '/_api/get-tests');
+ assert.isArray(getTests);
+ const functTests = getTests.filter((test) => {
+ return !!test.context.match(/Functional Tests/gi);
+ });
+ assert.isAtLeast(functTests.length, 6, 'At least 6 tests passed');
+ functTests.forEach((test) => {
+ assert.equal(test.state, 'passed', 'Tests in Passed State');
+ assert.isAtLeast(
+ test.assertions.length,
+ 1,
+ 'At least one assertion per test'
+ );
+ });
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
# --solutions--
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/issue-tracker.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/issue-tracker.md
index 2e44e8713bc..46e3eb91afb 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/issue-tracker.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/issue-tracker.md
@@ -1,63 +1,361 @@
---
id: 587d8249367417b2b2512c42
-title: 问题跟踪器
+title: Issue Tracker
challengeType: 4
-videoUrl: ''
+forumTopicId: 301569
dashedName: issue-tracker
---
# --description--
-构建一个功能类似于此的完整堆栈JavaScript应用程序： [https](https://protective-garage.glitch.me/) ： [//protective-garage.glitch.me/](https://protective-garage.glitch.me/) 。在这个项目上工作将涉及您在我们的入门项目上在Glitch上编写代码。完成此项目后，您可以将公共故障网址（到应用程序的主页）复制到此屏幕进行测试！您可以选择在另一个平台上编写项目，但必须公开显示我们的测试。使用[此链接](https://glitch.com/#!/import/github/freeCodeCamp/boilerplate-project-issuetracker/)在Glitch上启动此项目或在GitHub上克隆[此存储库](https://github.com/freeCodeCamp/boilerplate-project-issuetracker/) ！如果您使用Glitch，请记住将项目链接保存到安全的地方！
+Build a full stack JavaScript app that is functionally similar to this: <https://issue-tracker.freecodecamp.rocks/>. Working on this project will involve you writing your code using one of the following methods:
+
+- Clone [this GitHub repo](https://github.com/freeCodeCamp/boilerplate-project-issuetracker/) and complete your project locally.
+- Use [this repl.it starter project](https://repl.it/github/freeCodeCamp/boilerplate-project-issuetracker) to complete your project.
+- Use a site builder of your choice to complete the project. Be sure to incorporate all the files from our GitHub repo.
+
+When you are done, make sure a working demo of your project is hosted somewhere public. Then submit the URL to it in the `Solution Link` field. Optionally, also submit a link to your project's source code in the `GitHub Link` field.
+
+# --instructions--
+
+- Complete the necessary routes in `/routes/api.js`
+- Create all of the functional tests in `tests/2_functional-tests.js`
+- Copy the `sample.env` file to `.env` and set the variables appropriately
+- To run the tests uncomment `NODE_ENV=test` in your `.env` file
+- To run the tests in the console, use the command `npm run test`. To open the Repl.it console, press Ctrl+Shift+P (Cmd if on a Mac) and type "open shell"
+
+Write the following tests in `tests/2_functional-tests.js`:
+
+- Create an issue with every field: POST request to `/api/issues/{project}`
+- Create an issue with only required fields: POST request to `/api/issues/{project}`
+- Create an issue with missing required fields: POST request to `/api/issues/{project}`
+- View issues on a project: GET request to `/api/issues/{project}`
+- View issues on a project with one filter: GET request to `/api/issues/{project}`
+- View issues on a project with multiple filters: GET request to `/api/issues/{project}`
+- Update one field on an issue: PUT request to `/api/issues/{project}`
+- Update multiple fields on an issue: PUT request to `/api/issues/{project}`
+- Update an issue with missing `_id`: PUT request to `/api/issues/{project}`
+- Update an issue with no fields to update: PUT request to `/api/issues/{project}`
+- Update an issue with an invalid `_id`: PUT request to `/api/issues/{project}`
+- Delete an issue: DELETE request to `/api/issues/{project}`
+- Delete an issue with an invalid `_id`: DELETE request to `/api/issues/{project}`
+- Delete an issue with missing `_id`: DELETE request to `/api/issues/{project}`
# --hints--
-防止跨站点脚本（XSS）攻击。
+You can provide your own project, not the example URL.
```js
-
+(getUserInput) => {
+ assert(!/.*\/issue-tracker\.freecodecamp\.rocks/.test(getUserInput('url')));
+};
```
-我可以使用包含所需issue_title，issue_text，created_by和可选的assigned_to和status_text的表单数据POST / api / issues / {projectname}。
+You can send a `POST` request to `/api/issues/{projectname}` with form data containing the required fields `issue_title`, `issue_text`, `created_by`, and optionally `assigned_to` and `status_text`.
```js
-
+async (getUserInput) => {
+ try {
+ let test_data = {
+ issue_title: 'Faux Issue Title',
+ issue_text: 'Functional Test - Required Fields Only',
+ created_by: 'fCC'
+ };
+ const data = await $.post(
+ getUserInput('url') + '/api/issues/fcc-project',
+ test_data
+ );
+ assert.isObject(data);
+ assert.nestedInclude(data, test_data);
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-保存（和返回）的对象将包括所有这些字段（空白表示可选无输入），还包括created_on（日期/时间），updated_on（日期/时间），打开（布尔值，true表示打开，false表示关闭），和\_id。
+The `POST` request to `/api/issues/{projectname}` will return the created object, and must include all of the submitted fields. Excluded optional fields will be returned as empty strings. Additionally, include `created_on` (date/time), `updated_on` (date/time), `open` (boolean, `true` for open - default value, `false` for closed), and `_id`.
```js
-
+async (getUserInput) => {
+ try {
+ let test_data = {
+ issue_title: 'Faux Issue Title 2',
+ issue_text: 'Functional Test - Every field filled in',
+ created_by: 'fCC',
+ assigned_to: 'Chai and Mocha'
+ };
+ const data = await $.post(
+ getUserInput('url') + '/api/issues/fcc-project',
+ test_data
+ );
+ assert.isObject(data);
+ assert.nestedInclude(data, test_data);
+ assert.property(data, 'created_on');
+ assert.isNumber(Date.parse(data.created_on));
+ assert.property(data, 'updated_on');
+ assert.isNumber(Date.parse(data.updated_on));
+ assert.property(data, 'open');
+ assert.isBoolean(data.open);
+ assert.isTrue(data.open);
+ assert.property(data, '_id');
+ assert.isNotEmpty(data._id);
+ assert.property(data, 'status_text');
+ assert.isEmpty(data.status_text);
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-我可以使用id和对象中的任何字段PUT / api / issues / {projectname}，并使用值来对象表示对象。返回将“成功更新”或“无法更新”+ id。这应该始终更新updated_on。如果没有发送字段，则返回“未发送更新字段”。
+If you send a `POST` request to `/api/issues/{projectname}` without the required fields, returned will be the error `{ error: 'required field(s) missing' }`
```js
-
+async (getUserInput) => {
+ try {
+ let test_data = { created_by: 'fCC' };
+ const data = await $.post(getUserInput('url') + '/api/issues/fcc-project', {
+ created_by: 'fCC'
+ });
+ assert.isObject(data);
+ assert.property(data, 'error');
+ assert.equal(data.error, 'required field(s) missing');
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-我可以使用id DELETE / api / issues / {projectname}来完全删除问题。如果没有发送\_id，则返回“id error”，成功：“删除”+ id，失败：“无法删除”+ id。
+You can send a `GET` request to `/api/issues/{projectname}` for an array of all issues for that specific `projectname`, with all the fields present for each issue.
```js
-
+async (getUserInput) => {
+ try {
+ let test_data = { issue_text: 'Get Issues Test', created_by: 'fCC' };
+ const url =
+ getUserInput('url') +
+ '/api/issues/get_issues_test_' +
+ Date.now().toString().substring(7);
+ const data1 = await $.post(
+ url,
+ Object.assign(test_data, { issue_title: 'Faux Issue 1' })
+ );
+ assert.isObject(data1);
+ const data2 = await $.post(
+ url,
+ Object.assign(test_data, { issue_title: 'Faux Issue 2' })
+ );
+ assert.isObject(data2);
+ const data3 = await $.post(
+ url,
+ Object.assign(test_data, { issue_title: 'Faux Issue 3' })
+ );
+ assert.isObject(data3);
+ const getIssues = await $.get(url);
+ assert.isArray(getIssues);
+ assert.lengthOf(getIssues, 3);
+ let re = new RegExp('Faux Issue \\d');
+ getIssues.forEach((issue) => {
+ assert.property(issue, 'issue_title');
+ assert.match(issue.issue_title, re);
+ assert.property(issue, 'issue_text');
+ assert.property(issue, 'created_by');
+ assert.property(issue, 'assigned_to');
+ assert.property(issue, 'status_text');
+ assert.property(issue, 'open');
+ assert.property(issue, 'created_on');
+ assert.property(issue, 'updated_on');
+ assert.property(issue, '_id');
+ });
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-我可以GET / api / issues / {projectname}获取该特定项目中所有问题的数组，其中包含发布时返回的每个问题的所有信息。
+You can send a `GET` request to `/api/issues/{projectname}` and filter the request by also passing along any field and value as a URL query (ie. `/api/issues/{project}?open=false`). You can pass one or more field/value pairs at once.
```js
-
+async (getUserInput) => {
+ try {
+ let test_data = {
+ issue_title: 'To be Filtered',
+ issue_text: 'Filter Issues Test'
+ };
+ const url =
+ getUserInput('url') +
+ '/api/issues/get_issues_test_' +
+ Date.now().toString().substring(7);
+ const data1 = await $.post(
+ url,
+ Object.assign(test_data, { created_by: 'Alice', assigned_to: 'Bob' })
+ );
+ const data2 = await $.post(
+ url,
+ Object.assign(test_data, { created_by: 'Alice', assigned_to: 'Bob' })
+ );
+ const data3 = await $.post(
+ url,
+ Object.assign(test_data, { created_by: 'Alice', assigned_to: 'Eric' })
+ );
+ const data4 = await $.post(
+ url,
+ Object.assign(test_data, { created_by: 'Carol', assigned_to: 'Eric' })
+ );
+ const getSingle = await $.get(url + '?created_by=Alice');
+ assert.isArray(getSingle);
+ assert.lengthOf(getSingle, 3);
+ const getMultiple = await $.get(url + '?created_by=Alice&assigned_to=Bob');
+ assert.isArray(getMultiple);
+ assert.lengthOf(getMultiple, 2);
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-我可以通过传递查询中的任何字段和值来过滤我的获取请求（即。/ api / issues / {project}？open = false）。我可以传递尽可能多的字段/值。
+You can send a `PUT` request to `/api/issues/{projectname}` with an `_id` and one or more fields to update. On success, the `updated_on` field should be updated, and returned should be `{ result: 'successfully updated', '_id': _id }`.
```js
-
+async (getUserInput) => {
+ try {
+ let initialData = {
+ issue_title: 'Issue to be Updated',
+ issue_text: 'Functional Test - Put target',
+ created_by: 'fCC'
+ };
+ const url = getUserInput('url') + '/api/issues/fcc-project';
+ const itemToUpdate = await $.post(url, initialData);
+ const updateSucccess = await $.ajax({
+ url: url,
+ type: 'PUT',
+ data: { _id: itemToUpdate._id, issue_text: 'New Issue Text' }
+ });
+ assert.isObject(updateSucccess);
+ assert.deepEqual(updateSucccess, {
+ result: 'successfully updated',
+ _id: itemToUpdate._id
+ });
+ const getUpdatedId = await $.get(url + '?_id=' + itemToUpdate._id);
+ assert.isArray(getUpdatedId);
+ assert.isObject(getUpdatedId[0]);
+ assert.isAbove(
+ Date.parse(getUpdatedId[0].updated_on),
+ Date.parse(getUpdatedId[0].created_on)
+ );
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-所有11项功能测试都已完成并通过。
+When the `PUT` request sent to `/api/issues/{projectname}` does not include an `_id`, the return value is `{ error: 'missing _id' }`.
```js
+async (getUserInput) => {
+ try {
+ const url = getUserInput('url') + '/api/issues/fcc-project';
+ const badUpdate = await $.ajax({ url: url, type: 'PUT' });
+ assert.isObject(badUpdate);
+ assert.property(badUpdate, 'error');
+ assert.equal(badUpdate.error, 'missing _id');
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
+```
+When the `PUT` request sent to `/api/issues/{projectname}` does not include update fields, the return value is `{ error: 'no update field(s) sent', '_id': _id }`. On any other error, the return value is `{ error: 'could not update', '_id': _id }`.
+
+```js
+async (getUserInput) => {
+ try {
+ const url = getUserInput('url') + '/api/issues/fcc-project';
+ const badUpdate = await $.ajax({
+ url: url,
+ type: 'PUT',
+ data: { _id: '5f665eb46e296f6b9b6a504d' }
+ });
+ assert.deepEqual(badUpdate, {
+ error: 'no update field(s) sent',
+ _id: '5f665eb46e296f6b9b6a504d'
+ });
+ const badIdUpdate = await $.ajax({
+ url: url,
+ type: 'PUT',
+ data: { _id: '5f665eb46e296f6b9b6a504d', issue_text: 'New Issue Text' }
+ });
+ assert.deepEqual(badIdUpdate, {
+ error: 'could not update',
+ _id: '5f665eb46e296f6b9b6a504d'
+ });
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
+```
+
+You can send a `DELETE` request to `/api/issues/{projectname}` with an `_id` to delete an issue. If no `_id` is sent, the return value is `{ error: 'missing _id' }`. On success, the return value is `{ result: 'successfully deleted', '_id': _id }`. On failure, the return value is `{ error: 'could not delete', '_id': _id }`.
+
+```js
+async (getUserInput) => {
+ try {
+ let initialData = {
+ issue_title: 'Issue to be Deleted',
+ issue_text: 'Functional Test - Delete target',
+ created_by: 'fCC'
+ };
+ const url = getUserInput('url') + '/api/issues/fcc-project';
+ const itemToDelete = await $.post(url, initialData);
+ assert.isObject(itemToDelete);
+ const deleteSuccess = await $.ajax({
+ url: url,
+ type: 'DELETE',
+ data: { _id: itemToDelete._id }
+ });
+ assert.isObject(deleteSuccess);
+ assert.deepEqual(deleteSuccess, {
+ result: 'successfully deleted',
+ _id: itemToDelete._id
+ });
+ const noId = await $.ajax({ url: url, type: 'DELETE' });
+ assert.isObject(noId);
+ assert.deepEqual(noId, { error: 'missing _id' });
+ const badIdDelete = await $.ajax({
+ url: url,
+ type: 'DELETE',
+ data: { _id: '5f665eb46e296f6b9b6a504d', issue_text: 'New Issue Text' }
+ });
+ assert.isObject(badIdDelete);
+ assert.deepEqual(badIdDelete, {
+ error: 'could not delete',
+ _id: '5f665eb46e296f6b9b6a504d'
+ });
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
+```
+
+All 14 functional tests are complete and passing.
+
+```js
+async (getUserInput) => {
+ try {
+ const getTests = await $.get(getUserInput('url') + '/_api/get-tests');
+ assert.isArray(getTests);
+ assert.isAtLeast(getTests.length, 14, 'At least 14 tests passed');
+ getTests.forEach((test) => {
+ assert.equal(test.state, 'passed', 'Test in Passed State');
+ assert.isAtLeast(
+ test.assertions.length,
+ 1,
+ 'At least one assertion per test'
+ );
+ });
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
# --solutions--
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/metric-imperial-converter.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/metric-imperial-converter.md
index e53b7ab1fe0..a448bda36d3 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/metric-imperial-converter.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/metric-imperial-converter.md
@@ -1,93 +1,308 @@
---
id: 587d8249367417b2b2512c41
-title: 公制 - 英制转换器
+title: Metric-Imperial Converter
challengeType: 4
-videoUrl: ''
+forumTopicId: 301570
dashedName: metric-imperial-converter
---
# --description--
-构建一个功能类似于此的完整堆栈JavaScript应用程序： <https://incongruous-beard.glitch.me/>。在这个项目上工作将涉及您在我们的入门项目上在Glitch上编写代码。完成此项目后，您可以将公共故障网址（到应用程序的主页）复制到此屏幕进行测试！您可以选择在另一个平台上编写项目，但必须公开显示我们的测试。使用[此链接](https://glitch.com/#!/import/github/freeCodeCamp/boilerplate-project-metricimpconverter/)在Glitch上启动此项目或在GitHub上克隆[此存储库](https://github.com/freeCodeCamp/boilerplate-project-metricimpconverter/) ！如果您使用Glitch，请记住将项目链接保存到安全的地方！
+Build a full stack JavaScript app that is functionally similar to this: <https://metric-imperial-converter.freecodecamp.rocks/>. Working on this project will involve you writing your code using one of the following methods:
+
+- Clone [this GitHub repo](https://github.com/freeCodeCamp/boilerplate-project-metricimpconverter/) and complete your project locally.
+- Use [our repl.it starter project](https://repl.it/github/freeCodeCamp/boilerplate-project-metricimpconverter) to complete your project.
+- Use a site builder of your choice to complete the project. Be sure to incorporate all the files from our GitHub repo.
+
+When you are done, make sure a working demo of your project is hosted somewhere public. Then submit the URL to it in the `Solution Link` field. Optionally, also submit a link to your project's source code in the `GitHub Link` field.
+
+# --instructions--
+
+- Complete the necessary conversion logic in `/controllers/convertHandler.js`
+- Complete the necessary routes in `/routes/api.js`
+- Copy the `sample.env` file to `.env` and set the variables appropriately
+- To run the tests uncomment `NODE_ENV=test` in your `.env` file
+- To run the tests in the console, use the command `npm run test`. To open the Repl.it console, press Ctrl+Shift+P (Cmd if on a Mac) and type "open shell"
+
+Write the following tests in `tests/1_unit-tests.js`:
+
+- `convertHandler` should correctly read a whole number input.
+- `convertHandler` should correctly read a decimal number input.
+- `convertHandler` should correctly read a fractional input.
+- `convertHandler` should correctly read a fractional input with a decimal.
+- `convertHandler` should correctly return an error on a double-fraction (i.e. `3/2/3`).
+- `convertHandler` should correctly default to a numerical input of `1` when no numerical input is provided.
+- `convertHandler` should correctly read each valid input unit.
+- `convertHandler` should correctly return an error for an invalid input unit.
+- `convertHandler` should return the correct return unit for each valid input unit.
+- `convertHandler` should correctly return the spelled-out string unit for each valid input unit.
+- `convertHandler` should correctly convert `gal` to `L`.
+- `convertHandler` should correctly convert `L` to `gal`.
+- `convertHandler` should correctly convert `mi` to `km`.
+- `convertHandler` should correctly convert `km` to `mi`.
+- `convertHandler` should correctly convert `lbs` to `kg`.
+- `convertHandler` should correctly convert `kg` to `lbs`.
+
+Write the following tests in `tests/2_functional-tests.js`:
+
+- Convert a valid input such as `10L`: `GET` request to `/api/convert`.
+- Convert an invalid input such as `32g`: `GET` request to `/api/convert`.
+- Convert an invalid number such as `3/7.2/4kg`: `GET` request to `/api/convert`.
+- Convert an invalid number AND unit such as `3/7.2/4kilomegagram`: `GET` request to `/api/convert`.
+- Convert with no number such as `kg`: `GET` request to `/api/convert`.
# --hints--
-我将阻止客户端尝试猜测（嗅探）MIME类型。
+You can provide your own project, not the example URL.
```js
-
+getUserInput => {
+ assert(
+ !/.*\/metric-imperial-converter\.freecodecamp\.rocks/.test(
+ getUserInput('url')
+ )
+ );
+};
```
-我会阻止跨站点脚本（XSS）攻击。
-
+You can `GET` `/api/convert` with a single parameter containing an accepted number and unit and have it converted. (Hint: Split the input by looking for the index of the first character which will mark the start of the unit)
```js
-
```
-我可以使用包含可接受的数字和单位的单个参数进行GET / api / convert并将其转换。 （提示：通过查找将标记单元开头的第一个字符的索引来拆分输入）
+You can convert `'gal'` to `'L'` and vice versa. (1 gal to 3.78541 L)
```js
-
+async getUserInput => {
+ try {
+ const data1 = await $.get(getUserInput('url') + '/api/convert?input=1gal');
+ assert.equal(data1.returnNum, 3.78541);
+ assert.equal(data1.returnUnit, 'L');
+ const data2 = await $.get(getUserInput('url') + '/api/convert?input=10gal');
+ assert.equal(data2.returnNum, 37.8541);
+ assert.equal(data2.returnUnit, 'L');
+ const data3 = await $.get(getUserInput('url') + '/api/convert?input=1l');
+ assert.equal(data3.returnNum, 0.26417);
+ assert.equal(data3.returnUnit, 'gal');
+ const data4 = await $.get(getUserInput('url') + '/api/convert?input=10l');
+ assert.equal(data4.returnNum, 2.64172);
+ assert.equal(data4.returnUnit, 'gal');
+ } catch (xhr) {
+ throw new Error(xhr.responseText || xhr.message);
+ }
+};
```
-我可以将'gal'转换为'L'，反之亦然。 （1加仑到3.78541升）
+You can convert `'lbs'` to `'kg'` and vice versa. (1 lbs to 0.453592 kg)
```js
-
+async getUserInput => {
+ try {
+ const data1 = await $.get(getUserInput('url') + '/api/convert?input=1lbs');
+ assert.equal(data1.returnNum, 0.45359);
+ assert.equal(data1.returnUnit, 'kg');
+ const data2 = await $.get(getUserInput('url') + '/api/convert?input=10lbs');
+ assert.equal(data2.returnNum, 4.53592);
+ assert.equal(data2.returnUnit, 'kg');
+ const data3 = await $.get(getUserInput('url') + '/api/convert?input=1kg');
+ assert.equal(data3.returnNum, 2.20462);
+ assert.equal(data3.returnUnit, 'lbs');
+ const data4 = await $.get(getUserInput('url') + '/api/convert?input=10kg');
+ assert.equal(data4.returnNum, 22.04624);
+ assert.equal(data4.returnUnit, 'lbs');
+ } catch (xhr) {
+ throw new Error(xhr.responseText || xhr.message);
+ }
+};
```
-我可以将'lbs'转换为'kg'，反之亦然。 （1磅至0.453592千克）
+You can convert `'mi'` to `'km'` and vice versa. (1 mi to 1.60934 km)
```js
-
+async getUserInput => {
+ try {
+ const data1 = await $.get(getUserInput('url') + '/api/convert?input=1mi');
+ assert.equal(data1.returnNum, 1.60934);
+ assert.equal(data1.returnUnit, 'km');
+ const data2 = await $.get(getUserInput('url') + '/api/convert?input=10mi');
+ assert.equal(data2.returnNum, 16.0934);
+ assert.equal(data2.returnUnit, 'km');
+ const data3 = await $.get(getUserInput('url') + '/api/convert?input=1km');
+ assert.equal(data3.returnNum, 0.62137);
+ assert.equal(data3.returnUnit, 'mi');
+ const data4 = await $.get(getUserInput('url') + '/api/convert?input=10km');
+ assert.equal(data4.returnNum, 6.21373);
+ assert.equal(data4.returnUnit, 'mi');
+ } catch (xhr) {
+ throw new Error(xhr.responseText || xhr.message);
+ }
+};
```
-我可以将'mi'转换为'km'，反之亦然。 （1英里到1.60934公里）
+All incoming units should be accepted in both upper and lower case, but should be returned in both the `initUnit` and `returnUnit` in lower case, except for liter, which should be represented as an uppercase `'L'`.
```js
-
+async getUserInput => {
+ try {
+ const data1 = await $.get(getUserInput('url') + '/api/convert?input=1gal');
+ assert.equal(data1.initUnit, 'gal');
+ assert.equal(data1.returnUnit, 'L');
+ const data2 = await $.get(getUserInput('url') + '/api/convert?input=10L');
+ assert.equal(data2.initUnit, 'L');
+ assert.equal(data2.returnUnit, 'gal');
+ const data3 = await $.get(getUserInput('url') + '/api/convert?input=1l');
+ assert.equal(data3.initUnit, 'L');
+ assert.equal(data3.returnUnit, 'gal');
+ const data4 = await $.get(getUserInput('url') + '/api/convert?input=10KM');
+ assert.equal(data4.initUnit, 'km');
+ assert.equal(data4.returnUnit, 'mi');
+ } catch (xhr) {
+ throw new Error(xhr.responseText || xhr.message);
+ }
+};
```
-如果我的计量单位无效，则返回“无效单位”。
+If the unit of measurement is invalid, returned will be `'invalid unit'`.
```js
-
+async getUserInput => {
+ try {
+ const data = await $.get(getUserInput('url') + '/api/convert?input=1min');
+ assert(data.error === 'invalid unit' || data === 'invalid unit');
+ } catch (xhr) {
+ throw new Error(xhr.responseText || xhr.message);
+ }
+};
```
-如果我的号码无效，则返回“无效号码”。
+If the number is invalid, returned will be `'invalid number'`.
```js
-
+async getUserInput => {
+ try {
+ const data = await $.get(
+ getUserInput('url') + '/api/convert?input=1//2gal'
+ );
+ assert(data.error === 'invalid number' || data === 'invalid number');
+ } catch (xhr) {
+ throw new Error(xhr.responseText || xhr.message);
+ }
+};
```
-如果两者都无效，则返回“无效数字和单位”。
+If both the unit and number are invalid, returned will be `'invalid number and unit'`.
```js
-
+async getUserInput => {
+ try {
+ const data = await $.get(
+ getUserInput('url') + '/api/convert?input=1//2min'
+ );
+ assert(
+ data.error === 'invalid number and unit' ||
+ data === 'invalid number and unit'
+ );
+ } catch (xhr) {
+ throw new Error(xhr.responseText || xhr.message);
+ }
+};
```
-我可以在我的参数中使用分数，小数或两者（即5,1 / 2,2.5 / 6），但如果没有提供，则默认为1。
+You can use fractions, decimals or both in the parameter (ie. 5, 1/2, 2.5/6), but if nothing is provided it will default to 1.
```js
-
+async getUserInput => {
+ try {
+ const data1 = await $.get(getUserInput('url') + '/api/convert?input=mi');
+ assert.approximately(data1.initNum, 1, 0.001);
+ assert.approximately(data1.returnNum, 1.60934, 0.001);
+ assert.equal(data1.returnUnit, 'km');
+ const data2 = await $.get(getUserInput('url') + '/api/convert?input=1/5mi');
+ assert.approximately(data2.initNum, 1 / 5, 0.1);
+ assert.approximately(data2.returnNum, 0.32187, 0.001);
+ assert.equal(data2.returnUnit, 'km');
+ const data3 = await $.get(
+ getUserInput('url') + '/api/convert?input=1.5/7km'
+ );
+ assert.approximately(data3.initNum, 1.5 / 7, 0.001);
+ assert.approximately(data3.returnNum, 0.13315, 0.001);
+ assert.equal(data3.returnUnit, 'mi');
+ const data4 = await $.get(
+ getUserInput('url') + '/api/convert?input=3/2.7km'
+ );
+ assert.approximately(data4.initNum, 3 / 2.7, 0.001);
+ assert.approximately(data4.returnNum, 0.69041, 0.001);
+ assert.equal(data4.returnUnit, 'mi');
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-我的返回将包含initNum，initUnit，returnNum，returnUnit和字符串拼写单元，格式为“{initNum} {initial_Units}转换为{returnNum} {return_Units}”，结果舍入为字符串中的5位小数。
+Your return will consist of the `initNum`, `initUnit`, `returnNum`, `returnUnit`, and `string` spelling out units in the format `'{initNum} {initUnitString} converts to {returnNum} {returnUnitString}'` with the result rounded to 5 decimals.
```js
-
+async getUserInput => {
+ try {
+ const data = await $.get(getUserInput('url') + '/api/convert?input=2mi');
+ assert.equal(data.initNum, 2);
+ assert.equal(data.initUnit, 'mi');
+ assert.approximately(data.returnNum, 3.21868, 0.001);
+ assert.equal(data.returnUnit, 'km', 'returnUnit did not match');
+ assert.equal(data.string, '2 miles converts to 3.21868 kilometers');
+ } catch (xhr) {
+ throw new Error(xhr.responseText || xhr.message);
+ }
+};
```
-所有16个单元测试都已完成并通过。
+All 16 unit tests are complete and passing.
```js
-
+async getUserInput => {
+ try {
+ const getTests = await $.get(getUserInput('url') + '/_api/get-tests');
+ assert.isArray(getTests);
+ const unitTests = getTests.filter(test => {
+ return !!test.context.match(/Unit Tests/gi);
+ });
+ assert.isAtLeast(unitTests.length, 16, 'At least 16 tests passed');
+ unitTests.forEach(test => {
+ assert.equal(test.state, 'passed', 'Tests in Passed State');
+ assert.isAtLeast(
+ test.assertions.length,
+ 1,
+ 'At least one assertion per test'
+ );
+ });
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-所有5个功能测试都已完成并通过。
+All 5 functional tests are complete and passing.
```js
-
+async getUserInput => {
+ try {
+ const getTests = await $.get(getUserInput('url') + '/_api/get-tests');
+ assert.isArray(getTests);
+ const functTests = getTests.filter(test => {
+ return !!test.context.match(/Functional Tests/gi);
+ });
+ assert.isAtLeast(functTests.length, 5, 'At least 5 tests passed');
+ functTests.forEach(test => {
+ assert.equal(test.state, 'passed', 'Tests in Passed State');
+ assert.isAtLeast(
+ test.assertions.length,
+ 1,
+ 'At least one assertion per test'
+ );
+ });
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
# --solutions--
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/personal-library.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/personal-library.md
index 9799b943735..6cd072b4461 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/personal-library.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/personal-library.md
@@ -1,75 +1,213 @@
---
id: 587d824a367417b2b2512c43
-title: 个人图书馆
+title: Personal Library
challengeType: 4
-videoUrl: ''
+forumTopicId: 301571
dashedName: personal-library
---
# --description--
-构建一个功能类似于此的完整堆栈JavaScript应用程序： [https](https://fuzzy-mink.glitch.me/) ： [//spark-cathedral.glitch.me/](https://fuzzy-mink.glitch.me/) 。在这个项目上工作将涉及您在我们的入门项目上在Glitch上编写代码。完成此项目后，您可以将公共故障网址（到应用程序的主页）复制到此屏幕进行测试！您可以选择在另一个平台上编写项目，但必须公开显示我们的测试。使用[此链接](https://glitch.com/#!/import/github/freeCodeCamp/boilerplate-project-library/)在Glitch上启动此项目或在GitHub上克隆[此存储库](https://github.com/freeCodeCamp/boilerplate-project-library/) ！如果您使用Glitch，请记住将项目链接保存到安全的地方！
+Build a full stack JavaScript app that is functionally similar to this: <https://personal-library.freecodecamp.rocks/>. Working on this project will involve you writing your code using one of the following methods:
+
+- Clone [this GitHub repo](https://github.com/freeCodeCamp/boilerplate-project-library) and complete your project locally.
+- Use [our repl.it starter project](https://repl.it/github/freeCodeCamp/boilerplate-project-library)) to complete your project.
+- Use a site builder of your choice to complete the project. Be sure to incorporate all the files from our GitHub repo.
+
+When you are done, make sure a working demo of your project is hosted somewhere public. Then submit the URL to it in the `Solution Link` field. Optionally, also submit a link to your project's source code in the `GitHub Link` field.
+
+# --instructions--
+
+1. Add your MongoDB connection string to `.env` without quotes as `DB`
+ Example: `DB=mongodb://admin:pass@1234.mlab.com:1234/fccpersonallib`
+2. In your `.env` file set `NODE_ENV` to `test`, without quotes
+3. You need to create all routes within `routes/api.js`
+4. You will create all functional tests in `tests/2_functional-tests.js`
# --hints--
-我的网站上的任何内容都不会缓存在我的客户端中。
+You can provide your own project, not the example URL.
```js
-
+(getUserInput) => {
+ assert(
+ !/.*\/personal-library\.freecodecamp\.rocks/.test(getUserInput('url'))
+ );
+};
```
-标题会说该网站由'PHP 4.2.0'提供支持，即使它不是（作为安全措施）。
+You can send a <b>POST</b> request to `/api/books` with `title` as part of the form data to add a book. The returned response will be an object with the `title` and a unique `_id` as keys. If `title` is not included in the request, the returned response should be the string `missing required field title`.
```js
-
+async (getUserInput) => {
+ try {
+ let data1 = await $.post(getUserInput('url') + '/api/books', {
+ title: 'Faux Book 1'
+ });
+ assert.isObject(data1);
+ assert.property(data1, 'title');
+ assert.equal(data1.title, 'Faux Book 1');
+ assert.property(data1, '_id');
+ let data2 = await $.post(getUserInput('url') + '/api/books');
+ assert.isString(data2);
+ assert.equal(data2, 'missing required field title');
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-我可以将标题发布到/ api / books以添加书籍，并返回将是具有标题和唯一\_id的对象。
+You can send a <b>GET</b> request to `/api/books` and receive a JSON response representing all the books. The JSON response will be an array of objects with each object (book) containing `title`, `_id`, and `commentcount` properties.
```js
-
+async (getUserInput) => {
+ try {
+ let url = getUserInput('url') + '/api/books';
+ let a = $.post(url, { title: 'Faux Book A' });
+ let b = $.post(url, { title: 'Faux Book B' });
+ let c = $.post(url, { title: 'Faux Book C' });
+ Promise.all([a, b, c]).then(async () => {
+ let data = await $.get(url);
+ assert.isArray(data);
+ assert.isAtLeast(data.length, 3);
+ data.forEach((book) => {
+ assert.isObject(book);
+ assert.property(book, 'title');
+ assert.isString(book.title);
+ assert.property(book, '_id');
+ assert.property(book, 'commentcount');
+ assert.isNumber(book.commentcount);
+ });
+ });
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-我可以获取/ api / books来检索包含title，\_id和commentcount的所有书籍的数组。
+You can send a <b>GET</b> request to `/api/books/{_id}` to retrieve a single object of a book containing the properties `title`, `_id`, and a `comments` array (empty array if no comments present). If no book is found, return the string `no book exists`.
```js
-
+async (getUserInput) => {
+ try {
+ let url = getUserInput('url') + '/api/books';
+ let noBook = await $.get(url + '/5f665eb46e296f6b9b6a504d');
+ assert.isString(noBook);
+ assert.equal(noBook, 'no book exists');
+ let sampleBook = await $.post(url, { title: 'Faux Book Alpha' });
+ assert.isObject(sampleBook);
+ let bookId = sampleBook._id;
+ let bookQuery = await $.get(url + '/' + bookId);
+ assert.isObject(bookQuery);
+ assert.property(bookQuery, 'title');
+ assert.equal(bookQuery.title, 'Faux Book Alpha');
+ assert.property(bookQuery, 'comments');
+ assert.isArray(bookQuery.comments);
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-我可以使用/ api / books / {id}来检索包含\_title，\_id和注释数组的书的单个对象（如果没有注释，则为空数组）。
+You can send a <b>POST</b> request containing `comment` as the form body data to `/api/books/{_id}` to add a comment to a book. The returned response will be the books object similar to <b>GET</b> `/api/books/{_id}` request in an earlier test. If `comment` is not included in the request, return the string `missing required field comment`. If no book is found, return the string `no book exists`.
```js
-
+async (getUserInput) => {
+ try {
+ let url = getUserInput('url') + '/api/books';
+ let commentTarget = await $.post(url, { title: 'Notable Book' });
+ assert.isObject(commentTarget);
+ let bookId = commentTarget._id;
+ let bookCom1 = await $.post(url + '/' + bookId, {
+ comment: 'This book is fab!'
+ });
+ let bookCom2 = await $.post(url + '/' + bookId, {
+ comment: 'I did not care for it'
+ });
+ assert.isObject(bookCom2);
+ assert.property(bookCom2, '_id');
+ assert.property(bookCom2, 'title');
+ assert.property(bookCom2, 'comments');
+ assert.lengthOf(bookCom2.comments, 2);
+ bookCom2.comments.forEach((comment) => {
+ assert.isString(comment);
+ assert.oneOf(comment, ['This book is fab!', 'I did not care for it']);
+ });
+ let commentErr = await $.post(url + '/' + bookId);
+ assert.isString(commentErr);
+ assert.equal(commentErr, 'missing required field comment');
+ let failingComment = await $.post(url + '/5f665eb46e296f6b9b6a504d', {
+ comment: 'Never Seen Comment'
+ });
+ assert.isString(failingComment);
+ assert.equal(failingComment, 'no book exists');
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-我可以在/ api / books / {id}上发表评论，为书籍添加评论，并返回类似于get / api / books / {id}的书籍对象，包括新评论。
+You can send a <b>DELETE</b> request to `/api/books/{_id}` to delete a book from the collection. The returned response will be the string `delete successful` if successful. If no book is found, return the string `no book exists`.
```js
-
+async (getUserInput) => {
+ try {
+ let url = getUserInput('url') + '/api/books';
+ let deleteTarget = await $.post(url, { title: 'Deletable Book' });
+ assert.isObject(deleteTarget);
+ let bookId = deleteTarget._id;
+ let doDelete = await $.ajax({ url: url + '/' + bookId, type: 'DELETE' });
+ assert.isString(doDelete);
+ assert.equal(doDelete, 'delete successful');
+ let failingDelete = await $.ajax({
+ url: url + '/5f665eb46e296f6b9b6a504d',
+ type: 'DELETE'
+ });
+ assert.isString(failingDelete);
+ assert.equal(failingDelete, 'no book exists');
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-我可以删除/ api / books / {\_ id}从集合中删除一本书。如果成功，返回将“删除成功”。
+You can send a <b>DELETE</b> request to `/api/books` to delete all books in the database. The returned response will be the string `'complete delete successful` if successful.
```js
-
+async (getUserInput) => {
+ try {
+ const deleteAll = await $.ajax({
+ url: getUserInput('url') + '/api/books',
+ type: 'DELETE'
+ });
+ assert.isString(deleteAll);
+ assert.equal(deleteAll, 'complete delete successful');
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
-如果我试图要求一本不存在的书，我将被退回'没有书存在'。
+All 10 functional tests required are complete and passing.
```js
-
-```
-
-我可以向/ api / books发送删除请求以删除数据库中的所有书籍。如果成功，返回将是“完全删除成功”。
-
-```js
-
-```
-
-所有6项功能测试都是完整的并且通过。
-
-```js
-
+async (getUserInput) => {
+ try {
+ const getTests = await $.get(getUserInput('url') + '/_api/get-tests');
+ assert.isArray(getTests);
+ assert.isAtLeast(getTests.length, 10, 'At least 10 tests passed');
+ getTests.forEach((test) => {
+ assert.equal(test.state, 'passed', 'Test in Passed State');
+ assert.isAtLeast(
+ test.assertions.length,
+ 1,
+ 'At least one assertion per test'
+ );
+ });
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
# --solutions--
diff --git a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/sudoku-solver.md b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/sudoku-solver.md
index 5d9535cea45..dde3301c584 100644
--- a/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/sudoku-solver.md
+++ b/curriculum/challenges/chinese/06-quality-assurance/quality-assurance-projects/sudoku-solver.md
@@ -1,89 +1,351 @@
---
id: 5e601bf95ac9d0ecd8b94afd
+title: Sudoku Solver
challengeType: 4
dashedName: sudoku-solver
---
# --description--
-Build a full stack JavaScript app that is functionally similar to this: <https://bottlenose-eucalyptus.glitch.me/>.
+Build a full stack JavaScript app that is functionally similar to this: <https://sudoku-solver.freecodecamp.rocks/>. Working on this project will involve you writing your code using one of the following methods:
-Working on this project will involve you writing your code on Glitch on our starter project. After completing this project you can copy your public glitch url (to the homepage of your app) into this screen to test it! Optionally you may choose to write your project on another platform but it must be publicly visible for our testing.
+- Clone [this GitHub repo](https://github.com/freecodecamp/boilerplate-project-sudoku-solver) and complete your project locally.
+- Use [our repl.it starter project](https://repl.it/github/freeCodeCamp/boilerplate-project-sudoku-solver) to complete your project.
+- Use a site builder of your choice to complete the project. Be sure to incorporate all the files from our GitHub repo.
-Start this project on Glitch using [this link](https://glitch.com/edit/#!/remix/clone-from-repo?REPO_URL=https://github.com/freeCodeCamp/boilerplate-project-sudoku-solver/) or clone [this repository](https://github.com/freeCodeCamp/boilerplate-project-sudoku-solver/) on GitHub! If you use Glitch, remember to save the link to your project somewhere safe!
+When you are done, make sure a working demo of your project is hosted somewhere public. Then submit the URL to it in the `Solution Link` field. Optionally, also submit a link to your project's source code in the `GitHub Link` field.
+
+# --instructions--
+
+- All puzzle logic can go into `/controllers/sudoku-solver.js`
+ - The `validate` function should take a given puzzle string and check it to see if it has 81 valid characters for the input.
+ - The `check` functions should be validating against the *current* state of the board.
+ - The `solve` function should handle solving any given valid puzzle string, not just the test inputs and solutions. You are expected to write out the logic to solve this.
+- All routing logic can go into `/routes/api.js`
+- See the `puzzle-strings.js` file in `/controllers` for some sample puzzles your application should solve
+- To run the challenge tests on this page, set `NODE_ENV` to `test` without quotes in the `.env` file
+- To run the tests in the console, use the command `npm run test`. To open the Repl.it console, press Ctrl+Shift+P (Cmd if on a Mac) and type "open shell"
+
+Write the following tests in `tests/1_unit-tests.js`:
+
+- Logic handles a valid puzzle string of 81 characters
+- Logic handles a puzzle string with invalid characters (not 1-9 or `.`)
+- Logic handles a puzzle string that is not 81 characters in length
+- Logic handles a valid row placement
+- Logic handles an invalid row placement
+- Logic handles a valid column placement
+- Logic handles an invalid column placement
+- Logic handles a valid region (3x3 grid) placement
+- Logic handles an invalid region (3x3 grid) placement
+- Valid puzzle strings pass the solver
+- Invalid puzzle strings fail the solver
+- Solver returns the the expected solution for an incomplete puzzzle
+
+Write the following tests in `tests/2_functional-tests.js`
+
+- Solve a puzzle with valid puzzle string: POST request to `/api/solve`
+- Solve a puzzle with missing puzzle string: POST request to `/api/solve`
+- Solve a puzzle with invalid characters: POST request to `/api/solve`
+- Solve a puzzle with incorrect length: POST request to `/api/solve`
+- Solve a puzzle that cannot be solved: POST request to `/api/solve`
+- Check a puzzle placement with all fields: POST request to `/api/check`
+- Check a puzzle placement with single placement conflict: POST request to `/api/check`
+- Check a puzzle placement with multiple placement conflicts: POST request to `/api/check`
+- Check a puzzle placement with all placement conflicts: POST request to `/api/check`
+- Check a puzzle placement with missing required fields: POST request to `/api/check`
+- Check a puzzle placement with invalid characters: POST request to `/api/check`
+- Check a puzzle placement with incorrect length: POST request to `/api/check`
+- Check a puzzle placement with invalid placement coordinate: POST request to `/api/check`
+- Check a puzzle placement with invalid placement value: POST request to `/api/check`
# --hints--
-I can enter a sudoku puzzle by filling in the text area with either a number or period (.) to represent an empty cell.
+You should provide your own project, not the example URL.
```js
-
+(getUserInput) => {
+ const url = getUserInput('url');
+ assert(!/.*\/sudoku-solver\.freecodecamp\.rocks/.test(getUserInput('url')));
+};
```
-When a valid number is entered in the text area, the same number is applied to the correct cell of the sudoku grid.
+You can `POST` `/api/solve` with form data containing `puzzle` which will be a string containing a combination of numbers (1-9) and periods `.` to represent empty spaces. The returned object will contain a `solution` property with the solved puzzle.
```js
-
+async (getUserInput) => {
+ const input =
+ '..9..5.1.85.4....2432......1...69.83.9.....6.62.71...9......1945....4.37.4.3..6..';
+ const output =
+ '769235418851496372432178956174569283395842761628713549283657194516924837947381625';
+ const data = await fetch(getUserInput('url') + '/api/solve', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ puzzle: input })
+ });
+ const parsed = await data.json();
+ assert.property(parsed, 'solution');
+ assert.equal(parsed.solution, output);
+};
```
-I can enter a sudoku puzzle by adding numbers directly to the sudoku grid.
+If the object submitted to `/api/solve` is missing `puzzle`, the returned value will be `{ error: 'Required field missing' }`
```js
-
+async (getUserInput) => {
+ const input =
+ '..9..5.1.85.4....2432......1...69.83.9.....6.62.71...9......1945....4.37.4.3..6..';
+ const output = 'Required field missing';
+ const data = await fetch(getUserInput('url') + '/api/solve', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ notpuzzle: input })
+ });
+ const parsed = await data.json();
+ assert.property(parsed, 'error');
+ assert.equal(parsed.error, output);
+};
```
-When a valid number is entered in the sudoku grid, the same number appears in the correct position in the text area.
+If the puzzle submitted to `/api/solve` contains values which are not numbers or periods, the returned value will be `{ error: 'Invalid characters in puzzle' }`
```js
-
+async (getUserInput) => {
+ const input =
+ 'AA9..5.1.85.4....2432......1...69.83.9.....6.62.71...9......1945....4.37.4.3..6..';
+ const output = 'Invalid characters in puzzle';
+ const data = await fetch(getUserInput('url') + '/api/solve', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ puzzle: input })
+ });
+ const parsed = await data.json();
+ assert.property(parsed, 'error');
+ assert.equal(parsed.error, output);
+};
```
-The text area should only update the corresponding sudoku grid cell when a whole number between 1 and 9 is entered.
+If the puzzle submitted to `/api/solve` is greater or less than 81 characters, the returned value will be `{ error: 'Expected puzzle to be 81 characters long' }`
```js
-
+async (getUserInput) => {
+ const input =
+ '9..5.1.85.4....2432......1...69.83.9.....6.62.71...9......1945....4.37.4.3..6..';
+ const output = 'Expected puzzle to be 81 characters long';
+ const data = await fetch(getUserInput('url') + '/api/solve', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ puzzle: input })
+ });
+ const parsed = await data.json();
+ assert.property(parsed, 'error');
+ assert.equal(parsed.error, output);
+};
```
-The sudoku grid should only update the puzzle string in the text area when a whole number between 1 and 9 is entered into a cell.
+If the puzzle submitted to `/api/solve` is invalid or cannot be solved, the returned value will be `{ error: 'Puzzle cannot be solved' }`
```js
-
+async (getUserInput) => {
+ const input =
+ '9.9..5.1.85.4....2432......1...69.83.9.....6.62.71...9......1945....4.37.4.3..6..';
+ const output = 'Puzzle cannot be solved';
+ const data = await fetch(getUserInput('url') + '/api/solve', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ puzzle: input })
+ });
+ const parsed = await data.json();
+ assert.property(parsed, 'error');
+ assert.equal(parsed.error, output);
+};
```
-I can solve an incomplete puzzle by clicking the "Solve" button. When a solution is found, the sudoku grid and text area are automatically populated with the correct numbers for each cell in the grid or position in the text area.
+You can `POST` to `/api/check` an object containing `puzzle`, `coordinate`, and `value` where the `coordinate` is the letter A-I indicating the row, followed by a number 1-9 indicating the column, and `value` is a number from 1-9.
```js
-
+async (getUserInput) => {
+ const input =
+ '..9..5.1.85.4....2432......1...69.83.9.....6.62.71...9......1945....4.37.4.3..6..';
+ const coordinate = 'A1';
+ const value = '7';
+ const data = await fetch(getUserInput('url') + '/api/check', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ puzzle: input, coordinate, value })
+ });
+ const parsed = await data.json();
+ assert.property(parsed, 'valid');
+ assert.isTrue(parsed.valid);
+};
```
-This sudoku solver is not expected to be able to solve every incomplete puzzle. See `/public/puzzle-strings.js` for a list of puzzle strings it should be able to solve along with their solutions.
+The return value from the `POST` to `/api/check` will be an object containing a `valid` property, which is `true` if the number may be placed at the provided coordinate and `false` if the number may not. If false, the returned object will also contain a `conflict` property which is an array containing the strings `"row"`, `"column"`, and/or `"region"` depending on which makes the placement invalid.
```js
-
+async (getUserInput) => {
+ const input =
+ '..9..5.1.85.4....2432......1...69.83.9.....6.62.71...9......1945....4.37.4.3..6..';
+ const coordinate = 'A1';
+ const value = '1';
+ const conflict = ['row', 'column'];
+ const data = await fetch(getUserInput('url') + '/api/check', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ puzzle: input, coordinate, value })
+ });
+ const parsed = await data.json();
+ assert.property(parsed, 'valid');
+ assert.isFalse(parsed.valid);
+ assert.property(parsed, 'conflict');
+ assert.include(parsed.conflict, 'row');
+ assert.include(parsed.conflict, 'column');
+};
```
-If the puzzle is not 81 numbers or periods long, append the message "Error: Expected puzzle to be 81 characters long." to the `error-msg` `div` so the text appears in red.
+If the puzzle submitted to `/api/check` contains values which are not numbers or periods, the returned value will be `{ error: 'Invalid characters in puzzle' }`
```js
-
+async (getUserInput) => {
+ const input =
+ 'AA9..5.1.85.4....2432......1...69.83.9.....6.62.71...9......1945....4.37.4.3..6..';
+ const coordinate = 'A1';
+ const value = '1';
+ const output = 'Invalid characters in puzzle';
+ const data = await fetch(getUserInput('url') + '/api/check', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ puzzle: input, coordinate, value })
+ });
+ const parsed = await data.json();
+ assert.property(parsed, 'error');
+ assert.equal(parsed.error, output);
+};
```
-I can clear the text area and sudoku grid by clicking the "Clear" button.
+If the puzzle submitted to `/api/check` is greater or less than 81 characters, the returned value will be `{ error: 'Expected puzzle to be 81 characters long' }`
```js
-
+async (getUserInput) => {
+ const input =
+ '9..5.1.85.4....2432......1...69.83.9.....6.62.71...9......1945....4.37.4.3..6..';
+ const coordinate = 'A1';
+ const value = '1';
+ const output = 'Expected puzzle to be 81 characters long';
+ const data = await fetch(getUserInput('url') + '/api/check', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ puzzle: input, coordinate, value })
+ });
+ const parsed = await data.json();
+ assert.property(parsed, 'error');
+ assert.equal(parsed.error, output);
+};
```
-All 6 unit tests are complete and passing. See `/tests/1_unit-tests.js` for the expected behavior you should write tests for.
+If the object submitted to `/api/check` is missing `puzzle`, `coordinate` or `value`, the returned value will be `{ error: Required field(s) missing }`
```js
-
+async (getUserInput) => {
+ const input =
+ '..9..5.1.85.4....2432......1...69.83.9.....6.62.71...9......1945....4.37.4.3..6..';
+ const output = 'Required field(s) missing';
+ const data = await fetch(getUserInput('url') + '/api/check', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ puzzle: input })
+ });
+ const parsed = await data.json();
+ assert.property(parsed, 'error');
+ assert.equal(parsed.error, output);
+};
```
-All 4 functional tests are complete and passing. See `/tests/2_functional-tests.js` for the functionality you should write tests for.
+If the coordinate submitted to `api/check` does not point to an existing grid cell, the returned value will be `{ error: 'Invalid coordinate'}`
```js
+async (getUserInput) => {
+ const input =
+ '..9..5.1.85.4....2432......1...69.83.9.....6.62.71...9......1945....4.37.4.3..6..';
+ const output = 'Invalid coordinate';
+ const coordinate = 'XZ18';
+ const value = '7';
+ const data = await fetch(getUserInput('url') + '/api/check', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ puzzle: input, coordinate, value })
+ });
+ const parsed = await data.json();
+ assert.property(parsed, 'error');
+ assert.equal(parsed.error, output);
+};
+```
+If the `value` submitted to `/api/check` is not a number between 1 and 9, the returned values will be `{ error: 'Invalid value' }`
+
+```js
+async (getUserInput) => {
+ const input =
+ '..9..5.1.85.4....2432......1...69.83.9.....6.62.71...9......1945....4.37.4.3..6..';
+ const output = 'Invalid value';
+ const coordinate = 'A1';
+ const value = 'X';
+ const data = await fetch(getUserInput('url') + '/api/check', {
+ method: 'POST',
+ headers: { 'Content-Type': 'application/json' },
+ body: JSON.stringify({ puzzle: input, coordinate, value })
+ });
+ const parsed = await data.json();
+ assert.property(parsed, 'error');
+ assert.equal(parsed.error, output);
+};
+```
+
+All 12 unit tests are complete and passing. See `/tests/1_unit-tests.js` for the expected behavior you should write tests for.
+
+```js
+async (getUserInput) => {
+ try {
+ const getTests = await $.get(getUserInput('url') + '/_api/get-tests');
+ assert.isArray(getTests);
+ const units = getTests.filter((el) => el.context.includes('UnitTests'));
+ assert.isAtLeast(units.length, 12, 'At least 12 tests passed');
+ units.forEach((test) => {
+ assert.equal(test.state, 'passed', 'Test in Passed State');
+ assert.isAtLeast(
+ test.assertions.length,
+ 1,
+ 'At least one assertion per test'
+ );
+ });
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
+```
+
+All 14 functional tests are complete and passing. See `/tests/2_functional-tests.js` for the functionality you should write tests for.
+
+```js
+async (getUserInput) => {
+ try {
+ const getTests = await $.get(getUserInput('url') + '/_api/get-tests');
+ assert.isArray(getTests);
+ const funcs = getTests.filter((el) =>
+ el.context.includes('Functional Tests')
+ );
+ assert.isAtLeast(funcs.length, 14, 'At least 14 tests passed');
+ funcs.forEach((test) => {
+ assert.equal(test.state, 'passed', 'Test in Passed State');
+ assert.isAtLeast(
+ test.assertions.length,
+ 1,
+ 'At least one assertion per test'
+ );
+ });
+ } catch (err) {
+ throw new Error(err.responseText || err.message);
+ }
+};
```
# --solutions--
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/build-your-own-functions.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/build-your-own-functions.md
index 0a78065c880..a1797e50eb3 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/build-your-own-functions.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/build-your-own-functions.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f060b6c005b0e76f05b
+title: Build your own Functions
challengeType: 11
videoId: nLDychdBwUg
dashedName: build-your-own-functions
@@ -30,47 +31,35 @@ jane()
## --answers--
-Zap
-
+<pre>Zap
ABC
-
jane
-
fred
-
-jane
+jane</pre>
---
-Zap
-
+<pre>Zap
ABC
-
-Zap
+Zap</pre>
---
-ABC
-
+<pre>ABC
Zap
-
-jane
+jane</pre>
---
-ABC
-
+<pre>ABC
Zap
-
-ABC
+ABC</pre>
---
+<pre>Zap
Zap
-
-Zap
-
-Zap
+Zap</pre>
## --video-solution--
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/comparing-and-sorting-tuples.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/comparing-and-sorting-tuples.md
index c0c9260d927..011f26213ed 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/comparing-and-sorting-tuples.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/comparing-and-sorting-tuples.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f0b0b6c005b0e76f06d
+title: Comparing and Sorting Tuples
challengeType: 11
videoId: dZXzBXUxxCs
dashedName: comparing-and-sorting-tuples
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/conditional-execution.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/conditional-execution.md
index c2cd9705ae3..2e973cd5f62 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/conditional-execution.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/conditional-execution.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f050b6c005b0e76f058
+title: Conditional Execution
challengeType: 11
videoId: gz_IfIsZQtc
dashedName: conditional-execution
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/data-visualization-mailing-lists.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/data-visualization-mailing-lists.md
index 2e0dbcd47a1..b4030f10ff1 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/data-visualization-mailing-lists.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/data-visualization-mailing-lists.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f6a0b6c005b0e76f097
+title: 'Data Visualization: Mailing Lists'
challengeType: 11
videoId: RYdW660KkaQ
dashedName: data-visualization-mailing-lists
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/data-visualization-page-rank.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/data-visualization-page-rank.md
index f45087d8ee5..fd0b04359ae 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/data-visualization-page-rank.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/data-visualization-page-rank.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f6a0b6c005b0e76f096
+title: 'Data Visualization: Page Rank'
challengeType: 11
videoId: 6-w_qIUwaxU
dashedName: data-visualization-page-rank
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/dictionaries-and-loops.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/dictionaries-and-loops.md
index f5fb0acbeb8..e11c4968a0c 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/dictionaries-and-loops.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/dictionaries-and-loops.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f0a0b6c005b0e76f069
+title: Dictionaries and Loops
challengeType: 11
videoId: EEmekKiKG70
dashedName: dictionaries-and-loops
@@ -26,27 +27,23 @@ for key in counts:
## --answers--
+<pre>annie 42
+jan 100</pre>
+
+---
+
+<pre>chuck 1
annie 42
-
-jan 100
+jan 100</pre>
---
-chuck 1
-
-annie 42
-
-jan 100
+<pre>chuck 1</pre>
---
-chuck 1
-
----
-
-[Error]
+<pre>[Error]</pre>
## --video-solution--
1
-
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/dictionaries-common-applications.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/dictionaries-common-applications.md
index 9eb5f41ec1b..f1f0855d933 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/dictionaries-common-applications.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/dictionaries-common-applications.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f090b6c005b0e76f068
+title: 'Dictionaries: Common Applications'
challengeType: 11
videoId: f17xPfIXct0
dashedName: dictionaries-common-applications
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/files-as-a-sequence.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/files-as-a-sequence.md
index ca62e9c4e8d..dff23f6561c 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/files-as-a-sequence.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/files-as-a-sequence.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f080b6c005b0e76f063
+title: Files as a Sequence
challengeType: 11
videoId: cIA0EokbaHE
dashedName: files-as-a-sequence
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/intermediate-expressions.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/intermediate-expressions.md
index 09d9c4e1e02..06951beb680 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/intermediate-expressions.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/intermediate-expressions.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f050b6c005b0e76f057
+title: Intermediate Expressions
challengeType: 11
videoId: dKgUaIa5ATg
dashedName: intermediate-expressions
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/intermediate-strings.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/intermediate-strings.md
index 732ae7465d8..216b951cc5b 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/intermediate-strings.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/intermediate-strings.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f070b6c005b0e76f061
+title: Intermediate Strings
challengeType: 11
videoId: KgT_fYLXnyk
dashedName: intermediate-strings
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/introduction-elements-of-python.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/introduction-elements-of-python.md
index 95345121f4c..04796a35f1c 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/introduction-elements-of-python.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/introduction-elements-of-python.md
@@ -1,5 +1,6 @@
---
id: 5e6a54c358d3af90110a60a3
+title: 'Introduction: Elements of Python'
challengeType: 11
videoId: aRY_xjL35v0
dashedName: introduction-elements-of-python
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/introduction-hardware-achitecture.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/introduction-hardware-achitecture.md
index 9840c00801d..02afa120f1c 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/introduction-hardware-achitecture.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/introduction-hardware-achitecture.md
@@ -1,5 +1,6 @@
---
id: 5e6a54af58d3af90110a60a1
+title: 'Introduction: Hardware Architecture'
challengeType: 11
videoId: H6qtjRTfSog
dashedName: introduction-hardware-architecture
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/introduction-python-as-a-language.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/introduction-python-as-a-language.md
index 7aefd6f4188..a9ad7a9e2b4 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/introduction-python-as-a-language.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/introduction-python-as-a-language.md
@@ -1,5 +1,6 @@
---
id: 5e6a54ba58d3af90110a60a2
+title: 'Introduction: Python as a Language'
challengeType: 11
videoId: 0QeGbZNS_bY
dashedName: introduction-python-as-a-language
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/introduction-why-program.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/introduction-why-program.md
index 11b298b839f..f1549dbb070 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/introduction-why-program.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/introduction-why-program.md
@@ -1,5 +1,6 @@
---
id: 5e6a54a558d3af90110a60a0
+title: 'Introduction: Why Program?'
challengeType: 11
videoId: 3muQV-Im3Z0
dashedName: introduction-why-program
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/iterations-definite-loops.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/iterations-definite-loops.md
index 25b4c7c426b..6438238f031 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/iterations-definite-loops.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/iterations-definite-loops.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f070b6c005b0e76f05d
+title: 'Iterations: Definite Loops'
challengeType: 11
videoId: hiRTRAqNlpE
dashedName: iterations-definite-loops
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/iterations-loop-idioms.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/iterations-loop-idioms.md
index 3ad34088f4e..d0f5059a2ff 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/iterations-loop-idioms.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/iterations-loop-idioms.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f070b6c005b0e76f05e
+title: 'Iterations: Loop Idioms'
challengeType: 11
videoId: AelGAcoMXbI
dashedName: iterations-loop-idioms
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/iterations-more-patterns.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/iterations-more-patterns.md
index 186a6192d77..829fe700112 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/iterations-more-patterns.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/iterations-more-patterns.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f070b6c005b0e76f05f
+title: 'Iterations: More Patterns'
challengeType: 11
videoId: 9Wtqo6vha1M
dashedName: iterations-more-patterns
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/loops-and-iterations.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/loops-and-iterations.md
index bdb10acb412..5bf4c6df17d 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/loops-and-iterations.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/loops-and-iterations.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f060b6c005b0e76f05c
+title: Loops and Iterations
challengeType: 11
videoId: dLA-szNRnUY
dashedName: loops-and-iterations
@@ -22,35 +23,27 @@ while True:
## --answers--
-0
-
+<pre>0
1
-
-2
+2</pre>
---
-0
-
+<pre>0
1
-
2
-
-3
+3</pre>
---
-1
-
-2
+<pre>1
+2</pre>
---
-1
-
+<pre>1
2
-
-3
+3</pre>
## --video-solution--
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/make-a-relational-database.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/make-a-relational-database.md
index 140dba50d4d..36e76238d4c 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/make-a-relational-database.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/make-a-relational-database.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f170b6c005b0e76f08b
+title: Make a Relational Database
challengeType: 11
videoId: MQ5z4bdF92U
dashedName: make-a-relational-database
@@ -9,35 +10,35 @@ dashedName: make-a-relational-database
## --text--
-What SQL command would you use to retrieve all users that have the email address [quincy@freecodecamp.org](mailto:quincy@freecodecamp.org)?`mailto:quincy@freecodecamp.org`e to retrieve all users that have the email address [quincy@freecodecamp.org](mailto:quincy@freecodecamp.org)?`mailto:quincy@freecodecamp.org`have the email address [quincy@freecodecamp.org](mailto:quincy@freecodecamp.org)?`mailto:quincy@freecodecamp.org`@freecodecamp.org?
+What SQL command would you use to retrieve all users that have the email address `quincy@freecodecamp.org`?
## --answers--
-```
+```sql
SELECT Users WHERE email="quincy@freecodecamp.org"
```
---
-```
+```sql
SELECT Users WHERE email IS "quincy@freecodecamp.org"
```
---
-```
+```sql
SELECT ALL Users WHERE email="quincy@freecodecamp.org"
```
---
-```
+```sql
SELECT * FROM Users WHERE email IS "quincy@freecodecamp.org"
```
---
-```
+```sql
SELECT * FROM Users WHERE email="quincy@freecodecamp.org"
```
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/more-conditional-structures.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/more-conditional-structures.md
index 56285a41e06..135632d2b7d 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/more-conditional-structures.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/more-conditional-structures.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f060b6c005b0e76f059
+title: More Conditional Structures
challengeType: 11
videoId: HdL82tAZR20
dashedName: more-conditional-structures
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-protocol.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-protocol.md
index 20e7babdaa7..7196aa02da7 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-protocol.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-protocol.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f0c0b6c005b0e76f072
+title: Networking Protocol
challengeType: 11
videoId: c6vZGescaSc
dashedName: networking-protocol
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-text-processing.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-text-processing.md
index 53cb1c537c9..bf0c895949d 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-text-processing.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-text-processing.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f0c0b6c005b0e76f074
+title: 'Networking: Text Processing'
challengeType: 11
videoId: Pv_pJgVu8WI
dashedName: networking-text-processing
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-using-urllib-in-python.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-using-urllib-in-python.md
index 1d0062a68bb..a8f26e7a47e 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-using-urllib-in-python.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-using-urllib-in-python.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f0d0b6c005b0e76f075
+title: 'Networking: Using urllib in Python'
challengeType: 11
videoId: 7lFM1T_CxBs
dashedName: networking-using-urllib-in-python
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-web-scraping-with-python.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-web-scraping-with-python.md
index 750434f49de..ed31a3848ab 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-web-scraping-with-python.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-web-scraping-with-python.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f0d0b6c005b0e76f076
+title: 'Networking: Web Scraping with Python'
challengeType: 11
videoId: Uyioq2q4cEg
dashedName: networking-web-scraping-with-python
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-with-python.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-with-python.md
index 60d0f7414ad..b90462a16d5 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-with-python.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-with-python.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f0c0b6c005b0e76f071
+title: Networking with Python
challengeType: 11
videoId: _kJvneKVdNM
dashedName: networking-with-python
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-write-a-web-browser.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-write-a-web-browser.md
index e410cc5e2be..41fc6959e0e 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-write-a-web-browser.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/networking-write-a-web-browser.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f0c0b6c005b0e76f073
+title: 'Networking: Write a Web Browser'
challengeType: 11
videoId: zjyT9DaAjx4
dashedName: networking-write-a-web-browser
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/object-lifecycle.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/object-lifecycle.md
index b44721e3014..c85eabd257e 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/object-lifecycle.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/object-lifecycle.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f170b6c005b0e76f087
+title: Object Lifecycle
challengeType: 11
videoId: p1r3h_AMMIM
dashedName: object-lifecycle
@@ -32,39 +33,33 @@ q.party()
## --answers--
+<pre>
Quincy constructed
-
Miya constructed
-
Quincy party count 1
-
Miya party count 2
-
Quincy party count 3
+</pre>
---
+<pre>
Quincy constructed
-
Miya constructed
-
Quincy party count 1
-
Miya party count 1
-
Quincy party count 2
+</pre>
---
+<pre>
Quincy constructed
-
Quincy party count 1
-
Quincy party count 2
-
Miya constructed
-
Miya party count 1
+</pre>
## --video-solution--
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/objects-a-sample-class.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/objects-a-sample-class.md
index e392a534ce2..41d127dc0f4 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/objects-a-sample-class.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/objects-a-sample-class.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f160b6c005b0e76f086
+title: 'Objects: A Sample Class'
challengeType: 11
videoId: FiABKEuaSJ8
dashedName: objects-a-sample-class
@@ -25,27 +26,31 @@ an.party()
## --answers--
+<pre>
So far 1
-
So far 2
+</pre>
---
+<pre>
0
-
0
+</pre>
---
+<pre>
2
-
2
+</pre>
---
+<pre>
2
-
4
+</pre>
## --video-solution--
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/objects-inheritance.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/objects-inheritance.md
index 715e323476c..1de3c040129 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/objects-inheritance.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/objects-inheritance.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f170b6c005b0e76f088
+title: 'Objects: Inheritance'
challengeType: 11
videoId: FBL3alYrxRM
dashedName: objects-inheritance
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/python-dictionaries.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/python-dictionaries.md
index f7a5cd54f66..0c86c3bd097 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/python-dictionaries.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/python-dictionaries.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f090b6c005b0e76f067
+title: Python Dictionaries
challengeType: 11
videoId: dnzvfimrRMg
dashedName: python-dictionaries
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/python-functions.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/python-functions.md
index 0d51194fdad..2f4bb2bbd04 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/python-functions.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/python-functions.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f060b6c005b0e76f05a
+title: Python Functions
challengeType: 11
videoId: 3JGF-n3tDPU
dashedName: python-functions
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/python-lists.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/python-lists.md
index 3c9b5b49626..be78cd10c77 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/python-lists.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/python-lists.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f080b6c005b0e76f064
+title: Python Lists
challengeType: 11
videoId: Y0cvfDpYC_c
dashedName: python-lists
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/python-objects.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/python-objects.md
index d8593385a35..6082df58eba 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/python-objects.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/python-objects.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f160b6c005b0e76f085
+title: Python Objects
challengeType: 11
videoId: uJxGeTYy0us
dashedName: python-objects
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/reading-files.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/reading-files.md
index 677e00db267..9be11dc4f3c 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/reading-files.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/reading-files.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f080b6c005b0e76f062
+title: Reading Files
challengeType: 11
videoId: Fo1tW09KIwo
dashedName: reading-files
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/regular-expressions-matching-and-extracting-data.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/regular-expressions-matching-and-extracting-data.md
index 908ece91c0a..87a98241446 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/regular-expressions-matching-and-extracting-data.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/regular-expressions-matching-and-extracting-data.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f0b0b6c005b0e76f06f
+title: 'Regular Expressions: Matching and Extracting Data'
challengeType: 11
videoId: LaCZnTbQGkE
dashedName: regular-expressions-matching-and-extracting-data
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/regular-expressions-practical-applications.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/regular-expressions-practical-applications.md
index 1e4fd1bbefd..cf1d7b3825a 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/regular-expressions-practical-applications.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/regular-expressions-practical-applications.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f0b0b6c005b0e76f070
+title: 'Regular Expressions: Practical Applications'
challengeType: 11
videoId: xCjFU9G6x48
dashedName: regular-expressions-practical-applications
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/regular-expressions.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/regular-expressions.md
index 42c053f1cb1..0bff237a4a9 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/regular-expressions.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/regular-expressions.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f0b0b6c005b0e76f06e
+title: Regular Expressions
challengeType: 11
videoId: Yud_COr6pZo
dashedName: regular-expressions
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-database-design.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-database-design.md
index 5e768416158..21f07615fc3 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-database-design.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-database-design.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f180b6c005b0e76f08c
+title: Relational Database Design
challengeType: 11
videoId: AqdfbrpkbHk
dashedName: relational-database-design
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-databases-and-sqlite.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-databases-and-sqlite.md
index d311587faca..26732397d08 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-databases-and-sqlite.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-databases-and-sqlite.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f170b6c005b0e76f08a
+title: Relational Databases and SQLite
challengeType: 11
videoId: QlNod5-kFpA
dashedName: relational-databases-and-sqlite
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-databases-join-operation.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-databases-join-operation.md
index 19fba4f07ca..451e8b51079 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-databases-join-operation.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-databases-join-operation.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f180b6c005b0e76f08f
+title: 'Relational Databases: Join Operation'
challengeType: 11
videoId: jvDw3D9GKac
dashedName: relational-databases-join-operation
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-databases-many-to-many-relationships.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-databases-many-to-many-relationships.md
index 1019b6d3c7e..252eca3a880 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-databases-many-to-many-relationships.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-databases-many-to-many-relationships.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f190b6c005b0e76f090
+title: 'Relational Databases: Many-to-many Relationships'
challengeType: 11
videoId: z-SBYcvEQOc
dashedName: relational-databases-many-to-many-relationships
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-databases-relationship-building.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-databases-relationship-building.md
index 713d7fa6ef1..87e0754a625 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-databases-relationship-building.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/relational-databases-relationship-building.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f180b6c005b0e76f08e
+title: 'Relational Databases: Relationship Building'
challengeType: 11
videoId: CSbqczsHVnc
dashedName: relational-databases-relationship-building
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/representing-relationships-in-a-relational-database.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/representing-relationships-in-a-relational-database.md
index 45f71d34527..03b88fb6784 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/representing-relationships-in-a-relational-database.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/representing-relationships-in-a-relational-database.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f180b6c005b0e76f08d
+title: Representing Relationships in a Relational Database
challengeType: 11
videoId: '-orenCNdC2Q'
dashedName: representing-relationships-in-a-relational-database
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/strings-and-lists.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/strings-and-lists.md
index 28847672d79..51ab3b952b1 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/strings-and-lists.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/strings-and-lists.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f090b6c005b0e76f066
+title: Strings and Lists
challengeType: 11
videoId: lxcFa7ldCi0
dashedName: strings-and-lists
@@ -38,7 +39,7 @@ lar
---
-[lar@freecodecamp.org](mailto:lar@freecodecamp.org)`mailto:lar@freecodecamp.org`
+`lar@freecodecamp.org`
## --video-solution--
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/strings-in-python.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/strings-in-python.md
index 9845a9dfec7..2136025df66 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/strings-in-python.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/strings-in-python.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f070b6c005b0e76f060
+title: Strings in Python
challengeType: 11
videoId: LYZj207fKpQ
dashedName: strings-in-python
@@ -18,43 +19,39 @@ for n in "banana":
## --answers--
+<pre>
n
-
n
+</pre>
---
+<pre>
0
-
1
+</pre>
---
+<pre>
0
-
1
-
2
-
3
-
4
-
5
+</pre>
---
+<pre>
b
-
a
-
n
-
a
-
n
-
a
+</pre>
## --video-solution--
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/the-tuples-collection.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/the-tuples-collection.md
index df0210740d7..e11e5398ab7 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/the-tuples-collection.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/the-tuples-collection.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f0a0b6c005b0e76f06c
+title: The Tuples Collection
challengeType: 11
videoId: 3Lxpladfh2k
dashedName: the-tuples-collection
@@ -22,35 +23,35 @@ for (k,i) in d.items():
## --answers--
+<pre>
k i
-
k i
-
k i
+</pre>
---
+<pre>
quincy 0
-
beau 1
-
kris 2
+</pre>
---
+<pre>
quincy 1
-
beau 5
-
kris 9
+</pre>
---
+<pre>
1 quincy
-
5 beau
-
9 kris
+</pre>
## --video-solution--
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/using-web-services.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/using-web-services.md
index 8b25d44676f..5c2f7814c11 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/using-web-services.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/using-web-services.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f0e0b6c005b0e76f07a
+title: Using Web Services
challengeType: 11
videoId: oNl1OVDPGKE
dashedName: using-web-services
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/variables-expressions-and-statements.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/variables-expressions-and-statements.md
index b7b4ec7b9d1..50b6caa84bb 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/variables-expressions-and-statements.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/variables-expressions-and-statements.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f050b6c005b0e76f056
+title: 'Variables, Expressions, and Statements'
challengeType: 11
videoId: nELR-uyyrok
dashedName: variables-expressions-and-statements
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/visualizing-data-with-python.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/visualizing-data-with-python.md
index 21bcef72281..c258b56bce2 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/visualizing-data-with-python.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/visualizing-data-with-python.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f690b6c005b0e76f095
+title: Visualizing Data with Python
challengeType: 11
videoId: e3lydkH0prw
dashedName: visualizing-data-with-python
@@ -9,7 +10,7 @@ dashedName: visualizing-data-with-python
## --text--
-Most data needs to be **\_\_** before using it.
+Most data needs to be \_\_\_\_\_\_ before using it.
## --answers--
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-api-rate-limiting-and-security.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-api-rate-limiting-and-security.md
index b4ba5da9ad0..992602d0c69 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-api-rate-limiting-and-security.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-api-rate-limiting-and-security.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f150b6c005b0e76f080
+title: 'Web Services: API Rate Limiting and Security'
challengeType: 11
videoId: pI-g0lI8ngs
dashedName: web-services-api-rate-limiting-and-security
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-apis.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-apis.md
index b0bf2713d67..285380bd7b6 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-apis.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-apis.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f150b6c005b0e76f07f
+title: 'Web Services: APIs'
challengeType: 11
videoId: oUNn1psfBJg
dashedName: web-services-apis
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-json.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-json.md
index b57aef00566..336373054ca 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-json.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-json.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f140b6c005b0e76f07d
+title: 'Web Services: JSON'
challengeType: 11
videoId: ZJE-U56BppM
dashedName: web-services-json
@@ -15,13 +16,13 @@ What will the following code print?:
import json
data = '''
[
- { 'id' : '001',
- 'x' : '2',
- 'name' : 'Quincy'
+ { "id" : "001",
+ "x" : "2",
+ "name" : "Quincy"
} ,
- { 'id' : '009',
- 'x' : '7',
- 'name' : 'Mrugesh'
+ { "id" : "009",
+ "x" : "7",
+ "name" : "Mrugesh"
}
]
'''
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-service-oriented-approach.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-service-oriented-approach.md
index 7520eb2168a..614f5336a3e 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-service-oriented-approach.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-service-oriented-approach.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f140b6c005b0e76f07e
+title: 'Web Services: Service Oriented Approach'
challengeType: 11
videoId: muerlsCHExI
dashedName: web-services-service-oriented-approach
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-xml-schema.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-xml-schema.md
index 1f0b7841c02..8ccc87accbe 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-xml-schema.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-xml-schema.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f0e0b6c005b0e76f07c
+title: 'Web Services: XML Schema'
challengeType: 11
videoId: yWU9kTxW-nc
dashedName: web-services-xml-schema
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-xml.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-xml.md
index 6af5e3bc303..efda1068673 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-xml.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/web-services-xml.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f0e0b6c005b0e76f07b
+title: 'Web Services: XML'
challengeType: 11
videoId: _pZ0srbg7So
dashedName: web-services-xml
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/working-with-lists.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/working-with-lists.md
index 9299bd78e2e..dbeda928eee 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/working-with-lists.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/python-for-everybody/working-with-lists.md
@@ -1,5 +1,6 @@
---
id: 5e7b9f090b6c005b0e76f065
+title: Working with Lists
challengeType: 11
videoId: lCnHfTHkhbE
dashedName: working-with-lists
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/arithmetic-formatter.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/arithmetic-formatter.md
index 08d6e8e3877..c6f3482dba1 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/arithmetic-formatter.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/arithmetic-formatter.md
@@ -1,5 +1,6 @@
---
id: 5e44412c903586ffb414c94c
+title: Arithmetic Formatter
challengeType: 10
dashedName: arithmetic-formatter
---
@@ -8,14 +9,13 @@ dashedName: arithmetic-formatter
Create a function that receives a list of strings that are arithmetic problems and returns the problems arranged vertically and side-by-side.
-You can access [the full project description and starter code on repl.it](https://repl.it/@freeCodeCamp/fcc-arithmetic-arranger).
+You can access [the full project description and starter code on Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-arithmetic-formatter).
After going to that link, fork the project. Once you complete the project based on the instructions in 'README.md', submit your project link below.
We are still developing the interactive instructional part of the Python curriculum. For now, here are some videos on the freeCodeCamp.org YouTube channel that will teach you everything you need to know to complete this project:
-<ul>
- <li>
+<ul> <li>
<a href='https://www.freecodecamp.org/news/python-for-everybody/'>Python for Everybody Video Course</a> (14 hours)
</li>
<li>
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/budget-app.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/budget-app.md
index 6625903a39d..f67225f5341 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/budget-app.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/budget-app.md
@@ -1,5 +1,6 @@
---
id: 5e44413e903586ffb414c94e
+title: Budget App
challengeType: 10
dashedName: budget-app
---
@@ -8,7 +9,7 @@ dashedName: budget-app
Create a "Category" class that can be used to create different budget categories.
-You can access [the full project description and starter code on repl.it](https://repl.it/@freeCodeCamp/fcc-budget-app).
+You can access [the full project description and starter code on Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-budget-app).
After going to that link, fork the project. Once you complete the project based on the instructions in 'README.md', submit your project link below.
@@ -21,7 +22,7 @@ We are still developing the interactive instructional part of the Python curricu
<li>
<a href='https://www.freecodecamp.org/news/learn-python-basics-in-depth-video-course/'>Learn Python Video Course</a> (2 hours)
</li>
-<ul>
+</ul>
# --hints--
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/polygon-area-calculator.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/polygon-area-calculator.md
index 7d3bedb16ca..21c24f96526 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/polygon-area-calculator.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/polygon-area-calculator.md
@@ -1,5 +1,6 @@
---
id: 5e444147903586ffb414c94f
+title: Polygon Area Calculator
challengeType: 10
dashedName: polygon-area-calculator
---
@@ -8,7 +9,7 @@ dashedName: polygon-area-calculator
In this project you will use object oriented programming to create a Rectangle class and a Square class. The Square class should be a subclass of Rectangle and inherit methods and attributes.
-You can access [the full project description and starter code on repl.it](https://repl.it/@freeCodeCamp/fcc-shape-calculator).
+You can access [the full project description and starter code on Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-polygon-area-calculator).
After going to that link, fork the project. Once you complete the project based on the instructions in 'README.md', submit your project link below.
@@ -21,7 +22,7 @@ We are still developing the interactive instructional part of the Python curricu
<li>
<a href='https://www.freecodecamp.org/news/learn-python-basics-in-depth-video-course/'>Learn Python Video Course</a> (2 hours)
</li>
-<ul>
+</ul>
# --hints--
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/probability-calculator.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/probability-calculator.md
index 530898b4e83..58cf7485844 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/probability-calculator.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/probability-calculator.md
@@ -1,5 +1,6 @@
---
id: 5e44414f903586ffb414c950
+title: Probability Calculator
challengeType: 10
dashedName: probability-calculator
---
@@ -8,7 +9,7 @@ dashedName: probability-calculator
Write a program to determine the approximate probability of drawing certain balls randomly from a hat.
-You can access [the full project description and starter code on repl.it](https://repl.it/@freeCodeCamp/fcc-probability-calculator). After going to that link, fork the project. Once you complete the project based on the instructions in 'README.md', submit your project link below.
+You can access [the full project description and starter code on Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-probability-calculator). After going to that link, fork the project. Once you complete the project based on the instructions in 'README.md', submit your project link below.
We are still developing the interactive instructional part of the Python curriculum. For now, here are some videos on the freeCodeCamp.org YouTube channel that will teach you everything you need to know to complete this project:
@@ -19,7 +20,7 @@ We are still developing the interactive instructional part of the Python curricu
<li>
<a href='https://www.freecodecamp.org/news/learn-python-basics-in-depth-video-course/'>Learn Python Video Course</a> (2 hours)
</li>
-<ul>
+</ul>
# --hints--
diff --git a/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/time-calculator.md b/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/time-calculator.md
index bfbe5de6a25..2037f57ad7d 100644
--- a/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/time-calculator.md
+++ b/curriculum/challenges/chinese/07-scientific-computing-with-python/scientific-computing-with-python-projects/time-calculator.md
@@ -1,5 +1,6 @@
---
id: 5e444136903586ffb414c94d
+title: Time Calculator
challengeType: 10
dashedName: time-calculator
---
@@ -8,7 +9,7 @@ dashedName: time-calculator
Write a function named "add_time" that can add a duration to a start time and return the result.
-You can access [the full project description and starter code on repl.it](https://repl.it/@freeCodeCamp/fcc-time-calculator). After going to that link, fork the project. Once you complete the project based on the instructions in 'README.md', submit your project link below.
+You can access [the full project description and starter code on Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-time-calculator). After going to that link, fork the project. Once you complete the project based on the instructions in 'README.md', submit your project link below.
We are still developing the interactive instructional part of the Python curriculum. For now, here are some videos on the freeCodeCamp.org YouTube channel that will teach you everything you need to know to complete this project:
@@ -19,7 +20,7 @@ We are still developing the interactive instructional part of the Python curricu
<li>
<a href='https://www.freecodecamp.org/news/learn-python-basics-in-depth-video-course/'>Learn Python Video Course</a> (2 hours)
</li>
-<ul>
+</ul>
# --hints--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-analysis-example-a.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-analysis-example-a.md
index 813e73bbb4e..9a24560aa9a 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-analysis-example-a.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-analysis-example-a.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c14d
+title: Data Analysis Example A
challengeType: 11
videoId: nVAaxZ34khk
dashedName: data-analysis-example-a
@@ -7,9 +8,12 @@ dashedName: data-analysis-example-a
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/freecodecamp-pandas-real-life-example-24fa5bf8)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/FreeCodeCamp-Pandas-Real-Life-Example)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-analysis-example-b.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-analysis-example-b.md
index aeb6821802c..67f603c1296 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-analysis-example-b.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-analysis-example-b.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c14e
+title: Data Analysis Example B
challengeType: 11
videoId: 0kJz0q0pvgQ
dashedName: data-analysis-example-b
@@ -7,9 +8,12 @@ dashedName: data-analysis-example-b
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/freecodecamp-pandas-real-life-example-24fa5bf8)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/FreeCodeCamp-Pandas-Real-Life-Example)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
@@ -19,7 +23,7 @@ What does the `loc` method allow you to do?
## --answers--
-Retrieve a subset of rows and columns by supplying interger-location arguments.
+Retrieve a subset of rows and columns by supplying integer-location arguments.
---
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-cleaning-and-visualizations.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-cleaning-and-visualizations.md
index c1b7069494f..13cd90fb086 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-cleaning-and-visualizations.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-cleaning-and-visualizations.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c160
+title: Data Cleaning and Visualizations
challengeType: 11
videoId: mHjxzFS5_Z0
dashedName: data-cleaning-and-visualizations
@@ -7,9 +8,12 @@ dashedName: data-cleaning-and-visualizations
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/data-cleaning-rmotr-freecodecamp-fd76fa59)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/data-cleaning-rmotr-freecodecamp)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-cleaning-duplicates.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-cleaning-duplicates.md
index 52f39715a81..cf159cead12 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-cleaning-duplicates.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-cleaning-duplicates.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c15f
+title: Data Cleaning Duplicates
challengeType: 11
videoId: kj7QqjXhH6A
dashedName: data-cleaning-duplicates
@@ -7,9 +8,12 @@ dashedName: data-cleaning-duplicates
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/data-cleaning-rmotr-freecodecamp-fd76fa59)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/data-cleaning-rmotr-freecodecamp)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-cleaning-introduction.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-cleaning-introduction.md
index c1e2c7b704e..575be65eef4 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-cleaning-introduction.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-cleaning-introduction.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c15d
+title: Data Cleaning Introduction
challengeType: 11
videoId: ovYNhnltVxY
dashedName: data-cleaning-introduction
@@ -7,9 +8,12 @@ dashedName: data-cleaning-introduction
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/data-cleaning-rmotr-freecodecamp-fd76fa59)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/data-cleaning-rmotr-freecodecamp)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
@@ -32,25 +36,21 @@ print(s.notnull().sum())
---
-```
-0 True
+<pre>0 True
1 True
2 False
3 True
4 False
-dtype: bool
-```
+dtype: bool</pre>
---
-```
-0 False
+<pre>0 False
1 False
2 True
3 False
4 True
-dtype: bool
-```
+dtype: bool</pre>
## --video-solution--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-cleaning-with-dataframes.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-cleaning-with-dataframes.md
index dc83c45d6fc..887af41702d 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-cleaning-with-dataframes.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/data-cleaning-with-dataframes.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c15e
+title: Data Cleaning with DataFrames
challengeType: 11
videoId: sTMN_pdI6S0
dashedName: data-cleaning-with-dataframes
@@ -7,9 +8,12 @@ dashedName: data-cleaning-with-dataframes
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/data-cleaning-rmotr-freecodecamp-fd76fa59)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/data-cleaning-rmotr-freecodecamp)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
@@ -29,36 +33,36 @@ print(s)
## --answers--
-```
+<pre>
0 1.0
1 1.0
2 2.0
3 3.0
4 3.0
dtype: float64
-```
+</pre>
---
-```
+<pre>
0 NaN
1 1.0
2 2.0
3 2.0
4 3.0
dtype: float64
-```
+</pre>
---
-```
+<pre>
0 NaN
1 1.0
2 2.0
3 NaN
4 3.0
dtype: float64
-```
+</pre>
## --video-solution--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/how-to-use-jupyter-notebooks-intro.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/how-to-use-jupyter-notebooks-intro.md
index 3a494dc08a3..4ac7363d87d 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/how-to-use-jupyter-notebooks-intro.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/how-to-use-jupyter-notebooks-intro.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c14f
+title: How to use Jupyter Notebooks Intro
challengeType: 11
videoId: h8caJq2Bb9w
dashedName: how-to-use-jupyter-notebooks-intro
@@ -7,11 +8,13 @@ dashedName: how-to-use-jupyter-notebooks-intro
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/interactive-jupyterlab-tutorial-ac5fa63f)
-
-\- [Twitter Cheat Sheet](https://twitter.com/rmotr_com/status/1122176794696847361)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/ds-content-interactive-jupyterlab-tutorial)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
+- [Twitter Cheat Sheet](https://twitter.com/rmotr_com/status/1122176794696847361)
# --question--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/introduction-to-data-analysis.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/introduction-to-data-analysis.md
index 8ddaeb77ac0..a092ad93a88 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/introduction-to-data-analysis.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/introduction-to-data-analysis.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c14c
+title: Introduction to Data Analysis
challengeType: 11
videoId: VJrP2FUzKP0
dashedName: introduction-to-data-analysis
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/jupyter-notebooks-cells.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/jupyter-notebooks-cells.md
index 0a39943d4ff..137b9f53bd0 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/jupyter-notebooks-cells.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/jupyter-notebooks-cells.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c150
+title: Jupyter Notebooks Cells
challengeType: 11
videoId: 5PPegAs9aLA
dashedName: jupyter-notebooks-cells
@@ -7,31 +8,32 @@ dashedName: jupyter-notebooks-cells
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/interactive-jupyterlab-tutorial-ac5fa63f)
-
-\- [Twitter Cheat Sheet](https://twitter.com/rmotr_com/status/1122176794696847361)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/ds-content-interactive-jupyterlab-tutorial)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
+- [Twitter Cheat Sheet](https://twitter.com/rmotr_com/status/1122176794696847361)
# --question--
## --text--
-Which cells are responsible for rich display?
+What are the three main types of Jupyter Notebook Cell?
## --answers--
-Code Cells
+Code, Markdown, and Python
---
-Markdown Cells
+Code, Markdown, and Raw
---
-Raw Cells
+Markdown, Python, and Raw
## --video-solution--
-1
-
+2
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/jupyter-notebooks-importing-and-exporting-data.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/jupyter-notebooks-importing-and-exporting-data.md
index 5110e828c64..104edcae192 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/jupyter-notebooks-importing-and-exporting-data.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/jupyter-notebooks-importing-and-exporting-data.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c151
+title: Jupyter Notebooks Importing and Exporting Data
challengeType: 11
videoId: k1msxD3JIxE
dashedName: jupyter-notebooks-importing-and-exporting-data
@@ -7,11 +8,13 @@ dashedName: jupyter-notebooks-importing-and-exporting-data
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/interactive-jupyterlab-tutorial-ac5fa63f)
-
-\- [Twitter Cheat Sheet](https://twitter.com/rmotr_com/status/1122176794696847361)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/ds-content-interactive-jupyterlab-tutorial)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
+- [Twitter Cheat Sheet](https://twitter.com/rmotr_com/status/1122176794696847361)
# --question--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-algebra-and-size.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-algebra-and-size.md
index 0a21f04ca76..454f7c56465 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-algebra-and-size.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-algebra-and-size.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c157
+title: Numpy Algebra and Size
challengeType: 11
videoId: XAT97YLOKD8
dashedName: numpy-algebra-and-size
@@ -7,9 +8,12 @@ dashedName: numpy-algebra-and-size
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/freecodecamp-intro-to-numpy-6c285b74)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/freecodecamp-intro-to-numpy)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-arrays.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-arrays.md
index 27220bf8032..b5504e836e4 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-arrays.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-arrays.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c154
+title: Numpy Arrays
challengeType: 11
videoId: VDYVFHBL1AM
dashedName: numpy-arrays
@@ -7,9 +8,12 @@ dashedName: numpy-arrays
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/freecodecamp-intro-to-numpy-6c285b74)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/freecodecamp-intro-to-numpy)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
@@ -29,7 +33,7 @@ print(A[:, :2])
## --answers--
-```python
+```py
[['a' 'b']]
```
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-boolean-arrays.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-boolean-arrays.md
index 8cba8cac596..f271dff4b0f 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-boolean-arrays.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-boolean-arrays.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c156
+title: Numpy Boolean Arrays
challengeType: 11
videoId: N1ttsMmcVMM
dashedName: numpy-boolean-arrays
@@ -7,9 +8,12 @@ dashedName: numpy-boolean-arrays
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/freecodecamp-intro-to-numpy-6c285b74)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/freecodecamp-intro-to-numpy)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-introduction-a.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-introduction-a.md
index 0068522361c..fc38697f40e 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-introduction-a.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-introduction-a.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c152
+title: Numpy Introduction A
challengeType: 11
videoId: P-JjV6GBCmk
dashedName: numpy-introduction-a
@@ -7,9 +8,12 @@ dashedName: numpy-introduction-a
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/freecodecamp-intro-to-numpy-6c285b74)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/freecodecamp-intro-to-numpy)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
@@ -23,7 +27,7 @@ Often you won't work directly with Numpy.
---
-It's is extremely slow.
+It is extremely slow.
---
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-introduction-b.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-introduction-b.md
index b4be575ec95..a863e420741 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-introduction-b.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-introduction-b.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c153
+title: Numpy Introduction B
challengeType: 11
videoId: YIqgrNLAZkA
dashedName: numpy-introduction-b
@@ -7,9 +8,12 @@ dashedName: numpy-introduction-b
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/freecodecamp-intro-to-numpy-6c285b74)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/freecodecamp-intro-to-numpy)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-operations.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-operations.md
index 36d4193e26d..0072c840194 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-operations.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/numpy-operations.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c155
+title: Numpy Operations
challengeType: 11
videoId: eqSVcJbaPdk
dashedName: numpy-operations
@@ -7,9 +8,12 @@ dashedName: numpy-operations
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/freecodecamp-intro-to-numpy-6c285b74)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/freecodecamp-intro-to-numpy)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-conditional-selection-and-modifying-dataframes.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-conditional-selection-and-modifying-dataframes.md
index 38ba90c6e4f..3d17a73a558 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-conditional-selection-and-modifying-dataframes.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-conditional-selection-and-modifying-dataframes.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c15b
+title: Pandas Conditional Selection and Modifying DataFrames
challengeType: 11
videoId: BFlH0fN5xRQ
dashedName: pandas-conditional-selection-and-modifying-dataframes
@@ -7,9 +8,12 @@ dashedName: pandas-conditional-selection-and-modifying-dataframes
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/freecodecamp-intro-to-pandas-902ae59b)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/freecodecamp-intro-to-pandas)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
@@ -35,33 +39,33 @@ print(certificates_earned)
## --answers--
-```
+<pre>
Tom 13
Kris 11
Ahmad 9
Beau 7
Name: Longest streak, dtype: int64
-```
+</pre>
---
-```
+<pre>
Certificates Time (in months) Longest streak
Tom 8 16 13
Kris 2 5 11
Ahmad 5 9 9
Beau 6 12 7
-```
+</pre>
---
-```
+<pre>
Certificates Longest streak
Tom 8 13
Kris 2 11
Ahmad 5 9
Beau 6 7
-```
+</pre>
## --video-solution--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-creating-columns.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-creating-columns.md
index f9c378f0a4c..7d20f76c4ad 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-creating-columns.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-creating-columns.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c15c
+title: Pandas Creating Columns
challengeType: 11
videoId: _sSo2XZoB3E
dashedName: pandas-creating-columns
@@ -7,9 +8,12 @@ dashedName: pandas-creating-columns
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/freecodecamp-intro-to-pandas-902ae59b)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/freecodecamp-intro-to-pandas)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
@@ -17,13 +21,11 @@ More resources:
What code would add a "Certificates per month" column to the `certificates_earned` DataFrame like the one below?
-```
- Certificates Time (in months) Certificates per month
+<pre> Certificates Time (in months) Certificates per month
Tom 8 16 0.50
Kris 2 5 0.40
Ahmad 5 9 0.56
-Beau 6 12 0.50
-```
+Beau 6 12 0.50</pre>
## --answers--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-dataframes.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-dataframes.md
index dbdea2e3102..fd5eedb1768 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-dataframes.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-dataframes.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c15a
+title: Pandas DataFrames
challengeType: 11
videoId: 7SgFBYXaiH0
dashedName: pandas-dataframes
@@ -7,9 +8,12 @@ dashedName: pandas-dataframes
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/freecodecamp-intro-to-pandas-902ae59b)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/freecodecamp-intro-to-pandas)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
@@ -32,29 +36,29 @@ print(certificates_earned.iloc[2])
## --answers--
-```
+<pre>
Tom 16
Kris 5
Ahmad 9
Beau 12
Name: Time (in months), dtype: int64
-```
+</pre>
---
-```
+<pre>
Certificates 6
Time (in months) 12
Name: Beau, dtype: int64
-```
+</pre>
---
-```
+<pre>
Certificates 5
Time (in months) 9
Name: Ahmad, dtype: int64
-```
+</pre>
## --video-solution--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-indexing-and-conditional-selection.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-indexing-and-conditional-selection.md
index cb8df467271..1107f217f4e 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-indexing-and-conditional-selection.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-indexing-and-conditional-selection.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c159
+title: Pandas Indexing and Conditional Selection
challengeType: 11
videoId: '-ZOrgV_aA9A'
dashedName: pandas-indexing-and-conditional-selection
@@ -7,9 +8,12 @@ dashedName: pandas-indexing-and-conditional-selection
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/freecodecamp-intro-to-pandas-902ae59b)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/freecodecamp-intro-to-pandas)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
@@ -30,30 +34,30 @@ print(certificates_earned[certificates_earned > 5])
## --answers--
-```
+<pre>
Tom True
Kris False
Ahmad False
Beau True
dtype: int64
-```
+</pre>
---
-```
+<pre>
Tom 8
Ahmad 5
Beau 6
dtype: int64
-```
+</pre>
---
-```
+<pre>
Tom 8
Beau 6
dtype: int64
-```
+</pre>
## --video-solution--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-introduction.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-introduction.md
index 0fbc810348b..45a30bc050b 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-introduction.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/pandas-introduction.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c158
+title: Pandas Introduction
challengeType: 11
videoId: 0xACW-8cZU0
dashedName: pandas-introduction
@@ -7,9 +8,12 @@ dashedName: pandas-introduction
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/freecodecamp-intro-to-pandas-902ae59b)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/freecodecamp-intro-to-pandas)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/parsing-html-and-saving-data.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/parsing-html-and-saving-data.md
index 501c55be478..284a5eababf 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/parsing-html-and-saving-data.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/parsing-html-and-saving-data.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c164
+title: Parsing HTML and Saving Data
challengeType: 11
videoId: bJaqnTWQmb0
dashedName: parsing-html-and-saving-data
@@ -7,15 +8,12 @@ dashedName: parsing-html-and-saving-data
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Reading CSVs Notebook](https://notebooks.ai/rmotr-curriculum/rdp-reading-csv-and-txt-files-fb829f46)
-
-\- [Reading SQL](https://notebooks.ai/rmotr-curriculum/rdp-reading-data-from-relational-databases-2a3a889b)
-
-\- [Reading HTML](https://notebooks.ai/rmotr-curriculum/rdp-reading-html-tables-eb9cca73)
-
-\- [Reading Excel files](https://notebooks.ai/rmotr-curriculum/rdp-reading-excel-files-a6b99973)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/RDP-Reading-Data-with-Python-and-Pandas)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/python-functions-and-collections.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/python-functions-and-collections.md
index f6f9f4a81be..75a9a71e77f 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/python-functions-and-collections.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/python-functions-and-collections.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c166
+title: Python Functions and Collections
challengeType: 11
videoId: NzpU17ZVlUw
dashedName: python-functions-and-collections
@@ -7,9 +8,12 @@ dashedName: python-functions-and-collections
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/python-under-10-minutes-15addcb2)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/ds-content-python-under-10-minutes)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/python-introduction.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/python-introduction.md
index b516cfc3bd0..f67c83c389c 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/python-introduction.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/python-introduction.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c165
+title: Python Introduction
challengeType: 11
videoId: PrQV9JkLhb4
dashedName: python-introduction
@@ -7,9 +8,12 @@ dashedName: python-introduction
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/python-under-10-minutes-15addcb2)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/ds-content-python-under-10-minutes)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/python-iteration-and-modules.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/python-iteration-and-modules.md
index 49abd38b22f..4d4350768f2 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/python-iteration-and-modules.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/python-iteration-and-modules.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c167
+title: Python Iteration and Modules
challengeType: 11
videoId: XzosGWLafrY
dashedName: python-iteration-and-modules
@@ -7,9 +8,12 @@ dashedName: python-iteration-and-modules
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Notebook](https://notebooks.ai/rmotr-curriculum/python-under-10-minutes-15addcb2)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/ds-content-python-under-10-minutes)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/reading-data-csv-and-txt.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/reading-data-csv-and-txt.md
index f1a259dd0dc..18d6487a0a2 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/reading-data-csv-and-txt.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/reading-data-csv-and-txt.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c162
+title: Reading Data CSV and TXT
challengeType: 11
videoId: ViGEv0zOzUk
dashedName: reading-data-csv-and-txt
@@ -7,15 +8,12 @@ dashedName: reading-data-csv-and-txt
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Reading CSVs Notebook](https://notebooks.ai/rmotr-curriculum/rdp-reading-csv-and-txt-files-fb829f46)
-
-\- [Reading SQL](https://notebooks.ai/rmotr-curriculum/rdp-reading-data-from-relational-databases-2a3a889b)
-
-\- [Reading HTML](https://notebooks.ai/rmotr-curriculum/rdp-reading-html-tables-eb9cca73)
-
-\- [Reading Excel files](https://notebooks.ai/rmotr-curriculum/rdp-reading-excel-files-a6b99973)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/RDP-Reading-Data-with-Python-and-Pandas)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/reading-data-from-databases.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/reading-data-from-databases.md
index 2fbd44e4ca5..5502a07e7aa 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/reading-data-from-databases.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/reading-data-from-databases.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c163
+title: Reading Data from Databases
challengeType: 11
videoId: MtgXS1MofRw
dashedName: reading-data-from-databases
@@ -7,15 +8,12 @@ dashedName: reading-data-from-databases
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Reading CSVs Notebook](https://notebooks.ai/rmotr-curriculum/rdp-reading-csv-and-txt-files-fb829f46)
-
-\- [Reading SQL](https://notebooks.ai/rmotr-curriculum/rdp-reading-data-from-relational-databases-2a3a889b)
-
-\- [Reading HTML](https://notebooks.ai/rmotr-curriculum/rdp-reading-html-tables-eb9cca73)
-
-\- [Reading Excel files](https://notebooks.ai/rmotr-curriculum/rdp-reading-excel-files-a6b99973)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/RDP-Reading-Data-with-Python-and-Pandas)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/reading-data-introduction.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/reading-data-introduction.md
index fe9ea50beda..a16b2d01760 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/reading-data-introduction.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-course/reading-data-introduction.md
@@ -1,5 +1,6 @@
---
id: 5e9a093a74c4063ca6f7c161
+title: Reading Data Introduction
challengeType: 11
videoId: cDnt02BcHng
dashedName: reading-data-introduction
@@ -7,15 +8,12 @@ dashedName: reading-data-introduction
# --description--
+*Instead of using notebooks.ai like it shows in the video, you can use Google Colab instead.*
+
More resources:
-\- [Reading CSVs Notebook](https://notebooks.ai/rmotr-curriculum/rdp-reading-csv-and-txt-files-fb829f46)
-
-\- [Reading SQL](https://notebooks.ai/rmotr-curriculum/rdp-reading-data-from-relational-databases-2a3a889b)
-
-\- [Reading HTML](https://notebooks.ai/rmotr-curriculum/rdp-reading-html-tables-eb9cca73)
-
-\- [Reading Excel files](https://notebooks.ai/rmotr-curriculum/rdp-reading-excel-files-a6b99973)
+- [Notebooks on GitHub](https://github.com/ine-rmotr-curriculum/RDP-Reading-Data-with-Python-and-Pandas)
+- [How to open Notebooks from GitHub using Google Colab.](https://colab.research.google.com/github/googlecolab/colabtools/blob/master/notebooks/colab-github-demo.ipynb)
# --question--
@@ -23,13 +21,13 @@ More resources:
Given a file named `certificates.csv` with these contents:
-```
+<pre>
Name$Certificates$Time (in months)
Tom$8$16
Kris$2$5
Ahmad$5$9
Beau$6$12
-```
+</pre>
Fill in the blanks for the missing arguments below:
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/demographic-data-analyzer.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/demographic-data-analyzer.md
index 72a6b7e8a4b..66246f3d5d5 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/demographic-data-analyzer.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/demographic-data-analyzer.md
@@ -1,5 +1,6 @@
---
id: 5e46f7e5ac417301a38fb929
+title: Demographic Data Analyzer
challengeType: 10
dashedName: demographic-data-analyzer
---
@@ -8,7 +9,7 @@ dashedName: demographic-data-analyzer
In this challenge you must analyze demographic data using Pandas. You are given a dataset of demographic data that was extracted from the 1994 Census database.
-You can access [the full project description and starter code on repl.it](https://repl.it/@freeCodeCamp/fcc-demographic-data-analyzer).
+You can access [the full project description and starter code on Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-demographic-data-analyzer).
After going to that link, fork the project. Once you complete the project based on the instructions in 'README.md', submit your project link below.
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/mean-variance-standard-deviation-calculator.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/mean-variance-standard-deviation-calculator.md
index be6e980f47c..76fed744e5f 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/mean-variance-standard-deviation-calculator.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/mean-variance-standard-deviation-calculator.md
@@ -1,5 +1,6 @@
---
id: 5e46f7e5ac417301a38fb928
+title: Mean-Variance-Standard Deviation Calculator
challengeType: 10
dashedName: mean-variance-standard-deviation-calculator
---
@@ -8,7 +9,7 @@ dashedName: mean-variance-standard-deviation-calculator
Create a function that uses Numpy to output the mean, variance, and standard deviation of the rows, columns, and elements in a 3 x 3 matrix.
-You can access [the full project description and starter code on repl.it](https://repl.it/@freeCodeCamp/fcc-mean-var-std).
+You can access [the full project description and starter code on Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-mean-variance-standard-deviation-calculator).
After going to that link, fork the project. Once you complete the project based on the instructions in 'README.md', submit your project link below.
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/medical-data-visualizer.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/medical-data-visualizer.md
index 9004b2ffb30..4d7cfc2f138 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/medical-data-visualizer.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/medical-data-visualizer.md
@@ -1,5 +1,6 @@
---
id: 5e46f7f8ac417301a38fb92a
+title: Medical Data Visualizer
challengeType: 10
dashedName: medical-data-visualizer
---
@@ -8,7 +9,7 @@ dashedName: medical-data-visualizer
In this project, you will visualize and make calculations from medical examination data using matplotlib, seaborn, and pandas.
-You can access [the full project description and starter code on repl.it](https://repl.it/@freeCodeCamp/fcc-medical-data-visualizer).
+You can access [the full project description and starter code on Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-medical-data-visualizer).
After going to that link, fork the project. Once you complete the project based on the instructions in 'README.md', submit your project link below.
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/page-view-time-series-visualizer.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/page-view-time-series-visualizer.md
index d56da94399b..b33d2eeed6c 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/page-view-time-series-visualizer.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/page-view-time-series-visualizer.md
@@ -1,5 +1,6 @@
---
id: 5e46f802ac417301a38fb92b
+title: Page View Time Series Visualizer
challengeType: 10
dashedName: page-view-time-series-visualizer
---
@@ -8,7 +9,7 @@ dashedName: page-view-time-series-visualizer
For this project you will visualize time series data using a line chart, bar chart, and box plots. You will use Pandas, matplotlib, and seaborn to visualize a dataset containing the number of page views each day on the freeCodeCamp.org forum from 2016-05-09 to 2019-12-03. The data visualizations will help you understand the patterns in visits and identify yearly and monthly growth.
-You can access [the full project description and starter code on repl.it](https://repl.it/@freeCodeCamp/fcc-time-series-visualizer).
+You can access [the full project description and starter code on Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-page-view-time-series-visualizer).
After going to that link, fork the project. Once you complete the project based on the instructions in 'README.md', submit your project link below.
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/sea-level-predictor.md b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/sea-level-predictor.md
index 6f37bc680c5..58075caec0a 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/sea-level-predictor.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/data-analysis-with-python-projects/sea-level-predictor.md
@@ -1,5 +1,6 @@
---
id: 5e4f5c4b570f7e3a4949899f
+title: Sea Level Predictor
challengeType: 10
dashedName: sea-level-predictor
---
@@ -8,7 +9,7 @@ dashedName: sea-level-predictor
In this project, you will analyze a dataset of the global average sea level change since 1880. You will use the data to predict the sea level change through year 2050.
-You can access [the full project description and starter code on repl.it](https://repl.it/@freeCodeCamp/fcc-sea-level-predictor).
+You can access [the full project description and starter code on Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-sea-level-predictor).
After going to that link, fork the project. Once you complete the project based on the instructions in 'README.md', submit your project link below.
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/accessing-and-changing-elements,-rows,-columns.md b/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/accessing-and-changing-elements,-rows,-columns.md
index 95b3a750a28..3c833be5adb 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/accessing-and-changing-elements,-rows,-columns.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/accessing-and-changing-elements,-rows,-columns.md
@@ -1,5 +1,6 @@
---
id: 5e9a0a8e09c5df3cc3600ed4
+title: 'Accessing and Changing Elements, Rows, Columns'
challengeType: 11
videoId: v-7Y7koJ_N0
dashedName: accessing-and-changing-elements-rows-columns
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/basics-of-numpy.md b/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/basics-of-numpy.md
index 5dd0cd584b6..57b1924af7e 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/basics-of-numpy.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/basics-of-numpy.md
@@ -1,5 +1,6 @@
---
id: 5e9a0a8e09c5df3cc3600ed3
+title: Basics of Numpy
challengeType: 11
videoId: f9QrZrKQMLI
dashedName: basics-of-numpy
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/copying-arrays-warning.md b/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/copying-arrays-warning.md
index 4b98fbc12ef..afe7b8d83ea 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/copying-arrays-warning.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/copying-arrays-warning.md
@@ -1,5 +1,6 @@
---
id: 5e9a0a8e09c5df3cc3600ed7
+title: Copying Arrays Warning
challengeType: 11
videoId: iIoQ0_L0GvA
dashedName: copying-arrays-warning
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/initialize-array-problem.md b/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/initialize-array-problem.md
index 33fa3b1d2e8..34e8f276048 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/initialize-array-problem.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/initialize-array-problem.md
@@ -1,5 +1,6 @@
---
id: 5e9a0a8e09c5df3cc3600ed6
+title: Initialize Array Problem
challengeType: 11
videoId: 0jGfH8BPfOk
dashedName: initialize-array-problem
@@ -12,20 +13,33 @@ dashedName: initialize-array-problem
What is another way to produce the following array?
```py
-[[1. 1. 1. 1. 1.]
-[1. 0. 0. 0. 1.]
-[1. 0. 9. 0. 1.]
-[1. 0. 0. 0. 1.]
-[1. 1. 1. 1. 1.]]
+[[0. 0. 0. 0. 0. 0. 0.]
+[0. 1. 1. 1. 1. 1. 0.]
+[0. 1. 1. 1. 1. 1. 0.]
+[0. 1. 1. 5. 1. 1. 0.]
+[0. 1. 1. 1. 1. 1. 0.]
+[0. 1. 1. 1. 1. 1. 0.]
+[0. 0. 0. 0. 0. 0. 0.]]
```
## --answers--
```py
-output = np.ones((5, 5))
+output = np.ones((7, 7))
-z = np.zeros((3, 3))
-z[1, 1] = 9
+z = np.zeros((5, 5))
+z[2, 2] = 5
+
+output[1:1, -1:-1] = z
+```
+
+---
+
+```py
+output = np.zeros((7,7))
+
+z = np.ones((5, 5))
+z[2, 2] = 5
output[1:-1, 1:-1] = z
```
@@ -33,26 +47,15 @@ output[1:-1, 1:-1] = z
---
```py
-output = np.ones((5, 5))
+output = np.ones((7, 7))
-z = np.zeros((3, 3))
-z[1, 1] = 9
+z = np.zeros((5, 5))
+z[3, 3] = 5
-output[1:3, 1:3] = z
-```
-
----
-
-```py
-output = np.ones((5, 5))
-
-z = np.zeros((3, 3))
-z[1, 1] = 9
-
-output[4:1, 4:1] = z
+output[1:-1, 1:-1] = z
```
## --video-solution--
-1
+2
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/initializing-different-arrays.md b/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/initializing-different-arrays.md
index 44e16dfdf9a..a6957cc7e42 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/initializing-different-arrays.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/initializing-different-arrays.md
@@ -1,5 +1,6 @@
---
id: 5e9a0a8e09c5df3cc3600ed5
+title: Initializing Different Arrays
challengeType: 11
videoId: CEykdsKT4U4
dashedName: initializing-different-arrays
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/loading-data-and-advanced-indexing.md b/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/loading-data-and-advanced-indexing.md
index 00664a2c5e2..5afa96f450d 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/loading-data-and-advanced-indexing.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/loading-data-and-advanced-indexing.md
@@ -1,5 +1,6 @@
---
id: 5e9a0a8e09c5df3cc3600eda
+title: Loading Data and Advanced Indexing
challengeType: 11
videoId: tUdBZ7pF8Jg
dashedName: loading-data-and-advanced-indexing
@@ -11,10 +12,10 @@ dashedName: loading-data-and-advanced-indexing
Given a file named `data.txt` with these contents:
-```
+<pre>
29,97,32,100,45
15,88,5,75,22
-```
+</pre>
What code would produce the following array?
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/mathematics.md b/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/mathematics.md
index 9308d6af252..f3b1a3c1114 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/mathematics.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/mathematics.md
@@ -1,5 +1,6 @@
---
id: 5e9a0a8e09c5df3cc3600ed8
+title: Mathematics
challengeType: 11
videoId: 7txegvyhtVk
dashedName: mathematics
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/reorganizing-arrays.md b/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/reorganizing-arrays.md
index 422253b4e50..7e4903c8645 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/reorganizing-arrays.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/reorganizing-arrays.md
@@ -1,5 +1,6 @@
---
id: 5e9a0a8e09c5df3cc3600ed9
+title: Reorganizing Arrays
challengeType: 11
videoId: VNWAQbEM-C8
dashedName: reorganizing-arrays
diff --git a/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/what-is-numpy.md b/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/what-is-numpy.md
index 9823c9afb7d..03b2f65fc73 100644
--- a/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/what-is-numpy.md
+++ b/curriculum/challenges/chinese/08-data-analysis-with-python/numpy/what-is-numpy.md
@@ -1,5 +1,6 @@
---
id: 5e9a0a8e09c5df3cc3600ed2
+title: What is NumPy
challengeType: 11
videoId: 5Nwfs5Ej85Q
dashedName: what-is-numpy
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-projects/anonymous-message-board.md b/curriculum/challenges/chinese/09-information-security/information-security-projects/anonymous-message-board.md
index c149f65f55f..20970a8e212 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-projects/anonymous-message-board.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-projects/anonymous-message-board.md
@@ -1,84 +1,135 @@
---
id: 587d824a367417b2b2512c45
-title: 匿名留言板
+title: Anonymous Message Board
challengeType: 4
-videoUrl: ''
+forumTopicId: 301568
dashedName: anonymous-message-board
---
# --description--
-构建一个功能类似于此的完整堆栈JavaScript应用程序： <https://spiky-well-vein.glitch.me/> 。在这个项目上工作将涉及您在我们的入门项目上在Glitch上编写代码。完成此项目后，您可以将公共故障网址（到应用程序的主页）复制到此屏幕进行测试！您可以选择在另一个平台上编写项目，但必须公开显示我们的测试。使用[此链接](https://glitch.com/#!/import/github/freeCodeCamp/boilerplate-project-messageboard/)在Glitch上启动此项目或在GitHub上克隆[此存储库](https://github.com/freeCodeCamp/boilerplate-project-messageboard/) ！如果您使用Glitch，请记住将项目链接保存到安全的地方！
+Build a full stack JavaScript app that is functionally similar to this: <https://anonymous-message-board.freecodecamp.rocks/>.
+
+Working on this project will involve you writing your code using one of the following methods:
+
+- Clone [this GitHub repo](https://github.com/freeCodeCamp/boilerplate-project-messageboard/) and complete your project locally.
+- Use [our repl.it starter project](https://repl.it/github/freeCodeCamp/boilerplate-project-messageboard) to complete your project.
+- Use a site builder of your choice to complete the project. Be sure to incorporate all the files from our GitHub repo.
+
+When you are done, make sure a working demo of your project is hosted somewhere public. Then submit the URL to it in the `Solution Link` field. Optionally, also submit a link to your projects source code in the `GitHub Link` field.
+
+# --instructions--
+
+1. Set `NODE_ENV` to test without quotes when ready to write tests and DB to your databases connection string (in `.env`)
+2. Recommended to create controllers/handlers and handle routing in `routes/api.js`
+3. You will add any security features to `server.js`
+
+Write the following tests in `tests/2_functional-tests.js`:
+
+- Creating a new thread: POST request to `/api/threads/{board}`
+- Viewing the 10 most recent threads with 3 replies each: GET request to `/api/threads/{board}`
+- Deleting a thread with the incorrect password: DELETE request to `/api/threads/{board}` with an invalid `delete_password`
+- Deleting a thread with the correct password: DELETE request to `/api/threads/{board}` with a valid `delete_password`
+- Reporting a thread: PUT request to `/api/threads/{board}`
+- Creating a new reply: POST request to `/api/replies/{board}`
+- Viewing a single thread with all replies: GET request to `/api/replies/{board}`
+- Deleting a reply with the incorrect password: DELETE request to `/api/threads/{board}` with an invalid `delete_password`
+- Deleting a reply with the correct password: DELETE request to `/api/threads/{board}` with a valid `delete_password`
+- Reporting a reply: PUT request to `/api/replies/{board}`
# --hints--
-只允许您的网站加载到您自己网页上的iFrame中。
+You can provide your own project, not the example URL.
+
+```js
+(getUserInput) => {
+ assert(
+ !/.*\/anonymous-message-board\.freecodecamp\.rocks/.test(
+ getUserInput('url')
+ )
+ );
+};
+```
+
+Only allow your site to be loaded in an iFrame on your own pages.
+
+```js
+async (getUserInput) => {
+ const data = await fetch(getUserInput('url') + '/_api/app-info');
+ const parsed = await data.json();
+ assert.isTrue(parsed.headers['x-frame-options']?.includes('SAMEORIGIN'));
+};
+```
+
+Do not allow DNS prefetching.
+
+```js
+async (getUserInput) => {
+ const data = await fetch(getUserInput('url') + '/_api/app-info');
+ const parsed = await data.json();
+ assert.isTrue(parsed.headers['x-dns-prefetch-control']?.includes('off'));
+};
+```
+
+Only allow your site to send the referrer for your own pages.
+
+```js
+async (getUserInput) => {
+ const data = await fetch(getUserInput('url') + '/_api/app-info');
+ const parsed = await data.json();
+ assert.isTrue(parsed.headers['referrer-policy']?.includes('same-origin'));
+};
+```
+
+You can send a POST request to `/api/threads/{board}` with form data including `text` and `delete_password`. The saved database record will have at least the fields `_id`, `text`, `created_on`(date & time), `bumped_on`(date & time, starts same as `created_on`), `reported` (boolean), `delete_password`, & `replies` (array).
```js
```
-不允许DNS预取。
+You can send a POST request to `/api/replies/{board}` with form data including `text`, `delete_password`, & `thread_id`. This will update the `bumped_on` date to the comment's date. In the thread's `replies` array, an object will be saved with at least the properties `_id`, `text`, `created_on`, `delete_password`, & `reported`.
```js
```
-只允许您的网站为您自己的网页发送引荐来源。
+You can send a GET request to `/api/threads/{board}`. Returned will be an array of the most recent 10 bumped threads on the board with only the most recent 3 replies for each. The `reported` and `delete_password` fields will not be sent to the client.
```js
```
-我可以通过将表单数据文本和deletepassword*传递给/api/threads/ {board}将一个帖子发布到一个特定的留言板。（推荐res.redirect到电路板页面/ b / {board}）保存的至少是\_id，text ，createdon*（日期和时间），bumpedon*（日期和时间，与created_on相同），报告（布尔值），deletepassword*和回复（数组）。
+You can send a GET request to `/api/replies/{board}?thread_id={thread_id}`. Returned will be the entire thread with all its replies, also excluding the same fields from the client as the previous test.
```js
```
-我可以通过将表单数据文本，deletepassword*和threadid*传递给/ api / replies / {board}来对特定板上的线程发送回复，并且还会将bumped*on日期更新到注释日期。（推荐res.redirect到thread page / b / {board} / {thread_id}）在线程的回复数组中将保存\_id，text，createdon*，deletepassword\_，并报告。
+You can send a DELETE request to `/api/threads/{board}` and pass along the `thread_id` & `delete_password` to delete the thread. Returned will be the string `incorrect password` or `success`.
```js
```
-我可以在电路板上获取最近10个凸起线程的数组，其中最近只有来自/ api / threads / {board}的3个回复。 report和deletepasswords\_字段不会发送到客户端。
+You can send a DELETE request to `/api/replies/{board}` and pass along the `thread_id`, `reply_id`, & `delete_password`. Returned will be the string `incorrect password` or `success`. On success, the text of the `reply_id` will be changed to `[deleted]`.
```js
```
-我可以使用/ api / replies / {board}的所有回复获取整个帖子吗？thread_id = {thread_id}。同样隐藏客户端应该看到的相同字段。
+You can send a PUT request to `/api/threads/{board}` and pass along the `thread_id`. Returned will be the string `success`. The `reported` value of the `thread_id` will be changed to `true`.
```js
```
-如果我向/ api / threads / {board}发送DELETE请求并传递threadid*＆deletepassword*，我可以完全删除一个线程。 （文字回复将是“密码不正确”或“成功”）
+You can send a PUT request to `/api/replies/{board}` and pass along the `thread_id` & `reply_id`. Returned will be the string `success`. The `reported` value of the `reply_id` will be changed to `true`.
```js
```
-如果我向/ api / replies / {board}发送DELETE请求并传递threadid*，replyid*和deletepassword\_，我可以删除帖子（只是将文本更改为“[已删除]”而不是像线程一样完全删除）。 （文字回复将是“密码不正确”或“成功”）
-
-```js
-
-```
-
-我可以通过向/ api / threads / {board}发送PUT请求并传递threadid\_来报告一个线程并将其报告值更改为true。 （文字回复将是“成功”）
-
-```js
-
-```
-
-我可以通过向/ api / replies / {board}发送PUT请求并传递threadid*＆replyid*来报告回复并将其报告值更改为true。 （文字回复将是“成功”）
-
-```js
-
-```
-
-完整的功能测试，完全测试路线和通过。
+All 10 functional tests are complete and passing.
```js
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-projects/port-scanner.md b/curriculum/challenges/chinese/09-information-security/information-security-projects/port-scanner.md
index 53a3cbd7f2d..5c776ca7d07 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-projects/port-scanner.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-projects/port-scanner.md
@@ -1,6 +1,8 @@
---
id: 5e46f979ac417301a38fb932
+title: Port Scanner
challengeType: 10
+helpCategory: Python
dashedName: port-scanner
---
@@ -8,7 +10,7 @@ dashedName: port-scanner
Create a port scanner using Python.
-You can access [the full project description and starter code on repl.it](https://repl.it/@freeCodeCamp/fcc-port-scanner).
+You can access [the full project description and starter code on Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-port-scanner).
After going to that link, fork the project. Once you complete the project based on the instructions in 'README.md', submit your project link below.
@@ -21,7 +23,7 @@ We are still developing the interactive instructional part of the Python curricu
<li>
<a href='https://www.freecodecamp.org/news/learn-python-basics-in-depth-video-course/'>Learn Python Video Course</a> (2 hours)
</li>
-<ul>
+</ul>
# --hints--
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-projects/secure-real-time-multiplayer-game.md b/curriculum/challenges/chinese/09-information-security/information-security-projects/secure-real-time-multiplayer-game.md
index 2256f3ff2d2..3ced001df35 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-projects/secure-real-time-multiplayer-game.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-projects/secure-real-time-multiplayer-game.md
@@ -1,19 +1,34 @@
---
id: 5e601c775ac9d0ecd8b94aff
+title: Secure Real Time Multiplayer Game
challengeType: 4
dashedName: secure-real-time-multiplayer-game
---
# --description--
-Develop a 2D real time multiplayer game using the HTML Canvas API and [Socket.io](https://socket.io/) that is functionally similar to this: <https://thread-valley-lipstick.glitch.me/>.
+Develop a 2D real time multiplayer game using the HTML Canvas API and [Socket.io](https://socket.io/) that is functionally similar to this: <https://secure-real-time-multiplayer-game.freecodecamp.rocks/>. Working on this project will involve you writing your code using one of the following methods:
-Working on this project will involve you writing your code on Glitch on our starter project. After completing this project you can copy your public glitch url (to the homepage of your app) into this screen to test it! Optionally you may choose to write your project on another platform but it must be publicly visible for our testing.
+- Clone [this GitHub repo](https://github.com/freeCodeCamp/boilerplate-project-secure-real-time-multiplayer-game/) and complete your project locally.
+- Use [our repl.it starter project](https://repl.it/github/freeCodeCamp/boilerplate-project-secure-real-time-multiplayer-game) to complete your project.
+- Use a site builder of your choice to complete the project. Be sure to incorporate all the files from our GitHub repo.
-Start this project on Glitch using [this link](https://glitch.com/edit/#!/remix/clone-from-repo?REPO_URL=https://github.com/freeCodeCamp/boilerplate-project-secure-real-time-multiplayer-game/) or clone [this repository](https://github.com/freeCodeCamp/boilerplate-project-secure-real-time-multiplayer-game) on GitHub! If you use Glitch, remember to save the link to your project somewhere safe!
+When you are done, make sure a working demo of your project is hosted somewhere public. Then submit the URL to it in the `Solution Link` field. Optionally, also submit a link to your project's source code in the `GitHub Link` field.
# --hints--
+You can provide your own project, not the example URL.
+
+```js
+(getUserInput) => {
+ assert(
+ !/.*\/secure-real-time-multiplayer-game\.freecodecamp\.rocks/.test(
+ getUserInput('url')
+ )
+ );
+};
+```
+
Multiple players can connect to a server and play.
```js
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-projects/sha-1-password-cracker.md b/curriculum/challenges/chinese/09-information-security/information-security-projects/sha-1-password-cracker.md
index 71ed376b44d..8a79bb1b7c4 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-projects/sha-1-password-cracker.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-projects/sha-1-password-cracker.md
@@ -1,6 +1,8 @@
---
id: 5e46f983ac417301a38fb933
+title: SHA-1 Password Cracker
challengeType: 10
+helpCategory: Python
dashedName: sha-1-password-cracker
---
@@ -8,7 +10,7 @@ dashedName: sha-1-password-cracker
For this project you will learn about the importance of good security by creating a password cracker to figure out passwords that were hashed using SHA-1.
-You can access [the full project description and starter code on repl.it](https://repl.it/@freeCodeCamp/fcc-brute-force-password-cracker).
+You can access [the full project description and starter code on Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-SHA-1-password-cracker).
After going to that link, fork the project. Once you complete the project based on the instructions in 'README.md', submit your project link below.
@@ -21,7 +23,7 @@ We are still developing the interactive instructional part of the Python curricu
<li>
<a href='https://www.freecodecamp.org/news/learn-python-basics-in-depth-video-course/'>Learn Python Video Course</a> (2 hours)
</li>
-<ul>
+</ul>
# --hints--
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-projects/stock-price-checker.md b/curriculum/challenges/chinese/09-information-security/information-security-projects/stock-price-checker.md
index 4bbe859f94a..ce584c58d58 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-projects/stock-price-checker.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-projects/stock-price-checker.md
@@ -1,57 +1,126 @@
---
id: 587d824a367417b2b2512c44
-title: 股票价格检查
+title: Stock Price Checker
challengeType: 4
-videoUrl: ''
+forumTopicId: 301572
dashedName: stock-price-checker
---
# --description--
-构建一个功能类似于此的完整堆栈JavaScript应用程序： [https](https://sphenoid-crater.glitch.me/) ： [//giant-chronometer.glitch.me/](https://sphenoid-crater.glitch.me/) 。在这个项目上工作将涉及您在我们的入门项目上在Glitch上编写代码。完成此项目后，您可以将公共故障网址（到应用程序的主页）复制到此屏幕进行测试！您可以选择在另一个平台上编写项目，但必须公开显示我们的测试。使用[此链接](https://glitch.com/#!/import/github/freeCodeCamp/boilerplate-project-stockchecker/)在Glitch上启动此项目或在GitHub上克隆[此存储库](https://github.com/freeCodeCamp/boilerplate-project-stockchecker/) ！如果您使用Glitch，请记住将项目链接保存到安全的地方！
+Build a full stack JavaScript app that is functionally similar to this: <https://stock-price-checker.freecodecamp.rocks/>.
+
+Since all reliable stock price APIs require an API key, we've built a workaround. Use <https://stock-price-checker-proxy.freecodecamp.rocks/> to get up-to-date stock price information without needing to sign up for your own key.
+
+Working on this project will involve you writing your code using one of the following methods:
+
+- Clone [this GitHub repo](https://github.com/freeCodeCamp/boilerplate-project-stockchecker/) and complete your project locally.
+- Use [our repl.it starter project](https://repl.it/github/freeCodeCamp/boilerplate-project-stockchecker) to complete your project.
+- Use a site builder of your choice to complete the project. Be sure to incorporate all the files from our GitHub repo.
+
+When you are done, make sure a working demo of your project is hosted somewhere public. Then submit the URL to it in the `Solution Link` field. Optionally, also submit a link to your projects source code in the `GitHub Link` field.
+
+# --instructions--
+
+1. SET `NODE_ENV` to `test` without quotes and set `DB` to your MongoDB connection string
+2. Complete the project in `routes/api.js` or by creating a handler/controller
+3. You will add any security features to `server.js`
+4. You will create all of the functional tests in `tests/2_functional-tests.js`
+
+Write the following tests in `tests/2_functional-tests.js`:
+
+- Viewing one stock: GET request to `/api/stock-prices/`
+- Viewing one stock and liking it: GET request to `/api/stock-prices/`
+- Viewing the same stock and liking it again: GET request to `/api/stock-prices/`
+- Viewing two stocks: GET request to `/api/stock-prices/`
+- Viewing two stocks and liking them: GET request to `/api/stock-prices/`
# --hints--
-将内容安全策略设置为仅允许从服务器加载脚本和css。
+You can provide your own project, not the example URL.
+
+```js
+(getUserInput) => {
+ assert(
+ !/.*\/stock-price-checker\.freecodecamp\.rocks/.test(getUserInput('url'))
+ );
+};
+```
+
+You should set the content security policies to only allow loading of scripts and CSS from your server.
+
+```js
+async (getUserInput) => {
+ const data = await fetch(getUserInput('url') + '/_api/app-info');
+ const parsed = await data.json();
+ assert.isTrue(
+ parsed.headers['content-security-policy'].includes("script-src 'self'")
+ );
+ assert.isTrue(
+ parsed.headers['content-security-policy'].includes("style-src 'self'")
+ );
+};
+```
+
+You can send a `GET` request to `/api/stock-prices` with form data containing a NASDAQ stock ticker. Returned will be an object with the property `stockData`.
+
+```js
+async (getUserInput) => {
+ const data = await fetch(
+ getUserInput('url') + '/api/stock-prices?stock=GOOG'
+ );
+ const parsed = await data.json();
+ assert.property(parsed, 'stockData');
+};
+```
+
+The `stockData` property includes the `stock` ticker as a string, the `price` as a number, and `likes` as a number.
+
+```js
+async (getUserInput) => {
+ const data = await fetch(
+ getUserInput('url') + '/api/stock-prices?stock=GOOG'
+ );
+ const parsed = await data.json();
+ const ticker = parsed.stockData;
+ assert.typeOf(ticker.price, 'number');
+ assert.typeOf(ticker.likes, 'number');
+ assert.typeOf(ticker.stock, 'string');
+};
+```
+
+You can also pass along a `like` field as `true` (boolean) to have your like added to the stock(s). Only 1 like per IP should be accepted.
```js
```
-我可以使用包含纳斯达克股票代码的表格数据获得/ api /股票价格并收回一个对象stockData。
+If you pass along 2 stocks, the returned value will be an array with information about both stocks. Instead of `likes`, it will display `rel_likes` (the difference between the likes on both stocks) for both `stockData` objects.
```js
-
+async (getUserInput) => {
+ const data = await fetch(
+ getUserInput('url') + '/api/stock-prices?stock=GOOG&stock=MSFT'
+ );
+ const parsed = await data.json();
+ const ticker = parsed.stockData;
+ assert.typeOf(ticker, 'array');
+ assert.property(ticker[0], 'rel_likes');
+ assert.property(ticker[1], 'rel_likes');
+};
```
-在stockData中，我可以看到股票（字符串，股票代码），价格（字符串格式的小数）和喜欢（int）。
+All 5 functional tests are complete and passing.
```js
-
-```
-
-我也可以传递像true（boolean）这样的字段来将我的喜欢添加到股票中。每个IP应该只接受1个。
-
-```js
-
-```
-
-如果我传递2只股票，则返回对象将是一个包含股票信息的数组。而不是喜欢，它将显示两者上的rel_likes（两只股票之间的差异）。
-
-```js
-
-```
-
-获得当前价格的一个好方法是使用以下外部API（用您的股票替换“GOOG”）：https：//finance.google.com/finance/info？q = NASDAQ％3AOGOOG
-
-```js
-
-```
-
-所有5个功能测试都已完成并通过。
-
-```js
-
+async (getUserInput) => {
+ const tests = await fetch(getUserInput('url') + '/_api/get-tests');
+ const parsed = await tests.json();
+ assert.isTrue(parsed.length >= 5);
+ parsed.forEach((test) => {
+ assert.equal(test.state, 'passed');
+ });
+};
```
# --solutions--
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/ask-browsers-to-access-your-site-via-https-only-with-helmet.hsts.md b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/ask-browsers-to-access-your-site-via-https-only-with-helmet.hsts.md
index 23cec353ef9..6bf78cfcccb 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/ask-browsers-to-access-your-site-via-https-only-with-helmet.hsts.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/ask-browsers-to-access-your-site-via-https-only-with-helmet.hsts.md
@@ -1,6 +1,6 @@
---
id: 587d8248367417b2b2512c3c
-title: 要求浏览器通过HTTPS访问您的站点仅限于使用helmet.hsts（）
+title: Ask Browsers to Access Your Site via HTTPS Only with helmet.hsts()
challengeType: 2
forumTopicId: 301573
dashedName: ask-browsers-to-access-your-site-via-https-only-with-helmet-hsts
@@ -8,19 +8,19 @@ dashedName: ask-browsers-to-access-your-site-via-https-only-with-helmet-hsts
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-infosec) 的基础上进行开发。你也可以从 [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-infosec), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/).
-HTTP 严格安全传输（HSTS）是一个能帮助我们抵御协议 [降级攻击](https://en.wikipedia.org/wiki/Downgrade_attack) 和 [Cookie 挟持](https://en.wikipedia.org/wiki/Session_hijacking) 的 WEB 安全协议。如果你的网站能够通过 HTTPS 访问，那么你就可以让你的用户避免使用不安全的 HTTP 协议访问。为此，你只需要在 header 中设置 Strict-Transport-Security 字段，以此来告诉浏览器在今后的一段时间内使用 HTTPS 来请求网站内容。这一设置对首次请求之后所以的后续请求都适用。
+HTTP Strict Transport Security (HSTS) is a web security policy which helps to protect websites against protocol downgrade attacks and cookie hijacking. If your website can be accessed via HTTPS you can ask user’s browsers to avoid using insecure HTTP. By setting the header Strict-Transport-Security, you tell the browsers to use HTTPS for the future requests in a specified amount of time. This will work for the requests coming after the initial request.
# --instructions--
-通过 `helmet.hsts()`，网站会在未来的90天内使用 HTTPS。我们还可以传入配置对象 `{maxAge: timeInSeconds, force: true}`。Repl.it 默认已经开启 hsts，但你仍然可以通过添加 `{force: true}` 来覆盖它。我们会拦截 Glitch 请求的 header 来进行此挑战的测试，然后恢复此项配置。
+Configure `helmet.hsts()` to use HTTPS for the next 90 days. Pass the config object `{maxAge: timeInSeconds, force: true}`. You can create a variable `ninetyDaysInSeconds = 90*24*60*60;` to use for the `timeInSeconds`. Repl.it already has hsts enabled. To override its settings you need to set the field "force" to true in the config object. We will intercept and restore the Repl.it header, after inspecting it for testing.
-注意: 配置 HTTPS 需要域名以及 SSL/TSL 证书。
+Note: Configuring HTTPS on a custom website requires the acquisition of a domain, and a SSL/TLS Certificate.
# --hints--
-应正确加载 helmet.hsts() 中间件
+helmet.hsts() middleware should be mounted correctly
```js
(getUserInput) =>
@@ -35,7 +35,7 @@ HTTP 严格安全传输（HSTS）是一个能帮助我们抵御协议 [降级攻
);
```
-应将 maxAge 设置为 7776000 ms（90 天）
+maxAge should be equal to 7776000 s (90 days)
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/avoid-inferring-the-response-mime-type-with-helmet.nosniff.md b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/avoid-inferring-the-response-mime-type-with-helmet.nosniff.md
index 5366fdf9de7..2bbd3bf02e4 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/avoid-inferring-the-response-mime-type-with-helmet.nosniff.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/avoid-inferring-the-response-mime-type-with-helmet.nosniff.md
@@ -1,6 +1,6 @@
---
id: 587d8248367417b2b2512c3a
-title: 使用 helment.noSniff() 来避免推断响应的 MIME 类型
+title: Avoid Inferring the Response MIME Type with helmet.noSniff()
challengeType: 2
forumTopicId: 301574
dashedName: avoid-inferring-the-response-mime-type-with-helmet-nosniff
@@ -8,13 +8,15 @@ dashedName: avoid-inferring-the-response-mime-type-with-helmet-nosniff
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-infosec) 的基础上进行开发。你也可以从 [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-infosec), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/). Browsers can use content or MIME sniffing to override response `Content-Type` headers to guess and process the data using an implicit content type. While this can be convenient in some scenarios, it can also lead to some dangerous attacks. This middleware sets the X-Content-Type-Options header to `nosniff`, instructing the browser to not bypass the provided `Content-Type`.
-浏览器可以通过 content 或者 MIME 嗅探来判断不同的响应内容。这两个的优先级比 header 中的 Content-Type 高。这在一些情况下非常方便，但也会造成一定的安全风险。我们可以通过中间件来把 header 中的 X-Content-Type-Options 字段设置为 nosniff，这样浏览器就不会绕过 header 中的 Content-Type 了。
+# --instructions--
+
+Use the `helmet.noSniff()` method on your server.
# --hints--
-应正确加载 helmet.noSniff() 中间件
+helmet.noSniff() middleware should be mounted correctly
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/configure-helmet-using-the-parent-helmet-middleware.md b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/configure-helmet-using-the-parent-helmet-middleware.md
index 730769bbad0..6865d3896c2 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/configure-helmet-using-the-parent-helmet-middleware.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/configure-helmet-using-the-parent-helmet-middleware.md
@@ -1,6 +1,6 @@
---
id: 587d8249367417b2b2512c40
-title: 使用 helmet() 中间件来配置 Helmet
+title: Configure Helmet Using the ‘parent’ helmet() Middleware
challengeType: 2
forumTopicId: 301575
dashedName: configure-helmet-using-the-parent-helmet-middleware
@@ -8,32 +8,32 @@ dashedName: configure-helmet-using-the-parent-helmet-middleware
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-infosec) 的基础上进行开发。你也可以从 [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-infosec), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/).
-`app.use(helmet())` 会自动加载除了 `noCache()` 和 `contentSecurityPolicy()` 以外的，上面所有提到的中间件。但如果需要的话，我们也可以手动加入这两个中间件。通过修改配置对象，你还可以启用或禁用其它中间件。
+`app.use(helmet())` will automatically include all the middleware introduced above, except `noCache()`, and `contentSecurityPolicy()`, but these can be enabled if necessary. You can also disable or configure any other middleware individually, using a configuration object.
-<h3>示例：</h3>
+**Example:**
```js
app.use(helmet({
- frameguard: { // 配置
+ frameguard: { // configure
action: 'deny'
},
- contentSecurityPolicy: { // 启用并配置
+ contentSecurityPolicy: { // enable and configure
directives: {
defaultSrc: ["self"],
styleSrc: ['style.com'],
}
},
- dnsPrefetchControl: false // 禁用
+ dnsPrefetchControl: false // disable
}))
```
-为了方便教学和测试，我们会一个一个地引入中间件。但在实际项目中，通过使用父级的 `helmet()` 来实现是最清晰和简洁的。
+We introduced each middleware separately for teaching purposes and for ease of testing. Using the ‘parent’ `helmet()` middleware is easy to implement in a real project.
# --hints--
-没有测试—这是一个介绍关卡
+no tests - it's a descriptive challenge
```js
assert(true);
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/disable-client-side-caching-with-helmet.nocache.md b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/disable-client-side-caching-with-helmet.nocache.md
index 08bd3c2ffaf..21cf373400d 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/disable-client-side-caching-with-helmet.nocache.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/disable-client-side-caching-with-helmet.nocache.md
@@ -1,6 +1,6 @@
---
id: 587d8249367417b2b2512c3e
-title: 使用 helment.noCache() 禁用客户端缓存
+title: Disable Client-Side Caching with helmet.noCache()
challengeType: 2
forumTopicId: 301576
dashedName: disable-client-side-caching-with-helmet-nocache
@@ -8,13 +8,17 @@ dashedName: disable-client-side-caching-with-helmet-nocache
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-infosec) 的基础上进行开发。你也可以从 [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-infosec), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/).
-如果你为你的网站发布了一个更新，此时你一定想要用户看到最新的版本。为此，你可以通过禁用浏览器缓存来实现。而且，这个功能在开发环境中也非常有用。但另一方面，缓存可以为你的网站带来性能方面的提升，因此你应该只在必要的时候禁用缓存。
+If you are releasing an update for your website, and you want the users to always download the newer version, you can (try to) disable caching on client’s browser. It can be useful in development too. Caching has performance benefits, which you will lose, so only use this option when there is a real need.
+
+# --instructions--
+
+Use the `helmet.noCache()` method on your server.
# --hints--
-应正确加载 helmet.noCache() 中间件
+helmet.noCache() middleware should be mounted correctly
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/disable-dns-prefetching-with-helmet.dnsprefetchcontrol.md b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/disable-dns-prefetching-with-helmet.dnsprefetchcontrol.md
index df8553f7a3f..5d212b6dbe5 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/disable-dns-prefetching-with-helmet.dnsprefetchcontrol.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/disable-dns-prefetching-with-helmet.dnsprefetchcontrol.md
@@ -1,6 +1,6 @@
---
id: 587d8248367417b2b2512c3d
-title: 使用 helmet.dnsPrefetchControl() 禁用 DNS 预获取
+title: Disable DNS Prefetching with helmet.dnsPrefetchControl()
challengeType: 2
forumTopicId: 301577
dashedName: disable-dns-prefetching-with-helmet-dnsprefetchcontrol
@@ -8,13 +8,17 @@ dashedName: disable-dns-prefetching-with-helmet-dnsprefetchcontrol
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-infosec) 的基础上进行开发。你也可以从 [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-infosec), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/).
-为了提高性能，大部分浏览器都会为页面上的链接预先加载 DNS 记录。这样当用户点击一个链接的时候浏览器已经知道其 IP 地址了。但这也会造成 DNS 服务的过度使用（如果你有一个百万用户数量级的大型网站）、隐私问题（窃听者可以借此推测出你在访问哪个页面）、页面统计数据准确性（有些没访问过的链接会被标记成已访问）。如果你对安全性要求比较高，你应该禁用 DNS 预加载。当然，这样做会让你损失一些性能。
+To improve performance, most browsers prefetch DNS records for the links in a page. In that way the destination ip is already known when the user clicks on a link. This may lead to over-use of the DNS service (if you own a big website, visited by millions people…), privacy issues (one eavesdropper could infer that you are on a certain page), or page statistics alteration (some links may appear visited even if they are not). If you have high security needs you can disable DNS prefetching, at the cost of a performance penalty.
+
+# --instructions--
+
+Use the `helmet.dnsPrefetchControl()` method on your server.
# --hints--
-应正确加载 helmet.dnsPrefetchControl() 中间件
+helmet.dnsPrefetchControl() middleware should be mounted correctly
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/hash-and-compare-passwords-asynchronously.md b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/hash-and-compare-passwords-asynchronously.md
index aeaf39a64cf..5d6126daa7c 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/hash-and-compare-passwords-asynchronously.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/hash-and-compare-passwords-asynchronously.md
@@ -1,6 +1,6 @@
---
id: 58a25bcff9fc0f352b528e7d
-title: 哈希和异步比较密码
+title: Hash and Compare Passwords Asynchronously
challengeType: 2
forumTopicId: 301578
dashedName: hash-and-compare-passwords-asynchronously
@@ -8,21 +8,21 @@ dashedName: hash-and-compare-passwords-asynchronously
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-infosec) 的基础上进行开发。你也可以从 [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-bcrypt), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-bcrypt/).
-由于哈希运算会占用很大计算机资源并且会耗费比较多时间，因此比较推荐的做法是异步调用哈希算法，这样就不会因此阻止其它连接或请求了。异步调用哈希方法非常简单，只需要：
+As hashing is designed to be computationally intensive, it is recommended to do so asynchronously on your server as to avoid blocking incoming connections while you hash. All you have to do to hash a password asynchronous is call
```js
bcrypt.hash(myPlaintextPassword, saltRounds, (err, hash) => {
- /* 在数据库中存储 hash */
+ /*Store hash in your db*/
});
```
# --instructions--
-把这段哈希方法添加到你的服务器（我们已经定义好这个方法的变量给你直接使用了），然后你可以尝试在控制台输出。之后，我们通常需要把哈希的结果保存到数据库。
+Add this hashing function to your server(we've already defined the variables used in the function for you to use) and log it to the console for you to see! At this point you would normally save the hash to your database.
-当你需要对比用户输入的值是否和之前哈希过的值一样的时候，只需要调用对比函数：
+Now when you need to figure out if a new input is the same data as the hash you would just use the compare function.
```js
bcrypt.compare(myPlaintextPassword, hash, (err, res) => {
@@ -30,7 +30,7 @@ bcrypt.compare(myPlaintextPassword, hash, (err, res) => {
});
```
-当控制台输出生成的哈希并在对比的回调中输出结果后，我们就可以将其添加到现有的哈希函数中。控制台中会首先输出一个哈希结果，然后输出 `true`。如果将比较函数中的 "myPlaintextPassword" 更改为 "someOtherPlaintextPassword"，则比较的结果应显示 `false`。
+Add this into your existing hash function(since you need to wait for the hash to complete before calling the compare function) after you log the completed hash and log 'res' to the console within the compare. You should see in the console a hash then 'true' is printed! If you change 'myPlaintextPassword' in the compare function to 'someOtherPlaintextPassword' then it should say false.
```js
bcrypt.hash('passw0rd!', 13, (err, hash) => {
@@ -43,11 +43,11 @@ bcrypt.hash('passw0rd!', 13, (err, hash) => {
```
-请在完成挑战后提交你的页面。
+Submit your page when you think you've got it right.
# --hints--
-应异步地进行哈希并正确地执行对比
+Async hash should be generated and correctly compared.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/hash-and-compare-passwords-synchronously.md b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/hash-and-compare-passwords-synchronously.md
index 75c293965e2..c03109f16ac 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/hash-and-compare-passwords-synchronously.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/hash-and-compare-passwords-synchronously.md
@@ -1,6 +1,6 @@
---
id: 58a25bcff9fc0f352b528e7e
-title: 哈希和同步比较密码
+title: Hash and Compare Passwords Synchronously
challengeType: 2
forumTopicId: 301579
dashedName: hash-and-compare-passwords-synchronously
@@ -8,31 +8,31 @@ dashedName: hash-and-compare-passwords-synchronously
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-infosec) 的基础上进行开发。你也可以从 [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-bcrypt), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-bcrypt/).
-同步执行哈希运算是非常简单的，但这会在哈希计算量大并且次数多的情况下造成延迟。如果想要同步执行哈希运算，只需要这样调用：
+Hashing synchronously is just as easy to do but can cause lag if using it server side with a high cost or with hashing done very often. Hashing with this method is as easy as calling
```js
var hash = bcrypt.hashSync(myPlaintextPassword, saltRounds);
```
-把同步哈希的方法添加到你的代码并在控制台输出。和之前一样，我们已经为你定义好了你需要使用的变量，你不需要做任何改动。你可能注意到，即使你对同一个密码进行哈希运算，同步哈希和异步哈希的结果也是不一样的。这是因为每次哈希的盐都是随机生成的，这种随机反映在前 22 个字符的不同上。 当你需要对比用户输入的值是否和之前哈希过的值一样的时候，只需要调用对比函数：
+Add this method of hashing to your code and then log the result to the console. Again, the variables used are already defined in the server so you won't need to adjust them. You may notice even though you are hashing the same password as in the async function, the result in the console is different- this is due to the salt being randomly generated each time as seen by the first 22 characters in the third string of the hash. Now to compare a password input with the new sync hash, you would use the compareSync method:
```js
var result = bcrypt.compareSync(myPlaintextPassword, hash);
```
-返回的结果为 `true` 或 `false`
+with the result being a boolean true or false.
# --instructions--
-请添加这个方法，并把结果输出到控制台，以此来验证同步哈希操作是否成功。
+Add the function in and log the result to the console to see it working.
-请在完成挑战后提交你的页面。
+Submit your page when you think you've got it right.
# --hints--
-应同步地进行哈希并正确地执行对比
+Sync hash should be generated and correctly compared.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/hide-potentially-dangerous-information-using-helmet.hidepoweredby.md b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/hide-potentially-dangerous-information-using-helmet.hidepoweredby.md
index 2fd826a5dd7..630302ea383 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/hide-potentially-dangerous-information-using-helmet.hidepoweredby.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/hide-potentially-dangerous-information-using-helmet.hidepoweredby.md
@@ -1,6 +1,6 @@
---
id: 587d8247367417b2b2512c37
-title: 使用 helmet.hidePoweredBy() 隐藏潜在的危险信息
+title: Hide Potentially Dangerous Information Using helmet.hidePoweredBy()
challengeType: 2
forumTopicId: 301580
dashedName: hide-potentially-dangerous-information-using-helmet-hidepoweredby
@@ -8,13 +8,13 @@ dashedName: hide-potentially-dangerous-information-using-helmet-hidepoweredby
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-infosec) 的基础上进行开发。你也可以从 [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-infosec), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/).
-如果黑客发现你的网站是用 Express 搭建的，那么他们就可以利用 Express 或 Node 现存的漏洞来攻击你的网站。`X-Powered-By: Express` 默认情况下会添加到所有响应的 header。不过 `helmet.hidePoweredBy()` 中间件可以帮你把 header 中的 X-Powered-By 字段移除。你甚至可以把它设置成其它的值来骗过黑客，比如 `app.use(helmet.hidePoweredBy({ setTo: 'PHP 4.2.0' }))`
+Hackers can exploit known vulnerabilities in Express/Node if they see that your site is powered by Express. X-Powered-By: Express is sent in every request coming from Express by default. The `helmet.hidePoweredBy()` middleware will remove the X-Powered-By header. You can also explicitly set the header to something else, to throw people off. e.g. `app.use(helmet.hidePoweredBy({ setTo: 'PHP 4.2.0' }))`
# --hints--
-应正确地加载 helmet.hidePoweredBy() 中间件
+helmet.hidePoweredBy() middleware should be mounted correctly
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/install-and-require-helmet.md b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/install-and-require-helmet.md
index 0a91cf2b48e..0a58f1d5971 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/install-and-require-helmet.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/install-and-require-helmet.md
@@ -1,6 +1,6 @@
---
id: 587d8247367417b2b2512c36
-title: 安装和引入 Helmet
+title: Install and Require Helmet
challengeType: 2
forumTopicId: 301581
dashedName: install-and-require-helmet
@@ -8,17 +8,19 @@ dashedName: install-and-require-helmet
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-infosec) 的基础上进行开发。你也可以从 [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-infosec), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/).
-Helmet 通过配置不同的 HTTP header 信息来使你的 Express 应用更加安全。
+Helmet helps you secure your Express apps by setting various HTTP headers.
# --instructions--
-安装版本号为 `3.21.3` 的 Helmet，并引入它。
+All your code for these lessons goes in the `myApp.js` file between the lines of code we have started you off with. Do not change or delete the code we have added for you.
+
+Install Helmet version `3.21.3`, then require it.
# --hints--
-`package.json` 中应存在版本号为 `3.21.3` 的 `helmet` 依赖项
+`helmet` version `3.21.3` should be in `package.json`
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/mitigate-the-risk-of-clickjacking-with-helmet.frameguard.md b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/mitigate-the-risk-of-clickjacking-with-helmet.frameguard.md
index 80783bd2b97..1759d2e7271 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/mitigate-the-risk-of-clickjacking-with-helmet.frameguard.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/mitigate-the-risk-of-clickjacking-with-helmet.frameguard.md
@@ -1,6 +1,6 @@
---
id: 587d8247367417b2b2512c38
-title: 使用 helmet.frameguard() 降低点击劫持的风险
+title: Mitigate the Risk of Clickjacking with helmet.frameguard()
challengeType: 2
forumTopicId: 301582
dashedName: mitigate-the-risk-of-clickjacking-with-helmet-frameguard
@@ -8,17 +8,19 @@ dashedName: mitigate-the-risk-of-clickjacking-with-helmet-frameguard
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-infosec) 的基础上进行开发。你也可以从 [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-infosec), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/).
-黑客可能会不经过你的允许，把你的页面嵌套在 \``或者`</code> 标签里，用以实现“点击劫持”。点击劫持是一种视觉上的欺骗手段，让用户误以为自己在与所看到的网页交互。通过 iframe，黑客可以在你的页面上添加一个透明的“层”，然后把自己的恶意代码放在一个用户看不到的按钮中。这样一来，你网站的执行环境就被黑客设置成了他想要的效果。helmet 中间件可以设置 header 中的 X-Frame-Options 字段，这样我们就能设置哪些人才可以通过 frame/iframe 引入我们的页面了。这个配置有三个选项：DENY、SAMEORIGIN、ALLOW-FROM。 在这个挑战中，我们的应用不需要被 iframe 引用。 </section>\`
+Your page could be put in a `<frame>` or `<iframe>` without your consent. This can result in clickjacking attacks, among other things. Clickjacking is a technique of tricking a user into interacting with a page different from what the user thinks it is. This can be obtained executing your page in a malicious context, by mean of iframing. In that context a hacker can put a hidden layer over your page. Hidden buttons can be used to run bad scripts. This middleware sets the X-Frame-Options header. It restricts who can put your site in a frame. It has three modes: DENY, SAMEORIGIN, and ALLOW-FROM.
+
+We don’t need our app to be framed.
# --instructions--
-请调用 `helmet.frameguard()`，并传入配置对象 `{action: 'deny'}`。
+Use `helmet.frameguard()` passing with the configuration object `{action: 'deny'}`.
# --hints--
-应正确加载 helmet.frameguard() 中间件
+helmet.frameguard() middleware should be mounted correctly
```js
(getUserInput) =>
@@ -36,7 +38,7 @@ dashedName: mitigate-the-risk-of-clickjacking-with-helmet-frameguard
);
```
-应将 helmet.frameguard() 中的 action 值设置为 'DENY'
+helmet.frameguard() 'action' should be set to 'DENY'
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/mitigate-the-risk-of-cross-site-scripting-xss-attacks-with-helmet.xssfilter.md b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/mitigate-the-risk-of-cross-site-scripting-xss-attacks-with-helmet.xssfilter.md
index e051c95dfbe..298aac04f08 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/mitigate-the-risk-of-cross-site-scripting-xss-attacks-with-helmet.xssfilter.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/mitigate-the-risk-of-cross-site-scripting-xss-attacks-with-helmet.xssfilter.md
@@ -1,6 +1,7 @@
---
id: 587d8247367417b2b2512c39
-title: 使用 helmet.xssFilter() 降低跨站点脚本（XSS）攻击的风险
+title: >-
+ Mitigate the Risk of Cross Site Scripting (XSS) Attacks with helmet.xssFilter()
challengeType: 2
forumTopicId: 301583
dashedName: mitigate-the-risk-of-cross-site-scripting-xss-attacks-with-helmet-xssfilter
@@ -8,21 +9,23 @@ dashedName: mitigate-the-risk-of-cross-site-scripting-xss-attacks-with-helmet-xs
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-infosec) 的基础上进行开发。你也可以从 [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-infosec), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/).
-跨站脚本攻击（XSS）是一种比较常见的攻击手段，通过给页面注入恶意脚本来获取用户的 session cookie 和密码等信息。
+Cross-site scripting (XSS) is a frequent type of attack where malicious scripts are injected into vulnerable pages, with the purpose of stealing sensitive data like session cookies, or passwords.
-防止这类型的攻击也非常简单：“永远不要相信用户的输入”。作为一名开发人员，你一定要对所有外部的输入进行审查，其中包括来自表单、GET 请求的 URL、POST 请求主体的内容等。审核的意思是，你需要查找并给有潜在风险的字符进行编码，比如 < 和 >
+The basic rule to lower the risk of an XSS attack is simple: “Never trust user’s input”. As a developer you should always sanitize all the input coming from the outside. This includes data coming from forms, GET query urls, and even from POST bodies. Sanitizing means that you should find and encode the characters that may be dangerous e.g. <, >.
-现代浏览器中已经存在有效的策略来应对这一类问题。通常是通过配置 HTTP 请求的 header 来实现。
+Modern browsers can help mitigating the risk by adopting better software strategies. Often these are configurable via http headers.
-HTTP header 的 X-XSS-Protection 字段就可以为我们提供最基本的保护。浏览器通过启发式过滤法则来检测有潜在风险的注入脚本，如果这个 header 字段被启用，浏览器就会改变脚本中存在风险的代码，从而使恶意代码不再生效。
+The X-XSS-Protection HTTP header is a basic protection. The browser detects a potential injected script using a heuristic filter. If the header is enabled, the browser changes the script code, neutralizing it. It still has limited support.
-但是浏览器对这个功能的支持十分有限。
+# --instructions--
+
+Use `helmet.xssFilter()` to sanitize input sent to your server.
# --hints--
-应正确加载 helmet.xssFilter() 中间件
+helmet.xssFilter() middleware should be mounted correctly
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/prevent-ie-from-opening-untrusted-html-with-helmet.ienoopen.md b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/prevent-ie-from-opening-untrusted-html-with-helmet.ienoopen.md
index b8723e47876..1a05e0ba5c3 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/prevent-ie-from-opening-untrusted-html-with-helmet.ienoopen.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/prevent-ie-from-opening-untrusted-html-with-helmet.ienoopen.md
@@ -1,6 +1,6 @@
---
id: 587d8248367417b2b2512c3b
-title: 使用 helment.ieNoOpen() 防止 IE 打开不受信任的 HTML
+title: Prevent IE from Opening Untrusted HTML with helmet.ieNoOpen()
challengeType: 2
forumTopicId: 301584
dashedName: prevent-ie-from-opening-untrusted-html-with-helmet-ienoopen
@@ -8,13 +8,17 @@ dashedName: prevent-ie-from-opening-untrusted-html-with-helmet-ienoopen
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-infosec) 的基础上进行开发。你也可以从 [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-infosec), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/).
-有些网站会下载不安全的 HTML 文件，某些版本的 IE 默认情况下还会在你网站的作用域下打开这些 HTML 文件。换句话说，这些不安全的 HTML 页面可以在你的网站做恶意行为。我们可以通过中间件来设置 header 中的 X-Download-Options 字段，让它的值为 noopen。这样就可以防止 IE 在不信任的网站下执行下载的文件。
+Some web applications will serve untrusted HTML for download. Some versions of Internet Explorer by default open those HTML files in the context of your site. This means that an untrusted HTML page could start doing bad things in the context of your pages. This middleware sets the X-Download-Options header to noopen. This will prevent IE users from executing downloads in the trusted site’s context.
+
+# --instructions--
+
+Use the `helmet.ieNoOpen()` method on your server.
# --hints--
-应正确加载 helmet.ieNoOpen() 中间件
+helmet.ieNoOpen() middleware should be mounted correctly
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/set-a-content-security-policy-with-helmet.contentsecuritypolicy.md b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/set-a-content-security-policy-with-helmet.contentsecuritypolicy.md
index 47fa32c564b..72aae680a93 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/set-a-content-security-policy-with-helmet.contentsecuritypolicy.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/set-a-content-security-policy-with-helmet.contentsecuritypolicy.md
@@ -1,6 +1,6 @@
---
id: 587d8249367417b2b2512c3f
-title: 使用 helment.contentSecurityPolicy() 设置内容安全策略
+title: Set a Content Security Policy with helmet.contentSecurityPolicy()
challengeType: 2
forumTopicId: 301585
dashedName: set-a-content-security-policy-with-helmet-contentsecuritypolicy
@@ -8,21 +8,21 @@ dashedName: set-a-content-security-policy-with-helmet-contentsecuritypolicy
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-infosec) 的基础上进行开发。你也可以从 [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/) 上克隆。
+As a reminder, this project is being built upon the following starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-infosec), or cloned from [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/).
-在这个挑战中，我们要重点讨论现代浏览器中一种有效并且能大幅度减轻安全风险和很多种类型常见攻击的安全防护。通过配置内容安全策略，你可以防止很多类型的脚本恶意注入。这会让你的应用远离 XSS 漏洞、恶意追踪、恶意 frames 和很多其他攻击。CSP 通过配置资源白名单来避免这些问题。你可以给任何一种类型的页面资源（脚本、样式文件、字体、frames、媒体文件等）做这个配置。它支持很多指令，所以网站管理员可以做细致的控制。更多详情请参考 HTML 5 Rocks 和 KeyCDN。不幸的是，一些旧的浏览器不支持 CSP。
+This challenge highlights one promising new defense that can significantly reduce the risk and impact of many type of attacks in modern browsers. By setting and configuring a Content Security Policy you can prevent the injection of anything unintended into your page. This will protect your app from XSS vulnerabilities, undesired tracking, malicious frames, and much more. CSP works by defining an allowed list of content sources which are trusted. You can configure them for each kind of resource a web page may need (scripts, stylesheets, fonts, frames, media, and so on…). There are multiple directives available, so a website owner can have a granular control. See HTML 5 Rocks, KeyCDN for more details. Unfortunately CSP is unsupported by older browser.
-默认的指令很容易受到攻击, 所以设置 defaultSrc 指令作为降级方案很重要。Helmet 同时支持 defaultSrc 和 default-src 命名规范。降级方案可以应用在大部分指令上。
+By default, directives are wide open, so it’s important to set the defaultSrc directive as a fallback. Helmet supports both defaultSrc and default-src naming styles. The fallback applies for most of the unspecified directives.
# --instructions--
-在这个练习中，我们使用 `helmet.contentSecurityPolicy()` 并配置 `defaultSrc` 为 `["self"]`（允许的资源列表必须在一个数组当中）。这样做表示只信任自己网站的域名。另外，配置 `scriptSrc` 指令可以限制脚本只能本网站域名上或者信任的域名 'trusted-cdn.com' 上下载。
+In this exercise, use `helmet.contentSecurityPolicy()`, and configure it setting the `defaultSrc directive` to `["self"]` (the list of allowed sources must be in an array), in order to trust only your website address by default. Set also the `scriptSrc` directive so that you will allow scripts to be downloaded from your website, and from the domain 'trusted-cdn.com'.
-提示: 在 `self` 关键词中, 单引号也是关键词的一部分, 所以你应该用双引号来包起它才能正常工作。
+Hint: in the `self` keyword, the single quotes are part of the keyword itself, so it needs to be enclosed in double quotes to be working.
# --hints--
-应正确加载 helmet.csp() 中间件
+helmet.csp() middleware should be mounted correctly
```js
(getUserInput) =>
@@ -36,7 +36,7 @@ dashedName: set-a-content-security-policy-with-helmet-contentsecuritypolicy
);
```
-你的 csp 配置不正确，defaultSrc 应为 ["'self'"] 并且 scriptSrc 应为 ["'self'", 'trusted-cdn.com']
+Your csp config is not correct. defaultSrc should be ["'self'"] and scriptSrc should be ["'self'", 'trusted-cdn.com']
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/understand-bcrypt-hashes.md b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/understand-bcrypt-hashes.md
index 6ad89782baf..6828dbe0f0f 100644
--- a/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/understand-bcrypt-hashes.md
+++ b/curriculum/challenges/chinese/09-information-security/information-security-with-helmetjs/understand-bcrypt-hashes.md
@@ -1,6 +1,6 @@
---
id: 58a25bcef9fc0f352b528e7c
-title: 了解 BCrypt 的哈希加密
+title: Understand BCrypt Hashes
challengeType: 2
forumTopicId: 301586
dashedName: understand-bcrypt-hashes
@@ -8,21 +8,23 @@ dashedName: understand-bcrypt-hashes
# --description--
-请注意，本项目在 [这个 Repl.it 项目](https://repl.it/github/freeCodeCamp/boilerplate-infosec) 的基础上进行开发。你也可以从 [GitHub](https://github.com/freeCodeCamp/boilerplate-infosec/) 上克隆。
+For the following challenges, you will be working with a new starter project that is different from the previous one. You can find the new starter project on [Repl.it](https://repl.it/github/freeCodeCamp/boilerplate-bcrypt), or clone it from [GitHub](https://github.com/freeCodeCamp/boilerplate-bcrypt/).
-BCrypt 哈希非常安全。哈希其实就是原数据的加密形式，结果唯一；通过把原始数据放进哈希算法然后返回固定长度的字符串实现。为了让这个过程更加复杂和安全，你还可以加你的哈希添加 *盐*。加盐其实就是给在哈希算法执行前给源数据添加随机的字符串数据，这会让破解哈希结果更加复杂
+BCrypt hashes are very secure. A hash is basically a fingerprint of the original data- always unique. This is accomplished by feeding the original data into an algorithm and returning a fixed length result. To further complicate this process and make it more secure, you can also *salt* your hash. Salting your hash involves adding random data to the original data before the hashing process which makes it even harder to crack the hash.
-BCrypt 的哈希结果会是这样 `$2a$13$ZyprE5MRw2Q3WpNOGZWGbeG7ADUre1Q8QO.uUUtcbqloU0yvzavOm`。它遵循固定的结构，前面一小节 `$2a` 说明了该哈希采用什么算法。下一部分 `$13` 定义了 *cost*。cost 就是生成此哈希结果所用到的资源数量级；它是 2^cost 的对数，可以表说单位时间内数据放入哈希算法的次数。举个例子，cost 为 10 的意思就是说你能够在一个普通的计算机上每秒钟计算出 10 个密码的哈希结果。然而，如果 cost 为 15，那计算每个哈希结果就要 3 秒钟。再举例子，如果 cost 是 31，那每次哈希运算需要话费好几天才能完成。通常 cost 为 12 的哈希运算就已经足够安全。哈希结果的最后一部分 `$ZyprE5MRw2Q3WpNOGZWGbeG7ADUre1Q8QO.uUUtcbqloU0yvzavOm` 看起来像是由随机数字、点和字母组成的字符串，但实际上它有两部分内容。前面 22 个字符是加入的盐，剩下的就是加密过的密码！
+BCrypt hashes will always looks like `$2a$13$ZyprE5MRw2Q3WpNOGZWGbeG7ADUre1Q8QO.uUUtcbqloU0yvzavOm` which does have a structure. The first small bit of data `$2a` is defining what kind of hash algorithm was used. The next portion `$13` defines the *cost*. Cost is about how much power it takes to compute the hash. It is on a logarithmic scale of 2^cost and determines how many times the data is put through the hashing algorithm. For example, at a cost of 10 you are able to hash 10 passwords a second on an average computer, however at a cost of 15 it takes 3 seconds per hash... and to take it further, at a cost of 31 it would takes multiple days to complete a hash. A cost of 12 is considered very secure at this time. The last portion of your hash `$ZyprE5MRw2Q3WpNOGZWGbeG7ADUre1Q8QO.uUUtcbqloU0yvzavOm`, looks like one large string of numbers, periods, and letters but it is actually two separate pieces of information. The first 22 characters is the salt in plain text, and the rest is the hashed password!
# --instructions--
-要开始使用 BCrypt, 只需将它添加到你的依赖列表，然后在你的服务器引入 'bcrypt'。
+To begin using BCrypt, add it as a dependency in your project and require it as 'bcrypt' in your server.
-请在完成挑战后提交你的页面。
+Add all your code for these lessons in the `server.js` file between the code we have started you off with. Do not change or delete the code we have added for you.
+
+Submit your page when you think you've got it right.
# --hints--
-应添加 BCyrpt 至依赖列表
+BCrypt should be a dependency.
```js
(getUserInput) =>
@@ -41,7 +43,7 @@ BCrypt 的哈希结果会是这样 `$2a$13$ZyprE5MRw2Q3WpNOGZWGbeG7ADUre1Q8QO.uU
);
```
-应正确引入 BCrypt
+BCrypt should be properly required.
```js
(getUserInput) =>
diff --git a/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/creating-a-tcp-client.md b/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/creating-a-tcp-client.md
index 40bfd739da5..a808410376c 100644
--- a/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/creating-a-tcp-client.md
+++ b/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/creating-a-tcp-client.md
@@ -1,5 +1,6 @@
---
id: 5ea9997bbec2e9bc47e94db0
+title: Creating a TCP Client
challengeType: 11
videoId: ugYfJNTawks
dashedName: creating-a-tcp-client
diff --git a/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/developing-a-banner-grabber.md b/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/developing-a-banner-grabber.md
index b11fc954bac..88f5f3fcfe1 100644
--- a/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/developing-a-banner-grabber.md
+++ b/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/developing-a-banner-grabber.md
@@ -1,5 +1,6 @@
---
id: 5ea9997bbec2e9bc47e94db3
+title: Developing a Banner Grabber
challengeType: 11
videoId: CeGW761BIsA
dashedName: developing-a-banner-grabber
diff --git a/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/developing-a-port-scanner.md b/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/developing-a-port-scanner.md
index cc067a6a240..a35002583eb 100644
--- a/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/developing-a-port-scanner.md
+++ b/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/developing-a-port-scanner.md
@@ -1,5 +1,6 @@
---
id: 5ea9997bbec2e9bc47e94db4
+title: Developing a Port Scanner
challengeType: 11
videoId: z_qkqZS7KZ4
dashedName: developing-a-port-scanner
diff --git a/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/developing-an-nmap-scanner-part-1.md b/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/developing-an-nmap-scanner-part-1.md
index 1cdd896f68a..8f26492699b 100644
--- a/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/developing-an-nmap-scanner-part-1.md
+++ b/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/developing-an-nmap-scanner-part-1.md
@@ -1,5 +1,6 @@
---
id: 5ea9997bbec2e9bc47e94db1
+title: Developing an Nmap Scanner part 1
challengeType: 11
videoId: jYk9XaGoAnk
dashedName: developing-an-nmap-scanner-part-1
diff --git a/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/developing-an-nmap-scanner-part-2.md b/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/developing-an-nmap-scanner-part-2.md
index 207a8247bda..ab3a65968f7 100644
--- a/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/developing-an-nmap-scanner-part-2.md
+++ b/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/developing-an-nmap-scanner-part-2.md
@@ -1,5 +1,6 @@
---
id: 5ea9997bbec2e9bc47e94db2
+title: Developing an Nmap Scanner part 2
challengeType: 11
videoId: a98PscnUsTg
dashedName: developing-an-nmap-scanner-part-2
diff --git a/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/introduction-and-setup.md b/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/introduction-and-setup.md
index b901a00dc98..c243ea29541 100644
--- a/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/introduction-and-setup.md
+++ b/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/introduction-and-setup.md
@@ -1,5 +1,6 @@
---
id: 5ea9997bbec2e9bc47e94dae
+title: Introduction and Setup
challengeType: 11
videoId: XeQ7ZKtb998
dashedName: introduction-and-setup
diff --git a/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/understanding-sockets-and-creating-a-tcp-server.md b/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/understanding-sockets-and-creating-a-tcp-server.md
index 7e8669782a1..7a0f528118f 100644
--- a/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/understanding-sockets-and-creating-a-tcp-server.md
+++ b/curriculum/challenges/chinese/09-information-security/python-for-penetration-testing/understanding-sockets-and-creating-a-tcp-server.md
@@ -1,5 +1,6 @@
---
id: 5ea9997bbec2e9bc47e94daf
+title: Understanding Sockets and Creating a TCP Server
challengeType: 11
videoId: F1QI9tNuDQg
dashedName: understanding-sockets-and-creating-a-tcp-server
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/find-the-symmetric-difference.md b/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/find-the-symmetric-difference.md
index f3ae8e6fe19..f5fa4e3e8c3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/find-the-symmetric-difference.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/find-the-symmetric-difference.md
@@ -1,78 +1,84 @@
---
id: a3f503de51cf954ede28891d
-title: 找到对称差异
+title: Find the Symmetric Difference
challengeType: 5
-videoUrl: ''
+forumTopicId: 301611
dashedName: find-the-symmetric-difference
---
# --description--
-创建一个带有两个或更多数组的函数，并返回所提供数组的<dfn>对称差</dfn> （ `△`或`⊕` ）数组。给定两个集合（例如集合`A = {1, 2, 3}`并且集合`B = {2, 3, 4}` ），两个集合的数学术语“对称差异”是在任一集合中的元素集合。两组，但两者都没有（ `A △ B = C = {1, 4}` ）。对于你所采取的每一个额外的对称差异（比如在集合`D = {2, 3}` ），你应该得到具有两个集合中的任何一个但不是两个集合的元素的集合（ `C △ D = {1, 4} △ {2, 3} = {1, 2, 3, 4}` ）。结果数组必须仅包含唯一值（ *不重复* ）。如果卡住，请记得使用[Read-Search-Ask](https://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck-coding/19514) 。尝试配对程序。编写自己的代码。
+The mathematical term <dfn>symmetric difference</dfn> (`△` or `⊕`) of two sets is the set of elements which are in either of the two sets but not in both. For example, for sets `A = {1, 2, 3}` and `B = {2, 3, 4}`, `A △ B = {1, 4}`.
+
+Symmetric difference is a binary operation, which means it operates on only two elements. So to evaluate an expression involving symmetric differences among *three* elements (`A △ B △ C`), you must complete one operation at a time. Thus, for sets `A` and `B` above, and `C = {2, 3}`, `A △ B △ C = (A △ B) △ C = {1, 4} △ {2, 3} = {1, 2, 3, 4}`.
+
+# --instructions--
+
+Create a function that takes two or more arrays and returns an array of their symmetric difference. The returned array must contain only unique values (*no duplicates*).
# --hints--
-`sym([1, 2, 3], [5, 2, 1, 4])`应返回`[3, 4, 5]` 。
+`sym([1, 2, 3], [5, 2, 1, 4])` should return `[3, 4, 5]`.
```js
assert.sameMembers(sym([1, 2, 3], [5, 2, 1, 4]), [3, 4, 5]);
```
-`sym([1, 2, 3], [5, 2, 1, 4])`应仅包含三个元素。
+`sym([1, 2, 3], [5, 2, 1, 4])` should contain only three elements.
```js
assert.equal(sym([1, 2, 3], [5, 2, 1, 4]).length, 3);
```
-`sym([1, 2, 3, 3], [5, 2, 1, 4])`应该返回`[3, 4, 5]` 。
+`sym([1, 2, 3, 3], [5, 2, 1, 4])` should return `[3, 4, 5]`.
```js
assert.sameMembers(sym([1, 2, 3, 3], [5, 2, 1, 4]), [3, 4, 5]);
```
-`sym([1, 2, 3, 3], [5, 2, 1, 4])`应仅包含三个元素。
+`sym([1, 2, 3, 3], [5, 2, 1, 4])` should contain only three elements.
```js
assert.equal(sym([1, 2, 3, 3], [5, 2, 1, 4]).length, 3);
```
-`sym([1, 2, 3], [5, 2, 1, 4, 5])`应返回`[3, 4, 5]` 。
+`sym([1, 2, 3], [5, 2, 1, 4, 5])` should return `[3, 4, 5]`.
```js
assert.sameMembers(sym([1, 2, 3], [5, 2, 1, 4, 5]), [3, 4, 5]);
```
-`sym([1, 2, 3], [5, 2, 1, 4, 5])`应仅包含三个元素。
+`sym([1, 2, 3], [5, 2, 1, 4, 5])` should contain only three elements.
```js
assert.equal(sym([1, 2, 3], [5, 2, 1, 4, 5]).length, 3);
```
-`sym([1, 2, 5], [2, 3, 5], [3, 4, 5])`应该返回`[1, 4, 5]`
+`sym([1, 2, 5], [2, 3, 5], [3, 4, 5])` should return `[1, 4, 5]`
```js
assert.sameMembers(sym([1, 2, 5], [2, 3, 5], [3, 4, 5]), [1, 4, 5]);
```
-`sym([1, 2, 5], [2, 3, 5], [3, 4, 5])`应仅包含三个元素。
+`sym([1, 2, 5], [2, 3, 5], [3, 4, 5])` should contain only three elements.
```js
assert.equal(sym([1, 2, 5], [2, 3, 5], [3, 4, 5]).length, 3);
```
-`sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5])`应该返回`[1, 4, 5]` 。
+`sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5])` should return `[1, 4, 5]`.
```js
assert.sameMembers(sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5]), [1, 4, 5]);
```
-`sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5])`应仅包含三个元素。
+`sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5])` should contain only three elements.
```js
assert.equal(sym([1, 1, 2, 5], [2, 2, 3, 5], [3, 4, 5, 5]).length, 3);
```
-`sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3])`应该返回`[2, 3, 4, 6, 7]` `sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3])` `[2, 3, 4, 6, 7]` 。
+`sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3])` should return `[2, 3, 4, 6, 7]`.
```js
assert.sameMembers(
@@ -81,7 +87,7 @@ assert.sameMembers(
);
```
-`sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3])`应仅包含五个元素。
+`sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3])` should contain only five elements.
```js
assert.equal(
@@ -90,7 +96,7 @@ assert.equal(
);
```
-`sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3], [5, 3, 9, 8], [1])`应该返回`[1, 2, 4, 5, 6, 7, 8, 9]` 。
+`sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3], [5, 3, 9, 8], [1])` should return `[1, 2, 4, 5, 6, 7, 8, 9]`.
```js
assert.sameMembers(
@@ -106,7 +112,7 @@ assert.sameMembers(
);
```
-`sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3], [5, 3, 9, 8], [1])`应该只包含八个元素。
+`sym([3, 3, 3, 2, 5], [2, 1, 5, 7], [3, 4, 6, 6], [1, 2, 3], [5, 3, 9, 8], [1])` should contain only eight elements.
```js
assert.equal(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-bubble-sort.md b/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-bubble-sort.md
index 9651125779f..6cf8150f25e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-bubble-sort.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-bubble-sort.md
@@ -1,25 +1,30 @@
---
id: 8d5123c8c441eddfaeb5bdef
-title: 实施冒泡排序
+title: Implement Bubble Sort
challengeType: 1
-videoUrl: ''
+forumTopicId: 301612
dashedName: implement-bubble-sort
---
# --description--
-这是排序算法的几个挑战中的第一个。给定一组未排序的项目，我们希望能够返回已排序的数组。我们将看到几种不同的方法来实现这一点，并学习这些不同方法之间的一些权衡。虽然大多数现代语言都有这样的操作的内置排序方法，但了解一些常见的基本方法并了解如何实现它们仍然很重要。在这里我们将看到冒泡排序。冒泡排序方法从未排序数组的开头开始，并向末端“冒泡”未排序的值，遍历数组直到它完全排序。它通过比较相邻的项目并在它们出现故障时交换它们来完成此操作。该方法继续循环遍历数组，直到没有发生交换，此时数组被排序。该方法需要通过阵列进行多次迭代，并且对于平均和最差情况，具有二次时间复杂度。虽然简单，但在大多数情况下通常都是不切实际的。 **说明：** 编写一个函数`bubbleSort` ，它将整数数组作为输入，并按从最小到最大的排序顺序返回这些整数的数组。 **注意：**
-我们从幕后调用这个功能;我们使用的测试数组在编辑器中被注释掉了。尝试记录`array`以查看您的排序算法！
+This is the first of several challenges on sorting algorithms. Given an array of unsorted items, we want to be able to return a sorted array. We will see several different methods to do this and learn some tradeoffs between these different approaches. While most modern languages have built-in sorting methods for operations like this, it is still important to understand some of the common basic approaches and learn how they can be implemented.
+
+Here we will see bubble sort. The bubble sort method starts at the beginning of an unsorted array and 'bubbles up' unsorted values towards the end, iterating through the array until it is completely sorted. It does this by comparing adjacent items and swapping them if they are out of order. The method continues looping through the array until no swaps occur at which point the array is sorted.
+
+This method requires multiple iterations through the array and for average and worst cases has quadratic time complexity. While simple, it is usually impractical in most situations.
+
+**Instructions:** Write a function `bubbleSort` which takes an array of integers as input and returns an array of these integers in sorted order from least to greatest.
# --hints--
-`bubbleSort`是一个功能。
+`bubbleSort` should be a function.
```js
assert(typeof bubbleSort == 'function');
```
-`bubbleSort`返回一个已排序的数组（从最小到最大）。
+`bubbleSort` should return a sorted array (least to greatest).
```js
assert(
@@ -47,7 +52,7 @@ assert(
);
```
-`bubbleSort`返回一个除订单外没有变化的数组。
+`bubbleSort` should return an array that is unchanged except for order.
```js
assert.sameMembers(
@@ -74,10 +79,10 @@ assert.sameMembers(
);
```
-`bubbleSort`不应使用内置的`.sort()`方法。
+`bubbleSort` should not use the built-in `.sort()` method.
```js
-assert.strictEqual(code.search(/\.sort\(/), -1);
+assert(isBuiltInSortUsed());
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-insertion-sort.md b/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-insertion-sort.md
index ddc447f9c07..ce4a857bea2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-insertion-sort.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-insertion-sort.md
@@ -1,25 +1,26 @@
---
id: 587d8259367417b2b2512c86
-title: 实现插入排序
+title: Implement Insertion Sort
challengeType: 1
-videoUrl: ''
+forumTopicId: 301613
dashedName: implement-insertion-sort
---
# --description--
-我们将看到的下一个排序方法是插入排序。此方法通过在列表的开头构建排序数组来工作。它以第一个元素开始排序数组。然后它检查下一个元素并将其向后交换到已排序的数组，直到它处于排序位置。它继续遍历列表并将新项目向后交换到已排序的部分，直到它到达结尾。该算法在平均和最差情况下具有二次时间复杂度。 **说明：** 编写一个函数`insertionSort` ，它将一个整数数组作为输入，并按照从最小到最大的排序顺序返回这些整数的数组。 **注意：**
-我们从幕后调用这个功能;我们使用的测试数组在编辑器中被注释掉了。尝试记录`array`以查看您的排序算法！
+The next sorting method we'll look at is insertion sort. This method works by building up a sorted array at the beginning of the list. It begins the sorted array with the first element. Then it inspects the next element and swaps it backwards into the sorted array until it is in sorted position. It continues iterating through the list and swapping new items backwards into the sorted portion until it reaches the end. This algorithm has quadratic time complexity in the average and worst cases.
+
+**Instructions:** Write a function `insertionSort` which takes an array of integers as input and returns an array of these integers in sorted order from least to greatest.
# --hints--
-`insertionSort`是一个函数。
+`insertionSort` should be a function.
```js
assert(typeof insertionSort == 'function');
```
-`insertionSort`返回一个排序数组（从最小到最大）。
+`insertionSort` should return a sorted array (least to greatest).
```js
assert(
@@ -47,7 +48,7 @@ assert(
);
```
-`insertionSort`返回一个除订单外没有变化的数组。
+`insertionSort` should return an array that is unchanged except for order.
```js
assert.sameMembers(
@@ -74,10 +75,10 @@ assert.sameMembers(
);
```
-`insertionSort`不应使用内置的`.sort()`方法。
+`insertionSort` should not use the built-in `.sort()` method.
```js
-assert.strictEqual(code.search(/\.sort\(/), -1);
+assert(isBuiltInSortUsed());
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-merge-sort.md b/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-merge-sort.md
index dcc50c40b0f..ab1dca43817 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-merge-sort.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-merge-sort.md
@@ -1,25 +1,34 @@
---
id: 587d825c367417b2b2512c8f
-title: 实现合并排序
+title: Implement Merge Sort
challengeType: 1
-videoUrl: ''
+forumTopicId: 301614
dashedName: implement-merge-sort
---
# --description--
-另一种非常常见的中间排序算法是合并排序。像快速排序一样，合并排序也使用分而治之的递归方法对数组进行排序。它利用了这样一个事实：只要每个数组首先排序，就可以相对容易地对两个数组进行排序。但是我们只从一个数组作为输入开始，那么我们如何从中获得两个排序的数组呢？好吧，我们可以递归地将原始输入分成两部分，直到我们到达具有一个项目的数组的基本情况。单项数组是自然排序的，因此我们可以开始组合。这个组合将展开拆分原始数组的递归调用，最终生成所有元素的最终排序数组。然后，合并排序的步骤是： **1）** 递归地将输入数组拆分为一半，直到产生仅具有一个元素的子数组。 **2）将** 每个排序的子数组合并在一起以产生最终的排序数组。合并排序是一种有效的排序方法，时间复杂度为*O（nlog（n））* 。该算法很受欢迎，因为它性能高且易于实现。顺便说一句，这将是我们在此处介绍的最后一种排序算法。但是，稍后在关于树数据结构的部分中，我们将描述堆排序，这是另一种在其实现中需要二进制堆的有效排序方法。 **说明：** 编写一个函数`mergeSort` ，它以整数数组作为输入，并按从最小到最大的排序顺序返回这些整数的数组。实现这一点的一个好方法是编写一个函数，例如`merge` ，它负责合并两个已排序的数组，另一个函数，例如`mergeSort` ，它负责递归，生成单项数组以提供给merge。祝你好运！ **注意：**
-我们从幕后调用这个功能;我们使用的测试数组在编辑器中被注释掉了。尝试记录`array`以查看您的排序算法！
+Another common intermediate sorting algorithm is merge sort. Like quick sort, merge sort also uses a divide-and-conquer, recursive methodology to sort an array. It takes advantage of the fact that it is relatively easy to sort two arrays as long as each is sorted in the first place. But we'll start with only one array as input, so how do we get to two sorted arrays from that? Well, we can recursively divide the original input in two until we reach the base case of an array with one item. A single-item array is naturally sorted, so then we can start combining. This combination will unwind the recursive calls that split the original array, eventually producing a final sorted array of all the elements. The steps of merge sort, then, are:
+
+**1)** Recursively split the input array in half until a sub-array with only one element is produced.
+
+**2)** Merge each sorted sub-array together to produce the final sorted array.
+
+Merge sort is an efficient sorting method, with time complexity of *O(nlog(n))*. This algorithm is popular because it is performant and relatively easy to implement.
+
+As an aside, this will be the last sorting algorithm we cover here. However, later in the section on tree data structures we will describe heap sort, another efficient sorting method that requires a binary heap in its implementation.
+
+**Instructions:** Write a function `mergeSort` which takes an array of integers as input and returns an array of these integers in sorted order from least to greatest. A good way to implement this is to write one function, for instance `merge`, which is responsible for merging two sorted arrays, and another function, for instance `mergeSort`, which is responsible for the recursion that produces single-item arrays to feed into merge. Good luck!
# --hints--
-`mergeSort`是一个函数。
+`mergeSort` should be a function.
```js
assert(typeof mergeSort == 'function');
```
-`mergeSort`返回一个排序数组（从最小到最大）。
+`mergeSort` should return a sorted array (least to greatest).
```js
assert(
@@ -47,7 +56,7 @@ assert(
);
```
-`mergeSort`返回一个除订单外没有变化的数组。
+`mergeSort` should return an array that is unchanged except for order.
```js
assert.sameMembers(
@@ -74,10 +83,10 @@ assert.sameMembers(
);
```
-`mergeSort`不应使用内置的`.sort()`方法。
+`mergeSort` should not use the built-in `.sort()` method.
```js
-assert.strictEqual(code.search(/\.sort\(/), -1);
+assert(isBuiltInSortUsed());
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-quick-sort.md b/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-quick-sort.md
index 699b639048b..53ac770885b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-quick-sort.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-quick-sort.md
@@ -1,25 +1,28 @@
---
id: 587d825a367417b2b2512c89
-title: 实施快速排序
+title: Implement Quick Sort
challengeType: 1
-videoUrl: ''
+forumTopicId: 301615
dashedName: implement-quick-sort
---
# --description--
-在这里，我们将继续讨论中间排序算法：快速排序。快速排序是对数组进行排序的一种有效的，递归的分而治之的方法。在此方法中，在原始数组中选择了一个数据透视值。然后将该数组分成两个小于和大于数值的子数组。然后，我们在两个子阵列上结合递归调用快速排序算法的结果。这一直持续到达到空或单项数组的基本情况，我们返回。递归调用的展开将返回已排序的数组。快速排序是一种非常有效的排序方法，平均提供*O（nlog（n））*性能。它也相对容易实现。这些属性使其成为一种流行且有用的排序方法。 **说明：** 编写一个函数`quickSort` ，它将整数数组作为输入，并按从最小到最大的排序顺序返回这些整数的数组。虽然枢轴值的选择很重要，但任何支点都可以用于我们的目的。为简单起见，可以使用第一个或最后一个元素。 **注意：**
-我们从幕后调用这个功能;我们使用的测试数组在编辑器中被注释掉了。尝试记录`array`以查看您的排序算法！
+Here we will move on to an intermediate sorting algorithm: quick sort. Quick sort is an efficient, recursive divide-and-conquer approach to sorting an array. In this method, a pivot value is chosen in the original array. The array is then partitioned into two subarrays of values less than and greater than the pivot value. We then combine the result of recursively calling the quick sort algorithm on both sub-arrays. This continues until the base case of an empty or single-item array is reached, which we return. The unwinding of the recursive calls return us the sorted array.
+
+Quick sort is a very efficient sorting method, providing *O(nlog(n))* performance on average. It is also relatively easy to implement. These attributes make it a popular and useful sorting method.
+
+**Instructions:** Write a function `quickSort` which takes an array of integers as input and returns an array of these integers in sorted order from least to greatest. While the choice of the pivot value is important, any pivot will do for our purposes here. For simplicity, the first or last element could be used.
# --hints--
-`quickSort`是一个功能。
+`quickSort` should be a function.
```js
assert(typeof quickSort == 'function');
```
-`quickSort`返回一个排序数组（从最小到最大）。
+`quickSort` should return a sorted array (least to greatest).
```js
assert(
@@ -47,7 +50,7 @@ assert(
);
```
-`quickSort`返回一个除订单外没有变化的数组。
+`quickSort([1,4,2,8,345,123,43,32,5643,63,123,43,2,55,1,234,92])` should return an array that is unchanged except for order.
```js
assert.sameMembers(
@@ -74,10 +77,10 @@ assert.sameMembers(
);
```
-`quickSort`不应使用内置的`.sort()`方法。
+`quickSort` should not use the built-in `.sort()` method.
```js
-assert.strictEqual(code.search(/\.sort\(/), -1);
+assert(isBuiltInSortUsed());
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-selection-sort.md b/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-selection-sort.md
index 1a19a8447ec..1a1b0820904 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-selection-sort.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/implement-selection-sort.md
@@ -1,25 +1,26 @@
---
id: 587d8259367417b2b2512c85
-title: 实施选择排序
+title: Implement Selection Sort
challengeType: 1
-videoUrl: ''
+forumTopicId: 301616
dashedName: implement-selection-sort
---
# --description--
-这里我们将实现选择排序。选择排序的工作原理是选择列表中的最小值并使用列表中的第一个值进行交换。然后它从第二个位置开始，选择剩余列表中的最小值，并将其与第二个元素交换。它继续遍历列表并交换元素，直到它到达列表的末尾。现在列表已排序。在所有情况下，选择排序都具有二次时间复杂度。 **说明** ：编写一个函数`selectionSort` ，它将一个整数数组作为输入，并按照从最小到最大的排序顺序返回这些整数的数组。 **注意：**
-我们从幕后调用这个功能;我们使用的测试数组在编辑器中被注释掉了。尝试记录`array`以查看您的排序算法！
+Here we will implement selection sort. Selection sort works by selecting the minimum value in a list and swapping it with the first value in the list. It then starts at the second position, selects the smallest value in the remaining list, and swaps it with the second element. It continues iterating through the list and swapping elements until it reaches the end of the list. Now the list is sorted. Selection sort has quadratic time complexity in all cases.
+
+**Instructions**: Write a function `selectionSort` which takes an array of integers as input and returns an array of these integers in sorted order from least to greatest.
# --hints--
-`selectionSort`是一个函数。
+`selectionSort` should be a function.
```js
assert(typeof selectionSort == 'function');
```
-`selectionSort`返回一个排序数组（从最小到最大）。
+`selectionSort` should return a sorted array (least to greatest).
```js
assert(
@@ -47,7 +48,7 @@ assert(
);
```
-`selectionSort`返回一个除订单外没有变化的数组。
+`selectionSort` should return an array that is unchanged except for order.
```js
assert.sameMembers(
@@ -74,10 +75,10 @@ assert.sameMembers(
);
```
-`selectionSort`不应使用内置的`.sort()`方法。
+`selectionSort` should not use the built-in `.sort()` method.
```js
-assert.strictEqual(code.search(/\.sort\(/), -1);
+assert(isBuiltInSortUsed());
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/inventory-update.md b/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/inventory-update.md
index e17a27ee74f..e21c765a5f9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/inventory-update.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/inventory-update.md
@@ -1,18 +1,18 @@
---
id: a56138aff60341a09ed6c480
-title: 库存更新
+title: Inventory Update
challengeType: 5
-videoUrl: ''
+forumTopicId: 16019
dashedName: inventory-update
---
# --description--
-比较并更新存储在2D阵列中的库存与新交付的第二个2D阵列。更新当前现有库存物料数量（在`arr1` ）。如果找不到商品，请将新商品和数量添加到库存数组中。返回的库存数组应按项目的字母顺序排列。如果卡住，请记得使用[Read-Search-Ask](https://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck-coding/19514) 。尝试配对程序。编写自己的代码。
+Compare and update the inventory stored in a 2D array against a second 2D array of a fresh delivery. Update the current existing inventory item quantities (in `arr1`). If an item cannot be found, add the new item and quantity into the inventory array. The returned inventory array should be in alphabetical order by item.
# --hints--
-函数`updateInventory`应该返回一个数组。
+The function `updateInventory` should return an array.
```js
assert.isArray(
@@ -33,7 +33,7 @@ assert.isArray(
);
```
-`updateInventory([[21, "Bowling Ball"], [2, "Dirty Sock"], [1, "Hair Pin"], [5, "Microphone"]], [[2, "Hair Pin"], [3, "Half-Eaten Apple"], [67, "Bowling Ball"], [7, "Toothpaste"]])`应该返回一个长度为6的数组。
+`updateInventory([[21, "Bowling Ball"], [2, "Dirty Sock"], [1, "Hair Pin"], [5, "Microphone"]], [[2, "Hair Pin"], [3, "Half-Eaten Apple"], [67, "Bowling Ball"], [7, "Toothpaste"]])` should return an array with a length of 6.
```js
assert.equal(
@@ -55,7 +55,7 @@ assert.equal(
);
```
-`updateInventory([[21, "Bowling Ball"], [2, "Dirty Sock"], [1, "Hair Pin"], [5, "Microphone"]], [[2, "Hair Pin"], [3, "Half-Eaten Apple"], [67, "Bowling Ball"], [7, "Toothpaste"]])`应返回`[[88, "Bowling Ball"], [2, "Dirty Sock"], [3, "Hair Pin"], [3, "Half-Eaten Apple"], [5, "Microphone"], [7, "Toothpaste"]]` 。
+`updateInventory([[21, "Bowling Ball"], [2, "Dirty Sock"], [1, "Hair Pin"], [5, "Microphone"]], [[2, "Hair Pin"], [3, "Half-Eaten Apple"], [67, "Bowling Ball"], [7, "Toothpaste"]])` should return `[[88, "Bowling Ball"], [2, "Dirty Sock"], [3, "Hair Pin"], [3, "Half-Eaten Apple"], [5, "Microphone"], [7, "Toothpaste"]]`.
```js
assert.deepEqual(
@@ -84,7 +84,7 @@ assert.deepEqual(
);
```
-`updateInventory([[21, "Bowling Ball"], [2, "Dirty Sock"], [1, "Hair Pin"], [5, "Microphone"]], [])`应该返回`[[21, "Bowling Ball"], [2, "Dirty Sock"], [1, "Hair Pin"], [5, "Microphone"]]` 。
+`updateInventory([[21, "Bowling Ball"], [2, "Dirty Sock"], [1, "Hair Pin"], [5, "Microphone"]], [])` should return `[[21, "Bowling Ball"], [2, "Dirty Sock"], [1, "Hair Pin"], [5, "Microphone"]]`.
```js
assert.deepEqual(
@@ -106,7 +106,7 @@ assert.deepEqual(
);
```
-`updateInventory([], [[2, "Hair Pin"], [3, "Half-Eaten Apple"], [67, "Bowling Ball"], [7, "Toothpaste"]])`应该返回`[[67, "Bowling Ball"], [2, "Hair Pin"], [3, "Half-Eaten Apple"], [7, "Toothpaste"]]` 。
+`updateInventory([], [[2, "Hair Pin"], [3, "Half-Eaten Apple"], [67, "Bowling Ball"], [7, "Toothpaste"]])` should return `[[67, "Bowling Ball"], [2, "Hair Pin"], [3, "Half-Eaten Apple"], [7, "Toothpaste"]]`.
```js
assert.deepEqual(
@@ -128,7 +128,7 @@ assert.deepEqual(
);
```
-`updateInventory([[0, "Bowling Ball"], [0, "Dirty Sock"], [0, "Hair Pin"], [0, "Microphone"]], [[1, "Hair Pin"], [1, "Half-Eaten Apple"], [1, "Bowling Ball"], [1, "Toothpaste"]])`应返回`[[1, "Bowling Ball"], [0, "Dirty Sock"], [1, "Hair Pin"], [1, "Half-Eaten Apple"], [0, "Microphone"], [1, "Toothpaste"]]` 。
+`updateInventory([[0, "Bowling Ball"], [0, "Dirty Sock"], [0, "Hair Pin"], [0, "Microphone"]], [[1, "Hair Pin"], [1, "Half-Eaten Apple"], [1, "Bowling Ball"], [1, "Toothpaste"]])` should return `[[1, "Bowling Ball"], [0, "Dirty Sock"], [1, "Hair Pin"], [1, "Half-Eaten Apple"], [0, "Microphone"], [1, "Toothpaste"]]`.
```js
assert.deepEqual(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/no-repeats-please.md b/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/no-repeats-please.md
index 6ecd1e43190..7fc964e03fc 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/no-repeats-please.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/no-repeats-please.md
@@ -1,72 +1,74 @@
---
id: a7bf700cd123b9a54eef01d5
-title: 请不要重复
+title: No Repeats Please
challengeType: 5
-videoUrl: ''
+forumTopicId: 16037
dashedName: no-repeats-please
---
# --description--
-返回没有重复连续字母的提供字符串的总排列数。假设提供的字符串中的所有字符都是唯一的。例如， `aab`应该返回2，因为它总共有6个排列（ `aab` ， `aab` ， `aba` ， `aba` ， `baa` ， `baa` ），但只有2个（ `aba`和`aba` ）没有相同的字母（在这种情况下为`a` ）重复。如果卡住，请记得使用[Read-Search-Ask](https://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck-coding/19514) 。尝试配对程序。编写自己的代码。
+Return the number of total permutations of the provided string that don't have repeated consecutive letters. Assume that all characters in the provided string are each unique.
+
+For example, `aab` should return 2 because it has 6 total permutations (`aab`, `aab`, `aba`, `aba`, `baa`, `baa`), but only 2 of them (`aba` and `aba`) don't have the same letter (in this case `a`) repeating.
# --hints--
-`permAlone("aab")`应返回一个数字。
+`permAlone("aab")` should return a number.
```js
assert.isNumber(permAlone('aab'));
```
-`permAlone("aab")`应返回2。
+`permAlone("aab")` should return 2.
```js
assert.strictEqual(permAlone('aab'), 2);
```
-`permAlone("aaa")`应该返回0。
+`permAlone("aaa")` should return 0.
```js
assert.strictEqual(permAlone('aaa'), 0);
```
-`permAlone("aabb")`应该返回8。
+`permAlone("aabb")` should return 8.
```js
assert.strictEqual(permAlone('aabb'), 8);
```
-`permAlone("abcdefa")`应返回3600。
+`permAlone("abcdefa")` should return 3600.
```js
assert.strictEqual(permAlone('abcdefa'), 3600);
```
-`permAlone("abfdefa")`应返回2640。
+`permAlone("abfdefa")` should return 2640.
```js
assert.strictEqual(permAlone('abfdefa'), 2640);
```
-`permAlone("zzzzzzzz")`应该返回0。
+`permAlone("zzzzzzzz")` should return 0.
```js
assert.strictEqual(permAlone('zzzzzzzz'), 0);
```
-`permAlone("a")`应返回1。
+`permAlone("a")` should return 1.
```js
assert.strictEqual(permAlone('a'), 1);
```
-`permAlone("aaab")`应该返回0。
+`permAlone("aaab")` should return 0.
```js
assert.strictEqual(permAlone('aaab'), 0);
```
-`permAlone("aaabb")`应该返回12。
+`permAlone("aaabb")` should return 12.
```js
assert.strictEqual(permAlone('aaabb'), 12);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/pairwise.md b/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/pairwise.md
index b07c76b95af..215b2fda0aa 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/pairwise.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/algorithms/pairwise.md
@@ -1,50 +1,64 @@
---
id: a3f503de51cfab748ff001aa
-title: 成对
+title: Pairwise
challengeType: 5
-videoUrl: ''
+forumTopicId: 301617
dashedName: pairwise
---
# --description--
-给定一个数组`arr` ，找到其总和等于第二个参数`arg`元素对，并返回它们的索引之和。您可以使用具有相同数字元素但索引不同的多个对。每对应使用尽可能低的可用指数。一旦元素被使用，它就不能被重用来与另一个元素配对。例如， `pairwise([1, 1, 2], 3)`使用indice 0处的1而不是indice 1处的1创建一对`[2, 1]` ，因为0 + 2 <1 + 2。例如， `pairwise([7, 9, 11, 13, 15], 20)`返回`6` 。总和为20的对是`[7, 13]`和`[9, 11]` 。然后我们可以用它们的索引和值写出数组。
+Given an array `arr`, find element pairs whose sum equal the second argument `arg` and return the sum of their indices.
-| **指数** | 0 | 1 | 2 | 3 | 4 |
-| ------ | - | - | -- | -- | -- |
-| 值 | 7 | 9 | 11 | 13 | 15 |
+You may use multiple pairs that have the same numeric elements but different indices. Each pair should use the lowest possible available indices. Once an element has been used it cannot be reused to pair with another element. For instance, `pairwise([1, 1, 2], 3)` creates a pair `[2, 1]` using the 1 at index 0 rather than the 1 at index 1, because 0+2 < 1+2.
-下面我们将采用相应的索引并添加它们。 7 + 13 = 20→指数0 + 3 = 3
-9 + 11 = 20→指数1 + 2 = 3
-3 + 3 = 6→返回`6`如果卡住，请记住使用[Read-Search-Ask](https://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck-coding/19514) 。尝试配对程序。编写自己的代码。
+For example `pairwise([7, 9, 11, 13, 15], 20)` returns `6`. The pairs that sum to 20 are `[7, 13]` and `[9, 11]`. We can then write out the array with their indices and values.
+
+<div style='margin-left: 2em;'>
+
+| Index | 0 | 1 | 2 | 3 | 4 |
+| ----- | - | - | -- | -- | -- |
+| Value | 7 | 9 | 11 | 13 | 15 |
+
+</div>
+
+Below we'll take their corresponding indices and add them.
+
+<div style='margin-left: 2em;'>
+
+7 + 13 = 20 → Indices 0 + 3 = 3
+9 + 11 = 20 → Indices 1 + 2 = 3
+3 + 3 = 6 → Return `6`
+
+</div>
# --hints--
-`pairwise([1, 4, 2, 3, 0, 5], 7)`应该返回11。
+`pairwise([1, 4, 2, 3, 0, 5], 7)` should return 11.
```js
assert.deepEqual(pairwise([1, 4, 2, 3, 0, 5], 7), 11);
```
-`pairwise([1, 3, 2, 4], 4)`应该返回1。
+`pairwise([1, 3, 2, 4], 4)` should return 1.
```js
assert.deepEqual(pairwise([1, 3, 2, 4], 4), 1);
```
-`pairwise([1, 1, 1], 2)`应该返回1。
+`pairwise([1, 1, 1], 2)` should return 1.
```js
assert.deepEqual(pairwise([1, 1, 1], 2), 1);
```
-`pairwise([0, 0, 0, 0, 1, 1], 1)`应返回10。
+`pairwise([0, 0, 0, 0, 1, 1], 1)` should return 10.
```js
assert.deepEqual(pairwise([0, 0, 0, 0, 1, 1], 1), 10);
```
-`pairwise([], 100)`应该返回0。
+`pairwise([], 100)` should return 0.
```js
assert.deepEqual(pairwise([], 100), 0);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/add-a-new-element-to-a-binary-search-tree.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/add-a-new-element-to-a-binary-search-tree.md
index 926e376ab32..77f653d0f40 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/add-a-new-element-to-a-binary-search-tree.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/add-a-new-element-to-a-binary-search-tree.md
@@ -1,18 +1,34 @@
---
id: 587d8257367417b2b2512c7b
-title: 将新元素添加到二叉搜索树
+title: Add a New Element to a Binary Search Tree
challengeType: 1
-videoUrl: ''
+forumTopicId: 301618
dashedName: add-a-new-element-to-a-binary-search-tree
---
# --description--
-现在我们已经了解了基础知识，让我们编写一个更复杂的方法。在此挑战中，我们将创建一个向二叉搜索树添加新值的方法。该方法应该被称为`add` ，它应该接受一个整数值来添加到树中。注意保持二叉搜索树的不变量：每个左子项中的值应小于或等于父值，并且每个右子项中的值应大于或等于父值。在这里，让我们这样做，以便我们的树不能容纳重复的值。如果我们尝试添加已存在的值，则该方法应返回`null` 。否则，如果添加成功，则应返回`undefined` 。提示：树是自然递归的数据结构！
+This series of challenges will introduce the tree data structure. Trees are an important and versatile data structure in computer science. Of course, their name comes from the fact that when visualized they look much like the trees we are familiar with in the natural world. A tree data structure begins with one node, typically referred to as the root, and from here branches out to additional nodes, each of which may have more child nodes, and so on and so forth. The data structure is usually visualized with the root node at the top; you can think of it as a natural tree flipped upside down.
+
+First, let's describe some common terminology we will encounter with trees. The root node is the top of the tree. Data points in the tree are called nodes. Nodes with branches leading to other nodes are referred to as the parent of the node the branch leads to (the child). Other more complicated familial terms apply as you might expect. A subtree refers to all the descendants of a particular node, branches may be referred to as edges, and leaf nodes are nodes at the end of the tree that have no children. Finally, note that trees are inherently recursive data structures. That is, any children of a node are parents of their own subtree, and so on. The recursive nature of trees is important to understand when designing algorithms for common tree operations.
+
+To begin, we will discuss a particular type of a tree, the binary tree. In fact, we will actually discuss a particular binary tree, a binary search tree. Let's describe what this means. While the tree data structure can have any number of branches at a single node, a binary tree can only have two branches for every node. Furthermore, a binary search tree is ordered with respect to the child subtrees, such that the value of each node in the left subtree is less than or equal to the value of the parent node, and the value of each node in the right subtree is greater than or equal to the value of the parent node. It's very helpful to visualize this relationship in order to understand it better:
+
+<div style='width: 100%; display: flex; justify-content: center; align-items: center;'><img style='width: 100%; max-width: 350px; background-color: var(--gray-05);' src='https://user-images.githubusercontent.com/18563015/32136009-1e665d98-bbd6-11e7-9133-63184f9f9182.png'></div>
+
+Now this ordered relationship is very easy to see. Note that every value to the left of 8, the root node, is less than 8, and every value to the right is greater than 8. Also notice that this relationship applies to each of the subtrees as well. For example, the first left child is a subtree. 3 is the parent node, and it has exactly two child nodes — by the rules governing binary search trees, we know without even looking that the left child of this node (and any of its children) will be less than 3, and the right child (and any of its children) will be greater than 3 (but also less than the structure's root value), and so on.
+
+Binary search trees are very common and useful data structures because they provide logarithmic time in the average case for several common operations such as lookup, insertion, and deletion.
+
+# --instructions--
+
+We'll start simple. We've defined the skeleton of a binary search tree structure here in addition to a function to create nodes for our tree. Observe that each node may have a left and right value. These will be assigned child subtrees if they exist. In our binary search tree, you will create a method to add new values to our binary search tree. The method should be called `add` and it should accept an integer value to add to the tree. Take care to maintain the invariant of a binary search tree: the value in each left child should be less than or equal to the parent value, and the value in each right child should be greater than or equal to the parent value. Here, let's make it so our tree cannot hold duplicate values. If we try to add a value that already exists, the method should return `null`. Otherwise, if the addition is successful, `undefined` should be returned.
+
+**Hint:** trees are naturally recursive data structures!
# --hints--
-存在`BinarySearchTree`数据结构。
+The `BinarySearchTree` data structure should exist.
```js
assert(
@@ -26,7 +42,7 @@ assert(
);
```
-二叉搜索树有一个名为`add`的方法。
+The binary search tree should have a method called `add`.
```js
assert(
@@ -42,7 +58,7 @@ assert(
);
```
-add方法根据二叉搜索树规则添加元素。
+The add method should add elements according to the binary search tree rules.
```js
assert(
@@ -71,7 +87,7 @@ assert(
);
```
-添加已存在的元素将返回`null`
+Adding an element that already exists should return `null`.
```js
assert(
@@ -128,7 +144,7 @@ function Node(value) {
function BinarySearchTree() {
this.root = null;
// Only change code below this line
-
+
// Only change code above this line
}
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/add-elements-at-a-specific-index-in-a-linked-list.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/add-elements-at-a-specific-index-in-a-linked-list.md
index b064be4f486..3ca9c8c0f4c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/add-elements-at-a-specific-index-in-a-linked-list.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/add-elements-at-a-specific-index-in-a-linked-list.md
@@ -1,18 +1,22 @@
---
id: 587d8252367417b2b2512c67
-title: 在链接列表中的特定索引处添加元素
+title: Add Elements at a Specific Index in a Linked List
challengeType: 1
-videoUrl: ''
+forumTopicId: 301619
dashedName: add-elements-at-a-specific-index-in-a-linked-list
---
# --description--
-让我们创建一个addAt（index，element）方法，在给定的索引处添加一个元素。就像我们如何删除给定索引处的元素一样，我们需要在遍历链表时跟踪currentIndex。当currentIndex与给定索引匹配时，我们需要重新分配上一个节点的下一个属性以引用新添加的节点。并且新节点应该引用currentIndex中的下一个节点。回到康加线的例子，一个新人想加入这条线，但他想加入中间。你处于中间位置，所以你要把手从你前面的人身上移开。新人走过去，把手放在你曾经牵过手的那个人身上，现在你已经掌握了新人。说明创建addAt（index，element）方法，该方法在给定索引处添加元素。如果无法添加元素，则返回false。注意请记住检查给定索引是否为负数或者是否长于链接列表的长度。
+Let's create a addAt(index,element) method that adds an element at a given index. Just like how we remove elements at a given index, we need to keep track of the currentIndex as we traverse the linked list. When the currentIndex matches the given index, we would need to reassign the previous node's next property to reference the new added node. And the new node should reference the next node in the currentIndex. Returning to the conga line example, a new person wants to join the line, but he wants to join in the middle. You are in the middle of the line, so you take your hands off of the person ahead of you. The new person walks over and puts his hands on the person you once had hands on, and you now have your hands on the new person.
+
+# --instructions--
+
+Create an `addAt(index,element)` method that adds an element at a given index. Return false if an element could not be added. **Note:** Remember to check if the given index is a negative or is longer than the length of the linked list.
# --hints--
-当给定索引为0时， `addAt`方法应重新分配`head`到新节点。
+Your `addAt` method should reassign `head` to the new node when the given index is 0.
```js
assert(
@@ -26,7 +30,7 @@ assert(
);
```
-对于添加到链接列表的每个新节点， `addAt`方法应该将链表的长度增加一。
+Your `addAt` method should increase the length of the linked list by one for each new node added to the linked list.
```js
assert(
@@ -40,7 +44,7 @@ assert(
);
```
-如果无法添加节点，则`addAt`方法应返回`false` 。
+Your `addAt` method should return `false` if a node was unable to be added.
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/adjacency-list.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/adjacency-list.md
index 2419338a92d..72868a02d7d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/adjacency-list.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/adjacency-list.md
@@ -1,48 +1,50 @@
---
id: 587d8256367417b2b2512c77
-title: 邻接名单
+title: Adjacency List
challengeType: 1
-videoUrl: ''
+forumTopicId: 301620
dashedName: adjacency-list
---
# --description--
-图表可以以不同方式表示。这里我们描述一种方法，称为<dfn>邻接列表</dfn> 。邻接列表本质上是项目符号列表，其中左侧是节点，右侧列出它所连接的所有其他节点。下面是邻接列表的表示。
+Graphs can be represented in different ways. Here we describe one way, which is called an <dfn>adjacency list</dfn>. An adjacency list is essentially a bulleted list where the left side is the node and the right side lists all the other nodes it's connected to. Below is a representation of an adjacency list.
-> Node1：Node2，Node3
-> Node2：Node1
-> Node3：Node1
+<blockquote>Node1: Node2, Node3<br>Node2: Node1<br>Node3: Node1</blockquote>
-以上是无向图，因为`Node1`连接到`Node2`和`Node3` ，并且该信息与`Node2`和`Node3`显示的连接一致。有向图的邻接列表意味着列表的每一行都显示方向。如果指示上述内容，那么`Node2: Node1`将意味着有向边缘从`Node2`指向`Node1` 。我们可以将上面的无向图表示为邻接列表，方法是将其放在JavaScript对象中。
+Above is an undirected graph because `Node1` is connected to `Node2` and `Node3`, and that information is consistent with the connections `Node2` and `Node3` show. An adjacency list for a directed graph would mean each row of the list shows direction. If the above was directed, then `Node2: Node1` would mean there the directed edge is pointing from `Node2` towards `Node1`. We can represent the undirected graph above as an adjacency list by putting it within a JavaScript object.
-> var undirectedG = {
-> 节点1：\[“Node2”，“Node3”]，
-> Node2：\[“Node1”]，
-> Node3：\[“Node1”]
-> };
+```js
+var undirectedG = {
+ Node1: ["Node2", "Node3"],
+ Node2: ["Node1"],
+ Node3: ["Node1"]
+};
+```
-这也可以更简单地表示为一个数组，其中节点只有数字而不是字符串标签。
+This can also be more simply represented as an array where the nodes just have numbers rather than string labels.
-> var undirectedGArr = \[
-> \[1,2]，＃Node1
-> \[0]，＃Node2
-> \[0] #Node3
-> ]。
+```js
+var undirectedGArr = [
+ [1, 2], // Node1
+ [0], // Node2
+ [0] // Node3
+];
+```
# --instructions--
-创建一个社交网络作为无向图，其中有4个节点/人名为`James` ， `Jill` ， `Jenny`和`Jeff` 。詹姆斯和杰夫，吉尔和珍妮以及杰夫和珍妮之间存在边缘/关系。
+Create a social network as an undirected graph with 4 nodes/people named `James`, `Jill`, `Jenny`, and `Jeff`. There are edges/relationships between James and Jeff, Jill and Jenny, and Jeff and Jenny.
# --hints--
-`undirectedAdjList`应该只包含四个节点。
+`undirectedAdjList` should only contain four nodes.
```js
assert(Object.keys(undirectedAdjList).length === 4);
```
-`Jeff`和`James`之间应该有优势。
+There should be an edge between `Jeff` and `James`.
```js
assert(
@@ -51,7 +53,7 @@ assert(
);
```
-`Jill`和`Jenny`之间应该有一个优势。
+There should be an edge between `Jill` and `Jenny`.
```js
assert(
@@ -60,7 +62,7 @@ assert(
);
```
-`Jeff`和`Jenny`之间应该有优势。
+There should be an edge between `Jeff` and `Jenny`.
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/adjacency-matrix.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/adjacency-matrix.md
index b17b97c7780..a852fb9b089 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/adjacency-matrix.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/adjacency-matrix.md
@@ -1,56 +1,54 @@
---
id: 587d8256367417b2b2512c78
-title: 邻接矩阵
+title: Adjacency Matrix
challengeType: 1
-videoUrl: ''
+forumTopicId: 301621
dashedName: adjacency-matrix
---
# --description--
-表示图形的另一种方法是将其置于<dfn>邻接矩阵中</dfn> 。 <dfn>邻接矩阵</dfn>是二维（2D）阵列，其中每个嵌套数组具有与外部数组相同数量的元素。换句话说，它是数字的矩阵或网格，其中数字代表边缘。零意味着没有边缘或关系。
+Another way to represent a graph is to put it in an <dfn>adjacency matrix</dfn>. An <dfn>adjacency matrix</dfn> is a two-dimensional (2D) array where each nested array has the same number of elements as the outer array. In other words, it is a matrix or grid of numbers, where the numbers represent the edges.
-> 1 2 3
->
-> * * *
->
-> 1 | 0 1 1
-> 2 | 1 0 0
-> 3 | 1 0 0
+**Note**: The numbers to the top and left of the matrix are just labels for the nodes. Inside the matrix, ones mean there exists an edge between the vertices (nodes) representing the row and column. Finally, zeros mean there is no edge or relationship.
-上面是一个非常简单的无向图，其中有三个节点，第一个节点连接到第二个和第三个节点。 **注意** ：矩阵顶部和左侧的数字只是节点的标签。下面是同一件事的JavaScript实现。
+<pre>
+ 1 2 3
+ \------
+1 | 0 1 1
+2 | 1 0 0
+3 | 1 0 0
+</pre>
-> var adjMat = \[
-> \[0,1,1]，
-> \[1,0,0]，
-> \[1,0,0]
-> ]。
+Above is a very simple, undirected graph where you have three nodes, where the first node is connected to the second and third node. Below is a JavaScript implementation of the same thing.
-与邻接列表不同，矩阵的每个“行”必须具有与图中的节点相同数量的元素。这里我们有一个三乘三矩阵，这意味着我们的图中有三个节点。有向图看起来很相似。下面是第一节点具有指向第二节点的边缘，然后第二节点具有指向第三节点的边缘的图。
+```js
+var adjMat = [
+ [0, 1, 1],
+ [1, 0, 0],
+ [1, 0, 0]
+];
+```
-> var adjMatDirected = \[
-> \[0,1,0]，
-> \[0,0,1]，
-> \[0,0,0]
-> ]。
+Unlike an adjacency list, each "row" of the matrix has to have the same number of elements as nodes in the graph. Here we have a three by three matrix, which means we have three nodes in our graph. A directed graph would look similar. Below is a graph where the first node has an edge pointing toward the second node, and then the second node has an edge pointing to the third node.
-图形的边缘也可以有权
+```js
+var adjMatDirected = [
+ [0, 1, 0],
+ [0, 0, 1],
+ [0, 0, 0]
+];
+```
-<dfn>重</dfn>
-
- 。到目前为止，我们有
-
-<dfn>未加权的</dfn>
-
-边缘，只有存在和缺少边是二进制（ `0`或`1` ）。根据您的应用，您可以拥有不同的重量。
+Graphs can also have <dfn>weights</dfn> on their edges. So far, we have <dfn>unweighted</dfn> edges where just the presence and lack of edge is binary (`0` or `1`). You can have different weights depending on your application.
# --instructions--
-创建具有五个节点的无向图的邻接矩阵。该矩阵应该是多维数组。这五个节点在第一和第四节点，第一和第三节点，第三和第五节点以及第四和第五节点之间具有关系。所有边缘权重都是一个。
+Create an adjacency matrix of an undirected graph with five nodes. This matrix should be in a multi-dimensional array. These five nodes have relationships between the first and fourth node, the first and third node, the third and fifth node, and the fourth and fifth node. All edge weights are one.
# --hints--
-`undirectedAdjList`应该只包含五个节点。
+`undirectedAdjList` should only contain five nodes.
```js
assert(
@@ -65,25 +63,25 @@ assert(
);
```
-第一个和第四个节点之间应该有一条边。
+There should be an edge between the first and fourth node.
```js
assert(adjMatUndirected[0][3] === 1 && adjMatUndirected[3][0] === 1);
```
-第一个和第三个节点之间应该有一条边。
+There should be an edge between the first and third node.
```js
assert(adjMatUndirected[0][2] === 1 && adjMatUndirected[2][0] === 1);
```
-第三个和第五个节点之间应该有一条边。
+There should be an edge between the third and fifth node.
```js
assert(adjMatUndirected[2][4] === 1 && adjMatUndirected[4][2] === 1);
```
-第四个和第五个节点之间应该有一条边。
+There should be an edge between the fourth and fifth node.
```js
assert(adjMatUndirected[3][4] === 1 && adjMatUndirected[4][3] === 1);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/breadth-first-search.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/breadth-first-search.md
index 95cd3539720..12070241354 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/breadth-first-search.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/breadth-first-search.md
@@ -1,22 +1,38 @@
---
id: 587d825c367417b2b2512c90
-title: 广度优先搜索
+title: Breadth-First Search
challengeType: 1
-videoUrl: ''
+forumTopicId: 301622
dashedName: breadth-first-search
---
# --description--
-到目前为止，我们已经学会了创建图表表示的不同方法。现在怎么办？一个自然的问题是图中任何两个节点之间的距离是多少？输入<dfn>图遍历算法</dfn> 。 <dfn>遍历算法</dfn>是遍历或访问图中节点的算法。一种遍历算法是广度优先搜索算法。该算法从一个节点开始，首先访问一个边缘的所有邻居，然后继续访问它们的每个邻居。在视觉上，这就是算法正在做的事情。 要实现此算法，您需要输入图形结构和要启动的节点。首先，您需要了解距起始节点的距离。这个你想要开始你所有的距离最初一些大的数字，如`Infinity` 。这为从起始节点无法访问节点的情况提供了参考。接下来，您将要从开始节点转到其邻居。这些邻居是一个边缘，此时你应该添加一个距离单位到你要跟踪的距离。最后，有助于实现广度优先搜索算法的重要数据结构是队列。这是一个数组，您可以在其中添加元素到一端并从另一端删除元素。这也称为<dfn>FIFO</dfn>或<dfn>先进先出</dfn>数据结构。
+So far, we've learned different ways of creating representations of graphs. What now? One natural question to have is what are the distances between any two nodes in the graph? Enter <dfn>graph traversal algorithms</dfn>.
+
+<dfn>Traversal algorithms</dfn> are algorithms to traverse or visit nodes in a graph. One type of traversal algorithm is the breadth-first search algorithm.
+
+This algorithm starts at one node and visits all its neighbors that are one edge away. It then goes on to visit each of their neighbors and so on until all nodes have been reached.
+
+An important data structure that will help implement the breadth-first search algorithm is the queue. This is an array where you can add elements to one end and remove elements from the other end. This is also known as a <dfn>FIFO</dfn> or <dfn>First-In-First-Out</dfn> data structure.
+
+Visually, this is what the algorithm is doing. 
+
+The grey shading represents a node getting added into the queue and the black shading represents a node getting removed from the queue. See how every time a node gets removed from the queue (node turns black), all their neighbors get added into the queue (node turns grey).
+
+To implement this algorithm, you'll need to input a graph structure and a node you want to start at.
+
+First, you'll want to be aware of the distances from, or number of edges away from, the start node. You'll want to start all your distances with some large number, like `Infinity`. This prevents counting issues for when a node may not be reachable from your start node. Next, you'll want to go from the start node to its neighbors. These neighbors are one edge away and at this point you should add one unit of distance to the distances you're keeping track of.
# --instructions--
-编写一个函数`bfs()` ，它将邻接矩阵图（二维数组）和节点标签根作为参数。节点标签只是`0`到`n - 1`之间节点的整数值，其中`n`是图中节点的总数。您的函数将输出JavaScript对象键值对与节点及其与根的距离。如果无法到达节点，则其距离应为`Infinity` 。
+Write a function `bfs()` that takes an adjacency matrix graph (a two-dimensional array) and a node label root as parameters. The node label will just be the integer value of the node between `0` and `n - 1`, where `n` is the total number of nodes in the graph.
+
+Your function will output a JavaScript object key-value pairs with the node and its distance from the root. If the node could not be reached, it should have a distance of `Infinity`.
# --hints--
-输入图`[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]` ，起始节点为`1`应该返回`{0: 1, 1: 0, 2: 1, 3: 2}`
+The input graph `[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]` with a start node of `1` should return `{0: 1, 1: 0, 2: 1, 3: 2}`
```js
assert(
@@ -33,7 +49,7 @@ assert(
);
```
-输入图`[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 0], [0, 0, 0, 0]]` ，起始节点为`1`应该返回`{0: 1, 1: 0, 2: 1, 3: Infinity}`
+The input graph `[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 0], [0, 0, 0, 0]]` with a start node of `1` should return `{0: 1, 1: 0, 2: 1, 3: Infinity}`
```js
assert(
@@ -50,7 +66,7 @@ assert(
);
```
-输入图`[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]` ，起始节点为`0`应该返回`{0: 0, 1: 1, 2: 2, 3: 3}`
+The input graph `[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]` with a start node of `0` should return `{0: 0, 1: 1, 2: 2, 3: 3}`
```js
assert(
@@ -67,7 +83,7 @@ assert(
);
```
-起始节点为`0`的输入图`[[0, 1], [1, 0]]`应返回`{0: 0, 1: 1}`
+The input graph `[[0, 1], [1, 0]]` with a start node of `0` should return `{0: 0, 1: 1}`
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/check-if-an-element-is-present-in-a-binary-search-tree.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/check-if-an-element-is-present-in-a-binary-search-tree.md
index acec9a3f5da..03daa65cb23 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/check-if-an-element-is-present-in-a-binary-search-tree.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/check-if-an-element-is-present-in-a-binary-search-tree.md
@@ -1,18 +1,22 @@
---
id: 587d8257367417b2b2512c7c
-title: 检查二进制搜索树中是否存在元素
+title: Check if an Element is Present in a Binary Search Tree
challengeType: 1
-videoUrl: ''
+forumTopicId: 301623
dashedName: check-if-an-element-is-present-in-a-binary-search-tree
---
# --description--
-现在我们对二进制搜索树有了一般意义，让我们更详细地讨论它。二进制搜索树为平均情况下的查找，插入和删除的常见操作提供对数时间，并且在最坏情况下提供线性时间。为什么是这样？这些基本操作中的每一个都要求我们在树中找到一个项目（或者在插入的情况下找到它应该去的地方），并且由于每个父节点处的树结构，我们向左或向右分支并且有效地排除了一半的大小剩下的树。这使得搜索与树中节点数的对数成比例，这在平均情况下为这些操作创建对数时间。好的，但最坏的情况呢？那么，可考虑从以下值建构一棵树，将它们从左至右： `10` ， `12` ， `17` ， `25` 。根据我们的规则二叉搜索树，我们将增加`12`到右侧`10` ， `17` ，以这样的权利，以及`25`到这一权利。现在我们的树类似于一个链表，并且遍历它以找到`25`将要求我们以线性方式遍历所有项目。因此，在最坏的情况下，线性时间。这里的问题是树是不平衡的。我们将更多地了解这在以下挑战中意味着什么。说明：在此挑战中，我们将为树创建一个实用程序。编写一个方法`isPresent` ，它接受一个整数值作为输入，并在二叉搜索树中返回该值是否存在的布尔值。
+Now that we have a general sense of what a binary search tree is let's talk about it in a little more detail. Binary search trees provide logarithmic time for the common operations of lookup, insertion, and deletion in the average case, and linear time in the worst case. Why is this? Each of those basic operations requires us to find an item in the tree (or in the case of insertion to find where it should go) and because of the tree structure at each parent node we are branching left or right and effectively excluding half the size of the remaining tree. This makes the search proportional to the logarithm of the number of nodes in the tree, which creates logarithmic time for these operations in the average case. Ok, but what about the worst case? Well, consider constructing a tree from the following values, adding them left to right: `10`, `12`, `17`, `25`. Following our rules for a binary search tree, we will add `12` to the right of `10`, `17` to the right of this, and `25` to the right of this. Now our tree resembles a linked list and traversing it to find `25` would require us to traverse all the items in linear fashion. Hence, linear time in the worst case. The problem here is that the tree is unbalanced. We'll look a little more into what this means in the following challenges.
+
+# --instructions--
+
+In this challenge, we will create a utility for our tree. Write a method `isPresent` which takes an integer value as input and returns a boolean value for the presence or absence of that value in the binary search tree.
# --hints--
-存在`BinarySearchTree`数据结构。
+The `BinarySearchTree` data structure should exist.
```js
assert(
@@ -26,7 +30,7 @@ assert(
);
```
-二叉搜索树有一个名为`isPresent`的方法。
+The binary search tree should have a method called `isPresent`.
```js
assert(
@@ -42,7 +46,7 @@ assert(
);
```
-`isPresent`方法正确检查添加到树中的元素是否存在。
+The `isPresent` method should correctly check for the presence or absence of elements added to the tree.
```js
assert(
@@ -70,7 +74,7 @@ assert(
);
```
-`isPresent`处理树为空的情况。
+`isPresent` should handle cases where the tree is empty.
```js
assert(
@@ -141,7 +145,7 @@ function Node(value) {
function BinarySearchTree() {
this.root = null;
// Only change code below this line
-
+
// Only change code above this line
}
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/check-if-binary-search-tree.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/check-if-binary-search-tree.md
index ddaf38f2318..42081a1355d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/check-if-binary-search-tree.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/check-if-binary-search-tree.md
@@ -89,7 +89,7 @@ function BinarySearchTree() {
}
function isBinarySearchTree(tree) {
// Only change code below this line
-
+
// Only change code above this line
}
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-circular-queue.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-circular-queue.md
index 57acd6f252f..16ff688fabf 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-circular-queue.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-circular-queue.md
@@ -1,42 +1,62 @@
---
id: 587d8255367417b2b2512c75
-title: 创建循环队列
+title: Create a Circular Queue
challengeType: 1
-videoUrl: ''
+forumTopicId: 301625
dashedName: create-a-circular-queue
---
# --description--
-在此挑战中，您将创建一个循环队列。循环队列基本上是一个写入集合末尾的队列，然后开始在集合开头写自己。这种类型的数据结构在某些情况下具有一些有用的应用。例如，循环队列可用于流媒体。队列填满后，新媒体数据就会开始覆盖旧数据。说明这个概念的一个好方法是使用数组：
+In this challenge you will be creating a Circular Queue. A circular queue is a queue that writes to the end of a collection then begins overwriting itself at the beginning of the collection. This type of data structure is useful in certain situations. For example, a circular queue can be used for streaming media. Once the queue is full, new media data will overwrite old data.
-> \[1,2,3,4,5]
-> ^读@ 0
-> ^写@ 0
+A good way to illustrate this concept is with an array of length `5`:
-这里的读写都在`0` 。现在队列获得3个新记录`a` ， `b`和`c` 。我们的队列现在看起来像：
+```js
+[null, null, null, null, null]
+ ^Read @ 0
+ ^Write @ 0
+```
-> \[a，b，c，4,5]
-> ^读@ 0
-> ^写@ 3
+Here the read and write are both at position `0`. Now the queue gets 3 new records `a`, `b`, and `c`. Our queue now looks like:
-当读头读取时，它可以删除值或保留它们：
+```js
+[a, b, c, null, null]
+ ^Read @ 0
+ ^Write @ 3
+```
-> \[null，null，null，4,5]
-> ^阅读@ 3
-> ^写@ 3
+As the read head reads, it can remove values or keep them:
-一旦写入到达数组的末尾，它就会循环回到开头：
+```js
+[null, null, null, null, null]
+ ^Read @ 3
+ ^Write @ 3
+```
-> \[f，null，null，d，e]
-> ^阅读@ 3
-> ^写@ 1
+Now we write the values `d`, `e`, and `f` to the queue. Once the write reaches the end of the array it loops back to the beginning:
-这种方法需要恒定的内存量，但允许处理更大尺寸的文件。说明：在此挑战中，我们将实现循环队列。循环队列应提供`enqueue`和`dequeue`方法，允许您读取和写入队列。类本身也应该接受一个整数，您可以使用该整数在创建队列时指定队列的大小。我们已经在代码编辑器中为您编写了此类的起始版本。将项目排入队列时，写入指针应向前推进，并在到达队列末尾时循环回到开头。同样，当您使项目出列时，读指针应向前推进。不应允许写指针移过读指针（我们的类不会让你覆盖你还没有读过的数据），并且读指针不能超过你写的数据。此外，如果成功，则`enqueue`方法应返回您入`enqueue`的项，否则返回`null` 。类似地，当你使一个项目出列时，它应该被返回，如果你不能出列，你应该返回`null` 。
+```js
+[f, null, null, d, e]
+ ^Read @ 3
+ ^Write @ 1
+```
+
+This approach requires a constant amount of memory but allows files of a much larger size to be processed.
+
+# --instructions--
+
+In this challenge we will implement a circular queue. The circular queue should provide `enqueue` and `dequeue` methods which allow you to read from and write to the queue. The class itself should also accept an integer argument which you can use to specify the size of the queue when created. We've written the starting version of this class for you in the code editor.
+
+When you enqueue items to the queue, the write pointer should advance forward and loop back to the beginning once it reaches the end of the queue. The `enqueue` method should return the item you enqueued if it is successful; otherwise it will return `null`.
+
+Likewise, the read pointer should advance forward as you dequeue items. When you dequeue an item, that item should be returned. If you cannot dequeue an item, you should return `null`.
+
+The write pointer should not be allowed to move past the read pointer (our class won't let you overwrite data you haven't read yet) and the read pointer should not be able to advance past data you have written.
# --hints--
-`enqueue`方法将项添加到循环队列。
+The `enqueue` method should add items to the circular queue.
```js
assert(
@@ -51,7 +71,7 @@ assert(
);
```
-您不能通过读指针将项排入队列。
+You should not enqueue items past the read pointer.
```js
assert(
@@ -69,7 +89,7 @@ assert(
);
```
-`dequeue`方法使队列中的项目出列。
+The `dequeue` method should dequeue items from the queue.
```js
assert(
@@ -85,7 +105,7 @@ assert(
);
```
-项目出列后，其队列中的位置应重置为`null` 。
+After an item is dequeued, its position in the queue should be reset to `null`.
```js
assert(
@@ -102,7 +122,7 @@ assert(
);
```
-尝试通过写指针出列队列返回`null`并且不会使写指针前进。
+Trying to dequeue past the write pointer should return `null` and does not advance the write pointer.
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-doubly-linked-list.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-doubly-linked-list.md
index c02a0c5789a..16c2e122fa7 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-doubly-linked-list.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-doubly-linked-list.md
@@ -1,22 +1,26 @@
---
id: 587d825a367417b2b2512c87
-title: 创建双向链接列表
+title: Create a Doubly Linked List
challengeType: 1
-videoUrl: ''
+forumTopicId: 301626
dashedName: create-a-doubly-linked-list
---
# --description--
-到目前为止，我们创建的所有链接列表都是单链表。在这里，我们将创建一个<dfn>双向链表</dfn> 。顾名思义，双向链表中的节点引用了列表中的下一个和上一个节点。这允许我们在两个方向上遍历列表，但它还需要使用更多内存，因为每个节点必须包含对列表中前一个节点的附加引用。
+All of the linked lists we've created so far are singly linked lists. Here, we'll create a <dfn>doubly linked list</dfn>. As the name implies, nodes in a doubly linked list have references to the next and previous node in the list.
+
+This allows us to traverse the list in both directions but it also requires more memory to be used because every node must contain an additional reference to the previous node in the list.
# --instructions--
-我们提供了一个`Node`对象并启动了我们的`DoublyLinkedList` 。让我们将两个方法添加到名为`add` and `remove`双向链表`remove` 。 `add`方法应该将给定元素添加到列表中，而`remove`方法应该删除列表中所有出现的给定元素。编写这些方法时要小心处理任何可能的边缘情况，例如删除第一个或最后一个元素。此外，删除空列表中的任何项应返回`null` 。
+We've provided a `Node` object and started our `DoublyLinkedList`. Let's add two methods to our doubly linked list called `add` and `remove`. The `add` method should add the given element to the list while the `remove` method should remove all occurrences of a given element in the list.
+
+Be careful to handle any possible edge cases when writing these methods, such as deletions for the first or last element. Also, removing any item on an empty list should return `null`.
# --hints--
-存在DoublyLinkedList数据结构。
+The DoublyLinkedList data structure should exist.
```js
assert(
@@ -30,7 +34,7 @@ assert(
);
```
-DoublyLinkedList有一个名为add的方法。
+The DoublyLinkedList should have a method called add.
```js
assert(
@@ -47,7 +51,7 @@ assert(
);
```
-DoublyLinkedList有一个名为remove的方法。
+The DoublyLinkedList should have a method called remove.
```js
assert(
@@ -64,7 +68,7 @@ assert(
);
```
-从空列表中删除项目将返回null。
+Removing an item from an empty list should return null.
```js
assert(
@@ -78,7 +82,7 @@ assert(
);
```
-add方法将项添加到列表中。
+The add method should add items to the list.
```js
assert(
@@ -95,7 +99,7 @@ assert(
);
```
-每个节点都跟踪前一个节点。
+Each node should keep track of the previous node.
```js
assert(
@@ -112,7 +116,7 @@ assert(
);
```
-可以从列表中删除第一个项目。
+The first item should be removable from the list.
```js
assert(
@@ -130,7 +134,7 @@ assert(
);
```
-最后一项可以从列表中删除。
+The last item should be removable from the list.
```js
assert(
@@ -156,7 +160,7 @@ assert(
DoublyLinkedList.prototype = Object.assign(
DoublyLinkedList.prototype,
{
-
+
print() {
if (this.head == null) {
return null;
@@ -200,7 +204,7 @@ var DoublyLinkedList = function() {
this.head = null;
this.tail = null;
// Only change code below this line
-
+
// Only change code above this line
};
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-hash-table.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-hash-table.md
index 3c27f082ae7..65b68904323 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-hash-table.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-hash-table.md
@@ -1,18 +1,28 @@
---
id: 587d825b367417b2b2512c8e
-title: 创建一个哈希表
+title: Create a Hash Table
challengeType: 1
-videoUrl: ''
+forumTopicId: 301627
dashedName: create-a-hash-table
---
# --description--
-在这个挑战中，我们将学习哈希表。哈希表用于实现关联数组或键值对的映射，就像我们刚刚研究的对象和地图一样。例如，JavaScript对象可以实现为哈希表（其实际实现将取决于它运行的环境）。哈希表的工作方式是它接受一个键输入并以确定的方式将此键散列到某个数值。然后将该数值用作存储相关值的实际键。然后，如果您尝试再次访问相同的密钥，则散列函数将处理密钥，返回相同的数字结果，然后将用于查找关联的值。这平均提供非常有效的O（1）查找时间。散列表可以实现为具有散列函数的数组，从而生成指定范围内的数组索引。在这种方法中，数组大小的选择很重要，散列函数也是如此。例如，如果散列函数为两个不同的键生成相同的值，该怎么办？这称为碰撞。处理冲突的一种方法是仅将两个键值对存储在该索引处。然后，在查找其中任何一个时，您将不得不遍历项目桶以找到您要查找的密钥。良好的散列函数可最大限度地减少冲突，从而保持有效的搜索时间。在这里，我们不会关注散列或散列表实现的细节，我们将尝试大致了解它们的工作原理。说明：让我们创建哈希表的基本功能。我们已经创建了一个天真的散列函数供您使用。您可以将字符串值传递给函数哈希，它将返回一个哈希值，您可以将其用作存储键。在this.collection对象中根据此散列值存储项目。创建这三种方法：添加，删除和查找。第一个应该接受一个键值对来添加到哈希表。第二个应该在传递密钥时删除键值对。第三个应该接受一个键并返回相关的值，如果该键不存在则返回null。请务必编写代码以解决冲突！
+In this challenge we will learn about hash tables. A Hash table is used to implement associative arrays, or mappings of key-value pairs, like the objects and Maps we have just been studying. A JavaScript object could be implemented as a hash table, for instance (its actual implementation will depend on the environment it's running in). The way a hash table works is that it takes a key input and hashes this key in a deterministic way to some numerical value. This numerical value is then used as the actual key the associated value is stored by. Then, if you try to access the same key again, the hashing function will process the key, return the same numerical result, which will then be used to look up the associated value. This provides very efficient O(1) lookup time on average.
+
+Hash tables can be implemented as arrays with hash functions producing array indices within a specified range. In this method, the choice of the array size is important, as is the hashing function. For instance, what if the hashing function produces the same value for two different keys? This is called a collision. One way to handle collisions is to just store both key-value pairs at that index. Then, upon lookup of either, you would have to iterate through the bucket of items to find the key you are looking for. A good hashing function will minimize collisions to maintain efficient search time.
+
+Here, we won't be concerned with the details of hashing or hash table implementation, we will just try to get a general sense of how they work.
+
+# --instructions--
+
+Let's create the basic functionality of a hash table. We've created a naive hashing function for you to use. You can pass a string value to the function `hash` and it will return a hashed value you can use as a key for storage. Store items based on this hashed value in the `this.collection` object. Create these three methods: `add`, `remove`, and `lookup`. The first should accept a key value pair to add to the hash table. The second should remove a key-value pair when passed a key. The third should accept a key and return the associated value or `null` if the key is not present.
+
+Be sure to write your code to account for collisions!
# --hints--
-存在HashTable数据结构。
+The HashTable data structure should exist.
```js
assert(
@@ -26,7 +36,7 @@ assert(
);
```
-HashTable有一个add方法。
+The HashTable should have an add method.
```js
assert(
@@ -40,7 +50,7 @@ assert(
);
```
-HashTable有一个删除方法。
+The HashTable should have a remove method.
```js
assert(
@@ -54,7 +64,7 @@ assert(
);
```
-HashTable有一个查找方法。
+The HashTable should have a lookup method.
```js
assert(
@@ -68,7 +78,7 @@ assert(
);
```
-add方法添加键值对，lookup方法返回与给定键关联的值。
+The add method should add key value pairs and the lookup method should return the values associated with a given key.
```js
assert(
@@ -83,32 +93,25 @@ assert(
);
```
-remove方法接受一个键作为输入，并删除关联的键值对。
+The remove method should accept a key as input and should remove the associated key value pair.
```js
assert(
(function () {
var test = false;
+ var hashValue = hash('key');
if (typeof HashTable !== 'undefined') {
test = new HashTable();
}
+ test.add = addMethodSolution;
test.add('key', 'value');
test.remove('key');
- test.lookup = function (key) {
- var theHash = hash(key);
- if (this.collection.hasOwnProperty(theHash)[key]) {
- return this.collection[theHash][key];
- }
- return null;
- };
- var lookup = test.lookup('key');
- test.lookup = null;
- return lookup === null;
+ return !test.collection.hasOwnProperty(hashValue);
})()
);
```
-使用哈希函数添加项目。
+Items should be added using the hash function.
```js
assert(
@@ -126,7 +129,7 @@ assert(
);
```
-哈希表处理冲突。
+The hash table should handle collisions.
```js
assert(
@@ -187,7 +190,7 @@ var hash = string => {
var HashTable = function() {
this.collection = {};
// Only change code below this line
-
+
// Only change code above this line
};
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-linked-list-class.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-linked-list-class.md
index b157522f456..0c768cf0a28 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-linked-list-class.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-linked-list-class.md
@@ -1,22 +1,32 @@
---
id: 587d8251367417b2b2512c62
-title: 创建链接列表类
+title: Create a Linked List Class
challengeType: 1
-videoUrl: ''
+forumTopicId: 301628
dashedName: create-a-linked-list-class
---
# --description--
-让我们创建一个`linked list`类。每个链接列表都应该从一些基本属性开始： `head` （列表中的第一项）和`length` （列表中的项目数）。有时您会看到链接列表的实现，其中包含列表的最后一个元素的`tail` ，但是现在我们将坚持使用这两个元素。每当我们向链表添加元素时，我们的`length`属性应该加1。我们想要一种方法将项添加到链表中，因此我们要创建的第一种方法是`add`方法。如果我们的列表为空，那么向链表添加一个元素就足够了：我们只是将该元素包装在`Node`类中，然后将该节点分配给链表的`head` 。但是如果我们的名单已经有一个或多个成员呢？我们如何在列表中添加元素？回想一下，链表中的每个节点都有一个`next`属性。要将节点添加到列表，请在列表中找到最后一个节点，并将该节点的最后`next`属性指向新节点。 （提示：当节点的`next`属性为`null`时，您知道已到达链表的末尾。）
+Let's create a `linked list` class. Every linked list should start out with a few basic properties: a `head` (the first item in your list) and a `length` (number of items in your list). Sometimes you'll see implementations of linked lists that incorporate a `tail` for the last element of the list, but for now we'll just stick with these two. Whenever we add an element to the linked list, our `length` property should be incremented by one.
+
+We'll want to have a way to add items to our linked list, so the first method we'll want to create is the `add` method.
+
+If our list is empty, adding an element to our linked list is straightforward enough: we just wrap that element in a `Node` class, and we assign that node to the `head` of our linked list.
+
+But what if our list already has one or more members? How do we add an element to the list? Recall that each node in a linked list has a `next` property. To add a node to the list, find the last node in the list, and point that last node's `next` property at our new node. (Hint: you know you've reached the end of a linked list when a node's `next` property is `null`.)
# --instructions--
-编写一个add方法，将您推送到链接列表的第一个节点分配给`head` ;之后，每当添加一个节点时，每个节点都应该由前一个节点的`next`属性引用。注意每次将元素添加到链接列表时，列表的`length`应增加1。
+Write an add method that assigns the first node you push to the linked list to the `head`; after that, whenever adding a node, every node should be referenced by the previous node's `next` property.
+
+Note
+
+Your list's `length` should increase by one every time an element is added to the linked list.
# --hints--
-您的`LinkedList`类应该有一个`add`方法。
+Your `LinkedList` class should have a `add` method.
```js
assert(
@@ -27,7 +37,7 @@ assert(
);
```
-您的`LinkedList`类应该为添加的第一个节点分配`head` 。
+Your `LinkedList` class should assign `head` to the first node added.
```js
assert(
@@ -39,7 +49,7 @@ assert(
);
```
-`LinkedList`类中的上一个`node`应该引用创建的最新节点。
+The previous `node` in your `LinkedList` class should have reference to the newest node created.
```js
assert(
@@ -52,7 +62,7 @@ assert(
);
```
-`LinkedList`类的`size`应该等于`LinkedList`的节点数量。
+The `size` of your `LinkedList` class should equal the amount of nodes in the linked list.
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-map-data-structure.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-map-data-structure.md
index b7f2c7928d4..0c32270e722 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-map-data-structure.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-map-data-structure.md
@@ -1,40 +1,32 @@
---
id: 8d5823c8c441eddfaeb5bdef
-title: 创建地图数据结构
+title: Create a Map Data Structure
challengeType: 1
-videoUrl: ''
+forumTopicId: 301629
dashedName: create-a-map-data-structure
---
# --description--
-接下来的几个挑战将涵盖地图和哈希表。映射是存储键值对的数据结构。在JavaScript中，我们可以将它们作为对象使用。地图可根据键值快速查找存储的项目，是非常常见且有用的数据结构。说明：让我们开始创建自己的地图。因为JavaScript对象提供了比我们在此处编写的任何内容更有效的地图结构，所以这主要是作为学习练习。但是，JavaScript对象仅向我们提供某些操作。如果我们想定义自定义操作怎么办？使用此处提供的`Map`对象作为JavaScript `object`的包装器。在Map对象上创建以下方法和操作：
+The next few challenges will cover maps and hash tables. Maps are data structures that store key-value pairs. In JavaScript, these are available to us as objects. Maps provide rapid lookup of stored items based on key values and are very common and useful data structures.
-- `add`接受要添加到地图的`key, value`对。
+# --instructions--
-- `remove`接受一个键并删除关联的`key, value`对
+Let's get some practice creating our own map. Because JavaScript objects provide a much more efficient map structure than anything we could write here, this is intended primarily as a learning exercise. However, JavaScript objects only provide us with certain operations. What if we wanted to define custom operations? Use the `Map` object provided here as a wrapper around a JavaScript `object`. Create the following methods and operations on the Map object:
-- `get`接受一个`key`并返回存储的`value`
-
-- 如果密钥存在`has`接受`key`并返回
-
- <dfn>true，</dfn>
-
- 否则返回
-
- <dfn>false</dfn>
-
- 。
-
-- `values`返回地图中所有`values`的数组
-
-- `size`返回地图中的项目数
-
-- `clear`清空地图
+<ul>
+<li><code>add</code> accepts a <code>key, value</code> pair to add to the map.</li>
+<li><code>remove</code> accepts a key and removes the associated <code>key, value</code> pair</li>
+<li><code>get</code> accepts a <code>key</code> and returns the stored <code>value</code></li>
+<li><code>has</code> accepts a <code>key</code> and returns <dfn>true</dfn> if the key exists or <dfn>false</dfn> if it doesn't.</li>
+<li><code>values</code> returns an array of all the values in the map</li>
+<li><code>size</code> returns the number of items in the map</li>
+<li><code>clear</code> empties the map</li>
+</ul>
# --hints--
-存在地图数据结构。
+The Map data structure should exist.
```js
assert(
@@ -48,7 +40,7 @@ assert(
);
```
-Map对象具有以下方法：add，remove，get，has，values，clear和size。
+The Map object should have the following methods: add, remove, get, has, values, clear, and size.
```js
assert(
@@ -70,7 +62,7 @@ assert(
);
```
-add方法将项添加到地图中。
+The add method should add items to the map.
```js
assert(
@@ -87,7 +79,7 @@ assert(
);
```
-has方法对于添加的项返回true，对于缺少的项返回false。
+The has method should return true for added items and false for absent items.
```js
assert(
@@ -102,7 +94,7 @@ assert(
);
```
-get方法接受键作为输入并返回关联的值。
+The get method should accept keys as input and should return the associated values.
```js
assert(
@@ -117,7 +109,7 @@ assert(
);
```
-values方法将存储在映射中的所有值作为数组中的字符串返回。
+The values method should return all the values stored in the map as strings in an array.
```js
assert(
@@ -139,7 +131,7 @@ assert(
);
```
-clear方法清空映射，size方法返回映射中存在的项目数。
+The clear method should empty the map and the size method should return the number of items present in the map.
```js
assert(
@@ -166,7 +158,7 @@ assert(
var Map = function() {
this.collection = {};
// Only change code below this line
-
+
// Only change code above this line
};
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-priority-queue-class.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-priority-queue-class.md
index 63f8dad51b3..f7f70a8b32c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-priority-queue-class.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-priority-queue-class.md
@@ -1,18 +1,30 @@
---
id: 587d8255367417b2b2512c74
-title: 创建优先级队列类
+title: Create a Priority Queue Class
challengeType: 1
-videoUrl: ''
+forumTopicId: 301630
dashedName: create-a-priority-queue-class
---
# --description--
-在此挑战中，您将创建一个优先级队列。优先级队列是一种特殊类型的队列，其中项目可能具有指定其优先级的附加信息。这可以简单地用整数表示。项目优先级将覆盖确定序列项目已出列的放置顺序。如果具有较高优先级的项目在具有较低优先级的项目之后排队，则较高优先级项目将在所有其他项目之前出列。例如，让我们假设我们有一个包含三个项目的优先级队列： `[['kitten', 2], ['dog', 2], ['rabbit', 2]]`这里第二个值（整数）表示项目优先级。如果我们将优先级为`1` `['human', 1]`排入队列（假设优先级较低，则优先级较低），那么它将成为第一个出列的项目。该集合将是这样的： `[['human', 1], ['kitten', 2], ['dog', 2], ['rabbit', 2]]` 。我们已经开始在代码编辑器中编写`PriorityQueue` 。您需要添加一个`enqueue`方法来添加具有优先级的项目，一个用于删除项目的`dequeue`方法，一个用于返回队列中项目数量的`size`方法，一个用于返回队列`front`元素的`front`方法，以及最后一个`isEmpty`方法，如果队列为空则返回`true` ，否则返回`false` 。入`enqueue`应接受上面显示格式的项目（ `['human', 1]` ），其中`1`表示优先级。 `dequeue`应该只返回当前项目，而不是其优先级。
+In this challenge you will be creating a Priority Queue. A Priority Queue is a special type of Queue in which items may have additional information which specifies their priority. This could be simply represented with an integer. Item priority will override placement order in determining the sequence items are dequeued. If an item with a higher priority is enqueued after items with lower priority, the higher priority item will be dequeued before all the others.
+
+For instance, let’s imagine we have a priority queue with three items:
+
+`[['kitten', 2], ['dog', 2], ['rabbit', 2]]`
+
+Here the second value (an integer) represents item priority. If we enqueue `['human', 1]` with a priority of `1` (assuming lower priorities are given precedence) it would then be the first item to be dequeued. The collection would look like this:
+
+`[['human', 1], ['kitten', 2], ['dog', 2], ['rabbit', 2]]`.
+
+We’ve started writing a `PriorityQueue` in the code editor. You will need to add an `enqueue` method for adding items with a priority, a `dequeue` method for removing and returning items, a `size` method to return the number of items in the queue, a `front` method to return the element at the front of the queue, and finally an `isEmpty` method that will return `true` if the queue is empty or `false` if it is not.
+
+The `enqueue` should accept items with the format shown above (`['human', 1]`) where `1` represents the priority. `dequeue` and `front` should return only the item's name, not its priority.
# --hints--
-您的`Queue`类应该有一个`enqueue`方法。
+Your `PriorityQueue` class should have a `enqueue` method.
```js
assert(
@@ -23,7 +35,7 @@ assert(
);
```
-您的`Queue`类应该有一个`dequeue`方法。
+Your `PriorityQueue` class should have a `dequeue` method.
```js
assert(
@@ -34,7 +46,7 @@ assert(
);
```
-您的`Queue`类应该有一个`size`方法。
+Your `PriorityQueue` class should have a `size` method.
```js
assert(
@@ -45,7 +57,18 @@ assert(
);
```
-您的`Queue`类应该有一个`isEmpty`方法。
+Your `PriorityQueue` class should have a `front` method.
+
+```js
+assert(
+ (function () {
+ var test = new PriorityQueue();
+ return typeof test.front === 'function';
+ })()
+);
+```
+
+Your `PriorityQueue` class should have an `isEmpty` method.
```js
assert(
@@ -56,7 +79,7 @@ assert(
);
```
-当项目入队和出列时，您的PriorityQueue应使用`size`方法正确跟踪当前项目数。
+Your `PriorityQueue` class should correctly keep track of the current number of items using the `size` method as items are enqueued and dequeued.
```js
assert(
@@ -75,7 +98,34 @@ assert(
);
```
-当队列为空时， `isEmpty`方法应该返回`true` 。
+The `front` method should return the correct item at the front of the queue as items are enqueued and dequeued.
+
+```js
+assert(
+ (function () {
+ var test = new PriorityQueue();
+ test.enqueue(['David Brown', 2]);
+ var front1 = test.front();
+ test.enqueue(['Jon Snow', 1]);
+ var front2 = test.front();
+ test.dequeue();
+ test.enqueue(['A', 3]);
+ var front3 = test.front();
+ test.enqueue(['B', 3]);
+ test.enqueue(['C', 3]);
+ test.dequeue();
+ var front4 = test.front();
+ return (
+ front1 === 'David Brown' &&
+ front2 === 'Jon Snow' &&
+ front3 === 'David Brown' &&
+ front4 === 'A'
+ );
+ })()
+);
+```
+
+The `isEmpty` method should return `true` when the queue is empty.
```js
assert(
@@ -91,7 +141,7 @@ assert(
);
```
-优先级队列应该在具有较低优先级的项之前返回具有较高优先级的项，否则以先进先出顺序返回项。
+The priority queue should return items with a higher priority before items with a lower priority and return items in first-in-first-out order otherwise.
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-queue-class.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-queue-class.md
index e0960f57450..306b90e0c08 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-queue-class.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-queue-class.md
@@ -1,18 +1,26 @@
---
id: 587d8250367417b2b2512c60
-title: 创建队列类
+title: Create a Queue Class
challengeType: 1
-videoUrl: ''
+forumTopicId: 301631
dashedName: create-a-queue-class
---
# --description--
-与堆栈一样，队列是元素的集合。但与堆栈不同，队列遵循FIFO（先入先出）原则。添加到队列的元素将被推送到队列的尾部或末尾，并且只允许删除队列前面的元素。我们可以使用数组来表示队列，但就像堆栈一样，我们希望限制我们对队列的控制量。队列类的两个主要方法是enqueue和dequeue方法。 enqueue方法将元素推送到队列的尾部，dequeue方法移除并返回队列前面的元素。其他有用的方法是front，size和isEmpty方法。说明编写一个将元素推送到队列尾部的入队方法，一个删除并返回前面元素的出列方法，一个让我们看到前面元素的前方法，一个显示长度的大小方法，以及一个isEmpty方法检查队列是否为空。
+Like stacks, queues are a collection of elements. But unlike stacks, queues follow the FIFO (First-In First-Out) principle. Elements added to a queue are pushed to the tail, or the end, of the queue, and only the element at the front of the queue is allowed to be removed.
+
+We could use an array to represent a queue, but just like stacks, we want to limit the amount of control we have over our queues.
+
+The two main methods of a queue class is the enqueue and the dequeue method. The enqueue method pushes an element to the tail of the queue, and the dequeue method removes and returns the element at the front of the queue. Other useful methods are the front, size, and isEmpty methods.
+
+# --instructions--
+
+Write an `enqueue` method that pushes an element to the tail of the queue, a `dequeue` method that removes and returns the front element, a `front` method that lets us see the front element, a `size` method that shows the length, and an `isEmpty` method to check if the queue is empty.
# --hints--
-您的`Queue`类应该有一个`enqueue`方法。
+Your `Queue` class should have a `enqueue` method.
```js
assert(
@@ -23,7 +31,7 @@ assert(
);
```
-您的`Queue`类应该有一个`dequeue`方法。
+Your `Queue` class should have a `dequeue` method.
```js
assert(
@@ -34,7 +42,7 @@ assert(
);
```
-您的`Queue`类应该有一个`front`方法。
+Your `Queue` class should have a `front` method.
```js
assert(
@@ -45,7 +53,7 @@ assert(
);
```
-您的`Queue`类应该有一个`size`方法。
+Your `Queue` class should have a `size` method.
```js
assert(
@@ -56,7 +64,7 @@ assert(
);
```
-您的`Queue`类应该有一个`isEmpty`方法。
+Your `Queue` class should have an `isEmpty` method.
```js
assert(
@@ -67,7 +75,7 @@ assert(
);
```
-`dequeue`方法应该删除并返回队列的前端元素
+The `dequeue` method should remove and return the front element of the queue
```js
assert(
@@ -80,7 +88,7 @@ assert(
);
```
-`front`方法应该返回队列的front元素的值
+The `front` method should return value of the front element of the queue
```js
assert(
@@ -93,7 +101,7 @@ assert(
);
```
-`size`方法应该返回队列的长度
+The `size` method should return the length of the queue
```js
assert(
@@ -105,7 +113,7 @@ assert(
);
```
-如果队列中有元素，则`isEmpty`方法应返回`false`
+The `isEmpty` method should return `false` if there are elements in the queue
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-set-class.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-set-class.md
index da27e8e4bc5..d818be6a807 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-set-class.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-set-class.md
@@ -1,43 +1,47 @@
---
id: 8d1323c8c441eddfaeb5bdef
-title: 创建一个Set类
+title: Create a Set Class
challengeType: 1
-videoUrl: ''
+forumTopicId: 301632
dashedName: create-a-set-class
---
# --description--
-在接下来的几个练习中，我们将创建一个函数来模拟一个名为“Set”的数据结构。 Set类似于数组，但不能包含重复值。 Set的典型用途是简单地检查项目是否存在。这可以用对象实现，例如：
+In this exercise we are going to create a class named `Set` to emulate an abstract data structure called "set". A set is like an array, but it cannot contain duplicate values. The typical use for a set is to simply check for the presence of an item. We can see how the ES6 `Set` object works in the example below:
-> var set = new Object（）;
-> set.foo = true;
-> //看看我们的集合中是否存在foo：
-> console.log（set.foo）// true
+```js
+const set1 = new Set([1, 2, 3, 5, 5, 2, 0]);
+console.log(set1);
+// output: {1, 2, 3, 5, 0}
+console.log(set1.has(1));
+// output: true
+console.log(set1.has(6));
+// output: false
+```
-在接下来的几个练习中，我们将从头开始构建一个全功能的Set。对于本练习，只要该值中尚不存在该值，就创建一个将值添加到set集合的函数。例如：
+First, we will create an add method that adds a value to our set collection as long as the value does not already exist in the set. Then we will create a remove method that removes a value from the set collection if it already exists. And finally, we will create a size method that returns the number of elements inside the set collection.
-> this.add = function（element）{
-> //一些代码来为集合添加值
-> }
+# --instructions--
-如果成功添加该值，则该函数应返回`true`否则返回`false` 。
+Create an `add` method that adds a unique value to the set collection and returns `true` if the value was successfully added and `false` otherwise.
+
+Create a `remove` method that accepts a value and checks if it exists in the set. If it does, then this method should remove it from the set collection, and return `true`. Otherwise, it should return `false`. Create a `size` method that returns the size of the set collection.
# --hints--
-您的`Set`类应该有一个`add`方法。
+Your `Set` class should have an `add` method.
```js
assert(
(function () {
var test = new Set();
return typeof test.add === 'function';
- })(),
- 'Your <code>Set</code> class should have an <code>add</code> method.'
+ })()
);
```
-您的`add`方法不应添加重复值。
+Your `add` method should not add duplicate values.
```js
assert(
@@ -48,12 +52,11 @@ assert(
test.add('a');
var vals = test.values();
return vals[0] === 'a' && vals[1] === 'b' && vals.length === 2;
- })(),
- 'Your <code>add</code> method should not add duplicate values.'
+ })()
);
```
-成功添加值后， `add`方法应返回`true` 。
+Your `add` method should return `true` when a value has been successfully added.
```js
assert(
@@ -61,12 +64,11 @@ assert(
var test = new Set();
var result = test.add('a');
return result != undefined && result === true;
- })(),
- 'Your <code>add</code> method should return <code>true</code> when a value has been successfully added.'
+ })()
);
```
-`add`重复值时， `add`方法应返回`false` 。
+Your `add` method should return `false` when a duplicate value is added.
```js
assert(
@@ -75,8 +77,73 @@ assert(
test.add('a');
var result = test.add('a');
return result != undefined && result === false;
+ })()
+);
+```
+
+Your `Set` class should have a `remove` method.
+
+```js
+assert(
+ (function () {
+ var test = new Set();
+ return typeof test.remove === 'function';
+ })()
+);
+```
+
+Your `remove` method should only remove items that are present in the set.
+
+```js
+assert.deepEqual(
+ (function () {
+ var test = new Set();
+ test.add('a');
+ test.add('b');
+ test.remove('c');
+ return test.values();
})(),
- 'Your <code>add</code> method should return <code>false</code> when a duplicate value is added.'
+ ['a', 'b']
+);
+```
+
+Your `remove` method should remove the given item from the set.
+
+```js
+assert(
+ (function () {
+ var test = new Set();
+ test.add('a');
+ test.add('b');
+ test.remove('a');
+ var vals = test.values();
+ return vals[0] === 'b' && vals.length === 1;
+ })()
+);
+```
+
+Your `Set` class should have a `size` method.
+
+```js
+assert(
+ (function () {
+ var test = new Set();
+ return typeof test.size === 'function';
+ })()
+);
+```
+
+The `size` method should return the number of elements in the collection.
+
+```js
+assert(
+ (function () {
+ var test = new Set();
+ test.add('a');
+ test.add('b');
+ test.remove('a');
+ return test.size() === 1;
+ })()
);
```
@@ -103,7 +170,7 @@ class Set {
}
// Only change code below this line
-
+
// Only change code above this line
}
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-stack-class.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-stack-class.md
index 5ac5bbc03b3..a26dec74f65 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-stack-class.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-stack-class.md
@@ -1,18 +1,22 @@
---
id: 587d8250367417b2b2512c5f
-title: 创建一个堆栈类
+title: Create a Stack Class
challengeType: 1
-videoUrl: ''
+forumTopicId: 301633
dashedName: create-a-stack-class
---
# --description--
-在上一节中，我们讨论了堆栈是什么以及如何使用数组来表示堆栈。在本节中，我们将创建自己的堆栈类。虽然您可以使用数组来创建堆栈，但有时最好限制我们对堆栈的控制量。除了`push`和`pop`方法之外，堆栈还有其他有用的方法。让我们为我们的堆栈类添加一个`peek` ， `isEmpty`和`clear`方法。说明编写一个`push`方法，将元素推送到堆栈顶部，一个`pop`方法删除堆栈顶部的元素，一个`peek`堆栈中第一个元素的`peek`方法，一个`isEmpty`方法，用于检查是否存在stack是空的，是一个`clear`堆栈中所有元素的方法。通常堆栈没有这个，但我们添加了一个控制台记录集合的`print`助手方法。
+In the last section, we talked about what a stack is and how we can use an array to represent a stack. In this section, we will be creating our own stack class. Although you can use arrays to create stacks, sometimes it is best to limit the amount of control we have with our stacks. Apart from the `push` and `pop` method, stacks have other useful methods. Let's add a `peek`, `isEmpty`, and `clear` method to our stack class.
+
+# --instructions--
+
+Write a `push` method that pushes an element to the top of the stack, a `pop` method that removes and returns the element on the top of the stack, a `peek` method that looks at the top element in the stack, an `isEmpty` method that checks if the stack is empty, and a `clear` method that removes all elements from the stack. Normally stacks don't have this, but we've added a `print` helper method that console logs the collection.
# --hints--
-你的`Stack`类应该有一个`push`方法。
+Your `Stack` class should have a `push` method.
```js
assert(
@@ -23,7 +27,7 @@ assert(
);
```
-你的`Stack`类应该有一个`pop`方法。
+Your `Stack` class should have a `pop` method.
```js
assert(
@@ -34,7 +38,7 @@ assert(
);
```
-你的`Stack`类应该有一个`peek`方法。
+Your `Stack` class should have a `peek` method.
```js
assert(
@@ -45,7 +49,7 @@ assert(
);
```
-您的`Stack`类应该有一个`isEmpty`方法。
+Your `Stack` class should have a `isEmpty` method.
```js
assert(
@@ -56,7 +60,7 @@ assert(
);
```
-你的`Stack`类应该有一个`clear`方法。
+Your `Stack` class should have a `clear` method.
```js
assert(
@@ -67,7 +71,7 @@ assert(
);
```
-`peek`方法应该返回堆栈的顶部元素
+The `peek` method should return the top element of the stack
```js
assert(
@@ -79,7 +83,7 @@ assert(
);
```
-`pop`方法应该删除并返回堆栈的顶部元素
+The `pop` method should remove and return the top element of the stack
```js
assert(
@@ -91,7 +95,7 @@ assert(
);
```
-如果堆栈不包含任何元素，则`isEmpty`方法应返回true
+The `isEmpty` method should return true if a stack does not contain any elements
```js
assert(
@@ -102,7 +106,7 @@ assert(
);
```
-`clear`方法应该从堆栈中删除所有元素
+The `clear` method should remove all element from the stack
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-trie-search-tree.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-trie-search-tree.md
index 4a3b6ed355a..fb838c883e0 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-trie-search-tree.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-a-trie-search-tree.md
@@ -1,18 +1,22 @@
---
id: 587d8259367417b2b2512c84
-title: 创建Trie搜索树
+title: Create a Trie Search Tree
challengeType: 1
-videoUrl: ''
+forumTopicId: 301634
dashedName: create-a-trie-search-tree
---
# --description--
-在这里，我们将继续从二叉搜索树开始，看看另一种称为trie的树结构。 trie是一种常用于保存字符串的有序搜索树，或者更通用的关联数组或其中键是字符串的动态数据集。当许多键具有重叠前缀时，它们非常擅长存储数据集，例如，字典中的所有单词。与二叉树不同，节点不与实际值相关联。相反，节点的路径表示特定的键。例如，如果我们想将字符串代码存储在trie中，我们将有四个节点，每个字母对应一个节点：c - o - d - e。然后，通过所有这些节点的路径将创建代码作为字符串 - 该路径是我们存储的密钥。然后，如果我们想要添加字符串编码，它将在d之后分支之前共享前三个代码节点。通过这种方式，可以非常紧凑地存储大型数据集。此外，搜索可以非常快，因为它实际上限于您存储的字符串的长度。此外，与二叉树不同，节点可以存储任意数量的子节点。正如您可能从上面的示例中猜到的那样，一些元数据通常存储在保存密钥结尾的节点上，以便在以后的遍历中仍可以检索密钥。例如，如果我们在上面的示例中添加了代码，我们需要某种方式来知道代码中的e代表先前输入的密钥的结尾。否则，当我们添加代码时，这些信息将会丢失。说明：让我们创建一个存储单词的trie。它将通过add方法接受单词并将它们存储在trie数据结构中。它还允许我们查询给定字符串是否是带有isWord方法的单词，并使用print方法检索输入到trie中的所有单词。 isWord应该返回一个布尔值，print应该将所有这些单词的数组作为字符串值返回。为了让我们验证这个数据结构是否正确实现，我们为树中的每个节点提供了一个Node结构。每个节点都是一个具有keys属性的对象，该属性是JavaScript Map对象。这将保存作为每个节点的有效密钥的各个字母。我们还在节点上创建了一个end属性，如果节点表示单词的终止，则可以将其设置为true。
+Here we will move on from binary search trees and take a look at another type of tree structure called a trie. A trie is an ordered search tree commonly used to hold strings, or more generically associative arrays or dynamic datasets in which the keys are strings. They are very good at storing sets of data when many keys will have overlapping prefixes, for example, all the words in a dictionary. Unlike a binary tree, nodes are not associated with actual values. Instead, the path to a node represents a specific key. For instance, if we wanted to store the string code in a trie, we would have four nodes, one for each letter: c — o — d — e. Following that path through all these nodes will then create code as a string — that path is the key we stored. Then, if we wanted to add the string coding, it would share the first three nodes of code before branching away after the d. In this way, large datasets can be stored very compactly. In addition, search can be very quick because it is effectively limited to the length of the string you are storing. Furthermore, unlike binary trees a node can store any number of child nodes. As you might have guessed from the above example, some metadata is commonly stored at nodes that hold the end of a key so that on later traversals that key can still be retrieved. For instance, if we added codes in our example above we would need some way to know that the e in code represents the end of a key that was previously entered. Otherwise, this information would effectively be lost when we add codes.
+
+# --instructions--
+
+Let's create a trie to store words. It will accept words through an `add` method and store these in a trie data structure. It will also allow us to query if a given string is a word with an `isWord` method, and retrieve all the words entered into the trie with a `print` method. `isWord` should return a boolean value and print should return an array of all these words as string values. In order for us to verify that this data structure is implemented correctly, we've provided a `Node` structure for each node in the tree. Each node will be an object with a `keys` property which is a JavaScript Map object. This will hold the individual letters that are valid keys of each node. We've also created an `end` property on the nodes that can be set to `true` if the node represents the termination of a word.
# --hints--
-Trie有一个add方法。
+The Trie should have an add method.
```js
assert(
@@ -28,7 +32,7 @@ assert(
);
```
-Trie有一种打印方法。
+The Trie should have a print method.
```js
assert(
@@ -44,7 +48,7 @@ assert(
);
```
-Trie有一个isWord方法。
+The Trie should have an isWord method.
```js
assert(
@@ -60,7 +64,7 @@ assert(
);
```
-print方法将添加到trie的所有项目作为数组中的字符串返回。
+The print method should return all items added to the trie as strings in an array.
```js
assert(
@@ -89,7 +93,7 @@ assert(
);
```
-isWord方法仅对添加到trie的单词返回true，对所有其他单词返回false。
+The isWord method should return true only for words added to the trie and false for all other words.
```js
assert(
@@ -135,7 +139,7 @@ var Node = function() {
};
var Trie = function() {
// Only change code below this line
-
+
// Only change code above this line
};
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-an-es6-javascript-map.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-an-es6-javascript-map.md
index 72b3de8caed..96746bab118 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-an-es6-javascript-map.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-an-es6-javascript-map.md
@@ -1,24 +1,28 @@
---
id: 587d825b367417b2b2512c8d
-title: 创建ES6 JavaScript地图
+title: Create an ES6 JavaScript Map
challengeType: 1
-videoUrl: ''
+forumTopicId: 301635
dashedName: create-an-es6-javascript-map
---
# --description--
-新版本的JavaScript为我们提供了一个内置的Map对象，它提供了我们在上一次挑战中手工编写的大部分功能。此Map对象虽然与常规JavaScript对象类似，但它提供了一些普通对象缺少的有用功能。例如，ES6 Map跟踪添加到其中的项目的插入顺序。以下是其方法的更完整概述： `.has(key)`基于键的存在返回true或false `.get(key)`返回与键相关联的值`.set(key, value)`设置新键，值对`.delete(key)`删除一个键，值对`.clear()`删除所有键值对`.entries()`以插入顺序返回所有键的数组`.values()`返回插入中所有值的数组order说明：定义一个JavaScript Map对象并为其分配一个名为myMap的变量。添加密钥，值对`freeCodeCamp` ， `Awesome!`它。
+The new version of JavaScript provides us with a built-in Map object which provides much of the functionality we wrote by hand in the last challenge. This Map object, although similar to regular JavaScript objects, provides some useful functionality that normal objects lack. For example, an ES6 Map tracks the insertion order of items that are added to it. Here is a more complete overview of its methods: `.has(key)` returns true or false based on the presence of a key `.get(key)` returns the value associated with a key `.set(key, value)` sets a new key, value pair `.delete(key)` removes a key, value pair `.clear()` removes all key, value pairs `.entries()` returns an array of all the keys in insertion order `.values()` returns an array of all the values in insertion order
+
+# --instructions--
+
+Define a JavaScript Map object and assign to it a variable called myMap. Add the key, value pair `freeCodeCamp`, `Awesome!` to it.
# --hints--
-myMap对象存在。
+The myMap object should exist.
```js
assert(typeof myMap === 'object');
```
-myMap包含键值对`freeCodeCamp` ， `Awesome!` 。
+myMap should contain the key value pair `freeCodeCamp`, `Awesome!`.
```js
assert(myMap.get('freeCodeCamp') === 'Awesome!');
@@ -27,7 +31,6 @@ assert(myMap.get('freeCodeCamp') === 'Awesome!');
# --seed--
## --seed-contents--
-
```js
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-and-add-to-sets-in-es6.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-and-add-to-sets-in-es6.md
index 224ce7ca96d..0f6b957518f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-and-add-to-sets-in-es6.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/create-and-add-to-sets-in-es6.md
@@ -1,30 +1,48 @@
---
id: 587d8254367417b2b2512c70
-title: 在ES6中创建和添加集
+title: Create and Add to Sets in ES6
challengeType: 1
-videoUrl: ''
+forumTopicId: 301636
dashedName: create-and-add-to-sets-in-es6
---
# --description--
-既然您已经完成了ES5，那么您将在ES6中执行类似的操作。这将相当容易。 ES6包含一个内置的数据结构`Set`现在包含了您手动编写的许多操作。我们来看看：创建一个新的空集： `var set = new Set();`您可以使用值创建一个集合： `var set = new Set(1);`您可以使用数组创建一个集合： `var set = new Set([1, 2, 3]);`创建集合后，可以使用`add`方法添加所需的值：
+Now that you have worked through ES5, you are going to perform something similar in ES6. This will be considerably easier. ES6 contains a built-in data structure `Set` so many of the operations you wrote by hand are now included for you. Let's take a look:
-> var set = new Set（\[1,2,3]）;
-> set.add（\[4,5,6]）;
+To create a new empty set:
-提醒一下，集合是一种不能包含重复值的数据结构：
+`var set = new Set();`
-> var set = new Set（\[1,2,3,1,2,3]）;
-> // set仅包含\[1,2,3]
+You can create a set with a value:
+
+`var set = new Set(1);`
+
+You can create a set with an array:
+
+`var set = new Set([1, 2, 3]);`
+
+Once you have created a set, you can add the values you wish using the `add` method:
+
+```js
+var set = new Set([1, 2, 3]);
+set.add([4, 5, 6]);
+```
+
+As a reminder, a set is a data structure that cannot contain duplicate values:
+
+```js
+var set = new Set([1, 2, 3, 1, 2, 3]);
+// set contains [1, 2, 3] only
+```
# --instructions--
-在本练习中，返回一个具有以下值的集合： `1, 2, 3, 'Taco', 'Cat', 'Awesome'`
+For this exercise, return a set with the following values: `1, 2, 3, 'Taco', 'Cat', 'Awesome'`
# --hints--
-您的`Set`应该只包含值`1, 2, 3, Taco, Cat, Awesome` 。
+Your `Set` should only contain the values `1, 2, 3, Taco, Cat, Awesome`.
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/delete-a-leaf-node-in-a-binary-search-tree.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/delete-a-leaf-node-in-a-binary-search-tree.md
index f6888093b58..43d58845d26 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/delete-a-leaf-node-in-a-binary-search-tree.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/delete-a-leaf-node-in-a-binary-search-tree.md
@@ -1,18 +1,22 @@
---
id: 587d8258367417b2b2512c80
-title: 删除二进制搜索树中的叶节点
+title: Delete a Leaf Node in a Binary Search Tree
challengeType: 1
-videoUrl: ''
+forumTopicId: 301637
dashedName: delete-a-leaf-node-in-a-binary-search-tree
---
# --description--
-这是我们将在二叉搜索树中实现更难操作的三个挑战中的第一个：删除。删除很困难，因为删除节点会破坏树中的链接。必须仔细重新建立这些链接以确保维护二叉树结构。对于某些删除，这意味着必须重新排列树。通常，在尝试删除节点时，您将遇到以下三种情况之一：叶节点：要删除的目标没有子节点。一个孩子：要删除的目标只有一个孩子。两个子节点：要删除的目标有两个子节点。删除叶节点很简单，我们只需删除它。删除具有一个子节点的节点也相对容易，我们只需删除它并将其父节点链接到我们删除的节点的子节点。但是，删除具有两个子节点的节点更加困难，因为这会创建两个需要重新连接到父树的子节点。我们将在第三个挑战中看到如何处理这个案例。此外，在处理删除时，您需要注意一些边缘情况。如果树是空的怎么办？如果要删除的节点是根节点怎么办？如果树中只有两个元素怎么办？现在，让我们处理第一种删除叶节点的情况。说明：在我们的二叉树上创建一个名为`remove` 。我们将在这里为我们的删除操作构建逻辑。首先，您需要在remove中创建一个函数，该函数在当前树中找到我们尝试删除的节点。如果树中不存在该节点，则`remove`应返回`null` 。现在，如果目标节点是没有子节点的叶节点，则应将其父节点引用设置为`null` 。这有效地从树中删除节点。为此，您必须跟踪我们尝试删除的节点的父节点。创建一种跟踪目标节点具有的子节点数的方法也很有用，因为这将确定我们的删除属于哪种情况。我们将在下一次挑战中处理第二和第三个案例。祝你好运！
+This is the first of three challenges where we will implement a more difficult operation in binary search trees: deletion. Deletion is difficult because removing nodes breaks links in the tree. These links must be carefully reestablished to ensure the binary tree structure is maintained. For some deletions, this means the tree must be rearranged. In general, you will encounter one of three cases when trying to delete a node: Leaf Node: The target to delete has zero children. One Child: The target to delete only has one child. Two Children: The target to delete has two child nodes. Removing a leaf node is easy, we simply remove it. Deleting a node with one child is also relatively easy, we simply remove it and link its parent to child of the node we deleted. Removing a node with two children is more difficult, however, because this creates two child nodes that need to be reconnected to the parent tree. We'll see how to deal with this case in the third challenge. Additionally, you need to be mindful of some edge cases when handling deletion. What if the tree is empty? What if the node to delete is the root node? What if there are only two elements in the tree? For now, let's handle the first case where we delete a leaf node.
+
+# --instructions--
+
+Create a method on our binary tree called `remove`. We'll build the logic for our deletion operation in here. First, you'll want to create a function within remove that finds the node we are trying to delete in the current tree. If the node is not present in the tree, `remove` should return `null`. Now, if the target node is a leaf node with no children, then the parent reference to it should be set to `null`. This effectively deletes the node from the tree. To do this, you will have to keep track of the parent of the node we are trying to delete as well. It will also be useful to create a way to track the number of children the target node has, as this will determine which case our deletion falls under. We will handle the second and third cases in the next challenges. Good luck!
# --hints--
-存在`BinarySearchTree`数据结构。
+The `BinarySearchTree` data structure should exist.
```js
assert(
@@ -26,7 +30,7 @@ assert(
);
```
-二叉搜索树有一个名为`remove`的方法。
+The binary search tree should have a method called `remove`.
```js
assert(
@@ -42,7 +46,7 @@ assert(
);
```
-尝试删除不存在的元素将返回`null` 。
+Trying to remove an element that does not exist should return `null`.
```js
assert(
@@ -61,7 +65,7 @@ assert(
);
```
-如果根节点没有子节点，则删除它会将根节点设置为`null` 。
+If the root node has no children, deleting it should set the root to `null`.
```js
assert(
@@ -82,7 +86,7 @@ assert(
);
```
-`remove`方法从树中删除叶节点
+The `remove` method should remove leaf nodes from the tree.
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/delete-a-node-with-one-child-in-a-binary-search-tree.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/delete-a-node-with-one-child-in-a-binary-search-tree.md
index f69987caf2d..db24b0b29ee 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/delete-a-node-with-one-child-in-a-binary-search-tree.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/delete-a-node-with-one-child-in-a-binary-search-tree.md
@@ -1,18 +1,22 @@
---
id: 587d8258367417b2b2512c81
-title: 在二叉搜索树中删除具有一个子节点的节点
+title: Delete a Node with One Child in a Binary Search Tree
challengeType: 1
-videoUrl: ''
+forumTopicId: 301638
dashedName: delete-a-node-with-one-child-in-a-binary-search-tree
---
# --description--
-现在我们可以删除叶子节点，让我们继续第二种情况：删除一个子节点。对于这种情况，假设我们有一棵树，其中包含以下节点1 - 2 - 3，其中1是根。要删除2，我们只需要在1到3中做出正确的引用。更一般地说，为了删除只有一个子节点的节点，我们将该节点的父引用作为树中的下一个节点。说明：我们在`remove`方法中提供了一些代码，用于完成上一次挑战中的任务。我们找到要删除的目标及其父节点，并定义目标节点具有的子节点数。让我们在这里为仅有一个子节点的目标节点添加下一个案例。在这里，我们必须确定单个子节点是树中的左或右分支，然后在父节点中设置正确的引用以指向此节点。另外，让我们考虑目标是根节点的情况（这意味着父节点将为`null` ）。只要通过测试，请随意用自己的代码替换所有入门代码。
+Now that we can delete leaf nodes let's move on to the second case: deleting a node with one child. For this case, say we have a tree with the following nodes 1 — 2 — 3 where 1 is the root. To delete 2, we simply need to make the right reference in 1 point to 3. More generally to delete a node with only one child, we make that node's parent reference the next node in the tree.
+
+# --instructions--
+
+We've provided some code in our `remove` method that accomplishes the tasks from the last challenge. We find the target to delete and its parent and define the number of children the target node has. Let's add the next case here for target nodes with only one child. Here, we'll have to determine if the single child is a left or right branch in the tree and then set the correct reference in the parent to point to this node. In addition, let's account for the case where the target is the root node (this means the parent node will be `null`). Feel free to replace all the starter code with your own as long as it passes the tests.
# --hints--
-存在`BinarySearchTree`数据结构。
+The `BinarySearchTree` data structure should exist.
```js
assert(
@@ -26,7 +30,7 @@ assert(
);
```
-二叉搜索树有一个名为`remove`的方法。
+The binary search tree should have a method called `remove`.
```js
assert(
@@ -42,7 +46,7 @@ assert(
);
```
-尝试删除不存在的元素将返回`null` 。
+Trying to remove an element that does not exist should return `null`.
```js
assert(
@@ -61,7 +65,7 @@ assert(
);
```
-如果根节点没有子节点，则删除它会将根节点设置为`null` 。
+If the root node has no children, deleting it should set the root to `null`.
```js
assert(
@@ -82,7 +86,7 @@ assert(
);
```
-`remove`方法从树中删除叶节点
+The `remove` method should remove leaf nodes from the tree.
```js
assert(
@@ -110,7 +114,7 @@ assert(
);
```
-`remove`方法删除具有一个子节点的节点。
+The `remove` method should remove nodes with one child.
```js
assert(
@@ -136,7 +140,7 @@ assert(
);
```
-删除具有两个节点的树中的根将第二个节点设置为根。
+Removing the root in a tree with two nodes should set the second to be the root.
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/delete-a-node-with-two-children-in-a-binary-search-tree.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/delete-a-node-with-two-children-in-a-binary-search-tree.md
index 85f46647be7..aab61e9c37f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/delete-a-node-with-two-children-in-a-binary-search-tree.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/delete-a-node-with-two-children-in-a-binary-search-tree.md
@@ -1,18 +1,22 @@
---
id: 587d8258367417b2b2512c82
-title: 在二叉搜索树中删除具有两个子节点的节点
+title: Delete a Node with Two Children in a Binary Search Tree
challengeType: 1
-videoUrl: ''
+forumTopicId: 301639
dashedName: delete-a-node-with-two-children-in-a-binary-search-tree
---
# --description--
-删除具有两个子节点的节点是最难实现的。删除这样的节点会生成两个不再连接到原始树结构的子树。我们如何重新连接它们？一种方法是在目标节点的右子树中找到最小值，并用该值替换目标节点。以这种方式选择替换确保它大于左子树中的每个节点，它成为新的父节点，但也小于右子树中的每个节点，它成为新的父节点。完成此替换后，必须从右子树中删除替换节点。即使这个操作也很棘手，因为替换可能是一个叶子，或者它本身可能是一个右子树的父亲。如果是叶子，我们必须删除其父对它的引用。否则，它必须是目标的正确子项。在这种情况下，我们必须用替换值替换目标值，并使目标引用替换的右子。说明：让我们通过处理第三种情况来完成我们的`remove`方法。我们为前两种情况再次提供了一些代码。现在添加一些代码来处理具有两个子节点的目标节点。任何边缘情况要注意？如果树只有三个节点怎么办？完成后，这将完成二进制搜索树的删除操作。干得好，这是一个非常难的问题！
+Removing nodes that have two children is the hardest case to implement. Removing a node like this produces two subtrees that are no longer connected to the original tree structure. How can we reconnect them? One method is to find the smallest value in the right subtree of the target node and replace the target node with this value. Selecting the replacement in this way ensures that it is greater than every node in the left subtree it becomes the new parent of but also less than every node in the right subtree it becomes the new parent of. Once this replacement is made the replacement node must be removed from the right subtree. Even this operation is tricky because the replacement may be a leaf or it may itself be the parent of a right subtree. If it is a leaf we must remove its parent's reference to it. Otherwise, it must be the right child of the target. In this case, we must replace the target value with the replacement value and make the target reference the replacement's right child.
+
+# --instructions--
+
+Let's finish our `remove` method by handling the third case. We've provided some code again for the first two cases. Add some code now to handle target nodes with two children. Any edge cases to be aware of? What if the tree has only three nodes? Once you are finished this will complete our deletion operation for binary search trees. Nice job, this is a pretty hard problem!
# --hints--
-存在`BinarySearchTree`数据结构。
+The `BinarySearchTree` data structure should exist.
```js
assert(
@@ -26,7 +30,7 @@ assert(
);
```
-二叉搜索树有一个名为`remove`的方法。
+The binary search tree should have a method called `remove`.
```js
assert(
@@ -42,7 +46,7 @@ assert(
);
```
-尝试删除不存在的元素将返回`null` 。
+Trying to remove an element that does not exist should return `null`.
```js
assert(
@@ -58,7 +62,7 @@ assert(
);
```
-如果根节点没有子节点，则删除它会将根节点设置为`null` 。
+If the root node has no children, deleting it should set the root to `null`.
```js
assert(
@@ -76,7 +80,7 @@ assert(
);
```
-`remove`方法从树中删除叶节点
+The `remove` method should remove leaf nodes from the tree.
```js
assert(
@@ -103,7 +107,7 @@ assert(
);
```
-`remove`方法删除具有一个子节点的节点。
+The `remove` method should remove nodes with one child.
```js
assert(
@@ -129,7 +133,7 @@ assert(
);
```
-删除具有两个节点的树中的根将第二个节点设置为根。
+Removing the root in a tree with two nodes should set the second to be the root.
```js
assert(
@@ -151,7 +155,7 @@ assert(
);
```
-`remove`方法在保留二叉搜索树结构的同时删除具有两个子节点的节点。
+The `remove` method should remove nodes with two children while maintaining the binary search tree structure.
```js
assert(
@@ -208,7 +212,7 @@ assert(
);
```
-可以在三个节点的树上删除根。
+The root should be removable on a tree of three nodes.
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/depth-first-search.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/depth-first-search.md
index 89acbab5bf3..8fedd91cc51 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/depth-first-search.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/depth-first-search.md
@@ -1,26 +1,38 @@
---
id: 587d825d367417b2b2512c96
-title: 深度优先搜索
+title: Depth-First Search
challengeType: 1
-videoUrl: ''
+forumTopicId: 301640
dashedName: depth-first-search
---
# --description--
-与<dfn>广度优先搜索</dfn>类似，这里我们将学习另一种称为<dfn>深度优先搜索的</dfn>图遍历算法。广度优先搜索搜索远离源节点的增量边长度，而<dfn>深度优先搜索</dfn>首先尽可能地沿着边缘路径向下<dfn>搜索</dfn> 。一旦到达路径的一端，搜索将回溯到具有未访问边缘路径的最后一个节点并继续搜索。在视觉上，这就是算法正在做的事情，其中顶部节点是搜索的起始点。
+Similar to <dfn>breadth-first search</dfn>, here we will learn about another graph traversal algorithm called <dfn>depth-first search</dfn>.
-<img class='img-responsive' src='https://camo.githubusercontent.com/aaad9e39961daf34d967c616edeb50abf3bf1235/68747470733a2f2f75706c6f61642e77696b696d656469612e6f72672f77696b6970656469612f636f6d6d6f6e732f372f37662f44657074682d46697273742d5365617263682e676966'>
+Whereas the breadth-first search searches incremental edge lengths away from the source node, <dfn>depth-first search</dfn> first goes down a path of edges as far as it can.
-该算法的简单输出是可从给定节点到达的节点列表。因此，在实施此算法时，您需要跟踪您访问的节点。
+Once it reaches one end of a path, the search will backtrack to the last node with an un-visited edge path and continue searching.
+
+The animation below shows how the algorithm works. The algorithm starts with the top node and visits the nodes in the numbered order.
+
+<img class='img-responsive' src='https://camo.githubusercontent.com/aaad9e39961daf34d967c616edeb50abf3bf1235/68747470733a2f2f75706c6f61642e77696b696d656469612e6f72672f77696b6970656469612f636f6d6d6f6e732f372f37662f44657074682d46697273742d5365617263682e676966' />
+
+Notice how, unlike breadth-first search, every time a node is visited, it doesn't visit all of its neighbors. Instead, it first visits one of its neighbors and continues down that path until there are no more nodes to be visited on that path.
+
+To implement this algorithm, you'll want to use a stack. A stack is an array where the last element added is the first to be removed. This is also known as a <dfn>Last-In-First-Out</dfn> data structure. A stack is helpful in depth-first search algorithms because, as we add neighbors to the stack, we want to visit the most recently added neighbors first and remove them from the stack.
+
+A simple output of this algorithm is a list of nodes which are reachable from a given node. Therefore, you'll also want to keep track of the nodes you visit.
# --instructions--
-编写一个函数`dfs()` ，它将无向，邻接矩阵`graph`和节点标签`root`作为参数。节点标签将只是`0`和`n - 1`之间节点的数值，其中`n`是图中节点的总数。您的函数应输出从`root`可到达的所有节点的数组。
+Write a function `dfs()` that takes an undirected, adjacency matrix `graph` and a node label `root` as parameters. The node label will just be the numeric value of the node between `0` and `n - 1`, where `n` is the total number of nodes in the graph.
+
+Your function should output an array of all nodes reachable from `root`.
# --hints--
-输入图`[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]` ，起始节点为`1`应返回一个数组`0` ， `1` ， `2` ，和`3` 。
+The input graph `[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]` with a start node of `1` should return an array with `0`, `1`, `2`, and `3`.
```js
assert.sameMembers(
@@ -37,7 +49,7 @@ assert.sameMembers(
);
```
-输入图`[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]` ，起始节点为`1`应该返回一个包含四个元素的数组。
+The input graph `[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]` with a start node of `1` should return an array with four elements.
```js
assert(
@@ -53,7 +65,7 @@ assert(
);
```
-输入图`[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 0], [0, 0, 0, 0]]` ，起始节点为`3`应该返回一个`3`的数组。
+The input graph `[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 0], [0, 0, 0, 0]]` with a start node of `3` should return an array with `3`.
```js
assert.sameMembers(
@@ -70,7 +82,7 @@ assert.sameMembers(
);
```
-输入图`[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 0], [0, 0, 0, 0]]` ，起始节点为`3`应该返回一个包含一个元素的数组。
+The input graph `[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 0], [0, 0, 0, 0]]` with a start node of `3` should return an array with one element.
```js
assert(
@@ -86,7 +98,7 @@ assert(
);
```
-输入图`[[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]` ，起始节点为`3`应该返回一个`2`和`3`的数组。
+The input graph `[[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]` with a start node of `3` should return an array with `2` and `3`.
```js
assert.sameMembers(
@@ -103,7 +115,7 @@ assert.sameMembers(
);
```
-输入图`[[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]` ，起始节点为`3`应该返回一个包含两个元素的数组。
+The input graph `[[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]` with a start node of `3` should return an array with two elements.
```js
assert(
@@ -119,7 +131,7 @@ assert(
);
```
-输入图`[[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]` ，起始节点为`0`应该返回一个`0`和`1`的数组。
+The input graph `[[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]` with a start node of `0` should return an array with `0` and `1`.
```js
assert.sameMembers(
@@ -136,7 +148,7 @@ assert.sameMembers(
);
```
-输入图`[[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]` ，起始节点为`0`应该返回一个包含两个元素的数组。
+The input graph `[[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]` with a start node of `0` should return an array with two elements.
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/find-the-minimum-and-maximum-height-of-a-binary-search-tree.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/find-the-minimum-and-maximum-height-of-a-binary-search-tree.md
index cd73d0e9c02..eb71f5ea47f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/find-the-minimum-and-maximum-height-of-a-binary-search-tree.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/find-the-minimum-and-maximum-height-of-a-binary-search-tree.md
@@ -1,18 +1,26 @@
---
id: 587d8257367417b2b2512c7d
-title: 找到二叉搜索树的最小和最大高度
+title: Find the Minimum and Maximum Height of a Binary Search Tree
challengeType: 1
-videoUrl: ''
+forumTopicId: 301641
dashedName: find-the-minimum-and-maximum-height-of-a-binary-search-tree
---
# --description--
-在最后一个挑战中，我们描述了树可能变得不平衡的情景。为了理解平衡的概念，让我们看看另一个树属性：高度。树中的高度表示从根节点到任何给定叶节点的距离。高度分支的树结构中的不同路径可以具有不同的高度，但是对于给定的树，将具有最小和最大高度。如果树是平衡的，则这些值最多相差一个。这意味着在平衡树中，所有叶节点都存在于同一级别中，或者如果它们不在同一级别内，则它们最多相隔一个级别。平衡的属性对于树很重要，因为它决定了树操作的效率。正如我们在上一次挑战中所解释的那样，我们面临严重不平衡树木的最坏情况时间复杂性。自平衡树通常用于在具有动态数据集的树中解决此问题。这些的常见例子包括AVL树，红黑树和B树。这些树都包含额外的内部逻辑，当插入或删除创建不平衡状态时，它会重新平衡树。注意：与height相似的属性是depth，它指的是给定节点距根节点的距离。说明：为我们的二叉树编写两种方法： `findMinHeight`和`findMaxHeight` 。这些方法应分别返回给定二叉树内最小和最大高度的整数值。如果节点为空，请为其指定高度`-1` （这是基本情况）。最后，添加第三个方法`isBalanced` ，它返回`true`或`false`具体取决于树是否平衡。您可以使用刚才编写的前两种方法来确定这一点。
+In the last challenge we described a scenario in which a tree could become unbalanced. To understand the concept of balance, let's take a look at another tree property: height. Height in a tree represents the distance from the root node to any given leaf node. Different paths in a highly branched tree structure may have different heights, but for a given tree there will be a minimum and maximum height. If the tree is balanced, these values will differ at most by one. This means that in a balanced tree, all the leaf nodes exist within the same level, or if they are not within the same level they are at most one level apart.
+
+The property of balance is important for trees because it is what determines the efficiency of tree operations. As we explained in the last challenge, we face worst case time complexity for heavily unbalanced trees. Self-balancing trees are commonly used to account for this issue in trees with dynamic data sets. Common examples of these include AVL trees, red-black trees, and B-trees. These trees all contain additional internal logic which re-balance the tree when insertions or deletions create a state of imbalance.
+
+**Note:** A similar property to height is depth, which refers to how far a given node is from the root node.
+
+# --instructions--
+
+Write two methods for our binary tree: `findMinHeight` and `findMaxHeight`. These methods should return an integer value for the minimum and maximum height within a given binary tree, respectively. If the node is empty let's assign it a height of `-1` (that's the base case). Finally, add a third method `isBalanced` which returns `true` or `false` depending on whether the tree is balanced or not. You can use the first two methods you just wrote to determine this.
# --hints--
-存在`BinarySearchTree`数据结构。
+The `BinarySearchTree` data structure should exist.
```js
assert(
@@ -26,7 +34,7 @@ assert(
);
```
-二叉搜索树有一个名为`findMinHeight`的方法。
+The binary search tree should have a method called `findMinHeight`.
```js
assert(
@@ -42,7 +50,7 @@ assert(
);
```
-二叉搜索树有一个名为`findMaxHeight`的方法。
+The binary search tree should have a method called `findMaxHeight`.
```js
assert(
@@ -58,7 +66,7 @@ assert(
);
```
-二叉搜索树有一个名为`isBalanced`的方法。
+The binary search tree should have a method called `isBalanced`.
```js
assert(
@@ -74,7 +82,7 @@ assert(
);
```
-`findMinHeight`方法返回树的最小高度。
+The `findMinHeight` method should return the minimum height of the tree.
```js
assert(
@@ -101,7 +109,7 @@ assert(
);
```
-`findMaxHeight`方法返回树的最大高度。
+The `findMaxHeight` method should return the maximum height of the tree.
```js
assert(
@@ -128,7 +136,7 @@ assert(
);
```
-空树返回高度`-1` 。
+An empty tree should return a height of `-1`.
```js
assert(
@@ -147,7 +155,7 @@ assert(
);
```
-如果树是平衡二叉搜索树，则`isBalanced`方法返回true。
+The `isBalanced` method should return `false` if the tree is an unbalanced binary search tree.
```js
assert(
@@ -169,7 +177,33 @@ assert(
test.add(45);
test.add(73);
test.add(8);
- return !test.isBalanced();
+ return test.isBalanced() === false;
+ })()
+);
+```
+
+The `isBalanced` method should return `true` if the tree is a balanced binary search tree.
+
+```js
+assert(
+ (function () {
+ var test = false;
+ if (typeof BinarySearchTree !== 'undefined') {
+ test = new BinarySearchTree();
+ } else {
+ return false;
+ }
+ if (typeof test.isBalanced !== 'function') {
+ return false;
+ }
+ test.add(10);
+ test.add(3);
+ test.add(22);
+ test.add(1);
+ test.add(4);
+ test.add(17);
+ test.add(32);
+ return test.isBalanced() === true;
})()
);
```
@@ -227,7 +261,7 @@ function Node(value) {
function BinarySearchTree() {
this.root = null;
// Only change code below this line
-
+
// Only change code above this line
}
```
@@ -244,7 +278,7 @@ function Node(value) {
function BinarySearchTree() {
this.root = null;
// Only change code below this line
-
+
// Only change code above this line
this.findMinHeight = function(root = this.root) {
// empty tree.
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/find-the-minimum-and-maximum-value-in-a-binary-search-tree.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/find-the-minimum-and-maximum-value-in-a-binary-search-tree.md
index 30720fe407f..23077f9d938 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/find-the-minimum-and-maximum-value-in-a-binary-search-tree.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/find-the-minimum-and-maximum-value-in-a-binary-search-tree.md
@@ -1,20 +1,18 @@
---
id: 587d8256367417b2b2512c7a
-title: 在二叉搜索树中查找最小值和最大值
+title: Find the Minimum and Maximum Value in a Binary Search Tree
challengeType: 1
-videoUrl: ''
+forumTopicId: 301642
dashedName: find-the-minimum-and-maximum-value-in-a-binary-search-tree
---
# --description--
-这一系列挑战将介绍树数据结构。树木是计算机科学中重要且通用的数据结构。当然，他们的名字来源于这样一个事实：当他们看到它们时，它们看起来很像我们在自然界中熟悉的树木。树数据结构以一个节点（通常称为根）开始，并从此处分支到其他节点，每个节点可以具有更多子节点，依此类推。数据结构通常以顶部的根节点可视化;你可以把它想象成一棵倒置的天然树。首先，让我们描述一下我们将在树上遇到的一些常用术语。根节点是树的顶部。树中的数据点称为节点。具有通向其他节点的分支的节点被称为分支通向的节点的父节点（子节点）。其他更复杂的家庭术语适用于您所期望的。子树是指特定节点的所有后代，分支可以称为边，而叶节点是树末端没有子节点的节点。最后，请注意树本质上是递归数据结构。也就是说，节点的任何子节点都是其子树的父节点，依此类推。在为常见树操作设计算法时，树的递归性质非常重要。首先，我们将讨论一种特定类型的树，即二叉树。实际上，我们实际上将讨论一个特定的二叉树，一个二叉搜索树。让我们来描述这意味着什么。虽然树数据结构可以在单个节点上具有任意数量的分支，但是二叉树对于每个节点只能具有两个分支。此外，针对子子树排序二叉搜索树，使得左子树中的每个节点的值小于或等于父节点的值，并且右子树中的每个节点的值是大于或等于父节点的值。
-
-现在这个有序的关系很容易看到。请注意，根节点8左侧的每个值都小于8，右侧的每个值都大于8.还要注意，此关系也适用于每个子树。例如，第一个左子项是子树。 3是父节点，它有两个子节点 - 通过控制二进制搜索树的规则，我们知道甚至没有看到这个节点的左子节点（及其任何子节点）将小于3，右边child（及其任何子级）将大于3（但也小于结构的根值），依此类推。二进制搜索树是非常常见且有用的数据结构，因为它们在几种常见操作（例如查找，插入和删除）的平均情况下提供对数时间。说明：我们将从简单开始。除了为树创建节点的函数之外，我们还在这里定义了二叉搜索树结构的骨架。观察每个节点可能具有左右值。如果它们存在，将为它们分配子子树。在我们的二叉搜索树中，定义两个方法， `findMin`和`findMax` 。这些方法应返回二叉搜索树中保存的最小值和最大值（不用担心现在向树中添加值，我们在后台添加了一些值）。如果遇到困难，请反思二进制搜索树必须为true的不变量：每个左子树小于或等于其父树，每个右子树大于或等于其父树。我们还要说我们的树只能存储整数值。如果树为空，则任一方法都应返回`null` 。
+In this challenge you will define two methods, `findMin` and `findMax`. These methods should return the minimum and maximum value held in the binary search tree (don't worry about adding values to the tree for now, we have added some in the background). If you get stuck, reflect on the invariant that must be true for binary search trees: each left subtree is less than or equal to its parent and each right subtree is greater than or equal to its parent. Let's also say that our tree can only store integer values. If the tree is empty, either method should return `null`.
# --hints--
-存在`BinarySearchTree`数据结构。
+The `BinarySearchTree` data structure should exist.
```js
assert(
@@ -28,7 +26,7 @@ assert(
);
```
-二叉搜索树有一个名为`findMin`的方法。
+The binary search tree should have a method called `findMin`.
```js
assert(
@@ -44,7 +42,7 @@ assert(
);
```
-二叉搜索树有一个名为`findMax`的方法。
+The binary search tree should have a method called `findMax`.
```js
assert(
@@ -60,7 +58,7 @@ assert(
);
```
-`findMin`方法返回二叉搜索树中的最小值。
+The `findMin` method should return the minimum value in the binary search tree.
```js
assert(
@@ -87,7 +85,7 @@ assert(
);
```
-`findMax`方法返回二叉搜索树中的最大值。
+The `findMax` method should return the maximum value in the binary search tree.
```js
assert(
@@ -114,7 +112,7 @@ assert(
);
```
-`findMin`和`findMax`方法为空树返回`null` 。
+The `findMin` and `findMax` methods should return `null` for an empty tree.
```js
assert(
@@ -189,7 +187,7 @@ function Node(value) {
function BinarySearchTree() {
this.root = null;
// Only change code below this line
-
+
// Only change code above this line
}
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/implement-heap-sort-with-a-min-heap.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/implement-heap-sort-with-a-min-heap.md
index fedea5e8bbe..158817a64db 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/implement-heap-sort-with-a-min-heap.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/implement-heap-sort-with-a-min-heap.md
@@ -1,18 +1,24 @@
---
id: 587d825b367417b2b2512c8c
-title: 用最小堆实现堆排序
+title: Implement Heap Sort with a Min Heap
challengeType: 1
-videoUrl: ''
+forumTopicId: 301643
dashedName: implement-heap-sort-with-a-min-heap
---
# --description--
-现在我们可以添加和删除元素，让我们看看堆可用于的一些应用程序。堆通常用于实现优先级队列，因为它们始终将最大值或最小值的项存储在第一个位置。此外，它们还用于实现称为堆排序的排序算法。我们将在这里看到如何做到这一点。堆排序使用最小堆，与最大堆相反。最小堆始终将最小值的元素存储在根位置。堆排序通过获取未排序的数组，将数组中的每个项目添加到最小堆中，然后将最小堆中的每个项目提取到新数组中。最小堆结构确保新数组将包含至少最大顺序的原始项。这是最有效的排序算法之一，具有O（nlog（n））的平均和最差情况性能。说明：让我们用最小堆实现堆排序。您可以在此处调整最大堆代码。使用insert，remove和sort方法创建一个MinHeap对象。 sort方法应返回最小堆中从最小到最大排序的所有元素的数组。
+Now that we can add and remove elements let's see some of the applications heaps can be used for. Heaps are commonly used to implement priority queues because they always store an item of greatest or least value in first position. In addition, they are used to implement a sorting algorithm called heap sort. We'll see how to do this here. Heap sort uses a min heap, the reverse of a max heap. A min heap always stores the element of least value in the root position.
+
+Heap sort works by taking an unsorted array, adding each item in the array into a min heap, and then extracting every item out of the min heap into a new array. The min heap structure ensures that the new array will contain the original items in least to greatest order. This is one of the most efficient sorting algorithms with average and worst case performance of O(nlog(n)).
+
+# --instructions--
+
+Let's implement heap sort with a min heap. Feel free to adapt your max heap code here. Create an object `MinHeap` with `insert`, `remove`, and `sort` methods. The `sort` method should return an array of all the elements in the min heap sorted from smallest to largest.
# --hints--
-存在MinHeap数据结构。
+The MinHeap data structure should exist.
```js
assert(
@@ -26,7 +32,7 @@ assert(
);
```
-MinHeap有一个名为insert的方法。
+MinHeap should have a method called insert.
```js
assert(
@@ -42,7 +48,7 @@ assert(
);
```
-MinHeap有一个名为remove的方法。
+MinHeap should have a method called remove.
```js
assert(
@@ -58,7 +64,7 @@ assert(
);
```
-MinHeap有一个名为sort的方法。
+MinHeap should have a method called sort.
```js
assert(
@@ -74,28 +80,31 @@ assert(
);
```
-sort方法返回一个数组，其中包含按排序顺序添加到最小堆的所有项。
+The sort method should return an array containing all items added to the min heap in sorted order.
```js
assert(
- (function () {
- var test = false;
- if (typeof MinHeap !== 'undefined') {
- test = new MinHeap();
- } else {
+ (() => {
+ if (typeof MinHeap === 'undefined') {
return false;
}
- test.insert(3);
- test.insert(12);
- test.insert(5);
- test.insert(10);
- test.insert(1);
- test.insert(27);
- test.insert(42);
- test.insert(57);
- test.insert(5);
- var result = test.sort();
- return isSorted(result);
+
+ const heap = new MinHeap();
+ const arr = createRandomArray(25);
+
+ for (let i of arr) {
+ heap.insert(i);
+ }
+
+ const result = heap.sort();
+ arr.sort((a, b) => a - b);
+
+ for (let i = 0; i < arr.length; i++) {
+ if (arr[i] !== result[i]) {
+ return false;
+ }
+ }
+ return true;
})()
);
```
@@ -116,14 +125,14 @@ function createRandomArray(size = 5){
let a = new Array(size);
for(let i = 0; i < size; i++)
a[i] = Math.floor(Math.random() * 100);
-
+
return a;
}
const array = createRandomArray(25);
var MinHeap = function() {
// Only change code below this line
-
+
// Only change code above this line
};
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/incidence-matrix.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/incidence-matrix.md
index 15b3307b8c7..068b629cdd8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/incidence-matrix.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/incidence-matrix.md
@@ -1,69 +1,60 @@
---
id: 587d8256367417b2b2512c79
-title: 发生率矩阵
+title: Incidence Matrix
challengeType: 1
-videoUrl: ''
+forumTopicId: 301644
dashedName: incidence-matrix
---
# --description--
-表示图形的另一种方式是将其置于<dfn>关联矩阵中。</dfn> <dfn>入射矩阵</dfn>是二维（2D）阵列。一般而言，关联矩阵在其两个维度之间涉及两个不同类别的对象。这种矩阵类似于邻接矩阵。但是，行和列在这里意味着其他东西。在图表中，我们有边缘和节点。这些将是我们的“两类不同的对象”。该矩阵将使行为节点，列为边。这意味着我们可以拥有不均匀的行数和列数。每列将代表一个独特的边缘。此外，每个边连接两个节点。要显示两个节点之间存在边缘，您将在特定列的两行中放置1。下面是一个3节点图，节点1和节点3之间有一条边。
+Yet another way to represent a graph is to put it in an <dfn>incidence matrix.</dfn>
-> 1
->
-> * * *
->
-> 1 | 1
-> 2 | 0
-> 3 | 1
+An <dfn>incidence matrix</dfn> is a two-dimensional (2D) array. Generally speaking, an incidence matrix relates two different classes of objects between its two dimensions. This kind of matrix is similar to an adjacency matrix. However, the rows and columns mean something else here.
-以下是具有4个边和4个节点的`incidence matrix`的示例。请记住，列是边，行是节点本身。
+In graphs, we have edges and nodes. These will be our "two different classes of objects". This matrix will have the rows be the nodes and columns be the edges. This means that we can have an uneven number of rows and columns.
-> 1 2 3 4
->
-> * * *
->
-> 1 | 0 1 1 1
-> 2 | 1 1 0 0
-> 3 | 1 0 0 1
-> 4 | 0 0 1 0
+Each column will represent a unique edge. Also, each edge connects two nodes. To show that there is an edge between two nodes, you will put a 1 in the two rows of a particular column. Below is a 3 node graph with one edge between node 1 and node 3.
-下面是同一件事的JavaScript实现。
+<blockquote> 1<br> ---<br>1 | 1<br>2 | 0<br>3 | 1</blockquote>
-> var incMat = \[
-> \[0,1,1,1]，
-> \[1,1,0,0]，
-> \[1,0,0,1]，
-> \[0,0,1,0]
-> ]。
+Here is an example of an `incidence matrix` with 4 edges and 4 nodes. Remember, the columns are the edges and rows are the nodes themselves.
-要制作有向图，请使用`-1`表示离开特定节点的边，使用`1`作为边进入节点。
+<blockquote> 1 2 3 4<br> --------<br>1 | 0 1 1 1<br>2 | 1 1 0 0<br>3 | 1 0 0 1<br>4 | 0 0 1 0</blockquote>
-> var incMatDirected = \[
-> \[0，-1,1，-1]，
-> \[-1,1,0,0]，
-> \[1,0,0,1]，
-> \[0,0，-1,0]
-> ]。
+Below is a JavaScript implementation of the same thing.
-图形的边缘也可以有权
+```js
+var incMat = [
+ [0, 1, 1, 1],
+ [1, 1, 0, 0],
+ [1, 0, 0, 1],
+ [0, 0, 1, 0]
+];
+```
-<dfn>重</dfn>
+To make a directed graph, use `-1` for an edge leaving a particular node and `1` for an edge entering a node.
- 。到目前为止，我们有
+```js
+var incMatDirected = [
+ [ 0, -1, 1, -1],
+ [-1, 1, 0, 0],
+ [ 1, 0, 0, 1],
+ [ 0, 0, -1, 0]
+];
+```
-<dfn>未加权的</dfn>
-
-边缘，只有存在和缺少边是二进制（ `0`或`1` ）。根据您的应用，您可以拥有不同的重量。不同的权重表示为大于1的数字。
+Graphs can also have <dfn>weights</dfn> on their edges. So far, we have <dfn>unweighted</dfn> edges where just the presence and lack of edge is binary (`0` or `1`). You can have different weights depending on your application. A different weight is represented as numbers greater than 1.
# --instructions--
-创建具有五个节点和四个边的无向图的关联矩阵。该矩阵应该是多维数组。这五个节点在关系之后具有关系。第一边缘在第一和第二节点之间。第二个边缘位于第二个和第三个节点之间。第三个边缘位于第三个和第五个节点之间。并且四个边缘在第四和第二节点之间。所有边权重均为1，边缘顺序很重要。
+Create an incidence matrix of an undirected graph with five nodes and four edges. This matrix should be in a multi-dimensional array.
+
+These five nodes have the following relationships. The first edge is between the first and second node. The second edge is between the second and third node. The third edge is between the third and fifth node. The fourth edge is between the fourth and second node. All edge weights are one and the edge order matters.
# --hints--
-`incMatUndirected`应该只包含五个节点。
+`incMatUndirected` should only contain five nodes.
```js
assert(
@@ -78,25 +69,25 @@ assert(
);
```
-第一个和第二个节点之间应该有第一条边。
+There should be a first edge between the first and second node.
```js
assert(incMatUndirected[0][0] === 1 && incMatUndirected[1][0] === 1);
```
-第二个和第三个节点之间应该有第二条边。
+There should be a second edge between the second and third node.
```js
assert(incMatUndirected[1][1] === 1 && incMatUndirected[2][1] === 1);
```
-第三个和第五个节点之间应该有第三条边。
+There should be a third edge between the third and fifth node.
```js
assert(incMatUndirected[2][2] === 1 && incMatUndirected[4][2] === 1);
```
-第二个和第四个节点之间应该有第四条边。
+There should be a fourth edge between the second and fourth node.
```js
assert(incMatUndirected[1][3] === 1 && incMatUndirected[3][3] === 1);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/insert-an-element-into-a-max-heap.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/insert-an-element-into-a-max-heap.md
index d48c31cedb6..2d20974121c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/insert-an-element-into-a-max-heap.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/insert-an-element-into-a-max-heap.md
@@ -1,18 +1,50 @@
---
id: 587d825a367417b2b2512c8a
-title: 将元素插入最大堆
+title: Insert an Element into a Max Heap
challengeType: 1
-videoUrl: ''
+forumTopicId: 301703
dashedName: insert-an-element-into-a-max-heap
---
# --description--
-现在我们将继续讨论另一个树数据结构，即二进制堆。二进制堆是部分有序的二叉树，它满足堆属性。 heap属性指定父节点和子节点之间的关系。您可能有一个最大堆，其中所有父节点都大于或等于其子节点，或者最小堆，其中反向为真。二进制堆也是完整的二叉树。这意味着树的所有级别都被完全填充，如果最后一级被部分填充，则从左到右填充。虽然二进制堆可以实现为具有包含左和右引用的节点的树结构，但是根据堆属性的部分排序允许我们用数组表示堆。父子关系是我们感兴趣的，通过简单的算术，我们可以计算任何父节点的子节点和任何子节点的父节点。例如，考虑二进制最小堆的数组表示： `[ 6, 22, 30, 37, 63, 48, 42, 76 ]`根节点是第一个元素，6。它的子节点是22和30.如果我们看在这些值的数组索引之间的关系中，对于索引i，子项为2 \* i + 1和2 \* i + 2.同样，索引0处的元素是索引1和2处的这两个子项的父项。通常，我们可以在任何索引处找到节点的父节点，其中包含以下内容：（i - 1）/ 2.当二叉树增长到任意大小时，这些模式将成立。最后，我们可以稍微调整一下，通过跳过数组中的第一个元素，使这个算法更容易。这样做会为给定索引i处的任何元素创建以下关系：示例数组表示形式： `[ null, 6, 22, 30, 37, 63, 48, 42, 76 ]`元素的左子项：i \* 2元素的右子项：i \* 2 + 1一个元素的父元素：i / 2一旦你绕过数学运算，使用数组表示非常有用，因为使用这个算法可以快速确定节点位置，因为你不需要内存使用量减少维护对子节点的引用。说明：这里我们将创建一个最大堆。首先创建一个insert方法，将元素添加到堆中。在插入期间，始终保持堆属性非常重要。对于最大堆，这意味着根元素应始终在树中具有最大值，并且所有父节点应该大于其子节点。对于堆的数组实现，这通常分三步完成：将新元素添加到数组的末尾。如果元素大于其父元素，请切换它们。继续切换，直到新元素小于其父元素或到达树的根。最后，添加一个print方法，该方法返回已添加到堆中的所有项的数组。
+Now we will move on to another tree data structure, the binary heap. A binary heap is a partially ordered binary tree which satisfies the heap property. The heap property specifies a relationship between parent and child nodes. You may have a max heap, in which all parent nodes are greater than or equal to their child nodes, or a min heap, in which the reverse is true. Binary heaps are also complete binary trees. This means that all levels of the tree are fully filled and if the last level is partially filled it is filled from left to right.
+
+While binary heaps may be implemented as tree structures with nodes that contain left and right references, the partial ordering according to the heap property allows us to represent the heap with an array. The parent-children relationship is what we're interested in and with simple arithmetic we can compute the children of any parent and the parent of any child node.
+
+For instance, consider this array representation of a binary min heap:
+
+`[ 6, 22, 30, 37, 63, 48, 42, 76 ]`
+
+The root node is the first element, `6`. Its children are `22` and `30`. If we look at the relationship between the array indices of these values, for index `i` the children are `2 * i + 1` and `2 * i + 2`. Similarly, the element at index `0` is the parent of these two children at indices `1` and `2`. More generally, we can find the parent of a node at any index with the following: `Math.floor((i - 1) / 2)`. These patterns will hold true as the binary tree grows to any size. Finally, we can make a slight adjustment to make this arithmetic even easier by skipping the first element in the array. Doing this creates the following relationship for any element at a given index `i`:
+
+Example array representation:
+
+`[ null, 6, 22, 30, 37, 63, 48, 42, 76 ]`
+
+An element's left child: `i * 2`
+
+An element's right child: `i * 2 + 1`
+
+An element's parent: `Math.floor(i / 2)`
+
+Once you wrap your head around the math, using an array representation is very useful because node locations can be quickly determined with this arithmetic and memory usage is diminished because you don't need to maintain references to child nodes.
+
+# --instructions--
+
+Instructions: Here we will create a max heap. Start by just creating an `insert` method which adds elements to our heap. During insertion, it is important to always maintain the heap property. For a max heap this means the root element should always have the greatest value in the tree and all parent nodes should be greater than their children. For an array implementation of a heap, this is typically accomplished in three steps:
+
+<ol>
+ <li>Add the new element to the end of the array.</li>
+ <li>If the element is larger than its parent, switch them.</li>
+ <li>Continue switching until the new element is either smaller than its parent or you reach the root of the tree.</li>
+</ol>
+
+Finally, add a `print` method which returns an array of all the items that have been added to the heap.
# --hints--
-存在MaxHeap数据结构。
+The MaxHeap data structure should exist.
```js
assert(
@@ -26,7 +58,7 @@ assert(
);
```
-MaxHeap有一个名为insert的方法。
+MaxHeap should have a method called insert.
```js
assert(
@@ -42,7 +74,7 @@ assert(
);
```
-MaxHeap有一个名为print的方法。
+MaxHeap should have a method called print.
```js
assert(
@@ -58,7 +90,7 @@ assert(
);
```
-insert方法根据max heap属性添加元素。
+The insert method should add elements according to the max heap property.
```js
assert(
@@ -87,7 +119,7 @@ assert(
```js
var MaxHeap = function() {
// Only change code below this line
-
+
// Only change code above this line
};
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/invert-a-binary-tree.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/invert-a-binary-tree.md
index e697f6633aa..8be38b0d5d8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/invert-a-binary-tree.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/invert-a-binary-tree.md
@@ -1,18 +1,18 @@
---
id: 587d8259367417b2b2512c83
-title: 反转二叉树
+title: Invert a Binary Tree
challengeType: 1
-videoUrl: ''
+forumTopicId: 301704
dashedName: invert-a-binary-tree
---
# --description--
-这里我们将创建一个反转二叉树的函数。给定二叉树，我们希望生成一个新树，它等效于该树的镜像。与原始树的inorder遍历相比，在倒置树上运行inorder遍历将以相反的顺序探索节点。在我们的二叉树上编写一个名为`invert`的方法。调用此方法应该反转当前树结构。理想情况下，我们希望在线性时间内就地执行此操作。也就是说，我们只访问每个节点一次，我们在不使用任何额外内存的情况下修改现有的树结构。祝你好运！
+Here will we create a function to invert a binary tree. Given a binary tree, we want to produce a new tree that is equivalently the mirror image of this tree. Running an inorder traversal on an inverted tree will explore the nodes in reverse order when compared to the inorder traversal of the original tree. Write a method to do this called `invert` on our binary tree. Calling this method should invert the current tree structure. Ideally, we would like to do this in-place in linear time. That is, we only visit each node once and we modify the existing tree structure as we go, without using any additional memory. Good luck!
# --hints--
-存在`BinarySearchTree`数据结构。
+The `BinarySearchTree` data structure should exist.
```js
assert(
@@ -26,7 +26,7 @@ assert(
);
```
-二叉搜索树有一个名为`invert`的方法。
+The binary search tree should have a method called `invert`.
```js
assert(
@@ -42,7 +42,7 @@ assert(
);
```
-`invert`方法正确地反转树结构。
+The `invert` method should correctly invert the tree structure.
```js
assert(
@@ -70,7 +70,7 @@ assert(
);
```
-反转空树返回`null` 。
+Inverting an empty tree should return `null`.
```js
assert(
@@ -160,7 +160,7 @@ function Node(value) {
function BinarySearchTree() {
this.root = null;
// Only change code below this line
-
+
// Only change code above this line
}
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/learn-how-a-stack-works.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/learn-how-a-stack-works.md
index 21e03755bd1..19ea0d3c265 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/learn-how-a-stack-works.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/learn-how-a-stack-works.md
@@ -1,40 +1,50 @@
---
id: 587d8250367417b2b2512c5e
-title: 了解堆栈的工作原理
+title: Learn how a Stack Works
challengeType: 1
-videoUrl: ''
+forumTopicId: 301705
dashedName: learn-how-a-stack-works
---
# --description--
-你可能熟悉桌子上的一摞书。您可能已使用文本编辑器的撤消功能。您也可能习惯按手机上的后退按钮返回应用中的上一个视图。你知道他们都有什么共同之处吗？它们都以某种方式存储数据，以便您可以向后遍历。堆栈中最顶层的书是最后放在那里的书。如果您从堆栈的顶部删除该书，则会显示在最后一本书之前放置的书籍，依此类推。如果你考虑一下，在上面的所有例子中，你都会获得<dfn>Last-In-First-Out</dfn>服务。我们将尝试使用我们的代码来模仿它。该数据存储方案称为<dfn>堆栈</dfn> 。特别是，我们必须实现将JavaScript对象推送到堆栈顶部的`push()`方法;和`pop()`方法，它删除当前位于堆栈顶部的JavaScript对象。
+You are probably familiar with stack of books on your table. You have likely used the undo feature of a text editor. You are also probably used to hitting the back button on your phone to go back to the previous view in your app.
+
+You know what they all have in common? They all store the data in a way so that you can traverse backwards.
+
+The topmost book in the stack was the one that was put there last. If you remove that book from your stack's top, you would expose the book that was put there before the last book and so on.
+
+If you think about it, in all the above examples, you are getting <dfn>Last-In-First-Out</dfn> type of service. We will try to mimic this with our code.
+
+This data storage scheme is called a <dfn>Stack</dfn>. In particular, we would have to implement the `push()` method that pushes JavaScript objects at the top of the stack; and `pop()` method, that removes the JavaScript object that's at the top of the stack at the current moment.
# --instructions--
-这里我们有一堆作为数组表示的家庭作业： `"BIO12"`位于基础， `"PSY44"`位于堆栈的顶部。修改给定的数组，并使用上面提到的JavaScript方法将其视为`stack` 。从堆栈中删除顶部元素`"PSY44"` 。然后添加`"CS50"`作为堆栈的新顶部元素。
+Here we have a stack of homework assignments represented as an array: `"BIO12"` is at the base, and `"PSY44"` is at the top of the stack.
+
+Modify the given array and treat it like a `stack` using the JavaScript methods mentioned above. Remove the top element `"PSY44"` from the stack. Then add `"CS50"` to be the new top element of the stack.
# --hints--
-`homeworkStack`应该只包含4个元素。
+`homeworkStack` should only contain 4 elements.
```js
assert(homeworkStack.length === 4);
```
-`homeworkStack`的最后一个元素应该是`"CS50"` 。
+The last element in `homeworkStack` should be `"CS50"`.
```js
assert(homeworkStack[3] === 'CS50');
```
-`homeworkStack`不应包含`"PSY44"` 。
+`homeworkStack` should not contain `"PSY44"`.
```js
assert(homeworkStack.indexOf('PSY44') === -1);
```
-不应更改`homeworkStack`的初始声明。
+The initial declaration of the `homeworkStack` should not be changed.
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/perform-a-difference-on-two-sets-of-data.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/perform-a-difference-on-two-sets-of-data.md
index bde78cab38b..1c793291283 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/perform-a-difference-on-two-sets-of-data.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/perform-a-difference-on-two-sets-of-data.md
@@ -1,18 +1,20 @@
---
id: 587d8254367417b2b2512c6e
-title: 对两组数据执行差异
+title: Perform a Difference on Two Sets of Data
challengeType: 1
-videoUrl: ''
+forumTopicId: 301706
dashedName: perform-a-difference-on-two-sets-of-data
---
# --description--
-在本练习中，我们将对两组数据进行区分。我们将在我们的`Set`数据结构上创建一个名为`difference` 。集合的差异应比较两组并返回第一组中不存在的项目。此方法应将另一个`Set`作为参数，并返回两个集的`difference` 。例如，如果`setA = ['a','b','c']`和`setB = ['a','b','d','e']` ，则setA和setB的差异为： `setA.difference(setB) = ['c']` 。
+In this exercise we are going to perform a difference on 2 sets of data. We will create a method on our `Set` data structure called `difference`. A difference of sets should compare two sets and return the items present in the first set that are absent in the second. This method should take another `Set` as an argument and return the `difference` of the two sets.
+
+For example, if `setA = ['a','b','c']` and `setB = ['a','b','d','e']`, then the difference of setA and setB is: `setA.difference(setB) = ['c']`.
# --hints--
-你的`Set`类应该有一个`difference`方法。
+Your `Set` class should have a `difference` method.
```js
assert(
@@ -23,7 +25,7 @@ assert(
);
```
-收回了适当的收藏
+Your `difference` method should return the proper collection.
```js
assert(
@@ -122,7 +124,7 @@ class Set {
return newSet;
}
// Only change code below this line
-
+
// Only change code above this line
}
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/perform-a-subset-check-on-two-sets-of-data.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/perform-a-subset-check-on-two-sets-of-data.md
index 3b6820b7dc8..ea0caa1b3ed 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/perform-a-subset-check-on-two-sets-of-data.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/perform-a-subset-check-on-two-sets-of-data.md
@@ -1,29 +1,31 @@
---
id: 587d8254367417b2b2512c6f
-title: 对两组数据执行子集检查
+title: Perform a Subset Check on Two Sets of Data
challengeType: 1
-videoUrl: ''
+forumTopicId: 301707
dashedName: perform-a-subset-check-on-two-sets-of-data
---
# --description--
-在本练习中，我们将对2组数据执行子集测试。我们将在我们的`Set`数据结构上创建一个名为`subset` 。这将比较第一组与第二组，如果第一组完全包含在第二组中，则它将返回true。例如，如果`setA = ['a','b']`和`setB = ['a','b','c','d']` ，则setA和setB的子集为： `setA.subset(setB)`应该是`true` 。
+In this exercise, we are going to perform a subset test on 2 sets of data. We will create a method on our `Set` data structure called `isSubsetOf`. This will compare the first set against the second, and if the first set is fully contained within the second, it will return `true`.
+
+For example, if `setA = ['a','b']` and `setB = ['a','b','c','d']`, then `setA` is a subset of `setB`, so `setA.isSubsetOf(setB)` should return `true`.
# --hints--
-你的`Set`类应该有一个`union`方法。
+Your `Set` class should have a `isSubsetOf` method.
```js
assert(
(function () {
var test = new Set();
- return typeof test.subset === 'function';
+ return typeof test.isSubsetOf === 'function';
})()
);
```
-第一个Set（）包含在第二个Set中
+The first Set() should be contained in the second Set
```js
assert(
@@ -35,13 +37,13 @@ assert(
setB.add('c');
setB.add('a');
setB.add('d');
- var subsetSetAB = setA.subset(setB);
- return subsetSetAB === true;
+ var aIsSubsetOfB = setA.isSubsetOf(setB);
+ return aIsSubsetOfB === true;
})()
);
```
-`["a", "b"].subset(["a", "b", "c", "d"])`应该返回`true` “）
+`['a', 'b'].isSubsetOf(['a', 'b', 'c', 'd'])` should return `true`
```js
assert(
@@ -54,13 +56,13 @@ assert(
setB.add('b');
setB.add('c');
setB.add('d');
- var subsetSetAB = setA.subset(setB);
- return subsetSetAB === true;
+ var aIsSubsetOfB = setA.isSubsetOf(setB);
+ return aIsSubsetOfB === true;
})()
);
```
-`["a", "b", "c"].subset(["a", "b"])`应返回`false` “）
+`['a', 'b', 'c'].isSubsetOf(['a', 'b'])` should return `false`
```js
assert(
@@ -72,26 +74,26 @@ assert(
setA.add('c');
setB.add('a');
setB.add('b');
- var subsetSetAB = setA.subset(setB);
- return subsetSetAB === false;
+ var aIsSubsetOfB = setA.isSubsetOf(setB);
+ return aIsSubsetOfB === false;
})()
);
```
-`[].subset([])`应该返回`true`
+`[].isSubsetOf([])` should return `true`
```js
assert(
(function () {
var setA = new Set();
var setB = new Set();
- var subsetSetAB = setA.subset(setB);
- return subsetSetAB === true;
+ var aIsSubsetOfB = setA.isSubsetOf(setB);
+ return aIsSubsetOfB === true;
})()
);
```
-`["a", "b"].subset(["c", "d"])`应返回`false` “）
+`['a', 'b'].isSubsetOf(['c', 'd'])` should return `false`
```js
assert(
@@ -102,8 +104,8 @@ assert(
setA.add('b');
setB.add('c');
setB.add('d');
- var subsetSetAB = setA.subset(setB);
- return subsetSetAB === false;
+ var aIsSubsetOfB = setA.isSubsetOf(setB);
+ return aIsSubsetOfB === false;
})()
);
```
@@ -198,7 +200,7 @@ class Set {
return newSet;
}
// Only change code below this line
-
+
// Only change code above this line
}
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/perform-a-union-on-two-sets.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/perform-a-union-on-two-sets.md
index 6bc5d35c412..04ecfb767f8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/perform-a-union-on-two-sets.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/perform-a-union-on-two-sets.md
@@ -1,18 +1,20 @@
---
id: 587d8253367417b2b2512c6c
-title: 在两个集上执行联合
+title: Perform a Union on Two Sets
challengeType: 1
-videoUrl: ''
+forumTopicId: 301708
dashedName: perform-a-union-on-two-sets
---
# --description--
-在本练习中，我们将对两组数据执行联合。我们将在我们的`Set`数据结构上创建一个名为`union` 。此方法应将另一个`Set`作为参数，并返回两个集合的`union`集，不包括任何重复值。例如，如果`setA = ['a','b','c']`和`setB = ['a','b','d','e']` ，则setA和setB的并集为： `setA.union(setB) = ['a', 'b', 'c', 'd', 'e']` 。
+In this exercise we are going to perform a union on two sets of data. We will create a method on our `Set` data structure called `union`. This method should take another `Set` as an argument and return the `union` of the two sets, excluding any duplicate values.
+
+For example, if `setA = ['a','b','c']` and `setB = ['a','b','d','e']`, then the union of setA and setB is: `setA.union(setB) = ['a', 'b', 'c', 'd', 'e']`.
# --hints--
-你的`Set`类应该有一个`union`方法。
+Your `Set` class should have a `union` method.
```js
assert(
@@ -23,7 +25,7 @@ assert(
);
```
-收回了适当的收藏
+The union of a Set containing values ["a", "b", "c"] and a Set containing values ["c", "d"] should return a new Set containing values ["a", "b", "c", "d"].
```js
assert(
@@ -65,12 +67,12 @@ class Set {
}
// This method will return all the values in the set
values() {
- return Object.keys(this.dictionary);
+ return Object.values(this.dictionary);
}
// This method will add an element to the set
add(element) {
if (!this.has(element)) {
- this.dictionary[element] = true;
+ this.dictionary[element] = element;
this.length++;
return true;
}
@@ -111,12 +113,12 @@ class Set {
}
values() {
- return Object.keys(this.dictionary);
+ return Object.values(this.dictionary);
}
add(element) {
if (!this.has(element)) {
- this.dictionary[element] = true;
+ this.dictionary[element] = element;
this.length++;
return true;
}
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/perform-an-intersection-on-two-sets-of-data.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/perform-an-intersection-on-two-sets-of-data.md
index 78b991e7e60..a108cb3b4ee 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/perform-an-intersection-on-two-sets-of-data.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/perform-an-intersection-on-two-sets-of-data.md
@@ -1,18 +1,20 @@
---
id: 587d8253367417b2b2512c6d
-title: 在两组数据上执行交集
+title: Perform an Intersection on Two Sets of Data
challengeType: 1
-videoUrl: ''
+forumTopicId: 301709
dashedName: perform-an-intersection-on-two-sets-of-data
---
# --description--
-在本练习中，我们将对两组数据执行交集。我们将在我们的`Set`数据结构上创建一个名为`intersection` 。集合的交集表示两个或更多集合共有的所有值。此方法应将另一个`Set`作为参数，并返回两个集合的`intersection` 。例如，如果`setA = ['a','b','c']`和`setB = ['a','b','d','e']` ，则setA和setB的交集为： `setA.intersection(setB) = ['a', 'b']` 。
+In this exercise we are going to perform an intersection on 2 sets of data. We will create a method on our `Set` data structure called `intersection`. An intersection of sets represents all values that are common to two or more sets. This method should take another `Set` as an argument and return the `intersection` of the two sets.
+
+For example, if `setA = ['a','b','c']` and `setB = ['a','b','d','e']`, then the intersection of setA and setB is: `setA.intersection(setB) = ['a', 'b']`.
# --hints--
-您的`Set`类应该有一个`intersection`方法。
+Your `Set` class should have a `intersection` method.
```js
assert(
@@ -23,7 +25,7 @@ assert(
);
```
-收回了适当的收藏
+The proper collection should be returned.
```js
assert(
@@ -99,7 +101,7 @@ class Set {
return newSet;
}
// Only change code below this line
-
+
// Only change code above this line
}
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/remove-an-element-from-a-max-heap.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/remove-an-element-from-a-max-heap.md
index 9eb2ddbf2d3..858073f2112 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/remove-an-element-from-a-max-heap.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/remove-an-element-from-a-max-heap.md
@@ -1,18 +1,28 @@
---
id: 587d825b367417b2b2512c8b
-title: 从最大堆中删除元素
+title: Remove an Element from a Max Heap
challengeType: 1
-videoUrl: ''
+forumTopicId: 301710
dashedName: remove-an-element-from-a-max-heap
---
# --description--
-现在我们可以向堆中添加元素，让我们看看如何删除元素。删除和插入元素都需要类似的逻辑。在最大堆中，您通常需要删除最大值，因此这只需要从树的根中提取它。这将破坏我们树的堆属性，因此我们必须以某种方式重新建立它。通常，对于最大堆，这可以通过以下方式完成：将堆中的最后一个元素移动到根位置。如果root的子节点大于它，则将root与较大值的子节点交换。继续交换，直到父级大于两个子级，或者到达树中的最后一级。说明：向我们的最大堆添加一个名为remove的方法。此方法应返回已添加到最大堆的最大值，并将其从堆中删除。它还应该重新排序堆，以便保持堆属性。删除元素后，堆中剩余的下一个最大元素应该成为根。此处再次添加插入方法。
+Now that we can add elements to our heap let's see how we can remove elements. Removing and inserting elements both require similar logic. In a max heap you will usually want to remove the greatest value, so this involves simply extracting it from the root of our tree. This will break the heap property of our tree, so we must reestablish it in some way. Typically, for a max heap this is done in the following way:
+
+<ol>
+ <li>Move the last element in the heap into the root position.</li>
+ <li>If either child of the root is greater than it, swap the root with the child of greater value.</li>
+ <li>Continue swapping until the parent is greater than both children or you reach the last level in the tree.</li>
+</ol>
+
+# --instructions--
+
+Instructions: Add a method to our max heap called `remove`. This method should return the greatest value that has been added to our max heap and remove it from the heap. It should also reorder the heap so the heap property is maintained. After removing an element, the next greatest element remaining in the heap should become the root.
# --hints--
-存在MaxHeap数据结构。
+The MaxHeap data structure should exist.
```js
assert(
@@ -26,7 +36,7 @@ assert(
);
```
-MaxHeap有一个名为print的方法。
+MaxHeap should have a method called print.
```js
assert(
@@ -42,7 +52,7 @@ assert(
);
```
-MaxHeap有一个名为insert的方法。
+MaxHeap should have a method called insert.
```js
assert(
@@ -58,7 +68,7 @@ assert(
);
```
-MaxHeap有一个名为remove的方法。
+MaxHeap should have a method called remove.
```js
assert(
@@ -74,7 +84,7 @@ assert(
);
```
-remove方法从最大堆中删除最大元素，同时保持最大堆属性。
+The remove method should remove the greatest element from the max heap while maintaining the max heap property.
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/remove-elements-from-a-linked-list-by-index.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/remove-elements-from-a-linked-list-by-index.md
index 022c53458b6..9a8e59f77fc 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/remove-elements-from-a-linked-list-by-index.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/remove-elements-from-a-linked-list-by-index.md
@@ -1,34 +1,41 @@
---
id: 587d8251367417b2b2512c65
-title: 按索引从链接列表中删除元素
+title: Remove Elements from a Linked List by Index
challengeType: 1
-videoUrl: ''
+forumTopicId: 301711
dashedName: remove-elements-from-a-linked-list-by-index
---
# --description--
-在我们继续讨论另一个数据结构之前，让我们先了解链接列表的最后几点练习。让我们编写一个`removeAt`方法，删除给定`index`处的`element` 。该方法应该称为`removeAt(index)` 。要删除某个`index`处的`element` ，我们需要在沿着链表移动时保持每个节点的运行计数。用于遍历链表的元素的常用技术涉及<dfn>“转轮”</dfn>或“哨兵”，它们“指向”代码所比较的节点。在我们的情况下，开始于`head`我们的名单中，我们先从一个`currentIndex`始于变量`0` 。对于我们传递的每个节点， `currentIndex`应该增加1。就像我们的`remove(element)`方法一样，当我们在removeAt（index）方法中删除节点时，我们需要注意不要孤立列表的其余部分。我们通过确保引用已删除节点的节点具有对下一节点的引用来保持节点连续。
+Before we move on to another data structure, let's get a couple of last bits of practice with linked lists.
+
+Let's write a `removeAt` method that removes the `element` at a given `index`. The method should be called `removeAt(index)`. To remove an `element` at a certain `index`, we'll need to keep a running count of each node as we move along the linked list.
+
+A common technique used to iterate through the elements of a linked list involves a <dfn>'runner'</dfn>, or sentinel, that 'points' at the nodes that your code is comparing. In our case, starting at the `head` of our list, we start with a `currentIndex` variable that starts at `0`. The `currentIndex` should increment by one for each node we pass.
+
+Just like our `remove(element)` method, which [we covered in a previous lesson](/learn/coding-interview-prep/data-structures/remove-elements-from-a-linked-list), we need to be careful not to orphan the rest of our list when we remove the node in our `removeAt(index)` method. We keep our nodes contiguous by making sure that the node that has reference to the removed node has a reference to the next node.
# --instructions--
-编写`removeAt(index)`方法，删除并返回给定`index`处的节点。如果给定`index`为负数，或者大于或等于链表`length` ，则该方法应返回`null` 。注意请记住保持`currentIndex`计数。
+Write a `removeAt(index)` method that removes and returns a node at a given `index`. The method should return `null` if the given `index` is either negative, or greater than or equal to the `length` of the linked list.
+
+**Note:** Remember to keep count of the `currentIndex`.
# --hints--
-您的`LinkedList`类应该有一个`removeAt`方法。
+Your `LinkedList` class should have a `removeAt` method.
```js
assert(
(function () {
var test = new LinkedList();
return typeof test.removeAt === 'function';
- })(),
- 'Your <code>LinkedList</code> class should have a <code>removeAt</code> method.'
+ })()
);
```
-您的`removeAt`方法应该减少链表的`length`
+Your `removeAt` method should reduce the `length` of the linked list by one.
```js
assert(
@@ -39,12 +46,43 @@ assert(
test.add('kitten');
test.removeAt(1);
return test.size() === 2;
- })(),
- 'Your <code>removeAt</code> method should reduce the <code>length</code> of the linked list'
+ })()
);
```
-您的`removeAt`方法还应该返回已删除节点的元素。
+Your `removeAt` method should remove the element at the specified index from the linked list.
+
+```js
+assert(
+ (function () {
+ var test = new LinkedList();
+ test.add('cat');
+ test.add('dog');
+ test.add('kitten');
+ test.add('bird');
+ test.removeAt(1);
+ return (
+ JSON.stringify(test.head()) ===
+ '{"element":"cat","next":{"element":"kitten","next":{"element":"bird","next":null}}}'
+ );
+ })()
+);
+```
+
+When only one element is present in the linked list, your `removeAt` method should remove and return the element at specified index, and reduce the length of the linked list.
+
+```js
+assert(
+ (function () {
+ var test = new LinkedList();
+ test.add('cat');
+ var removedItem = test.removeAt(0);
+ return test.head() === null && test.size() === 0 && removedItem === 'cat';
+ })()
+);
+```
+
+Your `removeAt` method should return the element of the removed node.
```js
assert(
@@ -54,12 +92,11 @@ assert(
test.add('dog');
test.add('kitten');
return test.removeAt(1) === 'dog';
- })(),
- 'Your <code>removeAt</code> method should also return the element of the removed node.'
+ })()
);
```
-如果给定索引小于`0`则`removeAt`方法也应返回`null`
+Your `removeAt` method should return `null` and the linked list should not change if the given index is less than `0`.
```js
assert(
@@ -68,13 +105,17 @@ assert(
test.add('cat');
test.add('dog');
test.add('kitten');
- return test.removeAt(-1) === null;
- })(),
- 'Your <code>removeAt</code> method should also return <code>null</code> if the given index is less than <code>0</code>'
+ var removedItem = test.removeAt(-1);
+ return (
+ removedItem === null &&
+ JSON.stringify(test.head()) ===
+ '{"element":"cat","next":{"element":"dog","next":{"element":"kitten","next":null}}}'
+ );
+ })()
);
```
-如果给定索引等于或大于链表的`length` ，则`removeAt`方法也应返回`null` 。
+Your `removeAt` method should return `null` and the linked list should not change if the given index is greater than or equal to the `length` of the list.
```js
assert(
@@ -83,9 +124,13 @@ assert(
test.add('cat');
test.add('dog');
test.add('kitten');
- return test.removeAt(3) === null;
- })(),
- 'Your <code>removeAt</code> method should also return <code>null</code> if the given index is equal or more than the <code>length</code> of the linked list.'
+ var removedItem = test.removeAt(3);
+ return (
+ removedItem === null &&
+ JSON.stringify(test.head()) ===
+ '{"element":"cat","next":{"element":"dog","next":{"element":"kitten","next":null}}}'
+ );
+ })()
);
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/remove-elements-from-a-linked-list.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/remove-elements-from-a-linked-list.md
index 277d34a7d35..a202036ff08 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/remove-elements-from-a-linked-list.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/remove-elements-from-a-linked-list.md
@@ -1,34 +1,41 @@
---
id: 587d8251367417b2b2512c63
-title: 从链接列表中删除元素
+title: Remove Elements from a Linked List
challengeType: 1
-videoUrl: ''
+forumTopicId: 301712
dashedName: remove-elements-from-a-linked-list
---
# --description--
-链接列表的任何实现所需的下一个重要方法是`remove`方法。此方法应将要删除的元素作为参数，然后搜索列表以查找并删除包含该元素的节点。每当我们从链表中删除一个节点时，重要的是我们不要意外地孤立列表的其余部分。回想一下，每个节点的`next`属性都指向列表中跟随它的节点。如果我们删除中间元素，比如说，我们要确保我们从该元素的前一个节点的`next`属性到中间元素的`next`属性（这是列表中的下一个节点）的连接！这可能听起来真的很混乱，所以让我们回到康加线的例子，这样我们就有了一个很好的概念模型。想象自己在康加舞线上，直接在你面前的人离开了这条线。刚离开生产线的人不再将手放在任何人身上 - 而且你不再把手放在离开的人身上。你向前走，把你的手放在你看到的下一个人身上。如果我们要删除的元素是`head`元素，我们将`head`重新分配给链表的第二个节点。
+The next important method that any implementation of a linked list will need is a `remove` method. This method should take the element we want to remove as an argument, and then search the list to find and remove the node that contains that element.
+
+Whenever we remove a node from a linked list, it's important that we don't accidentally orphan the rest of the list in doing so. Recall that every node's `next` property points to the node that follows it in the list. If we're removing the middle element, say, we'll want to make sure that we have a connection from that element's previous node's `next` property to the middle element's `next` property (which is the next node in the list!)
+
+This might sound really confusing, so let's return to the conga line example so we have a good conceptual model. Picture yourself in a conga line, and the person directly in front of you leaves the line. The person who just left the line no longer has her hands on anyone in line--and you no longer have your hands on the person that left. You step forward and put your hands on next person you see.
+
+If the element we wish to remove is the `head` element, we reassign the `head` to the second node of the linked list.
# --instructions--
-编写一个`remove`方法，该方法接受一个元素并将其从链表中删除。注意每次从链接列表中删除元素时，列表的`length`应减少一。
+Write a `remove` method that takes an element and removes it from the linked list.
+
+**Note:** The `length` of the list should decrease by one every time an element is removed from the linked list.
# --hints--
-您的`LinkedList`类应该有一个`remove`方法。
+Your `LinkedList` class should have a `remove` method.
```js
assert(
(function () {
var test = new LinkedList();
return typeof test.remove === 'function';
- })(),
- 'Your <code>LinkedList</code> class should have a <code>remove</code> method.'
+ })()
);
```
-`remove`第一个节点时， `remove`方法应重新分配`head`到第二个节点。
+Your `remove` method should reassign `head` to the second node when the first node is removed.
```js
assert(
@@ -38,12 +45,11 @@ assert(
test.add('dog');
test.remove('cat');
return test.head().element === 'dog';
- })(),
- 'Your <code>remove</code> method should reassign <code>head</code> to the second node when the first node is removed.'
+ })()
);
```
-对于每个删除的节点，您的`remove`方法应该将链表的`length`减少一个。
+Your `remove` method should decrease the `length` of the linked list by one for every node removed.
```js
assert(
@@ -51,14 +57,31 @@ assert(
var test = new LinkedList();
test.add('cat');
test.add('dog');
+ test.add('hamster');
test.remove('cat');
- return test.size() === 1;
- })(),
- 'Your <code>remove</code> method should decrease the <code>length</code> of the linked list by one for every node removed.'
+ test.remove('fish');
+ return test.size() === 2;
+ })()
);
```
-您的`remove`方法应该将已删除节点的上`next`节点的引用重新分配给已删除节点的`next`引用。
+Your `remove` method should reassign the reference of the previous node of the removed node to the removed node's `next` reference.
+
+```js
+assert(
+ (function () {
+ var test = new LinkedList();
+ test.add('cat');
+ test.add('dog');
+ test.add('snake');
+ test.add('kitten');
+ test.remove('snake');
+ return test.head().next.next.element === 'kitten';
+ })()
+);
+```
+
+Your `remove` method should not change the linked list if the element does not exist in the linked list.
```js
assert(
@@ -67,10 +90,12 @@ assert(
test.add('cat');
test.add('dog');
test.add('kitten');
- test.remove('dog');
- return test.head().next.element === 'kitten';
- })(),
- 'Your <code>remove</code> method should reassign the reference of the previous node of the removed node to the removed node's <code>next</code> reference.'
+ test.remove('elephant');
+ return (
+ JSON.stringify(test.head()) ===
+ '{"element":"cat","next":{"element":"dog","next":{"element":"kitten","next":null}}}'
+ );
+ })()
);
```
@@ -169,7 +194,7 @@ function LinkedList() {
previous = currentNode;
currentNode = currentNode.next;
}
-
+
if (currentNode.next === null && currentNode.element !== element) {
return;
}
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/remove-items-from-a-set-in-es6.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/remove-items-from-a-set-in-es6.md
index 6118f6f7285..a3c2f3b97e7 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/remove-items-from-a-set-in-es6.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/remove-items-from-a-set-in-es6.md
@@ -1,27 +1,35 @@
---
id: 587d8254367417b2b2512c71
-title: 从ES6中的集中删除项目
+title: Remove items from a set in ES6
challengeType: 1
-videoUrl: ''
+forumTopicId: 301713
dashedName: remove-items-from-a-set-in-es6
---
# --description--
-让我们使用`delete`方法练习从ES6集中`delete` 。首先，创建一个ES6 Set `var set = new Set([1,2,3]);`现在使用`delete`方法从Set中删除一个项目。
+Let's practice removing items from an ES6 Set using the `delete` method.
-> set.delete（1）;
-> console.log（\[... set]）//应该返回\[2,3]
->
-> >
+First, create an ES6 Set:
+
+`var set = new Set([1,2,3]);`
+
+Now remove an item from your Set with the `delete` method.
+
+```js
+set.delete(1);
+console.log([...set]) // should return [ 2, 3 ]
+```
# --instructions--
-现在，创建一个整数为1,2,3,4和5的集合。删除值2和5，然后返回集合。
+Now, create a set with the integers 1, 2, 3, 4, & 5.
+
+Remove the values 2 and 5, and then return the set.
# --hints--
-您的集应包含值1,3和4
+Your Set should contain the values 1, 3, & 4
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/reverse-a-doubly-linked-list.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/reverse-a-doubly-linked-list.md
index 8cb1f2ac34a..58edfaa2a50 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/reverse-a-doubly-linked-list.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/reverse-a-doubly-linked-list.md
@@ -1,18 +1,18 @@
---
id: 587d825a367417b2b2512c88
-title: 反转双重链接列表
+title: Reverse a Doubly Linked List
challengeType: 1
-videoUrl: ''
+forumTopicId: 301714
dashedName: reverse-a-doubly-linked-list
---
# --description--
-让我们为我们的双向链表创建一个名为reverse的方法，它可以反转列表。一旦执行该方法，头部应指向前一个尾部，尾部应指向前一个头部。现在，如果我们从头到尾遍历列表，我们应该以与原始列表相反的顺序来满足节点。尝试反转空列表应返回null。
+Let's create one more method for our doubly linked list called reverse which reverses the list in place. Once the method is executed the head should point to the previous tail and the tail should point to the previous head. Now, if we traverse the list from head to tail we should meet the nodes in a reverse order compared to the original list. Trying to reverse an empty list should return null.
# --hints--
-存在DoublyLinkedList数据结构。
+The DoublyLinkedList data structure should exist.
```js
assert(
@@ -26,24 +26,7 @@ assert(
);
```
-DoublyLinkedList有一个名为add的方法。
-
-```js
-assert(
- (function () {
- var test = false;
- if (typeof DoublyLinkedList !== 'undefined') {
- test = new DoublyLinkedList();
- }
- if (test.add == undefined) {
- return false;
- }
- return typeof test.add == 'function';
- })()
-);
-```
-
-DoublyLinkedList有一个名为reverse的方法。
+The DoublyLinkedList should have a method called reverse.
```js
assert(
@@ -60,7 +43,7 @@ assert(
);
```
-反转空列表将返回null。
+Reversing an empty list should return null.
```js
assert(
@@ -74,7 +57,7 @@ assert(
);
```
-反向方法反转列表。
+The reverse method should reverse the list.
```js
assert(
@@ -86,13 +69,15 @@ assert(
test.add(58);
test.add(61);
test.add(32);
+ test.add(95);
+ test.add(41);
test.reverse();
- return test.print().join('') == '326158';
+ return test.print().join('') == '4195326158';
})()
);
```
-当列表反转时，正确维护下一个和上一个引用。
+The next and previous references should be correctly maintained when a list is reversed.
```js
assert(
@@ -104,8 +89,10 @@ assert(
test.add(11);
test.add(22);
test.add(33);
+ test.add(44);
+ test.add(55);
test.reverse();
- return test.printReverse().join('') == '112233';
+ return test.printReverse().join('') == '1122334455';
})()
);
```
@@ -178,7 +165,7 @@ var DoublyLinkedList = function() {
this.head = null;
this.tail = null;
// Only change code below this line
-
+
// Only change code above this line
};
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/search-within-a-linked-list.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/search-within-a-linked-list.md
index 466f57ab29f..03361689fef 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/search-within-a-linked-list.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/search-within-a-linked-list.md
@@ -1,22 +1,62 @@
---
id: 587d8251367417b2b2512c64
-title: 在链接列表中搜索
+title: Search within a Linked List
challengeType: 1
-videoUrl: ''
+forumTopicId: 301715
dashedName: search-within-a-linked-list
---
# --description--
-让我们为链表类添加一些更有用的方法。如果我们可以判断我们的列表是否为空，那么它是否有用，就像我们的`Stack`和`Queue`类一样？我们还应该能够在链表中找到特定元素。遍历数据结构是你想要进行大量练习的东西！让我们创建一个`indexOf`方法，该方法将`element`作为参数，并在链表中返回该元素的`index` 。如果在链接列表中找不到该元素，则返回`-1` 。让我们实现一个相反的方法：一个`elementAt`方法，它将`index`作为参数并返回给定`index`处的`element` 。如果未找到任何`element` ，则返回`undefined` 。
+Let's add a few more useful methods to our linked list class. Wouldn't it be useful if we could tell if our list was empty or not, as with our `Stack` and `Queue` classes?
+
+We should also be able to find specific elements in our linked list. Traversing through data structures is something you'll want to get a lot of practice with! Let's create an `indexOf` method that takes an `element` as an argument, and returns that element's `index` in the linked list. If the element is not found in the linked list, return `-1`.
+
+Let's also implement a method that does the opposite: an `elementAt` method that takes an `index` as an argument and returns the `element` at the given `index`. If no `element` is found, return `undefined`.
# --instructions--
-编写一个检查链表是否为空的`isEmpty`方法，返回给定元素`index`的`indexOf`方法，以及返回给定`index.`处`element`的`elementAt` `index.`
+Write an `isEmpty` method that checks if the linked list is empty, an `indexOf` method that returns the `index` of a given element, and an `elementAt` that returns an `element` at a given `index.`
# --hints--
-您的`LinkedList`类应该有一个`indexOf`方法。
+Your `LinkedList` class should have an `isEmpty` method.
+
+```js
+assert(
+ (function () {
+ var test = new LinkedList();
+ return typeof test.isEmpty === 'function';
+ })()
+);
+```
+
+Your `isEmpty` method should return `false` when there is at least one element in linked list.
+
+```js
+assert(
+ (function () {
+ var test = new LinkedList();
+ test.add('cat');
+ test.add('dog');
+ test.add('kitten');
+ return test.isEmpty() === false;
+ })()
+);
+```
+
+Your `isEmpty` method should return `true` when there are no elements in linked list.
+
+```js
+assert(
+ (function () {
+ var test = new LinkedList();
+ return test.isEmpty() === true;
+ })()
+);
+```
+
+Your `LinkedList` class should have an `indexOf` method.
```js
assert(
@@ -27,7 +67,35 @@ assert(
);
```
-您的`LinkedList`类应该有一个`elementAt`方法。
+Your `indexOf` method should return the index of a given element found in linked list.
+
+```js
+assert(
+ (function () {
+ var test = new LinkedList();
+ test.add('cat');
+ test.add('dog');
+ test.add('kitten');
+ return test.indexOf('cat') === 0;
+ })()
+);
+```
+
+Your `indexOf` method should return `-1` if the given element is not found in linked list
+
+```js
+assert(
+ (function () {
+ var test = new LinkedList();
+ test.add('cat');
+ test.add('dog');
+ test.add('kitten');
+ return test.indexOf('pony') === -1;
+ })()
+);
+```
+
+Your `LinkedList` class should have an `elementAt` method.
```js
assert(
@@ -38,35 +106,7 @@ assert(
);
```
-您的`size`方法应返回链表的长度
-
-```js
-assert(
- (function () {
- var test = new LinkedList();
- test.add('cat');
- test.add('dog');
- test.add('kitten');
- return test.size() === 3;
- })()
-);
-```
-
-`indexOf`方法应该返回给定元素的索引。
-
-```js
-assert(
- (function () {
- var test = new LinkedList();
- test.add('cat');
- test.add('dog');
- test.add('kitten');
- return test.indexOf('kitten') === 2;
- })()
-);
-```
-
-您的`elementAt`方法应该返回给定索引处的元素。
+Your `elementAt` method should return the element found at a given index in linked list.
```js
assert(
@@ -80,6 +120,20 @@ assert(
);
```
+Your `elementAt` method should return `undefined` if the given element is not found at a given index in linked list.
+
+```js
+assert(
+ (function () {
+ var test = new LinkedList();
+ test.add('cat');
+ test.add('dog');
+ test.add('kitten');
+ return test.elementAt(5) === undefined;
+ })()
+);
+```
+
# --seed--
## --seed-contents--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/typed-arrays.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/typed-arrays.md
index 805486af799..2433c0e2be6 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/typed-arrays.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/typed-arrays.md
@@ -1,81 +1,75 @@
---
id: 587d8253367417b2b2512c6a
-title: 键入的数组
+title: Typed Arrays
challengeType: 1
-videoUrl: ''
+forumTopicId: 301716
dashedName: typed-arrays
---
# --description--
-数组是可以容纳许多不同元素的JavaScript对象。 `var complexArr = [1, 5, "2", "Word", {"name": "James"}];`基本上后台发生的事情是您的浏览器会自动为该阵列提供适当的内存空间。如果添加或删除数据，它也会根据需要进行更改。但是，在高性能和不同元素类型的世界中，有时您需要更具体地了解为阵列提供多少内存。 <dfn>类型化数组</dfn>是这个问题的答案。您现在可以说要为阵列提供多少内存。下面是可用的不同类型数组的基本概述，以及该数组中每个元素的大小（以字节为单位）。
+Arrays are JavaScript objects that can hold a lot of different elements.
-| 类型 | 每个元素大小以字节为单位 |
-| ------------------- | ------------ |
-| `Int8Array` | 1 |
-| `Uint8Array` | 1 |
-| `Uint8ClampedArray` | 1 |
-| `Int16Array` | 2 |
-| `Uint16Array` | 2 |
-| `Int32Array` | 4 |
-| `Uint32Array` | 4 |
-| `Float32Array` | 4 |
-| `Float64Array` | 8 |
+`var complexArr = [1, 5, "2", "Word", {"name": "James"}];`
-创建这种类型的数组有两种方法。一种方法是直接创建它。下面是如何创建一个3长度的`Int16Array` 。
+Basically what happens in the background is that your browser will automatically give the right amount of memory space for that array. It will also change as needed if you add or remove data.
-> var i8 = new Int16Array（3）;
-> 的console.log（I8）;
-> //返回\[0,0,0]
+However, in the world of high performance and different element types, sometimes you need to be more specific on how much memory is given to an array.
-您还可以创建一个
+<dfn>Typed arrays</dfn> are the answer to this problem. You are now able to say how much memory you want to give an array. Below is a basic overview of the different types of arrays available and the size in bytes for each element in that array.
-<dfn>缓冲区</dfn>
+<table class='table table-striped'><tbody><tr><th>Type</th><th>Each element size in bytes</th></tr><tr><td><code>Int8Array</code></td><td>1</td></tr><tr><td><code>Uint8Array</code></td><td>1</td></tr><tr><td><code>Uint8ClampedArray</code></td><td>1</td></tr><tr><td><code>Int16Array</code></td><td>2</td></tr><tr><td><code>Uint16Array</code></td><td>2</td></tr><tr><td><code>Int32Array</code></td><td>4</td></tr><tr><td><code>Uint32Array</code></td><td>4</td></tr><tr><td><code>Float32Array</code></td><td>4</td></tr><tr><td><code>Float64Array</code></td><td>8</td></tr></tbody></table>
-来分配您希望数组占用多少数据（以字节为单位）。 **注意**
-要使用缓冲区创建类型化数组，需要将字节数分配为上面列出的字节的倍数。
+There are two ways in creating these kind of arrays. One way is to create it directly. Below is how to create a 3 length `Int16Array`.
-> //以不同方式创建相同的Int16Array数组
-> var byteSize = 6; //需要是2的倍数
-> var buffer = new ArrayBuffer（byteSize）;
-> var i8View = new Int16Array（buffer）;
-> buffer.byteLength; //返回6
-> i8View.byteLength; //返回6
-> 的console.log（i8View）; //返回\[0,0,0]
+```js
+var i8 = new Int16Array(3);
+console.log(i8);
+// Returns [0, 0, 0]
+```
-<dfn>缓冲区</dfn>
+You can also create a <dfn>buffer</dfn> to assign how much data (in bytes) you want the array to take up. **Note**
+To create typed arrays using buffers, you need to assign the number of bytes to be a multiple of the bytes listed above.
-是仅承载数据的通用对象。您无法正常访问它们。要访问它们，您需要先创建一个
+```js
+// Create same Int16Array array differently
+var byteSize = 6; // Needs to be multiple of 2
+var buffer = new ArrayBuffer(byteSize);
+var i8View = new Int16Array(buffer);
+buffer.byteLength; // Returns 6
+i8View.byteLength; // Returns 6
+console.log(i8View); // Returns [0, 0, 0]
+```
-<dfn>视图</dfn>
+<dfn>Buffers</dfn> are general purpose objects that just carry data. You cannot access them normally. To access them, you need to first create a <dfn>view</dfn>.
- 。
+```js
+i8View[0] = 42;
+console.log(i8View); // Returns [42, 0, 0]
+```
-> i8View \[0] = 42;
-> 的console.log（i8View）; //返回\[42,0,0]
-
-**注意**
-类型化数组没有传统数组所具有的某些方法，如`.pop()`或`.push()` 。类型化数组也会失败`Array.isArray()` ，它会检查某些内容是否为数组。虽然更简单，但对于不太复杂的JavaScript引擎来说，这可能是一个优势。
+**Note**
+Typed arrays do not have some of the methods traditional arrays have such as `.pop()` or `.push()`. Typed arrays also fail `Array.isArray()` that checks if something is an array. Although simpler, this can be an advantage for less-sophisticated JavaScript engines to implement them.
# --instructions--
-首先创建一个64字节的`buffer` 。然后创建一个`Int32Array`类型数组，其中包含一个名为`i32View`的视图。
+First create a `buffer` that is 64-bytes. Then create a `Int32Array` typed array with a view of it called `i32View`.
# --hints--
-您的`buffer`应该是64字节大。
+Your `buffer` should be 64 bytes large.
```js
assert(buffer.byteLength === 64);
```
-您的缓冲区的`i32View`视图应该是64字节大。
+Your `i32View` view of your buffer should be 64 bytes large.
```js
assert(i32View.byteLength === 64);
```
-您的缓冲区的`i32View`视图应为16个元素长。
+Your `i32View` view of your buffer should be 16 elements long.
```js
assert(i32View.length === 16);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/use-.has-and-.size-on-an-es6-set.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/use-.has-and-.size-on-an-es6-set.md
index 06d5f7ffe57..63c09ecd329 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/use-.has-and-.size-on-an-es6-set.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/use-.has-and-.size-on-an-es6-set.md
@@ -1,22 +1,34 @@
---
id: 587d8255367417b2b2512c72
-title: 在ES6集上使用.has和.size
+title: Use .has and .size on an ES6 Set
challengeType: 1
-videoUrl: ''
+forumTopicId: 301717
dashedName: use--has-and--size-on-an-es6-set
---
# --description--
-让我们看一下ES6 Set对象上可用的.has和.size方法。首先，创建一个ES6 Set `var set = new Set([1,2,3]);` .has方法将检查该值是否包含在集合中。 `var hasTwo = set.has(2);` .size方法将返回一个表示Set `var howBig = set.size;`大小的整数`var howBig = set.size;`
+Let's look at the .has and .size methods available on the ES6 Set object.
+
+First, create an ES6 Set
+
+`var set = new Set([1,2,3]);`
+
+The .has method will check if the value is contained within the set.
+
+`var hasTwo = set.has(2);`
+
+The .size method will return an integer representing the size of the Set
+
+`var howBig = set.size;`
# --instructions--
-在本练习中，我们将数组和值传递给checkSet（）函数。您的函数应该从数组参数创建ES6集。查找该集是否包含value参数。找到集合的大小。并在数组中返回这两个值。
+In this exercise we will pass an array and a value to the checkSet() function. Your function should create an ES6 set from the array argument. Find if the set contains the value argument. Find the size of the set. And return those two values in an array.
# --hints--
-`checkSet([4, 5, 6], 3)`应该返回[false，3]
+`checkSet([4, 5, 6], 3)` should return [ false, 3 ]
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/use-breadth-first-search-in-a-binary-search-tree.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/use-breadth-first-search-in-a-binary-search-tree.md
index 2ffe6c5dc0b..b02f10043d3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/use-breadth-first-search-in-a-binary-search-tree.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/use-breadth-first-search-in-a-binary-search-tree.md
@@ -1,18 +1,24 @@
---
id: 587d8258367417b2b2512c7f
-title: 在二叉搜索树中使用广度优先搜索
+title: Use Breadth First Search in a Binary Search Tree
challengeType: 1
-videoUrl: ''
+forumTopicId: 301718
dashedName: use-breadth-first-search-in-a-binary-search-tree
---
# --description--
-这里我们将介绍另一种树遍历方法：广度优先搜索。与上一次挑战中的深度优先搜索方法相比，广度优先搜索在继续进入下一级别之前探索树中给定级别中的所有节点。通常，队列在广度优先搜索算法的设计中用作辅助数据结构。在此方法中，我们首先将根节点添加到队列中。然后我们开始一个循环，我们将队列中的第一个项目出列，将其添加到一个新数组，然后检查它们的子子树。如果它的孩子不是空的，他们每个都被排队。此过程将继续，直到队列为空。说明：让我们在树中创建一个名为`levelOrder`的广度优先搜索方法。此方法应返回一个包含所有树节点值的数组，并以广度优先的方式进行探索。确保返回数组中的值，而不是节点本身。应从左到右遍历一个级别。接下来，让我们编写一个名为`reverseLevelOrder`的类似方法，它在每个级别执行相同的搜索，但是反向（从右到左）。
+Here we will introduce another tree traversal method: breadth-first search. In contrast to the depth-first search methods from the last challenge, breadth-first search explores all the nodes in a given level within a tree before continuing on to the next level. Typically, queues are utilized as helper data structures in the design of breadth-first search algorithms.
+
+In this method, we start by adding the root node to a queue. Then we begin a loop where we dequeue the first item in the queue, add it to a new array, and then inspect both its child subtrees. If its children are not null, they are each enqueued. This process continues until the queue is empty.
+
+# --instructions--
+
+Let's create a breadth-first search method in our tree called `levelOrder`. This method should return an array containing the values of all the tree nodes, explored in a breadth-first manner. Be sure to return the values in the array, not the nodes themselves. A level should be traversed from left to right. Next, let's write a similar method called `reverseLevelOrder` which performs the same search but in the reverse direction (right to left) at each level.
# --hints--
-存在`BinarySearchTree`数据结构。
+The `BinarySearchTree` data structure should exist.
```js
assert(
@@ -26,7 +32,7 @@ assert(
);
```
-二叉搜索树有一个名为`levelOrder`的方法。
+The binary search tree should have a method called `levelOrder`.
```js
assert(
@@ -42,7 +48,7 @@ assert(
);
```
-二叉搜索树有一个名为`reverseLevelOrder`的方法。
+The binary search tree should have a method called `reverseLevelOrder`.
```js
assert(
@@ -58,7 +64,7 @@ assert(
);
```
-`levelOrder`方法返回按级别顺序探索的树节点值的数组。
+The `levelOrder` method should return an array of the tree node values explored in level order.
```js
assert(
@@ -88,7 +94,7 @@ assert(
);
```
-`reverseLevelOrder`方法返回以反向级别顺序探索的树节点值的数组。
+The `reverseLevelOrder` method should return an array of the tree node values explored in reverse level order.
```js
assert(
@@ -118,7 +124,7 @@ assert(
);
```
-`levelOrder`方法为空树返回`null` 。
+The `levelOrder` method should return `null` for an empty tree.
```js
assert(
@@ -137,7 +143,7 @@ assert(
);
```
-`reverseLevelOrder`方法为空树返回`null` 。
+The `reverseLevelOrder` method should return `null` for an empty tree.
```js
assert(
@@ -208,7 +214,7 @@ function Node(value) {
function BinarySearchTree() {
this.root = null;
// Only change code below this line
-
+
// Only change code above this line
}
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/use-depth-first-search-in-a-binary-search-tree.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/use-depth-first-search-in-a-binary-search-tree.md
index 31d90eb9a13..c03fb750777 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/use-depth-first-search-in-a-binary-search-tree.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/use-depth-first-search-in-a-binary-search-tree.md
@@ -1,18 +1,22 @@
---
id: 587d8257367417b2b2512c7e
-title: 在二叉搜索树中使用深度优先搜索
+title: Use Depth First Search in a Binary Search Tree
challengeType: 1
-videoUrl: ''
+forumTopicId: 301719
dashedName: use-depth-first-search-in-a-binary-search-tree
---
# --description--
-我们知道如何在二叉搜索树中搜索特定值。但是，如果我们只想探索整棵树呢？或者，如果我们没有有序树，我们只需要搜索一个值？这里我们将介绍一些可用于探索树数据结构的树遍历方法。首先是深度优先搜索。在深度优先搜索中，在搜索继续到另一个子树之前，尽可能深地探索给定子树。有三种方法可以完成：按顺序：从最左边的节点开始搜索，到最右边的节点结束。预购：在树叶前探索所有的根。下订单：在根之前探索所有的叶子。您可能会猜到，您可以选择不同的搜索方法，具体取决于树存储的数据类型以及您要查找的内容。对于二叉搜索树，inorder遍历以排序顺序返回节点。说明：这里我们将在二叉搜索树上创建这三种搜索方法。深度优先搜索是一种固有的递归操作，只要子节点存在，它就会继续探索更多的子树。一旦理解了这个基本概念，您就可以简单地重新排列探索节点和子树的顺序，以生成上述三个搜索中的任何一个。例如，在后序搜索中，我们希望在开始返回任何节点本身之前一直递归到叶节点，而在预订搜索中，我们希望首先返回节点，然后继续递归在树下。在我们的树上定义`inorder` ， `preorder`和`postorder`方法。这些方法中的每一个都应该返回表示树遍历的项数组。确保返回数组中每个节点的整数值，而不是节点本身。最后，如果树为空，则返回`null` 。
+We know how to search a binary search tree for a specific value. But what if we just want to explore the entire tree? Or what if we don't have an ordered tree and we need to just search for a value? Here we will introduce some tree traversal methods which can be used to explore tree data structures. First up is depth-first search. In depth-first search, a given subtree is explored as deeply as possible before the search continues on to another subtree. There are three ways this can be done: In-order: Begin the search at the left-most node and end at the right-most node. Pre-order: Explore all the roots before the leaves. Post-order: Explore all the leaves before the roots. As you may guess, you may choose different search methods depending on what type of data your tree is storing and what you are looking for. For a binary search tree, an inorder traversal returns the nodes in sorted order.
+
+# --instructions--
+
+Here we will create these three search methods on our binary search tree. Depth-first search is an inherently recursive operation which continues to explore further subtrees so long as child nodes are present. Once you understand this basic concept, you can simply rearrange the order in which you explore the nodes and subtrees to produce any of the three searches above. For example, in post-order search we would want to recurse all the way to a leaf node before we begin to return any of the nodes themselves, whereas in pre-order search we would want to return the nodes first, and then continue recursing down the tree. Define `inorder`, `preorder`, and `postorder` methods on our tree. Each of these methods should return an array of items which represent the tree traversal. Be sure to return the integer values at each node in the array, not the nodes themselves. Finally, return `null` if the tree is empty.
# --hints--
-存在`BinarySearchTree`数据结构。
+The `BinarySearchTree` data structure should exist.
```js
assert(
@@ -26,7 +30,7 @@ assert(
);
```
-二叉搜索树有一个名为`inorder`的方法。
+The binary search tree should have a method called `inorder`.
```js
assert(
@@ -42,7 +46,7 @@ assert(
);
```
-二叉搜索树有一个名为`preorder`的方法。
+The binary search tree should have a method called `preorder`.
```js
assert(
@@ -58,7 +62,7 @@ assert(
);
```
-二叉搜索树有一个名为`postorder`的方法。
+The binary search tree should have a method called `postorder`.
```js
assert(
@@ -74,7 +78,7 @@ assert(
);
```
-`inorder`方法返回由inorder遍历产生的节点值数组。
+The `inorder` method should return an array of the node values that result from an inorder traversal.
```js
assert(
@@ -104,7 +108,7 @@ assert(
);
```
-`preorder`方法返回由前序遍历产生的节点值数组。
+The `preorder` method should return an array of the node values that result from a preorder traversal.
```js
assert(
@@ -134,7 +138,7 @@ assert(
);
```
-`postorder`方法返回由后序遍历产生的节点值数组。
+The `postorder` method should return an array of the node values that result from a postorder traversal.
```js
assert(
@@ -164,7 +168,7 @@ assert(
);
```
-`inorder`方法为空树返回`null` 。
+The `inorder` method should return `null` for an empty tree.
```js
assert(
@@ -183,7 +187,7 @@ assert(
);
```
-`preorder`方法为空树返回`null` 。
+The `preorder` method should return `null` for an empty tree.
```js
assert(
@@ -202,7 +206,7 @@ assert(
);
```
-`postorder`方法为空树返回`null` 。
+The `postorder` method should return `null` for an empty tree.
```js
assert(
@@ -274,7 +278,7 @@ function Node(value) {
function BinarySearchTree() {
this.root = null;
// Only change code below this line
-
+
// Only change code above this line
}
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/use-spread-and-notes-for-es5-set-integration.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/use-spread-and-notes-for-es5-set-integration.md
index 7e60c7c4c81..79ba7313b8c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/use-spread-and-notes-for-es5-set-integration.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/use-spread-and-notes-for-es5-set-integration.md
@@ -1,26 +1,34 @@
---
id: 587d8255367417b2b2512c73
-title: 使用Spread和Notes进行ES5 Set（）集成
+title: Use Spread and Notes for ES5 Set() Integration
challengeType: 1
-videoUrl: ''
+forumTopicId: 301720
dashedName: use-spread-and-notes-for-es5-set-integration
---
# --description--
-你还记得ES6传播运营商`...` ？ `...`可以在ES6中获取可迭代对象并将它们转换为数组。让我们创建一个Set，并检查传播函数。
+Do you remember the ES6 spread operator `...`?
-> var set = new Set（\[1,2,3]）;
-> var setToArr = \[... set]
-> console.log（setToArr）//返回\[1,2,3]
+`...` can take iterable objects in ES6 and turn them into arrays.
+
+Let's create a Set, and check out the spread function.
+
+```js
+var set = new Set([1,2,3]);
+var setToArr = [...set]
+console.log(setToArr) // returns [ 1, 2, 3 ]
+```
# --instructions--
-在本练习中，我们将set对象传递给`checkSet`函数。它应该返回一个包含Set值的数组。现在你已经成功学会了如何使用ES6 `Set()`对象，干得好！
+In this exercise we will pass a set object to the `checkSet` function. It should return an array containing the values of the Set.
+
+Now you've successfully learned how to use the ES6 `Set()` object, good job!
# --hints--
-您的套装已正确退回！
+`checkSet(new Set([1,2,3,4,5,6,7])` should return `[1, 2, 3, 4, 5, 6, 7]`.
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/work-with-nodes-in-a-linked-list.md b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/work-with-nodes-in-a-linked-list.md
index d84fd4e0f50..c746995d5db 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/work-with-nodes-in-a-linked-list.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/work-with-nodes-in-a-linked-list.md
@@ -1,28 +1,32 @@
---
id: 587d8251367417b2b2512c61
-title: 使用链接列表中的节点
+title: Work with Nodes in a Linked List
challengeType: 1
-videoUrl: ''
+forumTopicId: 301721
dashedName: work-with-nodes-in-a-linked-list
---
# --description--
-您将在计算机科学中遇到的另一个常见数据结构是<dfn>链表</dfn> 。链表是数据元素的线性集合，称为“节点”，每个数据元素指向下一个。链表中的每个<dfn>节点都</dfn>包含两个关键信息： `element`本身和对下一个`node`的引用。想象一下你在康加舞线上。你的手掌握在线下的下一个人身上，你身后的人就会抓住你。你可以直接看到这个人，但是他们阻挡了前方其他人的视线。一个节点就像一个康加舞线上的人：他们知道自己是谁，他们只能看到下一个人，但他们并不知道前方或后方的其他人。
+Another common data structure you'll run into in computer science is the <dfn>linked list</dfn>. A linked list is a linear collection of data elements, called 'nodes', each of which points to the next. Each <dfn>node</dfn> in a linked list contains two key pieces of information: the `element` itself, and a reference to the next `node`.
+
+Imagine that you are in a conga line. You have your hands on the next person in the line, and the person behind you has their hands on you. You can see the person straight ahead of you, but they are blocking the view of the other people ahead in line. A node is just like a person in a conga line: they know who they are and they can only see the next person in line, but they are not aware of the other people ahead or behind them.
# --instructions--
-在我们的代码编辑器中，我们创建了两个节点， `Kitten`和`Puppy` ，我们手动将`Kitten`节点连接到`Puppy`节点。创建`Cat`和`Dog`节点并手动将它们添加到该行。
+In our code editor, we've created two nodes, `Kitten` and `Puppy`, and we've manually connected the `Kitten` node to the `Puppy` node.
+
+Create a `Cat` and `Dog` node and manually add them to the line.
# --hints--
-您的`Puppy`节点应该具有对`Cat`节点的引用。
+Your `Puppy` node should have a reference to a `Cat` node.
```js
assert(Puppy.next.element === 'Cat');
```
-您的`Cat`节点应该具有对`Dog`节点的引用。
+Your `Cat` node should have a reference to a `Dog` node.
```js
assert(Cat.next.element === 'Dog');
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-1-multiples-of-3-and-5.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-1-multiples-of-3-and-5.md
index 243277e6d99..ff256d48a47 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-1-multiples-of-3-and-5.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-1-multiples-of-3-and-5.md
@@ -1,43 +1,49 @@
---
id: 5900f36e1000cf542c50fe80
-title: 问题1：3和5的倍数
+title: 'Problem 1: Multiples of 3 and 5'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301722
dashedName: problem-1-multiples-of-3-and-5
---
# --description--
-如果我们列出所有10以下是3或5的倍数的自然数，我们会得到3，5，6和9。这些倍数的总和是23。
+If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
-求出所有在`number`以下的3或5的倍数的总和。
+Find the sum of all the multiples of 3 or 5 below the provided parameter value `number`.
# --hints--
-`multiplesOf3and5(1000)`应该返回233168。
+`multiplesOf3and5(10)` should return a number.
```js
-assert.strictEqual(multiplesOf3and5(1000), 233168);
+assert(typeof multiplesOf3and5(10) === 'number');
```
-`multiplesOf3and5(49)`应该返回543。
+`multiplesOf3and5(49)` should return 543.
```js
assert.strictEqual(multiplesOf3and5(49), 543);
```
-`multiplesOf3and5(19564)`应该返回89301183。
+`multiplesOf3and5(1000)` should return 233168.
```js
-assert.strictEqual(multiplesOf3and5(19564), 89301183);
+assert.strictEqual(multiplesOf3and5(1000), 233168);
```
-您的函数未使用我们的测试值返回正确的结果。
+`multiplesOf3and5(8456)` should return 16687353.
```js
assert.strictEqual(multiplesOf3and5(8456), 16687353);
```
+`multiplesOf3and5(19564)` should return 89301183.
+
+```js
+assert.strictEqual(multiplesOf3and5(19564), 89301183);
+```
+
# --seed--
## --seed-contents--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-10-summation-of-primes.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-10-summation-of-primes.md
index 03f3f9fedfa..d1194e13f33 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-10-summation-of-primes.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-10-summation-of-primes.md
@@ -1,36 +1,44 @@
---
id: 5900f3761000cf542c50fe89
-title: 问题10：素数的总和
+title: 'Problem 10: Summation of primes'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301723
dashedName: problem-10-summation-of-primes
---
# --description--
-低于10的素数之和为2 + 3 + 5 + 7 = 17.求出n以下所有素数的总和。
+The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
+
+Find the sum of all the primes below `n`.
# --hints--
-`primeSummation(17)`应该返回41。
+`primeSummation(17)` should return a number.
+
+```js
+assert(typeof primeSummation(17) === 'number');
+```
+
+`primeSummation(17)` should return 41.
```js
assert.strictEqual(primeSummation(17), 41);
```
-`primeSummation(2001)`应该返回277050。
+`primeSummation(2001)` should return 277050.
```js
assert.strictEqual(primeSummation(2001), 277050);
```
-`primeSummation(140759)`应该返回873608362。
+`primeSummation(140759)` should return 873608362.
```js
assert.strictEqual(primeSummation(140759), 873608362);
```
-`primeSummation(2000000)`应返回142913828922。
+`primeSummation(2000000)` should return 142913828922.
```js
assert.strictEqual(primeSummation(2000000), 142913828922);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-100-arranged-probability.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-100-arranged-probability.md
index f94cdb91d42..7a5d2fd453d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-100-arranged-probability.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-100-arranged-probability.md
@@ -1,21 +1,31 @@
---
id: 5900f3d01000cf542c50fee3
-title: 问题100：安排概率
+title: 'Problem 100: Arranged probability'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301724
dashedName: problem-100-arranged-probability
---
# --description--
-如果一个盒子包含21个彩色光盘，由15个蓝色光盘和6个红色光盘组成，随机拍摄两张光盘，可以看出拍摄两张蓝色光盘的概率，P（BB）=（15/21） ）×（14/20）= 1/2。下一个这样的安排，其中有50％的机会随机拍摄两张蓝色光盘，是一个包含八十五个蓝色光盘和三十五个红色光盘的盒子。通过找到第一个包含总共超过1012 = 1,000,000,000,000个光盘的布置，确定该盒子将包含的蓝色光盘的数量。
+If a box contains twenty-one coloured discs, composed of fifteen blue discs and six red discs, and two discs were taken at random, it can be seen that the probability of taking two blue discs, P(BB) = (15/21)×(14/20) = 1/2.
+
+The next such arrangement, for which there is exactly 50% chance of taking two blue discs at random, is a box containing eighty-five blue discs and thirty-five red discs.
+
+By finding the first arrangement to contain over 10<sup>12</sup> = 1,000,000,000,000 discs in total, determine the number of blue discs that the box would contain.
# --hints--
-`euler100()`应该返回756872327473。
+`arrangedProbability()` should return a number.
```js
-assert.strictEqual(euler100(), 756872327473);
+assert(typeof arrangedProbability() === 'number');
+```
+
+`arrangedProbability()` should return 756872327473.
+
+```js
+assert.strictEqual(arrangedProbability(), 756872327473);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-101-optimum-polynomial.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-101-optimum-polynomial.md
index 2c6293c4d4a..75731144759 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-101-optimum-polynomial.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-101-optimum-polynomial.md
@@ -1,22 +1,32 @@
---
id: 5900f3d21000cf542c50fee4
-title: 问题101：最佳多项式
+title: 'Problem 101: Optimum polynomial'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301725
dashedName: problem-101-optimum-polynomial
---
# --description--
-如果我们被给出序列的前k个项，则不可能肯定地说下一个项的值，因为存在无限多个可以对序列建模的多项式函数。举个例子，让我们考虑一下立方体数字的顺序。这由生成函数定义，un = n3：1,8,27,64,125,216 ......假设我们只给出了该序列的前两个项。根据“简单就是最好”的原则，我们应该假设一个线性关系，并预测下一个项为15（公共差异7）。即使我们被提出前三个术语，按照相同的简单原则，也应假设二次关系。我们将OP（k，n）定义为序列的前k个项的最佳多项式生成函数的第n项。应该清楚的是，OP（k，n）将准确地生成n≤k的序列项，并且可能第一个不正确的项（FIT）将是OP（k，k + 1）;在这种情况下，我们将其称为坏OP（BOP）。作为一个基础，如果我们只给出第一个序列项，那么假设恒定是最明智的;也就是说，对于n≥2，OP（1，n）= u1。因此，我们获得了立方序列的以下OP：
+If we are presented with the first k terms of a sequence it is impossible to say with certainty the value of the next term, as there are infinitely many polynomial functions that can model the sequence.
-OP（1，n）= 11 1,1,1,1 ...... OP（2，n）= 7n-6 1,8,15，...... OP（3，n）= 6n2-11n + 6 1,8,27,58，... OP（4，n）= n3 1,8,27,64,125，......
+As an example, let us consider the sequence of cube numbers. This is defined by the generating function, un = n3: 1, 8, 27, 64, 125, 216, ...
-显然，对于k≥4，不存在BOP。通过考虑BOP产生的FIT之和（以红色表示），我们得到1 + 15 + 58 = 74.考虑下面的十度多项式生成函数：un = 1 - n + n2 - n3 + n4 - n5 + n6 - n7 + n8 - n9 + n10求BOP的FIT之和。
+Suppose we were only given the first two terms of this sequence. Working on the principle that "simple is best" we should assume a linear relationship and predict the next term to be 15 (common difference 7). Even if we were presented with the first three terms, by the same principle of simplicity, a quadratic relationship should be assumed.
+
+We shall define OP(k, n) to be the nth term of the optimum polynomial generating function for the first k terms of a sequence. It should be clear that OP(k, n) will accurately generate the terms of the sequence for n ≤ k, and potentially the first incorrect term (FIT) will be OP(k, k+1); in which case we shall call it a bad OP (BOP).
+
+As a basis, if we were only given the first term of sequence, it would be most sensible to assume constancy; that is, for n ≥ 2, OP(1, n) = u1.
+
+Hence we obtain the following OPs for the cubic sequence:
+
+OP(1, n) = 1 1, 1, 1, 1, ... OP(2, n) = 7n−6 1, 8, 15, ... OP(3, n) = 6n2−11n+6 1, 8, 27, 58, ... OP(4, n) = n3 1, 8, 27, 64, 125, ...
+
+Clearly no BOPs exist for k ≥ 4. By considering the sum of FITs generated by the BOPs (indicated in red above), we obtain 1 + 15 + 58 = 74. Consider the following tenth degree polynomial generating function: un = 1 − n + n2 − n3 + n4 − n5 + n6 − n7 + n8 − n9 + n10 Find the sum of FITs for the BOPs.
# --hints--
-`euler101()`应该返回37076114526。
+`euler101()` should return 37076114526.
```js
assert.strictEqual(euler101(), 37076114526);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-102-triangle-containment.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-102-triangle-containment.md
index 694866044ab..3f11a537f1c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-102-triangle-containment.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-102-triangle-containment.md
@@ -1,18 +1,30 @@
---
id: 5900f3d21000cf542c50fee5
-title: 问题102：三角形遏制
+title: 'Problem 102: Triangle containment'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301726
dashedName: problem-102-triangle-containment
---
# --description--
-在笛卡尔平面上随机绘制三个不同的点，其中-1000≤x，y≤1000，从而形成三角形。考虑以下两个三角形：A（-340,495），B（-153，-910），C（835，-947）X（-175,41），Y（-421，-714），Z（574， - 645）可以验证三角形ABC包含原点，而三角形XYZ不包含原点。使用triangles.txt（右键单击并将“Save Link / Target As ...”保存），包含一千个“随机”三角形坐标的27K文本文件，找到内部包含原点的三角形数。注意：文件中的前两个示例表示上面给出的示例中的三角形。
+Three distinct points are plotted at random on a Cartesian plane, for which -1000 ≤ x, y ≤ 1000, such that a triangle is formed.
+
+Consider the following two triangles:
+
+A(-340,495), B(-153,-910), C(835,-947)
+
+X(-175,41), Y(-421,-714), Z(574,-645)
+
+It can be verified that triangle ABC contains the origin, whereas triangle XYZ does not.
+
+Using triangles.txt (right click and 'Save Link/Target As...'), a 27K text file containing the coordinates of one thousand "random" triangles, find the number of triangles for which the interior contains the origin.
+
+NOTE: The first two examples in the file represent the triangles in the example given above.
# --hints--
-`euler102()`应该返回228。
+`euler102()` should return 228.
```js
assert.strictEqual(euler102(), 228);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-103-special-subset-sums-optimum.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-103-special-subset-sums-optimum.md
index 4518a984d58..1ba5808084f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-103-special-subset-sums-optimum.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-103-special-subset-sums-optimum.md
@@ -1,18 +1,34 @@
---
id: 5900f3d61000cf542c50fee7
-title: 问题103：特殊子集和：最佳
+title: 'Problem 103: Special subset sums: optimum'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301727
dashedName: problem-103-special-subset-sums-optimum
---
# --description--
-设S（A）表示大小为n的集合A中的元素之和。如果对于任何两个非空的不相交子集B和C，我们将其称为特殊和集合，以下属性为真：S（B）≠S（C）;也就是说，子集的总和不能相等。如果B包含的元素多于C，则S（B）> S（C）。如果对于给定的n，S（A）被最小化，我们将其称为最优的特殊和集。下面给出前五个最佳特殊和集。 n = 1：{1} n = 2：{1,2} n = 3：{2,3,4} n = 4：{3,5,6,7} n = 5：{6,9,11 ，12,13}似乎对于给定的最优集合，A = {a1，a2，...，an}，下一个最优集合的形式为B = {b，a1 + b，a2 + b，... ..，an + b}，其中b是前一行的“中间”元素。通过应用这个“规则”，我们期望n = 6的最优集合为A = {11,17,20,22,23,24}，其中S（A）= 117.但是，这不是最佳集合因为我们仅应用算法来提供接近最优的集合。 n = 6的最佳集合是A = {11,18,19,20,22,25}，其中S（A）= 115并且对应的集合字符串：111819202225。假设A是n =的最优特殊和集合7，找到它的设置字符串。注意：此问题与问题105和问题106有关。
+Let S(A) represent the sum of elements in set A of size n. We shall call it a special sum set if for any two non-empty disjoint subsets, B and C, the following properties are true:
+
+S(B) ≠ S(C); that is, sums of subsets cannot be equal.
+
+If B contains more elements than C then S(B) > S(C).
+
+If S(A) is minimised for a given n, we shall call it an optimum special sum set. The first five optimum special sum sets are given below.
+
+n = 1: {1}n = 2: {1, 2}n = 3: {2, 3, 4}n = 4: {3, 5, 6, 7}n = 5: {6, 9, 11, 12, 13}
+
+It seems that for a given optimum set, A = {a1, a2, ... , an}, the next optimum set is of the form B = {b, a1+b, a2+b, ... ,an+b}, where b is the "middle" element on the previous row.
+
+By applying this "rule" we would expect the optimum set for n = 6 to be A = {11, 17, 20, 22, 23, 24}, with S(A) = 117. However, this is not the optimum set, as we have merely applied an algorithm to provide a near optimum set. The optimum set for n = 6 is A = {11, 18, 19, 20, 22, 25}, with S(A) = 115 and corresponding set string: 111819202225.
+
+Given that A is an optimum special sum set for n = 7, find its set string.
+
+NOTE: This problem is related to Problem 105 and Problem 106.
# --hints--
-`euler103()`应该返回20313839404245。
+`euler103()` should return 20313839404245.
```js
assert.strictEqual(euler103(), 20313839404245);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-104-pandigital-fibonacci-ends.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-104-pandigital-fibonacci-ends.md
index 74da553a323..625840f94aa 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-104-pandigital-fibonacci-ends.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-104-pandigital-fibonacci-ends.md
@@ -1,18 +1,24 @@
---
id: 5900f3d51000cf542c50fee6
-title: 问题104：Pandigital Fibonacci结束
+title: 'Problem 104: Pandigital Fibonacci ends'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301728
dashedName: problem-104-pandigital-fibonacci-ends
---
# --description--
-Fibonacci序列由递归关系定义：Fn = Fn-1 + Fn-2，其中F1 = 1且F2 = 1.事实证明，包含113位数字的F541是第一个斐波那契数字，其中最后九个数字是1-9 pandigital（包含所有数字1到9，但不一定按顺序）。 F2749包含575个数字，是第一个斐波那契数字，前九个数字是1-9 pandigital。鉴于Fk是第一个斐波纳契数，前九个数字和后九个数字是1-9 pandigital，找到k。
+The Fibonacci sequence is defined by the recurrence relation:
+
+Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
+
+It turns out that F541, which contains 113 digits, is the first Fibonacci number for which the last nine digits are 1-9 pandigital (contain all the digits 1 to 9, but not necessarily in order). And F2749, which contains 575 digits, is the first Fibonacci number for which the first nine digits are 1-9 pandigital.
+
+Given that Fk is the first Fibonacci number for which the first nine digits AND the last nine digits are 1-9 pandigital, find k.
# --hints--
-`euler104()`应返回329468。
+`euler104()` should return 329468.
```js
assert.strictEqual(euler104(), 329468);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-105-special-subset-sums-testing.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-105-special-subset-sums-testing.md
index 59aff4a17f9..28d4d6ac863 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-105-special-subset-sums-testing.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-105-special-subset-sums-testing.md
@@ -1,18 +1,28 @@
---
id: 5900f3d61000cf542c50fee8
-title: 问题105：特殊子集总和：测试
+title: 'Problem 105: Special subset sums: testing'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301729
dashedName: problem-105-special-subset-sums-testing
---
# --description--
-设S（A）表示大小为n的集合A中的元素之和。如果对于任何两个非空的不相交子集B和C，我们将其称为特殊和集合，以下属性为真：S（B）≠S（C）;也就是说，子集的总和不能相等。如果B包含的元素多于C，则S（B）> S（C）。例如，{81,88,75,42,87,84,86,65}不是一个特殊的和集，因为65 + 87 + 88 = 75 + 81 + 84，而{157,150,164,119,79 ，159,161,139,158}满足所有可能的子集对组合的规则，并且S（A）= 1286.使用sets.txt（右键单击并“将链接/目标另存为...”），4K文本文件包含七到十二个元素的一百个集合（上面给出的两个例子是文件中的前两个集合），识别所有特殊的和集，A1，A2，...，Ak，并找到S的值（ A1）+ S（A2）+ ... + S（Ak）。注意：此问题与问题103和问题106有关。
+Let S(A) represent the sum of elements in set A of size n. We shall call it a special sum set if for any two non-empty disjoint subsets, B and C, the following properties are true:
+
+S(B) ≠ S(C); that is, sums of subsets cannot be equal.
+
+If B contains more elements than C then S(B) > S(C).
+
+For example, {81, 88, 75, 42, 87, 84, 86, 65} is not a special sum set because 65 + 87 + 88 = 75 + 81 + 84, whereas {157, 150, 164, 119, 79, 159, 161, 139, 158} satisfies both rules for all possible subset pair combinations and S(A) = 1286.
+
+Using sets.txt (right click and "Save Link/Target As..."), a 4K text file with one-hundred sets containing seven to twelve elements (the two examples given above are the first two sets in the file), identify all the special sum sets, A1, A2, ..., Ak, and find the value of S(A1) + S(A2) + ... + S(Ak).
+
+NOTE: This problem is related to Problem 103 and Problem 106.
# --hints--
-`euler105()`应返回73702。
+`euler105()` should return 73702.
```js
assert.strictEqual(euler105(), 73702);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-106-special-subset-sums-meta-testing.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-106-special-subset-sums-meta-testing.md
index 872a6ff1f2c..bb1db2e2607 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-106-special-subset-sums-meta-testing.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-106-special-subset-sums-meta-testing.md
@@ -1,18 +1,30 @@
---
id: 5900f3d71000cf542c50fee9
-title: 问题106：特殊子集和：元测试
+title: 'Problem 106: Special subset sums: meta-testing'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301730
dashedName: problem-106-special-subset-sums-meta-testing
---
# --description--
-设S（A）表示大小为n的集合A中的元素之和。如果对于任何两个非空的不相交子集B和C，我们将其称为特殊和集合，以下属性为真：S（B）≠S（C）;也就是说，子集的总和不能相等。如果B包含的元素多于C，则S（B）> S（C）。对于这个问题，我们假设给定的集合包含n个严格增加的元素，并且它已经满足第二个规则。令人惊讶的是，在可以从n = 4的集合中获得的25个可能的子集对中，仅需要对这些对中的1个进行相等性测试（第一规则）。类似地，当n = 7时，仅需要测试966个子集对中的70个。对于n = 12，可以获得多少261625个子集对需要进行相等性测试？注意：此问题与问题103和问题105有关。
+Let S(A) represent the sum of elements in set A of size n. We shall call it a special sum set if for any two non-empty disjoint subsets, B and C, the following properties are true:
+
+S(B) ≠ S(C); that is, sums of subsets cannot be equal.
+
+If B contains more elements than C then S(B) > S(C).
+
+For this problem we shall assume that a given set contains n strictly increasing elements and it already satisfies the second rule.
+
+Surprisingly, out of the 25 possible subset pairs that can be obtained from a set for which n = 4, only 1 of these pairs need to be tested for equality (first rule). Similarly, when n = 7, only 70 out of the 966 subset pairs need to be tested.
+
+For n = 12, how many of the 261625 subset pairs that can be obtained need to be tested for equality?
+
+NOTE: This problem is related to Problem 103 and Problem 105.
# --hints--
-`euler106()`应返回21384。
+`euler106()` should return 21384.
```js
assert.strictEqual(euler106(), 21384);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-107-minimal-network.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-107-minimal-network.md
index 22e7634752f..f51dfe801d6 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-107-minimal-network.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-107-minimal-network.md
@@ -1,22 +1,22 @@
---
id: 5900f3d91000cf542c50feea
-title: 问题107：最小网络
+title: 'Problem 107: Minimal network'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301731
dashedName: problem-107-minimal-network
---
# --description--
-以下无向网络由七个顶点和十二个边组成，总权重为243。
+The following undirected network consists of seven vertices and twelve edges with a total weight of 243.
-相同的网络可以由下面的矩阵表示。 ABCDEFG A-161221 --- B16--1720-- C12--28-31- D211728-181923 E-20-18--11 F-3119--27 G --- 231127-但是，有可能通过删除一些边缘来优化网络，并仍然确保网络上的所有点保持连接。实现最大节省的网络如下所示。它的权重为93，比原始网络节省了243 - 93 = 150。
+The same network can be represented by the matrix below. ABCDEFG A-161221--- B16--1720-- C12--28-31- D211728-181923 E-20-18--11 F--3119--27 G---231127- However, it is possible to optimise the network by removing some edges and still ensure that all points on the network remain connected. The network which achieves the maximum saving is shown below. It has a weight of 93, representing a saving of 243 − 93 = 150 from the original network.
-使用network.txt（右键单击并“保存链接/目标为...”），一个6K文本文件包含一个具有四十个顶点的网络，并以矩阵形式给出，找到通过删除冗余边缘可以实现的最大节省确保网络保持连接。
+Using network.txt (right click and 'Save Link/Target As...'), a 6K text file containing a network with forty vertices, and given in matrix form, find the maximum saving which can be achieved by removing redundant edges whilst ensuring that the network remains connected.
# --hints--
-`euler107()`应该返回259679。
+`euler107()` should return 259679.
```js
assert.strictEqual(euler107(), 259679);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-108-diophantine-reciprocals-i.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-108-diophantine-reciprocals-i.md
index 662a9402518..d216cfa9fd2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-108-diophantine-reciprocals-i.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-108-diophantine-reciprocals-i.md
@@ -1,20 +1,28 @@
---
id: 5900f3d91000cf542c50feeb
-title: 问题108：丢番图互惠I
+title: 'Problem 108: Diophantine Reciprocals I'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301732
dashedName: problem-108-diophantine-reciprocals-i
---
# --description--
-在下面的等式中，x，y和n是正整数。 1 / `x` + 1 / `y` = 1 / `n`对于`n` = 4，恰好有三种不同的解：1/5 + 1/20 = 1/4
+In the following equation x, y, and n are positive integers.
+
+1/`x` + 1/`y` = 1/`n`
+
+For `n` = 4 there are exactly three distinct solutions:
+
+1/5 + 1/20 = 1/4
1/6 + 1/12 = 1/4
-1/8 + 1/8 = 1/4不同解的数量超过一千的`n的`最小值是多少？
+1/8 + 1/8 = 1/4
+
+What is the least value of `n` for which the number of distinct solutions exceeds one-thousand?
# --hints--
-diophantineOne `diophantineOne()`应返回180180。
+`diophantineOne()` should return 180180.
```js
assert.strictEqual(diophantineOne(), 180180);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-109-darts.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-109-darts.md
index fcfa7c4cecc..e7a580b45b3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-109-darts.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-109-darts.md
@@ -1,16 +1,16 @@
---
id: 5900f3db1000cf542c50feec
-title: 问题109：飞镖
+title: 'Problem 109: Darts'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301733
dashedName: problem-109-darts
---
# --description--
-在飞镖游戏中，玩家在目标板上投掷三个飞镖，该目标板被分成二十个相等大小的部分，编号为一到二十。
+In the game of darts a player throws three darts at a target board which is split into twenty equal sized sections numbered one to twenty.
-飞镖的得分由飞镖着陆区域的数量决定。红色/绿色外圈外的飞镖着陆得分为零。该戒指内的黑色和奶油色区域代表单个分数。然而，红色/绿色外圈和中圈分别得分为双倍和高音。在董事会的中心是两个同心圆，称为公牛区或公牛眼。外面的公牛价值25点，内部公牛是双倍，价值50点。规则有很多变化，但在最流行的游戏中，玩家将以301或501的分数开始，并且第一个将他们的总跑动总数减少到零的玩家是胜利者。然而，玩“双打”系统是正常的，这意味着玩家必须在他们的最终飞镖上获得双倍（包括在棋盘中央的双靶心）才能获胜;任何其他飞镖都会将他们的跑动总数减少到一个或更低意味着这组三个飞镖的得分是“半身像”。当玩家能够完成他们当前的分数时，它被称为“结账”，最高结账时间是170：T20 T20 D25（两个高音20和双牛）。有6种截然不同的结账方式，分数为6：
+The score of a dart is determined by the number of the region that the dart lands in. A dart landing outside the red/green outer ring scores zero. The black and cream regions inside this ring represent single scores. However, the red/green outer ring and middle ring score double and treble scores respectively. At the centre of the board are two concentric circles called the bull region, or bulls-eye. The outer bull is worth 25 points and the inner bull is a double, worth 50 points. There are many variations of rules but in the most popular game the players will begin with a score 301 or 501 and the first player to reduce their running total to zero is a winner. However, it is normal to play a "doubles out" system, which means that the player must land a double (including the double bulls-eye at the centre of the board) on their final dart to win; any other dart that would reduce their running total to one or lower means the score for that set of three darts is "bust". When a player is able to finish on their current score it is called a "checkout" and the highest checkout is 170: T20 T20 D25 (two treble 20s and double bull). There are exactly eleven distinct ways to checkout on a score of 6:
D3
@@ -24,11 +24,11 @@ S4 D1
S1 S1 D2 S1 T1 D1 S1 S3 D1 D1 D1 D1 D1 S2 D1 S2 S2 D1
-请注意，D1 D2被认为与D2 D1不同，因为它们在不同的双打上完成。但是，组合S1 T1 D1被认为与T1 S1 D1相同。此外，我们不会在考虑组合时包含未命中数;例如，D3与0 D3和0 0 D3相同。令人难以置信的是，共有42336种不同的检查方式。玩家以低于100的分数结账有多少种不同的方式？
+Note that D1 D2 is considered different to D2 D1 as they finish on different doubles. However, the combination S1 T1 D1 is considered the same as T1 S1 D1. In addition we shall not include misses in considering combinations; for example, D3 is the same as 0 D3 and 0 0 D3. Incredibly there are 42336 distinct ways of checking out in total. How many distinct ways can a player checkout with a score less than 100?
# --hints--
-`euler109()`应返回38182。
+`euler109()` should return 38182.
```js
assert.strictEqual(euler109(), 38182);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-11-largest-product-in-a-grid.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-11-largest-product-in-a-grid.md
index a7b4ad7494e..d7610e3ac39 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-11-largest-product-in-a-grid.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-11-largest-product-in-a-grid.md
@@ -1,71 +1,62 @@
---
id: 5900f3781000cf542c50fe8a
-title: 问题11：网格中最大的产品
+title: 'Problem 11: Largest product in a grid'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301734
dashedName: problem-11-largest-product-in-a-grid
---
# --description--
-在下面的20×20网格中，沿对角线的四个数字标记为红色。
+In the 20×20 grid below, four numbers along a diagonal line have been marked in red.
-08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
+<div style='text-align: center;'>
+ 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08<br>
+ 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00<br>
+ 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65<br>
+ 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91<br>
+ 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80<br>
+ 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50<br>
+ 32 98 81 28 64 23 67 10 <span style='color: red'><b>26</b></span> 38 40 67 59 54 70 66 18 38 64 70<br>
+ 67 26 20 68 02 62 12 20 95 <span style='color: red'><b>63</b></span> 94 39 63 08 40 91 66 49 94 21<br>
+ 24 55 58 05 66 73 99 26 97 17 <span style='color: red'><b>78</b></span> 78 96 83 14 88 34 89 63 72<br>
+ 21 36 23 09 75 00 76 44 20 45 35 <span style='color: red'><b>14</b></span> 00 61 33 97 34 31 33 95<br>
+ 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92<br>
+ 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57<br>
+ 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58<br>
+ 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40<br>
+ 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66<br>
+ 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69<br>
+ 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36<br>
+ 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16<br>
+ 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54<br>
+ 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48<br>
+</div>
-49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
+The product of these numbers is 26 × 63 × 78 × 14 = 1788696.
-81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
-
-52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
-
-22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
-
-24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
-
-32 98 81 28 64 23 67 10 **26** 38 40 67 59 54 70 66 18 38 64 70
-
-67 26 20 68 02 62 12 20 95 **63** 94 39 63 08 40 91 66 49 94 21
-
-24 55 58 05 66 73 99 26 97 17 **78** 78 96 83 14 88 34 89 63 72
-
-21 36 23 09 75 00 76 44 20 45 35 **14** 00 61 33 97 34 31 33 95
-
-78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
-
-16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
-
-86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
-
-19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
-
-04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
-
-88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
-
-04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
-
-20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
-
-20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
-
-01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
-
-这些数字的乘积是26×63×78×14 = 1788696.在给定的`arr`网格中，相同方向（上，下，左，右或对角）的四个相邻数字的最大乘积是多少？
+What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in a given `arr` grid?
# --hints--
-`largestGridProduct(grid)`应该返回70600674。
+`largestGridProduct(testGrid)` should return a number.
```js
-assert.strictEqual(largestGridProduct(grid), 70600674);
+assert(typeof largestGridProduct(testGrid) === 'number');
```
-`largestGridProduct(testGrid)`应该返回14169081。
+`largestGridProduct(testGrid)` should return 14169081.
```js
assert.strictEqual(largestGridProduct(testGrid), 14169081);
```
+`largestGridProduct(grid)` should return 70600674.
+
+```js
+assert.strictEqual(largestGridProduct(grid), 70600674);
+```
+
# --seed--
## --seed-contents--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-110-diophantine-reciprocals-ii.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-110-diophantine-reciprocals-ii.md
index 43455e5d1a4..c50c81a1647 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-110-diophantine-reciprocals-ii.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-110-diophantine-reciprocals-ii.md
@@ -1,18 +1,24 @@
---
id: 5900f3db1000cf542c50feed
-title: 问题110：丢番图互惠II
+title: 'Problem 110: Diophantine Reciprocals II'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301735
dashedName: problem-110-diophantine-reciprocals-ii
---
# --description--
-在下面的等式中，x，y和n是正整数。 1 / `x` + 1 / `y` = 1 / `n`可以证实，当`n` = 1260时，存在113个不同的解，并且这是`n的`最小值，其中不同解的总数超过100。 `n的`最小值是多少，不同解决方案的数量超过四百万？
+In the following equation x, y, and n are positive integers.
+
+1/`x` + 1/`y` = 1/`n`
+
+It can be verified that when `n` = 1260 there are 113 distinct solutions and this is the least value of `n` for which the total number of distinct solutions exceeds one hundred.
+
+What is the least value of `n` for which the number of distinct solutions exceeds four million?
# --hints--
-`diophantineTwo`应该返回9350130049860600。
+`diophantineTwo()` should return 9350130049860600.
```js
assert.strictEqual(diophantineTwo(), 9350130049860600);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-111-primes-with-runs.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-111-primes-with-runs.md
index a20e2aa90d1..920085b1f32 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-111-primes-with-runs.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-111-primes-with-runs.md
@@ -1,22 +1,30 @@
---
id: 5900f3db1000cf542c50feee
-title: 问题111：运行的Primes
+title: 'Problem 111: Primes with runs'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301736
dashedName: problem-111-primes-with-runs
---
# --description--
-考虑到包含重复数字的4位数素数，很明显它们不能全部相同：1111可被11整除，2222可被22整除，依此类推。但是有九个4位数的素数包含三个素数：1117,1151,1171,1181,1511,1811,2111,4111,8111我们将说M（n，d）表示n-的最大重复数位数。数字素数，其中d是重复数字，N（n，d）表示这些素数的数量，S（n，d）表示这些素数的总和。所以M（4,1）= 3是4位数的素数的最大重复位数，其中一个是重复的数字，有N（4,1）= 9个这样的素数，这些素数的总和是S （4,1）= 22275.事实证明，对于d = 0，只能使M（4,0）= 2个重复数字，但是存在N（4,0）= 13个这样的情况。同样地，我们获得了4位素数的以下结果。
+Considering 4-digit primes containing repeated digits it is clear that they cannot all be the same: 1111 is divisible by 11, 2222 is divisible by 22, and so on. But there are nine 4-digit primes containing three ones:
-数字，d M（4，d）N（4，d）S（4，d）0 2 13 67061 1 3 9 22275 2 3 1 2221 3 3 12 46214 4 3 2 8888 5 3 1 5557 6 3 1 6661 7 3 9 57863 8 3 1 8887 9 3 7 48073
+1117, 1151, 1171, 1181, 1511, 1811, 2111, 4111, 8111
-对于d = 0到9，所有S（4，d）的总和是273700.求所有S（10，d）的总和。
+We shall say that M(n, d) represents the maximum number of repeated digits for an n-digit prime where d is the repeated digit, N(n, d) represents the number of such primes, and S(n, d) represents the sum of these primes.
+
+So M(4, 1) = 3 is the maximum number of repeated digits for a 4-digit prime where one is the repeated digit, there are N(4, 1) = 9 such primes, and the sum of these primes is S(4, 1) = 22275. It turns out that for d = 0, it is only possible to have M(4, 0) = 2 repeated digits, but there are N(4, 0) = 13 such cases.
+
+In the same way we obtain the following results for 4-digit primes.
+
+Digit, d M(4, d) N(4, d) S(4, d) 0 2 13 67061 1 3 9 22275 2 3 1 2221 3 3 12 46214 4 3 2 8888 5 3 1 5557 6 3 1 6661 7 3 9 57863 8 3 1 8887 9 3 7 48073
+
+For d = 0 to 9, the sum of all S(4, d) is 273700. Find the sum of all S(10, d).
# --hints--
-`euler111()`应返回612407567715。
+`euler111()` should return 612407567715.
```js
assert.strictEqual(euler111(), 612407567715);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-112-bouncy-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-112-bouncy-numbers.md
index 3e82355620e..cd998d5bdeb 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-112-bouncy-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-112-bouncy-numbers.md
@@ -1,18 +1,28 @@
---
id: 5900f3dd1000cf542c50feef
-title: 问题112：有弹性的数字
+title: 'Problem 112: Bouncy numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301738
dashedName: problem-112-bouncy-numbers
---
# --description--
-如果左边的数字没有超过数字，则从左到右工作，称为递增数字;例如，134468。同样，如果右边的数字没有超过数字，则称为递减数字;例如，66420。我们将调用一个既不增加也不减少“有弹性”数的正整数;例如，155349。显然，不会有任何低于一百的弹性数字，但只有超过一千（525）的数字超过一半是有弹性的。事实上，有弹性数字首次达到50％的最小数量是538.令人惊讶的是，有弹性的数字变得越来越普遍，当我们达到21780时，有弹性数字的比例等于90％。找出有弹性数字的比例正好为99％的最小数字。
+Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, 134468.
+
+Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.
+
+We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, 155349.
+
+Clearly there cannot be any bouncy numbers below one-hundred, but just over half of the numbers below one-thousand (525) are bouncy. In fact, the least number for which the proportion of bouncy numbers first reaches 50% is 538.
+
+Surprisingly, bouncy numbers become more and more common and by the time we reach 21780 the proportion of bouncy numbers is equal to 90%.
+
+Find the least number for which the proportion of bouncy numbers is exactly 99%.
# --hints--
-`euler112()`应返回1587000。
+`euler112()` should return 1587000.
```js
assert.strictEqual(euler112(), 1587000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-113-non-bouncy-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-113-non-bouncy-numbers.md
index 2ec7c774c22..bc28697a24f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-113-non-bouncy-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-113-non-bouncy-numbers.md
@@ -1,18 +1,26 @@
---
id: 5900f3dd1000cf542c50fef0
-title: 问题113：非弹性数字
+title: 'Problem 113: Non-bouncy numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301739
dashedName: problem-113-non-bouncy-numbers
---
# --description--
-如果左边的数字没有超过数字，则从左到右工作，称为递增数字;例如，134468。同样，如果右边的数字没有超过数字，则称为递减数字;例如，66420。我们将调用一个既不增加也不减少“有弹性”数的正整数;例如，随着n的增加，低于n的弹性数的比例增加，使得只有12951个数字低于一百万个而不是有弹性而只有277032个非有弹性的数字低于1010.一个数量低于一个googol（10100） ）是不是有弹性？
+Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, 134468.
+
+Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.
+
+We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, 155349.
+
+As n increases, the proportion of bouncy numbers below n increases such that there are only 12951 numbers below one-million that are not bouncy and only 277032 non-bouncy numbers below 1010.
+
+How many numbers below a googol (10100) are not bouncy?
# --hints--
-`euler113()`应该返回51161058134250。
+`euler113()` should return 51161058134250.
```js
assert.strictEqual(euler113(), 51161058134250);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-114-counting-block-combinations-i.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-114-counting-block-combinations-i.md
index e21e53fae21..c9ec66c9aab 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-114-counting-block-combinations-i.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-114-counting-block-combinations-i.md
@@ -1,20 +1,20 @@
---
id: 5900f3e01000cf542c50fef2
-title: 问题114：计数块组合I
+title: 'Problem 114: Counting block combinations I'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301740
dashedName: problem-114-counting-block-combinations-i
---
# --description--
-测量七个单元长度的行具有红色块，其上放置有最小长度为三个单元，使得任何两个红色块（允许为不同长度）由至少一个黑色方块隔开。有七种方法可以做到这一点。
+A row measuring seven units in length has red blocks with a minimum length of three units placed on it, such that any two red blocks (which are allowed to be different lengths) are separated by at least one black square. There are exactly seventeen ways of doing this.
-一行可以测量多达50个长度的行数？注意：虽然上面的示例不适合这种可能性，但通常允许混合块大小。例如，在一个长度为8个单位的行上，您可以使用红色（3），黑色（1）和红色（4）。
+How many ways can a row measuring fifty units in length be filled? NOTE: Although the example above does not lend itself to the possibility, in general it is permitted to mix block sizes. For example, on a row measuring eight units in length you could use red (3), black (1), and red (4).
# --hints--
-`euler114()`应该返回16475640049。
+`euler114()` should return 16475640049.
```js
assert.strictEqual(euler114(), 16475640049);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-115-counting-block-combinations-ii.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-115-counting-block-combinations-ii.md
index bbd1b0c9311..069a8a2c384 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-115-counting-block-combinations-ii.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-115-counting-block-combinations-ii.md
@@ -1,18 +1,30 @@
---
id: 5900f3df1000cf542c50fef1
-title: 问题115：计数块组合II
+title: 'Problem 115: Counting block combinations II'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301741
dashedName: problem-115-counting-block-combinations-ii
---
# --description--
-注意：这是问题114的更难的版本。测量n个单位长度的行具有红色块，其上放置最小长度为m个单位，使得任何两个红色块（允许不同长度）被分开至少有一个黑色方块。让fill-count函数F（m，n）表示可以填充行的方式的数量。例如，F（3,29）= 673135和F（3,30）= 1089155.也就是说，对于m = 3，可以看出n = 30是填充计数函数首次超过的最小值一百万。同样，对于m = 10，可以验证F（10,56）= 880711和F（10,57）= 1148904，因此n = 57是填充计数函数首次超过的最小值一百万。对于m = 50，找到填充计数函数首先超过一百万的n的最小值。
+NOTE: This is a more difficult version of Problem 114.
+
+A row measuring n units in length has red blocks with a minimum length of m units placed on it, such that any two red blocks (which are allowed to be different lengths) are separated by at least one black square.
+
+Let the fill-count function, F(m, n), represent the number of ways that a row can be filled.
+
+For example, F(3, 29) = 673135 and F(3, 30) = 1089155.
+
+That is, for m = 3, it can be seen that n = 30 is the smallest value for which the fill-count function first exceeds one million.
+
+In the same way, for m = 10, it can be verified that F(10, 56) = 880711 and F(10, 57) = 1148904, so n = 57 is the least value for which the fill-count function first exceeds one million.
+
+For m = 50, find the least value of n for which the fill-count function first exceeds one million.
# --hints--
-`euler115()`应返回168。
+`euler115()` should return 168.
```js
assert.strictEqual(euler115(), 168);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-116-red-green-or-blue-tiles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-116-red-green-or-blue-tiles.md
index b837f71e29e..1f61195dea8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-116-red-green-or-blue-tiles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-116-red-green-or-blue-tiles.md
@@ -1,24 +1,26 @@
---
id: 5900f3e01000cf542c50fef3
-title: 问题116：红色，绿色或蓝色瓷砖
+title: 'Problem 116: Red, green or blue tiles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301742
dashedName: problem-116-red-green-or-blue-tiles
---
# --description--
-一排五个黑色正方形瓷砖是用彩色椭圆形瓷砖代替的，其中红色（长度为2），绿色（长度为3）或蓝色（长度为4）。如果选择红色瓷砖，则可以通过七种方式完成此操作。
+A row of five black square tiles is to have a number of its tiles replaced with coloured oblong tiles chosen from red (length two), green (length three), or blue (length four).
-如果选择绿色瓷砖，有三种方法。
+If red tiles are chosen there are exactly seven ways this can be done.
-如果选择蓝色瓷砖，有两种方法。
+If green tiles are chosen there are three ways.
-假设颜色不能混合，则有7 + 3 + 2 = 12种方式替换长度为五个单位的行中的黑色瓷砖。如果不能混合颜色并且必须使用至少一个彩色瓷砖，那么连续测量五十个单位的黑色瓷砖有多少种不同的方式可以替换？注意：这与问题117有关。
+And if blue tiles are chosen there are two ways.
+
+Assuming that colours cannot be mixed there are 7 + 3 + 2 = 12 ways of replacing the black tiles in a row measuring five units in length. How many different ways can the black tiles in a row measuring fifty units in length be replaced if colours cannot be mixed and at least one coloured tile must be used? NOTE: This is related to Problem 117.
# --hints--
-`euler116()`应该返回20492570929。
+`euler116()` should return 20492570929.
```js
assert.strictEqual(euler116(), 20492570929);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-117-red-green-and-blue-tiles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-117-red-green-and-blue-tiles.md
index 5174b57e2dd..80f46d12016 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-117-red-green-and-blue-tiles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-117-red-green-and-blue-tiles.md
@@ -1,20 +1,20 @@
---
id: 5900f3e21000cf542c50fef4
-title: 问题117：红色，绿色和蓝色瓷砖
+title: 'Problem 117: Red, green, and blue tiles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301743
dashedName: problem-117-red-green-and-blue-tiles
---
# --description--
-使用黑色正方形瓷砖和椭圆形瓷砖的组合，选自：测量两个单位的红色瓷砖，测量三个单位的绿色瓷砖和测量四个单位的蓝色瓷砖，可以以十五种不同的方式平铺一个长度为五个单位的行。
+Using a combination of black square tiles and oblong tiles chosen from: red tiles measuring two units, green tiles measuring three units, and blue tiles measuring four units, it is possible to tile a row measuring five units in length in exactly fifteen different ways.
-一行测量五十个单位长度的平铺方式有多少？注意：这与问题116有关。
+How many ways can a row measuring fifty units in length be tiled? NOTE: This is related to Problem 116.
# --hints--
-`euler117()`应该返回100808458960497。
+`euler117()` should return 100808458960497.
```js
assert.strictEqual(euler117(), 100808458960497);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-118-pandigital-prime-sets.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-118-pandigital-prime-sets.md
index c1b9653cf7d..8bcd4b5dcd4 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-118-pandigital-prime-sets.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-118-pandigital-prime-sets.md
@@ -1,18 +1,20 @@
---
id: 5900f3e21000cf542c50fef5
-title: 问题118：Pandigital prime set
+title: 'Problem 118: Pandigital prime sets'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301744
dashedName: problem-118-pandigital-prime-sets
---
# --description--
-使用所有数字1到9并自由地连接它们以形成十进制整数，可以形成不同的集合。有趣的是，集合{2,5,47,89,631}，属于它的所有元素都是素数。包含每个数字1到9的多少个不同的集合只包含主要元素？
+Using all of the digits 1 through 9 and concatenating them freely to form decimal integers, different sets can be formed. Interestingly with the set {2,5,47,89,631}, all of the elements belonging to it are prime.
+
+How many distinct sets containing each of the digits one through nine exactly once contain only prime elements?
# --hints--
-`euler118()`应返回44680。
+`euler118()` should return 44680.
```js
assert.strictEqual(euler118(), 44680);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-119-digit-power-sum.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-119-digit-power-sum.md
index 8bd1d35bfe5..40853a13145 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-119-digit-power-sum.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-119-digit-power-sum.md
@@ -1,18 +1,24 @@
---
id: 5900f3e41000cf542c50fef6
-title: 问题119：数字功率总和
+title: 'Problem 119: Digit power sum'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301745
dashedName: problem-119-digit-power-sum
---
# --description--
-数字512很有意思，因为它等于它的数字加到某个幂的总和：5 + 1 + 2 = 8，而83 = 512.具有这个属性的数字的另一个例子是614656 = 284.我们将定义一个成为这个序列的第n个术语，并坚持一个数字必须包含至少两个数字才能得到一个总和。给出a2 = 512和a10 = 614656。找到a30。
+The number 512 is interesting because it is equal to the sum of its digits raised to some power: 5 + 1 + 2 = 8, and 83 = 512. Another example of a number with this property is 614656 = 284.
+
+We shall define an to be the nth term of this sequence and insist that a number must contain at least two digits to have a sum.
+
+You are given that a2 = 512 and a10 = 614656.
+
+Find a30.
# --hints--
-`euler119()`应该返回248155780267521。
+`euler119()` should return 248155780267521.
```js
assert.strictEqual(euler119(), 248155780267521);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-12-highly-divisible-triangular-number.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-12-highly-divisible-triangular-number.md
index ff05afe0cd8..b14276a81e7 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-12-highly-divisible-triangular-number.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-12-highly-divisible-triangular-number.md
@@ -1,62 +1,64 @@
---
id: 5900f3781000cf542c50fe8b
-title: 问题12：高度可分的三角数
+title: 'Problem 12: Highly divisible triangular number'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301746
dashedName: problem-12-highly-divisible-triangular-number
---
# --description--
-通过添加自然数生成三角数的序列。所以第7个三角形数字是1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.前十个术语是：
+The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:
-1,3,6,10,15,21,28,36,45,55 ......
+<div style='text-align: center;'>1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...</div>
-让我们列出前七个三角形数字的因子：
+Let us list the factors of the first seven triangle numbers:
-**1：** 1
+<div style='padding-left: 4em;'><b>1:</b> 1</div>
+<div style='padding-left: 4em;'><b>3:</b> 1, 3</div>
+<div style='padding-left: 4em;'><b>6:</b> 1, 2, 3, 6</div>
+<div style='padding-left: 4em;'><b>10:</b> 1, 2, 5, 10</div>
+<div style='padding-left: 4em;'><b>15:</b> 1, 3, 5, 15</div>
+<div style='padding-left: 4em;'><b>21:</b> 1, 3, 7, 21</div>
+<div style='padding-left: 4em;'><b>28:</b> 1, 2, 4, 7, 14, 28</div>
-**3：** 1,3
+We can see that 28 is the first triangle number to have over five divisors.
-**6：** 1,2,3,6
-
-**10：** 1,2,5,10
-
-**15：** 1,3,5,15
-
-**21：** 1,3,7,21
-
-**28：** 1,2,4,7,14,28
-
-我们可以看到28是第一个超过五个除数的三角形数。超过`n`除数的第一个三角形数的值是多少？
+What is the value of the first triangle number to have over `n` divisors?
# --hints--
-`divisibleTriangleNumber(5)`应该返回28。
+`divisibleTriangleNumber(5)` should return a number.
+
+```js
+assert(typeof divisibleTriangleNumber(5) === 'number');
+```
+
+`divisibleTriangleNumber(5)` should return 28.
```js
assert.strictEqual(divisibleTriangleNumber(5), 28);
```
-`divisibleTriangleNumber(23)`应该返回630。
+`divisibleTriangleNumber(23)` should return 630.
```js
assert.strictEqual(divisibleTriangleNumber(23), 630);
```
-divisibleTriangleNumber `divisibleTriangleNumber(167)`应该返回1385280。
+`divisibleTriangleNumber(167)` should return 1385280.
```js
assert.strictEqual(divisibleTriangleNumber(167), 1385280);
```
-divisibleTriangleNumber `divisibleTriangleNumber(374)`应该返回17907120。
+`divisibleTriangleNumber(374)` should return 17907120.
```js
assert.strictEqual(divisibleTriangleNumber(374), 17907120);
```
-divisibleTriangleNumber `divisibleTriangleNumber(500)`应该返回76576500。
+`divisibleTriangleNumber(500)` should return 76576500.
```js
assert.strictEqual(divisibleTriangleNumber(500), 76576500);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-120-square-remainders.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-120-square-remainders.md
index 498f306ef15..6bcf398e782 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-120-square-remainders.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-120-square-remainders.md
@@ -1,18 +1,22 @@
---
id: 5900f3e41000cf542c50fef7
-title: 问题120：方形剩余部分
+title: 'Problem 120: Square remainders'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301747
dashedName: problem-120-square-remainders
---
# --description--
-设r是（a-1）n +（a + 1）n除以a2时的余数。例如，如果a = 7且n = 3，则r = 42：63 + 83 =728≡42mod 49.并且当n变化时，r也将变化，但是对于a = 7，结果是rmax = 42。对于3≤a≤1000，找到Σrmax。
+Let r be the remainder when (a−1)n + (a+1)n is divided by a2.
+
+For example, if a = 7 and n = 3, then r = 42: 63 + 83 = 728 ≡ 42 mod 49. And as n varies, so too will r, but for a = 7 it turns out that rmax = 42.
+
+For 3 ≤ a ≤ 1000, find ∑ rmax.
# --hints--
-`euler120()`应返回333082500。
+`euler120()` should return 333082500.
```js
assert.strictEqual(euler120(), 333082500);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-121-disc-game-prize-fund.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-121-disc-game-prize-fund.md
index 2ad88f79420..1cc716c41ca 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-121-disc-game-prize-fund.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-121-disc-game-prize-fund.md
@@ -1,18 +1,24 @@
---
id: 5900f3e51000cf542c50fef8
-title: 问题121：光盘游戏奖金
+title: 'Problem 121: Disc game prize fund'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301748
dashedName: problem-121-disc-game-prize-fund
---
# --description--
-一个包包含一个红色圆盘和一个蓝色圆盘。在有机会的游戏中，玩家随机拍摄光盘并记录其颜色。每次转动后，光盘返回到包中，添加一个额外的红色光盘，随机拍摄另一张光盘。如果玩家在游戏结束时拍摄的红色光盘多于红色光盘，则玩家可以支付1英镑玩游戏并获胜。如果游戏进行了四轮比赛，那么玩家获胜的概率恰好是11/120，因此银行家在这场比赛中应该分配的最高奖金将是10英镑，然后才会产生损失。请注意，任何支付都将是一个整数磅，并且还包括为玩游戏而支付的原始£1，因此在给出的示例中，玩家实际上赢得了9英镑。找到应该分配给单个游戏的最大奖金，其中玩15个回合。
+A bag contains one red disc and one blue disc. In a game of chance a player takes a disc at random and its colour is noted. After each turn the disc is returned to the bag, an extra red disc is added, and another disc is taken at random.
+
+The player pays £1 to play and wins if they have taken more blue discs than red discs at the end of the game.
+
+If the game is played for four turns, the probability of a player winning is exactly 11/120, and so the maximum prize fund the banker should allocate for winning in this game would be £10 before they would expect to incur a loss. Note that any payout will be a whole number of pounds and also includes the original £1 paid to play the game, so in the example given the player actually wins £9.
+
+Find the maximum prize fund that should be allocated to a single game in which fifteen turns are played.
# --hints--
-`euler121()`应返回2269。
+`euler121()` should return 2269.
```js
assert.strictEqual(euler121(), 2269);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-122-efficient-exponentiation.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-122-efficient-exponentiation.md
index 4c2f63d409c..7da1b4274e3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-122-efficient-exponentiation.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-122-efficient-exponentiation.md
@@ -1,18 +1,32 @@
---
id: 5900f3e61000cf542c50fef9
-title: 问题122：有效取幂
+title: 'Problem 122: Efficient exponentiation'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301749
dashedName: problem-122-efficient-exponentiation
---
# --description--
-最简单的计算n15的方法需要十四次乘法：n×n×...×n = n15但是使用“二进制”方法可以在六次乘法中计算它：n×n = n2n2×n2 = n4n4×n4 = n8n8 ×n4 = n12n12×n2 = n14n14×n = n15然而，只能在五次乘法中计算它：n×n = n2n2×n = n3n3×n3 = n6n6×n6 = n12n12×n3 = n15我们将定义m （k）是计算nk的最小乘法数;例如m（15）= 5.对于1≤k≤200，找到Σm（k）。
+The most naive way of computing n15 requires fourteen multiplications:
+
+n × n × ... × n = n15
+
+But using a "binary" method you can compute it in six multiplications:
+
+n × n = n2n2 × n2 = n4n4 × n4 = n8n8 × n4 = n12n12 × n2 = n14n14 × n = n15
+
+However it is yet possible to compute it in only five multiplications:
+
+n × n = n2n2 × n = n3n3 × n3 = n6n6 × n6 = n12n12 × n3 = n15
+
+We shall define m(k) to be the minimum number of multiplications to compute nk; for example m(15) = 5.
+
+For 1 ≤ k ≤ 200, find ∑ m(k).
# --hints--
-`euler122()`应返回1582。
+`euler122()` should return 1582.
```js
assert.strictEqual(euler122(), 1582);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-123-prime-square-remainders.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-123-prime-square-remainders.md
index 6ff5a1277c3..cb7b7afda10 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-123-prime-square-remainders.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-123-prime-square-remainders.md
@@ -1,18 +1,24 @@
---
id: 5900f3e71000cf542c50fefa
-title: 问题123：素数正方形余数
+title: 'Problem 123: Prime square remainders'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301750
dashedName: problem-123-prime-square-remainders
---
# --description--
-令pn为第n个素数：2,3,5,7,11 ......，并且当r（pn-1）n +（pn + 1）n除以pn2时，令r为余数。例如，当n = 3时，p3 = 5，并且43 + 63 =280≡5mod 25.余数首先超过109的n的最小值是7037.求出余数首次超过的n的最小值1010。
+Let pn be the nth prime: 2, 3, 5, 7, 11, ..., and let r be the remainder when (pn−1)n + (pn+1)n is divided by pn2.
+
+For example, when n = 3, p3 = 5, and 43 + 63 = 280 ≡ 5 mod 25.
+
+The least value of n for which the remainder first exceeds 109 is 7037.
+
+Find the least value of n for which the remainder first exceeds 1010.
# --hints--
-`euler123()`应该返回21035。
+`euler123()` should return 21035.
```js
assert.strictEqual(euler123(), 21035);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-124-ordered-radicals.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-124-ordered-radicals.md
index ea4981cf727..1dfc34dc9b3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-124-ordered-radicals.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-124-ordered-radicals.md
@@ -1,18 +1,22 @@
---
id: 5900f3e81000cf542c50fefb
-title: 问题124：有序的激进分子
+title: 'Problem 124: Ordered radicals'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301751
dashedName: problem-124-ordered-radicals
---
# --description--
-n，rad（n）的基数是n的不同素因子的乘积。例如，504 = 23×32×7，因此rad（504）= 2×3×7 = 42.如果我们计算1≤n≤10的rad（n），则在rad（n）上对它们进行排序，并进行排序如果激进值相等，我们得到：未分类
+The radical of n, rad(n), is the product of the distinct prime factors of n. For example, 504 = 23 × 32 × 7, so rad(504) = 2 × 3 × 7 = 42.
-排序n rad（n）
+If we calculate rad(n) for 1 ≤ n ≤ 10, then sort them on rad(n), and sorting on n if the radical values are equal, we get:
-n rad（n）k 11
+Unsorted
+
+Sorted n rad(n)
+
+n rad(n) k 11
111 22
@@ -32,11 +36,11 @@ n rad（n）k 11
779 1010
-101010令E（k）为有序n列中的第k个元素;例如，E（4）= 8且E（6）= 9.如果rad（n）按1≤n≤100000排序，则找到E（10000）。
+101010 Let E(k) be the kth element in the sorted n column; for example, E(4) = 8 and E(6) = 9. If rad(n) is sorted for 1 ≤ n ≤ 100000, find E(10000).
# --hints--
-`euler124()`应返回21417。
+`euler124()` should return 21417.
```js
assert.strictEqual(euler124(), 21417);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-125-palindromic-sums.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-125-palindromic-sums.md
index b4cdf648fc5..460f5e2a237 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-125-palindromic-sums.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-125-palindromic-sums.md
@@ -1,18 +1,22 @@
---
id: 5900f3e91000cf542c50fefc
-title: 问题125：回文总和
+title: 'Problem 125: Palindromic sums'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301752
dashedName: problem-125-palindromic-sums
---
# --description--
-回文数595很有意思，因为它可以写成连续正方形的总和：62 + 72 + 82 + 92 + 102 + 112 + 122.正好有11个以下的回文可以写成连续的平方和，并且这些回文的总和是4164.注意，没有包括1 = 02 + 12，因为该问题涉及正整数的平方。找到小于108的所有数字的总和，这些数字都是回文并且可以写为连续正方形的总和。
+The palindromic number 595 is interesting because it can be written as the sum of consecutive squares: 62 + 72 + 82 + 92 + 102 + 112 + 122.
+
+There are exactly eleven palindromes below one-thousand that can be written as consecutive square sums, and the sum of these palindromes is 4164. Note that 1 = 02 + 12 has not been included as this problem is concerned with the squares of positive integers.
+
+Find the sum of all the numbers less than 108 that are both palindromic and can be written as the sum of consecutive squares.
# --hints--
-`euler125()`应该返回2906969179。
+`euler125()` should return 2906969179.
```js
assert.strictEqual(euler125(), 2906969179);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-126-cuboid-layers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-126-cuboid-layers.md
index edbfced8aaf..fe9b4a6aa80 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-126-cuboid-layers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-126-cuboid-layers.md
@@ -1,20 +1,20 @@
---
id: 5900f3ea1000cf542c50fefd
-title: 问题126：长方体层
+title: 'Problem 126: Cuboid layers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301753
dashedName: problem-126-cuboid-layers
---
# --description--
-覆盖尺寸为3 x 2 x 1的长方体上每个可见面的最小立方体数量为22。
+The minimum number of cubes to cover every visible face on a cuboid measuring 3 x 2 x 1 is twenty-two.
-如果我们在这个固体上添加第二层，则需要四十六个立方体来覆盖每个可见面，第三层需要七十八个立方体，第四层需要一百一十八个立方体来覆盖每个可见面。然而，尺寸为5 x 1 x 1的长方体上的第一层也需要22个立方体;类似地，尺寸为5 x 3 x 1,7 x 2 x 1和11 x 1 x 1的长方体上的第一层都包含四十六个立方体。我们将定义C（n）来表示在其一个层中包含n个立方体的长方体的数量。因此，C（22）= 2，C（46）= 4，C（78）= 5，并且C（118）= 8.结果，154是n的最小值，其中C（n）= 10。找到n的最小值，其中C（n）= 1000。
+If we then add a second layer to this solid it would require forty-six cubes to cover every visible face, the third layer would require seventy-eight cubes, and the fourth layer would require one-hundred and eighteen cubes to cover every visible face. However, the first layer on a cuboid measuring 5 x 1 x 1 also requires twenty-two cubes; similarly the first layer on cuboids measuring 5 x 3 x 1, 7 x 2 x 1, and 11 x 1 x 1 all contain forty-six cubes. We shall define C(n) to represent the number of cuboids that contain n cubes in one of its layers. So C(22) = 2, C(46) = 4, C(78) = 5, and C(118) = 8. It turns out that 154 is the least value of n for which C(n) = 10. Find the least value of n for which C(n) = 1000.
# --hints--
-`euler126()`应返回18522。
+`euler126()` should return 18522.
```js
assert.strictEqual(euler126(), 18522);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-127-abc-hits.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-127-abc-hits.md
index f83bd37c5ad..665e17c0a74 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-127-abc-hits.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-127-abc-hits.md
@@ -1,18 +1,42 @@
---
id: 5900f3ec1000cf542c50fefe
-title: 问题127：abc-hits
+title: 'Problem 127: abc-hits'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301754
dashedName: problem-127-abc-hits
---
# --description--
-n，rad（n）的基数是n的不同素因子的乘积。例如，504 = 23×32×7，因此rad（504）= 2×3×7 = 42.如果出现以下情况，我们将正整数（a，b，c）的三元组定义为abc-hit：GCD（ a，b）= GCD（a，c）= GCD（b，c）= 1 a <ba + b = c rad（abc）<c例如，（5,27,32）是abc-hit，因为：GCD（5,27）= GCD（5,32）= GCD（27,32）= 1 5 <27 5 + 27 = 32 rad（4320）= 30 <32事实证明abc-hits是非常罕见的c <1000只有31次abc命中，Σc= 12523。查找Σc表示c <120000。
+The radical of n, rad(n), is the product of distinct prime factors of n. For example, 504 = 23 × 32 × 7, so rad(504) = 2 × 3 × 7 = 42.
+
+We shall define the triplet of positive integers (a, b, c) to be an abc-hit if:
+
+GCD(a, b) = GCD(a, c) = GCD(b, c) = 1
+
+a < b
+
+a + b = c
+
+rad(abc) < c
+
+For example, (5, 27, 32) is an abc-hit, because:
+
+GCD(5, 27) = GCD(5, 32) = GCD(27, 32) = 1
+
+5 < 27
+
+5 + 27 = 32
+
+rad(4320) = 30 < 32
+
+It turns out that abc-hits are quite rare and there are only thirty-one abc-hits for c < 1000, with ∑c = 12523.
+
+Find ∑c for c < 120000.
# --hints--
-`euler127()`应该返回18407904。
+`euler127()` should return 18407904.
```js
assert.strictEqual(euler127(), 18407904);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-128-hexagonal-tile-differences.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-128-hexagonal-tile-differences.md
index bbac86239e0..7bdd6242348 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-128-hexagonal-tile-differences.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-128-hexagonal-tile-differences.md
@@ -1,20 +1,22 @@
---
id: 5900f3ec1000cf542c50feff
-title: 问题128：六边形瓷砖差异
+title: 'Problem 128: Hexagonal tile differences'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301755
dashedName: problem-128-hexagonal-tile-differences
---
# --description--
-编号为1的六边形瓷砖由六个六边形瓷砖环围绕，从“12点钟”开始，并以逆时针方向对瓷砖2至7进行编号。新环以相同的方式添加，下一个环编号为8至19,20至37,38至61，依此类推。下图显示了前三个环。
+A hexagonal tile with number 1 is surrounded by a ring of six hexagonal tiles, starting at "12 o'clock" and numbering the tiles 2 to 7 in an anti-clockwise direction.
-通过找到tile n与其六个邻居中的每一个之间的差异，我们将PD（n）定义为那些作为素数的差异的数量。例如，在瓦片8周围顺时针工作，差异是12,29,11,6,1和13.因此PD（8）= 3.以相同的方式，瓦片17周围的差异是1,17,16,1 ，11和10，因此PD（17）= 2.可以证明PD（n）的最大值是3.如果PD（n）= 3的所有瓦片按升序列出以形成一个序列，第10个瓦片将是271.按此顺序找到第2000个瓦片。
+New rings are added in the same fashion, with the next rings being numbered 8 to 19, 20 to 37, 38 to 61, and so on. The diagram below shows the first three rings.
+
+By finding the difference between tile n and each of its six neighbours we shall define PD(n) to be the number of those differences which are prime. For example, working clockwise around tile 8 the differences are 12, 29, 11, 6, 1, and 13. So PD(8) = 3. In the same way, the differences around tile 17 are 1, 17, 16, 1, 11, and 10, hence PD(17) = 2. It can be shown that the maximum value of PD(n) is 3. If all of the tiles for which PD(n) = 3 are listed in ascending order to form a sequence, the 10th tile would be 271. Find the 2000th tile in this sequence.
# --hints--
-`euler128()`应该返回14516824220。
+`euler128()` should return 14516824220.
```js
assert.strictEqual(euler128(), 14516824220);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-129-repunit-divisibility.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-129-repunit-divisibility.md
index 8694a9e131e..0d8f6c2916c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-129-repunit-divisibility.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-129-repunit-divisibility.md
@@ -1,18 +1,24 @@
---
id: 5900f3ef1000cf542c50ff01
-title: 问题129：重新划分可分性
+title: 'Problem 129: Repunit divisibility'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301756
dashedName: problem-129-repunit-divisibility
---
# --description--
-完全由1组成的数字称为repunit。我们将R（k）定义为长度k的重新定位;例如，R（6）= 111111.假设n是正整数且GCD（n，10）= 1，则可以证明总是存在一个值k，其中R（k）可被n整除让A（n）成为k的最小值;例如，A（7）= 6且A（41）= 5.A（n）首先超过10的n的最小值是17.求出A（n）首先超过1的n的最小值 - 百万。
+A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.
+
+Given that n is a positive integer and GCD(n, 10) = 1, it can be shown that there always exists a value, k, for which R(k) is divisible by n, and let A(n) be the least such value of k; for example, A(7) = 6 and A(41) = 5.
+
+The least value of n for which A(n) first exceeds ten is 17.
+
+Find the least value of n for which A(n) first exceeds one-million.
# --hints--
-`euler129()`应该返回1000023。
+`euler129()` should return 1000023.
```js
assert.strictEqual(euler129(), 1000023);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-13-large-sum.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-13-large-sum.md
index fa2dfc9a717..05775850cff 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-13-large-sum.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-13-large-sum.md
@@ -1,26 +1,133 @@
---
id: 5900f37a1000cf542c50fe8c
-title: 问题13：大笔金额
+title: 'Problem 13: Large sum'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301757
dashedName: problem-13-large-sum
---
# --description--
-计算出以下一百个50位数字之和的前十位数。
+Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.
-37107287533902102798797998220837590246510135740250 46376937677490009712648124896970078050417018260538 74324986199524741059474233309513058123726617309629 91942213363574161572522430563301811072406154908250 23067588207539346171171980310421047513778063246676 89261670696623633820136378418383684178734361726757 28112879812849979408065481931592621691275889832738 44274228917432520321923589422876796487670272189318 47451445736001306439091167216856844588711603153276 70386486105843025439939619828917593665686757934951 62176457141856560629502157223196586755079324193331 64906352462741904929101432445813822663347944758178 92575867718337217661963751590579239728245598838407 58203565325359399008402633568948830189458628227828 80181199384826282014278194139940567587151170094390 35398664372827112653829987240784473053190104293586 86515506006295864861532075273371959191420517255829 71693888707715466499115593487603532921714970056938 54370070576826684624621495650076471787294438377604 5328265410875682844319119063469 4037855217779295145 36123272525000296071075082563815656710885258350721 45876576172410976447339110607218265236877223636045 17423706905851860660448207621209813287860733969412 81142660418086830619328460811191061556940512689692 51934325451728388641918047049293215058642563049483 62467221648435076201727918039944693004732956340691 15732444386908125794514089057706229429197107928209 55037687525678773091862540744969844508330393682126 18336384825330154686196124348767681297534375946515 80386287592878490201521685554828717201219257766954 78182833757993103614740356856449095527097864797581 16726320100436897842553539920931837441497806860984 48403098129077791799088218795327364475675590848030 87086987551392711854517078544161852424320693150332 59959406895756536782107074926966537676326235447210 69793950679652694742597709739166693763042633987085 41052684708299085211399427365734116182760315001271 65378607361501080857009149939512557028198746004375 35829035317434717326932123578154982629742552737307 94953759765 105305946966067683156574377167401875275 88902802571733229619176668713819931811048770190271 25267680276078003013678680992525463401061632866526 36270218540497705585629946580636237993140746255962 24074486908231174977792365466257246923322810917141 91430288197103288597806669760892938638285025333403 34413065578016127815921815005561868836468420090470 23053081172816430487623791969842487255036638784583 11487696932154902810424020138335124462181441773470 63783299490636259666498587618221225225512486764533 67720186971698544312419572409913959008952310058822 95548255300263520781532296796249481641953868218774 76085327132285723110424803456124867697064507995236 37774242535411291684276865538926205024910326572967 23701913275725675285653248258265463092207058596522 29798860272258331913126375147341994889534765745501 18495701454879288984856827726077713721403798879715 38298203783031473527721580348144513491373226651381 34829543829199918180278916522431027392251122869539 409579530664052326325380441000596549391598 79593635 29746152185502371307642255121183693803580388584903 41698116222072977186158236678424689157993532961922 62467957194401269043877107275048102390895523597457 23189706772547915061505504953922979530901129967519 86188088225875314529584099251203829009407770775672 11306739708304724483816533873502340845647058077308 82959174767140363198008187129011875491310547126581 97623331044818386269515456334926366572897563400500 42846280183517070527831839425882145521227251250327 55121603546981200581762165212827652751691296897789 32238195734329339946437501907836945765883352399886 75506164965184775180738168837861091527357929701337 62177842752192623401942399639168044983993173312731 32924185707147349566916674687634660915035914677504 99518671430235219628894890102423325116913619626622 73267460800591547471830798392868535206946944540724 76841822524674417161514036427982273348055556214818 97142617910342598647204516893989422179826088076852 87783646182799346313767754307809363333018982642090 1084880252167467088321 5120185883543223812876952786 71329612474782464538636993009049310363619763878039 62184073572399794223406235393808339651327408011116 66627891981488087797941876876144230030984490851411 60661826293682836764744779239180335110989069790714 85786944089552990653640447425576083659976645795096 66024396409905389607120198219976047599490197230297 64913982680032973156037120041377903785566085089252 16730939319872750275468906903707539413042652315011 94809377245048795150954100921645863754710598436791 78639167021187492431995700641917969777599028300699 15368713711936614952811305876380278410754449733078 40789923115535562561142322423255033685442488917353 44889911501440648020369068063960672322193204149535 41503128880339536053299340368006977710650566631954 81234880673210146739058568557934581403627822703280 82616570773948327592232845941706525094512325230608 22918802058777319719839450180888072429661980811197 77158542502016545090413245809786882778948721859617 72107838435069186155435662884062257473692284509516 20 849603980134001723930671666823555245252804609722 53503534226472524250874054075591789781264330331690
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+</div>
# --hints--
-`largeSum(testNums)`应该返回8348422521。
+`largeSum(testNums)` should return a number.
+
+```js
+assert(typeof largeSum(testNums) === 'number');
+```
+
+`largeSum(testNums)` should return 8348422521.
```js
assert.strictEqual(largeSum(testNums), 8348422521);
```
-`largeSum(fiftyDigitNums)`应返回5537376230。
+`largeSum(fiftyDigitNums)` should return 5537376230.
```js
assert.strictEqual(largeSum(fiftyDigitNums), 5537376230);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-130-composites-with-prime-repunit-property.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-130-composites-with-prime-repunit-property.md
index 1f64300ef36..0f3c2c7a3cc 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-130-composites-with-prime-repunit-property.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-130-composites-with-prime-repunit-property.md
@@ -1,18 +1,26 @@
---
id: 5900f3ee1000cf542c50ff00
-title: 问题130：具有主要repunit属性的复合材料
+title: 'Problem 130: Composites with prime repunit property'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301758
dashedName: problem-130-composites-with-prime-repunit-property
---
# --description--
-完全由1组成的数字称为repunit。我们将R（k）定义为长度k的重新定位;例如，R（6）= 111111.假设n是正整数且GCD（n，10）= 1，则可以证明总是存在一个值k，其中R（k）可被n整除让A（n）成为k的最小值;例如，A（7）= 6和A（41）= 5.对于所有素数，p> 5，p-1可以被A（p）整除。例如，当p = 41时，A（41）= 5，并且40可被5整除。但是，也有罕见的复合值，这也是正确的;前五个例子是91,259,451,481和703.找到n的前25个复合值之和，其中GCD（n，10）= 1，n-1可被A（n）整除。
+A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.
+
+Given that n is a positive integer and GCD(n, 10) = 1, it can be shown that there always exists a value, k, for which R(k) is divisible by n, and let A(n) be the least such value of k; for example, A(7) = 6 and A(41) = 5.
+
+You are given that for all primes, p > 5, that p − 1 is divisible by A(p). For example, when p = 41, A(41) = 5, and 40 is divisible by 5.
+
+However, there are rare composite values for which this is also true; the first five examples being 91, 259, 451, 481, and 703.
+
+Find the sum of the first twenty-five composite values of n for whichGCD(n, 10) = 1 and n − 1 is divisible by A(n).
# --hints--
-`euler130()`应返回149253。
+`euler130()` should return 149253.
```js
assert.strictEqual(euler130(), 149253);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-131-prime-cube-partnership.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-131-prime-cube-partnership.md
index a60228c6838..729f8689fba 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-131-prime-cube-partnership.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-131-prime-cube-partnership.md
@@ -1,18 +1,24 @@
---
id: 5900f3ef1000cf542c50ff02
-title: 问题131：Prime立方体伙伴关系
+title: 'Problem 131: Prime cube partnership'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301759
dashedName: problem-131-prime-cube-partnership
---
# --description--
-存在一些素数值p，其中存在正整数n，使得表达式n3 + n2p是完美的立方体。例如，当p = 19时，83 + 82×19 = 123.最令人惊讶的是，对于具有此属性的每个素数，n的值是唯一的，并且在100之下只有四个这样的素数。一百万以下的素数有多少这个非凡的财产？
+There are some prime values, p, for which there exists a positive integer, n, such that the expression n3 + n2p is a perfect cube.
+
+For example, when p = 19, 83 + 82×19 = 123.
+
+What is perhaps most surprising is that for each prime with this property the value of n is unique, and there are only four such primes below one-hundred.
+
+How many primes below one million have this remarkable property?
# --hints--
-`euler131()`应该返回173。
+`euler131()` should return 173.
```js
assert.strictEqual(euler131(), 173);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-132-large-repunit-factors.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-132-large-repunit-factors.md
index 8a641ae102a..d71927659ba 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-132-large-repunit-factors.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-132-large-repunit-factors.md
@@ -1,18 +1,22 @@
---
id: 5900f3f11000cf542c50ff03
-title: 问题132：大的重新安置因素
+title: 'Problem 132: Large repunit factors'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301760
dashedName: problem-132-large-repunit-factors
---
# --description--
-完全由1组成的数字称为repunit。我们将R（k）定义为长度k的重新定位。例如，R（10）= 1111111111 = 11×41×271×9091，并且这些素因子的总和是9414.求出R（109）的前40个素因子的总和。
+A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k.
+
+For example, R(10) = 1111111111 = 11×41×271×9091, and the sum of these prime factors is 9414.
+
+Find the sum of the first forty prime factors of R(109).
# --hints--
-`euler132()`应返回843296。
+`euler132()` should return 843296.
```js
assert.strictEqual(euler132(), 843296);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-133-repunit-nonfactors.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-133-repunit-nonfactors.md
index 5d17a655aae..649dd7f1f33 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-133-repunit-nonfactors.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-133-repunit-nonfactors.md
@@ -1,18 +1,24 @@
---
id: 5900f3f21000cf542c50ff04
-title: 问题133：重新计算非因素
+title: 'Problem 133: Repunit nonfactors'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301761
dashedName: problem-133-repunit-nonfactors
---
# --description--
-完全由1组成的数字称为repunit。我们将R（k）定义为长度k的重新定位;例如，R（6）= 111111.让我们考虑R（10n）形式的重新组合。虽然R（10），R（100）或R（1000）不能被17整除，但R（10000）可被17整除。但是没有n的值，R（10n）将除以19。事实上，值得注意的是，11,17,41和73是低于100的唯一四个素数，可以是R（10n）的因子。求出十万以下所有素数的总和，这些素数永远不会是R（10n）的因子。
+A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.
+
+Let us consider repunits of the form R(10n).
+
+Although R(10), R(100), or R(1000) are not divisible by 17, R(10000) is divisible by 17. Yet there is no value of n for which R(10n) will divide by 19. In fact, it is remarkable that 11, 17, 41, and 73 are the only four primes below one-hundred that can be a factor of R(10n).
+
+Find the sum of all the primes below one-hundred thousand that will never be a factor of R(10n).
# --hints--
-`euler133()`应返回453647705。
+`euler133()` should return 453647705.
```js
assert.strictEqual(euler133(), 453647705);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-134-prime-pair-connection.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-134-prime-pair-connection.md
index cb8930908b6..a21b71ea01a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-134-prime-pair-connection.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-134-prime-pair-connection.md
@@ -1,18 +1,22 @@
---
id: 5900f3f21000cf542c50ff05
-title: 问题134：素对对连接
+title: 'Problem 134: Prime pair connection'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301762
dashedName: problem-134-prime-pair-connection
---
# --description--
-考虑连续的素数p1 = 19和p2 = 23.可以证实1219是最小的数字，使得最后的数字由p1形成，同时也可以被p2整除。实际上，除了p1 = 3和p2 = 5之外，对于每对连续质数，p2> p1，存在n的值，其中最后的数字由p1形成，n可以被p2整除。设S是n的这些值中最小的。找到每对连续质数的ΣS，其中5≤p1≤1000000。
+Consider the consecutive primes p1 = 19 and p2 = 23. It can be verified that 1219 is the smallest number such that the last digits are formed by p1 whilst also being divisible by p2.
+
+In fact, with the exception of p1 = 3 and p2 = 5, for every pair of consecutive primes, p2 > p1, there exist values of n for which the last digits are formed by p1 and n is divisible by p2. Let S be the smallest of these values of n.
+
+Find ∑ S for every pair of consecutive primes with 5 ≤ p1 ≤ 1000000.
# --hints--
-`euler134()`应该返回18613426663617120。
+`euler134()` should return 18613426663617120.
```js
assert.strictEqual(euler134(), 18613426663617120);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-135-same-differences.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-135-same-differences.md
index de7d846d71d..f0de3b5c589 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-135-same-differences.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-135-same-differences.md
@@ -1,18 +1,24 @@
---
id: 5900f3f31000cf542c50ff06
-title: 问题135：同样的差异
+title: 'Problem 135: Same differences'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301763
dashedName: problem-135-same-differences
---
# --description--
-给定正整数，x，y和z是算术级数的连续项，正整数n的最小值，其中等式x2-y2-z2 = n，恰好有两个解是n = 27：342 - 272 - 202 = 122 - 92 - 62 = 27事实证明，n = 1155是具有正好十个解的最小值。 n不到一百万的多少个值有十个不同的解？
+Given the positive integers, x, y, and z, are consecutive terms of an arithmetic progression, the least value of the positive integer, n, for which the equation, x2 − y2 − z2 = n, has exactly two solutions is n = 27:
+
+342 − 272 − 202 = 122 − 92 − 62 = 27
+
+It turns out that n = 1155 is the least value which has exactly ten solutions.
+
+How many values of n less than one million have exactly ten distinct solutions?
# --hints--
-`euler135()`应该返回4989。
+`euler135()` should return 4989.
```js
assert.strictEqual(euler135(), 4989);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-136-singleton-difference.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-136-singleton-difference.md
index 14b7a06d35d..34fda8a4eeb 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-136-singleton-difference.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-136-singleton-difference.md
@@ -1,18 +1,24 @@
---
id: 5900f3f51000cf542c50ff07
-title: 问题136：单身人士差异
+title: 'Problem 136: Singleton difference'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301764
dashedName: problem-136-singleton-difference
---
# --description--
-正整数x，y和z是算术级数的连续项。假设n是一个正整数，当n = 20时，方程x2 - y2 - z2 = n恰好有一个解：132 - 102 - 72 = 20实际上有二十五个n低于一百的值，其中方程有一个独特的解决方案。 n小于五千万的有多少个值只有一个解？
+The positive integers, x, y, and z, are consecutive terms of an arithmetic progression. Given that n is a positive integer, the equation, x2 − y2 − z2 = n, has exactly one solution when n = 20:
+
+132 − 102 − 72 = 20
+
+In fact there are twenty-five values of n below one hundred for which the equation has a unique solution.
+
+How many values of n less than fifty million have exactly one solution?
# --hints--
-`euler136()`应返回2544559。
+`euler136()` should return 2544559.
```js
assert.strictEqual(euler136(), 2544559);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-137-fibonacci-golden-nuggets.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-137-fibonacci-golden-nuggets.md
index 5f1dce4dd0c..2863f66aa1a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-137-fibonacci-golden-nuggets.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-137-fibonacci-golden-nuggets.md
@@ -1,26 +1,34 @@
---
id: 5900f3f51000cf542c50ff08
-title: 问题137：斐波那契金块
+title: 'Problem 137: Fibonacci golden nuggets'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301765
dashedName: problem-137-fibonacci-golden-nuggets
---
# --description--
-考虑无穷多项式系列AF（x）= xF1 + x2F2 + x3F3 + ...，其中Fk是斐波纳契数列中的第k项：1,1,2,3,5,8，...;也就是说，Fk = Fk-1 + Fk-2，F1 = 1且F2 = 1.对于这个问题，我们将对x的值感兴趣，其中AF（x）是正整数。令人惊讶的是AF（1/2）=（1/2）.1 +（1/2）2.1 +（1/2）3.2 +（1/2）4.3 +（1/2）5.5 + ......
+Consider the infinite polynomial series AF(x) = xF1 + x2F2 + x3F3 + ..., where Fk is the kth term in the Fibonacci sequence: 1, 1, 2, 3, 5, 8, ... ; that is, Fk = Fk−1 + Fk−2, F1 = 1 and F2 = 1.
-= 1/2 + 1/4 + 2/8 + 3/16 + 5/32 + ......
+For this problem we shall be interested in values of x for which AF(x) is a positive integer.
-= 2前五个自然数的x的相应值如下所示。
+Surprisingly AF(1/2)
-xAF（x）√2-111/ 22（√13-2）/ 33（√89-5）/ 84（√34-3）/ 55
+=
-如果x是理性的，我们将AF（x）称为金块，因为它们变得越来越稀少;例如，第10个金块是74049690.找到第15个金块。
+(1/2).1 + (1/2)2.1 + (1/2)3.2 + (1/2)4.3 + (1/2)5.5 + ...
+
+= 1/2 + 1/4 + 2/8 + 3/16 + 5/32 + ...
+
+= 2 The corresponding values of x for the first five natural numbers are shown below.
+
+xAF(x) √2−11 1/22 (√13−2)/33 (√89−5)/84 (√34−3)/55
+
+We shall call AF(x) a golden nugget if x is rational, because they become increasingly rarer; for example, the 10th golden nugget is 74049690. Find the 15th golden nugget.
# --hints--
-`euler137()`应该返回1120149658760。
+`euler137()` should return 1120149658760.
```js
assert.strictEqual(euler137(), 1120149658760);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-138-special-isosceles-triangles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-138-special-isosceles-triangles.md
index ef272bef15b..5675e1f938a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-138-special-isosceles-triangles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-138-special-isosceles-triangles.md
@@ -1,20 +1,20 @@
---
id: 5900f3f61000cf542c50ff09
-title: 问题138：特殊的等腰三角形
+title: 'Problem 138: Special isosceles triangles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301766
dashedName: problem-138-special-isosceles-triangles
---
# --description--
-考虑具有基本长度，b = 16和腿，L = 17的等腰三角形。
+Consider the isosceles triangle with base length, b = 16, and legs, L = 17.
-通过使用毕达哥拉斯定理，可以看出三角形的高度h =√（172-82）= 15，比基本长度小1。当b = 272且L = 305时，我们得到h = 273，这比基本长度多一个，这是第二个最小的等腰三角形，具有h = b±1的性质。找到12个最小等腰的ΣL h = b±1且b，L为正整数的三角形。
+By using the Pythagorean theorem it can be seen that the height of the triangle, h = √(172 − 82) = 15, which is one less than the base length. With b = 272 and L = 305, we get h = 273, which is one more than the base length, and this is the second smallest isosceles triangle with the property that h = b ± 1. Find ∑ L for the twelve smallest isosceles triangles for which h = b ± 1 and b, L are positive integers.
# --hints--
-`euler138()`应该返回1118049290473932。
+`euler138()` should return 1118049290473932.
```js
assert.strictEqual(euler138(), 1118049290473932);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-139-pythagorean-tiles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-139-pythagorean-tiles.md
index f769895f46b..800c0ccbab2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-139-pythagorean-tiles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-139-pythagorean-tiles.md
@@ -1,20 +1,22 @@
---
id: 5900f3f71000cf542c50ff0a
-title: 问题139：毕达哥拉斯瓷砖
+title: 'Problem 139: Pythagorean tiles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301767
dashedName: problem-139-pythagorean-tiles
---
# --description--
-设（a，b，c）表示具有整数长边的直角三角形的三个边。可以将四个这样的三角形放在一起以形成长度为c的正方形。例如，（3,4,5）三角形可以放在一起形成一个5乘5的正方形，中间有一个1个洞，可以看到5乘5的正方形可以用二十五个平铺1个方格。
+Let (a, b, c) represent the three sides of a right angle triangle with integral length sides. It is possible to place four such triangles together to form a square with length c.
-但是，如果使用（5,12,13）三角形，则孔将按7乘7测量，并且这些不能用于平铺13乘13平方。鉴于直角三角形的周长小于一亿，有多少毕达哥拉斯三角形允许这样的平铺？
+For example, (3, 4, 5) triangles can be placed together to form a 5 by 5 square with a 1 by 1 hole in the middle and it can be seen that the 5 by 5 square can be tiled with twenty-five 1 by 1 squares.
+
+However, if (5, 12, 13) triangles were used then the hole would measure 7 by 7 and these could not be used to tile the 13 by 13 square. Given that the perimeter of the right triangle is less than one-hundred million, how many Pythagorean triangles would allow such a tiling to take place?
# --hints--
-`euler139()`应该返回10057761。
+`euler139()` should return 10057761.
```js
assert.strictEqual(euler139(), 10057761);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-14-longest-collatz-sequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-14-longest-collatz-sequence.md
index d3d159983d9..807f362857a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-14-longest-collatz-sequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-14-longest-collatz-sequence.md
@@ -1,57 +1,73 @@
---
id: 5900f37a1000cf542c50fe8d
-title: 问题14：最长的Collatz序列
+title: 'Problem 14: Longest Collatz sequence'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301768
dashedName: problem-14-longest-collatz-sequence
---
# --description--
-为正整数集定义以下迭代序列：
+The following iterative sequence is defined for the set of positive integers:
-`n` → `n` / 2（ `n`是偶数）
+<div style='padding-left: 4em;'><var>n</var> → <var>n</var>/2 (<var>n</var> is even)</div>
-`n` →3 `n` + 1（ `n`为奇数）
+<div style='padding-left: 4em;'><var>n</var> → 3<var>n</var> + 1 (<var>n</var> is odd)</div>
-使用上面的规则并从13开始，我们生成以下序列：
+Using the rule above and starting with 13, we generate the following sequence:
-13→40→20→10→5→16→8→4→2→1
+<div style='text-align: center;'>13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1</div>
-可以看出，该序列（从13开始并在1结束）包含10个项。虽然尚未证实（Collatz问题），但是认为所有起始数字都在1处结束。在给定`limit`下，哪个起始数产生最长链？注意：一旦链条启动，条款允许超过一百万。
+It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
+
+Which starting number, under the given `limit`, produces the longest chain?
+
+**Note:** Once the chain starts the terms are allowed to go above one million.
# --hints--
-`longestCollatzSequence(14)`应该返回9。
+`longestCollatzSequence(14)` should return a number.
+
+```js
+assert(typeof longestCollatzSequence(14) === 'number');
+```
+
+`longestCollatzSequence(14)` should return 9.
```js
assert.strictEqual(longestCollatzSequence(14), 9);
```
-`longestCollatzSequence(5847)`应返回3711。
+`longestCollatzSequence(5847)` should return 3711.
```js
assert.strictEqual(longestCollatzSequence(5847), 3711);
```
-`longestCollatzSequence(46500)`应返回35655。
+`longestCollatzSequence(46500)` should return 35655.
```js
assert.strictEqual(longestCollatzSequence(46500), 35655);
```
-`longestCollatzSequence(54512)`应返回52527。
+`longestCollatzSequence(54512)` should return 52527.
```js
assert.strictEqual(longestCollatzSequence(54512), 52527);
```
-`longestCollatzSequence(1000000)`应返回837799。
+`longestCollatzSequence(100000)` should return 77031.
```js
assert.strictEqual(longestCollatzSequence(100000), 77031);
```
+`longestCollatzSequence(1000000)` should return 837799.
+
+```js
+assert.strictEqual(longestCollatzSequence(1000000), 837799);
+```
+
# --seed--
## --seed-contents--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-140-modified-fibonacci-golden-nuggets.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-140-modified-fibonacci-golden-nuggets.md
index b9e47e16866..01664be7aef 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-140-modified-fibonacci-golden-nuggets.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-140-modified-fibonacci-golden-nuggets.md
@@ -1,22 +1,26 @@
---
id: 5900f3fa1000cf542c50ff0c
-title: 问题140：改进的斐波那契金块
+title: 'Problem 140: Modified Fibonacci golden nuggets'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301769
dashedName: problem-140-modified-fibonacci-golden-nuggets
---
# --description--
-考虑无穷多项式系列AG（x）= xG1 + x2G2 + x3G3 + ...，其中Gk是二阶递归关系的第k项，Gk = Gk-1 + Gk-2，G1 = 1，G2 = 4;也就是说，1,4,5,9,14,23 ......对于这个问题，我们将关注x的值，其中AG（x）是正整数。前五个自然数的x的相应值如下所示。
+Consider the infinite polynomial series AG(x) = xG1 + x2G2 + x3G3 + ..., where Gk is the kth term of the second order recurrence relation Gk = Gk−1 + Gk−2, G1 = 1 and G2 = 4; that is, 1, 4, 5, 9, 14, 23, ... .
-xAG（x）（√5-1）/ 41 2/52（√22-2）/ 63（√137-5）/ 144 1/25
+For this problem we shall be concerned with values of x for which AG(x) is a positive integer.
-如果x是理性的，我们将称AG（x）为金块，因为它们变得越来越稀少;例如，第20个金块是211345365.找到前30个金块的总和。
+The corresponding values of x for the first five natural numbers are shown below.
+
+xAG(x) (√5−1)/41 2/52 (√22−2)/63 (√137−5)/144 1/25
+
+We shall call AG(x) a golden nugget if x is rational, because they become increasingly rarer; for example, the 20th golden nugget is 211345365. Find the sum of the first thirty golden nuggets.
# --hints--
-`euler140()`应该返回5673835352990。
+`euler140()` should return 5673835352990.
```js
assert.strictEqual(euler140(), 5673835352990);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-141-investigating-progressive-numbers-n-which-are-also-square.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-141-investigating-progressive-numbers-n-which-are-also-square.md
index b324b398d2b..4bcd7e069a6 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-141-investigating-progressive-numbers-n-which-are-also-square.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-141-investigating-progressive-numbers-n-which-are-also-square.md
@@ -1,18 +1,26 @@
---
id: 5900f3f91000cf542c50ff0b
-title: 问题141：调查渐进数n，它们也是正方形
+title: 'Problem 141: Investigating progressive numbers, n, which are also square'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301770
dashedName: problem-141-investigating-progressive-numbers-n-which-are-also-square
---
# --description--
-正整数n除以d，商和余数分别为q和r。另外，d，q和r是几何序列中的连续正整数项，但不一定是该顺序。例如，58除以6具有商9和余数4.还可以看出，4,6,9是几何序列中的连续项（公共比率3/2）。我们将这样的数字称为n，进步。一些渐进的数字，例如9和10404 = 1022，恰好也是完美的正方形。所有渐进完美正方形的总和低于十万是124657.找到所有渐进完美正方形的总和低于一万亿（1012）。
+A positive integer, n, is divided by d and the quotient and remainder are q and r respectively. In addition d, q, and r are consecutive positive integer terms in a geometric sequence, but not necessarily in that order.
+
+For example, 58 divided by 6 has quotient 9 and remainder 4. It can also be seen that 4, 6, 9 are consecutive terms in a geometric sequence (common ratio 3/2).
+
+We will call such numbers, n, progressive.
+
+Some progressive numbers, such as 9 and 10404 = 1022, happen to also be perfect squares. The sum of all progressive perfect squares below one hundred thousand is 124657.
+
+Find the sum of all progressive perfect squares below one trillion (1012).
# --hints--
-`euler141()`应该返回878454337159。
+`euler141()` should return 878454337159.
```js
assert.strictEqual(euler141(), 878454337159);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-142-perfect-square-collection.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-142-perfect-square-collection.md
index c4141d72111..03d921eac9a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-142-perfect-square-collection.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-142-perfect-square-collection.md
@@ -1,18 +1,18 @@
---
id: 5900f3fa1000cf542c50ff0d
-title: 问题142：完美的方形集合
+title: 'Problem 142: Perfect Square Collection'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301771
dashedName: problem-142-perfect-square-collection
---
# --description--
-找到具有整数x> y> z> 0的最小x + y + z，使得x + y，x-y，x + z，x-z，y + z，y-z都是完美正方形。
+Find the smallest x + y + z with integers x > y > z > 0 such that x + y, x − y, x + z, x − z, y + z, y − z are all perfect squares.
# --hints--
-`euler142()`应返回1006193。
+`euler142()` should return 1006193.
```js
assert.strictEqual(euler142(), 1006193);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-143-investigating-the-torricelli-point-of-a-triangle.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-143-investigating-the-torricelli-point-of-a-triangle.md
index ffd7b11cb14..fd11d827b7d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-143-investigating-the-torricelli-point-of-a-triangle.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-143-investigating-the-torricelli-point-of-a-triangle.md
@@ -1,20 +1,24 @@
---
id: 5900f3fc1000cf542c50ff0e
-title: 问题143：研究三角形的Torricelli点
+title: 'Problem 143: Investigating the Torricelli point of a triangle'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301772
dashedName: problem-143-investigating-the-torricelli-point-of-a-triangle
---
# --description--
-设ABC为三角形，所有内角均小于120度。设X为三角形内的任意点，并使XA = p，XC = q，XB = r。 Fermat挑战Torricelli找到X的位置，使p + q + r最小化。 Torricelli能够证明，如果在三角形ABC的每一侧构造等边三角形AOB，BNC和AMC，则AOB，BNC和AMC的外接圆将在三角形内的单个点T处相交。此外，他证明了T，称为Torricelli / Fermat点，最小化p + q + r。更值得注意的是，可以证明，当总和最小化时，AN = BM = CO = p + q + r，并且AN，BM和CO也在T处相交。
+Let ABC be a triangle with all interior angles being less than 120 degrees. Let X be any point inside the triangle and let XA = p, XC = q, and XB = r.
-如果总和最小化并且a，b，c，p，q和r都是正整数，我们将称三角形ABC为Torricelli三角形。例如，a = 399，b = 455，c = 511是Torricelli三角形的示例，其中p + q + r = 784.找到Torricelli三角形的p + q +r≤120000的所有不同值的总和。
+Fermat challenged Torricelli to find the position of X such that p + q + r was minimised.
+
+Torricelli was able to prove that if equilateral triangles AOB, BNC and AMC are constructed on each side of triangle ABC, the circumscribed circles of AOB, BNC, and AMC will intersect at a single point, T, inside the triangle. Moreover he proved that T, called the Torricelli/Fermat point, minimises p + q + r. Even more remarkable, it can be shown that when the sum is minimised, AN = BM = CO = p + q + r and that AN, BM and CO also intersect at T.
+
+If the sum is minimised and a, b, c, p, q and r are all positive integers we shall call triangle ABC a Torricelli triangle. For example, a = 399, b = 455, c = 511 is an example of a Torricelli triangle, with p + q + r = 784. Find the sum of all distinct values of p + q + r ≤ 120000 for Torricelli triangles.
# --hints--
-`euler143()`应返回30758397。
+`euler143()` should return 30758397.
```js
assert.strictEqual(euler143(), 30758397);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-144-investigating-multiple-reflections-of-a-laser-beam.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-144-investigating-multiple-reflections-of-a-laser-beam.md
index b3ecd756d22..88d8b1e5a8d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-144-investigating-multiple-reflections-of-a-laser-beam.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-144-investigating-multiple-reflections-of-a-laser-beam.md
@@ -1,22 +1,26 @@
---
id: 5900f3fc1000cf542c50ff0f
-title: 问题144：研究激光束的多次反射
+title: 'Problem 144: Investigating multiple reflections of a laser beam'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301773
dashedName: problem-144-investigating-multiple-reflections-of-a-laser-beam
---
# --description--
-在激光物理学中，“白色单元”是镜像系统，其充当激光束的延迟线。光束进入细胞，在镜子上反弹，最终恢复原状。我们将考虑的特定白色单元是椭圆形，其等式为4x2 + y2 = 100。顶部对应于-0.01≤x≤+ 0.01的部分缺失，允许光线通过孔进入和退出。
+In laser physics, a "white cell" is a mirror system that acts as a delay line for the laser beam. The beam enters the cell, bounces around on the mirrors, and eventually works its way back out.
-此问题中的光束从白色单元外部的点（0.0,10.1）开始，光束首先在（1.4，-9.6）处撞击镜子。每次激光束撞击椭圆的表面时，它遵循通常的反射定律“入射角等于反射角”。也就是说，入射光束和反射光束都与入射点处的法线形成相同的角度。在左图中，红线表示激光束与白色单元壁之间的前两个接触点;蓝线表示在第一次反弹入射点处与椭圆相切的直线。给定椭圆的任意点（x，y）处的切线的斜率m为：m = -4x / y法线为垂直于入射点处的该切线。右侧的动画显示了光束的前10个反射。
+The specific white cell we will be considering is an ellipse with the equation 4x2 + y2 = 100
-在离开之前，光束到达白细胞内表面的次数是多少？
+The section corresponding to −0.01 ≤ x ≤ +0.01 at the top is missing, allowing the light to enter and exit through the hole.
+
+The light beam in this problem starts at the point (0.0,10.1) just outside the white cell, and the beam first impacts the mirror at (1.4,-9.6). Each time the laser beam hits the surface of the ellipse, it follows the usual law of reflection "angle of incidence equals angle of reflection." That is, both the incident and reflected beams make the same angle with the normal line at the point of incidence. In the figure on the left, the red line shows the first two points of contact between the laser beam and the wall of the white cell; the blue line shows the line tangent to the ellipse at the point of incidence of the first bounce.The slope m of the tangent line at any point (x,y) of the given ellipse is: m = −4x/yThe normal line is perpendicular to this tangent line at the point of incidence. The animation on the right shows the first 10 reflections of the beam.
+
+How many times does the beam hit the internal surface of the white cell before exiting?
# --hints--
-`euler144()`应返回354。
+`euler144()` should return 354.
```js
assert.strictEqual(euler144(), 354);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-145-how-many-reversible-numbers-are-there-below-one-billion.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-145-how-many-reversible-numbers-are-there-below-one-billion.md
index 08e8f0876af..0ec3679a6f9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-145-how-many-reversible-numbers-are-there-below-one-billion.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-145-how-many-reversible-numbers-are-there-below-one-billion.md
@@ -1,22 +1,22 @@
---
id: 5900f3fd1000cf542c50ff10
-title: 问题145：有多少可逆数字低于十亿？
+title: 'Problem 145: How many reversible numbers are there below one-billion?'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301774
dashedName: problem-145-how-many-reversible-numbers-are-there-below-one-billion
---
# --description--
-一些正整数n具有sum \[n + reverse（n）]完全由奇数（十进制）数字组成的属性。例如，36 + 63 = 99和409 + 904 = 1313.我们将这些数字称为可逆的;所以36,63,409和904是可逆的。 n或反向（n）中不允许前导零。
+Some positive integers n have the property that the sum \[ n + reverse(n) ] consists entirely of odd (decimal) digits. For instance, 36 + 63 = 99 and 409 + 904 = 1313. We will call such numbers reversible; so 36, 63, 409, and 904 are reversible. Leading zeroes are not allowed in either n or reverse(n).
-有一千个可逆数字低于一千。
+There are 120 reversible numbers below one-thousand.
-有多少可逆数字低于十亿（109）？
+How many reversible numbers are there below one-billion (109)?
# --hints--
-`euler145()`应该返回608720。
+`euler145()` should return 608720.
```js
assert.strictEqual(euler145(), 608720);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-146-investigating-a-prime-pattern.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-146-investigating-a-prime-pattern.md
index 0781a2e8c34..ddfec6766b2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-146-investigating-a-prime-pattern.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-146-investigating-a-prime-pattern.md
@@ -1,20 +1,20 @@
---
id: 5900f3fe1000cf542c50ff11
-title: 问题146：调查素数模式
+title: 'Problem 146: Investigating a Prime Pattern'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301775
dashedName: problem-146-investigating-a-prime-pattern
---
# --description--
-整数n2 + 1，n2 + 3，n2 + 7，n2 + 9，n2 + 13和n2 + 27为其连续质数的最小正整数n为10。所有小于100万的整数n的总和 是1242490。
+The smallest positive integer n for which the numbers n2+1, n2+3, n2+7, n2+9, n2+13, and n2+27 are consecutive primes is 10. The sum of all such integers n below one-million is 1242490.
-所有小于1.5亿的整数n的总和是多少？
+What is the sum of all such integers n below 150 million?
# --hints--
-`euler146()`应该返回676333270。
+`euler146()` should return 676333270.
```js
assert.strictEqual(euler146(), 676333270);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-147-rectangles-in-cross-hatched-grids.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-147-rectangles-in-cross-hatched-grids.md
index d420d03e572..3b422811f69 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-147-rectangles-in-cross-hatched-grids.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-147-rectangles-in-cross-hatched-grids.md
@@ -1,24 +1,24 @@
---
id: 5900f3ff1000cf542c50ff12
-title: 问题147：交叉阴影网格中的矩形
+title: 'Problem 147: Rectangles in cross-hatched grids'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301776
dashedName: problem-147-rectangles-in-cross-hatched-grids
---
# --description--
-在3x2交叉阴影网格中，如图所示，总共37个不同的矩形可以位于该网格内。
+In a 3x2 cross-hatched grid, a total of 37 different rectangles could be situated within that grid as indicated in the sketch.
-有5个小于3x2的网格，垂直和水平尺寸很重要，即1x1,2x1,3x1,1x2和2x2。如果它们中的每一个都是交叉阴影线，则可以在这些较小的网格中放置以下数量的不同矩形：1x1：1 2x1：4 3x1：8 1x2：4 2x2：18
+There are 5 grids smaller than 3x2, vertical and horizontal dimensions being important, i.e. 1x1, 2x1, 3x1, 1x2 and 2x2. If each of them is cross-hatched, the following number of different rectangles could be situated within those smaller grids: 1x1: 1 2x1: 4 3x1: 8 1x2: 4 2x2: 18
-将它们添加到3x2网格中的37个，总共72个不同的矩形可以位于3x2和更小的网格中。
+Adding those to the 37 of the 3x2 grid, a total of 72 different rectangles could be situated within 3x2 and smaller grids.
-在47x43和更小的网格中可以放置多少个不同的矩形？
+How many different rectangles could be situated within 47x43 and smaller grids?
# --hints--
-`euler147()`应该返回846910284。
+`euler147()` should return 846910284.
```js
assert.strictEqual(euler147(), 846910284);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-148-exploring-pascals-triangle.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-148-exploring-pascals-triangle.md
index 84990a4491b..13babe893b0 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-148-exploring-pascals-triangle.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-148-exploring-pascals-triangle.md
@@ -1,74 +1,34 @@
---
id: 5900f4021000cf542c50ff14
-title: 问题148：探索帕斯卡的三角形
+title: 'Problem 148: Exploring Pascal''s triangle'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301777
dashedName: problem-148-exploring-pascals-triangle
---
# --description--
-我们可以很容易地验证Pascal三角形的前七行中的所有条目都不能被7整除：
+We can easily verify that none of the entries in the first seven rows of Pascal's triangle are divisible by 7:
-1
+<pre>
+ 1
+ 1 1
+ 1 2 1
+ 1 3 3 1
+ 1 4 6 4 1
+ 1 5 10 10 5 1
+1 6 15 20 15 6 1
+</pre>
-1
+However, if we check the first one hundred rows, we will find that only 2361 of the 5050 entries are not divisible by 7.
-1
+# --instructions--
-1
-
-2
-
-1
-
-1
-
-3
-
-3
-
-1
-
-1
-
-4
-
-6
-
-4
-
-1
-
-1
-
-五
-
-10
-
-10
-
-五
-
-1 1
-
-6
-
-15
-
-20
-
-15
-
-6
-
-1然而，如果我们检查前100行，我们会发现5050个条目中只有2361个不能被7整除。
-
-找到Pascal三角形的前十亿（109）行中不能被7整除的条目数。
+Find the number of entries which are not divisible by 7 in the first one billion (10<sup>9</sup>) rows of Pascal's triangle.
# --hints--
-`euler148()`应该返回2129970655314432。
+`euler148()` should return 2129970655314432.
```js
assert.strictEqual(euler148(), 2129970655314432);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-149-searching-for-a-maximum-sum-subsequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-149-searching-for-a-maximum-sum-subsequence.md
index a20b8b0a08a..c76d5370c24 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-149-searching-for-a-maximum-sum-subsequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-149-searching-for-a-maximum-sum-subsequence.md
@@ -1,32 +1,32 @@
---
id: 5900f4021000cf542c50ff13
-title: 问题149：搜索最大和子序列
+title: 'Problem 149: Searching for a maximum-sum subsequence'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301778
dashedName: problem-149-searching-for-a-maximum-sum-subsequence
---
# --description--
-查看下表，可以很容易地验证任意方向（水平，垂直，对角或反对角）上相邻数字的最大和为16（= 8 + 7 + 1）。
+Looking at the table below, it is easy to verify that the maximum possible sum of adjacent numbers in any direction (horizontal, vertical, diagonal or anti-diagonal) is 16 (= 8 + 7 + 1).
−25329−6513273−18−4 8
-现在，让我们重复搜索，但范围更大：
+Now, let us repeat the search, but on a much larger scale:
-首先，使用称为“滞后斐波那契生成器”的特定形式生成四百万个伪随机数：
+First, generate four million pseudo-random numbers using a specific form of what is known as a "Lagged Fibonacci Generator":
-对于1≤k≤55，sk = \[100003 − 200003k + 300007k3]（模1000000）− 500000。 对于56≤k≤4000000，sk = \[sk-24 + sk-55 + 1000000]（模1000000）− 500000。
+For 1 ≤ k ≤ 55, sk = \[100003 − 200003k + 300007k3] (modulo 1000000) − 500000. For 56 ≤ k ≤ 4000000, sk = \[sk−24 + sk−55 + 1000000] (modulo 1000000) − 500000.
-因此，s10 = -393027，s100 = 86613。
+Thus, s10 = −393027 and s100 = 86613.
-然后，将s的项排列在2000×2000表中，使用前2000个数字（顺序）填充第一行，使用后2000个数字填充第二行，依此类推。
+The terms of s are then arranged in a 2000×2000 table, using the first 2000 numbers to fill the first row (sequentially), the next 2000 numbers to fill the second row, and so on.
-最后，在任何方向（水平，垂直，对角线或反对角线）上找到（任意数量）相邻项的最大和。
+Finally, find the greatest sum of (any number of) adjacent entries in any direction (horizontal, vertical, diagonal or anti-diagonal).
# --hints--
-`euler149()`应该返回52852124。
+`euler149()` should return 52852124.
```js
assert.strictEqual(euler149(), 52852124);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-15-lattice-paths.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-15-lattice-paths.md
index 2799a5a1034..225a31ee93c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-15-lattice-paths.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-15-lattice-paths.md
@@ -1,32 +1,40 @@
---
id: 5900f37b1000cf542c50fe8e
-title: 问题15：格子路径
+title: 'Problem 15: Lattice paths'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301780
dashedName: problem-15-lattice-paths
---
# --description--
-从2×2网格的左上角开始，只能向右和向下移动，右下角有6条路线。 
+Starting in the top left corner of a 2×2 grid, and only being able to move to the right and down, there are exactly 6 routes to the bottom right corner.
-通过给定的`gridSize`有多少这样的路由？
+<img class="img-responsive center-block" alt="a diagram of 6 2 by 2 grids showing all the routes to the bottom right corner" src="https://cdn-media-1.freecodecamp.org/project-euler/1Atixoj.gif" style="background-color: white; padding: 10px;" />
+
+How many such routes are there through a given `gridSize`?
# --hints--
-`latticePaths(4)`应该返回70。
+`latticePaths(4)` should return a number.
+
+```js
+assert(typeof latticePaths(4) === 'number');
+```
+
+`latticePaths(4)` should return 70.
```js
assert.strictEqual(latticePaths(4), 70);
```
-`latticePaths(9)`应该返回48620。
+`latticePaths(9)` should return 48620.
```js
assert.strictEqual(latticePaths(9), 48620);
```
-`latticePaths(20)`应该返回137846528820。
+`latticePaths(20)` should return 137846528820.
```js
assert.strictEqual(latticePaths(20), 137846528820);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-150-searching-a-triangular-array-for-a-sub-triangle-having-minimum-sum.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-150-searching-a-triangular-array-for-a-sub-triangle-having-minimum-sum.md
index a312ea4c2fc..fa659c2f46e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-150-searching-a-triangular-array-for-a-sub-triangle-having-minimum-sum.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-150-searching-a-triangular-array-for-a-sub-triangle-having-minimum-sum.md
@@ -1,34 +1,37 @@
---
id: 5900f4031000cf542c50ff15
-title: 问题150：在三角形阵列中搜索具有最小和的子三角形
+title: >-
+ Problem 150: Searching a triangular array for a sub-triangle having minimum-sum
challengeType: 5
-videoUrl: ''
+forumTopicId: 301781
dashedName: problem-150-searching-a-triangular-array-for-a-sub-triangle-having-minimum-sum
---
# --description--
-在正整数和负整数的三角形阵列中，我们希望找到一个子三角形，使得它包含的数字之和尽可能小。在下面的示例中，可以很容易地验证标记的三角形满足具有-42的总和的条件。
+In a triangular array of positive and negative integers, we wish to find a sub-triangle such that the sum of the numbers it contains is the smallest possible.
-我们希望制作一个包含一千行的三角形数组，因此我们使用一种随机数生成器（称为线性同余生成器）生成5009个伪随机数sk，范围为±219，如下所示：t：= 0
+In the example below, it can be easily verified that the marked triangle satisfies this condition having a sum of −42.
-对于k = 1到k = 500500：
+We wish to make such a triangular array with one thousand rows, so we generate 500500 pseudo-random numbers sk in the range ±219, using a type of random number generator (known as a Linear Congruential Generator) as follows: t := 0
-t：=（615949 \* t + 797807）modulo 220 sk：= t-219因此：s1 = 273519，s2 = -153582，s3 = 450905等我们的三角形数组然后使用伪随机数形成：
+for k = 1 up to k = 500500:
+
+t := (615949\*t + 797807) modulo 220 sk := t−219 Thus: s1 = 273519, s2 = −153582, s3 = 450905 etc Our triangular array is then formed using the pseudo-random numbers thus:
s1 s2 s3 s4 s5 s6
-s7 s8 s9 s10 ......
+s7 s8 s9 s10 ...
-子三角形可以从数组的任何元素开始，并在我们喜欢的范围内向下延伸（从下一行直接接收它下面的两个元素，之后直接从该行下面的三个元素，依此类推）。
+Sub-triangles can start at any element of the array and extend down as far as we like (taking-in the two elements directly below it from the next row, the three elements directly below from the row after that, and so on).
-“三角形的总和”定义为它包含的所有元素的总和。
+The "sum of a sub-triangle" is defined as the sum of all the elements it contains.
-找到可能的最小子三角形和。
+Find the smallest possible sub-triangle sum.
# --hints--
-`euler150()`应返回-271248680。
+`euler150()` should return -271248680.
```js
assert.strictEqual(euler150(), -271248680);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-151-paper-sheets-of-standard-sizes-an-expected-value-problem.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-151-paper-sheets-of-standard-sizes-an-expected-value-problem.md
index 02b13b43cc7..e75e5a0fdfe 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-151-paper-sheets-of-standard-sizes-an-expected-value-problem.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-151-paper-sheets-of-standard-sizes-an-expected-value-problem.md
@@ -1,20 +1,26 @@
---
id: 5900f4031000cf542c50ff16
-title: 问题151：标准尺寸的纸张：期望值问题
+title: 'Problem 151: Paper sheets of standard sizes: an expected-value problem'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301782
dashedName: problem-151-paper-sheets-of-standard-sizes-an-expected-value-problem
---
# --description--
-印刷车间每周运行16批（工作），每批需要一张A5尺寸的特殊颜色纸。每个星期一早上，工头打开一个新信封，里面装着一张大小为A1的特殊纸张。他继续把它切成两半，从而得到两张A2尺寸的纸。然后，他将其中一个切成两半，得到两张A3尺寸，依此类推，直到他获得了本周第一批所需的A5尺寸纸张。所有未使用的纸张都放回信封中。
+A printing shop runs 16 batches (jobs) every week and each batch requires a sheet of special colour-proofing paper of size A5.
-在每个后续批次开始时，他随机从信封中取出一张纸。如果它的大小为A5，他会使用它。如果它更大，他会重复“切成两半”的程序，直到他有他需要的东西，任何剩余的床单总是放回信封里。排除本周的第一批和最后一批，找到工头在信封中找到一张纸的预期次数（每周）。使用格式x.xxxxxx将您的答案四舍五入到小数点后六位。
+Every Monday morning, the foreman opens a new envelope, containing a large sheet of the special paper with size A1.
+
+He proceeds to cut it in half, thus getting two sheets of size A2. Then he cuts one of them in half to get two sheets of size A3 and so on until he obtains the A5-size sheet needed for the first batch of the week.
+
+All the unused sheets are placed back in the envelope.
+
+At the beginning of each subsequent batch, he takes from the envelope one sheet of paper at random. If it is of size A5, he uses it. If it is larger, he repeats the 'cut-in-half' procedure until he has what he needs and any remaining sheets are always placed back in the envelope. Excluding the first and last batch of the week, find the expected number of times (during each week) that the foreman finds a single sheet of paper in the envelope. Give your answer rounded to six decimal places using the format x.xxxxxx .
# --hints--
-<code>euler151()</ code>应该返回0.464399。
+`euler151()` should return 0.464399.
```js
assert.strictEqual(euler151(), 0.464399);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-152-writing-one-half-as-a-sum-of-inverse-squares.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-152-writing-one-half-as-a-sum-of-inverse-squares.md
index c06a3dbe880..1acd9f0652e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-152-writing-one-half-as-a-sum-of-inverse-squares.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-152-writing-one-half-as-a-sum-of-inverse-squares.md
@@ -1,22 +1,22 @@
---
id: 5900f4041000cf542c50ff17
-title: 问题152：将一半写为倒数平方和
+title: 'Problem 152: Writing one half as a sum of inverse squares'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301783
dashedName: problem-152-writing-one-half-as-a-sum-of-inverse-squares
---
# --description--
-有几种方法可以使用不同的整数将数字1/2写成反平方和。
+There are several ways to write the number 1/2 as a sum of inverse squares using distinct integers.
-例如，可以使用数字{2,3,4,5,7,12,15,20,28,35}：
+For instance, the numbers {2,3,4,5,7,12,15,20,28,35} can be used:
-实际上，仅使用2到45之间的整数（包括2和45），完全可以通过三种方式来实现，其余两种方式是：{2,3,4,6,7,9,10,20,28,35,36,45 }和{2,3,4,6,7,9,12,15,28,30,35,36,45}。 使用2到80之间（含2和80）的不同整数，有多少种方法可以将数字1/2写成反平方和？
+In fact, only using integers between 2 and 45 inclusive, there are exactly three ways to do it, the remaining two being: {2,3,4,6,7,9,10,20,28,35,36,45} and {2,3,4,6,7,9,12,15,28,30,35,36,45}. How many ways are there to write the number 1/2 as a sum of inverse squares using distinct integers between 2 and 80 inclusive?
# --hints--
-<code>euler152()</ code>应该返回301。
+`euler152()` should return 301.
```js
assert.strictEqual(euler152(), 301);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-153-investigating-gaussian-integers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-153-investigating-gaussian-integers.md
index 16649ab014a..b1eaa92c44b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-153-investigating-gaussian-integers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-153-investigating-gaussian-integers.md
@@ -1,50 +1,50 @@
---
id: 5900f4051000cf542c50ff18
-title: 问题153：调查高斯整数
+title: 'Problem 153: Investigating Gaussian Integers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301784
dashedName: problem-153-investigating-gaussian-integers
---
# --description--
-众所周知，方程x2 = -1没有实数x的解。
+As we all know the equation x2=-1 has no solutions for real x.
-然而，如果我们引入虚数i，则该等式具有两个解：x = i且x = -i。
+If we however introduce the imaginary number i this equation has two solutions: x=i and x=-i.
-如果我们更进一步，等式（x-3）2 = -4有两个复数解：x = 3 + 2i和x = 3-2i。 x = 3 + 2i和x = 3-2i被称为彼此的复共轭。
+If we go a step further the equation (x-3)2=-4 has two complex solutions: x=3+2i and x=3-2i. x=3+2i and x=3-2i are called each others' complex conjugate.
-形式a + bi的数字称为复数。
+Numbers of the form a+bi are called complex numbers.
-通常，+ bi和a-bi是彼此的复共轭。高斯整数是复数a + bi，使得a和b都是整数。
+In general a+bi and a−bi are each other's complex conjugate. A Gaussian Integer is a complex number a+bi such that both a and b are integers.
-常规整数也是高斯整数（b = 0）。
+The regular integers are also Gaussian integers (with b=0).
-为了将它们与b≠0的高斯整数区分开来，我们称这样的整数为“有理整数”。
+To distinguish them from Gaussian integers with b ≠ 0 we call such integers "rational integers."
-如果结果也是高斯整数，则高斯整数称为有理整数n的除数。
+A Gaussian integer is called a divisor of a rational integer n if the result is also a Gaussian integer.
-例如，如果我们将5除以1 + 2i，我们可以通过以下方式简化：
+If for example we divide 5 by 1+2i we can simplify in the following manner:
-通过1 + 2i的复共轭乘以分子和分母：1-2i。
+Multiply numerator and denominator by the complex conjugate of 1+2i: 1−2i.
-结果是。
+The result is .
-所以1 + 2i是5的除数。
+So 1+2i is a divisor of 5.
-请注意，1 + i不是5的除数，因为。
+Note that 1+i is not a divisor of 5 because .
-还要注意，如果高斯整数（a + bi）是有理整数n的除数，则其复共轭（a-bi）也是n的除数。实际上，5有六个除数，使得实部是正的：{1,1 + 2i，1 - 2i，2 + i，2 - i，5}。
+Note also that if the Gaussian Integer (a+bi) is a divisor of a rational integer n, then its complex conjugate (a−bi) is also a divisor of n. In fact, 5 has six divisors such that the real part is positive: {1, 1 + 2i, 1 − 2i, 2 + i, 2 − i, 5}.
-以下是前五个正整数的所有除数的表：
+The following is a table of all of the divisors for the first five positive rational integers:
-n高斯整数除数，具有正实数partSum s（n）
+n Gaussian integer divisors with positive real partSum s(n) of these
-divisors111 21,1 + i，1-i，25 31,34 41,1 + i，1-i，2,2 + 2i，2-2i，413 51,1 + 2i，1-2i，2 + i， 2-i，512对于具有正实部的除数，那么，我们有：。对于1≤n≤105，Σs（n）= 17924657155。什么是Σs（n）1≤n≤108？
+divisors111 21, 1+i, 1-i, 25 31, 34 41, 1+i, 1-i, 2, 2+2i, 2-2i,413 51, 1+2i, 1-2i, 2+i, 2-i, 512 For divisors with positive real parts, then, we have: . For 1 ≤ n ≤ 105, ∑ s(n)=17924657155. What is ∑ s(n) for 1 ≤ n ≤ 108?
# --hints--
-`euler153()`应该返回17971254122360636。
+`euler153()` should return 17971254122360636.
```js
assert.strictEqual(euler153(), 17971254122360636);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-154-exploring-pascals-pyramid.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-154-exploring-pascals-pyramid.md
index d51b418b36b..8d8abe0dfcd 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-154-exploring-pascals-pyramid.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-154-exploring-pascals-pyramid.md
@@ -1,20 +1,20 @@
---
id: 5900f4071000cf542c50ff19
-title: 问题154：探索帕斯卡的金字塔
+title: 'Problem 154: Exploring Pascal''s pyramid'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301785
dashedName: problem-154-exploring-pascals-pyramid
---
# --description--
-使用球形球构造三角形金字塔，使得每个球恰好位于下一个较低水平的三个球上。
+A triangular pyramid is constructed using spherical balls so that each ball rests on exactly three balls of the next lower level.
-然后，我们计算从顶点到每个位置的路径数量：路径从顶点开始并向下前进到当前位置正下方的三个球体中的任何一个。因此，到达某个位置的路径数是紧接在其上方的数字的总和（取决于位置，在其上方最多有三个数字）。结果是Pascal的金字塔，每个级别n的数字是三项式展开（x + y + z）n的系数。 （x + y + z）200000的扩展中有多少个系数是1012的倍数？
+Then, we calculate the number of paths leading from the apex to each position: A path starts at the apex and progresses downwards to any of the three spheres directly below the current position. Consequently, the number of paths to reach a certain position is the sum of the numbers immediately above it (depending on the position, there are up to three numbers above it). The result is Pascal's pyramid and the numbers at each level n are the coefficients of the trinomial expansion (x + y + z)n. How many coefficients in the expansion of (x + y + z)200000 are multiples of 1012?
# --hints--
-`euler154()`应该返回479742450。
+`euler154()` should return 479742450.
```js
assert.strictEqual(euler154(), 479742450);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-155-counting-capacitor-circuits.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-155-counting-capacitor-circuits.md
index e05abd8b9a4..404c223c6a4 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-155-counting-capacitor-circuits.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-155-counting-capacitor-circuits.md
@@ -1,24 +1,24 @@
---
id: 5900f4081000cf542c50ff1a
-title: 问题155：计算电容器电路
+title: 'Problem 155: Counting Capacitor Circuits'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301786
dashedName: problem-155-counting-capacitor-circuits
---
# --description--
-电路仅使用相同值C的相同电容器。
+An electric circuit uses exclusively identical capacitors of the same value C.
-电容器可以串联或并联连接以形成子单元，子单元然后可以与其他电容器或其他子单元串联或并联连接以形成更大的子单元，以此类推直到最终电路。使用这个简单的程序和多达n个相同的电容器，我们可以制造具有一系列不同总电容的电路。例如，使用最多n = 3个电容器，每个电容器为60 F，我们可以获得以下7个不同的总电容值：
+The capacitors can be connected in series or in parallel to form sub-units, which can then be connected in series or in parallel with other capacitors or other sub-units to form larger sub-units, and so on up to a final circuit. Using this simple procedure and up to n identical capacitors, we can make circuits having a range of different total capacitances. For example, using up to n=3 capacitors of 60 F each, we can obtain the following 7 distinct total capacitance values:
-如果我们用D（n）表示当使用多达n个等值电容器时我们可以获得的不同总电容值的数量和上述简单程序，我们得到：D（1）= 1，D（2）= 3 ，D（3）= 7 ...求D（18）。提醒：当并联电容器C1，C2等时，总电容为CT = C1 + C2 + ......，
+If we denote by D(n) the number of distinct total capacitance values we can obtain when using up to n equal-valued capacitors and the simple procedure described above, we have: D(1)=1, D(2)=3, D(3)=7 ... Find D(18). Reminder : When connecting capacitors C1, C2 etc in parallel, the total capacitance is CT = C1 + C2 +...,
-而当它们串联连接时，总电容由下式给出：
+whereas when connecting them in series, the overall capacitance is given by:
# --hints--
-`euler155()`应返回3857447。
+`euler155()` should return 3857447.
```js
assert.strictEqual(euler155(), 3857447);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-156-counting-digits.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-156-counting-digits.md
index 1885525df3f..3ab75c957bc 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-156-counting-digits.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-156-counting-digits.md
@@ -1,32 +1,32 @@
---
id: 5900f4091000cf542c50ff1b
-title: 问题156：计数数字
+title: 'Problem 156: Counting Digits'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301787
dashedName: problem-156-counting-digits
---
# --description--
-从零开始，自然数字在基数10中写下，如下所示：
+Starting from zero the natural numbers are written down in base 10 like this:
-0 1 2 3 4 5 6 7 8 9 10 11 12 ....
+0 1 2 3 4 5 6 7 8 9 10 11 12....
-考虑数字d = 1。在我们写下每个数字n后，我们将更新已发生的数字并将此数字称为f（n，1）。那么f（n，1）的第一个值如下：
+Consider the digit d=1. After we write down each number n, we will update the number of ones that have occurred and call this number f(n,1). The first values for f(n,1), then, are as follows:
-nf（n，1）00 11 21 31 41 51 61 71 81 91 102 114 125
+nf(n,1) 00 11 21 31 41 51 61 71 81 91 102 114 125
-请注意，f（n，1）永远不等于3。
+Note that f(n,1) never equals 3.
-因此，等式f（n，1）= n的前两个解是n = 0并且n = 1。下一个解决方案是n = 199981。以相同的方式，函数f（n，d）给出在写入数字n之后已经写下的总位数d。
+So the first two solutions of the equation f(n,1)=n are n=0 and n=1. The next solution is n=199981. In the same manner the function f(n,d) gives the total number of digits d that have been written down after the number n has been written.
-实际上，对于每个数字d≠0,0是方程f（n，d）= n的第一个解。设s（d）是f（n，d）= n的所有解的总和。
+In fact, for every digit d ≠ 0, 0 is the first solution of the equation f(n,d)=n. Let s(d) be the sum of all the solutions for which f(n,d)=n.
-你得到s（1）= 22786974071。找到Σs（d）的1≤d≤9。注意：如果对于某些n，对于多于一个d的值，f（n，d）= n，对于d的每个值，再次计算n的这个值。 F（N，d）= N。
+You are given that s(1)=22786974071. Find ∑ s(d) for 1 ≤ d ≤ 9. Note: if, for some n, f(n,d)=n for more than one value of d this value of n is counted again for every value of d for which f(n,d)=n.
# --hints--
-`euler156`应该返回21295121502550。
+`euler156()` should return 21295121502550.
```js
assert.strictEqual(euler156(), 21295121502550);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-157-solving-the-diophantine-equation.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-157-solving-the-diophantine-equation.md
index 4a195d325ca..fa1e3270516 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-157-solving-the-diophantine-equation.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-157-solving-the-diophantine-equation.md
@@ -1,64 +1,62 @@
---
id: 5900f4091000cf542c50ff1c
-title: 问题157：解决丢番图方程
+title: 'Problem 157: Solving the diophantine equation'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301788
dashedName: problem-157-solving-the-diophantine-equation
---
# --description--
-491/5000
+Consider the diophantine equation 1/a+1/b= p/10n with a, b, p, n positive integers and a ≤ b.
-考虑具有a，b，p，n个正整数且a≤b的双色子方程1 / a + 1 / b = p / 10n。
+For n=1 this equation has 20 solutions that are listed below:
-对于n = 1，此等式有20个解决方案，如下所示：
+1/1+1/1=20/10
-1/1 + 1/1 = 20/10
+1/1+1/2=15/10
-1/1 + 1/2 = 15/10
+1/1+1/5=12/10
-1/1 + 1/5 = 12/10
+1/1+1/10=11/10
-1/1 + 1/10 = 11/10
+1/2+1/2=10/10
-1/2 + 1/2 = 10/10
+1/2+1/5=7/10
-1/2 + 1/5 = 7/10
+1/2+1/10=6/10
-1/2 + 1/10 = 6/10
+1/3+1/6=5/10
-1/3 + 1/6 = 5/10
+1/3+1/15=4/10
-1/3 + 1/15 = 4/10
+1/4+1/4=5/10
-1/4 + 1/4 = 5/10
+1/4+1/20=3/10
-1/4 + 1/20 = 3/10
+1/5+1/5=4/10
-1/5 + 1/5 = 4/10
+1/5+1/10=3/10
-1/5 + 1/10 = 3/10
+1/6+1/30=2/10
-1/6 + 1/30 = 2/10
+1/10+1/10=2/10
-1/10 + 1/10 = 2/10
+1/11+1/110=1/10
-1/11 + 1/110 = 1/10
+1/12+1/60=1/10
-1/12 + 1/60 = 1/10
+1/14+1/35=1/10
-1/14 + 1/35 = 1/10
+1/15+1/30=1/10
-1/15 + 1/30 = 1/10
+1/20+1/20=1/10
-1/20 + 1/20 = 1/10
-
-对于1≤n≤9，该方程有多少个解？
+How many solutions has this equation for 1 ≤ n ≤ 9?
# --hints--
-`euler157()`应返回53490。
+`euler157()` should return 53490.
```js
assert.strictEqual(euler157(), 53490);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-158-exploring-strings-for-which-only-one-character-comes-lexicographically-after-its-neighbour-to-the-left.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-158-exploring-strings-for-which-only-one-character-comes-lexicographically-after-its-neighbour-to-the-left.md
index 6226d9a2b43..6b979cb0d7f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-158-exploring-strings-for-which-only-one-character-comes-lexicographically-after-its-neighbour-to-the-left.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-158-exploring-strings-for-which-only-one-character-comes-lexicographically-after-its-neighbour-to-the-left.md
@@ -1,19 +1,34 @@
---
id: 5900f40a1000cf542c50ff1d
-title: 问题158：探索在其邻居之后只有一个字符按字典顺序出现的字符串
+title: >-
+ Problem 158: Exploring strings for which only one character comes lexicographically after its neighbour to the left
challengeType: 5
-videoUrl: ''
+forumTopicId: 301789
dashedName: >-
problem-158-exploring-strings-for-which-only-one-character-comes-lexicographically-after-its-neighbour-to-the-left
---
# --description--
-从字母表的26个字母中取三个不同的字母，可以形成长度为3的字符串。例如'abc'，'hat'和'zyx'。当我们研究这三个例子时，我们看到对于'abc'，两个字符在其左边的邻居之后以字典方式出现。对于“帽子”，只有一个字符在其左边的邻居之后按字典顺序排列。对于'zyx'，在左边的邻居之后，字典上有零个字符。总共有10400个长度为3的字符串，其中一个字符在其左边的邻居之后按字典顺序排列。我们现在考虑字母表中n≤26个不同字符的字符串。对于每个n，p（n）是长度为n的字符串的数量，正好一个字符在其左边的邻居之后按字典顺序排列。 p（n）的最大值是多少？
+Taking three different letters from the 26 letters of the alphabet, character strings of length three can be formed.
+
+Examples are 'abc', 'hat' and 'zyx'.
+
+When we study these three examples we see that for 'abc' two characters come lexicographically after its neighbour to the left.
+
+For 'hat' there is exactly one character that comes lexicographically after its neighbour to the left. For 'zyx' there are zero characters that come lexicographically after its neighbour to the left.
+
+In all there are 10400 strings of length 3 for which exactly one character comes lexicographically after its neighbour to the left.
+
+We now consider strings of n ≤ 26 different characters from the alphabet.
+
+For every n, p(n) is the number of strings of length n for which exactly one character comes lexicographically after its neighbour to the left.
+
+What is the maximum value of p(n)?
# --hints--
-`euler158()`应该返回409511334375。
+`euler158()` should return 409511334375.
```js
assert.strictEqual(euler158(), 409511334375);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-159-digital-root-sums-of-factorisations.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-159-digital-root-sums-of-factorisations.md
index 50383e2a11d..b5e3b4945bd 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-159-digital-root-sums-of-factorisations.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-159-digital-root-sums-of-factorisations.md
@@ -1,22 +1,24 @@
---
id: 5900f40c1000cf542c50ff1e
-title: 问题159：因子的数字根和
+title: 'Problem 159: Digital root sums of factorisations'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301790
dashedName: problem-159-digital-root-sums-of-factorisations
---
# --description--
-复合数可以通过许多不同的方式考虑。例如，不包括乘以一，24可以用7种不同的方式考虑：
+A composite number can be factored many different ways.
+
+For instance, not including multiplication by one, 24 can be factored in 7 distinct ways:
24 = 2x2x2x3 24 = 2x3x4 24 = 2x2x6 24 = 4x6 24 = 3x8 24 = 2x12 24 = 24
-回想一下，基数为10的数字的数字根是通过将该数字的数字加在一起而得到的，并重复该过程，直到到达的数字小于10.因此，467的数字根是8。应将数字根和（DRS）称为我们数字的各个因子的数字根的总和。下图显示了所有DRS值24.因子分解数字根Sum2x2x2x3 92x3x4 92x2x6 104x6 103x8 112x12 524 6 24的最大数字根和为11.函数mdrs（n）给出n的最大数字根和。所以mdrs（24）= 11。找到Σmdrs（n）为1 <n <1,000,000。
+Recall that the digital root of a number, in base 10, is found by adding together the digits of that number, and repeating that process until a number is arrived at that is less than 10. Thus the digital root of 467 is 8. We shall call a Digital Root Sum (DRS) the sum of the digital roots of the individual factors of our number. The chart below demonstrates all of the DRS values for 24. FactorisationDigital Root Sum2x2x2x3 92x3x4 92x2x6 104x6 103x8 112x12 524 6The maximum Digital Root Sum of 24 is 11. The function mdrs(n) gives the maximum Digital Root Sum of n. So mdrs(24)=11. Find ∑mdrs(n) for 1 < n < 1,000,000.
# --hints--
-`euler159()`应返回14489159。
+`euler159()` should return 14489159.
```js
assert.strictEqual(euler159(), 14489159);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-16-power-digit-sum.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-16-power-digit-sum.md
index 0a9dac314fe..b71b4800826 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-16-power-digit-sum.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-16-power-digit-sum.md
@@ -1,30 +1,38 @@
---
id: 5900f37d1000cf542c50fe8f
-title: 问题16：电源数字总和
+title: 'Problem 16: Power digit sum'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301791
dashedName: problem-16-power-digit-sum
---
# --description--
-2 <sup>15<!-- sup-->= 32768，其位数之和为3 + 2 + 7 + 6 + 8 = 26。 2 <sup><code>指数<!-- code--><!-- sup-->的数字总和是多少？</code></sup></sup>
+2<sup>15</sup> = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
+
+What is the sum of the digits of the number 2<sup><code>exponent</code></sup>?
# --hints--
-`powerDigitSum(15)`应该返回26。
+`powerDigitSum(15)` should return a number.
+
+```js
+assert(typeof powerDigitSum(15) === 'number');
+```
+
+`powerDigitSum(15)` should return 26.
```js
assert.strictEqual(powerDigitSum(15), 26);
```
-`powerDigitSum(128)`应该返回166。
+`powerDigitSum(128)` should return 166.
```js
assert.strictEqual(powerDigitSum(128), 166);
```
-`powerDigitSum(1000)`应返回1366。
+`powerDigitSum(1000)` should return 1366.
```js
assert.strictEqual(powerDigitSum(1000), 1366);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-160-factorial-trailing-digits.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-160-factorial-trailing-digits.md
index 656ceb6da5d..c7319ba84c3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-160-factorial-trailing-digits.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-160-factorial-trailing-digits.md
@@ -1,18 +1,28 @@
---
id: 5900f40d1000cf542c50ff1f
-title: 问题160：因子尾随数字
+title: 'Problem 160: Factorial trailing digits'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301794
dashedName: problem-160-factorial-trailing-digits
---
# --description--
-对于任何N，让f（N）为N！中尾随零之前的最后五位数。例如，9！ = 362880所以f（9）= 36288 10！ = 3628800所以f（10）= 36288 20！ = 2432902008176640000所以f（20）= 17664查找f（1,000,000,000,000）
+For any N, let f(N) be the last five digits before the trailing zeroes in N!.
+
+For example,
+
+9! = 362880 so f(9)=36288
+
+10! = 3628800 so f(10)=36288
+
+20! = 2432902008176640000 so f(20)=17664
+
+Find f(1,000,000,000,000)
# --hints--
-`euler160()`应返回16576。
+`euler160()` should return 16576.
```js
assert.strictEqual(euler160(), 16576);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-161-triominoes.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-161-triominoes.md
index d1a9da60dd7..ea369bae60d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-161-triominoes.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-161-triominoes.md
@@ -1,24 +1,26 @@
---
id: 5900f40d1000cf542c50ff20
-title: 问题161：Triominoes
+title: 'Problem 161: Triominoes'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301795
dashedName: problem-161-triominoes
---
# --description--
-三角形是由三个通过边缘连接的正方形组成的形状。有两种基本形式：
+A triomino is a shape consisting of three squares joined via the edges.
-如果考虑所有可能的方向，则有六个：
+There are two basic forms:
-任何n×m网格的nxm可以被3整除，可以用三角形平铺。如果我们考虑通过反射或从另一个平铺旋转获得的倾斜不同，有41种方式可以使用三角形平铺2乘9的网格：
+If all possible orientations are taken into account there are six:
-有多少种方式可以通过三角形以这种方式平铺9乘12的网格？
+Any n by m grid for which nxm is divisible by 3 can be tiled with triominoes. If we consider tilings that can be obtained by reflection or rotation from another tiling as different there are 41 ways a 2 by 9 grid can be tiled with triominoes:
+
+In how many ways can a 9 by 12 grid be tiled in this way by triominoes?
# --hints--
-`euler161()`应该返回20574308184277972。
+`euler161()` should return 20574308184277972.
```js
assert.strictEqual(euler161(), 20574308184277972);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-162-hexadecimal-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-162-hexadecimal-numbers.md
index c1c5e3600f8..53a6bcd7722 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-162-hexadecimal-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-162-hexadecimal-numbers.md
@@ -1,18 +1,32 @@
---
id: 5900f40e1000cf542c50ff21
-title: 问题162：十六进制数
+title: 'Problem 162: Hexadecimal numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301796
dashedName: problem-162-hexadecimal-numbers
---
# --description--
-在十六进制数中，系统使用16个不同的数字表示数字：0,1,2,3,4,5,6,7,8,9，A，B，C，D，E，F写入时的十六进制数AF在十进制数系统中等于10x16 + 15 = 175。在3位十六进制数10A，1A0，A10和A01中，数字0,1和A都存在。就像用十进制数写的数字一样，我们写十六进制数而不带前导零。包含最多十六个十六进制数字的十六进制数是多少，所有数字0,1和A至少出现一次？将答案作为十六进制数字给出。 （A，B，C，D，E和F大写，没有任何前导或尾随代码将数字标记为十六进制且没有前导零，例如1A3F而不是：1a3f而不是0x1a3f而不是$ 1A3F而不是＃1A3F而不是0000001A3F）
+In the hexadecimal number system numbers are represented using 16 different digits:
+
+0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
+
+The hexadecimal number AF when written in the decimal number system equals 10x16+15=175.
+
+In the 3-digit hexadecimal numbers 10A, 1A0, A10, and A01 the digits 0,1 and A are all present.
+
+Like numbers written in base ten we write hexadecimal numbers without leading zeroes.
+
+How many hexadecimal numbers containing at most sixteen hexadecimal digits exist with all of the digits 0,1, and A present at least once?
+
+Give your answer as a hexadecimal number.
+
+(A,B,C,D,E and F in upper case, without any leading or trailing code that marks the number as hexadecimal and without leading zeroes , e.g. 1A3F and not: 1a3f and not 0x1a3f and not $1A3F and not #1A3F and not 0000001A3F)
# --hints--
-`euler162()` 应该返回3D58725572C62302。
+`euler162()` should return 3D58725572C62302.
```js
assert.strictEqual(euler162(), '3D58725572C62302');
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-163-cross-hatched-triangles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-163-cross-hatched-triangles.md
index ec3fd3bcc0c..72ed1158b19 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-163-cross-hatched-triangles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-163-cross-hatched-triangles.md
@@ -1,20 +1,20 @@
---
id: 5900f40f1000cf542c50ff22
-title: 问题163：阴影线三角形
+title: 'Problem 163: Cross-hatched triangles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301797
dashedName: problem-163-cross-hatched-triangles
---
# --description--
-考虑等边三角形，其中从每个顶点到相对侧的中间绘制直线，例如在下面草图中的1号三角形中。
+Consider an equilateral triangle in which straight lines are drawn from each vertex to the middle of the opposite side, such as in the size 1 triangle in the sketch below.
-现在可以在该三角形中观察到具有不同形状或大小或方向或位置的十六个三角形。使用大小为1的三角形作为构建块，可以形成更大的三角形，例如上面草图中的大小为2的三角形。现在可以在该尺寸2三角形中观察到具有不同形状或尺寸或方向或位置的一百四十个三角形。可以观察到，2号三角形包含4个1号三角形构造块。大小为3的三角形将包含9个大小为1的三角形构建块，因此大小为n的三角形将包含n2个大小为1的三角形构建块。如果我们将T（n）表示为大小为n的三角形中存在的三角形的数量，则T（1）= 16 T（2）= 104 Find T（36）。
+Sixteen triangles of either different shape or size or orientation or location can now be observed in that triangle. Using size 1 triangles as building blocks, larger triangles can be formed, such as the size 2 triangle in the above sketch. One-hundred and four triangles of either different shape or size or orientation or location can now be observed in that size 2 triangle. It can be observed that the size 2 triangle contains 4 size 1 triangle building blocks. A size 3 triangle would contain 9 size 1 triangle building blocks and a size n triangle would thus contain n2 size 1 triangle building blocks. If we denote T(n) as the number of triangles present in a triangle of size n, then T(1) = 16 T(2) = 104 Find T(36).
# --hints--
-`euler163()`应返回343047。
+`euler163()` should return 343047.
```js
assert.strictEqual(euler163(), 343047);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-164-numbers-for-which-no-three-consecutive-digits-have-a-sum-greater-than-a-given-value.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-164-numbers-for-which-no-three-consecutive-digits-have-a-sum-greater-than-a-given-value.md
index 65441b3267b..8082f00bff3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-164-numbers-for-which-no-three-consecutive-digits-have-a-sum-greater-than-a-given-value.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-164-numbers-for-which-no-three-consecutive-digits-have-a-sum-greater-than-a-given-value.md
@@ -1,19 +1,20 @@
---
id: 5900f4111000cf542c50ff23
-title: 问题164：没有三个连续数字的总和大于给定值的数字
+title: >-
+ Problem 164: Numbers for which no three consecutive digits have a sum greater than a given value
challengeType: 5
-videoUrl: ''
+forumTopicId: 301798
dashedName: >-
problem-164-numbers-for-which-no-three-consecutive-digits-have-a-sum-greater-than-a-given-value
---
# --description--
-存在多少20个数字n（没有任何前导零），使得n的三个连续数字的总和不大于9？
+How many 20 digit numbers n (without any leading zero) exist such that no three consecutive digits of n have a sum greater than 9?
# --hints--
-`euler164()`应该返回378158756814587。
+`euler164()` should return 378158756814587.
```js
assert.strictEqual(euler164(), 378158756814587);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-165-intersections.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-165-intersections.md
index 4ca468809d9..a98373c3384 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-165-intersections.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-165-intersections.md
@@ -1,20 +1,26 @@
---
id: 5900f4111000cf542c50ff24
-title: 问题165：交叉口
+title: 'Problem 165: Intersections'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301799
dashedName: problem-165-intersections
---
# --description--
-段由其两个端点唯一定义。通过考虑平面几何中的两个线段，存在三种可能性：段具有零点，一个点或无限多个共同点。此外，当两个段恰好具有一个共同点时，可能是该公共点是任一段或两者的端点的情况。如果两个段的公共点不是任一段的端点，则它是两个段的内点。如果T是L1和L2的唯一公共点，则我们将两个段L1和L2的公共点T称为L1和L2的真实交点，并且T是两个段的内点。
+A segment is uniquely defined by its two endpoints. By considering two line segments in plane geometry there are three possibilities:
-考虑三个段L1，L2和L3：L1：（27,44）到（12,32）L2：（46,53）到（17,62）L3：（46,70）到（22,40）可以证实线段L2和L3具有真实的交叉点。我们注意到，作为L3的终点之一：（22,40）位于L1上，这不被认为是真正的交点。 L1和L2没有共同点。因此，在三个线段中，我们找到一个真正的交叉点。现在让我们对5000个线段进行相同的操作。为此，我们使用所谓的“Blum Blum Shub”伪随机数生成器生成20000个数字。 s0 = 290797 sn + 1 = sn×sn（modulo 50515093）tn = sn（modulo 500）为了创建每个线段，我们使用四个连续的数字tn。也就是说，第一个线段由下式给出：（t1，t2）到（t3，t4）根据上述发生器计算的前四个数字应该是：27,144,12和232.因此第一个线段是（ 27,144）至（12,232）。在5000个线段中发现了多少个不同的真实交叉点？
+the segments have zero points, one point, or infinitely many points in common.
+
+Moreover when two segments have exactly one point in common it might be the case that that common point is an endpoint of either one of the segments or of both. If a common point of two segments is not an endpoint of either of the segments it is an interior point of both segments.
+
+We will call a common point T of two segments L1 and L2 a true intersection point of L1 and L2 if T is the only common point of L1 and L2 and T is an interior point of both segments.
+
+Consider the three segments L1, L2, and L3: L1: (27, 44) to (12, 32) L2: (46, 53) to (17, 62) L3: (46, 70) to (22, 40) It can be verified that line segments L2 and L3 have a true intersection point. We note that as the one of the end points of L3: (22,40) lies on L1 this is not considered to be a true point of intersection. L1 and L2 have no common point. So among the three line segments, we find one true intersection point. Now let us do the same for 5000 line segments. To this end, we generate 20000 numbers using the so-called "Blum Blum Shub" pseudo-random number generator. s0 = 290797 sn+1 = sn×sn (modulo 50515093) tn = sn (modulo 500) To create each line segment, we use four consecutive numbers tn. That is, the first line segment is given by: (t1, t2) to (t3, t4) The first four numbers computed according to the above generator should be: 27, 144, 12 and 232. The first segment would thus be (27,144) to (12,232). How many distinct true intersection points are found among the 5000 line segments?
# --hints--
-`euler165()`应该返回2868868。
+`euler165()` should return 2868868.
```js
assert.strictEqual(euler165(), 2868868);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-166-criss-cross.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-166-criss-cross.md
index ddd71ac7697..9f1db5cea49 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-166-criss-cross.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-166-criss-cross.md
@@ -1,26 +1,26 @@
---
id: 5900f4131000cf542c50ff25
-title: 问题166：克里斯十字架
+title: 'Problem 166: Criss Cross'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301800
dashedName: problem-166-criss-cross
---
# --description--
-4x4网格填充数字d，0≤d≤9。
+A 4x4 grid is filled with digits d, 0 ≤ d ≤ 9.
-可以看出，在网格中
+It can be seen that in the grid
6 3 3 0 5 0 4 3 0 7 1 4 1 2 4 5
-每行和每列的总和值为12.此外，每个对角线的总和也是12。
+the sum of each row and each column has the value 12. Moreover the sum of each diagonal is also 12.
-在多少种方法中，您可以使用数字d，0≤d≤9填充4x4网格，以便每行，每列和两个对角线具有相同的总和？
+In how many ways can you fill a 4x4 grid with the digits d, 0 ≤ d ≤ 9 so that each row, each column, and both diagonals have the same sum?
# --hints--
-`euler166()`应返回7130034。
+`euler166()` should return 7130034.
```js
assert.strictEqual(euler166(), 7130034);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-167-investigating-ulam-sequences.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-167-investigating-ulam-sequences.md
index 5a409efa5bc..6f30b4b2729 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-167-investigating-ulam-sequences.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-167-investigating-ulam-sequences.md
@@ -1,18 +1,28 @@
---
id: 5900f4141000cf542c50ff26
-title: 问题167：研究Ulam序列
+title: 'Problem 167: Investigating Ulam sequences'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301801
dashedName: problem-167-investigating-ulam-sequences
---
# --description--
-对于两个正整数a和b，Ulam序列U（a，b）由U（a，b）1 = a，U（a，b）2 = b定义，对于k> 2，U（a，b） ）k是大于U（a，b）（k-1）的最小整数，它可以用一种方式写成U（a，b）的两个不同的先前成员的总和。例如，序列U（1,2）以1,2,3 = 1 + 2,4 = 1 + 3,6 = 2 + 4,8 = 2 + 6,11 = 3 + 8开始; 5不属于它，因为5 = 1 + 4 = 2 + 3有两个表示作为前两个成员的总和，同样7 = 1 + 6 = 3 + 4.找到ΣU（2,2n + 1）k为2≤n≤10，其中k = 1011。
+For two positive integers a and b, the Ulam sequence U(a,b) is defined by U(a,b)1 = a, U(a,b)2 = b and for k > 2,
+
+U(a,b)k is the smallest integer greater than U(a,b)(k-1) which can be written in exactly one way as the sum of two distinct previous members of U(a,b).
+
+For example, the sequence U(1,2) begins with
+
+1, 2, 3 = 1 + 2, 4 = 1 + 3, 6 = 2 + 4, 8 = 2 + 6, 11 = 3 + 8;
+
+5 does not belong to it because 5 = 1 + 4 = 2 + 3 has two representations as the sum of two previous members, likewise 7 = 1 + 6 = 3 + 4.
+
+Find ∑U(2,2n+1)k for 2 ≤ n ≤10, where k = 1011.
# --hints--
-`euler167()`应该返回3916160068885。
+`euler167()` should return 3916160068885.
```js
assert.strictEqual(euler167(), 3916160068885);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-168-number-rotations.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-168-number-rotations.md
index f29553ba6d6..9ad5f78e64d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-168-number-rotations.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-168-number-rotations.md
@@ -1,18 +1,24 @@
---
id: 5900f4151000cf542c50ff27
-title: 问题168：数字轮换
+title: 'Problem 168: Number Rotations'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301802
dashedName: problem-168-number-rotations
---
# --description--
-考虑数字142857.我们可以通过将最后一个数字（7）移动到它的前面来右旋这个数字，给我们714285.可以验证714285 = 5×142857。这证明了142857的一个不寻常的属性：它是右旋的除数。找到具有此属性的所有整数n，10 <n <10100之和的最后5位数。
+Consider the number 142857. We can right-rotate this number by moving the last digit (7) to the front of it, giving us 714285.
+
+It can be verified that 714285=5×142857.
+
+This demonstrates an unusual property of 142857: it is a divisor of its right-rotation.
+
+Find the last 5 digits of the sum of all integers n, 10 < n < 10100, that have this property.
# --hints--
-`euler168()`应返回`euler168()` 。
+`euler168()` should return 59206.
```js
assert.strictEqual(euler168(), 59206);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-169-exploring-the-number-of-different-ways-a-number-can-be-expressed-as-a-sum-of-powers-of-2.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-169-exploring-the-number-of-different-ways-a-number-can-be-expressed-as-a-sum-of-powers-of-2.md
index 9b62b318865..e0bf8ea804d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-169-exploring-the-number-of-different-ways-a-number-can-be-expressed-as-a-sum-of-powers-of-2.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-169-exploring-the-number-of-different-ways-a-number-can-be-expressed-as-a-sum-of-powers-of-2.md
@@ -1,19 +1,32 @@
---
id: 5900f4151000cf542c50ff28
-title: 问题169：探索数字可以表示为2的幂之和的不同方式的数量
+title: >-
+ Problem 169: Exploring the number of different ways a number can be expressed as a sum of powers of 2
challengeType: 5
-videoUrl: ''
+forumTopicId: 301803
dashedName: >-
problem-169-exploring-the-number-of-different-ways-a-number-can-be-expressed-as-a-sum-of-powers-of-2
---
# --description--
-将f（0）= 1和f（n）定义为不同方式的数量n可以表示为2的整数幂之和，使用每个幂不超过两次。例如，f（10）= 5因为有五种不同的表达方式10：1 + 1 + 8 1 + 1 + 4 + 41 + 1 + 2 + 2 + 4 2 + 4 + 4 2 + 8什么是f （1025）？
+Define f(0)=1 and f(n) to be the number of different ways n can be expressed as a sum of integer powers of 2 using each power no more than twice.
+
+For example, f(10)=5 since there are five different ways to express 10:
+
+1 + 1 + 8
+
+1 + 1 + 4 + 41 + 1 + 2 + 2 + 4
+
+2 + 4 + 4
+
+2 + 8
+
+What is f(1025)?
# --hints--
-`euler169()`应该返回178653872807。
+`euler169()` should return 178653872807.
```js
assert.strictEqual(euler169(), 178653872807);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-17-number-letter-counts.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-17-number-letter-counts.md
index eacd7697702..2b096d1d78b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-17-number-letter-counts.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-17-number-letter-counts.md
@@ -1,30 +1,40 @@
---
id: 5900f37d1000cf542c50fe90
-title: 问题17：数字字母计数
+title: 'Problem 17: Number letter counts'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301804
dashedName: problem-17-number-letter-counts
---
# --description--
-如果数字1到5用文字写出：一，二，三，四，五，则总共使用3 + 3 + 5 + 4 + 4 = 19个字母。如果从1到包含`limit`所有数字都用文字写出，那么会使用多少个字母？ **注意：** 不要计算空格或连字符。例如，342（三百四十二）包含23个字母，115（一百一十五）包含20个字母。在写出数字时使用“和”符合英国的用法。
+If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
+
+If all the numbers from 1 to given `limit` inclusive were written out in words, how many letters would be used?
+
+**Note:** Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.
# --hints--
-`numberLetterCounts(5)`应返回19。
+`numberLetterCounts(5)` should return a number.
+
+```js
+assert(typeof numberLetterCounts(5) === 'number');
+```
+
+`numberLetterCounts(5)` should return 19.
```js
assert.strictEqual(numberLetterCounts(5), 19);
```
-`numberLetterCounts(150)`应该返回1903。
+`numberLetterCounts(150)` should return 1903.
```js
assert.strictEqual(numberLetterCounts(150), 1903);
```
-`numberLetterCounts(1000)`应该返回21124。
+`numberLetterCounts(1000)` should return 21124.
```js
assert.strictEqual(numberLetterCounts(1000), 21124);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-170-find-the-largest-0-to-9-pandigital-that-can-be-formed-by-concatenating-products.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-170-find-the-largest-0-to-9-pandigital-that-can-be-formed-by-concatenating-products.md
index 8a64702485e..7a7e20558b5 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-170-find-the-largest-0-to-9-pandigital-that-can-be-formed-by-concatenating-products.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-170-find-the-largest-0-to-9-pandigital-that-can-be-formed-by-concatenating-products.md
@@ -1,27 +1,28 @@
---
id: 5900f4161000cf542c50ff29
-title: 问题170：找到可以通过连接产品形成的最大的0到9个pandigital
+title: >-
+ Problem 170: Find the largest 0 to 9 pandigital that can be formed by concatenating products
challengeType: 5
-videoUrl: ''
+forumTopicId: 301805
dashedName: >-
problem-170-find-the-largest-0-to-9-pandigital-that-can-be-formed-by-concatenating-products
---
# --description--
-取数字6并乘以1273和9854：
+Take the number 6 and multiply it by each of 1273 and 9854:
-6×1273 = 7638 6×9854 = 59124
+6 × 1273 = 7638 6 × 9854 = 59124
-通过连接这些产品，我们得到1到9 pandigital 763859124.我们将763859124称为“6和（1273,9854）”的连接产品。另请注意，输入数字612739854的串联也是1到9 pandigital。
+By concatenating these products we get the 1 to 9 pandigital 763859124. We will call 763859124 the "concatenated product of 6 and (1273,9854)". Notice too, that the concatenation of the input numbers, 612739854, is also 1 to 9 pandigital.
-对于0到9个pandigital数字也可以这样做。
+The same can be done for 0 to 9 pandigital numbers.
-什么是具有两个或更多其他整数的整数的0到9 pandigital 10位连接产品，这样输入数字的串联也是0到9 pandigital 10位数字？
+What is the largest 0 to 9 pandigital 10-digit concatenated product of an integer with two or more other integers, such that the concatenation of the input numbers is also a 0 to 9 pandigital 10-digit number?
# --hints--
-`euler170()`应返回9857164023。
+`euler170()` should return 9857164023.
```js
assert.strictEqual(euler170(), 9857164023);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-171-finding-numbers-for-which-the-sum-of-the-squares-of-the-digits-is-a-square.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-171-finding-numbers-for-which-the-sum-of-the-squares-of-the-digits-is-a-square.md
index a9fd5ee9c45..2b355e111f4 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-171-finding-numbers-for-which-the-sum-of-the-squares-of-the-digits-is-a-square.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-171-finding-numbers-for-which-the-sum-of-the-squares-of-the-digits-is-a-square.md
@@ -1,19 +1,28 @@
---
id: 5900f4181000cf542c50ff2a
-title: 问题171：找到数字的平方和为正方形的数字
+title: >-
+ Problem 171: Finding numbers for which the sum of the squares of the digits is a square
challengeType: 5
-videoUrl: ''
+forumTopicId: 301806
dashedName: >-
problem-171-finding-numbers-for-which-the-sum-of-the-squares-of-the-digits-is-a-square
---
# --description--
-对于正整数n，令f（n）为n的数字（在基数10中）的平方和，例如f（3）= 32 = 9，f（25）= 22 + 52 = 4 + 25 = 29，f（442）= 42 + 42 + 22 = 16 + 16 + 4 = 36找到所有n的总和的最后九位数，0 <n <1020，使得f（n）是一个完美的平方。
+For a positive integer n, let f(n) be the sum of the squares of the digits (in base 10) of n, e.g.
+
+f(3) = 32 = 9,
+
+f(25) = 22 + 52 = 4 + 25 = 29,
+
+f(442) = 42 + 42 + 22 = 16 + 16 + 4 = 36
+
+Find the last nine digits of the sum of all n, 0 < n < 1020, such that f(n) is a perfect square.
# --hints--
-`euler171()`应该返回142989277。
+`euler171()` should return 142989277.
```js
assert.strictEqual(euler171(), 142989277);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-172-investigating-numbers-with-few-repeated-digits.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-172-investigating-numbers-with-few-repeated-digits.md
index daf0142ee1b..a25ab3224fd 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-172-investigating-numbers-with-few-repeated-digits.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-172-investigating-numbers-with-few-repeated-digits.md
@@ -1,18 +1,18 @@
---
id: 5900f4181000cf542c50ff2b
-title: 问题172：调查重复数字很少的数字
+title: 'Problem 172: Investigating numbers with few repeated digits'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301807
dashedName: problem-172-investigating-numbers-with-few-repeated-digits
---
# --description--
-有多少个18位数字n（没有前导零），n中没有数字出现超过三次？
+How many 18-digit numbers n (without leading zeros) are there such that no digit occurs more than three times in n?
# --hints--
-`euler172()`应返回227485267000992000。
+`euler172()` should return 227485267000992000.
```js
assert.strictEqual(euler172(), 227485267000992000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-173-using-up-to-one-million-tiles-how-many-different-hollow-square-laminae-can-be-formed.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-173-using-up-to-one-million-tiles-how-many-different-hollow-square-laminae-can-be-formed.md
index 424608b02c9..700b710e805 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-173-using-up-to-one-million-tiles-how-many-different-hollow-square-laminae-can-be-formed.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-173-using-up-to-one-million-tiles-how-many-different-hollow-square-laminae-can-be-formed.md
@@ -1,21 +1,22 @@
---
id: 5900f41a1000cf542c50ff2c
-title: 问题173：使用多达一百万个瓷砖可以形成多少个不同的“空心”方形薄片？
+title: >-
+ Problem 173: Using up to one million tiles how many different "hollow" square laminae can be formed?
challengeType: 5
-videoUrl: ''
+forumTopicId: 301808
dashedName: >-
problem-173-using-up-to-one-million-tiles-how-many-different-hollow-square-laminae-can-be-formed
---
# --description--
-我们将方形薄片定义为具有方形“孔”的方形轮廓，使得该形状具有垂直和水平对称性。例如，使用正好三十二平方的瓷砖，我们可以形成两个不同的方形薄片：
+We shall define a square lamina to be a square outline with a square "hole" so that the shape possesses vertical and horizontal symmetry. For example, using exactly thirty-two square tiles we can form two different square laminae:
-使用一百个瓷砖，并且不一定一次使用所有瓷砖，可以形成四十一个不同的方形薄片。使用多达一百万个瓷砖可以形成多少个不同的方形薄片？
+With one-hundred tiles, and not necessarily using all of the tiles at one time, it is possible to form forty-one different square laminae. Using up to one million tiles how many different square laminae can be formed?
# --hints--
-`euler173()`应该返回1572729。
+`euler173()` should return 1572729.
```js
assert.strictEqual(euler173(), 1572729);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-174-counting-the-number-of-hollow-square-laminae-that-can-form-one-two-three-...-distinct-arrangements.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-174-counting-the-number-of-hollow-square-laminae-that-can-form-one-two-three-...-distinct-arrangements.md
index 648de5a8d66..13277919381 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-174-counting-the-number-of-hollow-square-laminae-that-can-form-one-two-three-...-distinct-arrangements.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-174-counting-the-number-of-hollow-square-laminae-that-can-form-one-two-three-...-distinct-arrangements.md
@@ -1,21 +1,24 @@
---
id: 5900f41a1000cf542c50ff2d
-title: 问题174：计算可以形成一个，两个，三个......不同排列的“空心”方形薄片的数量
+title: >-
+ Problem 174: Counting the number of "hollow" square laminae that can form one, two, three, ... distinct arrangements
challengeType: 5
-videoUrl: ''
+forumTopicId: 301809
dashedName: >-
problem-174-counting-the-number-of-hollow-square-laminae-that-can-form-one-two-three-----distinct-arrangements
---
# --description--
-我们将方形薄片定义为具有方形“孔”的方形轮廓，使得该形状具有垂直和水平对称性。给定八个瓷砖，可以仅以一种方式形成薄层：3x3正方形，中间有1x1个孔。但是，使用32个瓷砖可以形成两个不同的薄片。
+We shall define a square lamina to be a square outline with a square "hole" so that the shape possesses vertical and horizontal symmetry.
-如果t表示使用的瓦片数，我们将说t = 8是类型L（1）并且t = 32是类型L（2）。令N（n）为t≤1000000的数，使得t为L（n）型;例如，N（15）= 832.对于1≤n≤10，ΣN（n）是多少？
+Given eight tiles it is possible to form a lamina in only one way: 3x3 square with a 1x1 hole in the middle. However, using thirty-two tiles it is possible to form two distinct laminae.
+
+If t represents the number of tiles used, we shall say that t = 8 is type L(1) and t = 32 is type L(2). Let N(n) be the number of t ≤ 1000000 such that t is type L(n); for example, N(15) = 832. What is ∑ N(n) for 1 ≤ n ≤ 10?
# --hints--
-`euler174()`应该返回209566。
+`euler174()` should return 209566.
```js
assert.strictEqual(euler174(), 209566);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-175-fractions-involving-the-number-of-different-ways-a-number-can-be-expressed-as-a-sum-of-powers-of-2.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-175-fractions-involving-the-number-of-different-ways-a-number-can-be-expressed-as-a-sum-of-powers-of-2.md
index 4a93260b33a..cd3abbf807a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-175-fractions-involving-the-number-of-different-ways-a-number-can-be-expressed-as-a-sum-of-powers-of-2.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-175-fractions-involving-the-number-of-different-ways-a-number-can-be-expressed-as-a-sum-of-powers-of-2.md
@@ -1,23 +1,24 @@
---
id: 5900f41c1000cf542c50ff2e
-title: 问题175：涉及不同方式的数量的分数数字可以表示为2的幂的总和
+title: >-
+ Problem 175: Fractions involving the number of different ways a number can be expressed as a sum of powers of 2
challengeType: 5
-videoUrl: ''
+forumTopicId: 301810
dashedName: >-
problem-175-fractions-involving-the-number-of-different-ways-a-number-can-be-expressed-as-a-sum-of-powers-of-2
---
# --description--
-将f（0）= 1和f（n）定义为将n作为2的幂之和进行写入的方式的数量，其中没有功率发生超过两次。
+Define f(0)=1 and f(n) to be the number of ways to write n as a sum of powers of 2 where no power occurs more than twice.
-例如，f（10）= 5因为有五种不同的表达方式10:10 = 8 + 2 = 8 + 1 + 1 = 4 + 4 + 2 = 4 + 2 + 2 + 1 + 1 = 4 + 4 + 1 + 1
+For example, f(10)=5 since there are five different ways to express 10:10 = 8+2 = 8+1+1 = 4+4+2 = 4+2+2+1+1 = 4+4+1+1
-可以证明，对于每个分数p / q（p> 0，q> 0），存在至少一个整数n，使得f（n）/ f（n-1）= p / q。例如，f（n）/ f（n-1）= 13/17的最小n是241. 241的二进制扩展是11110001.从最高有效位到最低有效位读取这个二进制数有4个1，3个零和1个。我们将字符串4,3,1称为缩短的二进制扩展241.找到最小n的缩短二进制扩展，其中f（n）/ f（n-1）= 123456789/987654321。以逗号分隔的整数给出答案，没有任何空格。
+It can be shown that for every fraction p/q (p>0, q>0) there exists at least one integer n such that f(n)/f(n-1)=p/q. For instance, the smallest n for which f(n)/f(n-1)=13/17 is 241. The binary expansion of 241 is 11110001. Reading this binary number from the most significant bit to the least significant bit there are 4 one's, 3 zeroes and 1 one. We shall call the string 4,3,1 the Shortened Binary Expansion of 241. Find the Shortened Binary Expansion of the smallest n for which f(n)/f(n-1)=123456789/987654321. Give your answer as comma separated integers, without any whitespaces.
# --hints--
-`euler175()`应该返回1,13717420,8。
+`euler175()` should return 1, 13717420, 8.
```js
assert.strictEqual(euler175(), 1, 13717420, 8);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-176-right-angled-triangles-that-share-a-cathetus.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-176-right-angled-triangles-that-share-a-cathetus.md
index 716548a6a70..c3b6c090fed 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-176-right-angled-triangles-that-share-a-cathetus.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-176-right-angled-triangles-that-share-a-cathetus.md
@@ -1,18 +1,20 @@
---
id: 5900f41c1000cf542c50ff2f
-title: 问题176：共用一个cathetus的直角三角形
+title: 'Problem 176: Right-angled triangles that share a cathetus'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301811
dashedName: problem-176-right-angled-triangles-that-share-a-cathetus
---
# --description--
-具有侧面（9,12,15），（12,16,20），（5,12,13）和（12,35,37）的四个直角三角形都具有相等的一个短边（导管）可以证明，没有其他整数侧直角三角形存在，其中一个导管等于12.找到最小整数，可以是恰好47547个不同整数侧直角三角形的cathetus的长度。
+The four right-angled triangles with sides (9,12,15), (12,16,20), (5,12,13) and (12,35,37) all have one of the shorter sides (catheti) equal to 12. It can be shown that no other integer sided right-angled triangle exists with one of the catheti equal to 12.
+
+Find the smallest integer that can be the length of a cathetus of exactly 47547 different integer sided right-angled triangles.
# --hints--
-`euler176()`应返回96818198400000。
+`euler176()` should return 96818198400000.
```js
assert.strictEqual(euler176(), 96818198400000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-177-integer-angled-quadrilaterals.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-177-integer-angled-quadrilaterals.md
index bed2a88b238..2e391146546 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-177-integer-angled-quadrilaterals.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-177-integer-angled-quadrilaterals.md
@@ -1,20 +1,20 @@
---
id: 5900f41e1000cf542c50ff30
-title: 问题177：整数角四边形
+title: 'Problem 177: Integer angled Quadrilaterals'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301812
dashedName: problem-177-integer-angled-quadrilaterals
---
# --description--
-设ABCD为凸四边形，对角线为AC和BD。在每个顶点处，对角线与两侧中的每一侧形成一个角度，从而产生八个角度角。
+Let ABCD be a convex quadrilateral, with diagonals AC and BD. At each vertex the diagonal makes an angle with each of the two sides, creating eight corner angles.
-例如，在顶点A处，两个角度是CAD，CAB。我们称之为四边形，当以“整角四边形”度数度量时，所有八个角度角都具有整数值。整数角度四边形的示例是正方形，其中所有八个角度角都是45°。另一个例子是DAC = 20°，BAC = 60°，ABD = 50°，CBD = 30°，BCA = 40°，DCA = 30°，CDB = 80°，ADB = 50°。非相似整数角度四边形的总数是多少？注意：在计算中，如果计算出的角度在整数值的10-9范围内，则可以假定计算出的角度是整数。
+For example, at vertex A, the two angles are CAD, CAB. We call such a quadrilateral for which all eight corner angles have integer values when measured in degrees an "integer angled quadrilateral". An example of an integer angled quadrilateral is a square, where all eight corner angles are 45°. Another example is given by DAC = 20°, BAC = 60°, ABD = 50°, CBD = 30°, BCA = 40°, DCA = 30°, CDB = 80°, ADB = 50°. What is the total number of non-similar integer angled quadrilaterals? Note: In your calculations you may assume that a calculated angle is integral if it is within a tolerance of 10-9 of an integer value.
# --hints--
-`euler177()`应返回129325。
+`euler177()` should return 129325.
```js
assert.strictEqual(euler177(), 129325);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-178-step-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-178-step-numbers.md
index ba7325f749d..12bad9c5ada 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-178-step-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-178-step-numbers.md
@@ -1,20 +1,26 @@
---
id: 5900f41e1000cf542c50ff31
-title: 问题178：步骤编号
+title: 'Problem 178: Step Numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301813
dashedName: problem-178-step-numbers
---
# --description--
-考虑数字45656.可以看出，45656的每对连续数字的差值为1。每对连续数字的差值为1的数字称为步数。 pandigital数字包含从0到9的每个十进制数字至少一次。
+Consider the number 45656.
-有多少pandigital步数小于1040？
+It can be seen that each pair of consecutive digits of 45656 has a difference of one.
+
+A number for which every pair of consecutive digits has a difference of one is called a step number.
+
+A pandigital number contains every decimal digit from 0 to 9 at least once.
+
+How many pandigital step numbers less than 1040 are there?
# --hints--
-`euler178()`应该返回126461847755。
+`euler178()` should return 126461847755.
```js
assert.strictEqual(euler178(), 126461847755);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-179-consecutive-positive-divisors.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-179-consecutive-positive-divisors.md
index 6b3d080c876..1811def962b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-179-consecutive-positive-divisors.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-179-consecutive-positive-divisors.md
@@ -1,18 +1,18 @@
---
id: 5900f41f1000cf542c50ff32
-title: 问题179：连续的正向除数
+title: 'Problem 179: Consecutive positive divisors'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301814
dashedName: problem-179-consecutive-positive-divisors
---
# --description--
-求整数1 <n <107，其中n和n + 1具有相同的正除数。例如，14具有正除数1,2,7,14而15具有1,3,5,15。
+Find the number of integers 1 < n < 107, for which n and n + 1 have the same number of positive divisors. For example, 14 has the positive divisors 1, 2, 7, 14 while 15 has 1, 3, 5, 15.
# --hints--
-`euler179()`应返回986262。
+`euler179()` should return 986262.
```js
assert.strictEqual(euler179(), 986262);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-18-maximum-path-sum-i.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-18-maximum-path-sum-i.md
index e5cbe9b1a08..2ad9a7a7a6e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-18-maximum-path-sum-i.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-18-maximum-path-sum-i.md
@@ -1,17 +1,27 @@
---
id: 5900f37e1000cf542c50fe91
-title: 问题18：最大路径总和I.
+title: 'Problem 18: Maximum path sum I'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301815
dashedName: problem-18-maximum-path-sum-i
---
# --description--
-通过启动在低于该三角形的顶部和移动到相邻的数字下面的行中，最大总从上到下为23 **3**
-**7** 4
-2 **4** 6
-8 5 **9** 3也就是说，3 + 7 + 4 + 9 = 23.找到下面三角形从上到下的最大总数： 75
+By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
+
+<span style='display: block; text-align: center;'>
+ <strong style='color: red;'>3</strong><br>
+ <strong style='color: red;'>7</strong> 4<br>
+ 2 <strong style='color: red;'>4</strong> 6<br>
+ 8 5 <strong style='color: red;'>9</strong> 3
+</span>
+
+That is, 3 + 7 + 4 + 9 = 23.
+
+Find the maximum total from top to bottom of the triangle below:
+
+75
95 64
17 47 82
18 35 87 10
@@ -25,17 +35,25 @@ dashedName: problem-18-maximum-path-sum-i
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
-04 62 98 27 23 09 70 98 73 93 38 53 60 04 23 **注意：** 由于只有16384条路线，因此可以通过尝试每条路线来解决此问题。然而，问题67，与包含一百行的三角形是同样的挑战;它无法通过蛮力解决，需要一种聪明的方法！ ; O）
+04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
+
+**NOTE:** As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)
# --hints--
-`maximumPathSumI(testTriangle)`应该返回23。
+`maximumPathSumI(testTriangle)` should return a number.
+
+```js
+assert(typeof maximumPathSumI(testTriangle) === 'number');
+```
+
+`maximumPathSumI(testTriangle)` should return 23.
```js
assert.strictEqual(maximumPathSumI(testTriangle), 23);
```
-`maximumPathSumI(numTriangle)`应该返回1074。
+`maximumPathSumI(numTriangle)` should return 1074.
```js
assert.strictEqual(maximumPathSumI(numTriangle), 1074);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-180-rational-zeros-of-a-function-of-three-variables.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-180-rational-zeros-of-a-function-of-three-variables.md
index ed63edfbc4f..3b46677319d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-180-rational-zeros-of-a-function-of-three-variables.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-180-rational-zeros-of-a-function-of-three-variables.md
@@ -1,18 +1,34 @@
---
id: 5900f4201000cf542c50ff33
-title: 问题180：三个变量函数的有理零点
+title: 'Problem 180: Rational zeros of a function of three variables'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301816
dashedName: problem-180-rational-zeros-of-a-function-of-three-variables
---
# --description--
-对于任何整数n，考虑三个函数f1，n（x，y，z）= xn + 1 + yn + 1 - zn + 1f2，n（x，y，z）=（xy + yz + zx）\*（ xn-1 + yn-1-zn-1）f3，n（x，y，z）= xyz \*（xn-2 + yn-2-zn-2）及其组合fn（x，y，z）= f1，n（x，y，z）+ f2，n（x，y，z） - f3，n（x，y，z）如果是x，我们将（x，y，z）称为k阶的黄金三元组， y和z都是形式为a / b的有理数，0 <a <b≤k且存在（至少）一个整数n，因此fn（x，y，z）= 0.设s（x ，y，z）= x + y + z。设t = u / v是所有不同s（x，y，z）的所有黄金三元组（x，y，z）的总和。所有s（x，y，z）和t必须在减少形式。找到你+ v。
+For any integer n, consider the three functions
+
+f1,n(x,y,z) = xn+1 + yn+1 − zn+1f2,n(x,y,z) = (xy + yz + zx)\*(xn-1 + yn-1 − zn-1)f3,n(x,y,z) = xyz\*(xn-2 + yn-2 − zn-2)
+
+and their combination
+
+fn(x,y,z) = f1,n(x,y,z) + f2,n(x,y,z) − f3,n(x,y,z)
+
+We call (x,y,z) a golden triple of order k if x, y, and z are all rational numbers of the form a / b with
+
+0 < a < b ≤ k and there is (at least) one integer n, so that fn(x,y,z) = 0.
+
+Let s(x,y,z) = x + y + z.
+
+Let t = u / v be the sum of all distinct s(x,y,z) for all golden triples (x,y,z) of order 35. All the s(x,y,z) and t must be in reduced form.
+
+Find u + v.
# --hints--
-`euler180()`应该返回285196020571078980。
+`euler180()` should return 285196020571078980.
```js
assert.strictEqual(euler180(), 285196020571078980);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-181-investigating-in-how-many-ways-objects-of-two-different-colours-can-be-grouped.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-181-investigating-in-how-many-ways-objects-of-two-different-colours-can-be-grouped.md
index 9b68add7874..1aa8a4d0875 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-181-investigating-in-how-many-ways-objects-of-two-different-colours-can-be-grouped.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-181-investigating-in-how-many-ways-objects-of-two-different-colours-can-be-grouped.md
@@ -1,19 +1,26 @@
---
id: 5900f4231000cf542c50ff34
-title: 问题181：调查两种不同颜色的对象可以分组的方式
+title: >-
+ Problem 181: Investigating in how many ways objects of two different colours can be grouped
challengeType: 5
-videoUrl: ''
+forumTopicId: 301817
dashedName: >-
problem-181-investigating-in-how-many-ways-objects-of-two-different-colours-can-be-grouped
---
# --description--
-有三个黑色物体B和一个白色物体W它们可以用7种方式分组：（BBBW）（B，BBW）（B，B，BW）（B，B，B，W）（B，BB，W） （BBB，W）（BB，BW）在六十个黑色物体B和四十个白色物体W的分组方式中有多少种？
+Having three black objects B and one white object W they can be grouped in 7 ways like this:
+
+(BBBW)(B,BBW)(B,B,BW)(B,B,B,W)
+
+(B,BB,W)(BBB,W)(BB,BW)
+
+In how many ways can sixty black objects B and forty white objects W be thus grouped?
# --hints--
-`euler181()`应该返回83735848679360670。
+`euler181()` should return 83735848679360670.
```js
assert.strictEqual(euler181(), 83735848679360670);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-183-maximum-product-of-parts.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-183-maximum-product-of-parts.md
index 7c07bcd6c5d..0dcbca86fd5 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-183-maximum-product-of-parts.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-183-maximum-product-of-parts.md
@@ -1,32 +1,34 @@
---
id: 5900f4231000cf542c50ff36
-title: 问题183：零件的最大产品
+title: 'Problem 183: Maximum product of parts'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301819
dashedName: problem-183-maximum-product-of-parts
---
# --description--
-令N为正整数，并且将N分成k个相等的部分，r = N / k，使得N = r + r + ... + r。设P是这些部分的乘积，P = r×r×...×r = rk。
+Let N be a positive integer and let N be split into k equal parts, r = N/k, so that N = r + r + ... + r.
-例如，如果11被分成五个相等的部分，11 = 2.2 + 2.2 + 2.2 + 2.2 + 2.2，那么P = 2.25 = 51.53632。
+Let P be the product of these parts, P = r × r × ... × r = rk.
-对于给定的N值，设M（N）= Pmax。
+For example, if 11 is split into five equal parts, 11 = 2.2 + 2.2 + 2.2 + 2.2 + 2.2, then P = 2.25 = 51.53632.
-事实证明，N = 11的最大值是通过将11分成4个相等的部分得到的，这导致Pmax =（11/4）4;即，M（11）= 14641/256 = 57.19140625，这是终止小数。
+Let M(N) = Pmax for a given value of N.
-然而，对于N = 8，通过将其分成三个相等的部分来实现最大值，因此M（8）= 512/27，这是非终止小数。
+It turns out that the maximum for N = 11 is found by splitting eleven into four equal parts which leads to Pmax = (11/4)4; that is, M(11) = 14641/256 = 57.19140625, which is a terminating decimal.
-如果M（N）是非终止小数，则令D（N）= N，如果M（N）是终止小数，则D（N）= -N。
+However, for N = 8 the maximum is achieved by splitting it into three equal parts, so M(8) = 512/27, which is a non-terminating decimal.
-例如，5≤N≤100的ΣD（N）是2438。
+Let D(N) = N if M(N) is a non-terminating decimal and D(N) = -N if M(N) is a terminating decimal.
-求ΣD（N）为5≤N≤10000。
+For example, ΣD(N) for 5 ≤ N ≤ 100 is 2438.
+
+Find ΣD(N) for 5 ≤ N ≤ 10000.
# --hints--
-`euler183()`应该返回48861552。
+`euler183()` should return 48861552.
```js
assert.strictEqual(euler183(), 48861552);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-184-triangles-containing-the-origin.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-184-triangles-containing-the-origin.md
index d9a4a2dcfb5..dd5855935b1 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-184-triangles-containing-the-origin.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-184-triangles-containing-the-origin.md
@@ -1,22 +1,24 @@
---
id: 5900f4241000cf542c50ff37
-title: 问题184：包含原点的三角形
+title: 'Problem 184: Triangles containing the origin'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301820
dashedName: problem-184-triangles-containing-the-origin
---
# --description--
-考虑点（x，y）的集合Ir，其中半径为r的圆内部的整数坐标以原点为中心，即x2 + y2 <r2。对于半径为2，I2包含九个点（0,0），（1,0），（1,1），（0,1），（ - 1,1），（ - 1,0），（ -1，-1），（0，-1）和（1，-1）。在I2中有八个三角形具有全部三个顶点，其中包含内部的原点。其中两个如下所示，其他通过旋转从这些中获得。
+Consider the set Ir of points (x,y) with integer coordinates in the interior of the circle with radius r, centered at the origin, i.e. x2 + y2 < r2.
-对于半径为3，有360个三角形包含内部的原点并且所有顶点都在I3中，而对于I5，该数字是10600。
+For a radius of 2, I2 contains the nine points (0,0), (1,0), (1,1), (0,1), (-1,1), (-1,0), (-1,-1), (0,-1) and (1,-1). There are eight triangles having all three vertices in I2 which contain the origin in the interior. Two of them are shown below, the others are obtained from these by rotation.
-有多少个三角形包含内部的原点并且在I105中包含所有三个顶点？
+For a radius of 3, there are 360 triangles containing the origin in the interior and having all vertices in I3 and for I5 the number is 10600.
+
+How many triangles are there containing the origin in the interior and having all three vertices in I105?
# --hints--
-`euler184()`应返回1725323624056。
+`euler184()` should return 1725323624056.
```js
assert.strictEqual(euler184(), 1725323624056);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-185-number-mind.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-185-number-mind.md
index 0a24b57bc55..bd8a8539373 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-185-number-mind.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-185-number-mind.md
@@ -1,26 +1,28 @@
---
id: 5900f4251000cf542c50ff38
-title: 问题185：数字思维
+title: 'Problem 185: Number Mind'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301821
dashedName: problem-185-number-mind
---
# --description--
-游戏Number Mind是众所周知的游戏Master Mind的变种。而不是彩色钉，你必须猜测一个秘密的数字序列。在每次猜测之后，你只会告诉你猜对了多少个正确的数字。所以，如果序列是1234并且你猜到了2036，那么你会被告知你有一个正确的数字;但是，你不会被告知你在错误的地方也有另一个数字。
+The game Number Mind is a variant of the well known game Master Mind.
-例如，给出以下5位秘密序列的猜测，90342; 2正确70794; 0正确39458; 2正确34109; 1正确51545; 2正确12531; 1正确正确序列39542是唯一的。
+Instead of coloured pegs, you have to guess a secret sequence of digits. After each guess you're only told in how many places you've guessed the correct digit. So, if the sequence was 1234 and you guessed 2036, you'd be told that you have one correct digit; however, you would NOT be told that you also have another digit in the wrong place.
-基于以下猜测，
+For instance, given the following guesses for a 5-digit secret sequence, 90342 ;2 correct 70794 ;0 correct 39458 ;2 correct 34109 ;1 correct 51545 ;2 correct 12531 ;1 correct The correct sequence 39542 is unique.
-5616185650518293 2正确3847439647293047 1正确5855462940810587 3正确9742855507068353 3正确4296849643607543 3正确3174248439465858 1正确4513559094146117 2正确7890971548908067 3正确8157356344118483 1正确2615250744386899 2正确8690095851526254 3正确6375711915077050 1正确6913859173121360; 1更正6442889055042768; 2更正2321386104303845; 0更正2326509471271448; 2更正5251583379644322; 2更正1748270476758276; 3更正4895722652190306; 1更正3041631117224635; 3更正1841236454324589; 3更正2659862637316867; 2更正
+Based on the following guesses,
-找到唯一的16位秘密序列。
+5616185650518293 ;2 correct 3847439647293047 ;1 correct 5855462940810587 ;3 correct 9742855507068353 ;3 correct 4296849643607543 ;3 correct 3174248439465858 ;1 correct 4513559094146117 ;2 correct 7890971548908067 ;3 correct 8157356344118483 ;1 correct 2615250744386899 ;2 correct 8690095851526254 ;3 correct 6375711915077050 ;1 correct 6913859173121360 ;1 correct 6442889055042768 ;2 correct 2321386104303845 ;0 correct 2326509471271448 ;2 correct 5251583379644322 ;2 correct 1748270476758276 ;3 correct 4895722652190306 ;1 correct 3041631117224635 ;3 correct 1841236454324589 ;3 correct 2659862637316867 ;2 correct
+
+Find the unique 16-digit secret sequence.
# --hints--
-`euler185()`应该返回4640261571849533。
+`euler185()` should return 4640261571849533.
```js
assert.strictEqual(euler185(), 4640261571849533);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-186-connectedness-of-a-network.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-186-connectedness-of-a-network.md
index 515a80d2d28..119501f8331 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-186-connectedness-of-a-network.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-186-connectedness-of-a-network.md
@@ -1,28 +1,28 @@
---
id: 5900f4281000cf542c50ff39
-title: 问题186：网络的连通性
+title: 'Problem 186: Connectedness of a network'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301822
dashedName: problem-186-connectedness-of-a-network
---
# --description--
-以下是来自拥有100万用户的繁忙电话系统的记录：
+Here are the records from a busy telephone system with one million users:
-RecNrCallerCalled120000710005326001835004393600863701497 .........来电者的电话号码和记录n中的被叫号码是来电者（n）= S2n-1和被叫（n）= S2n，其中S1,2,3，...来来自“Lagged Fibonacci Generator”：
+RecNrCallerCalled120000710005326001835004393600863701497......... The telephone number of the caller and the called number in record n are Caller(n) = S2n-1 and Called(n) = S2n where S1,2,3,... come from the "Lagged Fibonacci Generator":
-对于1≤k≤55，Sk = \[100003 - 200003k + 300007k3]（模1000000）对于56≤k，Sk = \[Sk-24 + Sk-55]（模1000000）
+For 1 ≤ k ≤ 55, Sk = \[100003 - 200003k + 300007k3] (modulo 1000000) For 56 ≤ k, Sk = \[Sk-24 + Sk-55] (modulo 1000000)
-如果Caller（n）= Called（n），则假定用户误操作并且呼叫失败;否则通话成功。
+If Caller(n) = Called(n) then the user is assumed to have misdialled and the call fails; otherwise the call is successful.
-从记录的开头，我们说如果X调用Y，则任何一对用户X和Y都是朋友，反之亦然。类似地，如果X是Y的朋友并且Y是Z的朋友，则X是Z的朋友的朋友;等等更长的链条。
+From the start of the records, we say that any pair of users X and Y are friends if X calls Y or vice-versa. Similarly, X is a friend of a friend of Z if X is a friend of Y and Y is a friend of Z; and so on for longer chains.
-总理的电话号码是524287.经过多少次成功的电话，不计算误操作，99％的用户（包括PM）将成为总理的朋友，朋友的朋友等？
+The Prime Minister's phone number is 524287. After how many successful calls, not counting misdials, will 99% of the users (including the PM) be a friend, or a friend of a friend etc., of the Prime Minister?
# --hints--
-`euler186()`应返回2325629。
+`euler186()` should return 2325629.
```js
assert.strictEqual(euler186(), 2325629);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-187-semiprimes.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-187-semiprimes.md
index 40a6a26879b..043684a9d76 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-187-semiprimes.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-187-semiprimes.md
@@ -1,22 +1,22 @@
---
id: 5900f4291000cf542c50ff3a
-title: 问题187：半致
+title: 'Problem 187: Semiprimes'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301823
dashedName: problem-187-semiprimes
---
# --description--
-复合是包含至少两个素因子的数字。例如，15 = 3×5; 9 = 3×3; 12 = 2×2×3。
+A composite is a number containing at least two prime factors. For example, 15 = 3 × 5; 9 = 3 × 3; 12 = 2 × 2 × 3.
-在30以下有十种复合材料恰好包含两个，不一定是不同的主要因素：4,6,9,10,14,15,21,22,25,26。
+There are ten composites below thirty containing precisely two, not necessarily distinct, prime factors: 4, 6, 9, 10, 14, 15, 21, 22, 25, 26.
-有多少复合整数，n <108，恰好有两个，不一定是不同的素因子？
+How many composite integers, n < 108, have precisely two, not necessarily distinct, prime factors?
# --hints--
-`euler187()`应该返回17427258。
+`euler187()` should return 17427258.
```js
assert.strictEqual(euler187(), 17427258);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-188-the-hyperexponentiation-of-a-number.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-188-the-hyperexponentiation-of-a-number.md
index 1261f9ffddc..8c24a2f8e95 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-188-the-hyperexponentiation-of-a-number.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-188-the-hyperexponentiation-of-a-number.md
@@ -1,20 +1,24 @@
---
id: 5900f4291000cf542c50ff3b
-title: 问题188：数字的过度扩展
+title: 'Problem 188: The hyperexponentiation of a number'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301824
dashedName: problem-188-the-hyperexponentiation-of-a-number
---
# --description--
-由↑^ b或ba表示的正整数b的数字a的过度增强或分解通过以下方式递归地定义：a↑↑1 = a，a↑↑（k + 1）= a（a↑↑k ）。
+The hyperexponentiation or tetration of a number a by a positive integer b, denoted by a↑↑b or ba, is recursively defined by:
-因此我们有例如3↑↑2 = 33 = 27，因此3↑↑3 = 327 = 7625597484987和3↑↑4大致是103.6383346400240996 \* 10 ^ 12。查找1777↑↑1855的最后8位数字。
+a↑↑1 = a,
+
+a↑↑(k+1) = a(a↑↑k).
+
+Thus we have e.g. 3↑↑2 = 33 = 27, hence 3↑↑3 = 327 = 7625597484987 and 3↑↑4 is roughly 103.6383346400240996\*10^12. Find the last 8 digits of 1777↑↑1855.
# --hints--
-`euler188()`应该返回95962097。
+`euler188()` should return 95962097.
```js
assert.strictEqual(euler188(), 95962097);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-189-tri-colouring-a-triangular-grid.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-189-tri-colouring-a-triangular-grid.md
index 33831aabe10..4153d584425 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-189-tri-colouring-a-triangular-grid.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-189-tri-colouring-a-triangular-grid.md
@@ -1,26 +1,26 @@
---
id: 5900f4291000cf542c50ff3c
-title: 问题189：三角网格三色
+title: 'Problem 189: Tri-colouring a triangular grid'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301825
dashedName: problem-189-tri-colouring-a-triangular-grid
---
# --description--
-考虑以下64个三角形的配置：
+Consider the following configuration of 64 triangles:
-我们希望用三种颜色中的一种为每个三角形的内部着色：红色，绿色或蓝色，这样就不会有两个相邻的三角形具有相同的颜色。这种着色应称为有效。这里，如果两个三角形共享边缘，则称它们是相邻的。注意：如果它们只共享一个顶点，那么它们不是邻居。
+We wish to colour the interior of each triangle with one of three colours: red, green or blue, so that no two neighbouring triangles have the same colour. Such a colouring shall be called valid. Here, two triangles are said to be neighbouring if they share an edge. Note: if they only share a vertex, then they are not neighbours.
-例如，以下是上面网格的有效着色：
+For example, here is a valid colouring of the above grid:
-通过旋转或反射从着色C获得的着色C'被认为与C不同，除非两者相同。
+A colouring C' which is obtained from a colouring C by rotation or reflection is considered distinct from C unless the two are identical.
-上述配置有多少种不同的有效着色？
+How many distinct valid colourings are there for the above configuration?
# --hints--
-`euler189()`应该返回10834893628237824。
+`euler189()` should return 10834893628237824.
```js
assert.strictEqual(euler189(), 10834893628237824);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-19-counting-sundays.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-19-counting-sundays.md
index 82aefef0f30..c12d6d1c6b8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-19-counting-sundays.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-19-counting-sundays.md
@@ -1,40 +1,44 @@
---
id: 5900f37f1000cf542c50fe92
-title: 问题19：计算星期日
+title: 'Problem 19: Counting Sundays'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301827
dashedName: problem-19-counting-sundays
---
# --description--
-您将获得以下信息，但您可能更愿意为自己做一些研究。
+You are given the following information, but you may prefer to do some research for yourself.
-- 1900年1月1日是星期一。
-- 九月是三十天
- 四月，六月和十一月。
- 其余的都有三十一个，
- 仅拯救二月，
- 其中有二十八，风雨无阻。
- 在闰年，二十九岁。
-- 闰年发生在任何一年，可被4整除，但除非可被400整除，否则不会在一个世纪上。
-- 在二十世纪的第一个月（1901年1月1日至2000年12月31日），有多少个星期日下降？
+<ul>
+ <li>1 Jan 1900 was a Monday.</li>
+ <li>Thirty days has September,<br>April, June and November.<br>All the rest have thirty-one,<br>Saving February alone,<br>Which has twenty-eight, rain or shine.<br>And on leap years, twenty-nine.</li>
+ <li>A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.</li>
+</ul>
+
+How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?
# --hints--
-`countingSundays(1943, 1946)`应该返回6。
+`countingSundays(1943, 1946)` should return a number.
+
+```js
+assert(typeof countingSundays(1943, 1946) === 'number');
+```
+
+`countingSundays(1943, 1946)` should return 6.
```js
assert.strictEqual(countingSundays(1943, 1946), 6);
```
-`countingSundays(1995, 2000)`应该返回9。
+`countingSundays(1995, 2000)` should return 10.
```js
assert.strictEqual(countingSundays(1995, 2000), 10);
```
-`countingSundays(1901, 2000)`应该返回171。
+`countingSundays(1901, 2000)` should return 171.
```js
assert.strictEqual(countingSundays(1901, 2000), 171);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-190-maximising-a-weighted-product.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-190-maximising-a-weighted-product.md
index 7048e9601b3..143247df800 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-190-maximising-a-weighted-product.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-190-maximising-a-weighted-product.md
@@ -1,22 +1,22 @@
---
id: 5900f42b1000cf542c50ff3d
-title: 问题190：最大化加权产品
+title: 'Problem 190: Maximising a weighted product'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301828
dashedName: problem-190-maximising-a-weighted-product
---
# --description--
-令Sm =（x1，x2，...，xm）为正实数的m元组，其中x1 + x2 + ... + xm = m，其中Pm = x1 \* x22 \* ... \* xmm最大化。
+Let Sm = (x1, x2, ... , xm) be the m-tuple of positive real numbers with x1 + x2 + ... + xm = m for which Pm = x1 \* x22 \* ... \* xmm is maximised.
-例如，可以验证\[P10] = 4112（\[]是整数部分函数）。
+For example, it can be verified that \[P10] = 4112 (\[ ] is the integer part function).
-求Σ\[Pm]为2≤m≤15。
+Find Σ\[Pm] for 2 ≤ m ≤ 15.
# --hints--
-`euler190()`应该返回371048281。
+`euler190()` should return 371048281.
```js
assert.strictEqual(euler190(), 371048281);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-191-prize-strings.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-191-prize-strings.md
index 47bff4d76fc..bd93cdca4c2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-191-prize-strings.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-191-prize-strings.md
@@ -1,26 +1,26 @@
---
id: 5900f42b1000cf542c50ff3e
-title: 问题191：奖品串
+title: 'Problem 191: Prize Strings'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301829
dashedName: problem-191-prize-strings
---
# --description--
-特定学校为出勤和准时良好的儿童提供现金奖励。如果他们连续三天缺席或者不止一次缺席，那么他们将丧失奖品。
+A particular school offers cash rewards to children with good attendance and punctuality. If they are absent for three consecutive days or late on more than one occasion then they forfeit their prize.
-在n天期间，为每个孩子形成三个字符串，包括L's（晚），O（按时）和A'（缺席）。
+During an n-day period a trinary string is formed for each child consisting of L's (late), O's (on time), and A's (absent).
-虽然可以形成为期4天的八十三个三弦组，但恰好有四十三个弦可以获得奖品：
+Although there are eighty-one trinary strings for a 4-day period that can be formed, exactly forty-three strings would lead to a prize:
-OOOO OOOA OOOL OOAO OOAA OOAL OOLO OOLA OAOO OAOA OAOL OAAO OAAL OALO OALA OLOO OLOA OLAO OLAA AOOO AOOA AOOL AOAO AOAA AOO AOLA AOLA AAOO AAOA AAOL AALO AALA ALOO ALOA ALAO ALAA LOOO LOOA LOAO LOAA LAOO LAOA LAAO
+OOOO OOOA OOOL OOAO OOAA OOAL OOLO OOLA OAOO OAOA OAOL OAAO OAAL OALO OALA OLOO OLOA OLAO OLAA AOOO AOOA AOOL AOAO AOAA AOAL AOLO AOLA AAOO AAOA AAOL AALO AALA ALOO ALOA ALAO ALAA LOOO LOOA LOAO LOAA LAOO LAOA LAAO
-在30天内存在多少“奖品”字符串？
+How many "prize" strings exist over a 30-day period?
# --hints--
-`euler191()`应该返回1918080160。
+`euler191()` should return 1918080160.
```js
assert.strictEqual(euler191(), 1918080160);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-192-best-approximations.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-192-best-approximations.md
index cd0ccdf8415..6751cbffaf9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-192-best-approximations.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-192-best-approximations.md
@@ -1,24 +1,26 @@
---
id: 5900f42c1000cf542c50ff3f
-title: 问题192：最佳近似值
+title: 'Problem 192: Best Approximations'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301830
dashedName: problem-192-best-approximations
---
# --description--
-设x是实数。对于分母界限d，x的最佳近似是简化形式的有理数r / s，其中s≤d，使得任何比r更接近x的有理数具有大于d的分母：
+Let x be a real number.
-| P / QX | <| r / sx | ⇒q> d
+A best approximation to x for the denominator bound d is a rational number r/s in reduced form, with s ≤ d, such that any rational number which is closer to x than r/s has a denominator larger than d:
-例如，分母界限20的最佳近似值√13为18/5，分母界限30的最佳近似值为√13为101/28。
+|p/q-x| < |r/s-x| ⇒ q > d
-找到分母1012的最佳逼近的所有分母的总和为√n，其中n不是完美的正方形，1 <n≤100000。
+For example, the best approximation to √13 for the denominator bound 20 is 18/5 and the best approximation to √13 for the denominator bound 30 is 101/28.
+
+Find the sum of all denominators of the best approximations to √n for the denominator bound 1012, where n is not a perfect square and 1 < n ≤ 100000.
# --hints--
-`euler192()`应该返回57060635927998344。
+`euler192()` should return 57060635927998344.
```js
assert.strictEqual(euler192(), 57060635927998344);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-193-squarefree-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-193-squarefree-numbers.md
index f4909fe7742..1b9cc8d4900 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-193-squarefree-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-193-squarefree-numbers.md
@@ -1,20 +1,20 @@
---
id: 5900f42f1000cf542c50ff41
-title: 问题193：无广场数字
+title: 'Problem 193: Squarefree Numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301831
dashedName: problem-193-squarefree-numbers
---
# --description--
-如果没有素数的平方除n，则正整数n称为squarefree，因此1,2,3,5,6,7,10,11是无平方的，但不是4,8,9,12。
+A positive integer n is called squarefree, if no square of a prime divides n, thus 1, 2, 3, 5, 6, 7, 10, 11 are squarefree, but not 4, 8, 9, 12.
-250以下有多少平方数？
+How many squarefree numbers are there below 250?
# --hints--
-`euler193()`应该返回684465067343069。
+`euler193()` should return 684465067343069.
```js
assert.strictEqual(euler193(), 684465067343069);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-194-coloured-configurations.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-194-coloured-configurations.md
index bbbe9e5105e..947d65c7878 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-194-coloured-configurations.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-194-coloured-configurations.md
@@ -1,24 +1,28 @@
---
id: 5900f42f1000cf542c50ff40
-title: 问题194：彩色配置
+title: 'Problem 194: Coloured Configurations'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301832
dashedName: problem-194-coloured-configurations
---
# --description--
-考虑使用单位A：和B：构建的图形，其中单位沿着垂直边缘粘合，如图中所示。
+Consider graphs built with the units A:
-类型（a，b，c）的配置是由单元A和b单元B构成的图，其中图的顶点使用多达c种颜色着色，使得没有两个相邻顶点具有相同的颜色。上面的化合物图是类型（2,2,6）的配置的实例，实际上对于所有c≥4的类型（2,2，c）。
+and B: , where the units are glued along
-令N（a，b，c）为类型（a，b，c）的配置数。例如，N（1,0,3）= 24，N（0,2,4）= 92928，N（2,2,3）= 20736。
+the vertical edges as in the graph .
-找到N（25,75,1984）的最后8位数。
+A configuration of type (a,b,c) is a graph thus built of a units A and b units B, where the graph's vertices are coloured using up to c colours, so that no two adjacent vertices have the same colour. The compound graph above is an example of a configuration of type (2,2,6), in fact of type (2,2,c) for all c ≥ 4.
+
+Let N(a,b,c) be the number of configurations of type (a,b,c). For example, N(1,0,3) = 24, N(0,2,4) = 92928 and N(2,2,3) = 20736.
+
+Find the last 8 digits of N(25,75,1984).
# --hints--
-`euler194()`应该返回61190912。
+`euler194()` should return 61190912.
```js
assert.strictEqual(euler194(), 61190912);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-195-inscribed-circles-of-triangles-with-one-angle-of-60-degrees.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-195-inscribed-circles-of-triangles-with-one-angle-of-60-degrees.md
index 7f3a6e6f390..cd52202435f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-195-inscribed-circles-of-triangles-with-one-angle-of-60-degrees.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-195-inscribed-circles-of-triangles-with-one-angle-of-60-degrees.md
@@ -1,20 +1,28 @@
---
id: 5900f4311000cf542c50ff43
-title: 问题195：内切三角形的圆形，一个角度为60度
+title: 'Problem 195: Inscribed circles of triangles with one angle of 60 degrees'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301833
dashedName: problem-195-inscribed-circles-of-triangles-with-one-angle-of-60-degrees
---
# --description--
-让我们称一个整数边三角形，其中一个角度为60度，一个60度的三角形。设r是这样的60度三角形的内切圆的半径。有1234个60度三角形，其中r≤100。设T（n）是60度三角形的数量，其中r≤n，因此T（100）= 1234，T（1000）= 22767，和T（ 10000）= 359912。
+Let's call an integer sided triangle with exactly one angle of 60 degrees a 60-degree triangle.
-找到T（1053779）。
+Let r be the radius of the inscribed circle of such a 60-degree triangle.
+
+There are 1234 60-degree triangles for which r ≤ 100.
+
+Let T(n) be the number of 60-degree triangles for which r ≤ n, so
+
+T(100) = 1234, T(1000) = 22767, and T(10000) = 359912.
+
+Find T(1053779).
# --hints--
-`euler195()`应返回75085391。
+`euler195()` should return 75085391.
```js
assert.strictEqual(euler195(), 75085391);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-196-prime-triplets.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-196-prime-triplets.md
index 20df0663271..4ead4631ddc 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-196-prime-triplets.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-196-prime-triplets.md
@@ -1,34 +1,34 @@
---
id: 5900f4301000cf542c50ff42
-title: 问题196：Prime三胞胎
+title: 'Problem 196: Prime triplets'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301834
dashedName: problem-196-prime-triplets
---
# --description--
-以下列方式从所有正整数构建三角形：
+Build a triangle from all positive integers in the following way:
-1 2 3 4 5 6 7 8 9 1011 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 2829 30 31 32 33 34 35 3637 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66。 。 。
+1 2 3 4 5 6 7 8 9 1011 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 2829 30 31 32 33 34 35 3637 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 . . .
-每个正整数在三角形中最多有八个邻居。
+Each positive integer has up to eight neighbours in the triangle.
-如果三个素数中的一个具有另外两个作为三角形中的邻居，则一组三个素数被称为素数三元组。
+A set of three primes is called a prime triplet if one of the three primes has the other two as neighbours in the triangle.
-例如，在第二行中，素数2和3是某些素数三元组的元素。
+For example, in the second row, the prime numbers 2 and 3 are elements of some prime triplet.
-如果考虑第8行，它包含两个素数，它们是某些素数三元组的元素，即29和31.如果考虑第9行，它只包含一个素数，它是某个素数三元组的元素：37。
+If row 8 is considered, it contains two primes which are elements of some prime triplet, i.e. 29 and 31. If row 9 is considered, it contains only one prime which is an element of some prime triplet: 37.
-将S（n）定义为行n中素数的总和，它是任何素数三元组的元素。然后S（8）= 60并且S（9）= 37。
+Define S(n) as the sum of the primes in row n which are elements of any prime triplet. Then S(8)=60 and S(9)=37.
-你被给予S（10000）= 950007619。
+You are given that S(10000)=950007619.
-找到S（5678027）+ S（7208785）。
+Find S(5678027) + S(7208785).
# --hints--
-`euler196()`应返回322303240771079940。
+`euler196()` should return 322303240771079940.
```js
assert.strictEqual(euler196(), 322303240771079940);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-197-investigating-the-behaviour-of-a-recursively-defined-sequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-197-investigating-the-behaviour-of-a-recursively-defined-sequence.md
index d2a4a6ccb90..1e2f3fe30d6 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-197-investigating-the-behaviour-of-a-recursively-defined-sequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-197-investigating-the-behaviour-of-a-recursively-defined-sequence.md
@@ -1,20 +1,22 @@
---
id: 5900f4311000cf542c50ff44
-title: 问题197：研究递归定义的序列的行为
+title: 'Problem 197: Investigating the behaviour of a recursively defined sequence'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301835
dashedName: problem-197-investigating-the-behaviour-of-a-recursively-defined-sequence
---
# --description--
-给定函数f（x）=⌊230.403243784-x2⌋×10-9（⌊⌊是floor函数），序列un由u0 = -1和un + 1 = f（un）定义。
+Given is the function f(x) = ⌊230.403243784-x2⌋ × 10-9 ( ⌊ ⌋ is the floor-function),
-找到n + 1012的un + un + 1.在小数点后面给出9位数的答案。
+the sequence un is defined by u0 = -1 and un+1 = f(un).
+
+Find un + un+1 for n = 1012. Give your answer with 9 digits after the decimal point.
# --hints--
-`euler197()`应该返回1.710637717。
+`euler197()` should return 1.710637717.
```js
assert.strictEqual(euler197(), 1.710637717);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-198-ambiguous-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-198-ambiguous-numbers.md
index ceaf1de9d11..a955aa7c37e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-198-ambiguous-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-198-ambiguous-numbers.md
@@ -1,22 +1,22 @@
---
id: 5900f4331000cf542c50ff45
-title: 问题198：模糊数字
+title: 'Problem 198: Ambiguous Numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301836
dashedName: problem-198-ambiguous-numbers
---
# --description--
-对于分母界限d的实数x的最佳近似是有理数r / s（以简化形式），其中s≤d，因此任何比r更接近x的有理数p / q具有q> d。
+A best approximation to a real number x for the denominator bound d is a rational number r/s (in reduced form) with s ≤ d, so that any rational number p/q which is closer to x than r/s has q > d.
-通常，对于所有分母边界唯一确定对实数的最佳近似。但是，有一些例外，例如9/40对分母界限6具有两个最佳近似值1/4和1/5。如果至少有一个分母绑定为x拥有，我们将称为实数x不明确两个最佳近似值。显然，模糊的数字必然是理性的。
+Usually the best approximation to a real number is uniquely determined for all denominator bounds. However, there are some exceptions, e.g. 9/40 has the two best approximations 1/4 and 1/5 for the denominator bound 6. We shall call a real number x ambiguous, if there is at least one denominator bound for which x possesses two best approximations. Clearly, an ambiguous number is necessarily rational.
-有多少个模糊数x = p / q,0 <x <1/100，其分母q是否不超过108？
+How many ambiguous numbers x = p/q, 0 < x < 1/100, are there whose denominator q does not exceed 108?
# --hints--
-`euler198()`应该返回52374425。
+`euler198()` should return 52374425.
```js
assert.strictEqual(euler198(), 52374425);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-199-iterative-circle-packing.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-199-iterative-circle-packing.md
index 7bfb997e10c..0194bbb44d1 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-199-iterative-circle-packing.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-199-iterative-circle-packing.md
@@ -1,25 +1,39 @@
---
id: 5900f4341000cf542c50ff46
-title: 问题199：迭代圆包装
+title: 'Problem 199: Iterative Circle Packing'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301837
dashedName: problem-199-iterative-circle-packing
---
# --description--
-三个相等半径的圆放置在较大的圆内，使得每对圆彼此相切并且内圆不重叠。有四个未覆盖的“间隙”，它们将用更多的切圆迭代地填充。
+Three circles of equal radius are placed inside a larger circle such that each pair of circles is tangent to one another and the inner circles do not overlap. There are four uncovered "gaps" which are to be filled iteratively with more tangent circles.
-在每次迭代中，在每个间隙中放置最大尺寸的圆，这为下一次迭代创建了更多的间隙。经过3次迭代（如图）后，有108个间隙，未被圆圈覆盖的区域部分为0.06790342，四舍五入到小数点后8位。
+<img class="img-responsive center-block" alt="a diagram of non-overlapping concentric circles" src="https://cdn-media-1.freecodecamp.org/project-euler/199-circles-in-circles.gif" style="background-color: white; padding: 10px;" />
-10次迭代后，圆圈没有覆盖哪一部分区域？使用格式x.xxxxxxxx将您的答案四舍五入到小数点后八位。
+At each iteration, a maximally sized circle is placed in each gap, which creates more gaps for the next iteration. After 3 iterations (pictured), there are 108 gaps and the fraction of the area which is not covered by circles is 0.06790342, rounded to eight decimal places.
+
+What fraction of the area is not covered by circles after `n` iterations? Give your answer rounded to eight decimal places using the format x.xxxxxxxx .
# --hints--
-`euler199()`应该返回0.00396087。
+`iterativeCirclePacking(10)` should return a number.
```js
-assert.strictEqual(euler199(), 0.00396087);
+assert(typeof iterativeCirclePacking(10) === 'number');
+```
+
+`iterativeCirclePacking(10)` should return 0.00396087.
+
+```js
+assert.strictEqual(iterativeCirclePacking(10), 0.00396087);
+```
+
+`iterativeCirclePacking(3)` should return 0.06790342.
+
+```js
+assert.strictEqual(iterativeCirclePacking(3), 0.06790342);
```
# --seed--
@@ -46,7 +60,7 @@ function iterativeCirclePacking(n) {
a1 += 3 * getArea(k0, k1, k1, n);
a1 += getArea(k1, k1, k1, n);
let final = ((a0 - a1) / a0).toFixed(8);
-
+
return parseFloat(final);
function getArea(k1, k2, k3, depth) {
if (depth == 0) return 0.0;
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-2-even-fibonacci-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-2-even-fibonacci-numbers.md
index a1d690026be..ecf2a8aa73a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-2-even-fibonacci-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-2-even-fibonacci-numbers.md
@@ -1,69 +1,73 @@
---
id: 5900f36e1000cf542c50fe81
-title: 问题2：斐波那契数列中的偶数
+title: 'Problem 2: Even Fibonacci Numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301838
dashedName: problem-2-even-fibonacci-numbers
---
# --description--
-在斐波那契数列中，每一项都是前两项的和（第一项和第二项除外）。如果从1和2开始，前十项是：
+Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
<div style='text-align: center;'>1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...</div>
-求出斐波那契数列中值是偶数的项的和，至第`n`项（包括第`n`项）为止。
+By considering the terms in the Fibonacci sequence whose values do not exceed `n`, find the sum of the even-valued terms.
# --hints--
-`fiboEvenSum(10)`应该返回188。
+`fiboEvenSum(10)` should return a number.
```js
-assert.strictEqual(
- fiboEvenSum(10),
- 188,
- '<code>fiboEvenSum(10)</code> should return 188.'
-);
+assert(typeof fiboEvenSum(10) === 'number');
```
-`fiboEvenSum(23)`应该返回60696。
+Your function should return an `even` value.
```js
-assert.strictEqual(
- fiboEvenSum(23),
- 60696,
- '<code>fiboEvenSum(23)</code> should return 60696.'
-);
+assert.equal(fiboEvenSum(10) % 2 === 0, true);
```
-`fiboEvenSum(43)`应该返回1485607536。
+Your function should sum the even-valued Fibonacci numbers: `fiboEvenSum(8)` should return 10.
```js
-assert.strictEqual(
- fiboEvenSum(43),
- 1485607536,
- '<code>fiboEvenSum(43)</code> should return 1485607536.'
-);
+assert.strictEqual(fiboEvenSum(8), 10);
```
-您的函数未使用我们的测试值返回正确的结果。
+`fiboEvenSum(10)` should return 10.
```js
-assert.strictEqual(
- fiboEvenSum(18),
- 3382,
- 'Your function is not returning the correct result using our tests values.'
-);
+assert.strictEqual(fiboEvenSum(10), 10);
```
-您的函数应返回`even`数值。
+`fiboEvenSum(34)` should return 44.
```js
-assert.equal(
- fiboEvenSum(31) % 2 === 0,
- true,
- 'Your function should return an <code>even</code> value.'
-);
+assert.strictEqual(fiboEvenSum(34), 44);
+```
+
+`fiboEvenSum(60)` should return 44.
+
+```js
+assert.strictEqual(fiboEvenSum(60), 44);
+```
+
+`fiboEvenSum(1000)` should return 798.
+
+```js
+assert.strictEqual(fiboEvenSum(1000), 798);
+```
+
+`fiboEvenSum(100000)` should return 60696.
+
+```js
+assert.strictEqual(fiboEvenSum(100000), 60696);
+```
+
+`fiboEvenSum(4000000)` should return 4613732.
+
+```js
+assert.strictEqual(fiboEvenSum(4000000), 4613732);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-20-factorial-digit-sum.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-20-factorial-digit-sum.md
index 6fc1350d8a9..579ac97f6f9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-20-factorial-digit-sum.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-20-factorial-digit-sum.md
@@ -1,43 +1,53 @@
---
id: 5900f3801000cf542c50fe93
-title: 问题20：因子数字和
+title: 'Problem 20: Factorial digit sum'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301839
dashedName: problem-20-factorial-digit-sum
---
# --description--
-`n` ！意味着`n` ×（ `n` - 1）×...×3×2×1例如，10！ = 10×9×...×3×2×1 = 3628800，
-和数字10中的数字之和！是3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.找到数字之和`n` ！
+`n`! means `n` × (`n` − 1) × ... × 3 × 2 × 1
+
+For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
+and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
+
+Find the sum of the digits `n`!
# --hints--
-`sumFactorialDigits(10)`应该返回27。
+`sumFactorialDigits(10)` should return a number.
+
+```js
+assert(typeof sumFactorialDigits(10) === 'number');
+```
+
+`sumFactorialDigits(10)` should return 27.
```js
assert.strictEqual(sumFactorialDigits(10), 27);
```
-`sumFactorialDigits(25)`应该返回72。
+`sumFactorialDigits(25)` should return 72.
```js
assert.strictEqual(sumFactorialDigits(25), 72);
```
-`sumFactorialDigits(50)`应该返回216。
+`sumFactorialDigits(50)` should return 216.
```js
assert.strictEqual(sumFactorialDigits(50), 216);
```
-`sumFactorialDigits(75)`应该返回432。
+`sumFactorialDigits(75)` should return 432.
```js
assert.strictEqual(sumFactorialDigits(75), 432);
```
-`sumFactorialDigits(100)`应该返回648。
+`sumFactorialDigits(100)` should return 648.
```js
assert.strictEqual(sumFactorialDigits(100), 648);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-200-find-the-200th-prime-proof-sqube-containing-the-contiguous-sub-string-200.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-200-find-the-200th-prime-proof-sqube-containing-the-contiguous-sub-string-200.md
index f9eeb845ea3..c6458059ce1 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-200-find-the-200th-prime-proof-sqube-containing-the-contiguous-sub-string-200.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-200-find-the-200th-prime-proof-sqube-containing-the-contiguous-sub-string-200.md
@@ -1,25 +1,28 @@
---
id: 5900f4351000cf542c50ff47
-title: 问题200：找到包含连续子串“200”的第200个防爆大概
+title: >-
+ Problem 200: Find the 200th prime-proof sqube containing the contiguous sub-string "200"
challengeType: 5
-videoUrl: ''
+forumTopicId: 301840
dashedName: >-
problem-200-find-the-200th-prime-proof-sqube-containing-the-contiguous-sub-string-200
---
# --description--
-我们将一个sqube定义为一个形式的数字p2q3，其中p和q是不同的素数。例如，200 = 5223或120072949 = 232613。
+We shall define a sqube to be a number of the form, p2q3, where p and q are distinct primes.
-前五个角是72,108,200,392和500。
+For example, 200 = 5223 or 120072949 = 232613.
-有趣的是，200也是第一个不能改变任何单个数字来构成素数的数字;我们称这些数字为主要证据。包含连续子串“200”的下一个主要证明群是1992008。
+The first five squbes are 72, 108, 200, 392, and 500.
-找到包含连续子串“200”的第200个防质量的sqube。
+Interestingly, 200 is also the first number for which you cannot change any single digit to make a prime; we shall call such numbers, prime-proof. The next prime-proof sqube which contains the contiguous sub-string "200" is 1992008.
+
+Find the 200th prime-proof sqube containing the contiguous sub-string "200".
# --hints--
-`euler200()`应该返回229161792008。
+`euler200()` should return 229161792008.
```js
assert.strictEqual(euler200(), 229161792008);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-201-subsets-with-a-unique-sum.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-201-subsets-with-a-unique-sum.md
index c4723e9a2a1..045fd9b2b8b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-201-subsets-with-a-unique-sum.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-201-subsets-with-a-unique-sum.md
@@ -1,26 +1,28 @@
---
id: 5900f4361000cf542c50ff48
-title: 问题201：具有唯一总和的子集
+title: 'Problem 201: Subsets with a unique sum'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301841
dashedName: problem-201-subsets-with-a-unique-sum
---
# --description--
-对于任何数字集A，让sum（A）为A的元素之和。考虑集合B = {1,3,6,8,10,11}。 B的20个子集包含三个元素，它们的总和是：
+For any set A of numbers, let sum(A) be the sum of the elements of A.
-sum（{1,3,6}）= 10，sum（{1,3,8}）= 12，sum（{1,3,10}）= 14，sum（{1,3,11}）= 15，sum（{1,6,8}）= 15，sum（{1,6,10}）= 17，sum（{1,6,11}）= 18，sum（{1,8,10} ）= 19，sum（{1,8,11}）= 20，sum（{1,10,11}）= 22，sum（{3,6,8}）= 17，sum（{3,6， 10}）= 19，sum（{3,6,11}）= 20，sum（{3,8,10}）= 21，sum（{3,8,11}）= 22，sum（{3， 10,11}）= 24，sum（{6,8,10}）= 24，sum（{6,8,11}）= 25，sum（{6,10,11}）= 27，sum（{ 8,10,11}）= 29。
+Consider the set B = {1,3,6,8,10,11}. There are 20 subsets of B containing three elements, and their sums are:
-其中一些总和不止一次出现，有些则是独一无二的。对于集合A，设U（A，k）是A的k元素子集的唯一和的集合，在我们的例子中我们发现U（B，3）= {10,12,14,18,21,25 ，27,29}和和（U（B，3））= 156。
+sum({1,3,6}) = 10, sum({1,3,8}) = 12, sum({1,3,10}) = 14, sum({1,3,11}) = 15, sum({1,6,8}) = 15, sum({1,6,10}) = 17, sum({1,6,11}) = 18, sum({1,8,10}) = 19, sum({1,8,11}) = 20, sum({1,10,11}) = 22, sum({3,6,8}) = 17, sum({3,6,10}) = 19, sum({3,6,11}) = 20, sum({3,8,10}) = 21, sum({3,8,11}) = 22, sum({3,10,11}) = 24, sum({6,8,10}) = 24, sum({6,8,11}) = 25, sum({6,10,11}) = 27, sum({8,10,11}) = 29.
-现在考虑100个元素集S = {12,22，...，1002}。 S具有100891344545564193334812497256 50个元素子集。
+Some of these sums occur more than once, others are unique. For a set A, let U(A,k) be the set of unique sums of k-element subsets of A, in our example we find U(B,3) = {10,12,14,18,21,25,27,29} and sum(U(B,3)) = 156.
-确定所有整数的和，它们是S的50个元素子集中的一个的总和，即求和（U（S，50））。
+Now consider the 100-element set S = {12, 22, ... , 1002}. S has 100891344545564193334812497256 50-element subsets.
+
+Determine the sum of all integers which are the sum of exactly one of the 50-element subsets of S, i.e. find sum(U(S,50)).
# --hints--
-`euler201()`应返回115039000。
+`euler201()` should return 115039000.
```js
assert.strictEqual(euler201(), 115039000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-202-laserbeam.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-202-laserbeam.md
index d686fe45656..c42f68e4147 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-202-laserbeam.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-202-laserbeam.md
@@ -1,24 +1,24 @@
---
id: 5900f4371000cf542c50ff49
-title: 问题202：Laserbeam
+title: 'Problem 202: Laserbeam'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301843
dashedName: problem-202-laserbeam
---
# --description--
-三个镜子以等边三角形的形状排列，其反射表面指向内侧。在三角形的每个顶点处存在无穷小的间隙，激光束可以通过该间隙。
+Three mirrors are arranged in the shape of an equilateral triangle, with their reflective surfaces pointing inwards. There is an infinitesimal gap at each vertex of the triangle through which a laser beam may pass.
-标记顶点A，B和C.有两种方式可以使激光束进入顶点C，从11个表面反弹，然后通过相同的顶点退出：下面显示了一种方式;另一方面与此相反。
+Label the vertices A, B and C. There are 2 ways in which a laser beam may enter vertex C, bounce off 11 surfaces, then exit through the same vertex: one way is shown below; the other is the reverse of that.
-有80840种方式，激光束可以进入顶点C，从1000001表面反弹，然后通过相同的顶点退出。
+There are 80840 ways in which a laser beam may enter vertex C, bounce off 1000001 surfaces, then exit through the same vertex.
-激光束在顶点C进入的方式有多少，从12017639147表面反弹，然后通过相同的顶点退出？
+In how many ways can a laser beam enter at vertex C, bounce off 12017639147 surfaces, then exit through the same vertex?
# --hints--
-`euler202()`应返回1209002624。
+`euler202()` should return 1209002624.
```js
assert.strictEqual(euler202(), 1209002624);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-203-squarefree-binomial-coefficients.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-203-squarefree-binomial-coefficients.md
index d44230356a8..4ecf609e4b3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-203-squarefree-binomial-coefficients.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-203-squarefree-binomial-coefficients.md
@@ -1,26 +1,26 @@
---
id: 5900f4381000cf542c50ff4a
-title: 问题203：无平方二项系数
+title: 'Problem 203: Squarefree Binomial Coefficients'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301844
dashedName: problem-203-squarefree-binomial-coefficients
---
# --description--
-二项式系数nCk可以以三角形排列，Pascal的三角形，如下所示：
+The binomial coefficients nCk can be arranged in triangular form, Pascal's triangle, like this:
111121133114641151010511615201561172135352171 .........
-可以看出Pascal三角形的前八行包含十二个不同的数字：1,2,3,4,5,6,7,10,15,20,21和35。
+It can be seen that the first eight rows of Pascal's triangle contain twelve distinct numbers: 1, 2, 3, 4, 5, 6, 7, 10, 15, 20, 21 and 35.
-如果没有素数的平方除n，则正整数n称为squarefree。在Pascal三角形的前八行中的十二个不同数字中，除了4和20之外的所有数字都是无方形的。前八行中不同的无平方数字的总和为105。
+A positive integer n is called squarefree if no square of a prime divides n. Of the twelve distinct numbers in the first eight rows of Pascal's triangle, all except 4 and 20 are squarefree. The sum of the distinct squarefree numbers in the first eight rows is 105.
-找到Pascal三角形的前51行中不同的无平方数字的总和。
+Find the sum of the distinct squarefree numbers in the first 51 rows of Pascal's triangle.
# --hints--
-`euler203()`应该返回34029210557338。
+`euler203()` should return 34029210557338.
```js
assert.strictEqual(euler203(), 34029210557338);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-204-generalised-hamming-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-204-generalised-hamming-numbers.md
index 9b5a5627b3c..4d81e08eea0 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-204-generalised-hamming-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-204-generalised-hamming-numbers.md
@@ -1,26 +1,26 @@
---
id: 5900f4381000cf542c50ff4b
-title: 问题204：广义汉明数
+title: 'Problem 204: Generalised Hamming Numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301845
dashedName: problem-204-generalised-hamming-numbers
---
# --description--
-海明数是一个正数，其素数不大于5。
+A Hamming number is a positive number which has no prime factor larger than 5.
-因此，前几个汉明数是1、2、3、4、5、6、8、9、10、12、15。
+So the first few Hamming numbers are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15.
-有1105个汉明数字不超过108。
+There are 1105 Hamming numbers not exceeding 108.
-如果没有素数大于n的正数，我们将其称为n型广义汉明数。 因此，汉明数是类型5的广义汉明数。
+We will call a positive number a generalised Hamming number of type n, if it has no prime factor larger than n. Hence the Hamming numbers are the generalised Hamming numbers of type 5.
-有多少个类型为100的广义汉明数不超过109？
+How many generalised Hamming numbers of type 100 are there which don't exceed 109?
# --hints--
-`euler204()`应返回2944730。
+`euler204()` should return 2944730.
```js
assert.strictEqual(euler204(), 2944730);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-205-dice-game.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-205-dice-game.md
index cc429e8c8ca..41bf903b6d9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-205-dice-game.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-205-dice-game.md
@@ -1,22 +1,24 @@
---
id: 5900f4391000cf542c50ff4c
-title: 问题205：骰子游戏
+title: 'Problem 205: Dice Game'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301846
dashedName: problem-205-dice-game
---
# --description--
-彼得有九个四面（金字塔）骰子，每个骰子都有编号为1,2,3,4的面孔。科林有六个六面（立方体）骰子，每个骰子都有编号为1,2,3,4,5,6的面孔。
+Peter has nine four-sided (pyramidal) dice, each with faces numbered 1, 2, 3, 4.
-彼得和科林掷骰子并比较总数：总得分最高。如果总数相等，结果是平局。
+Colin has six six-sided (cubic) dice, each with faces numbered 1, 2, 3, 4, 5, 6.
-金字塔皮特击败立方科林的几率是多少？将您的答案四舍五入到0.abcdefg形式的七位小数
+Peter and Colin roll their dice and compare totals: the highest total wins. The result is a draw if the totals are equal.
+
+What is the probability that Pyramidal Pete beats Cubic Colin? Give your answer rounded to seven decimal places in the form 0.abcdefg
# --hints--
-`euler205()`应该返回0.5731441。
+`euler205()` should return 0.5731441.
```js
assert.strictEqual(euler205(), 0.5731441);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-206-concealed-square.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-206-concealed-square.md
index 7cbf4d7dd3c..eaedced890e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-206-concealed-square.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-206-concealed-square.md
@@ -1,18 +1,18 @@
---
id: 5900f43a1000cf542c50ff4d
-title: 问题206：隐藏的广场
+title: 'Problem 206: Concealed Square'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301847
dashedName: problem-206-concealed-square
---
# --description--
-找到唯一的正整数，其正方形的形式为1_2_3_4_5_6_7_8_9_0，其中每个“\_”是一个数字。
+Find the unique positive integer whose square has the form 1_2_3_4_5_6_7_8_9_0, where each “\_” is a single digit.
# --hints--
-`euler206()`应该返回1389019170。
+`euler206()` should return 1389019170.
```js
assert.strictEqual(euler206(), 1389019170);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-207-integer-partition-equations.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-207-integer-partition-equations.md
index 627a8bc851f..19f1d12c8d4 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-207-integer-partition-equations.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-207-integer-partition-equations.md
@@ -1,26 +1,28 @@
---
id: 5900f43c1000cf542c50ff4e
-title: 问题207：整数分区方程
+title: 'Problem 207: Integer partition equations'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301848
dashedName: problem-207-integer-partition-equations
---
# --description--
-对于一些正整数k，存在形式为4t = 2t + k的整数分区，其中4t，2t和k都是正整数而t是实数。
+For some positive integers k, there exists an integer partition of the form 4t = 2t + k,
-前两个这样的分区是41 = 21 + 2和41.5849625 ...... = 21.5849625 ... + 6。
+where 4t, 2t, and k are all positive integers and t is a real number.
-t也是整数的分区称为完美。对于任何m≥1，让P（m）为k≤m的完美分区的比例。因此P（6）= 1/2。
+The first two such partitions are 41 = 21 + 2 and 41.5849625... = 21.5849625... + 6.
-在下表中列出了一些P（m）P（5）= 1/1 P（10）= 1/2 P（15）= 2/3 P（20）= 1/2 P（25）= 1/2 P（30）= 2/5 ... P（180）= 1/4 P（185）= 3/13
+Partitions where t is also an integer are called perfect. For any m ≥ 1 let P(m) be the proportion of such partitions that are perfect with k ≤ m. Thus P(6) = 1/2.
-找到P（m）<1/12345的最小m
+In the following table are listed some values of P(m) P(5) = 1/1 P(10) = 1/2 P(15) = 2/3 P(20) = 1/2 P(25) = 1/2 P(30) = 2/5 ... P(180) = 1/4 P(185) = 3/13
+
+Find the smallest m for which P(m) < 1/12345
# --hints--
-`euler207()`应该返回44043947822。
+`euler207()` should return 44043947822.
```js
assert.strictEqual(euler207(), 44043947822);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-208-robot-walks.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-208-robot-walks.md
index c5aa6ce4ab9..36d5a8fff53 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-208-robot-walks.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-208-robot-walks.md
@@ -1,22 +1,22 @@
---
id: 5900f43f1000cf542c50ff51
-title: 问题208：机器人走路
+title: 'Problem 208: Robot Walks'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301849
dashedName: problem-208-robot-walks
---
# --description--
-机器人以一系列五分之一圆弧（72°）移动，每个步骤可自由选择顺时针或逆时针弧，但不能转动现场。
+A robot moves in a series of one-fifth circular arcs (72°), with a free choice of a clockwise or an anticlockwise arc for each step, but no turning on the spot.
-从北向南开始的70932条可能的闭合路径之一是
+One of 70932 possible closed paths of 25 arcs starting northward is
-鉴于机器人开始面向北方，在最终弧线之后，可以将多少70个弧长的行程返回到其起始位置？ （任何弧都可以多次遍历。）
+Given that the robot starts facing North, how many journeys of 70 arcs in length can it take that return it, after the final arc, to its starting position? (Any arc may be traversed multiple times.)
# --hints--
-`euler208()`应该返回331951449665644800。
+`euler208()` should return 331951449665644800.
```js
assert.strictEqual(euler208(), 331951449665644800);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-209-circular-logic.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-209-circular-logic.md
index 78b1316aa89..f7437447361 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-209-circular-logic.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-209-circular-logic.md
@@ -1,22 +1,24 @@
---
id: 5900f43e1000cf542c50ff4f
-title: 问题209：循环逻辑
+title: 'Problem 209: Circular Logic'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301850
dashedName: problem-209-circular-logic
---
# --description--
-k输入二进制真值表是从k个输入位（二进制数字，0 \[假]或1 \[真]）到1个输出位的映射。例如，逻辑AND和XOR函数的2输入二进制真值表是：
+A k-input binary truth table is a map from k input bits
-x y x AND y000010100111x y x XOR y000011101110多个6输入二进制真值表τ满足公式
+(binary digits, 0 \[false] or 1 \[true]) to 1 output bit. For example, the 2-input binary truth tables for the logical AND and XOR functions are:
-所有6位输入（a，b，c，τ，a，b，c，d，e，f）和τ（b，c，d，e，f，一个XOR（b和c））= 0 d，e，f）？
+x y x AND y000010100111x y x XOR y000011101110How many 6-input binary truth tables, τ, satisfy the formula
+
+τ(a, b, c, d, e, f) AND τ(b, c, d, e, f, a XOR (b AND c)) = 0 for all 6-bit inputs (a, b, c, d, e, f)?
# --hints--
-`euler209()`应该返回15964587728784。
+`euler209()` should return 15964587728784.
```js
assert.strictEqual(euler209(), 15964587728784);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-21-amicable-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-21-amicable-numbers.md
index b44a9aad063..58acfba70e1 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-21-amicable-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-21-amicable-numbers.md
@@ -1,36 +1,48 @@
---
id: 5900f3811000cf542c50fe94
-title: 问题21：友好的数字
+title: 'Problem 21: Amicable numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301851
dashedName: problem-21-amicable-numbers
---
# --description--
-设d（ `n` ）定义为`n`的适当除数之`和` （小于`n的`数均匀分成`n` ）。如果d（ `a` ）= `b`并且d（ `b` ）= `a` ，其中`a` ≠ `b` ，则`a`和`b`是友好对，并且`a`和`b`中的每`一个`被称为友好数字。例如，220的适当除数是1,2,4,5,10,11,20,22,44,55和110;因此d（220）= 284. 284的适当除数是1,2,4,71和142;所以d（284）= 220.评估`n`下所有友好数字的总和。
+Let d(`n`) be defined as the sum of proper divisors of `n` (numbers less than `n` which divide evenly into `n`).
+
+If d(`a`) = `b` and d(`b`) = `a`, where `a` ≠ `b`, then `a` and `b` are an amicable pair and each of `a` and `b` are called amicable numbers.
+
+For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
+
+Evaluate the sum of all the amicable numbers under `n`.
# --hints--
-`sumAmicableNum(1000)`应返回504。
+`sumAmicableNum(1000)` should return a number.
+
+```js
+assert(typeof sumAmicableNum(1000) === 'number');
+```
+
+`sumAmicableNum(1000)` should return 504.
```js
assert.strictEqual(sumAmicableNum(1000), 504);
```
-`sumAmicableNum(2000)`应该返回2898。
+`sumAmicableNum(2000)` should return 2898.
```js
assert.strictEqual(sumAmicableNum(2000), 2898);
```
-`sumAmicableNum(5000)`应该返回8442。
+`sumAmicableNum(5000)` should return 8442.
```js
assert.strictEqual(sumAmicableNum(5000), 8442);
```
-`sumAmicableNum(10000)`应返回31626。
+`sumAmicableNum(10000)` should return 31626.
```js
assert.strictEqual(sumAmicableNum(10000), 31626);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-210-obtuse-angled-triangles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-210-obtuse-angled-triangles.md
index b76da6e541d..1fc7d980330 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-210-obtuse-angled-triangles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-210-obtuse-angled-triangles.md
@@ -1,20 +1,26 @@
---
id: 5900f43e1000cf542c50ff50
-title: 问题210：钝角三角形
+title: 'Problem 210: Obtuse Angled Triangles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301852
dashedName: problem-210-obtuse-angled-triangles
---
# --description--
-考虑点（x，y）的集合S（r），其中整数坐标满足| x | + | y | ≤r。设O为点（0,0），C为点（r / 4，r / 4）。令N（r）为S（r）中的点B的数量，使得三角形OBC具有钝角，即最大角度α满足90°<α<180°。因此，例如，N（4）= 24并且N（8）= 100。
+Consider the set S(r) of points (x,y) with integer coordinates satisfying |x| + |y| ≤ r.
-什么是N（1,000,000,000）？
+Let O be the point (0,0) and C the point (r/4,r/4).
+
+Let N(r) be the number of points B in S(r), so that the triangle OBC has an obtuse angle, i.e. the largest angle α satisfies 90°<α<180°.
+
+So, for example, N(4)=24 and N(8)=100.
+
+What is N(1,000,000,000)?
# --hints--
-`euler210()`应该返回1598174770174689500。
+`euler210()` should return 1598174770174689500.
```js
assert.strictEqual(euler210(), 1598174770174689500);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-211-divisor-square-sum.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-211-divisor-square-sum.md
index d0b226f062e..bde41e4a79e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-211-divisor-square-sum.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-211-divisor-square-sum.md
@@ -1,18 +1,22 @@
---
id: 5900f43f1000cf542c50ff52
-title: 问题211：除数平方和
+title: 'Problem 211: Divisor Square Sum'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301853
dashedName: problem-211-divisor-square-sum
---
# --description--
-对于正整数n，令σ2（n）为其除数的平方和。例如，σ2（10）= 1 + 4 + 25 + 100 = 130.找到所有n的总和，0 <n <64,000,000，使得σ2（n）是完美的平方。
+For a positive integer n, let σ2(n) be the sum of the squares of its divisors. For example,
+
+σ2(10) = 1 + 4 + 25 + 100 = 130.
+
+Find the sum of all n, 0 < n < 64,000,000 such that σ2(n) is a perfect square.
# --hints--
-`euler211()`应返回1922364685。
+`euler211()` should return 1922364685.
```js
assert.strictEqual(euler211(), 1922364685);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-212-combined-volume-of-cuboids.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-212-combined-volume-of-cuboids.md
index 2c90da245af..901bbbc3d11 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-212-combined-volume-of-cuboids.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-212-combined-volume-of-cuboids.md
@@ -1,32 +1,32 @@
---
id: 5900f4411000cf542c50ff53
-title: 问题212：长方体的组合体积
+title: 'Problem 212: Combined Volume of Cuboids'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301854
dashedName: problem-212-combined-volume-of-cuboids
---
# --description--
-由参数{（x0，y0，z0），（dx，dy，dz）}指定的轴对齐长方体由所有点（X，Y，Z）组成，使得x0≤X≤x0+ dx，y0≤ Y≤y0+ dy且z0≤Z≤z0+ dz。长方体的体积是乘积，dx×dy×dz。长方体集合的总体积是它们的并集体积，如果任何长方体重叠，它将小于各个体积的总和。
+An axis-aligned cuboid, specified by parameters { (x0,y0,z0), (dx,dy,dz) }, consists of all points (X,Y,Z) such that x0 ≤ X ≤ x0+dx, y0 ≤ Y ≤ y0+dy and z0 ≤ Z ≤ z0+dz. The volume of the cuboid is the product, dx × dy × dz. The combined volume of a collection of cuboids is the volume of their union and will be less than the sum of the individual volumes if any cuboids overlap.
-设C1，...，C50000是50000轴对齐长方体的集合，使得Cn具有参数
+Let C1,...,C50000 be a collection of 50000 axis-aligned cuboids such that Cn has parameters
-x0 = S6n-5模10000y0 = S6n-4模10000z0 = S6n-3模10000dx = 1 +（S6n-2模399）dy = 1 +（S6n-1模399）dz = 1 +（S6n模399）
+x0 = S6n-5 modulo 10000y0 = S6n-4 modulo 10000z0 = S6n-3 modulo 10000dx = 1 + (S6n-2 modulo 399)dy = 1 + (S6n-1 modulo 399)dz = 1 + (S6n modulo 399)
-其中S1，...，S300000来自“Lagged Fibonacci Generator”：
+where S1,...,S300000 come from the "Lagged Fibonacci Generator":
-对于1≤k≤55，Sk = \[100003 - 200003k + 300007k3]（模1000000）对于56≤k，Sk = \[Sk-24 + Sk-55]（模1000000）
+For 1 ≤ k ≤ 55, Sk = \[100003 - 200003k + 300007k3] (modulo 1000000)For 56 ≤ k, Sk = \[Sk-24 + Sk-55] (modulo 1000000)
-因此，C1具有参数{（7,53,183），（94,369,56）}，C2具有参数{（2383,3563,5079），（42,212,344）}，等等。
+Thus, C1 has parameters {(7,53,183),(94,369,56)}, C2 has parameters {(2383,3563,5079),(42,212,344)}, and so on.
-前100个长方体C1，...，C100的总体积为723581599。
+The combined volume of the first 100 cuboids, C1,...,C100, is 723581599.
-所有50000个长方体的总体积是多少，C1，...，C50000？
+What is the combined volume of all 50000 cuboids, C1,...,C50000 ?
# --hints--
-`euler212()`应该返回328968937309。
+`euler212()` should return 328968937309.
```js
assert.strictEqual(euler212(), 328968937309);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-213-flea-circus.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-213-flea-circus.md
index 37d58d3109f..2fbc23393c1 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-213-flea-circus.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-213-flea-circus.md
@@ -1,20 +1,22 @@
---
id: 5900f4411000cf542c50ff54
-title: 问题213：跳蚤马戏团
+title: 'Problem 213: Flea Circus'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301855
dashedName: problem-213-flea-circus
---
# --description--
-一个30×30的正方形网格包含900个跳蚤，最初是每平方一个跳蚤。当响铃响起时，每个跳蚤随机跳到相邻的广场（通常有4种可能，除了网格边缘或角落处的跳蚤）。
+A 30×30 grid of squares contains 900 fleas, initially one flea per square.
-在响铃50次后，预计的未占用方格数量是多少？将您的答案四舍五入到小数点后六位。
+When a bell is rung, each flea jumps to an adjacent square at random (usually 4 possibilities, except for fleas on the edge of the grid or at the corners).
+
+What is the expected number of unoccupied squares after 50 rings of the bell? Give your answer rounded to six decimal places.
# --hints--
-`euler213()`应返回330.721154。
+`euler213()` should return 330.721154.
```js
assert.strictEqual(euler213(), 330.721154);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-214-totient-chains.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-214-totient-chains.md
index 86d48f8c655..754596cef69 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-214-totient-chains.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-214-totient-chains.md
@@ -1,26 +1,28 @@
---
id: 5900f4421000cf542c50ff55
-title: 问题214个欧拉链
+title: 'Problem 214: Totient Chains'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301856
dashedName: problem-214-totient-chains
---
# --description--
-设φ是欧拉的函数，即对于自然数n，φ（n）是k的数，1≤k≤n，gcd（k，n）= 1。
+Let φ be Euler's totient function, i.e. for a natural number n,
-通过迭代φ，每个正整数产生以1结尾的数字递减的链。例如，如果我们从5开始，则生成序列5,4,2,1。以下列出了长度为4的所有链条：
+φ(n) is the number of k, 1 ≤ k ≤ n, for which gcd(k,n) = 1.
-5,4,2,1 7,6,2,1 8,4,2,1 9,6,2,1 10,4,2,1 12,4,2,1 14,6,2,1 18 ，6,2,1
+By iterating φ, each positive integer generates a decreasing chain of numbers ending in 1. E.g. if we start with 5 the sequence 5,4,2,1 is generated. Here is a listing of all chains with length 4:
-这些链中只有两个以素数开头，它们的总和为12。
+5,4,2,1 7,6,2,1 8,4,2,1 9,6,2,1 10,4,2,1 12,4,2,1 14,6,2,1 18,6,2,1
-所有小于40000000的素数的总和是多少，它产生一个长度为25的链？
+Only two of these chains start with a prime, their sum is 12.
+
+What is the sum of all primes less than 40000000 which generate a chain of length 25?
# --hints--
-`euler214()`应该返回1677366278943。
+`euler214()` should return 1677366278943.
```js
assert.strictEqual(euler214(), 1677366278943);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-215-crack-free-walls.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-215-crack-free-walls.md
index 82caa3d4202..45f7f563e73 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-215-crack-free-walls.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-215-crack-free-walls.md
@@ -1,24 +1,24 @@
---
id: 5900f4431000cf542c50ff56
-title: 问题215：无裂缝墙
+title: 'Problem 215: Crack-free Walls'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301857
dashedName: problem-215-crack-free-walls
---
# --description--
-考虑用2×1和3×1砖（水平×垂直尺寸）建造墙壁的问题，使得为了额外的强度，水平相邻砖块之间的间隙从不在连续层中排列，即从不形成“运行”裂纹”。
+Consider the problem of building a wall out of 2×1 and 3×1 bricks (horizontal×vertical dimensions) such that, for extra strength, the gaps between horizontally-adjacent bricks never line up in consecutive layers, i.e. never form a "running crack".
-例如，由于以红色显示的运行裂缝，以下9×3墙是不可接受的：
+For example, the following 9×3 wall is not acceptable due to the running crack shown in red:
-有八种形成无裂纹9×3壁的方法，写成W（9,3）= 8。
+There are eight ways of forming a crack-free 9×3 wall, written W(9,3) = 8.
-计算W（32,10）。
+Calculate W(32,10).
# --hints--
-`euler215()`应该返回806844323190414。
+`euler215()` should return 806844323190414.
```js
assert.strictEqual(euler215(), 806844323190414);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-216-investigating-the-primality-of-numbers-of-the-form-2n2-1.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-216-investigating-the-primality-of-numbers-of-the-form-2n2-1.md
index 6fac07e2046..e7591cd46f5 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-216-investigating-the-primality-of-numbers-of-the-form-2n2-1.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-216-investigating-the-primality-of-numbers-of-the-form-2n2-1.md
@@ -1,20 +1,26 @@
---
id: 5900f4451000cf542c50ff57
-title: 问题216：调查2n2-1形式的数字的素数
+title: 'Problem 216: Investigating the primality of numbers of the form 2n2-1'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301858
dashedName: problem-216-investigating-the-primality-of-numbers-of-the-form-2n2-1
---
# --description--
-考虑形式为t（n）= 2n2-1的数字t（n），其中n> 1.第一个这样的数字是7,17,31,49,71,97,127和161.事实证明只有49 = 7 \* 7和161 = 7 \* 23不是素数。对于n≤10000，有2202个数字t（n）是素数。
+Consider numbers t(n) of the form t(n) = 2n2-1 with n > 1.
-对于n≤50,000,000，有多少个数t（n）是素数？
+The first such numbers are 7, 17, 31, 49, 71, 97, 127 and 161.
+
+It turns out that only 49 = 7\*7 and 161 = 7\*23 are not prime.
+
+For n ≤ 10000 there are 2202 numbers t(n) that are prime.
+
+How many numbers t(n) are prime for n ≤ 50,000,000 ?
# --hints--
-`euler216()`应该返回5437849。
+`euler216()` should return 5437849.
```js
assert.strictEqual(euler216(), 5437849);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-217-balanced-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-217-balanced-numbers.md
index ca6ad7d307b..625b4a6c6d2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-217-balanced-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-217-balanced-numbers.md
@@ -1,18 +1,26 @@
---
id: 5900f4461000cf542c50ff58
-title: 问题217：平衡数字
+title: 'Problem 217: Balanced Numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301859
dashedName: problem-217-balanced-numbers
---
# --description--
-具有k（十进制）数字的正整数称为平衡，如果其第一个⌈k/2⌉数字总和为与其最后的⌈k/2⌉数字相同的值，其中⌈x⌉，x的发音上限，是最小整数≥x，因此⌈π⌉= 4且⌈5⌉= 5.因此，例如，所有回文都是平衡的，如13722.设T（n）是所有平衡数小于10n的总和。因此：T（1）= 45，T（2）= 540，T（5）= 334795890.求T（47）mod 315
+A positive integer with k (decimal) digits is called balanced if its first ⌈k/2⌉ digits sum to the same value as its last ⌈k/2⌉ digits, where ⌈x⌉, pronounced ceiling of x, is the smallest integer ≥ x, thus ⌈π⌉ = 4 and ⌈5⌉ = 5.
+
+So, for example, all palindromes are balanced, as is 13722.
+
+Let T(n) be the sum of all balanced numbers less than 10n.
+
+Thus: T(1) = 45, T(2) = 540 and T(5) = 334795890.
+
+Find T(47) mod 315
# --hints--
-`euler217()`应该返回6273134。
+`euler217()` should return 6273134.
```js
assert.strictEqual(euler217(), 6273134);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-218-perfect-right-angled-triangles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-218-perfect-right-angled-triangles.md
index fc32210610e..db91d74874e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-218-perfect-right-angled-triangles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-218-perfect-right-angled-triangles.md
@@ -1,24 +1,30 @@
---
id: 5900f4461000cf542c50ff59
-title: 问题218：完美的直角三角形
+title: 'Problem 218: Perfect right-angled triangles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301860
dashedName: problem-218-perfect-right-angled-triangles
---
# --description--
-考虑具有边a = 7，b = 24和c = 25的直角三角形。这个三角形的面积是84，可以被完美的数字6和28整除。而且它是一个原始的直角三角形，因为gcd（a，b）= 1，gcd（b，c）= 1。 c也是一个完美的正方形。
+Consider the right angled triangle with sides a=7, b=24 and c=25.
-如果-it是一个原始的直角三角形，我们将称为直角三角形完美 - 斜边是一个完美的正方形
+The area of this triangle is 84, which is divisible by the perfect numbers 6 and 28.
-如果-it是一个完美的直角三角形，我们将称为直角三角形超完美，并且 - 区域是完美数字6和28的倍数。
+Moreover it is a primitive right angled triangle as gcd(a,b)=1 and gcd(b,c)=1.
-有多少c≤1016的完美直角三角形不是非常完美的？
+Also c is a perfect square.
+
+We will call a right angled triangle perfect if -it is a primitive right angled triangle -its hypotenuse is a perfect square
+
+We will call a right angled triangle super-perfect if -it is a perfect right angled triangle and -its area is a multiple of the perfect numbers 6 and 28.
+
+How many perfect right-angled triangles with c≤1016 exist that are not super-perfect?
# --hints--
-`euler218()`应该返回0。
+`euler218()` should return 0.
```js
assert.strictEqual(euler218(), 0);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-219-skew-cost-coding.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-219-skew-cost-coding.md
index 197364266d6..38d16eba872 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-219-skew-cost-coding.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-219-skew-cost-coding.md
@@ -1,26 +1,30 @@
---
id: 5900f4481000cf542c50ff5a
-title: 问题219：偏差成本编码
+title: 'Problem 219: Skew-cost coding'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301861
dashedName: problem-219-skew-cost-coding
---
# --description--
-设A和B为位串（0和1的序列）。如果A等于B的最左边长度（A）位，则A被称为B的前缀。例如，00110是001101001的前缀，但不是00111或100110的前缀。
+Let A and B be bit strings (sequences of 0's and 1's).
-大小为n的无前缀代码是n个不同位串的集合，因此没有字符串是任何其他字符串的前缀。例如，这是一个大小为6的无前缀代码：
+If A is equal to the leftmost length(A) bits of B, then A is said to be a prefix of B.
-0000,0001,001,01,10,11
+For example, 00110 is a prefix of 001101001, but not of 00111 or 100110.
-现在假设发送'0'位需要一分钱，而传输'1'需要4便士。然后，上面显示的无前缀代码的总成本是35便士，这恰好是所讨论的偏斜定价方案可能最便宜的。简而言之，我们写Cost（6）= 35。
+A prefix-free code of size n is a collection of n distinct bit strings such that no string is a prefix of any other. For example, this is a prefix-free code of size 6:
-什么是成本（109）？
+0000, 0001, 001, 01, 10, 11
+
+Now suppose that it costs one penny to transmit a '0' bit, but four pence to transmit a '1'. Then the total cost of the prefix-free code shown above is 35 pence, which happens to be the cheapest possible for the skewed pricing scheme in question. In short, we write Cost(6) = 35.
+
+What is Cost(109) ?
# --hints--
-`euler219()`应返回64564225042。
+`euler219()` should return 64564225042.
```js
assert.strictEqual(euler219(), 64564225042);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-22-names-scores.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-22-names-scores.md
index 5f14235a6c2..e93e40d6ef4 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-22-names-scores.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-22-names-scores.md
@@ -1,30 +1,40 @@
---
id: 5a51eabcad78bf416f316e2a
-title: 问题22：命名分数
+title: 'Problem 22: Names scores'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301862
dashedName: problem-22-names-scores
---
# --description--
-使用`names` ，包含超过五千个名字的数组，首先按字母顺序排序。然后计算每个名称的字母值，将该值乘以列表中的字母位置以获得名称分数。例如，当列表按字母顺序排序时，值为3 + 15 + 12 + 9 + 14 = 53的COLIN是列表中的第938个名称。因此，COLIN将获得938×53 = 49714的分数。文件中所有名称分数的总和是多少？
+Using `names`, an array defined in the background containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score.
+
+For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would obtain a score of 938 × 53 = 49714.
+
+What is the total of all the name scores in the array?
# --hints--
-`namesScores(test1)`应该返回791。
+`namesScores(test1)` should return a number.
+
+```js
+assert(typeof namesScores(test1) === 'number');
+```
+
+`namesScores(test1)` should return 791.
```js
assert.strictEqual(namesScores(test1), 791);
```
-`namesScores(test2)`应该返回1468。
+`namesScores(test2)` should return 1468.
```js
assert.strictEqual(namesScores(test2), 1468);
```
-`namesScores(names)`应返回871198282。
+`namesScores(names)` should return 871198282.
```js
assert.strictEqual(namesScores(names), 871198282);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-220-heighway-dragon.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-220-heighway-dragon.md
index 10096341265..1fccaef107f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-220-heighway-dragon.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-220-heighway-dragon.md
@@ -1,28 +1,28 @@
---
id: 5900f4481000cf542c50ff5b
-title: 问题220：高速公路龙
+title: 'Problem 220: Heighway Dragon'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301863
dashedName: problem-220-heighway-dragon
---
# --description--
-设D0为双字母串“Fa”。对于n≥1，通过字符串重写规则从Dn-1导出Dn：
+Let D0 be the two-letter string "Fa". For n≥1, derive Dn from Dn-1 by the string-rewriting rules:
-“a”→“aRbFR”“b”→“LFaLb”
+"a" → "aRbFR" "b" → "LFaLb"
-因此，D0 =“Fa”，D1 =“FaRbFR”，D2 =“FaRbFRRLFaLbFR”，等等。
+Thus, D0 = "Fa", D1 = "FaRbFR", D2 = "FaRbFRRLFaLbFR", and so on.
-这些字符串可以解释为计算机图形程序的指令，“F”表示“向前绘制一个单位”，“L”表示“向左转90度”，“R”表示“向右转90度”，和“a” “和”b“被忽略了。计算机光标的初始位置是（0,0），指向（0,1）。
+These strings can be interpreted as instructions to a computer graphics program, with "F" meaning "draw forward one unit", "L" meaning "turn left 90 degrees", "R" meaning "turn right 90 degrees", and "a" and "b" being ignored. The initial position of the computer cursor is (0,0), pointing up towards (0,1).
-然后Dn是一种奇特的绘画，被称为n阶的Heighway Dragon。例如，D10如下所示;将每个“F”计为一步，（18,16）处的突出显示点是500步后达到的位置。
+Then Dn is an exotic drawing known as the Heighway Dragon of order n. For example, D10 is shown below; counting each "F" as one step, the highlighted spot at (18,16) is the position reached after 500 steps.
-D50中1012步后光标的位置是什么？以x，y的形式给出答案，没有空格。
+What is the position of the cursor after 1012 steps in D50 ? Give your answer in the form x,y with no spaces.
# --hints--
-`euler220()`应返回139776,963904。
+`euler220()` should return 139776, 963904.
```js
assert.strictEqual(euler220(), 139776, 963904);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-221-alexandrian-integers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-221-alexandrian-integers.md
index baada948fa3..010d2bd57c9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-221-alexandrian-integers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-221-alexandrian-integers.md
@@ -1,44 +1,24 @@
---
id: 5900f4491000cf542c50ff5c
-title: 问题221：亚历山大整数
+title: 'Problem 221: Alexandrian Integers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301864
dashedName: problem-221-alexandrian-integers
---
# --description--
-如果存在整数p，q，r，我们将调用正整数A和“亚历山大整数”，这样：
+We shall call a positive integer A an "Alexandrian integer", if there exist integers p, q, r such that: A = p · q · r and 1/A = 1/p + 1/q + 1/r
-```
- A = p · q · r and 1A
-```
+<!-- TODO Use MathJax -->
-=
+For example, 630 is an Alexandrian integer (p = 5, q = −7, r = −18). In fact, 630 is the 6th Alexandrian integer, the first 6 Alexandrian integers being: 6, 42, 120, 156, 420 and 630.
-```
- 1p
-```
-
-\+
-
-```
- 1q
-```
-
-\+
-
-```
- 1r
-```
-
-例如，630是亚历山大整数（p = 5，q = -7，r = -18）。实际上，630是第6个亚历山大整数，前6个亚历山大整数是：6,42,120,156,420和630。
-
-找到第150000个亚历山大整数。
+Find the 150000th Alexandrian integer.
# --hints--
-`euler221()`应该返回1884161251122450。
+`euler221()` should return 1884161251122450.
```js
assert.strictEqual(euler221(), 1884161251122450);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-222-sphere-packing.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-222-sphere-packing.md
index dedaa3aba4a..423ce7ff8e6 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-222-sphere-packing.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-222-sphere-packing.md
@@ -1,20 +1,20 @@
---
id: 5900f44b1000cf542c50ff5d
-title: 问题222：球体包装
+title: 'Problem 222: Sphere Packing'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301865
dashedName: problem-222-sphere-packing
---
# --description--
-什么是最短管道的长度，内径为50毫米，可以完全包含21个半径为30毫米，31毫米，......，50毫米的球？
+What is the length of the shortest pipe, of internal radius 50mm, that can fully contain 21 balls of radii 30mm, 31mm, ..., 50mm?
-以微米（10-6米）四舍五入到最接近的整数给出答案。
+Give your answer in micrometres (10-6 m) rounded to the nearest integer.
# --hints--
-`euler222()`应该返回1590933。
+`euler222()` should return 1590933.
```js
assert.strictEqual(euler222(), 1590933);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-223-almost-right-angled-triangles-i.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-223-almost-right-angled-triangles-i.md
index faa448cf04c..28f8d8f2143 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-223-almost-right-angled-triangles-i.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-223-almost-right-angled-triangles-i.md
@@ -1,20 +1,20 @@
---
id: 5900f44b1000cf542c50ff5e
-title: 问题223：几乎是直角三角形I.
+title: 'Problem 223: Almost right-angled triangles I'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301866
dashedName: problem-223-almost-right-angled-triangles-i
---
# --description--
-如果边满足a2 + b2 = c2 + 1，让我们称一个整数边三角形，边a≤b≤c几乎不是锐角。
+Let us call an integer sided triangle with sides a ≤ b ≤ c barely acute if the sides satisfy a2 + b2 = c2 + 1.
-有多少个几乎没有锐角的三角形，周长≤25,000,000？
+How many barely acute triangles are there with perimeter ≤ 25,000,000?
# --hints--
-`euler223()`应该返回61614848。
+`euler223()` should return 61614848.
```js
assert.strictEqual(euler223(), 61614848);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-224-almost-right-angled-triangles-ii.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-224-almost-right-angled-triangles-ii.md
index 40acdfbb0d4..d9f557fd4a9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-224-almost-right-angled-triangles-ii.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-224-almost-right-angled-triangles-ii.md
@@ -1,20 +1,20 @@
---
id: 5900f44e1000cf542c50ff5f
-title: 问题224：几乎直角三角形II
+title: 'Problem 224: Almost right-angled triangles II'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301867
dashedName: problem-224-almost-right-angled-triangles-ii
---
# --description--
-如果边满足a2 + b2 = c2 - 1，让我们称一个边长为a≤b≤c的整数边三角形，几乎不会钝。
+Let us call an integer sided triangle with sides a ≤ b ≤ c barely obtuse if the sides satisfy a2 + b2 = c2 - 1.
-有多少个几乎没有钝角的三角形，周长≤75,000,000？
+How many barely obtuse triangles are there with perimeter ≤ 75,000,000?
# --hints--
-`euler224()`应返回4137330。
+`euler224()` should return 4137330.
```js
assert.strictEqual(euler224(), 4137330);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-225-tribonacci-non-divisors.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-225-tribonacci-non-divisors.md
index c384ae4dce2..ff5fb05c1a7 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-225-tribonacci-non-divisors.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-225-tribonacci-non-divisors.md
@@ -1,22 +1,24 @@
---
id: 5900f44e1000cf542c50ff60
-title: 问题225：Tribonacci非除数
+title: 'Problem 225: Tribonacci non-divisors'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301868
dashedName: problem-225-tribonacci-non-divisors
---
# --description--
-序列1,1,1,3,5,9,17,31,57,105,193,355,653,1201 ......由T1 = T2 = T3 = 1和Tn = Tn-1 + Tn定义-2 + Tn-3。
+The sequence 1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193, 355, 653, 1201 ...
-可以证明27不会划分该序列的任何项。实际上，27是具有该属性的第一个奇数。
+is defined by T1 = T2 = T3 = 1 and Tn = Tn-1 + Tn-2 + Tn-3.
-找到不分割上述序列的任何项的第124个奇数。
+It can be shown that 27 does not divide any terms of this sequence.In fact, 27 is the first odd number with this property.
+
+Find the 124th odd number that does not divide any terms of the above sequence.
# --hints--
-`euler225()`应该返回2009年。
+`euler225()` should return 2009.
```js
assert.strictEqual(euler225(), 2009);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-226-a-scoop-of-blancmange.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-226-a-scoop-of-blancmange.md
index 5a9cddab7ee..3aadacc50b8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-226-a-scoop-of-blancmange.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-226-a-scoop-of-blancmange.md
@@ -1,24 +1,24 @@
---
id: 5900f4511000cf542c50ff62
-title: 问题226：Blancmange的一个独家新闻
+title: 'Problem 226: A Scoop of Blancmange'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301869
dashedName: problem-226-a-scoop-of-blancmange
---
# --description--
-blancmange曲线是点（x，y）的集合，使得0≤x≤1，并且其中s（x）=从x到最接近的整数的距离。
+The blancmange curve is the set of points (x,y) such that 0 ≤ x ≤ 1 and ,where s(x) = the distance from x to the nearest integer.
-blancmange曲线下面积等于½，如下图所示为粉红色。
+The area under the blancmange curve is equal to ½, shown in pink in the diagram below.
-设C为中心（¼，½）和半径¼的圆，在图中以黑色显示。
+Let C be the circle with centre (¼,½) and radius ¼, shown in black in the diagram.
-blancmange曲线下面的哪个区域被C包围？将你的答案四舍五入到小数点后八位的形式为0.abcdefgh
+What area under the blancmange curve is enclosed by C?Give your answer rounded to eight decimal places in the form 0.abcdefgh
# --hints--
-`euler226()`应该返回0.11316017。
+`euler226()` should return 0.11316017.
```js
assert.strictEqual(euler226(), 0.11316017);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-227-the-chase.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-227-the-chase.md
index baac7fcd890..aff8e9b9340 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-227-the-chase.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-227-the-chase.md
@@ -1,22 +1,22 @@
---
id: 5900f44f1000cf542c50ff61
-title: 问题227：大通
+title: 'Problem 227: The Chase'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301870
dashedName: problem-227-the-chase
---
# --description--
-“追逐”是一个玩两个骰子和偶数玩家的游戏。
+"The Chase" is a game played with two dice and an even number of players.
-球员坐在桌子旁;游戏开始于两个相对的玩家，每个玩家有一个骰子。在每个回合中，两个玩家都会掷骰子。如果一名球员掷出一个1，他将骰子传给左边的邻居;如果他掷出一个6，他就把骰子传给右边的邻居;否则，他会为下一个回合保留死亡。游戏结束时，一个玩家在滚动并通过后都有两个骰子;那个玩家失败了。
+The players sit around a table; the game begins with two opposite players having one die each. On each turn, the two players with a die roll it. If a player rolls a 1, he passes the die to his neighbour on the left; if he rolls a 6, he passes the die to his neighbour on the right; otherwise, he keeps the die for the next turn. The game ends when one player has both dice after they have been rolled and passed; that player has then lost.
-在拥有100名玩家的游戏中，游戏持续的预期轮数是多少？将您的答案四舍五入到十位有效数字。
+In a game with 100 players, what is the expected number of turns the game lasts? Give your answer rounded to ten significant digits.
# --hints--
-`euler227()`应返回3780.618622。
+`euler227()` should return 3780.618622.
```js
assert.strictEqual(euler227(), 3780.618622);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-228-minkowski-sums.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-228-minkowski-sums.md
index 6197a4b4f74..3ee325d2be7 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-228-minkowski-sums.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-228-minkowski-sums.md
@@ -1,36 +1,38 @@
---
id: 5900f4511000cf542c50ff63
-title: 问题228：Minkowski Sums
+title: 'Problem 228: Minkowski Sums'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301871
dashedName: problem-228-minkowski-sums
---
# --description--
-设Sn是常规的n边多边形 - 或形状 - 其顶点
+<!-- TODO Use MathJax and re-write from projecteuler.net -->
-vk（k = 1,2，...，n）有坐标：
+Let Sn be the regular n-sided polygon – or shape – whose vertices
-```
- xk = cos( 2k-1/n ×180° ) yk = sin( 2k-1/n ×180° )
-```
+vk (k = 1,2,…,n) have coordinates:
-每个Sn都被解释为由周边和内部的所有点组成的填充形状。
+xk = cos( 2k-1/n ×180° )
-两个形状S和T的Minkowski和S + T是结果
+yk = sin( 2k-1/n ×180° )
-将S中的每个点添加到T中的每个点，其中以坐标方式执行点添加：
+Each Sn is to be interpreted as a filled shape consisting of all points on the perimeter and in the interior.
-（u，v）+（x，y）=（u + x，v + y）。
+The Minkowski sum, S+T, of two shapes S and T is the result of
-例如，S3和S4的总和是六边形，如下面粉红色所示：
+adding every point in S to every point in T, where point addition is performed coordinate-wise:
-S1864 + S1865 + ... + S1909有多少方面？
+(u, v) + (x, y) = (u+x, v+y).
+
+For example, the sum of S3 and S4 is the six-sided shape shown in pink below:
+
+How many sides does S1864 + S1865 + … + S1909 have?
# --hints--
-`euler228()`应返回86226。
+`euler228()` should return 86226.
```js
assert.strictEqual(euler228(), 86226);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-229-four-representations-using-squares.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-229-four-representations-using-squares.md
index 987366cd694..0431ab815bd 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-229-four-representations-using-squares.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-229-four-representations-using-squares.md
@@ -1,32 +1,32 @@
---
id: 5900f4521000cf542c50ff64
-title: 问题229：使用正方形的四个表示
+title: 'Problem 229: Four Representations using Squares'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301872
dashedName: problem-229-four-representations-using-squares
---
# --description--
-考虑数字3600.这是非常特别的，因为
+Consider the number 3600. It is very special, because
3600 = 482 + 362 3600 = 202 + 2×402 3600 = 302 + 3×302 3600 = 452 + 7×152
-类似地，我们发现88201 = 992 + 2802 = 2872 + 2×542 = 2832 + 3×522 = 1972 + 7×842。
+Similarly, we find that 88201 = 992 + 2802 = 2872 + 2×542 = 2832 + 3×522 = 1972 + 7×842.
-在1747年，欧拉证明哪些数字可以表示为两个正方形的总和。我们感兴趣的是数字n，它承认以下四种类型的表示：
+In 1747, Euler proved which numbers are representable as a sum of two squares. We are interested in the numbers n which admit representations of all of the following four types:
-n = a12 + b12n = a22 + 2 b22n = a32 + 3 b32n = a72 + 7 b72，
+n = a12 + b12n = a22 + 2 b22n = a32 + 3 b32n = a72 + 7 b72,
-其中ak和bk是正整数。
+where the ak and bk are positive integers.
-有75373这样的数字不超过107。
+There are 75373 such numbers that do not exceed 107.
-有多少这样的数字不超过2×109？
+How many such numbers are there that do not exceed 2×109?
# --hints--
-`euler229()`应该返回11325263。
+`euler229()` should return 11325263.
```js
assert.strictEqual(euler229(), 11325263);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-23-non-abundant-sums.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-23-non-abundant-sums.md
index 190e388702f..ce8b2262e4a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-23-non-abundant-sums.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-23-non-abundant-sums.md
@@ -1,36 +1,48 @@
---
id: 5900f3831000cf542c50fe96
-title: 问题23：非丰富的总和
+title: 'Problem 23: Non-abundant sums'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301873
dashedName: problem-23-non-abundant-sums
---
# --description--
-完美数字是一个数字，其正确除数的总和恰好等于数字。例如，28的适当除数之和为1 + 2 + 4 + 7 + 14 = 28，这意味着28是一个完美数。如果`n`的适当除数之和小于`n` ，则`n`被称为不足，如果该和超过`n`则称为`n` 。由于12是最小的有限数，1 + 2 + 3 + 4 + 6 = 16，可以写成两个有限数之和的最小数是24.通过数学分析，可以看出所有整数都大于28123可以写成两个数字的总和。然而，即使已知不能表示为两个充裕数的总和的最大数小于该限制，也不能通过分析进一步减小该上限。找出所有正整数<= `n`的总和，它不能写成两个丰富数字的总和。
+A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.
+
+A number `n` is called deficient if the sum of its proper divisors is less than `n` and it is called abundant if this sum exceeds `n`.
+
+As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.
+
+Find the sum of all positive integers <= `n` which cannot be written as the sum of two abundant numbers.
# --hints--
-`sumOfNonAbundantNumbers(10000)`应返回3731004。
+`sumOfNonAbundantNumbers(10000)` should return a number.
+
+```js
+assert(typeof sumOfNonAbundantNumbers(10000) === 'number');
+```
+
+`sumOfNonAbundantNumbers(10000)` should return 3731004.
```js
assert(sumOfNonAbundantNumbers(10000) === 3731004);
```
-`sumOfNonAbundantNumbers(15000)`应该返回4039939。
+`sumOfNonAbundantNumbers(15000)` should return 4039939.
```js
assert(sumOfNonAbundantNumbers(15000) === 4039939);
```
-`sumOfNonAbundantNumbers(20000)`应返回4159710。
+`sumOfNonAbundantNumbers(20000)` should return 4159710.
```js
assert(sumOfNonAbundantNumbers(20000) === 4159710);
```
-`sumOfNonAbundantNumbers(28123)`应该返回4179871。
+`sumOfNonAbundantNumbers(28123)` should return 4179871.
```js
assert(sumOfNonAbundantNumbers(28123) === 4179871);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-230-fibonacci-words.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-230-fibonacci-words.md
index fae079bacf7..16ea3131991 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-230-fibonacci-words.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-230-fibonacci-words.md
@@ -1,36 +1,36 @@
---
id: 5900f4531000cf542c50ff65
-title: 问题230：斐波纳契语
+title: 'Problem 230: Fibonacci Words'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301874
dashedName: problem-230-fibonacci-words
---
# --description--
-对于任何两个数字串A和B，我们将FA，B定义为序列（A，B，AB，BAB，ABBAB，...），其中每个术语是前两个术语的串联。
+For any two strings of digits, A and B, we define FA,B to be the sequence (A,B,AB,BAB,ABBAB,...) in which each term is the concatenation of the previous two.
-此外，我们将DA，B（n）定义为FA的第一项中的第n个数字，B包含至少n个数字。
+Further, we define DA,B(n) to be the nth digit in the first term of FA,B that contains at least n digits.
-例：
+Example:
-设A = 1415926535，B = 8979323846。我们希望找到DA，B（35）。
+Let A=1415926535, B=8979323846. We wish to find DA,B(35), say.
-FA，B的前几个术语是：1415926535 8979323846 14159265358979323846 897932384614159265358979323846 14159265358979323846897932384614159265358979323846
+The first few terms of FA,B are: 1415926535 8979323846 14159265358979323846 897932384614159265358979323846 14159265358979323846897932384614159265358979323846
-然后DA，B（35）是第五项中的第35位，即9。
+Then DA,B(35) is the 35th digit in the fifth term, which is 9.
-现在我们使用A小数点后面的前100位数字：14159265358979323846264338327950288419716939937510 58209749445923078164062862089986280348253421170679
+Now we use for A the first 100 digits of π behind the decimal point: 14159265358979323846264338327950288419716939937510 58209749445923078164062862089986280348253421170679
-对于B下一百个数字：
+and for B the next hundred digits:
-82148086513282306647093844609550582231725359408128 48111745028410270193852110555964462294895493038196。
+82148086513282306647093844609550582231725359408128 48111745028410270193852110555964462294895493038196 .
-求Σn= 0,1，...，17 10n×DA，B（（127 + 19n）×7n）。
+Find ∑n = 0,1,...,17 10n× DA,B((127+19n)×7n) .
# --hints--
-`euler230()`应返回850481152593119200。
+`euler230()` should return 850481152593119200.
```js
assert.strictEqual(euler230(), 850481152593119200);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-231-the-prime-factorisation-of-binomial-coefficients.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-231-the-prime-factorisation-of-binomial-coefficients.md
index 0a017e3e50f..c313225c729 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-231-the-prime-factorisation-of-binomial-coefficients.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-231-the-prime-factorisation-of-binomial-coefficients.md
@@ -1,20 +1,24 @@
---
id: 5900f4531000cf542c50ff66
-title: 问题231：二项式系数的素因子化
+title: 'Problem 231: The prime factorisation of binomial coefficients'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301875
dashedName: problem-231-the-prime-factorisation-of-binomial-coefficients
---
# --description--
-二项式系数10C3 = 120. 120 = 23×3×5 = 2×2×2×3×5，并且2 + 2 + 2 + 3 + 5 = 14.因此，素数因子分解中的项的总和为10C3是14岁。
+The binomial coefficient 10C3 = 120.
-找出素数因子分解中的术语总和为20000000C15000000。
+120 = 23 × 3 × 5 = 2 × 2 × 2 × 3 × 5, and 2 + 2 + 2 + 3 + 5 = 14.
+
+So the sum of the terms in the prime factorisation of 10C3 is 14.
+
+Find the sum of the terms in the prime factorisation of 20000000C15000000.
# --hints--
-`euler231()`应该返回7526965179680。
+`euler231()` should return 7526965179680.
```js
assert.strictEqual(euler231(), 7526965179680);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-232-the-race.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-232-the-race.md
index ebea8d8b32e..c109669bd8a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-232-the-race.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-232-the-race.md
@@ -1,24 +1,24 @@
---
id: 5900f4551000cf542c50ff67
-title: 问题232：比赛
+title: 'Problem 232: The Race'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301876
dashedName: problem-232-the-race
---
# --description--
-两名玩家分享一个无偏见的硬币并轮流玩“The Race”。在玩家1的回合中，他扔了一次硬币：如果它出现在Heads，他得分为1分;如果它出现Tails，他什么都没得分。在玩家2的回合中，她选择一个正整数T并将硬币投掷T次：如果它出现在所有的头上，她得到2T-1分;否则，她没有得分。玩家1排在第一位。获胜者是第一个达到100分或更多分。
+Two players share an unbiased coin and take it in turns to play "The Race". On Player 1's turn, he tosses the coin once: if it comes up Heads, he scores one point; if it comes up Tails, he scores nothing. On Player 2's turn, she chooses a positive integer T and tosses the coin T times: if it comes up all Heads, she scores 2T-1 points; otherwise, she scores nothing. Player 1 goes first. The winner is the first to 100 or more points.
-在每次转弯时，玩家2选择硬币投掷的数量T，以最大化她获胜的概率。
+On each turn Player 2 selects the number, T, of coin tosses that maximises the probability of her winning.
-玩家2获胜的概率是多少？
+What is the probability that Player 2 wins?
-将您的答案四舍五入到0.abcdefgh形式的八位小数。
+Give your answer rounded to eight decimal places in the form 0.abcdefgh .
# --hints--
-`euler232()`应返回0.83648556。
+`euler232()` should return 0.83648556.
```js
assert.strictEqual(euler232(), 0.83648556);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-233-lattice-points-on-a-circle.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-233-lattice-points-on-a-circle.md
index a69e8b3cc47..5903d7e8c95 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-233-lattice-points-on-a-circle.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-233-lattice-points-on-a-circle.md
@@ -1,20 +1,22 @@
---
id: 5900f4551000cf542c50ff68
-title: 问题233：圆上的格点
+title: 'Problem 233: Lattice points on a circle'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301877
dashedName: problem-233-lattice-points-on-a-circle
---
# --description--
-令f（N）是在通过（0,0），（N，0），（0，N）和（N，N）的圆上具有整数坐标的点的数量。可以证明f（10000）= 36。
+Let f(N) be the number of points with integer coordinates that are on a circle passing through (0,0), (N,0),(0,N), and (N,N).
-所有正整数N≤1011的总和是多少，f（N）= 420？
+It can be shown that f(10000) = 36.
+
+What is the sum of all positive integers N ≤ 1011 such that f(N) = 420 ?
# --hints--
-`euler233()`应该返回271204031455541300。
+`euler233()` should return 271204031455541300.
```js
assert.strictEqual(euler233(), 271204031455541300);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-234-semidivisible-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-234-semidivisible-numbers.md
index 9abf8b1ff22..edf39ca239d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-234-semidivisible-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-234-semidivisible-numbers.md
@@ -1,22 +1,26 @@
---
id: 5900f4571000cf542c50ff69
-title: 问题234：半透明数字
+title: 'Problem 234: Semidivisible numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301878
dashedName: problem-234-semidivisible-numbers
---
# --description--
-对于n≥4的整数，我们定义n的下素数根，用lps（n）表示，最大素数≤√n和n的上素数根，ups（n），作为最小素数≥ √N。因此，例如，lps（4）= 2 = ups（4），lps（1000）= 31，ups（1000）= 37.让我们称一个整数n≥4半可分，如果lps（n）和ups之一（n）除n，但不是两者。
+For an integer n ≥ 4, we define the lower prime square root of n, denoted by lps(n), as the largest prime ≤ √n and the upper prime square root of n, ups(n), as the smallest prime ≥ √n.
-不可超过15的半可分数之和为30，数字为8,10和12. 15不是半可分的，因为它是lps（15）= 3和ups（15）= 5的倍数。作为另一个例子，最多1000个的92个半可见数字的总和是34825。
+So, for example, lps(4) = 2 = ups(4), lps(1000) = 31, ups(1000) = 37.
-所有半数均不超过999966663333的总和是多少？
+Let us call an integer n ≥ 4 semidivisible, if one of lps(n) and ups(n) divides n, but not both.
+
+The sum of the semidivisible numbers not exceeding 15 is 30, the numbers are 8, 10 and 12. 15 is not semidivisible because it is a multiple of both lps(15) = 3 and ups(15) = 5. As a further example, the sum of the 92 semidivisible numbers up to 1000 is 34825.
+
+What is the sum of all semidivisible numbers not exceeding 999966663333 ?
# --hints--
-`euler234()`应该返回1259187438574927000。
+`euler234()` should return 1259187438574927000.
```js
assert.strictEqual(euler234(), 1259187438574927000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-235-an-arithmetic-geometric-sequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-235-an-arithmetic-geometric-sequence.md
index b76d7038f00..549fc0275de 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-235-an-arithmetic-geometric-sequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-235-an-arithmetic-geometric-sequence.md
@@ -1,22 +1,24 @@
---
id: 5900f4571000cf542c50ff6a
-title: 问题235：算术几何序列
+title: 'Problem 235: An Arithmetic Geometric sequence'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301879
dashedName: problem-235-an-arithmetic-geometric-sequence
---
# --description--
-给定是算术几何序列u（k）=（900-3k）rk-1。设s（n）=Σk= 1 ... nu（k）。
+Given is the arithmetic-geometric sequence u(k) = (900-3k)rk-1.
-找到r的值，其中s（5000）= - 600,000,000,000。
+Let s(n) = Σk=1...nu(k).
-将您的答案四舍五入到小数点后面的12位。
+Find the value of r for which s(5000) = -600,000,000,000.
+
+Give your answer rounded to 12 places behind the decimal point.
# --hints--
-`euler235()`应该返回1.002322108633。
+`euler235()` should return 1.002322108633.
```js
assert.strictEqual(euler235(), 1.002322108633);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-236-luxury-hampers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-236-luxury-hampers.md
index 0f6830915e8..01c7effbc3e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-236-luxury-hampers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-236-luxury-hampers.md
@@ -1,28 +1,28 @@
---
id: 5900f4591000cf542c50ff6b
-title: 问题236：豪华礼篮
+title: 'Problem 236: Luxury Hampers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301881
dashedName: problem-236-luxury-hampers
---
# --description--
-供应商'A'和'B'为奢侈品篮子市场提供以下数量的产品：
+Suppliers 'A' and 'B' provided the following numbers of products for the luxury hamper market:
-Product'A“B'Beluga Caviar5248640Christmas Cake13121888Gammon Joint26243776Vintage Port57603776Champagne Truffles39365664
+Product'A''B'Beluga Caviar5248640Christmas Cake13121888Gammon Joint26243776Vintage Port57603776Champagne Truffles39365664
-虽然供应商非常努力地将货物运送到完美的状态，但不可避免地会有一些变质 - 即产品变坏了。
+Although the suppliers try very hard to ship their goods in perfect condition, there is inevitably some spoilage - i.e. products gone bad.
-供应商使用两种类型的统计数据来比较他们的表现：每个供应商的五个每个产品的腐败率等于坏的产品数量除以供应的产品数量，依次为五个产品中的每一个。每个供应商的整体腐败率等于坏的产品总数除以该供应商提供的产品总数。令他们惊讶的是，供应商发现五种每种产品的腐败率都更差（更高） ）对于'B'而不是'A'由相同的因子（腐败率的比率），m> 1;然而，矛盾的是，'A'的整体腐败率比'B'更差，也是m倍。
+The suppliers compare their performance using two types of statistic:The five per-product spoilage rates for each supplier are equal to the number of products gone bad divided by the number of products supplied, for each of the five products in turn. The overall spoilage rate for each supplier is equal to the total number of products gone bad divided by the total number of products provided by that supplier.To their surprise, the suppliers found that each of the five per-product spoilage rates was worse (higher) for 'B' than for 'A' by the same factor (ratio of spoilage rates), m>1; and yet, paradoxically, the overall spoilage rate was worse for 'A' than for 'B', also by a factor of m.
-有三十五米> 1，这个惊人的结果可能会发生，其中最小的是1476/1475。
+There are thirty-five m>1 for which this surprising result could have occurred, the smallest of which is 1476/1475.
-m的最大可能值是多少？以u / v的形式将答案缩小到最低值。
+What's the largest possible value of m? Give your answer as a fraction reduced to its lowest terms, in the form u/v.
# --hints--
-`euler236()`应返回`euler236()` 。
+`euler236()` should return 123 / 59.
```js
assert.strictEqual(euler236(), 123 / 59);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-237-tours-on-a-4-x-n-playing-board.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-237-tours-on-a-4-x-n-playing-board.md
index ac9c50926ce..eec1076ec10 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-237-tours-on-a-4-x-n-playing-board.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-237-tours-on-a-4-x-n-playing-board.md
@@ -1,20 +1,30 @@
---
id: 5900f4591000cf542c50ff6c
-title: 问题237：在4 x n游戏板上游览
+title: 'Problem 237: Tours on a 4 x n playing board'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301882
dashedName: problem-237-tours-on-a-4-x-n-playing-board
---
# --description--
-设T（n）为4×n游戏板上的游览次数，使得：游览从左上角开始。巡视包括向上，向下，向左或向右一个方格的移动。游览访问每个广场一次。游览结束于左下角。该图显示了4×10板上的一次巡视：
+Let T(n) be the number of tours over a 4 × n playing board such that:
-T（10）是2329.什么是T（1012）模108？
+The tour starts in the top left corner.
+
+The tour consists of moves that are up, down, left, or right one square.
+
+The tour visits each square exactly once.
+
+The tour ends in the bottom left corner.
+
+The diagram shows one tour over a 4 × 10 board:
+
+T(10) is 2329. What is T(1012) modulo 108?
# --hints--
-`euler237()`应该返回15836928。
+`euler237()` should return 15836928.
```js
assert.strictEqual(euler237(), 15836928);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-238-infinite-string-tour.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-238-infinite-string-tour.md
index 0ba30ec3d59..4bb75276865 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-238-infinite-string-tour.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-238-infinite-string-tour.md
@@ -1,38 +1,38 @@
---
id: 5900f45b1000cf542c50ff6d
-title: 问题238：无限的字符串游览
+title: 'Problem 238: Infinite string tour'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301883
dashedName: problem-238-infinite-string-tour
---
# --description--
-使用“Blum Blum Shub”伪随机数生成器创建一系列数字：
+Create a sequence of numbers using the "Blum Blum Shub" pseudo-random number generator:
-s0 = 14025256 sn + 1 = sn2 mod 20300713
+s0 = 14025256 sn+1 = sn2 mod 20300713
-连接这些数字s0s1s2 ...以创建一个无限长度的字符串w。然后，w = 14025256741014958470038053646 ......
+Concatenate these numbers s0s1s2… to create a string w of infinite length. Then, w = 14025256741014958470038053646…
-对于正整数k，如果不存在w的子串，其中数字之和等于k，则p（k）被定义为零。如果w的至少一个子字符串存在且数字之和等于k，则我们定义p（k）= z，其中z是最早的这种子字符串的起始位置。
+For a positive integer k, if no substring of w exists with a sum of digits equal to k, p(k) is defined to be zero. If at least one substring of w exists with a sum of digits equal to k, we define p(k) = z, where z is the starting position of the earliest such substring.
-例如：
+For instance:
-子串1,14,1402 ......具有等于1,5,7，...的数字的总和，从位置1开始，因此p（1）= p（5）= p（7）= ...... = 1。
+The substrings 1, 14, 1402, … with respective sums of digits equal to 1, 5, 7, … start at position 1, hence p(1) = p(5) = p(7) = … = 1.
-子串4,402,4025，...各自的数字总和等于4,6,11 ......从位置2开始，因此p（4）= p（6）= p（11）= ...... = 2。
+The substrings 4, 402, 4025, … with respective sums of digits equal to 4, 6, 11, … start at position 2, hence p(4) = p(6) = p(11) = … = 2.
-子串02,0252，......各自的数字总和等于2,9，...从位置3开始，因此p（2）= p（9）= ...... = 3。
+The substrings 02, 0252, … with respective sums of digits equal to 2, 9, … start at position 3, hence p(2) = p(9) = … = 3.
-请注意，从位置3开始的子字符串025具有等于7的数字之和，但是存在较早的子字符串（从位置1开始），其数字之和等于7，因此p（7）= 1，而不是3。
+Note that substring 025 starting at position 3, has a sum of digits equal to 7, but there was an earlier substring (starting at position 1) with a sum of digits equal to 7, so p(7) = 1, not 3.
-我们可以验证，对于0 <k≤103，Σp（k）= 4742。
+We can verify that, for 0 < k ≤ 103, ∑ p(k) = 4742.
-求Σp（k），0 <k≤2·1015。
+Find ∑ p(k), for 0 < k ≤ 2·1015.
# --hints--
-`euler238()`应该返回9922545104535660。
+`euler238()` should return 9922545104535660.
```js
assert.strictEqual(euler238(), 9922545104535660);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-239-twenty-two-foolish-primes.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-239-twenty-two-foolish-primes.md
index f8d020a1adb..b68783fecf6 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-239-twenty-two-foolish-primes.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-239-twenty-two-foolish-primes.md
@@ -1,22 +1,22 @@
---
id: 5900f45c1000cf542c50ff6e
-title: 问题239：二十二个愚蠢的素数
+title: 'Problem 239: Twenty-two Foolish Primes'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301884
dashedName: problem-239-twenty-two-foolish-primes
---
# --description--
-编号为1到100的一组磁盘以随机顺序排列成一行。
+A set of disks numbered 1 through 100 are placed in a line in random order.
-我们有一个部分紊乱的概率是多少，这样才能确定22个素数盘远离它们的自然位置？ （任何数量的非主要磁盘也可以在其自然位置内或外找到。）
+What is the probability that we have a partial derangement such that exactly 22 prime number discs are found away from their natural positions? (Any number of non-prime disks may also be found in or out of their natural positions.)
-将答案四舍五入到小数点后面的12位，格式为0.abcdefghijkl。
+Give your answer rounded to 12 places behind the decimal point in the form 0.abcdefghijkl.
# --hints--
-`euler239()`应该返回0.001887854841。
+`euler239()` should return 0.001887854841.
```js
assert.strictEqual(euler239(), 0.001887854841);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-24-lexicographic-permutations.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-24-lexicographic-permutations.md
index 908feeb34ba..49998335d2e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-24-lexicographic-permutations.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-24-lexicographic-permutations.md
@@ -1,40 +1,46 @@
---
id: 5900f3841000cf542c50fe97
-title: 问题24：字典排列
+title: 'Problem 24: Lexicographic permutations'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301885
dashedName: problem-24-lexicographic-permutations
---
# --description--
-置换是对象的有序排列。例如，3124是数字1,2,3和4的一种可能的排列。如果所有排列都以数字或字母顺序列出，我们称之为词典顺序。字典排列0,1和2是：
+A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are:
-012 021 102 120 201 210
+<div style='text-align: center;'>012 021 102 120 201 210</div>
-数字0,1,2,3,4,5,6,7,8和9的第`n`个词典排列是什么？
+What is the `n`th lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
# --hints--
-`lexicographicPermutations(699999)`应该返回1938246570。
+`lexicographicPermutations(699999)` should return a number.
+
+```js
+assert(typeof lexicographicPermutations(699999) === 'number');
+```
+
+`lexicographicPermutations(699999)` should return 1938246570.
```js
assert(lexicographicPermutations(699999) == 1938246570);
```
-`lexicographicPermutations(899999)`应该返回2536987410。
+`lexicographicPermutations(899999)` should return 2536987410.
```js
assert(lexicographicPermutations(899999) == 2536987410);
```
-`lexicographicPermutations(900000)`应该返回2537014689。
+`lexicographicPermutations(900000)` should return 2537014689.
```js
assert(lexicographicPermutations(900000) == 2537014689);
```
-`lexicographicPermutations(999999)`应该返回2783915460。
+`lexicographicPermutations(999999)` should return 2783915460.
```js
assert(lexicographicPermutations(999999) == 2783915460);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-240-top-dice.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-240-top-dice.md
index dc705dc8992..392edfa3dda 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-240-top-dice.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-240-top-dice.md
@@ -1,28 +1,28 @@
---
id: 5900f45d1000cf542c50ff6f
-title: 问题240：顶部骰子
+title: 'Problem 240: Top Dice'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301887
dashedName: problem-240-top-dice
---
# --description--
-有1111种方式可以滚动五个6面骰子（编号为1到6的边），使前三个总和为15.一些例子是：
+There are 1111 ways in which five 6-sided dice (sides numbered 1 to 6) can be rolled so that the top three sum to 15. Some examples are:
-D1，D2，D3，D4，D5 = 4,3,6,3,5
+D1,D2,D3,D4,D5 = 4,3,6,3,5
-D1，D2，D3，D4，D5 = 4,3,3,5,6
+D1,D2,D3,D4,D5 = 4,3,3,5,6
-D1，D2，D3，D4，D5 = 3,3,3,6,6
+D1,D2,D3,D4,D5 = 3,3,3,6,6
-D1，D2，D3，D4，D5 = 6,6,3,3,3
+D1,D2,D3,D4,D5 = 6,6,3,3,3
-在多少个方面可以滚动20个12面骰子（编号为1到12的边），使前十个骰子总和达到70？
+In how many ways can twenty 12-sided dice (sides numbered 1 to 12) be rolled so that the top ten sum to 70?
# --hints--
-`euler240()`应该返回7448717393364182000。
+`euler240()` should return 7448717393364182000.
```js
assert.strictEqual(euler240(), 7448717393364182000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-241-perfection-quotients.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-241-perfection-quotients.md
index 40bd39dda54..b80b7872cb4 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-241-perfection-quotients.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-241-perfection-quotients.md
@@ -1,22 +1,22 @@
---
id: 5900f45d1000cf542c50ff70
-title: 问题241：完美商数
+title: 'Problem 241: Perfection Quotients'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301888
dashedName: problem-241-perfection-quotients
---
# --description--
-对于正整数n，令σ（n）为n的所有除数之和，因此例如σ（6）= 1 + 2 + 3 + 6 = 12。
+For a positive integer n, let σ(n) be the sum of all divisors of n, so e.g. σ(6) = 1 + 2 + 3 + 6 = 12.
-正如您所知，一个完美的数字是σ（n）= 2n的数字。
+A perfect number, as you probably know, is a number with σ(n) = 2n.
-让我们定义正整数asp（n）=σ（n）n的完美商。求出所有正整数n≤1018的和，其中p（n）的形式为k + 1/2，其中k是整数。
+Let us define the perfection quotient of a positive integer asp(n)= σ(n)n . Find the sum of all positive integers n ≤ 1018 for which p(n) has the form k + 1⁄2, where k is an integer.
# --hints--
-`euler241()`应该返回482316491800641150。
+`euler241()` should return 482316491800641150.
```js
assert.strictEqual(euler241(), 482316491800641150);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-242-odd-triplets.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-242-odd-triplets.md
index e5d6c79fc98..c8eb8d360c5 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-242-odd-triplets.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-242-odd-triplets.md
@@ -1,24 +1,24 @@
---
id: 5900f45f1000cf542c50ff71
-title: 问题242：奇数三胞胎
+title: 'Problem 242: Odd Triplets'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301889
dashedName: problem-242-odd-triplets
---
# --description--
-给定集合{1,2，...，n}，我们将f（n，k）定义为具有奇数元素之和的k元素子集的数量。例如，f（5,3）= 4，因为集合{1,2,3,4,5}有四个3元素子集具有奇数元素，即：{1,2,4}，{ 1,3,5}，{2,3,4}和{2,4,5}。
+Given the set {1,2,...,n}, we define f(n,k) as the number of its k-element subsets with an odd sum of elements. For example, f(5,3) = 4, since the set {1,2,3,4,5} has four 3-element subsets having an odd sum of elements, i.e.: {1,2,4}, {1,3,5}, {2,3,4} and {2,4,5}.
-当所有三个值n，k和f（n，k）都是奇数时，我们说它们产生奇数三元组\[n，k，f（n，k）]。
+When all three values n, k and f(n,k) are odd, we say that they make an odd-triplet \[n,k,f(n,k)].
-正好有五个奇数三元组，n≤10，即：\[1,1，f（1,1）= 1]，\[5,1，f（5,1）= 3]，\[5,5，f （5,5）= 1]，\[9,1，f（9,1）= 5]和\[9,9，f（9,9）= 1]。
+There are exactly five odd-triplets with n ≤ 10, namely: \[1,1,f(1,1) = 1], \[5,1,f(5,1) = 3], \[5,5,f(5,5) = 1], \[9,1,f(9,1) = 5] and \[9,9,f(9,9) = 1].
-n≤1012，有多少奇数三胞胎？
+How many odd-triplets are there with n ≤ 1012 ?
# --hints--
-`euler242()`应该返回997104142249036700。
+`euler242()` should return 997104142249036700.
```js
assert.strictEqual(euler242(), 997104142249036700);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-243-resilience.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-243-resilience.md
index e566e9c06b8..54c911233ab 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-243-resilience.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-243-resilience.md
@@ -1,22 +1,24 @@
---
id: 5900f4601000cf542c50ff73
-title: 问题243：恢复力
+title: 'Problem 243: Resilience'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301890
dashedName: problem-243-resilience
---
# --description--
-分子小于其分母的正分数称为适当分数。对于任何分母，d，将有d-1个适当的分数;例如，d = 12：1 / 12,2 / 12,3 / 12,4 / 12,5 / 12,6 / 12,7 / 12,8 / 12,9 / 12,10 / 12,11 / 12。
+A positive fraction whose numerator is less than its denominator is called a proper fraction.
-我们将称一个无法取消弹性部分的分数。此外，我们将分母R（d）的弹性定义为具有弹性的适当分数的比率;例如，R（12）= 4/11。事实上，d = 12是具有弹性R（d）<4/10的最小分母。
+For any denominator, d, there will be d−1 proper fractions; for example, with d = 12:1/12 , 2/12 , 3/12 , 4/12 , 5/12 , 6/12 , 7/12 , 8/12 , 9/12 , 10/12 , 11/12 .
-找到最小分母d，具有弹性R（d）<15499/94744。
+We shall call a fraction that cannot be cancelled down a resilient fraction. Furthermore we shall define the resilience of a denominator, R(d), to be the ratio of its proper fractions that are resilient; for example, R(12) = 4/11 . In fact, d = 12 is the smallest denominator having a resilience R(d) < 4/10 .
+
+Find the smallest denominator d, having a resilience R(d) < 15499/94744 .
# --hints--
-`euler243()`应该返回892371480。
+`euler243()` should return 892371480.
```js
assert.strictEqual(euler243(), 892371480);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-244-sliders.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-244-sliders.md
index 7f09925d243..dfc2ef5931d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-244-sliders.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-244-sliders.md
@@ -1,32 +1,34 @@
---
id: 5900f4601000cf542c50ff72
-title: 问题244：滑块
+title: 'Problem 244: Sliders'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301891
dashedName: problem-244-sliders
---
# --description--
-你可能知道游戏Fifteen Puzzle。在这里，我们有7个红色瓷砖和8个蓝色瓷砖，而不是编号瓷砖。移动由方块（左，右，上，下）的大写初始值表示，其中区块滑动，例如从配置（S）开始，通过序列LULUR我们到达配置（E）：
+You probably know the game Fifteen Puzzle. Here, instead of numbered tiles, we have seven red tiles and eight blue tiles.
-（S），（E）
+A move is denoted by the uppercase initial of the direction (Left, Right, Up, Down) in which the tile is slid, e.g. starting from configuration (S), by the sequence LULUR we reach the configuration (E):
-对于每个路径，其校验和由（伪代码）计算：
+(S), (E)
-checksum = 0 checksum =（checksum×243 + m1）mod 100 000 007 checksum =（checksum×243 + m2）mod 100 000 007 ... checksum =（checksum×243 + mn）mod 100 000 007其中mk是ASCII值移动序列中的第k个字母和移动的ASCII值为：
+For each path, its checksum is calculated by (pseudocode):
+
+checksum = 0 checksum = (checksum × 243 + m1) mod 100 000 007 checksum = (checksum × 243 + m2) mod 100 000 007 … checksum = (checksum × 243 + mn) mod 100 000 007 where mk is the ASCII value of the kth letter in the move sequence and the ASCII values for the moves are:
L76R82U85D68
-对于上面给出的序列LULUR，校验和将是19761398.现在，从配置（S）开始，找到达到配置（T）的所有最短路径。
+For the sequence LULUR given above, the checksum would be 19761398. Now, starting from configuration (S), find all shortest ways to reach configuration (T).
-（S），（T）
+(S), (T)
-具有最小长度的路径的所有校验和的总和是多少？
+What is the sum of all checksums for the paths having the minimal length?
# --hints--
-`euler244()`应该返回96356848。
+`euler244()` should return 96356848.
```js
assert.strictEqual(euler244(), 96356848);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-245-coresilience.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-245-coresilience.md
index d500830ea4a..47fc69a5fab 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-245-coresilience.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-245-coresilience.md
@@ -1,20 +1,20 @@
---
id: 5900f4621000cf542c50ff74
-title: 问题245：Coresilience
+title: 'Problem 245: Coresilience'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301892
dashedName: problem-245-coresilience
---
# --description--
-我们将称一个无法取消弹性部分的分数。此外，我们将分母R（d）的弹性定义为具有弹性的适当分数的比率;例如，R（12）= 4/11。
+We shall call a fraction that cannot be cancelled down a resilient fraction. Furthermore we shall define the resilience of a denominator, R(d), to be the ratio of its proper fractions that are resilient; for example, R(12) = 4⁄11.
-数d> 1的弹性则为φ（d）d-1，其中φ是欧拉的函数。我们进一步定义了n> 1的核心，即C（n）= n - φ（n）n - 1.素数p的核心是C（p）= 1p - 1.找出所有复合整数的总和1 <n≤2×1011，其中C（n）是单位分数。
+The resilience of a number d > 1 is then φ(d)d − 1 , where φ is Euler's totient function. We further define the coresilience of a number n > 1 as C(n)= n − φ(n)n − 1. The coresilience of a prime p is C(p) = 1p − 1. Find the sum of all composite integers 1 < n ≤ 2×1011, for which C(n) is a unit fraction.
# --hints--
-`euler245()`应该返回288084712410001。
+`euler245()` should return 288084712410001.
```js
assert.strictEqual(euler245(), 288084712410001);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-246-tangents-to-an-ellipse.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-246-tangents-to-an-ellipse.md
index b5023556abd..2d50a2e979d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-246-tangents-to-an-ellipse.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-246-tangents-to-an-ellipse.md
@@ -1,24 +1,26 @@
---
id: 5900f4621000cf542c50ff75
-title: 问题246：切线到椭圆
+title: 'Problem 246: Tangents to an ellipse'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301893
dashedName: problem-246-tangents-to-an-ellipse
---
# --description--
-椭圆的定义是：给定一个圆心c，其中心为M，半径为r，点G为d（G，M）
+A definition for an ellipse is:
-椭圆点的构造如下所示。
+Given a circle c with centre M and radius r and a point G such that d(G,M)
-给定的是M（-2000,1500）和G（8000,1500）。给定也是具有中心M和半径15000的圆c。与G和c等距的点的轨迹形成椭圆e。从外部的点P开始，绘制两个切线t1和t2到椭圆。让t1和t2接触椭圆的点为R和S.
+The construction of the points of the ellipse is shown below.
-对于多少格点P，角度RPS是否大于45度？
+Given are the points M(-2000,1500) and G(8000,1500). Given is also the circle c with centre M and radius 15000. The locus of the points that are equidistant from G and c form an ellipse e. From a point P outside e the two tangents t1 and t2 to the ellipse are drawn. Let the points where t1 and t2 touch the ellipse be R and S.
+
+For how many lattice points P is angle RPS greater than 45 degrees?
# --hints--
-`euler246()`应该返回810834388。
+`euler246()` should return 810834388.
```js
assert.strictEqual(euler246(), 810834388);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-247-squares-under-a-hyperbola.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-247-squares-under-a-hyperbola.md
index b7c252a6401..d9a5572eeac 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-247-squares-under-a-hyperbola.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-247-squares-under-a-hyperbola.md
@@ -1,24 +1,24 @@
---
id: 5900f4641000cf542c50ff76
-title: 问题247：双曲线下的正方形
+title: 'Problem 247: Squares under a hyperbola'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301894
dashedName: problem-247-squares-under-a-hyperbola
---
# --description--
-考虑受1≤x且0≤y≤1/ x约束的区域。
+Consider the region constrained by 1 ≤ x and 0 ≤ y ≤ 1/x.
-设S1是可以适应曲线的最大正方形。设S2是适合其余区域的最大正方形，依此类推。设Sn的指数是表示Sn左边的方格数和Sn以下的方格数的对（左下）。
+Let S1 be the largest square that can fit under the curve. Let S2 be the largest square that fits in the remaining area, and so on. Let the index of Sn be the pair (left, below) indicating the number of squares to the left of Sn and the number of squares below Sn.
-该图显示了一些用数字标记的方格。 S2左边有一个正方形，下面没有正方形，所以S2的索引是（1,0）。可以看出，S32的索引是（1,1），也是S50的索引。 50是Sn的指数为（1,1）的最大n。
+The diagram shows some such squares labelled by number. S2 has one square to its left and none below, so the index of S2 is (1,0). It can be seen that the index of S32 is (1,1) as is the index of S50. 50 is the largest n for which the index of Sn is (1,1).
-Sn指数为（3,3）的最大n是多少？
+What is the largest n for which the index of Sn is (3,3)?
# --hints--
-`euler247()`应该返回782252。
+`euler247()` should return 782252.
```js
assert.strictEqual(euler247(), 782252);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-248-numbers-for-which-eulers-totient-function-equals-13.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-248-numbers-for-which-eulers-totient-function-equals-13.md
index 6062ae41c2e..1641fd73a25 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-248-numbers-for-which-eulers-totient-function-equals-13.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-248-numbers-for-which-eulers-totient-function-equals-13.md
@@ -1,18 +1,20 @@
---
id: 5900f4651000cf542c50ff77
-title: 问题248：Euler的totient函数等于13的数字！
+title: 'Problem 248: Numbers for which Euler’s totient function equals 13!'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301895
dashedName: problem-248-numbers-for-which-eulers-totient-function-equals-13
---
# --description--
-φ（n）= 13的第一个数字n！是6227180929.找到第150,000个这样的数字。
+The first number n for which φ(n)=13! is 6227180929.
+
+Find the 150,000th such number.
# --hints--
-`euler248()`应返回23507044290。
+`euler248()` should return 23507044290.
```js
assert.strictEqual(euler248(), 23507044290);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-249-prime-subset-sums.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-249-prime-subset-sums.md
index 8953b199f62..71c8d1da197 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-249-prime-subset-sums.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-249-prime-subset-sums.md
@@ -1,18 +1,22 @@
---
id: 5900f4671000cf542c50ff79
-title: 问题249：Prime子集总和
+title: 'Problem 249: Prime Subset Sums'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301896
dashedName: problem-249-prime-subset-sums
---
# --description--
-令S = {2,3,5，...，4999}为小于5000的素数集。求出S的子集数，其元素之和为素数。输入最右边的16位数字作为答案。
+Let S = {2, 3, 5, ..., 4999} be the set of prime numbers less than 5000.
+
+Find the number of subsets of S, the sum of whose elements is a prime number.
+
+Enter the rightmost 16 digits as your answer.
# --hints--
-`euler249()`应该返回9275262564250418。
+`euler249()` should return 9275262564250418.
```js
assert.strictEqual(euler249(), 9275262564250418);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-25-1000-digit-fibonacci-number.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-25-1000-digit-fibonacci-number.md
index fc783c0c50b..ca7d277f447 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-25-1000-digit-fibonacci-number.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-25-1000-digit-fibonacci-number.md
@@ -1,138 +1,52 @@
---
id: 5900f3851000cf542c50fe98
-title: 问题25：1000位斐波纳契数
+title: 'Problem 25: 1000-digit Fibonacci number'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301897
dashedName: problem-25-1000-digit-fibonacci-number
---
# --description--
-Fibonacci序列由递归关系定义：
+The Fibonacci sequence is defined by the recurrence relation:
-F
+<div style='padding-left: 4em;'>F<sub>n</sub> = F<sub>n−1</sub> + F<sub>n−2</sub>, where F<sub>1</sub> = 1 and F<sub>2</sub> = 1.</div>
-<sub>n</sub>
+Hence the first 12 terms will be:
- = F
+<div style='padding-left: 4em; display: inline-grid; grid-template-rows: auto; row-gap: 7px;'><div>F<sub>1</sub> = 1</div><div>F<sub>2</sub> = 1</div><div>F<sub>3</sub> = 2</div><div>F<sub>4</sub> = 3</div><div>F<sub>5</sub> = 5</div><div>F<sub>6</sub> = 8</div><div>F<sub>7</sub> = 13</div><div>F<sub>8</sub> = 21</div><div>F<sub>9</sub> = 34</div><div>F<sub>10</sub> = 55</div><div>F<sub>11</sub> = 89</div><div>F<sub>12</sub> = 144</div></div>
-<sub>n-1</sub>
+The 12th term, F<sub>12</sub>, is the first term to contain three digits.
-- F
-
-<sub>n-2</sub>
-
- ，其中F
-
-<sub>1</sub>
-
- = 1且F
-
-<sub>2</sub>
-
- = 1。
-
-因此，前12个学期将是：
-
-F
-
-<sub>1</sub>
-
- = 1
-
-F
-
-<sub>2</sub>
-
- = 1
-
-F
-
-<sub>3</sub>
-
- = 2
-
-F
-
-<sub>4</sub>
-
- = 3
-
-F
-
-<sub>5</sub>
-
- = 5
-
-F
-
-<sub>6</sub>
-
- = 8
-
-F
-
-<sub>7</sub>
-
- = 13
-
-F
-
-<sub>8</sub>
-
- = 21
-
-F
-
-<sub>9</sub>
-
- = 34
-
-F
-
-<sub>10</sub>
-
- = 55
-
-F
-
-<sub>11</sub>
-
- = 89
-
-F
-
-<sub>12</sub>
-
- = 144
-
-第12个学期F
-
-<sub>12</sub>
-
-是第一个包含三位数的术语。 Fibonacci序列中包含`n个`数字的第一项的索引是多少？
+What is the index of the first term in the Fibonacci sequence to contain `n` digits?
# --hints--
-`digitFibonacci(5)`应该返回21。
+`digitFibonacci(5)` should return a number.
+
+```js
+assert(typeof digitFibonacci(5) === 'number');
+```
+
+`digitFibonacci(5)` should return 21.
```js
assert.strictEqual(digitFibonacci(5), 21);
```
-`digitFibonacci(10)`应该返回45。
+`digitFibonacci(10)` should return 45.
```js
assert.strictEqual(digitFibonacci(10), 45);
```
-`digitFibonacci(15)`应该返回69。
+`digitFibonacci(15)` should return 69.
```js
assert.strictEqual(digitFibonacci(15), 69);
```
-`digitFibonacci(20)`应该返回93。
+`digitFibonacci(20)` should return 93.
```js
assert.strictEqual(digitFibonacci(20), 93);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-250-250250.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-250-250250.md
index 720dfa83ae7..9c488260551 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-250-250250.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-250-250250.md
@@ -1,18 +1,18 @@
---
id: 5900f4661000cf542c50ff78
-title: 问题250：250250
+title: 'Problem 250: 250250'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301898
dashedName: problem-250-250250
---
# --description--
-找到{11,22,33，...，250250250250}的非空子集的数量，其元素的总和可被250整除。输入最右边的16位作为答案。
+Find the number of non-empty subsets of {11, 22, 33,..., 250250250250}, the sum of whose elements is divisible by 250. Enter the rightmost 16 digits as your answer.
# --hints--
-`euler250()`应该返回1425480602091519。
+`euler250()` should return 1425480602091519.
```js
assert.strictEqual(euler250(), 1425480602091519);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-251-cardano-triplets.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-251-cardano-triplets.md
index d423bb77a87..9023efb8182 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-251-cardano-triplets.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-251-cardano-triplets.md
@@ -1,24 +1,24 @@
---
id: 5900f4671000cf542c50ff7a
-title: 问题251：卡尔达诺三胞胎
+title: 'Problem 251: Cardano Triplets'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301899
dashedName: problem-251-cardano-triplets
---
# --description--
-如果满足条件，则正整数（a，b，c）的三元组称为Cardano三元组：
+A triplet of positive integers (a,b,c) is called a Cardano Triplet if it satisfies the condition:
-例如，（2,1,5）是Cardano Triplet。
+For example, (2,1,5) is a Cardano Triplet.
-存在149个Cardano三胞胎，其中a + b +c≤1000。
+There exist 149 Cardano Triplets for which a+b+c ≤ 1000.
-找出有多少卡尔达诺三胞胎存在，使a + b +c≤110,000,000。
+Find how many Cardano Triplets exist such that a+b+c ≤ 110,000,000.
# --hints--
-`euler251()`应该返回18946051。
+`euler251()` should return 18946051.
```js
assert.strictEqual(euler251(), 18946051);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-252-convex-holes.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-252-convex-holes.md
index 9f49c55b8c9..68251abd753 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-252-convex-holes.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-252-convex-holes.md
@@ -1,28 +1,28 @@
---
id: 5900f4691000cf542c50ff7b
-title: 问题252：凸孔
+title: 'Problem 252: Convex Holes'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301900
dashedName: problem-252-convex-holes
---
# --description--
-给定平面上的一组点，我们将凸孔定义为凸多边形，其具有作为顶点的任何给定点并且不包含其内部中的任何给定点（除了顶点之外，其他给定点可能位于在多边形的周长上）。
+Given a set of points on a plane, we define a convex hole to be a convex polygon having as vertices any of the given points and not containing any of the given points in its interior (in addition to the vertices, other given points may lie on the perimeter of the polygon).
-作为示例，下面的图像示出了一组二十个点和一些这样的凸孔。显示为红色七边形的凸孔具有等于1049694.5平方单位的面积，这是给定点集上的凸孔的最高可能区域。
+As an example, the image below shows a set of twenty points and a few such convex holes. The convex hole shown as a red heptagon has an area equal to 1049694.5 square units, which is the highest possible area for a convex hole on the given set of points.
-对于我们的例子，我们使用前20个点（T2k-1，T2k），对于k = 1,2，...，20，使用伪随机数生成器生成：
+For our example, we used the first 20 points (T2k−1, T2k), for k = 1,2,…,20, produced with the pseudo-random number generator:
-S0 = 290797 Sn + 1 = Sn2 mod 50515093 Tn =（Sn mod 2000）-1000
+S0 = 290797 Sn+1 = Sn2 mod 50515093 Tn = ( Sn mod 2000 ) − 1000
-即（527,144），（ - 488,732），（ - 454，-947），......
+i.e. (527, 144), (−488, 732), (−454, −947), …
-包含伪随机序列中前500个点的集合上凸孔的最大面积是多少？指定您的答案，包括小数点后的一位数。
+What is the maximum area for a convex hole on the set containing the first 500 points in the pseudo-random sequence? Specify your answer including one digit after the decimal point.
# --hints--
-`euler252()`应该返回104924。
+`euler252()` should return 104924.
```js
assert.strictEqual(euler252(), 104924);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-253-tidying-up.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-253-tidying-up.md
index 7071c8f82f2..78efca2bdc3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-253-tidying-up.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-253-tidying-up.md
@@ -1,32 +1,32 @@
---
id: 5900f4691000cf542c50ff7c
-title: 问题253：整理
+title: 'Problem 253: Tidying up'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301901
dashedName: problem-253-tidying-up
---
# --description--
-一个小孩子有一个“数字毛毛虫”，由四十个拼图块组成，每个拼图上有一个数字，当它们连成一条线时，按顺序显示数字1到40。
+A small child has a “number caterpillar” consisting of forty jigsaw pieces, each with one number on it, which, when connected together in a line, reveal the numbers 1 to 40 in order.
-每天晚上，孩子的父亲必须拿起散落在游戏室的毛毛虫的碎片。他随机拿起碎片并按正确顺序放置。当毛虫以这种方式建立时，它形成了逐渐融合在一起的不同部分。段的数量从零开始（没有放置件），通常增加到大约十一或十二，然后在完成单个段（放置所有件）之前往往再次下降。
+Every night, the child's father has to pick up the pieces of the caterpillar that have been scattered across the play room. He picks up the pieces at random and places them in the correct order. As the caterpillar is built up in this way, it forms distinct segments that gradually merge together. The number of segments starts at zero (no pieces placed), generally increases up to about eleven or twelve, then tends to drop again before finishing at a single segment (all pieces placed).
-例如：
+For example:
-片段放置所以Far121422936434554354 ......
+Piece Placed Segments So Far121422936434554354……
-设M是毛虫随机整理过程中遇到的最大段数。对于十件毛毛虫来说，每个M的可能性是多少
+Let M be the maximum number of segments encountered during a random tidy-up of the caterpillar. For a caterpillar of ten pieces, the number of possibilities for each M is
M Possibilities1512 2250912 31815264 41418112 5144000
-所以M的最可能值是3，平均值是385643/113400 = 3.400732，四舍五入到小数点后六位。
+so the most likely value of M is 3 and the average value is 385643⁄113400 = 3.400732, rounded to six decimal places.
-对于四十件毛毛虫而言，M最可能的价值是11;但是M的平均值是多少？将您的答案四舍五入到小数点后六位。
+The most likely value of M for a forty-piece caterpillar is 11; but what is the average value of M? Give your answer rounded to six decimal places.
# --hints--
-`euler253()`应返回11.492847。
+`euler253()` should return 11.492847.
```js
assert.strictEqual(euler253(), 11.492847);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-254-sums-of-digit-factorials.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-254-sums-of-digit-factorials.md
index db84985f922..028f20783ef 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-254-sums-of-digit-factorials.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-254-sums-of-digit-factorials.md
@@ -1,28 +1,28 @@
---
id: 5900f46b1000cf542c50ff7d
-title: 问题254：数字因子的总和
+title: 'Problem 254: Sums of Digit Factorials'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301902
dashedName: problem-254-sums-of-digit-factorials
---
# --description--
-将f（n）定义为n的数字的阶乘的总和。例如，f（342）= 3！ + 4！ + 2！ = 32。
+Define f(n) as the sum of the factorials of the digits of n. For example, f(342) = 3! + 4! + 2! = 32.
-将sf（n）定义为f（n）的数字之和。所以sf（342）= 3 + 2 = 5。
+Define sf(n) as the sum of the digits of f(n). So sf(342) = 3 + 2 = 5.
-将g（i）定义为最小的正整数n，使得sf（n）= i。虽然sf（342）是5，但sf（25）也是5，并且可以证实g（5）是25。
+Define g(i) to be the smallest positive integer n such that sf(n) = i. Though sf(342) is 5, sf(25) is also 5, and it can be verified that g(5) is 25.
-将sg（i）定义为g（i）的数字之和。所以sg（5）= 2 + 5 = 7。
+Define sg(i) as the sum of the digits of g(i). So sg(5) = 2 + 5 = 7.
-此外，可以证实g（20）是267并且1≤i≤20的Σsg（i）是156。
+Further, it can be verified that g(20) is 267 and ∑ sg(i) for 1 ≤ i ≤ 20 is 156.
-什么是Σsg（i）1≤i≤150？
+What is ∑ sg(i) for 1 ≤ i ≤ 150?
# --hints--
-`euler254()`应该返回8184523820510。
+`euler254()` should return 8184523820510.
```js
assert.strictEqual(euler254(), 8184523820510);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-255-rounded-square-roots.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-255-rounded-square-roots.md
index 5afbda444d3..8c6acc7113e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-255-rounded-square-roots.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-255-rounded-square-roots.md
@@ -1,30 +1,30 @@
---
id: 5900f46d1000cf542c50ff7f
-title: 问题255：圆角平方根
+title: 'Problem 255: Rounded Square Roots'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301903
dashedName: problem-255-rounded-square-roots
---
# --description--
-我们将正整数n的圆角平方根定义为n的平方根，四舍五入到最接近的整数。
+We define the rounded-square-root of a positive integer n as the square root of n rounded to the nearest integer.
-以下过程（基本上是Heron的方法适用于整数运算）找到n的舍入平方根：设d是数字n的位数。如果d为奇数，则设置x0 = 2×10（d-1）/ 2。如果d是偶数，则设置x0 = 7×10（d-2）/ 2。重复：
+The following procedure (essentially Heron's method adapted to integer arithmetic) finds the rounded-square-root of n: Let d be the number of digits of the number n. If d is odd, set x0 = 2×10(d-1)⁄2. If d is even, set x0 = 7×10(d-2)⁄2. Repeat:
-直到xk + 1 = xk。
+until xk+1 = xk.
-举个例子，让我们找到n = 4321.n的圆角平方根有4个数字，所以x0 = 7×10（4-2）/2 = 70.由于x2 = x1，我们在这里停止。因此，经过两次迭代后，我们发现4321的圆角平方根为66（实际平方根为65.7343137 ......）。
+As an example, let us find the rounded-square-root of n = 4321.n has 4 digits, so x0 = 7×10(4-2)⁄2 = 70. Since x2 = x1, we stop here. So, after just two iterations, we have found that the rounded-square-root of 4321 is 66 (the actual square root is 65.7343137…).
-使用此方法时所需的迭代次数非常低。例如，我们可以找到5位整数的圆角平方根（10,000≤n≤99,999），平均迭代次数为3.2102888889（平均值四舍五入到10位小数）。
+The number of iterations required when using this method is surprisingly low. For example, we can find the rounded-square-root of a 5-digit integer (10,000 ≤ n ≤ 99,999) with an average of 3.2102888889 iterations (the average value was rounded to 10 decimal places).
-使用上述过程，找到14位数字的圆角平方根（1013≤n<1014）所需的平均迭代次数是多少？将您的答案四舍五入到小数点后10位。
+Using the procedure described above, what is the average number of iterations required to find the rounded-square-root of a 14-digit number (1013 ≤ n < 1014)? Give your answer rounded to 10 decimal places.
-注意：符号⌊x⌋和⌈x⌉分别代表楼层功能和天花板功能。
+Note: The symbols ⌊x⌋ and ⌈x⌉ represent the floor function and ceiling function respectively.
# --hints--
-`euler255()`应返回4.447401118。
+`euler255()` should return 4.447401118.
```js
assert.strictEqual(euler255(), 4.447401118);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-256-tatami-free-rooms.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-256-tatami-free-rooms.md
index 26f1e34c422..7a55e36abec 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-256-tatami-free-rooms.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-256-tatami-free-rooms.md
@@ -1,34 +1,34 @@
---
id: 5900f46c1000cf542c50ff7e
-title: 问题256：无榻榻米房间
+title: 'Problem 256: Tatami-Free Rooms'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301904
dashedName: problem-256-tatami-free-rooms
---
# --description--
-榻榻米是长方形垫子，用于完全覆盖房间的地板，没有重叠。
+Tatami are rectangular mats, used to completely cover the floor of a room, without overlap.
-假设唯一可用的榻榻米尺寸为1×2，那么可以覆盖的房间的形状和大小显然存在一些限制。
+Assuming that the only type of available tatami has dimensions 1×2, there are obviously some limitations for the shape and size of the rooms that can be covered.
-对于这个问题，我们只考虑具有整数尺寸a，b和均匀尺寸s = a·b的矩形房间。我们使用术语“大小”来表示房间的地板表面积，并且 - 不失一般性 - 我们添加条件a≤b。
+For this problem, we consider only rectangular rooms with integer dimensions a, b and even size s = a·b. We use the term 'size' to denote the floor surface area of the room, and — without loss of generality — we add the condition a ≤ b.
-铺设榻榻米时要遵循一条规则：四个不同垫子的角落必须相遇。例如，考虑下面两个4×4房间的安排：
+There is one rule to follow when laying out tatami: there must be no points where corners of four different mats meet. For example, consider the two arrangements below for a 4×4 room:
-左边的排列是可以接受的，而右边的排列则不是：中间的红色“X”标志着四个榻榻米相遇的点。
+The arrangement on the left is acceptable, whereas the one on the right is not: a red "X" in the middle, marks the point where four tatami meet.
-由于这个规则，某些大小合适的房间不能覆盖榻榻米：我们称之为榻榻米房间。此外，我们将T（s）定义为大小为s的无榻榻米房间的数量。
+Because of this rule, certain even-sized rooms cannot be covered with tatami: we call them tatami-free rooms. Further, we define T(s) as the number of tatami-free rooms of size s.
-最小的无榻榻米房间尺寸为s = 70，尺寸为7×10。所有其他尺寸为s = 70的房间都可以铺上榻榻米;它们是：1×70,2×35和5×14。因此，T（70）= 1。
+The smallest tatami-free room has size s = 70 and dimensions 7×10. All the other rooms of size s = 70 can be covered with tatami; they are: 1×70, 2×35 and 5×14. Hence, T(70) = 1.
-类似地，我们可以验证T（1320）= 5，因为正好有5个无榻榻米的房间，大小为s = 1320：20×66,22×60,24×55,30×44和33×40。事实上，s = 1320是T（s）= 5的最小房间尺寸。
+Similarly, we can verify that T(1320) = 5 because there are exactly 5 tatami-free rooms of size s = 1320: 20×66, 22×60, 24×55, 30×44 and 33×40. In fact, s = 1320 is the smallest room-size s for which T(s) = 5.
-找到T（s）= 200的最小房间大小。
+Find the smallest room-size s for which T(s) = 200.
# --hints--
-`euler256()`应该返回85765680。
+`euler256()` should return 85765680.
```js
assert.strictEqual(euler256(), 85765680);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-257-angular-bisectors.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-257-angular-bisectors.md
index 61bc75cc163..629b1f0340e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-257-angular-bisectors.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-257-angular-bisectors.md
@@ -1,22 +1,26 @@
---
id: 5900f46e1000cf542c50ff80
-title: 问题257：角度平分器
+title: 'Problem 257: Angular Bisectors'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301905
dashedName: problem-257-angular-bisectors
---
# --description--
-给定是一个整数边三角形ABC，边a≤b≤c。 （AB = c，BC = a且AC = b）。三角形的角平分线在点E，F和G处与两侧相交（见下图）。
+Given is an integer sided triangle ABC with sides a ≤ b ≤ c.
-段EF，EG和FG将三角形ABC划分为四个较小的三角形：AEG，BFE，CGF和EFG。可以证明，对于这四个三角形中的每一个，比率区域（ABC）/面积（子三角形）是合理的。然而，存在这些比率中的一些或全部是积分的三角形。
+(AB = c, BC = a and AC = b).
-存在多少个周长≤100,000,000的三角形ABC，以便比率面积（ABC）/面积（AEG）是整数？
+The angular bisectors of the triangle intersect the sides at points E, F and G (see picture below).
+
+The segments EF, EG and FG partition the triangle ABC into four smaller triangles: AEG, BFE, CGF and EFG. It can be proven that for each of these four triangles the ratio area(ABC)/area(subtriangle) is rational. However, there exist triangles for which some or all of these ratios are integral.
+
+How many triangles ABC with perimeter≤100,000,000 exist so that the ratio area(ABC)/area(AEG) is integral?
# --hints--
-`euler257()`应该返回139012411。
+`euler257()` should return 139012411.
```js
assert.strictEqual(euler257(), 139012411);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-258-a-lagged-fibonacci-sequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-258-a-lagged-fibonacci-sequence.md
index 6ed5c1e4927..963b52807ec 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-258-a-lagged-fibonacci-sequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-258-a-lagged-fibonacci-sequence.md
@@ -1,20 +1,20 @@
---
id: 5900f46e1000cf542c50ff81
-title: 问题258：滞后的斐波那契序列
+title: 'Problem 258: A lagged Fibonacci sequence'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301906
dashedName: problem-258-a-lagged-fibonacci-sequence
---
# --description--
-序列定义为：
+A sequence is defined as:
-gk = 1，对于0≤k≤1999gk = gk-2000 + gk-1999，对于k≥2000。对于k = 1018，找到gk mod 20092010。
+gk = 1, for 0 ≤ k ≤ 1999 gk = gk-2000 + gk-1999, for k ≥ 2000. Find gk mod 20092010 for k = 1018.
# --hints--
-`euler258()`应该返回12747994。
+`euler258()` should return 12747994.
```js
assert.strictEqual(euler258(), 12747994);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-259-reachable-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-259-reachable-numbers.md
index d43086f8ac1..c9eab8e6efd 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-259-reachable-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-259-reachable-numbers.md
@@ -1,22 +1,22 @@
---
id: 5900f4701000cf542c50ff82
-title: 问题259：可达号码
+title: 'Problem 259: Reachable Numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301907
dashedName: problem-259-reachable-numbers
---
# --description--
-如果正整数可以由遵循以下规则的算术表达式产生，则称为正整数：
+A positive integer will be called reachable if it can result from an arithmetic expression obeying the following rules:
-按顺序使用数字1到9，每个数字恰好一次。可以连接任何连续的数字（例如，使用数字2,3和4，我们获得数字234）。只允许四种常用的二进制算术运算（加法，减法，乘法和除法）。每个操作可以使用任何次数，或者根本不使用。不允许一元减号。可以使用任意数量（可能嵌套的）括号来定义操作的顺序。例如，42是可达的，因为（1/23） *（（4* 5）-6）\*（78-9）= 42。
+Uses the digits 1 through 9, in that order and exactly once each. Any successive digits can be concatenated (for example, using the digits 2, 3 and 4 we obtain the number 234). Only the four usual binary arithmetic operations (addition, subtraction, multiplication and division) are allowed. Each operation can be used any number of times, or not at all. Unary minus is not allowed. Any number of (possibly nested) parentheses may be used to define the order of operations. For example, 42 is reachable, since (1/23) *((4*5)-6) \* (78-9) = 42.
-所有正可达整数的总和是多少？
+What is the sum of all positive reachable integers?
# --hints--
-`euler259()`应返回20101196798。
+`euler259()` should return 20101196798.
```js
assert.strictEqual(euler259(), 20101196798);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-26-reciprocal-cycles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-26-reciprocal-cycles.md
index 12871e07494..d6ee972f907 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-26-reciprocal-cycles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-26-reciprocal-cycles.md
@@ -1,90 +1,48 @@
---
id: 5900f3861000cf542c50fe99
-title: 问题26：互惠周期
+title: 'Problem 26: Reciprocal cycles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301908
dashedName: problem-26-reciprocal-cycles
---
# --description--
-单位分数在分子中包含1。给出分母2到10的单位分数的十进制表示：
+A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
-<sup><sub>二分之一</sub></sup>
+<div style='padding-left: 4em; display: inline-grid; grid-template-rows: auto; row-gap: 7px;'><div><sup>1</sup>/<sub>2</sub> = 0.5</div><div><sup>1</sup>/<sub>3</sub> = 0.(3)</div><div><sup>1</sup>/<sub>4</sub> = 0.25</div><div><sup>1</sup>/<sub>5</sub> = 0.2</div><div><sup>1</sup>/<sub>6</sub> = 0.1(6)</div><div><sup>1</sup>/<sub>7</sub> = 0.(142857)</div><div><sup>1</sup>/<sub>8</sub> = 0.125</div><div><sup>1</sup>/<sub>9</sub> = 0.(1)</div><div><sup>1</sup>/<sub>10</sub> = 0.1</div></div>
- = 0.5
+Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that <sup>1</sup>/<sub>7</sub> has a 6-digit recurring cycle.
-<sup><sub>三分之一</sub></sup>
-
- = 0（3）
-
-<sup><sub>四分之一</sub></sup>
-
- = 0.25
-
-<sup>的<sub>1/5</sub></sup>
-
- = 0.2
-
-<sup><sub>六分之一</sub></sup>
-
- = 0.1（6）
-
-<sup><sub>七分之一</sub></sup>
-
- = 0（142857）
-
-<sup><sub>八分之一</sub></sup>
-
- = 0.125
-
-<sup><sub>九分之一</sub></sup>
-
- = 0（1）
-
-<sup><sub>一十分之一</sub></sup>
-
- = 0.1
-
-其中0.1（6）表示0.166666 ...，并具有1位循环周期。可以看出，
-
-<sup>1</sup>
-
- /
-
-<sub>7</sub>
-
-具有6位循环周期。找到`d` < `n`的值，其中
-
-<sup>1</sup>
-
- /
-
-<sub>d</sub>
-
-包含其小数部分中最长的循环周期。
+Find the value of `d` < `n` for which <sup>1</sup>/<sub>d</sub> contains the longest recurring cycle in its decimal fraction part.
# --hints--
-`reciprocalCycles(700)`应该返回659。
+`reciprocalCycles(700)` should return a number.
+
+```js
+assert(typeof reciprocalCycles(700) === 'number');
+```
+
+`reciprocalCycles(700)` should return 659.
```js
assert(reciprocalCycles(700) == 659);
```
-`reciprocalCycles(800)`应该返回743。
+`reciprocalCycles(800)` should return 743.
```js
assert(reciprocalCycles(800) == 743);
```
-`reciprocalCycles(900)`应该返回887。
+`reciprocalCycles(900)` should return 887.
```js
assert(reciprocalCycles(900) == 887);
```
-`reciprocalCycles(1000)`应该返回983。
+`reciprocalCycles(1000)` should return 983.
```js
assert(reciprocalCycles(1000) == 983);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-260-stone-game.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-260-stone-game.md
index dedc5e3d352..55f0dccfd5d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-260-stone-game.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-260-stone-game.md
@@ -1,28 +1,30 @@
---
id: 5900f4701000cf542c50ff83
-title: 问题260：石头游戏
+title: 'Problem 260: Stone Game'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301909
dashedName: problem-260-stone-game
---
# --description--
-一场比赛由三堆石头和两名球员组成。在轮到时，玩家从堆中移除一个或多个宝石。但是，如果她从多个桩中取石头，她必须从每个选定的桩中移除相同数量的石头。
+A game is played with three piles of stones and two players.
-换句话说，玩家选择一些N> 0并从任何一个桩中移除：N个石头;或任意两桩（总共2N）中的每一块N石;或者从三个桩中的每一个（总共3N）中的N个石头。拿最后一块石头的玩家赢了比赛。
+At her turn, a player removes one or more stones from the piles. However, if she takes stones from more than one pile, she must remove the same number of stones from each of the selected piles.
-获胜配置是第一个玩家可以强制获胜的配置。例如，（0,0,13），（0,11,11）和（5,5,5）是获胜配置，因为第一个玩家可以立即移除所有宝石。
+In other words, the player chooses some N>0 and removes: N stones from any single pile; or N stones from each of any two piles (2N total); or N stones from each of the three piles (3N total). The player taking the last stone(s) wins the game.
-失败的配置是第二个玩家可以强制获胜的配置，无论第一个玩家做什么。例如，（0,1,2）和（1,3,3）正在失去配置：任何合法移动都会为第二个玩家留下获胜配置。
+A winning configuration is one where the first player can force a win. For example, (0,0,13), (0,11,11) and (5,5,5) are winning configurations because the first player can immediately remove all stones.
-考虑所有丢失的配置（xi，yi，zi），其中xi≤yi≤zi≤100。我们可以验证Σ（xi + yi + zi）= 173895。
+A losing configuration is one where the second player can force a win, no matter what the first player does. For example, (0,1,2) and (1,3,3) are losing configurations: any legal move leaves a winning configuration for the second player.
-求Σ（xi + yi + zi）其中（xi，yi，zi）在失去的配置范围内，xi≤yi≤zi≤1000。
+Consider all losing configurations (xi,yi,zi) where xi ≤ yi ≤ zi ≤ 100. We can verify that Σ(xi+yi+zi) = 173895 for these.
+
+Find Σ(xi+yi+zi) where (xi,yi,zi) ranges over the losing configurations with xi ≤ yi ≤ zi ≤ 1000.
# --hints--
-`euler260()`应该返回167542057。
+`euler260()` should return 167542057.
```js
assert.strictEqual(euler260(), 167542057);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-261-pivotal-square-sums.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-261-pivotal-square-sums.md
index 02f518bba2b..b265e7910b2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-261-pivotal-square-sums.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-261-pivotal-square-sums.md
@@ -1,22 +1,22 @@
---
id: 5900f4711000cf542c50ff84
-title: 问题261：Pivotal Square Sums
+title: 'Problem 261: Pivotal Square Sums'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301910
dashedName: problem-261-pivotal-square-sums
---
# --description--
-如果存在一对m> 0且n≥k的整数，则让我们称正整数ka square-pivot，使得直到k的（m + 1）个连续正方形的总和等于m个连续正方形的总和从（n + 1）开始：
+Let us call a positive integer k a square-pivot, if there is a pair of integers m > 0 and n ≥ k, such that the sum of the (m+1) consecutive squares up to k equals the sum of the m consecutive squares from (n+1) on:
-（km）2 + ... + k2 =（n + 1）2 + ... +（n + m）2。
+(k-m)2 + ... + k2 = (n+1)2 + ... + (n+m)2.
-一些小的方形枢轴是4：32 + 42 = 52 21：202 + 212 = 292 24：212 + 222 + 232 + 242 = 252 + 262 + 272 110：1082 + 1092 + 1102 = 1332 + 1342找到所有的总和不同的方形枢轴≤1010。
+Some small square-pivots are 4: 32 + 42 = 52 21: 202 + 212 = 292 24: 212 + 222 + 232 + 242 = 252 + 262 + 272 110: 1082 + 1092 + 1102 = 1332 + 1342Find the sum of all distinct square-pivots ≤ 1010.
# --hints--
-`euler261()`应该返回238890850232021。
+`euler261()` should return 238890850232021.
```js
assert.strictEqual(euler261(), 238890850232021);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-262-mountain-range.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-262-mountain-range.md
index 88e135d7cae..01cc16de4e1 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-262-mountain-range.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-262-mountain-range.md
@@ -1,28 +1,28 @@
---
id: 5900f4731000cf542c50ff85
-title: 问题262：山脉
+title: 'Problem 262: Mountain Range'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301911
dashedName: problem-262-mountain-range
---
# --description--
-以下等式表示山区的连续地形，给出任意点（x，y）处的高程h：
+The following equation represents the continuous topography of a mountainous region, giving the elevation h at any point (x,y):
-蚊子打算从A（200,200）飞到B（1400,1400），而不会离开0≤x，y≤1600的区域。
+A mosquito intends to fly from A(200,200) to B(1400,1400), without leaving the area given by 0 ≤ x, y ≤ 1600.
-由于中间的山脉，它首先直线上升到A'点，具有高度f。然后，在保持在相同高度f的同时，它在任何障碍物周围飞行，直到它到达B正上方的B'点。
+Because of the intervening mountains, it first rises straight up to a point A', having elevation f. Then, while remaining at the same elevation f, it flies around any obstacles until it arrives at a point B' directly above B.
-首先，确定fmin，它是允许从A到B的这种行程的最小恒定高度，同时保持在指定区域。然后，找到A'和B'之间最短路径的长度，同时以恒定高度fmin飞行。
+First, determine fmin which is the minimum constant elevation allowing such a trip from A to B, while remaining in the specified area. Then, find the length of the shortest path between A' and B', while flying at that constant elevation fmin.
-将该长度作为答案，四舍五入到小数点后三位。
+Give that length as your answer, rounded to three decimal places.
-注意：为方便起见，下面以适合大多数编程语言的形式重复上面显示的高程函数：h =（5000-0.005 *（x* x + y *y + x* y）+12.5 *（x + y））* exp（ - abs（0.000001 *（x* x + y *y）-0.0015* （x + y）+0.7））
+Note: For convenience, the elevation function shown above is repeated below, in a form suitable for most programming languages: h=( 5000-0.005*(x*x+y*y+x*y)+12.5*(x+y) )* exp( -abs(0.000001*(x*x+y*y)-0.0015*(x+y)+0.7) )
# --hints--
-`euler262()`应返回2531.205。
+`euler262()` should return 2531.205.
```js
assert.strictEqual(euler262(), 2531.205);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-263-an-engineers-dream-come-true.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-263-an-engineers-dream-come-true.md
index 16f44da8bdf..24fca67d821 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-263-an-engineers-dream-come-true.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-263-an-engineers-dream-come-true.md
@@ -1,28 +1,34 @@
---
id: 5900f4741000cf542c50ff86
-title: 问题263：工程师“梦想成真
+title: 'Problem 263: An engineers'' dream come true'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301912
dashedName: problem-263-an-engineers-dream-come-true
---
# --description--
-考虑数字6. 6的除数是：1,2,3和6.从1到6并且包括6的每个数可以写成6：1 = 1,2 = 2,3 =的不同除数之和1 + 2,4 = 1 + 3,5 = 2 + 3,6 = 6。如果从1到n并且包括n的每个数可以表示为n的不同除数之和，则数n被称为实数。
+Consider the number 6. The divisors of 6 are: 1,2,3 and 6.
-一对差异为6的连续素数被称为性感对（因为“性”是拉丁语中的“六”）。第一个性感的是（23,29）。
+Every number from 1 up to and including 6 can be written as a sum of distinct divisors of 6:
-我们偶尔会发现一对三对，这意味着三对连续的性感素数对，这样每对的第二个成员就是下一对的第一个成员。
+1=1, 2=2, 3=1+2, 4=1+3, 5=2+3, 6=6.
-我们将数字n称为：（n-9，n-3），（n-3，n + 3），（n + 3，n + 9）形成三对，数字n-8 ，n-4，n，n + 4和n + 8都是实用的，
+A number n is called a practical number if every number from 1 up to and including n can be expressed as a sum of distinct divisors of n.
-工程师的天堂。
+A pair of consecutive prime numbers with a difference of six is called a sexy pair (since "sex" is the Latin word for "six"). The first sexy pair is (23, 29).
-找出前四个工程师天堂的总和。
+We may occasionally find a triple-pair, which means three consecutive sexy prime pairs, such that the second member of each pair is the first member of the next pair.
+
+We shall call a number n such that : (n-9, n-3), (n-3,n+3), (n+3, n+9) form a triple-pair, and the numbers n-8, n-4, n, n+4 and n+8 are all practical,
+
+an engineers’ paradise.
+
+Find the sum of the first four engineers’ paradises.
# --hints--
-`euler263()`应该返回2039506520。
+`euler263()` should return 2039506520.
```js
assert.strictEqual(euler263(), 2039506520);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-264-triangle-centres.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-264-triangle-centres.md
index d2258c1499a..34be7dede4c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-264-triangle-centres.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-264-triangle-centres.md
@@ -1,24 +1,34 @@
---
id: 5900f4751000cf542c50ff87
-title: 问题264：三角中心
+title: 'Problem 264: Triangle Centres'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301913
dashedName: problem-264-triangle-centres
---
# --description--
-考虑所有三角形：格点上的所有顶点。原点O. Orthocentre在圆点H（5,0）处的圆环。有九个这样的三角形，其周长≤50。按周长的升序列出并显示，它们是：
+Consider all the triangles having:
-A（-4,3），B（5,0），C（4，-3）A（4,3），B（5,0），C（-4，-3）A（-3,4） ），B（5,0），C（3，-4）A（3,4），B（5,0），C（-3，-4）A（0,5），B（5,0） ），C（0，-5）A（1,8），B（8，-1），C（-4，-7）A（8,1），B（1，-8），C（ - 4,7）A（2,9），B（9，-2），C（-6，-7）A（9,2），B（2，-9），C（-6,7）
+All their vertices on lattice points.
-他们的周长总和，四舍五入到小数点后四位，是291.0089。
+Circumcentre at the origin O.
-查找周长≤105的所有此类三角形。输入其周长的总和四舍五入到小数点后四位。
+Orthocentre at the point H(5, 0).
+
+There are nine such triangles having a perimeter ≤ 50.
+
+Listed and shown in ascending order of their perimeter, they are:
+
+A(-4, 3), B(5, 0), C(4, -3) A(4, 3), B(5, 0), C(-4, -3) A(-3, 4), B(5, 0), C(3, -4) A(3, 4), B(5, 0), C(-3, -4) A(0, 5), B(5, 0), C(0, -5) A(1, 8), B(8, -1), C(-4, -7) A(8, 1), B(1, -8), C(-4, 7) A(2, 9), B(9, -2), C(-6, -7) A(9, 2), B(2, -9), C(-6, 7)
+
+The sum of their perimeters, rounded to four decimal places, is 291.0089.
+
+Find all such triangles with a perimeter ≤ 105. Enter as your answer the sum of their perimeters rounded to four decimal places.
# --hints--
-`euler264()`应该返回2816417.1055。
+`euler264()` should return 2816417.1055.
```js
assert.strictEqual(euler264(), 2816417.1055);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-265-binary-circles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-265-binary-circles.md
index 6ed2507df3e..0ed757aa65a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-265-binary-circles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-265-binary-circles.md
@@ -1,28 +1,28 @@
---
id: 5900f4761000cf542c50ff88
-title: 问题265：二进制圆圈
+title: 'Problem 265: Binary Circles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301914
dashedName: problem-265-binary-circles
---
# --description--
-2N二进制数字可以放在一个圆圈中，这样所有N位顺时针子序列都是不同的。
+2N binary digits can be placed in a circle so that all the N-digit clockwise subsequences are distinct.
-对于N = 3，两个这样的圆形布置是可能的，忽略旋转：
+For N=3, two such circular arrangements are possible, ignoring rotations:
-对于第一种布置，顺时针顺序的3位子序列是：000,001,010,101,011,111,110和100。
+For the first arrangement, the 3-digit subsequences, in clockwise order, are: 000, 001, 010, 101, 011, 111, 110 and 100.
-通过将以全零的子序列开始的二进制数字连接为最高有效位并顺时针进行，可以将每个循环排列编码为数字。因此，N = 3的两种排列表示为23和29：00010111 2 = 23 00011101 2 = 29
+Each circular arrangement can be encoded as a number by concatenating the binary digits starting with the subsequence of all zeros as the most significant bits and proceeding clockwise. The two arrangements for N=3 are thus represented as 23 and 29: 00010111 2 = 23 00011101 2 = 29
-调用S（N）唯一数值表示的总和，我们可以看到S（3）= 23 + 29 = 52。
+Calling S(N) the sum of the unique numeric representations, we can see that S(3) = 23 + 29 = 52.
-找到S（5）。
+Find S(5).
# --hints--
-`euler265()`应该返回209110240768。
+`euler265()` should return 209110240768.
```js
assert.strictEqual(euler265(), 209110240768);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-266-pseudo-square-root.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-266-pseudo-square-root.md
index fe22e857e25..18569d5b8e9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-266-pseudo-square-root.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-266-pseudo-square-root.md
@@ -1,20 +1,26 @@
---
id: 5900f4771000cf542c50ff89
-title: 问题266：伪平方根
+title: 'Problem 266: Pseudo Square Root'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301915
dashedName: problem-266-pseudo-square-root
---
# --description--
-12的除数是：1,2,3,4,6和12. 12的最大除数不超过12的平方根是3.我们将称为整数n的最大除数，它不超过n的伪平方根（PSR）的平方根。可以看出PSR（3102）= 47。
+The divisors of 12 are: 1,2,3,4,6 and 12.
-设p是190以下质数的乘积。求PSR（p）mod 1016。
+The largest divisor of 12 that does not exceed the square root of 12 is 3.
+
+We shall call the largest divisor of an integer n that does not exceed the square root of n the pseudo square root (PSR) of n.
+
+It can be seen that PSR(3102)=47.
+
+Let p be the product of the primes below 190. Find PSR(p) mod 1016.
# --hints--
-`euler266()`应该返回1096883702440585。
+`euler266()` should return 1096883702440585.
```js
assert.strictEqual(euler266(), 1096883702440585);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-267-billionaire.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-267-billionaire.md
index 7c3c4ebc2e0..0c308ff3ec8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-267-billionaire.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-267-billionaire.md
@@ -1,18 +1,28 @@
---
id: 5900f4771000cf542c50ff8a
-title: 问题267：亿万富翁
+title: 'Problem 267: Billionaire'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301916
dashedName: problem-267-billionaire
---
# --description--
-您将获得独特的投资机会。从1英镑的资本开始，你可以选择固定比例f来投注一个公平的硬币投掷1000次投掷。你的回报是对你的投注加倍，你失去了对尾巴的赌注。例如，如果f = 1/4，对于第一次投掷，你下注0.25英镑，如果头部出现你赢得0.5英镑，那么就有1.5英镑。然后你下注0.375英镑，如果第二次折腾是尾巴，你有1.125英镑。选择f以最大限度地提高在1,000次翻转后获得至少1,000,000,000英镑的机会，您成为亿万富翁的机会是多少？假设所有计算都是精确的（没有舍入），但是将答案四舍五入到小数点后面的12位数，形式为0.abcdefghijkl。
+You are given a unique investment opportunity.
+
+Starting with £1 of capital, you can choose a fixed proportion, f, of your capital to bet on a fair coin toss repeatedly for 1000 tosses.
+
+Your return is double your bet for heads and you lose your bet for tails.
+
+For example, if f = 1/4, for the first toss you bet £0.25, and if heads comes up you win £0.5 and so then have £1.5. You then bet £0.375 and if the second toss is tails, you have £1.125.
+
+Choosing f to maximize your chances of having at least £1,000,000,000 after 1,000 flips, what is the chance that you become a billionaire?
+
+All computations are assumed to be exact (no rounding), but give your answer rounded to 12 digits behind the decimal point in the form 0.abcdefghijkl.
# --hints--
-`euler267()`应返回0.999992836187。
+`euler267()` should return 0.999992836187.
```js
assert.strictEqual(euler267(), 0.999992836187);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-268-counting-numbers-with-at-least-four-distinct-prime-factors-less-than-100.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-268-counting-numbers-with-at-least-four-distinct-prime-factors-less-than-100.md
index 4808aa7151f..d11cb99eef8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-268-counting-numbers-with-at-least-four-distinct-prime-factors-less-than-100.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-268-counting-numbers-with-at-least-four-distinct-prime-factors-less-than-100.md
@@ -1,21 +1,22 @@
---
id: 5900f4791000cf542c50ff8b
-title: 问题268：计算至少有四个不同素因子小于100的数字
+title: >-
+ Problem 268: Counting numbers with at least four distinct prime factors less than 100
challengeType: 5
-videoUrl: ''
+forumTopicId: 301917
dashedName: >-
problem-268-counting-numbers-with-at-least-four-distinct-prime-factors-less-than-100
---
# --description--
-可以证实，有23个小于1000的正整数可被至少四个小于100的不同质数整除。
+It can be verified that there are 23 positive integers less than 1000 that are divisible by at least four distinct primes less than 100.
-找出少于1016的正整数可被至少四个小于100的不同质数整除。
+Find how many positive integers less than 1016 are divisible by at least four distinct primes less than 100.
# --hints--
-`euler268()`应该返回785478606870985。
+`euler268()` should return 785478606870985.
```js
assert.strictEqual(euler268(), 785478606870985);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-269-polynomials-with-at-least-one-integer-root.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-269-polynomials-with-at-least-one-integer-root.md
index cfa6e13fc8e..b4b26926105 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-269-polynomials-with-at-least-one-integer-root.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-269-polynomials-with-at-least-one-integer-root.md
@@ -1,24 +1,28 @@
---
id: 5900f4791000cf542c50ff8c
-title: 问题269：具有至少一个整数根的多项式
+title: 'Problem 269: Polynomials with at least one integer root'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301918
dashedName: problem-269-polynomials-with-at-least-one-integer-root
---
# --description--
-多项式P（x）的根或零是等式P（x）= 0的解。将Pn定义为其系数是n的数字的多项式。例如，P5703（x）= 5x3 + 7x2 + 3。
+A root or zero of a polynomial P(x) is a solution to the equation P(x) = 0.
-我们可以看到：Pn（0）是n的最后一位，Pn（1）是n的位数之和，Pn（10）是n本身。定义Z（k）作为正整数的数量，n ，不超过多项式Pn具有至少一个整数根的k。
+Define Pn as the polynomial whose coefficients are the digits of n.
-可以证实Z（100 000）是14696。
+For example, P5703(x) = 5x3 + 7x2 + 3.
-什么是Z（1016）？
+We can see that:Pn(0) is the last digit of n, Pn(1) is the sum of the digits of n, Pn(10) is n itself.Define Z(k) as the number of positive integers, n, not exceeding k for which the polynomial Pn has at least one integer root.
+
+It can be verified that Z(100 000) is 14696.
+
+What is Z(1016)?
# --hints--
-`euler269()`应该返回1311109198529286。
+`euler269()` should return 1311109198529286.
```js
assert.strictEqual(euler269(), 1311109198529286);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-27-quadratic-primes.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-27-quadratic-primes.md
index 119b01d77a0..a694ee2a595 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-27-quadratic-primes.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-27-quadratic-primes.md
@@ -1,40 +1,58 @@
---
id: 5900f3871000cf542c50fe9a
-title: 问题27：二次素数
+title: 'Problem 27: Quadratic primes'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301919
dashedName: problem-27-quadratic-primes
---
# --description--
-欧拉发现了显着的二次公式：$ n ^ 2 + n + 41 $事实证明，公式将为连续的整数值$ 0 \\ le n \\ le 39 $产生40个素数。但是，当$ n = 40时，40 ^ 2 + 40 + 41 = 40（40 + 1）+ 41 $可被41整除，当然$ n = 41时，41 ^ 2 + 41 + 41 $显然可以被整除41.发现了令人难以置信的公式$ n ^ 2 - 79n + 1601 $，它为连续值$ 0 \\ le n \\ le 79 $产生80个素数。系数-79和1601的乘积是-126479。考虑形式的二次方：
+Euler discovered the remarkable quadratic formula:
-$ n ^ 2 + an + b $，其中$ | a | <range $和$ | b | \\ le $ $其中$ | n | $是$ n $的模数/绝对值，例如$ | 11 | = 11 $和$ | -4 | = 4 $
+<div style='margin-left: 4em;'>$n^2 + n + 41$</div>
-找到系数的乘积，$ a $和$ b $，用于生成连续值$ n $的最大素数数的二次表达式，从$ n = 0 $开始。
+It turns out that the formula will produce 40 primes for the consecutive integer values $0 \\le n \\le 39$. However, when $n = 40, 40^2 + 40 + 41 = 40(40 + 1) + 41$ is divisible by 41, and certainly when $n = 41, 41^2 + 41 + 41$ is clearly divisible by 41.
+
+The incredible formula $n^2 - 79n + 1601$ was discovered, which produces 80 primes for the consecutive values $0 \\le n \\le 79$. The product of the coefficients, −79 and 1601, is −126479.
+
+Considering quadratics of the form:
+
+<div style='margin-left: 4em;'>
+ $n^2 + an + b$, where $|a| < range$ and $|b| \le range$<br>
+ where $|n|$ is the modulus/absolute value of $n$<br>
+ e.g. $|11| = 11$ and $|-4| = 4$<br>
+</div>
+
+Find the product of the coefficients, $a$ and $b$, for the quadratic expression that produces the maximum number of primes for consecutive values of $n$, starting with $n = 0$.
# --hints--
-`quadraticPrimes(200)`应返回-4925。
+`quadraticPrimes(200)` should return a number.
+
+```js
+assert(typeof quadraticPrimes(200) === 'number');
+```
+
+`quadraticPrimes(200)` should return -4925.
```js
assert(quadraticPrimes(200) == -4925);
```
-`quadraticPrimes(500)`应返回-18901。
+`quadraticPrimes(500)` should return -18901.
```js
assert(quadraticPrimes(500) == -18901);
```
-`quadraticPrimes(800)`应返回-43835。
+`quadraticPrimes(800)` should return -43835.
```js
assert(quadraticPrimes(800) == -43835);
```
-`quadraticPrimes(1000)`应返回-59231。
+`quadraticPrimes(1000)` should return -59231.
```js
assert(quadraticPrimes(1000) == -59231);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-270-cutting-squares.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-270-cutting-squares.md
index 8bc6e6a72be..5d490e3b17a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-270-cutting-squares.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-270-cutting-squares.md
@@ -1,20 +1,28 @@
---
id: 5900f47c1000cf542c50ff8e
-title: 问题270：切割方块
+title: 'Problem 270: Cutting Squares'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301920
dashedName: problem-270-cutting-squares
---
# --description--
-放置具有整数尺寸N×N的正方形纸，在原点处具有拐角，并且沿x轴和y轴具有两个侧面。然后，我们按照以下规则进行切割：我们只在位于正方形不同侧面的两个点之间进行直线切割，并且具有整数坐标。两个切口不能交叉，但是几个切口可以在相同的边界点处相遇。继续进行，直到不再进行合法削减。将任何反射或旋转计数为不同，我们将C（N）称为切割N×N平方的方式的数量。例如，C（1）= 2且C（2）= 30（如下所示）。
+A square piece of paper with integer dimensions N×N is placed with a corner at the origin and two of its sides along the x- and y-axes. Then, we cut it up respecting the following rules:
-什么是C（30）mod 108？
+We only make straight cuts between two points lying on different sides of the square, and having integer coordinates.
+
+Two cuts cannot cross, but several cuts can meet at the same border point.
+
+Proceed until no more legal cuts can be made.
+
+Counting any reflections or rotations as distinct, we call C(N) the number of ways to cut an N×N square. For example, C(1) = 2 and C(2) = 30 (shown below).
+
+What is C(30) mod 108 ?
# --hints--
-`euler270()`应该返回82282080。
+`euler270()` should return 82282080.
```js
assert.strictEqual(euler270(), 82282080);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-271-modular-cubes-part-1.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-271-modular-cubes-part-1.md
index 443c8c648b1..e9f9bb785cc 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-271-modular-cubes-part-1.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-271-modular-cubes-part-1.md
@@ -1,22 +1,22 @@
---
id: 5900f47b1000cf542c50ff8d
-title: 问题271：模块化立方体，第1部分
+title: 'Problem 271: Modular Cubes, part 1'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301921
dashedName: problem-271-modular-cubes-part-1
---
# --description--
-对于正数n，将S（n）定义为整数x的总和，为1
+For a positive number n, define S(n) as the sum of the integers x, for which 1
-当n = 91时，x有8个可能的值，即：9,16,22,29,53,74,79,81。因此，S（91）= 9 + 16 + 22 + 29 + 53 + 74 + 79 + 81 = 363。
+When n=91, there are 8 possible values for x, namely : 9, 16, 22, 29, 53, 74, 79, 81. Thus, S(91)=9+16+22+29+53+74+79+81=363.
-找到S（13082761331670030）。
+Find S(13082761331670030).
# --hints--
-`euler271()`应返回4617456485273130000。
+`euler271()` should return 4617456485273130000.
```js
assert.strictEqual(euler271(), 4617456485273130000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-272-modular-cubes-part-2.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-272-modular-cubes-part-2.md
index 6f242a2ea93..3cfb07b5b3f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-272-modular-cubes-part-2.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-272-modular-cubes-part-2.md
@@ -1,22 +1,22 @@
---
id: 5900f47d1000cf542c50ff8f
-title: 问题272：模块化立方体，第2部分
+title: 'Problem 272: Modular Cubes, part 2'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301922
dashedName: problem-272-modular-cubes-part-2
---
# --description--
-对于正数n，将C（n）定义为整数x的数量，为1
+For a positive number n, define C(n) as the number of the integers x, for which 1
-当n = 91时，x有8个可能的值，即：9,16,22,29,53,74,79,81。因此，C（91）= 8。
+When n=91, there are 8 possible values for x, namely : 9, 16, 22, 29, 53, 74, 79, 81. Thus, C(91)=8.
-求出正数n≤1011的总和，其中C（n）= 242。
+Find the sum of the positive numbers n≤1011 for which C(n)=242.
# --hints--
-`euler272()`应该返回8495585919506151000。
+`euler272()` should return 8495585919506151000.
```js
assert.strictEqual(euler272(), 8495585919506151000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-273-sum-of-squares.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-273-sum-of-squares.md
index f71caa82987..508c463ddd8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-273-sum-of-squares.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-273-sum-of-squares.md
@@ -1,20 +1,20 @@
---
id: 5900f47e1000cf542c50ff90
-title: 问题273：正方形的总和
+title: 'Problem 273: Sum of Squares'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301923
dashedName: problem-273-sum-of-squares
---
# --description--
-考虑以下形式的方程：a2 + b2 = N，0≤a≤b，a，b和N整数。
+Consider equations of the form: a2 + b2 = N, 0 ≤ a ≤ b, a, b and N integer.
-对于N = 65，有两种解决方案：a = 1，b = 8，a = 4，b = 7。我们将S（N）称为a2 + b2 = N，0≤a≤b，a，b和N整数的所有解的a的值之和。因此，S（65）= 1 + 4 = 5.找到ΣS（N），对于所有无平均N，只能被4k + 1形式的素数整除，其中4k + 1 <150。
+For N=65 there are two solutions: a=1, b=8 and a=4, b=7. We call S(N) the sum of the values of a of all solutions of a2 + b2 = N, 0 ≤ a ≤ b, a, b and N integer. Thus S(65) = 1 + 4 = 5. Find ∑S(N), for all squarefree N only divisible by primes of the form 4k+1 with 4k+1 < 150.
# --hints--
-`euler273()`应该返回2032447591196869000。
+`euler273()` should return 2032447591196869000.
```js
assert.strictEqual(euler273(), 2032447591196869000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-274-divisibility-multipliers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-274-divisibility-multipliers.md
index 97b29e0e104..7841b6ed8f3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-274-divisibility-multipliers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-274-divisibility-multipliers.md
@@ -1,30 +1,30 @@
---
id: 5900f47f1000cf542c50ff91
-title: 问题274：可分性乘数
+title: 'Problem 274: Divisibility Multipliers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301924
dashedName: problem-274-divisibility-multipliers
---
# --description--
-对于每个整数p> 1互质到10，有一个正的可分性乘数m <p，它对任何正整数n的后续函数保持p的可除性。
+For each integer p > 1 coprime to 10 there is a positive divisibility multiplier m < p which preserves divisibility by p for the following function on any positive integer, n:
-f（n）=（除了n的最后一位以外的所有数字）+（n的最后一位）\* m
+f(n) = (all but the last digit of n) + (the last digit of n) \* m
-也就是说，如果m是p的可分数乘数，则当且仅当n可被p整除时，f（n）可被p整除。
+That is, if m is the divisibility multiplier for p, then f(n) is divisible by p if and only if n is divisible by p.
-（当n远大于p时，f（n）将小于n，并且f的重复应用为p提供乘法可除性测试。）
+(When n is much larger than p, f(n) will be less than n and repeated application of f provides a multiplicative divisibility test for p.)
-例如，113的可分性乘数是34。
+For example, the divisibility multiplier for 113 is 34.
-f（76275）= 7627 + 5 *34 = 7797：76275和7797都可以被113f（12345）= 1234 + 5* 34 = 1404：12345和1404整除都不能被113整除
+f(76275) = 7627 + 5 *34 = 7797 : 76275 and 7797 are both divisible by 113f(12345) = 1234 + 5* 34 = 1404 : 12345 and 1404 are both not divisible by 113
-对于10和小于1000互质的素数的可除性乘数的总和是39517.对于10和小于107互质的素数的可除数乘数的总和是多少？
+The sum of the divisibility multipliers for the primes that are coprime to 10 and less than 1000 is 39517. What is the sum of the divisibility multipliers for the primes that are coprime to 10 and less than 107?
# --hints--
-`euler274()`应该返回1601912348822。
+`euler274()` should return 1601912348822.
```js
assert.strictEqual(euler274(), 1601912348822);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-275-balanced-sculptures.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-275-balanced-sculptures.md
index 9f63508f1ca..a65dc0151c8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-275-balanced-sculptures.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-275-balanced-sculptures.md
@@ -1,20 +1,30 @@
---
id: 5900f4801000cf542c50ff92
-title: 问题275：平衡雕塑
+title: 'Problem 275: Balanced Sculptures'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301925
dashedName: problem-275-balanced-sculptures
---
# --description--
-让我们如下定义n阶平衡雕塑：由n + 1个瓷砖组成的多边形，称为块（n个瓷砖）和底座（剩余瓷砖）;底座的中心位置（x = 0，y = 0）;块的y坐标大于零（所以底座是唯一的最低瓦片）;组合的所有块的质心具有等于零的x坐标。在对雕塑进行计数时，任何仅仅是关于y轴的反射的布置都不算是不同的。例如，6级平衡雕塑如下图所示;请注意，每对镜像（关于y轴）都算作一个雕塑：
+Let us define a balanced sculpture of order n as follows:
-有964个平衡雕塑，订单10和360505订单15.有多少平衡雕塑有18个订单？
+A polyomino made up of n+1 tiles known as the blocks (n tiles) and the plinth (remaining tile);
+
+the plinth has its centre at position (x = 0, y = 0);
+
+the blocks have y-coordinates greater than zero (so the plinth is the unique lowest tile);
+
+the centre of mass of all the blocks, combined, has x-coordinate equal to zero.
+
+When counting the sculptures, any arrangements which are simply reflections about the y-axis, are not counted as distinct. For example, the 18 balanced sculptures of order 6 are shown below; note that each pair of mirror images (about the y-axis) is counted as one sculpture:
+
+There are 964 balanced sculptures of order 10 and 360505 of order 15.How many balanced sculptures are there of order 18?
# --hints--
-`euler275()`应返回15030564。
+`euler275()` should return 15030564.
```js
assert.strictEqual(euler275(), 15030564);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-276-primitive-triangles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-276-primitive-triangles.md
index f1607395003..0e8392e57be 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-276-primitive-triangles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-276-primitive-triangles.md
@@ -1,18 +1,22 @@
---
id: 5900f4801000cf542c50ff93
-title: 问题276：原始三角形
+title: 'Problem 276: Primitive Triangles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301926
dashedName: problem-276-primitive-triangles
---
# --description--
-考虑具有整数边a，b和c的三角形，其中a≤b≤c。如果gcd（a，b，c）= 1，则整数边三角形（a，b，c）称为基元。存在多少个原始整数边三角形，周长不超过10 000 000？
+Consider the triangles with integer sides a, b and c with a ≤ b ≤ c.
+
+An integer sided triangle (a,b,c) is called primitive if gcd(a,b,c)=1.
+
+How many primitive integer sided triangles exist with a perimeter not exceeding 10 000 000?
# --hints--
-`euler276()`应该返回5777137137739633000。
+`euler276()` should return 5777137137739633000.
```js
assert.strictEqual(euler276(), 5777137137739633000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-277-a-modified-collatz-sequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-277-a-modified-collatz-sequence.md
index 09de613348e..1f081396c26 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-277-a-modified-collatz-sequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-277-a-modified-collatz-sequence.md
@@ -1,32 +1,32 @@
---
id: 5900f4811000cf542c50ff94
-title: 问题277：修改的Collatz序列
+title: 'Problem 277: A Modified Collatz sequence'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301927
dashedName: problem-277-a-modified-collatz-sequence
---
# --description--
-通过以下方式从起始值a1获得修改的整数Collatz序列：
+A modified Collatz sequence of integers is obtained from a starting value a1 in the following way:
-如果a可以被3整除，则a + 1 = an / 3.我们将此表示为一个大的向下步骤，“D”。
+an+1 = an/3 if an is divisible by 3. We shall denote this as a large downward step, "D".
-如果除以3得到1的余数，则a + 1 =（4an + 2）/ 3。我们将其称为向上步骤“U”。
+an+1 = (4an + 2)/3 if an divided by 3 gives a remainder of 1. We shall denote this as an upward step, "U".
-如果除以3得到余数为2，则a + 1 =（2an-1）/ 3.我们将这表示为一个小的向下步骤，“d”。
+an+1 = (2an - 1)/3 if an divided by 3 gives a remainder of 2. We shall denote this as a small downward step, "d".
-当某些a = 1时，序列终止。
+The sequence terminates when some an = 1.
-给定任何整数，我们可以列出步骤的顺序。例如，如果a1 = 231，则序列{an} = {231,77,51,17,11,7,10,14,9,3,1}对应于步骤“DdDddUUdDD”。
+Given any integer, we can list out the sequence of steps. For instance if a1=231, then the sequence {an}={231,77,51,17,11,7,10,14,9,3,1} corresponds to the steps "DdDddUUdDD".
-当然，还有其他序列以相同的序列“DdDddUUdDD ....”开头。例如，如果a1 = 1004064，则序列为DdDddUUdDDDdUDUUUdDdUUDDDUdDD。实际上，1004064是以序列DdDddUUdDD开头的最小可能a1> 106。
+Of course, there are other sequences that begin with that same sequence "DdDddUUdDD....". For instance, if a1=1004064, then the sequence is DdDddUUdDDDdUDUUUdDdUUDDDUdDD. In fact, 1004064 is the smallest possible a1 > 106 that begins with the sequence DdDddUUdDD.
-从序列“UDDDUdddDDUDDddDdDddDDUDDdUUDd”开始的最小a1> 1015是多少？
+What is the smallest a1 > 1015 that begins with the sequence "UDDDUdddDDUDDddDdDddDDUDDdUUDd"?
# --hints--
-`euler277()`应该返回1125977393124310。
+`euler277()` should return 1125977393124310.
```js
assert.strictEqual(euler277(), 1125977393124310);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-278-linear-combinations-of-semiprimes.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-278-linear-combinations-of-semiprimes.md
index 6361bc0724f..6e26a6a316f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-278-linear-combinations-of-semiprimes.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-278-linear-combinations-of-semiprimes.md
@@ -1,24 +1,24 @@
---
id: 5900f4831000cf542c50ff95
-title: 问题278：半正定的线性组合
+title: 'Problem 278: Linear Combinations of Semiprimes'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301928
dashedName: problem-278-linear-combinations-of-semiprimes
---
# --description--
-给定整数1 <a1 <a2 <... <an的值，考虑线性组合q1a1 + q2a2 + ... + qnan = b，仅使用整数值qk≥0。
+Given the values of integers 1 < a1 < a2 <... < an, consider the linear combination q1a1 + q2a2 + ... + qnan = b, using only integer values qk ≥ 0.
-注意，对于给定的ak集合，可能不是b的所有值都是可能的。例如，如果a1 = 5且a2 = 7，则没有q1≥0且q2≥0使得b可以是1,2,3,4,6,8,9,11,13,16,18或23 。
+Note that for a given set of ak, it may be that not all values of b are possible. For instance, if a1 = 5 and a2 = 7, there are no q1 ≥ 0 and q2 ≥ 0 such that b could be 1, 2, 3, 4, 6, 8, 9, 11, 13, 16, 18 or 23.
-事实上，23是a1 = 5和a2 = 7的b的最大不可能值。因此，我们称f（5,7）= 23.同样，可以证明f（6,10,15）= 29和f（14,22,77）= 195。
+In fact, 23 is the largest impossible value of b for a1 = 5 and a2 = 7. We therefore call f(5, 7) = 23. Similarly, it can be shown that f(6, 10, 15)=29 and f(14, 22, 77) = 195.
-找到Σf（p *q，p* r，q \* r），其中p，q和r是素数，p <q <r <5000。
+Find ∑ f(p*q,p*r,q\*r), where p, q and r are prime numbers and p < q < r < 5000.
# --hints--
-`euler278()`应该返回1228215747273908500。
+`euler278()` should return 1228215747273908500.
```js
assert.strictEqual(euler278(), 1228215747273908500);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-279-triangles-with-integral-sides-and-an-integral-angle.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-279-triangles-with-integral-sides-and-an-integral-angle.md
index ab90cb29dc9..9fc673b5ea4 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-279-triangles-with-integral-sides-and-an-integral-angle.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-279-triangles-with-integral-sides-and-an-integral-angle.md
@@ -1,18 +1,18 @@
---
id: 5900f4841000cf542c50ff96
-title: 问题279：具有整体边和整体角的三角形
+title: 'Problem 279: Triangles with integral sides and an integral angle'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301929
dashedName: problem-279-triangles-with-integral-sides-and-an-integral-angle
---
# --description--
-有多少个三角形有整体边，至少有一个整体角（以度为单位），周长不超过108？
+How many triangles are there with integral sides, at least one integral angle (measured in degrees), and a perimeter that does not exceed 108?
# --hints--
-`euler279()`应该返回416577688。
+`euler279()` should return 416577688.
```js
assert.strictEqual(euler279(), 416577688);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-28-number-spiral-diagonals.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-28-number-spiral-diagonals.md
index 62af21acaf2..a476199d1dc 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-28-number-spiral-diagonals.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-28-number-spiral-diagonals.md
@@ -1,50 +1,54 @@
---
id: 5900f3881000cf542c50fe9b
-title: 问题28：对角螺旋数
+title: 'Problem 28: Number spiral diagonals'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301930
dashedName: problem-28-number-spiral-diagonals
---
# --description--
-从数字1开始，沿顺时针方向向右移动，形成5 x 5螺旋，如下所示：
+Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:
-21 22 23 24 25
+<div style='padding-left: 4em;'>
+ <div style='color: red; display: inline;'>21</div> 22 23 24 <div style='color: red; display: inline;'>25</div><br>
+ 20 <div style='color: red; display: inline;'>7</div> 8 <div style='color: red; display: inline;'>9</div> 10<br>
+ 19 6 <div style='color: red; display: inline;'>1</div> 2 11<br>
+ 18 <div style='color: red; display: inline;'>5</div> 4 <div style='color: red; display: inline;'>3</div> 12<br>
+ <div style='color: red; display: inline;'>17</div> 16 15 14 <div style='color: red; display: inline;'>13</div><br>
+</div>
-20 7 8 9 10
+It can be verified that the sum of the numbers on the diagonals is 101.
-19 6 1 2 11
-
-18 5 4 3 12
-
-17 16 15 14 13
-
-可以验证对角线上的数字之和为101。
-
-以相同方式形成的n×n螺旋中的对角线上的数字的总和是多少？
+What is the sum of the numbers on the diagonals in an `n` by `n` spiral formed in the same way?
# --hints--
-`spiralDiagonals(101)`应该返回692101。
+`spiralDiagonals(101)` should return a number.
+
+```js
+assert(typeof spiralDiagonals(101) === 'number');
+```
+
+`spiralDiagonals(101)` should return 692101.
```js
assert(spiralDiagonals(101) == 692101);
```
-`spiralDiagonals(101)`应该返回18591725。
+`spiralDiagonals(303)` should return 18591725.
```js
assert(spiralDiagonals(303) == 18591725);
```
-`spiralDiagonals(101)`应该返回85986601。
+`spiralDiagonals(505)` should return 85986601.
```js
assert(spiralDiagonals(505) == 85986601);
```
-`spiralDiagonals(101)`应该返回669171001。
+`spiralDiagonals(1001)` should return 669171001.
```js
assert(spiralDiagonals(1001) == 669171001);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-280-ant-and-seeds.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-280-ant-and-seeds.md
index 5f3e693b4a6..642a7d802f9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-280-ant-and-seeds.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-280-ant-and-seeds.md
@@ -1,22 +1,22 @@
---
id: 5900f4841000cf542c50ff97
-title: 问题280：蚂蚁和种子
+title: 'Problem 280: Ant and seeds'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301931
dashedName: problem-280-ant-and-seeds
---
# --description--
-费力的蚂蚁在5x5的网格上随机行走。 步行从中央广场开始。 在每个步骤中，蚂蚁都会随机移动到相邻的正方形，而不会离开网格。 因此，根据蚂蚁的位置，每一步有2、3或4种可能的移动。
+A laborious ant walks randomly on a 5x5 grid. The walk starts from the central square. At each step, the ant moves to an adjacent square at random, without leaving the grid; thus there are 2, 3 or 4 possible moves at each step depending on the ant's position.
-步行开始时，将种子放在下排的每个正方形上。 当蚂蚁不携带种子并到达包含种子的下排的正方形时，它将开始携带种子。 蚂蚁会将种子放在最终到达的上一行的第一个空方格上。
+At the start of the walk, a seed is placed on each square of the lower row. When the ant isn't carrying a seed and reaches a square of the lower row containing a seed, it will start to carry the seed. The ant will drop the seed on the first empty square of the upper row it eventually reaches.
-在将所有种子都放到第一行之前，预期的步骤数是多少？ 将答案四舍五入到小数点后六位。
+What's the expected number of steps until all seeds have been dropped in the top row? Give your answer rounded to 6 decimal places.
# --hints--
-`euler280()`应该返回430.088247。
+`euler280()` should return 430.088247.
```js
assert.strictEqual(euler280(), 430.088247);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-281-pizza-toppings.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-281-pizza-toppings.md
index fb533ab7320..9ac6cf33a6f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-281-pizza-toppings.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-281-pizza-toppings.md
@@ -1,24 +1,24 @@
---
id: 5900f4861000cf542c50ff98
-title: 问题281：比萨馅料
+title: 'Problem 281: Pizza Toppings'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301932
dashedName: problem-281-pizza-toppings
---
# --description--
-您将得到一个比萨饼（完美的圆），该比萨饼已被切成等份的m·n，并且您想在每个切片上精确地放一个馅料。
+You are given a pizza (perfect circle) that has been cut into m·n equal pieces and you want to have exactly one topping on each slice.
-令f（m，n）表示在披萨上放置m种不同配料（m≥2）的方法，将每个配料恰好放在n个切片上（n≥1）。 反射被认为是不同的，旋转没有。
+Let f(m,n) denote the number of ways you can have toppings on the pizza with m different toppings (m ≥ 2), using each topping on exactly n slices (n ≥ 1). Reflections are considered distinct, rotations are not.
-因此，例如，f（2,1）= 1，f（2,2）= f（3,1）= 2和f（3,2）= 16.f（3,2）如下所示：
+Thus, for instance, f(2,1) = 1, f(2,2) = f(3,1) = 2 and f(3,2) = 16. f(3,2) is shown below:
-找出所有f（m，n）的总和，使f（m，n）≤1015。
+Find the sum of all f(m,n) such that f(m,n) ≤ 1015.
# --hints--
-`euler281()`应该返回1485776387445623。
+`euler281()` should return 1485776387445623.
```js
assert.strictEqual(euler281(), 1485776387445623);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-282-the-ackermann-function.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-282-the-ackermann-function.md
index 80017a67922..5abd58ec44b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-282-the-ackermann-function.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-282-the-ackermann-function.md
@@ -1,22 +1,22 @@
---
id: 5900f4861000cf542c50ff99
-title: 问题282：阿克曼函数
+title: 'Problem 282: The Ackermann function'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301933
dashedName: problem-282-the-ackermann-function
---
# --description--
-对于非负整数m，n，Ackermann函数A（m，n）定义如下：
+For non-negative integers m, n, the Ackermann function A(m, n) is defined as follows:
-例如A（1，0）= 2，A（2，2）= 7和A（3，4）= 125。
+For example A(1, 0) = 2, A(2, 2) = 7 and A(3, 4) = 125.
-找到A（n，n）并给出答案mod 148。
+Find A(n, n) and give your answer mod 148.
# --hints--
-`euler282()`应该返回1098988351。
+`euler282()` should return 1098988351.
```js
assert.strictEqual(euler282(), 1098988351);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-283-integer-sided-triangles-for-which-the-area--perimeter-ratio-is-integral.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-283-integer-sided-triangles-for-which-the-area--perimeter-ratio-is-integral.md
index bce865d64e1..e15776ec08f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-283-integer-sided-triangles-for-which-the-area--perimeter-ratio-is-integral.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-283-integer-sided-triangles-for-which-the-area--perimeter-ratio-is-integral.md
@@ -1,27 +1,28 @@
---
id: 5900f4881000cf542c50ff9a
-title: 问题283：面积*周长比为整数的整数边三角形
+title: >-
+ Problem 283: Integer sided triangles for which the area * perimeter ratio is integral
challengeType: 5
-videoUrl: ''
+forumTopicId: 301934
dashedName: >-
problem-283-integer-sided-triangles-for-which-the-area--perimeter-ratio-is-integral
---
# --description--
-考虑具有边6、8和10的三角形。可以看出，周长和面积都等于24。
+Consider the triangle with sides 6, 8 and 10. It can be seen that the perimeter and the area are both equal to 24.
-因此面积/周长比等于1。
+So the area/perimeter ratio is equal to 1.
-还要考虑具有边13、14和15的三角形。周长等于42，而面积等于84。
+Consider also the triangle with sides 13, 14 and 15. The perimeter equals 42 while the area is equal to 84.
-因此，对于该三角形，面积/周长比等于2。
+So for this triangle the area/perimeter ratio is equal to 2.
-求出面积/周长比等于不超过1000的正整数的所有整数边三角形的周长总和。
+Find the sum of the perimeters of all integer sided triangles for which the area/perimeter ratios are equal to positive integers not exceeding 1000.
# --hints--
-`euler283()`应该返回28038042525570324。
+`euler283()` should return 28038042525570324.
```js
assert.strictEqual(euler283(), 28038042525570324);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-284-steady-squares.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-284-steady-squares.md
index dcaeda68460..818483669ec 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-284-steady-squares.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-284-steady-squares.md
@@ -1,24 +1,24 @@
---
id: 5900f4891000cf542c50ff9b
-title: 问题284：平稳正方形
+title: 'Problem 284: Steady Squares'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301935
dashedName: problem-284-steady-squares
---
# --description--
-十进制编号系统中的3位数字376是具有特殊属性的数字的示例，该属性的平方以相同的数字结尾：3762 =141376。让我们将具有此属性的数字称为稳定平方。
+The 3-digit number 376 in the decimal numbering system is an example of numbers with the special property that its square ends with the same digits: 3762 = 141376. Let's call a number with this property a steady square.
-在其他编号系统中也可以看到稳定的正方形。 在基数14的编号系统中，三位数c37也是一个稳定的正方形：c372 = aa0c37，并且在同一编号系统中，其位数之和为c + 3 + 7 = 18。 字母a，b，c和d以类似于十六进制编号系统的方式分别用于10、11、12和13位数字。
+Steady squares can also be observed in other numbering systems. In the base 14 numbering system, the 3-digit number c37 is also a steady square: c372 = aa0c37, and the sum of its digits is c+3+7=18 in the same numbering system. The letters a, b, c and d are used for the 10, 11, 12 and 13 digits respectively, in a manner similar to the hexadecimal numbering system.
-对于1≤n≤9，在基数14的编号系统中，所有n位稳定正方形的位总和为2d8（十进制582）。 不允许以0开头的稳定正方形。
+For 1 ≤ n ≤ 9, the sum of the digits of all the n-digit steady squares in the base 14 numbering system is 2d8 (582 decimal). Steady squares with leading 0's are not allowed.
-在基数为14的编号系统中找到所有n位稳定正方形的位总和 1≤n≤10000（十进制），并在基数为14的系统中根据需要使用小写字母给出答案。
+Find the sum of the digits of all the n-digit steady squares in the base 14 numbering system for 1 ≤ n ≤ 10000 (decimal) and give your answer in the base 14 system using lower case letters where necessary.
# --hints--
-`euler284()`应该返回5a411d7b。
+`euler284()` should return 5a411d7b.
```js
assert.strictEqual(euler284(), '5a411d7b');
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-285-pythagorean-odds.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-285-pythagorean-odds.md
index dde6189b761..12ddabd7222 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-285-pythagorean-odds.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-285-pythagorean-odds.md
@@ -1,26 +1,26 @@
---
id: 5900f48a1000cf542c50ff9c
-title: 问题285：毕达哥拉斯赔率
+title: 'Problem 285: Pythagorean odds'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301936
dashedName: problem-285-pythagorean-odds
---
# --description--
-Albert选择一个正整数k，然后在区间\[0,1]中以均匀分布随机选择两个实数a，b。
+Albert chooses a positive integer k, then two real numbers a, b are randomly chosen in the interval \[0,1] with uniform distribution.
-然后计算和（k·a + 1）2 +（k·b + 1）2的平方根并四舍五入为最接近的整数。 如果结果等于k，他得分为k分； 否则他什么也没得分。
+The square root of the sum (k·a+1)2 + (k·b+1)2 is then computed and rounded to the nearest integer. If the result is equal to k, he scores k points; otherwise he scores nothing.
-例如，如果k ＝ 6，a ＝ 0.2，b ＝ 0.85，则（k·a + 1）2 ＋+（k·b + 1）2 ＝ 42.05。 42.05的平方根是6.484 ...，四舍五入到最接近的整数后，它变为6。 这等于k，因此他得到6分。
+For example, if k = 6, a = 0.2 and b = 0.85, then (k·a+1)2 + (k·b+1)2 = 42.05. The square root of 42.05 is 6.484... and when rounded to the nearest integer, it becomes 6. This is equal to k, so he scores 6 points.
-可以看出，如果他以k = 1，k = 2，...，k = 10进行10圈，则他的总得分的期望值四舍五入到小数点后五位。
+It can be shown that if he plays 10 turns with k = 1, k = 2, ..., k = 10, the expected value of his total score, rounded to five decimal places, is 10.20914.
-如果他以k = 1，k = 2，k = 3，...，k = 105进行105转，他的总得分的期望值是多少，四舍五入到小数点后五位？
+If he plays 105 turns with k = 1, k = 2, k = 3, ..., k = 105, what is the expected value of his total score, rounded to five decimal places?
# --hints--
-`euler285()`应该返回157055.80999。
+`euler285()` should return 157055.80999.
```js
assert.strictEqual(euler285(), 157055.80999);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-286-scoring-probabilities.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-286-scoring-probabilities.md
index 8fee2ed22ad..abb357c47a2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-286-scoring-probabilities.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-286-scoring-probabilities.md
@@ -1,24 +1,22 @@
---
id: 5900f48a1000cf542c50ff9d
-title: 问题286：计分概率
+title: 'Problem 286: Scoring probabilities'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301937
dashedName: problem-286-scoring-probabilities
---
# --description--
-447/5000
+Barbara is a mathematician and a basketball player. She has found that the probability of scoring a point when shooting from a distance x is exactly (1 - x/q), where q is a real constant greater than 50.
-芭芭拉（Barbara）是数学家和篮球运动员。 她发现从距离x射击时给一个点评分的概率正好是（1 -x / q），其中q是大于50的实常数。
+During each practice run, she takes shots from distances x = 1, x = 2, ..., x = 50 and, according to her records, she has precisely a 2 % chance to score a total of exactly 20 points.
-在每次练习中，她从x = 1，x = 2，...，x = 50的距离射击，并且根据她的记录，她恰好有2％的机会总得分为20分。
-
-找出q并将答案四舍五入到小数点后十位。
+Find q and give your answer rounded to 10 decimal places.
# --hints--
-`euler286()`应该返回52.6494571953。
+`euler286()` should return 52.6494571953.
```js
assert.strictEqual(euler286(), 52.6494571953);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-287-quadtree-encoding-a-simple-compression-algorithm.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-287-quadtree-encoding-a-simple-compression-algorithm.md
index 0751c755bc1..8d11a248e2f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-287-quadtree-encoding-a-simple-compression-algorithm.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-287-quadtree-encoding-a-simple-compression-algorithm.md
@@ -1,34 +1,34 @@
---
id: 5900f48b1000cf542c50ff9e
-title: 问题287：四叉树编码（一种简单的压缩算法）
+title: 'Problem 287: Quadtree encoding (a simple compression algorithm)'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301938
dashedName: problem-287-quadtree-encoding-a-simple-compression-algorithm
---
# --description--
-四叉树编码使我们能够将2N×2N黑白图像描述为比特序列（0和1）。这些序列应从左向右读取，如下所示：
+The quadtree encoding allows us to describe a 2N×2N black and white image as a sequence of bits (0 and 1). Those sequences are to be read from left to right like this:
-第一位处理完整的2N×2N区域；
+the first bit deals with the complete 2N×2N region;
-“ 0”表示拆分：
+"0" denotes a split:
-当前的2n×2n区域被分为4个子区域，尺寸为2n-1×2n-1，
+the current 2n×2n region is divided into 4 sub-regions of dimension 2n-1×2n-1,
-接下来的几位包含左上，右上，左下和右下子区域的描述-按此顺序；
+the next bits contains the description of the top left, top right, bottom left and bottom right sub-regions - in that order;
-“ 10”表示当前区域仅包含黑色像素；
+"10" indicates that the current region contains only black pixels;
-“ 11”表示当前区域仅包含白色像素。请考虑以下4×4图像（彩色标记表示可能发生分裂的位置）：
+"11" indicates that the current region contains only white pixels.Consider the following 4×4 image (colored marks denote places where a split can occur):
-该图像可以通过几个序列来描述，例如： 长度为30的“ 001010101001011111111010101101” 长度为16的“ 0100101111101110”是此图像的最小序列。
+This image can be described by several sequences, for example : "001010101001011111011010101010", of length 30, or "0100101111101110", of length 16, which is the minimal sequence for this image.
-对于正整数N，使用以下着色方案将DN定义为2N×2N图像： 坐标x = 0，y，= 0的像素对应于左下像素， 如果（x-2N-1）2 +（y-2N-1）2≤22N-2，则像素为黑色， 描述D24的最小序列的长度是多少？
+For a positive integer N, define DN as the 2N×2N image with the following coloring scheme: the pixel with coordinates x = 0, y = 0 corresponds to the bottom left pixel, if (x - 2N-1)2 + (y - 2N-1)2 ≤ 22N-2 then the pixel is black, otherwise the pixel is white.What is the length of the minimal sequence describing D24 ?
# --hints--
-`euler287()`应该返回313135496。
+`euler287()` should return 313135496.
```js
assert.strictEqual(euler287(), 313135496);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-288-an-enormous-factorial.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-288-an-enormous-factorial.md
index 3aebbc727d3..28a2e301ceb 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-288-an-enormous-factorial.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-288-an-enormous-factorial.md
@@ -1,28 +1,28 @@
---
id: 5900f48d1000cf542c50ff9f
-title: 问题288：一个巨大的阶乘
+title: 'Problem 288: An enormous factorial'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301939
dashedName: problem-288-an-enormous-factorial
---
# --description--
-对于任何素数p，数N（p，q）定义为
+For any prime p the number N(p,q) is defined by
-N（p，q）= ∑n = 0至q Tn \* pn，其中Tn由以下随机数生成器生成：
+N(p,q) = ∑n=0 to q Tn\*pn with Tn generated by the following random number generator:
-S0 = 290797 Sn + 1 = Sn2 mod 50515093 Tn = Sn mod p
+S0 = 290797 Sn+1 = Sn2 mod 50515093 Tn = Sn mod p
-令Nfac（p，q）为N（p，q）的阶乘。 令NF（p，q）为Nfac（p，q）中因子p的数量。
+Let Nfac(p,q) be the factorial of N(p,q). Let NF(p,q) be the number of factors p in Nfac(p,q).
-给出NF（3,10000）mod 320 = 624955285。
+You are given that NF(3,10000) mod 320=624955285.
-求NF（61,107）mod 6110
+Find NF(61,107) mod 6110
# --hints--
-`euler288()`应该返回605857431263982000。
+`euler288()` should return 605857431263982000.
```js
assert.strictEqual(euler288(), 605857431263982000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-289-eulerian-cycles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-289-eulerian-cycles.md
index 3de97857b59..fcac68664d9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-289-eulerian-cycles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-289-eulerian-cycles.md
@@ -1,28 +1,28 @@
---
id: 5900f48d1000cf542c50ffa0
-title: 问题289：欧拉循环
+title: 'Problem 289: Eulerian Cycles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301940
dashedName: problem-289-eulerian-cycles
---
# --description--
-令C（x，y）为穿过点（x，y），（x，y + 1），（x + 1，y）和（x + 1，y + 1）的圆。
+Let C(x,y) be a circle passing through the points (x, y), (x, y+1), (x+1, y) and (x+1, y+1).
-对于正整数m和n，令E（m，n）为由m·n个圆组成的配置： {C（x，y）：0≤x <m，0≤y <n，x和y是整数}
+For positive integers m and n, let E(m,n) be a configuration which consists of the m·n circles: { C(x,y): 0 ≤ x < m, 0 ≤ y < n, x and y are integers }
-E（m，n）上的欧拉循环是一条闭合路径，它恰好通过每个圆弧一次。 E（m，n）上可能有许多这样的路径，但是我们只对那些不会自交叉的路径感兴趣： 非相交路径仅在格点处触碰自身，但从未相交。
+An Eulerian cycle on E(m,n) is a closed path that passes through each arc exactly once. Many such paths are possible on E(m,n), but we are only interested in those which are not self-crossing: A non-crossing path just touches itself at lattice points, but it never crosses itself.
-下图显示了E（3,3）和一个欧拉非交叉路径的示例。
+The image below shows E(3,3) and an example of an Eulerian non-crossing path.
-令L（m，n）为E（m，n）上的欧拉非交叉路径数。 例如，L（1,2）= 2，L（2,2）= 37，L（3,3）= 104290。
+Let L(m,n) be the number of Eulerian non-crossing paths on E(m,n). For example, L(1,2) = 2, L(2,2) = 37 and L(3,3) = 104290.
-找出L（6,10）mod 1010。
+Find L(6,10) mod 1010.
# --hints--
-`euler289()`应该返回6567944538。
+`euler289()` should return 6567944538.
```js
assert.strictEqual(euler289(), 6567944538);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-29-distinct-powers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-29-distinct-powers.md
index 5f7a1c24685..28c05dfc85d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-29-distinct-powers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-29-distinct-powers.md
@@ -1,50 +1,57 @@
---
id: 5900f3891000cf542c50fe9c
-title: 问题29：不同的权力
+title: 'Problem 29: Distinct powers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301941
dashedName: problem-29-distinct-powers
---
# --description--
-考虑2≤a≤5和2≤b≤5的ab的所有整数组合：
+Consider all integer combinations of ab for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:
-22 = 4、23 = 8、24 = 16、25 = 32
+<div style='padding-left: 4em;'>
+ 2<sup>2</sup>=4, 2<sup>3</sup>=8, 2<sup>4</sup>=16, 2<sup>5</sup>=32 <br>
+ 3<sup>2</sup>=9, 3<sup>3</sup>=27, 3<sup>4</sup>=81, 3<sup>5</sup>=243 <br>
+ 4<sup>2</sup>=16, 4<sup>3</sup>=64, 4<sup>4</sup>=256, 4<sup>5</sup>=1024 <br>
+ 5<sup>2</sup>=25, 5<sup>3</sup>=125, 5<sup>4</sup>=625, 5<sup>5</sup>=3125 <br>
+</div>
-32 = 9、33 = 27、34 = 81、35 = 243
+If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:
-42 = 16、43 = 64、44 = 256、45 = 1024
+<div style='padding-left: 4em;'>
+ 4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
+</div>
-52 = 25、53 = 125、54 = 625、55 = 3125
-
-如果然后按数字顺序放置它们，并删除所有重复项，则会得到以下15个不同术语的序列：
-
-4，8，9，16，25，27，32，64，81，125，243，256，625，1024，3125
-
-对于2≤a≤n和2≤b≤n，由ab生成的序列中有多少个不同的项？
+How many distinct terms are in the sequence generated by `ab` for 2 ≤ `a` ≤ `n` and 2 ≤ `b` ≤ `n`?
# --hints--
-`distinctPowers(15)`应该返回177。
+`distinctPowers(15)` should return a number.
+
+```js
+assert(typeof distinctPowers(15) === 'number');
+```
+
+`distinctPowers(15)` should return 177.
```js
assert.strictEqual(distinctPowers(15), 177);
```
-`distinctPowers(20)`应该返回324。
+`distinctPowers(20)` should return 324.
```js
assert.strictEqual(distinctPowers(20), 324);
```
-`distinctPowers(25)`应该返回519。
+`distinctPowers(25)` should return 519.
```js
assert.strictEqual(distinctPowers(25), 519);
```
-`distinctPowers(30)`应该返回755。
+`distinctPowers(30)` should return 755.
```js
assert.strictEqual(distinctPowers(30), 755);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-290-digital-signature.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-290-digital-signature.md
index 7b4791e3a3a..e46dfedd830 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-290-digital-signature.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-290-digital-signature.md
@@ -1,14 +1,18 @@
---
id: 5900f48f1000cf542c50ffa1
-title: 问题290：数字签名
+title: 'Problem 290: Digital Signature'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301942
dashedName: problem-290-digital-signature
---
+# --description--
+
+How many integers 0 ≤ n < 1018 have the property that the sum of the digits of n equals the sum of digits of 137n?
+
# --hints--
-`euler290()`应该返回20444710234716470。
+`euler290()` should return 20444710234716470.
```js
assert.strictEqual(euler290(), 20444710234716470);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-291-panaitopol-primes.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-291-panaitopol-primes.md
index 7b8aeed6c9c..0c7d704d9f4 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-291-panaitopol-primes.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-291-panaitopol-primes.md
@@ -1,20 +1,20 @@
---
id: 5900f48f1000cf542c50ffa2
-title: 问题291：Panaitopol素数
+title: 'Problem 291: Panaitopol Primes'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301943
dashedName: problem-291-panaitopol-primes
---
# --description--
-如果对于某些正整数x和y，素数p称为Panaitopol素数。
+A prime number p is called a Panaitopol prime if for some positive integersx and y.
-找出多少个Panaitopol素数小于5×1015。
+Find how many Panaitopol primes are less than 5×1015.
# --hints--
-`euler291()`应该返回4037526。
+`euler291()` should return 4037526.
```js
assert.strictEqual(euler291(), 4037526);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-292-pythagorean-polygons.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-292-pythagorean-polygons.md
index c94fed7b9f2..446e8755315 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-292-pythagorean-polygons.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-292-pythagorean-polygons.md
@@ -1,28 +1,28 @@
---
id: 5900f4911000cf542c50ffa3
-title: 问题292：勾股多边形
+title: 'Problem 292: Pythagorean Polygons'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301944
dashedName: problem-292-pythagorean-polygons
---
# --description--
-我们将勾股定线多边形定义为具有以下特性的凸多边形：至少有三个顶点，
+We shall define a pythagorean polygon to be a convex polygon with the following properties:there are at least three vertices,
-没有三个顶点对齐，
+no three vertices are aligned,
-每个顶点都有整数坐标
+each vertex has integer coordinates,
-每个边都有整数长度。对于给定的整数n，将P（n）定义为周长≤n的不同毕达哥拉斯多边形的数量。
+each edge has integer length.For a given integer n, define P(n) as the number of distinct pythagorean polygons for which the perimeter is ≤ n.
-毕达哥拉斯多边形应该被认为是不同的，只要它们都不是另一个的翻译即可。
+Pythagorean polygons should be considered distinct as long as none is a translation of another.
-给出P（4）= 1，P（30）= 3655和P（60）= 891045。 找出P（120）。
+You are given that P(4) = 1, P(30) = 3655 and P(60) = 891045. Find P(120).
# --hints--
-`euler292()`应该返回3600060866。
+`euler292()` should return 3600060866.
```js
assert.strictEqual(euler292(), 3600060866);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-293-pseudo-fortunate-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-293-pseudo-fortunate-numbers.md
index f55d5ef46a6..18307e7b042 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-293-pseudo-fortunate-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-293-pseudo-fortunate-numbers.md
@@ -1,26 +1,26 @@
---
id: 5900f4931000cf542c50ffa4
-title: 问题293：伪幸运数
+title: 'Problem 293: Pseudo-Fortunate Numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301945
dashedName: problem-293-pseudo-fortunate-numbers
---
# --description--
-如果偶数正整数N是2的幂或它的不同质数因子是连续质数，则称该数为偶数。
+An even positive integer N will be called admissible, if it is a power of 2 or its distinct prime factors are consecutive primes.
-前十二个允许的数字是2,4,6,8,12,16,18,24,30,32,36,48。
+The first twelve admissible numbers are 2,4,6,8,12,16,18,24,30,32,36,48.
-如果允许N，则最小的整数M> 1，使得N + M为质数，将被称为N的伪幸运数。
+If N is admissible, the smallest integer M > 1 such that N+M is prime, will be called the pseudo-Fortunate number for N.
-例如，N = 630是可接受的，因为它是偶数，并且其不同的素数因子是连续的素数2,3,5和7。 631之后的下一个质数是641； 因此，630的伪幸运数为M = 11。 还可以看到，16的伪幸运数字是3。
+For example, N=630 is admissible since it is even and its distinct prime factors are the consecutive primes 2,3,5 and 7. The next prime number after 631 is 641; hence, the pseudo-Fortunate number for 630 is M=11. It can also be seen that the pseudo-Fortunate number for 16 is 3.
-找到所有小于109的允许数字N的所有唯一伪幸运数字之和。
+Find the sum of all distinct pseudo-Fortunate numbers for admissible numbers N less than 109.
# --hints--
-`euler293()`应该返回2209。
+`euler293()` should return 2209.
```js
assert.strictEqual(euler293(), 2209);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-294-sum-of-digits---experience-23.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-294-sum-of-digits---experience-23.md
index b86f5cec857..b7bc62a5f13 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-294-sum-of-digits---experience-23.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-294-sum-of-digits---experience-23.md
@@ -1,26 +1,26 @@
---
id: 5900f4931000cf542c50ffa5
-title: 问题294：数字总和-经验＃23
+title: 'Problem 294: Sum of digits - experience #23'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301946
dashedName: problem-294-sum-of-digits---experience-23
---
# --description--
-对于正整数k，将d（k）定义为通常的十进制表示形式中k的位数之和。
+For a positive integer k, define d(k) as the sum of the digits of k in its usual decimal representation.
-因此d（42）= 4 + 2 = 6。
+Thus d(42) = 4+2 = 6.
-对于正整数n，将S（n）定义为具有以下属性的正整数k <10n的数量： k可被23整除 d（k）= 23。
+For a positive integer n, define S(n) as the number of positive integers k < 10n with the following properties : k is divisible by 23 and d(k) = 23.
-给出S（9）= 263626和S（42）= 6377168878570056。
+You are given that S(9) = 263626 and S(42) = 6377168878570056.
-找到S（1112）并给出答案mod 109。
+Find S(1112) and give your answer mod 109.
# --hints--
-`euler294()`应该返回789184709。
+`euler294()` should return 789184709.
```js
assert.strictEqual(euler294(), 789184709);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-295-lenticular-holes.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-295-lenticular-holes.md
index 91d889fd597..70d1de6d608 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-295-lenticular-holes.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-295-lenticular-holes.md
@@ -1,36 +1,36 @@
---
id: 5900f4931000cf542c50ffa6
-title: 问题295：透镜孔
+title: 'Problem 295: Lenticular holes'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301947
dashedName: problem-295-lenticular-holes
---
# --description--
-如果满足以下条件，我们称两个圆包围的凸面为透镜孔：
+We call the convex area enclosed by two circles a lenticular hole if:
-两个圆的中心都在晶格点上。
+The centres of both circles are on lattice points.
-两个圆在两个不同的晶格点处相交。
+The two circles intersect at two distinct lattice points.
-被两个圆包围的凸区域的内部不包含任何晶格点。
+The interior of the convex area enclosed by both circles does not contain any lattice points.
-考虑一下圈子： C0：x2 + y2 = 25 C1：（x + 4）2+（y-4）2 = 1 C2：（x-12）2+（y-4）2 = 65
+Consider the circles: C0: x2+y2=25 C1: (x+4)2+(y-4)2=1 C2: (x-12)2+(y-4)2=65
-下图绘制了圆圈C0，C1和C2。
+The circles C0, C1 and C2 are drawn in the picture below.
-C0和C1以及C0和C2形成一个透镜孔。
+C0 and C1 form a lenticular hole, as well as C0 and C2.
-如果存在两个半径为r1和r2且形成一个透镜孔的圆，我们将一个有序正实数对（r1，r2）称为透镜对。 我们可以验证（1，5）和（5，√65）是以上示例的双凸透镜对。
+We call an ordered pair of positive real numbers (r1, r2) a lenticular pair if there exist two circles with radii r1 and r2 that form a lenticular hole. We can verify that (1, 5) and (5, √65) are the lenticular pairs of the example above.
-令L（N）为0 <r1≤r2≤N的不同双凸透镜对（r1，r2）的数量。 我们可以验证L（10）= 30和L（100）= 3442。
+Let L(N) be the number of distinct lenticular pairs (r1, r2) for which 0 < r1 ≤ r2 ≤ N. We can verify that L(10) = 30 and L(100) = 3442.
-求L（100 000）。
+Find L(100 000).
# --hints--
-`euler295()`应该返回4884650818。
+`euler295()` should return 4884650818.
```js
assert.strictEqual(euler295(), 4884650818);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-296-angular-bisector-and-tangent.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-296-angular-bisector-and-tangent.md
index bb63cc695ca..0d269dca3cc 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-296-angular-bisector-and-tangent.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-296-angular-bisector-and-tangent.md
@@ -1,22 +1,22 @@
---
id: 5900f4941000cf542c50ffa7
-title: 问题296：角平分线和切线
+title: 'Problem 296: Angular Bisector and Tangent'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301948
dashedName: problem-296-angular-bisector-and-tangent
---
# --description--
-给定一个整数边三角形ABC，其中BC≤AC≤AB.k是角度ACB的角平分线.m是C与ABC外接圆的切线.n是一条平行于m到B的线。
+Given is an integer sided triangle ABC with BC ≤ AC ≤ AB.k is the angular bisector of angle ACB.m is the tangent at C to the circumscribed circle of ABC.n is a line parallel to m through B.
-n和k的交集称为E。
+The intersection of n and k is called E.
-存在多少个周长不超过100000的三角形ABC，使得BE具有整数长度？
+How many triangles ABC with a perimeter not exceeding 100 000 exist such that BE has integral length?
# --hints--
-`euler296()`应该返回1137208419。
+`euler296()` should return 1137208419.
```js
assert.strictEqual(euler296(), 1137208419);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-297-zeckendorf-representation.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-297-zeckendorf-representation.md
index 33b874da00a..55d96b520c6 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-297-zeckendorf-representation.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-297-zeckendorf-representation.md
@@ -1,26 +1,26 @@
---
id: 5900f4951000cf542c50ffa8
-title: 问题297：Zeckendorf表示
+title: 'Problem 297: Zeckendorf Representation'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301949
dashedName: problem-297-zeckendorf-representation
---
# --description--
-斐波那契数列中的每个新项都是通过将前两个项相加而生成的。
+Each new term in the Fibonacci sequence is generated by adding the previous two terms.
-从1和2开始，前10个术语将是：1、2、3、5、8、13、21、34、55、89。
+Starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89.
-每个正整数都可以唯一地写为斐波纳契数列的非连续项之和。 例如，100 = 3 + 8 + 89。 这样的总和称为数字的Zeckendorf表示。
+Every positive integer can be uniquely written as a sum of nonconsecutive terms of the Fibonacci sequence. For example, 100 = 3 + 8 + 89. Such a sum is called the Zeckendorf representation of the number.
-对于任何n> 0的整数，令z（n）为n的Zeckendorf表示中的项数。 因此，z（5）= 1，z（14）= 2，z（100）= 3等。 另外，对于0 ＜n ＜106，∑ z（n）＝ 7894453。
+For any integer n>0, let z(n) be the number of terms in the Zeckendorf representation of n. Thus, z(5) = 1, z(14) = 2, z(100) = 3 etc. Also, for 0<n<106, ∑ z(n) = 7894453.
-求出∑ z（n）为0 < n < 1017。
+Find ∑ z(n) for 0<n<1017.
# --hints--
-`euler297()`应该返回2252639041804718000。
+`euler297()` should return 2252639041804718000.
```js
assert.strictEqual(euler297(), 2252639041804718000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-298-selective-amnesia.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-298-selective-amnesia.md
index 30bc64e6abc..a9c0c8ad7e0 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-298-selective-amnesia.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-298-selective-amnesia.md
@@ -1,24 +1,24 @@
---
id: 5900f4971000cf542c50ffa9
-title: 问题298：选择性健忘症
+title: 'Problem 298: Selective Amnesia'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301950
dashedName: problem-298-selective-amnesia
---
# --description--
-拉里（Larry）和罗宾（Robin）玩着一场记忆游戏，其中涉及一个随机数序列，介于1到10之间（含1和10），每次被召唤一个。每个玩家最多可以记住5个先前的数字。当被叫号码在玩家的记忆中时，该玩家将获得积分。如果不是，则播放器将被叫号码添加到他的记忆中，如果他的记忆已满，则删除另一个号码。
+Larry and Robin play a memory game involving of a sequence of random numbers between 1 and 10, inclusive, that are called out one at a time. Each player can remember up to 5 previous numbers. When the called number is in a player's memory, that player is awarded a point. If it's not, the player adds the called number to his memory, removing another number if his memory is full.
-两位选手都以空洞的回忆开始。两位玩家总是向他们的记忆中添加新的遗漏号码，但是在决定删除哪个号码时使用不同的策略： 拉里（Larry）的策略是删除最长时间内未拨打的电话。 Robin的策略是删除内存中时间最长的数字。
+Both players start with empty memories. Both players always add new missed numbers to their memory but use a different strategy in deciding which number to remove: Larry's strategy is to remove the number that hasn't been called in the longest time. Robin's strategy is to remove the number that's been in the memory the longest time.
-范例游戏：Turn 被叫号码 拉里的记忆 拉里的比分 罗宾的记忆 罗宾的分数 1个 1个 1个 0 1个 0 2 2 1,2 0 1,2 0 3 4 1,2,4 0 1,2,4 0 4 6 1,2,4,6 0 1,2,4,6 0 5 1个 1,2,4,6 1个 1,2,4,6 1个 6 8 1,2,4,6,8 1个 1,2,4,6,8 1个 7 10 1,4,6,8,10 1个 2,4,6,8,10 1个 8 2 1,2,6,8,10 1个 2,4,6,8,10 2 9 4 1,2,4,8,10 1个 2,4,6,8,10 3 10 1个 1,2,4,8,10 2 1,4,6,8,10 3
+Example game:Turn Callednumber Larry'smemory Larry'sscore Robin'smemory Robin'sscore 1 1 1 0 1 0 2 2 1,2 0 1,2 0 3 4 1,2,4 0 1,2,4 0 4 6 1,2,4,6 0 1,2,4,6 0 5 1 1,2,4,6 1 1,2,4,6 1 6 8 1,2,4,6,8 1 1,2,4,6,8 1 7 10 1,4,6,8,10 1 2,4,6,8,10 1 8 2 1,2,6,8,10 1 2,4,6,8,10 2 9 4 1,2,4,8,10 1 2,4,6,8,10 3 10 1 1,2,4,8,10 2 1,4,6,8,10 3
-用L表示拉里的分数，用R表示罗宾的分数，| L-R |的期望值是多少？ 50转后？使用x.xxxxxxxx格式将答案四舍五入到小数点后八位。
+Denoting Larry's score by L and Robin's score by R, what is the expected value of |L-R| after 50 turns? Give your answer rounded to eight decimal places using the format x.xxxxxxxx .
# --hints--
-`euler298()`应该返回1.76882294。
+`euler298()` should return 1.76882294.
```js
assert.strictEqual(euler298(), 1.76882294);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-299-three-similar-triangles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-299-three-similar-triangles.md
index fbbc58d5568..586b5890681 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-299-three-similar-triangles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-299-three-similar-triangles.md
@@ -1,28 +1,30 @@
---
id: 5900f4971000cf542c50ffaa
-title: 问题299：三个相似的三角形
+title: 'Problem 299: Three similar triangles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301951
dashedName: problem-299-three-similar-triangles
---
# --description--
-选择了四个具有整数坐标的点：A（a，0），B（b，0），C（0，c）和D（0，d），
+Four points with integer coordinates are selected:A(a, 0), B(b, 0), C(0, c) and D(0, d),
-其中0
+with 0 < a < b and 0 < c < d.
-仅当a = c时，容易证明三个三角形可以相似。
+Point P, also with integer coordinates, is chosen on the line AC so that the three triangles ABP, CDP and BDP are all similar.
-因此，给定a = c，我们正在寻找三元组（a，b，d），使得AC上至少存在一个点P（具有整数坐标），从而使三个三角形ABP，CDP和BDP都相似。
+It is easy to prove that the three triangles can be similar, only if a=c.
-例如，如果（a，b，d）＝（2，3，4），则可以容易地验证点P（1，1）满足上述条件。 请注意，三点式（2,3,4）和（2,4,3）被认为是截然不同的，尽管点P（1,1）对于两者而言是共同的。
+So, given that a=c, we are looking for triplets (a,b,d) such that at least one point P (with integer coordinates) exists on AC, making the three triangles ABP, CDP and BDP all similar.
-如果b + d <100，则存在92个不同的三元组（a，b，d），从而存在点P. 如果b + d <100000，则存在320471个不同的三元组（a，b，d），从而存在点P. 如果b + d <100000000，那么有几个不同的三元组（a，b，d）使得点P存在？
+For example, if (a,b,d)=(2,3,4), it can be easily verified that point P(1,1) satisfies the above condition. Note that the triplets (2,3,4) and (2,4,3) are considered as distinct, although point P(1,1) is common for both.
+
+If b+d < 100, there are 92 distinct triplets (a,b,d) such that point P exists. If b+d < 100 000, there are 320471 distinct triplets (a,b,d) such that point P exists. If b+d < 100 000 000, how many distinct triplets (a,b,d) are there such that point P exists?
# --hints--
-`euler299()`应该返回549936643。
+`euler299()` should return 549936643.
```js
assert.strictEqual(euler299(), 549936643);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-3-largest-prime-factor.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-3-largest-prime-factor.md
index a9dfaf6588a..0fbe58b9f1f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-3-largest-prime-factor.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-3-largest-prime-factor.md
@@ -1,50 +1,62 @@
---
id: 5900f36f1000cf542c50fe82
-title: 问题3：最大素数
+title: 'Problem 3: Largest prime factor'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301952
dashedName: problem-3-largest-prime-factor
---
# --description--
-13195的主要因子是5、7、13和29。
+The prime factors of 13195 are 5, 7, 13 and 29.
-给定<code>数字<!-- code-->的最大素数是多少？</code>
+What is the largest prime factor of the given `number`?
# --hints--
-`largestPrimeFactor(2)`应该返回2。
+`largestPrimeFactor(2)` should return a number.
+
+```js
+assert(typeof largestPrimeFactor(2) === 'number');
+```
+
+`largestPrimeFactor(2)` should return 2.
```js
assert.strictEqual(largestPrimeFactor(2), 2);
```
-`largestPrimeFactor(3)`应该返回3。
+`largestPrimeFactor(3)` should return 3.
```js
assert.strictEqual(largestPrimeFactor(3), 3);
```
-`largestPrimeFactor(5)`应该返回5。
+`largestPrimeFactor(5)` should return 5.
```js
assert.strictEqual(largestPrimeFactor(5), 5);
```
-`largestPrimeFactor(7)`应该返回7。
+`largestPrimeFactor(7)` should return 7.
```js
assert.strictEqual(largestPrimeFactor(7), 7);
```
-`largestPrimeFactor(13195)`应该返回29。
+`largestPrimeFactor(8)` should return 2.
+
+```js
+assert.strictEqual(largestPrimeFactor(8), 2);
+```
+
+`largestPrimeFactor(13195)` should return 29.
```js
assert.strictEqual(largestPrimeFactor(13195), 29);
```
-`largestPrimeFactor(600851475143)`应该返回6857。
+`largestPrimeFactor(600851475143)` should return 6857.
```js
assert.strictEqual(largestPrimeFactor(600851475143), 6857);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-30-digit-n-powers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-30-digit-n-powers.md
index e8a08282f48..1eb7fd409bb 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-30-digit-n-powers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-30-digit-n-powers.md
@@ -1,48 +1,54 @@
---
id: 5900f38a1000cf542c50fe9d
-title: 问题30：数字n次方
+title: 'Problem 30: Digit n powers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301953
dashedName: problem-30-digit-n-powers
---
# --description--
-令人惊讶的是，只有三个数字可以写为它们的数字的四次幂之和：
+Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:
-1634 = 1 <sup>4</sup> + 6 <sup>4</sup> + 3 <sup>4</sup> + 4 <sup>4</sup>
+<div style='margin-left: 4em;'>
+ 1634 = 1<sup>4</sup> + 6<sup>4</sup> + 3<sup>4</sup> + 4<sup>4</sup><br>
+ 8208 = 8<sup>4</sup> + 2<sup>4</sup> + 0<sup>4</sup> + 8<sup>4</sup><br>
+ 9474 = 9<sup>4</sup> + 4<sup>4</sup> + 7<sup>4</sup> + 4<sup>4</sup><br>
+</div>
-8208 = 8 <sup>4</sup> + 2 <sup>4</sup> + 0 <sup>4</sup> + 8 <sup>4</sup>
+As 1 = 1<sup>4</sup> is not a sum it is not included.
-9474 = 9 <sup>4</sup> + 4 <sup>4</sup> + 7 <sup>4</sup> + 4 <sup>4</sup>
+The sum of these numbers is 1634 + 8208 + 9474 = 19316.
-由于1 = 1 <sup>4</sup>不是总和，因此不包括在内。
-
-这些数字的总和为1634 + 8208 + 9474 = 19316。
-
-找出所有可以写为数字n次幂的数字之和。
+Find the sum of all the numbers that can be written as the sum of `n` powers of their digits.
# --hints--
-`digitnPowers(2)`应该返回0。
+`digitnPowers(2)` should return a number.
+
+```js
+assert(typeof digitnPowers(2) === 'number');
+```
+
+`digitnPowers(2)` should return 0.
```js
assert(digitnPowers(2) == 0);
```
-`digitnPowers(3)`应该返回1301。
+`digitnPowers(3)` should return 1301.
```js
assert(digitnPowers(3) == 1301);
```
-`digitnPowers(4)`应该返回19316。
+`digitnPowers(4)` should return 19316.
```js
assert(digitnPowers(4) == 19316);
```
-`digitnPowers(5)`应该返回443839。
+`digitnPowers(5)` should return 443839.
```js
assert(digitnPowers(5) == 443839);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-300-protein-folding.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-300-protein-folding.md
index 9d80bf971b7..52c5f272c6f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-300-protein-folding.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-300-protein-folding.md
@@ -1,30 +1,30 @@
---
id: 5900f49a1000cf542c50ffac
-title: 问题300：蛋白质折叠
+title: 'Problem 300: Protein folding'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301954
dashedName: problem-300-protein-folding
---
# --description--
-以一种非常简化的形式，我们可以将蛋白质视为由疏水（H）和极性（P）元素（例如HHPPHHHPHHPH。
+In a very simplified form, we can consider proteins as strings consisting of hydrophobic (H) and polar (P) elements, e.g. HHPPHHHPHHPH.
-对于这个问题，蛋白质的方向很重要。例如HPP被认为与PPH不同。因此，存在2n个由n个元素组成的不同蛋白质。
+For this problem, the orientation of a protein is important; e.g. HPP is considered distinct from PPH. Thus, there are 2n distinct proteins consisting of n elements.
-当人们自然地遇到这些弦时，它们总是以H-H接触点的数量尽可能多的方式折叠，因为这在能量上是有利的。 结果，H元素倾向于在内部积累，而P元素在外部。 天然蛋白质当然会在三个维度上折叠，但我们只会考虑蛋白质在两个维度上的折叠。
+When one encounters these strings in nature, they are always folded in such a way that the number of H-H contact points is as large as possible, since this is energetically advantageous. As a result, the H-elements tend to accumulate in the inner part, with the P-elements on the outside. Natural proteins are folded in three dimensions of course, but we will only consider protein folding in two dimensions.
-下图显示了我们的示例蛋白质可以折叠的两种可能方式（H-H接触点用红色圆点显示）。
+The figure below shows two possible ways that our example protein could be folded (H-H contact points are shown with red dots).
-左侧的折叠只有六个H-H接触点，因此永远不会自然发生。 另一方面，右侧的折叠具有9个H-H接触点，这对于此琴弦是最佳的。
+The folding on the left has only six H-H contact points, thus it would never occur naturally. On the other hand, the folding on the right has nine H-H contact points, which is optimal for this string.
-假设H和P元素同样可能出现在沿着字符串的任何位置，那么长度为8的随机蛋白质字符串最优折叠中H-H接触点的平均数量为850/28 = 3.3203125。
+Assuming that H and P elements are equally likely to occur in any position along the string, the average number of H-H contact points in an optimal folding of a random protein string of length 8 turns out to be 850 / 28=3.3203125.
-在最佳折叠长度为15的随机蛋白质串中，H-H接触点的平均数量是多少？ 要获得准确的结果，请使用尽可能多的小数位给出答案。
+What is the average number of H-H contact points in an optimal folding of a random protein string of length 15? Give your answer using as many decimal places as necessary for an exact result.
# --hints--
-`euler300()`应该返回8.0540771484375。
+`euler300()` should return 8.0540771484375.
```js
assert.strictEqual(euler300(), 8.0540771484375);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-301-nim.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-301-nim.md
index 16bed858d07..8197d8abf58 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-301-nim.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-301-nim.md
@@ -1,28 +1,33 @@
---
id: 5900f4991000cf542c50ffab
-title: 问题301：尼姆
+title: 'Problem 301: Nim'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301955
dashedName: problem-301-nim
---
# --description--
-尼姆（Nim）是一堆用石头堆砌而成的游戏，两名玩家轮流用它来清除任何堆石，直到没有石头为止。
+Nim is a game played with heaps of stones, where two players take it in turn to remove any number of stones from any heap until no stones remain.
-我们将考虑Nim的三堆普通播放版本，其工作方式如下：
+We'll consider the three-heap normal-play version of Nim, which works as follows:
-\-游戏开始时有三堆石头。 -玩家在回合时从任何一个堆中移出正数的石头。 -第一个无法移动（因为没有剩余的石头）的玩家输了。
+- At the start of the game there are three heaps of stones.
+- On his turn the player removes any positive number of stones from any single heap.
+- The first player unable to move (because no stones remain) loses.
-如果（n1，n2，n3）表示由大小为n1，n2和n3的堆组成的Nim位置，则存在一个简单函数X（n1，n2，n3）-您可以查找或尝试自己推断-返回值：如果采用完美策略，将要移动的玩家最终会输掉，则返回零；或非零，如果采用完美策略，将要移动的玩家最终会获胜。例如X（1,2,3）= 0，因为无论当前玩家做什么，他的对手都可以通过移动而留下两个相同大小的堆，而此时，当前玩家的每一步都可以被镜像他的对手，直到没有剩下的石头；因此当前玩家输了。为了显示：
+If (n1,n2,n3) indicates a Nim position consisting of heaps of size n1, n2 and n3 then there is a simple function X(n1,n2,n3) — that you may look up or attempt to deduce for yourself — that returns: zero if, with perfect strategy, the player about to move will eventually lose; or non-zero if, with perfect strategy, the player about to move will eventually win. For example X(1,2,3) = 0 because, no matter what the current player does, his opponent can respond with a move that leaves two heaps of equal size, at which point every move by the current player can be mirrored by his opponent until no stones remain; so the current player loses. To illustrate:
-\-当前玩家移至（1,2,1） -对手移动至（1,0,1） -当前玩家移至（0,0,1） -对手移动到（0,0,0），因此获胜。
+- current player moves to (1,2,1)
+- opponent moves to (1,0,1)
+- current player moves to (0,0,1)
+- opponent moves to (0,0,0), and so wins.
-对于多少个正整数n≤230，X（n，2n，3n）= 0？
+For how many positive integers n ≤ 230 does X(n,2n,3n) = 0 ?
# --hints--
-`euler301()`应该返回2178309。
+`euler301()` should return 2178309.
```js
assert.strictEqual(euler301(), 2178309);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-302-strong-achilles-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-302-strong-achilles-numbers.md
index 57da1b2bad3..78c112d626a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-302-strong-achilles-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-302-strong-achilles-numbers.md
@@ -1,30 +1,30 @@
---
id: 5900f49b1000cf542c50ffad
-title: 问题302：强阿喀琉斯数
+title: 'Problem 302: Strong Achilles Numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301956
dashedName: problem-302-strong-achilles-numbers
---
# --description--
-如果p2是n中每个素数p的n的因数，则正整数n是有力的。
+A positive integer n is powerful if p2 is a divisor of n for every prime factor p in n.
-如果n可以表示为另一个正整数的幂，则正整数n是理想幂。
+A positive integer n is a perfect power if n can be expressed as a power of another positive integer.
-如果n是强大的，但不是完美的幂，则正整数n是阿喀琉斯数。 例如，864和1800是阿喀琉斯数字：864 = 25·33和1800 = 23·32·52。
+A positive integer n is an Achilles number if n is powerful but not a perfect power. For example, 864 and 1800 are Achilles numbers: 864 = 25·33 and 1800 = 23·32·52.
-如果S和φ（S）都是阿喀琉斯数，我们将正整数S称为强阿喀琉斯数.1 例如，864是强致命弱点数：φ（864）= 288 = 25·32。 但是，1800不是强阿喀琉斯数字，因为：（1800）= 480 = 25·31·51。
+We shall call a positive integer S a Strong Achilles number if both S and φ(S) are Achilles numbers.1 For example, 864 is a Strong Achilles number: φ(864) = 288 = 25·32. However, 1800 isn't a Strong Achilles number because: φ(1800) = 480 = 25·31·51.
-低于104的有7个强弱点和低于108的656。
+There are 7 Strong Achilles numbers below 104 and 656 below 108.
-1018以下有多少个强阿喀琉斯数字？
+How many Strong Achilles numbers are there below 1018?
-1φ表示欧拉的拉伸函数。
+1 φ denotes Euler's totient function.
# --hints--
-`euler302()`应该返回1170060。
+`euler302()` should return 1170060.
```js
assert.strictEqual(euler302(), 1170060);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-303-multiples-with-small-digits.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-303-multiples-with-small-digits.md
index e3de16e29c4..1f9b36fad0d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-303-multiples-with-small-digits.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-303-multiples-with-small-digits.md
@@ -1,24 +1,24 @@
---
id: 5900f49b1000cf542c50ffae
-title: 问题303：小数的倍数
+title: 'Problem 303: Multiples with small digits'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301957
dashedName: problem-303-multiples-with-small-digits
---
# --description--
-对于正整数n，将f（n）定义为n的最小正整数，以10为底的数字只能使用≤2的数字。
+For a positive integer n, define f(n) as the least positive multiple of n that, written in base 10, uses only digits ≤ 2.
-因此，f（2）= 2，f（3）= 12，f（7）= 21，f（42）= 210，f（89）= 1121222。
+Thus f(2)=2, f(3)=12, f(7)=21, f(42)=210, f(89)=1121222.
-另外，。
+Also, .
-找 。
+Find .
# --hints--
-`euler303()`应该返回1111981904675169。
+`euler303()` should return 1111981904675169.
```js
assert.strictEqual(euler303(), 1111981904675169);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-304-primonacci.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-304-primonacci.md
index 606041ff553..aab56db5538 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-304-primonacci.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-304-primonacci.md
@@ -1,26 +1,26 @@
---
id: 5900f49d1000cf542c50ffaf
-title: 问题304：Primonacci
+title: 'Problem 304: Primonacci'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301958
dashedName: problem-304-primonacci
---
# --description--
-对于任何正整数n，函数next_prime（n）返回最小质数p，使得p> n。
+For any positive integer n the function next_prime(n) returns the smallest prime p such that p>n.
-序列a（n）定义为： 对于n> 1，a（1）= next_prime（1014）和a（n）= next_prime（a（n-1））。
+The sequence a(n) is defined by: a(1)=next_prime(1014) and a(n)=next_prime(a(n-1)) for n>1.
-斐波那契数列f（n）定义为： 对于n> 1，f（0）= 0，f（1）= 1和f（n）= f（n-1）+ f（n-2）。
+The fibonacci sequence f(n) is defined by: f(0)=0, f(1)=1 and f(n)=f(n-1)+f(n-2) for n>1.
-序列b（n）定义为f（a（n））。
+The sequence b(n) is defined as f(a(n)).
-求出∑b（n）为1≤n≤100000。 给出答案mod 1234567891011。
+Find ∑b(n) for 1≤n≤100 000. Give your answer mod 1234567891011.
# --hints--
-`euler304()`应该返回283988410192。
+`euler304()` should return 283988410192.
```js
assert.strictEqual(euler304(), 283988410192);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-305-reflexive-position.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-305-reflexive-position.md
index 555992fd870..c42f1440383 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-305-reflexive-position.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-305-reflexive-position.md
@@ -1,26 +1,26 @@
---
id: 5900f49d1000cf542c50ffb0
-title: 问题305：反身姿势
+title: 'Problem 305: Reflexive Position'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301959
dashedName: problem-305-reflexive-position
---
# --description--
-我们称S为（无限）字符串，该字符串是通过连接以10为底的连续正整数（从1开始）而形成的。
+Let's call S the (infinite) string that is made by concatenating the consecutive positive integers (starting from 1) written down in base 10.
-因此，S = 1234567891011121314151617181920212223242 ...
+Thus, S = 1234567891011121314151617181920212223242...
-很容易看出，任何数字在S中都会显示无数次。
+It's easy to see that any number will show up an infinite number of times in S.
-让我们将f（n）称为S中第n次出现n的起始位置。 例如，f（1）= 1，f（5）= 81，f（12）= 271和f（7780）= 111111365。
+Let's call f(n) the starting position of the nth occurrence of n in S. For example, f(1)=1, f(5)=81, f(12)=271 and f(7780)=111111365.
-求出∑f（3k）为1≤k≤13。
+Find ∑f(3k) for 1≤k≤13.
# --hints--
-`euler305()`应该返回18174995535140。
+`euler305()` should return 18174995535140.
```js
assert.strictEqual(euler305(), 18174995535140);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-306-paper-strip-game.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-306-paper-strip-game.md
index 55d1a5e5efe..d750803f3b2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-306-paper-strip-game.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-306-paper-strip-game.md
@@ -1,26 +1,26 @@
---
id: 5900f49f1000cf542c50ffb1
-title: 问题306：纸条游戏
+title: 'Problem 306: Paper-strip Game'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301960
dashedName: problem-306-paper-strip-game
---
# --description--
-以下游戏是组合博弈论的经典示例：
+The following game is a classic example of Combinatorial Game Theory:
-两名玩家从n个白色方块开始，轮流交替进行。 在每个回合中，玩家选择两个连续的白色方块并将其涂成黑色。 第一个无法移动的玩家输了。
+Two players start with a strip of n white squares and they take alternate turns. On each turn, a player picks two contiguous white squares and paints them black. The first player who cannot make a move loses.
-如果n = 1，则没有有效的移动，因此第一个玩家会自动失败。 如果n = 2，则只有一招有效，此后第二名玩家输了。 如果n = 3，则有两个有效的举动，但都留下第二个玩家输掉的情况。 如果n = 4，则第一个玩家有3个有效动作；她可以通过绘制两个中间方块来赢得比赛。 如果n = 5，则第一个玩家有四次有效移动（下面以红色显示）；但无论她做什么，第二名玩家（蓝色）都会获胜。
+If n = 1, there are no valid moves, so the first player loses automatically. If n = 2, there is only one valid move, after which the second player loses. If n = 3, there are two valid moves, but both leave a situation where the second player loses. If n = 4, there are three valid moves for the first player; she can win the game by painting the two middle squares. If n = 5, there are four valid moves for the first player (shown below in red); but no matter what she does, the second player (blue) wins.
-因此，对于1≤n≤5，有3个n值，第一位玩家可以对其施加强制胜利。 类似地，对于1≤n≤50，第一个玩家可以强制取胜的n值为40。
+So, for 1 ≤ n ≤ 5, there are 3 values of n for which the first player can force a win. Similarly, for 1 ≤ n ≤ 50, there are 40 values of n for which the first player can force a win.
-对于1≤n≤1 000 000，第一个玩家可以强制赢得多少n值？
+For 1 ≤ n ≤ 1 000 000, how many values of n are there for which the first player can force a win?
# --hints--
-`euler306()`应该返回852938。
+`euler306()` should return 852938.
```js
assert.strictEqual(euler306(), 852938);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-307-chip-defects.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-307-chip-defects.md
index 465fa73c197..eac5e7e113d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-307-chip-defects.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-307-chip-defects.md
@@ -1,22 +1,22 @@
---
id: 5900f4a01000cf542c50ffb2
-title: 问题307：芯片缺陷
+title: 'Problem 307: Chip Defects'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301961
dashedName: problem-307-chip-defects
---
# --description--
-k个缺陷随机分布在一家工厂生产的n个集成电路芯片中（在芯片上可以发现任何数量的缺陷，每个缺陷与其他缺陷无关）。
+k defects are randomly distributed amongst n integrated-circuit chips produced by a factory (any number of defects may be found on a chip and each defect is independent of the other defects).
-令p（k，n）表示存在至少3个缺陷的芯片的概率。 例如p（3.7）≈0.0204081633。
+Let p(k,n) represent the probability that there is a chip with at least 3 defects. For instance p(3,7) ≈ 0.0204081633.
-找到p（20,000，1 000 000）并以0.abcdefghij的形式将答案四舍五入到小数点后10位。
+Find p(20 000, 1 000 000) and give your answer rounded to 10 decimal places in the form 0.abcdefghij
# --hints--
-`euler307()`应该返回0.7311720251。
+`euler307()` should return 0.7311720251.
```js
assert.strictEqual(euler307(), 0.7311720251);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-308-an-amazing-prime-generating-automaton.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-308-an-amazing-prime-generating-automaton.md
index 6b6c896fb08..a0cfc2d8a96 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-308-an-amazing-prime-generating-automaton.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-308-an-amazing-prime-generating-automaton.md
@@ -1,26 +1,26 @@
---
id: 5900f4a11000cf542c50ffb3
-title: 问题308：惊人的生成素数的自动机
+title: 'Problem 308: An amazing Prime-generating Automaton'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301962
dashedName: problem-308-an-amazing-prime-generating-automaton
---
# --description--
-用编程语言Fractran编写的程序包含一个分数列表。
+A program written in the programming language Fractran consists of a list of fractions.
-Fractran虚拟机的内部状态是一个正整数，该整数最初设置为种子值。 Fractran程序的每次迭代都将状态整数乘以列表中的第一个分数，从而将其保留为整数。
+The internal state of the Fractran Virtual Machine is a positive integer, which is initially set to a seed value. Each iteration of a Fractran program multiplies the state integer by the first fraction in the list which will leave it an integer.
-例如，约翰·霍顿·康威（John Horton Conway）为黄金一代编写的Fractran程序之一，由以下14个部分组成：1791 ， 7885 ， 1951年 ， 2338 ， 2933 ， 7729 ， 9523 ， 7719 ， 117 ， 1113 ， 1311 ， 152 ， 17 ， 551 。 从种子整数2开始，程序的连续迭代产生以下序列： 15，825，725，1925，2275，425，...，68，4，30，...，136，8，60，...，544，32，240，...
+For example, one of the Fractran programs that John Horton Conway wrote for prime-generation consists of the following 14 fractions:1791 , 7885 , 1951 , 2338 , 2933 , 7729 , 9523 , 7719 , 117 , 1113 , 1311 , 152 , 17 , 551 . Starting with the seed integer 2, successive iterations of the program produce the sequence: 15, 825, 725, 1925, 2275, 425, ..., 68, 4, 30, ..., 136, 8, 60, ..., 544, 32, 240, ...
-此序列中出现的2的幂是22、23、25，... 可以证明，该序列中所有2的幂都具有质数指数，并且所有质数都以正确的顺序显示为2的幂指数！
+The powers of 2 that appear in this sequence are 22, 23, 25, ... It can be shown that all the powers of 2 in this sequence have prime exponents and that all the primes appear as exponents of powers of 2, in proper order!
-如果有人使用上面的Fractran程序来解决Project Euler问题7（找到第10001个素数），那么在该程序产生第210001个素数之前，将需要进行几次迭代？
+If someone uses the above Fractran program to solve Project Euler Problem 7 (find the 10001st prime), how many iterations would be needed until the program produces 210001st prime ?
# --hints--
-`euler308()`应该返回1539669807660924。
+`euler308()` should return 1539669807660924.
```js
assert.strictEqual(euler308(), 1539669807660924);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-309-integer-ladders.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-309-integer-ladders.md
index 2e1150caaa7..03d5603c1f9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-309-integer-ladders.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-309-integer-ladders.md
@@ -1,24 +1,24 @@
---
id: 5900f4a11000cf542c50ffb4
-title: 问题309：整数阶梯
+title: 'Problem 309: Integer Ladders'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301963
dashedName: problem-309-integer-ladders
---
# --description--
-在经典的“穿越梯子”问题中，我们得到了两个梯子的长度x和y，它们位于一条狭窄的水平街道的相对壁上。 我们还给出了两个梯子交叉的街道上方的高度h，并要求我们找到街道的宽度（w）。
+In the classic "Crossing Ladders" problem, we are given the lengths x and y of two ladders resting on the opposite walls of a narrow, level street. We are also given the height h above the street where the two ladders cross and we are asked to find the width of the street (w).
-在这里，我们仅关注所有四个变量均为正整数的实例。 例如，如果x = 70，y = 119，h = 30，我们可以计算出w = 56。
+Here, we are only concerned with instances where all four variables are positive integers. For example, if x = 70, y = 119 and h = 30, we can calculate that w = 56.
-实际上，对于x，y，h和0 <x <y <200的整数，只有五个三元组（x，y，h）产生w的整数解： （70、119、30），（74、182、21），（87、105、35），（100、116、35）和（119、175、40）。
+In fact, for integer values x, y, h and 0 < x < y < 200, there are only five triplets (x,y,h) producing integer solutions for w: (70, 119, 30), (74, 182, 21), (87, 105, 35), (100, 116, 35) and (119, 175, 40).
-对于整数值x，y，h和0 <x <y <1 000 000，有多少个三元组（x，y，h）产生w的整数解？
+For integer values x, y, h and 0 < x < y < 1 000 000, how many triplets (x,y,h) produce integer solutions for w?
# --hints--
-`euler309()`应该返回210139。
+`euler309()` should return 210139.
```js
assert.strictEqual(euler309(), 210139);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-31-coin-sums.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-31-coin-sums.md
index eefc3828f4e..0826a685f4c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-31-coin-sums.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-31-coin-sums.md
@@ -1,44 +1,50 @@
---
id: 5900f38b1000cf542c50fe9e
-title: 问题31：硬币总和
+title: 'Problem 31: Coin sums'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301965
dashedName: problem-31-coin-sums
---
# --description--
-在英格兰，货币由英镑，英镑和便士p组成，并且有八种普通硬币流通：
+In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:
-1p，2p，5p，10p，20p，50p，£1（100p）和£2（200p）。
+<div style='margin-left: 4em;'>1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).</div>
-可以通过以下方式赚取£2：
+It is possible to make £2 in the following way:
-1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p
+<div style='margin-left: 4em;'>1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p</div>
-使用任意数量的硬币可以制成多少种便士？
+How many different ways can `n` pence be made using any number of coins?
# --hints--
-`digitnPowers(50)`应该返回451。
+`coinSums(50)` should return a number.
+
+```js
+assert(typeof coinSums(50) === 'number');
+```
+
+`coinSums(50)` should return 451.
```js
assert(coinSums(50) == 451);
```
-`digitnPowers(100)`应该返回4563。
+`coinSums(100)` should return 4563.
```js
assert(coinSums(100) == 4563);
```
-`digitnPowers(150)`应该返回21873。
+`coinSums(150)` should return 21873.
```js
assert(coinSums(150) == 21873);
```
-`digitnPowers(200)`应该返回73682。
+`coinSums(200)` should return 73682.
```js
assert(coinSums(200) == 73682);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-310-nim-square.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-310-nim-square.md
index cbc8e1b2d25..c93f2fd5315 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-310-nim-square.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-310-nim-square.md
@@ -1,26 +1,26 @@
---
id: 5900f4a21000cf542c50ffb5
-title: 问题310：尼姆广场
+title: 'Problem 310: Nim Square'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301966
dashedName: problem-310-nim-square
---
# --description--
-爱丽丝和鲍勃玩游戏Nim Square。
+Alice and Bob play the game Nim Square.
-Nim Square就像普通的三堆普通游戏Nim一样，但是玩家只能从堆中取出一平方的石头。
+Nim Square is just like ordinary three-heap normal play Nim, but the players may only remove a square number of stones from a heap.
-三个堆中的石头数量由有序三元组（a，b，c）表示。
+The number of stones in the three heaps is represented by the ordered triple (a,b,c).
-如果0≤a≤b≤c≤29，则下一个玩家的失落位数为1160。
+If 0≤a≤b≤c≤29 then the number of losing positions for the next player is 1160.
-如果0≤a≤b≤c≤100000，则查找下一个玩家的失落位置数。
+Find the number of losing positions for the next player if 0≤a≤b≤c≤100 000.
# --hints--
-`euler310()`应该返回2586528661783。
+`euler310()` should return 2586528661783.
```js
assert.strictEqual(euler310(), 2586528661783);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-311-biclinic-integral-quadrilaterals.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-311-biclinic-integral-quadrilaterals.md
index c748eca9ab8..b5aee9d2b54 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-311-biclinic-integral-quadrilaterals.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-311-biclinic-integral-quadrilaterals.md
@@ -1,24 +1,28 @@
---
id: 5900f4a31000cf542c50ffb6
-title: 问题311：双斜积分四边形
+title: 'Problem 311: Biclinic Integral Quadrilaterals'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301967
dashedName: problem-311-biclinic-integral-quadrilaterals
---
# --description--
-ABCD是凸的整数边四边形，其1≤AB
+ABCD is a convex, integer sided quadrilateral with 1 ≤ AB < BC < CD < AD.
-例如，以下四边形是双斜积分四边形： AB = 19，BC = 29，CD = 37，AD = 43，BD = 48和AO = CO = 23。
+BD has integer length. O is the midpoint of BD. AO has integer length.
-令B（N）为满足AB2 + BC2 + CD2 + AD2≤N的不同双斜积分四边形ABCD的数量。 我们可以验证B（10000）= 49和B（1 000 000）= 38239。
+We'll call ABCD a biclinic integral quadrilateral if AO = CO ≤ BO = DO.
-求B（10000000000000）。
+For example, the following quadrilateral is a biclinic integral quadrilateral: AB = 19, BC = 29, CD = 37, AD = 43, BD = 48 and AO = CO = 23.
+
+Let B(N) be the number of distinct biclinic integral quadrilaterals ABCD that satisfy AB2+BC2+CD2+AD2 ≤ N. We can verify that B(10 000) = 49 and B(1 000 000) = 38239.
+
+Find B(10 000 000 000).
# --hints--
-`euler311()`应该返回2466018557。
+`euler311()` should return 2466018557.
```js
assert.strictEqual(euler311(), 2466018557);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-312-cyclic-paths-on-sierpiski-graphs.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-312-cyclic-paths-on-sierpiski-graphs.md
index abf43f6da44..b9083a73f0b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-312-cyclic-paths-on-sierpiski-graphs.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-312-cyclic-paths-on-sierpiski-graphs.md
@@ -1,26 +1,26 @@
---
id: 5900f4a51000cf542c50ffb7
-title: 问题312：Sierpiński图上的循环路径
+title: 'Problem 312: Cyclic paths on Sierpiński graphs'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301968
dashedName: problem-312-cyclic-paths-on-sierpiski-graphs
---
# --description--
-\-1阶（S1）的Sierpiński图是等边三角形。
+\- A Sierpiński graph of order-1 (S1) is an equilateral triangle.
-\-通过将Sn的三个副本放置在Sn上，从而使每对副本都有一个公共角，从而从Sn中获得Sn +1。
+\- Sn+1 is obtained from Sn by positioning three copies of Sn so that every pair of copies has one common corner.
-令C（n）为恰好一次通过Sn的所有顶点的循环数。 例如，C（3）= 8，因为可以在S3上绘制八个这样的循环，如下所示：
+Let C(n) be the number of cycles that pass exactly once through all the vertices of Sn. For example, C(3) = 8 because eight such cycles can be drawn on S3, as shown below:
-也可以验证： C（1）= C（2）= 1 C（5）= 71328803586048 C（10,000）mod 108 = 37652224 C（10,000）模138 = 617720485
+It can also be verified that : C(1) = C(2) = 1 C(5) = 71328803586048 C(10 000) mod 108 = 37652224 C(10 000) mod 138 = 617720485
-求C（C（C（10,000））mod 138。
+Find C(C(C(10 000))) mod 138.
# --hints--
-`euler312()`应该返回324681947。
+`euler312()` should return 324681947.
```js
assert.strictEqual(euler312(), 324681947);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-313-sliding-game.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-313-sliding-game.md
index 3af89ec341c..dd6ec7cfb3a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-313-sliding-game.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-313-sliding-game.md
@@ -1,24 +1,24 @@
---
id: 5900f4a61000cf542c50ffb8
-title: 问题313：滑行游戏
+title: 'Problem 313: Sliding game'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301969
dashedName: problem-313-sliding-game
---
# --description--
-在滑动游戏中，柜台可以水平或垂直滑入一个空白区域。 游戏的目的是将红色计数器从网格的左上角移至右下角。 该空间始终始于右下角。 例如，以下图片序列显示了如何在2 x 2网格上进行5次移动才能完成游戏。
+In a sliding game a counter may slide horizontally or vertically into an empty space. The objective of the game is to move the red counter from the top left corner of a grid to the bottom right corner; the space always starts in the bottom right corner. For example, the following sequence of pictures show how the game can be completed in five moves on a 2 by 2 grid.
-令S（m，n）代表完成游戏的最小移动次数。 例如，可以验证S（5.4）= 25。
+Let S(m,n) represent the minimum number of moves to complete the game on an m by n grid. For example, it can be verified that S(5,4) = 25.
-正好有5482个网格，其中S（m，n）= p2，其中p <100是质数。
+There are exactly 5482 grids for which S(m,n) = p2, where p < 100 is prime.
-S（m，n）= p2有多少个网格，其中p <106是素数？
+How many grids does S(m,n) = p2, where p < 106 is prime?
# --hints--
-`euler313()`应该返回2057774861813004。
+`euler313()` should return 2057774861813004.
```js
assert.strictEqual(euler313(), 2057774861813004);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-314-the-mouse-on-the-moon.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-314-the-mouse-on-the-moon.md
index d03a6060cac..78cd0f1cf65 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-314-the-mouse-on-the-moon.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-314-the-mouse-on-the-moon.md
@@ -1,26 +1,26 @@
---
id: 5900f4a71000cf542c50ffb9
-title: 问题314：月球上的老鼠
+title: 'Problem 314: The Mouse on the Moon'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301970
dashedName: problem-314-the-mouse-on-the-moon
---
# --description--
-月亮已经打开，可以免费获得土地，但是有一个陷阱。您必须在放样的土地周围建造一堵墙，在月球上建造一堵墙很昂贵。每个国家都被分配了500 m x 500 m平方的面积，但它们仅拥有其围墙所在的区域。 251001个帖子已放置在1米间隔的矩形网格中。墙必须是一系列闭合的直线，每条直线都在柱子之间延伸。
+The moon has been opened up, and land can be obtained for free, but there is a catch. You have to build a wall around the land that you stake out, and building a wall on the moon is expensive. Every country has been allotted a 500 m by 500 m square area, but they will possess only that area which they wall in. 251001 posts have been placed in a rectangular grid with 1 meter spacing. The wall must be a closed series of straight lines, each line running from post to post.
-当然，较大的国家/地区建造了2000 m的围墙，将整个250 000 m2的区域封闭起来。大芬维克公国（Duchy of Grand Fenwick）的预算较为紧张，并已要求您（他们的皇家程序员）计算出哪种形状会获得最佳的最大封闭面积/墙长比。
+The bigger countries of course have built a 2000 m wall enclosing the entire 250 000 m2 area. The Duchy of Grand Fenwick, has a tighter budget, and has asked you (their Royal Programmer) to compute what shape would get best maximum enclosed-area/wall-length ratio.
-您已经在一张纸上进行了一些初步的计算。 对于封闭25万平方米的2000米墙， 封闭区域/墙长比为125。 尽管不允许，但要弄清楚是否更好：如果在正方形区域内放置一个圆圈，使其接触四边，则该区域将等于π*2502 m2，周长将为π* 500 m，因此封闭区域/墙长比也将为125。
+You have done some preliminary calculations on a sheet of paper. For a 2000 meter wall enclosing the 250 000 m2 area the enclosed-area/wall-length ratio is 125. Although not allowed , but to get an idea if this is anything better: if you place a circle inside the square area touching the four sides the area will be equal to π*2502 m2 and the perimeter will be π*500 m, so the enclosed-area/wall-length ratio will also be 125.
-但是，如果从正方形切下边长为75 m，75 m和75√2m的四个三角形，则总面积将变为238750 m2，周长将变为1400 +300√2m。因此，得出的封闭区域/壁长比为130.87，这要好得多。
+However, if you cut off from the square four triangles with sides 75 m, 75 m and 75√2 m the total area becomes 238750 m2 and the perimeter becomes 1400+300√2 m. So this gives an enclosed-area/wall-length ratio of 130.87, which is significantly better.
-找到最大的封闭区域/墙长比。 以abc.defghijk的形式将答案四舍五入到小数点后8位。
+Find the maximum enclosed-area/wall-length ratio. Give your answer rounded to 8 places behind the decimal point in the form abc.defghijk.
# --hints--
-`euler314()`应该返回132.52756426。
+`euler314()` should return 132.52756426.
```js
assert.strictEqual(euler314(), 132.52756426);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-315-digital-root-clocks.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-315-digital-root-clocks.md
index 6adfc58c0da..38a2a69cca1 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-315-digital-root-clocks.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-315-digital-root-clocks.md
@@ -1,38 +1,38 @@
---
id: 5900f4a71000cf542c50ffba
-title: 问题315：数字根时钟
+title: 'Problem 315: Digital root clocks'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301971
dashedName: problem-315-digital-root-clocks
---
# --description--
-请Sam和Max将两个数字时钟转换为两个“数字根”时钟。
+Sam and Max are asked to transform two digital clocks into two "digital root" clocks.
-数字根时钟是一种逐步计算数字根的数字时钟。
+A digital root clock is a digital clock that calculates digital roots step by step.
-当时钟输入一个数字时，它将显示它，然后它将开始计算，显示所有中间值，直到得到结果为止。 例如，如果时钟被输入数字137，它将显示：“ 137”→“ 11”→“ 2”，然后它将变黑，等待下一个数字。
+When a clock is fed a number, it will show it and then it will start the calculation, showing all the intermediate values until it gets to the result. For example, if the clock is fed the number 137, it will show: "137" → "11" → "2" and then it will go black, waiting for the next number.
-每个数字均由一些亮段组成：三个水平段（顶部，中间，底部）和四个垂直段（左上，右上，左下，右下）。 数字“ 1”由垂直的右上角和右下角构成，数字“ 4”由中间的水平和垂直左上角，右上角和右下角组成。数字“ 8”将它们全部点亮。
+Every digital number consists of some light segments: three horizontal (top, middle, bottom) and four vertical (top-left, top-right, bottom-left, bottom-right). Number "1" is made of vertical top-right and bottom-right, number "4" is made by middle horizontal and vertical top-left, top-right and bottom-right. Number "8" lights them all.
-仅当打开/关闭分段时，时钟才消耗能量。 打开“ 2”将花费5个转换，而“ 7”将仅花费4个转换。
+The clocks consume energy only when segments are turned on/off. To turn on a "2" will cost 5 transitions, while a "7" will cost only 4 transitions.
-山姆和麦克斯建造了两个不同的时钟。
+Sam and Max built two different clocks.
-山姆的时钟喂食例如数字137：时钟显示“ 137”，然后关闭面板，然后打开下一个数字（“ 11”），然后再次关闭面板，最后打开最后一个数字（“ 2”）并在一段时间后关闭。 例如，对于数字137，Sam的时钟要求：“ 137” ： （2 + 5 + 4）×2 = 22个转换（“ 137”打开/关闭）。 “ 11” ： （2 + 2）×2 = 8个转换（“ 11”开/关）。 “ 2” ： （5）×2 = 10个转换（“ 2”打开/关闭）。
+Sam's clock is fed e.g. number 137: the clock shows "137", then the panel is turned off, then the next number ("11") is turned on, then the panel is turned off again and finally the last number ("2") is turned on and, after some time, off. For the example, with number 137, Sam's clock requires:"137" : (2 + 5 + 4) × 2 = 22 transitions ("137" on/off). "11" : (2 + 2) × 2 = 8 transitions ("11" on/off). "2" : (5) × 2 = 10 transitions ("2" on/off).
-总共进行了40次转换。
+For a grand total of 40 transitions.
-马克斯的时钟工作方式不同。除了关闭整个面板，它还足够聪明，可以仅关闭下一个数字不需要的那些段。 对于数字137，麦克斯的时钟要求：“ 137” ： 2 + 5 + 4 = 11个转换（“ 137”亮） 7个转换（关闭数字“ 11”不需要的段）。 “ 11” ： 0个转换（数字“ 11”已正确打开） 3个转换（关闭第一个“ 1”和第二个“ 1”的底部； 顶部与数字“ 2”相同）。 “ 2” ： 4个转换（打开其余段以获取“ 2”） 5个转换（关闭数字“ 2”）。
+Max's clock works differently. Instead of turning off the whole panel, it is smart enough to turn off only those segments that won't be needed for the next number. For number 137, Max's clock requires:"137" : 2 + 5 + 4 = 11 transitions ("137" on) 7 transitions (to turn off the segments that are not needed for number "11"). "11" : 0 transitions (number "11" is already turned on correctly) 3 transitions (to turn off the first "1" and the bottom part of the second "1"; the top part is common with number "2"). "2" : 4 transitions (to turn on the remaining segments in order to get a "2") 5 transitions (to turn off number "2").
-总共进行了30次转换。
+For a grand total of 30 transitions.
-当然，Max的时钟比Sam的时钟消耗更少的功率。 两个时钟都馈入A = 107和B = 2×107之间的所有素数。 求出Sam的时钟所需的跃迁总数与Max的时钟所需的跃迁总数之间的差。
+Of course, Max's clock consumes less power than Sam's one. The two clocks are fed all the prime numbers between A = 107 and B = 2×107. Find the difference between the total number of transitions needed by Sam's clock and that needed by Max's one.
# --hints--
-`euler315()`应该返回13625242。
+`euler315()` should return 13625242.
```js
assert.strictEqual(euler315(), 13625242);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-316-numbers-in-decimal-expansions.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-316-numbers-in-decimal-expansions.md
index 7e0d487ca37..4a2821b24e7 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-316-numbers-in-decimal-expansions.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-316-numbers-in-decimal-expansions.md
@@ -1,30 +1,30 @@
---
id: 5900f4a81000cf542c50ffbb
-title: 问题316：十进制扩展中的数字
+title: 'Problem 316: Numbers in decimal expansions'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301972
dashedName: problem-316-numbers-in-decimal-expansions
---
# --description--
-令p = p1 p2 p3 ...是无限数的随机数字序列，它们以相等的概率从{0,1,2,3,4,5,6,7,8,9}中选择。
+Let p = p1 p2 p3 ... be an infinite sequence of random digits, selected from {0,1,2,3,4,5,6,7,8,9} with equal probability.
-可以看出p对应于实数0.p1 p2 p3 ....
+It can be seen that p corresponds to the real number 0.p1 p2 p3 ....
-还可以看出，从间隔\[0,1）中选择一个随机实数等效于选择一个从{0,1,2,3,4,5,6,7,8， 9}的概率相等。
+It can also be seen that choosing a random real number from the interval \[0,1) is equivalent to choosing an infinite sequence of random digits selected from {0,1,2,3,4,5,6,7,8,9} with equal probability.
-对于具有d个十进制数字的任何正整数n，令k为最小索引，以使pk，pk + 1，... pk + d-1为n的十进制数字，顺序相同。 同样，令g（n）为k的期望值； 可以证明g（n）总是有限的，有趣的是，总是整数。
+For any positive integer n with d decimal digits, let k be the smallest index such that pk, pk+1, ...pk+d-1 are the decimal digits of n, in the same order. Also, let g(n) be the expected value of k; it can be proven that g(n) is always finite and, interestingly, always an integer number.
-例如，如果n = 535，则 对于p = 31415926535897 ....，我们得到k = 9 对于p = 355287143650049560000490848764084685354 ...，我们得到k = 36 等等，我们发现g（535）= 1008。
+For example, if n = 535, then for p = 31415926535897...., we get k = 9 for p = 355287143650049560000490848764084685354..., we get k = 36 etc and we find that g(535) = 1008.
-鉴于此，找到
+Given that , find
-注意：代表发言权功能。
+Note: represents the floor function.
# --hints--
-`euler316()`应该返回542934735751917760。
+`euler316()` should return 542934735751917760.
```js
assert.strictEqual(euler316(), 542934735751917760);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-317-firecracker.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-317-firecracker.md
index 12f6163faa2..bb528a225aa 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-317-firecracker.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-317-firecracker.md
@@ -1,22 +1,22 @@
---
id: 5900f4aa1000cf542c50ffbc
-title: 问题317：鞭炮
+title: 'Problem 317: Firecracker'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301973
dashedName: problem-317-firecracker
---
# --description--
-爆竹在离地面100 m的高度处爆炸。 它分为许多非常小的碎片，它们向各个方向移动。 它们都具有相同的初始速度20 m / s。
+A firecracker explodes at a height of 100 m above level ground. It breaks into a large number of very small fragments, which move in every direction; all of them have the same initial velocity of 20 m/s.
-我们假设碎片在没有空气阻力的情况下在g = 9.81 m / s2的均匀重力场中运动。
+We assume that the fragments move without air resistance, in a uniform gravitational field with g=9.81 m/s2.
-找到碎片到达地面之前通过的区域的体积（单位：m3）。 将答案四舍五入到小数点后四位。
+Find the volume (in m3) of the region through which the fragments move before reaching the ground. Give your answer rounded to four decimal places.
# --hints--
-`euler317()`应该返回1856532.8455。
+`euler317()` should return 1856532.8455.
```js
assert.strictEqual(euler317(), 1856532.8455);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-318-2011-nines.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-318-2011-nines.md
index a5f52ddc16d..1e8e9b8821e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-318-2011-nines.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-318-2011-nines.md
@@ -1,48 +1,48 @@
---
id: 5900f4ab1000cf542c50ffbd
-title: 问题318：2011个九
+title: 'Problem 318: 2011 nines'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301974
dashedName: problem-318-2011-nines
---
# --description--
-考虑实数√2+√3。
+Consider the real number √2+√3.
-当我们计算√2+√3的偶数幂时
+When we calculate the even powers of √2+√3
-我们得到：
+we get:
-（√2+√3）2 = 9.898979485566356 ...
+(√2+√3)2 = 9.898979485566356...
-（√2+√3）4 = 97.98979485566356 ...
+(√2+√3)4 = 97.98979485566356...
-（√2+√3）6 = 969.998969071069263 ...
+(√2+√3)6 = 969.998969071069263...
-（√2+√3）8 = 9601.99989585502907 ...
+(√2+√3)8 = 9601.99989585502907...
-（√2+√3）10 = 95049.999989479221 ...
+(√2+√3)10 = 95049.999989479221...
-（√2+√3）12 = 940897.9999989371855 ...
+(√2+√3)12 = 940897.9999989371855...
-（√2+√3）14 = 9313929.99999989263 ...
+(√2+√3)14 = 9313929.99999989263...
-（√2+√3）16 = 92198401.99999998915 ...
+(√2+√3)16 = 92198401.99999998915...
-这些幂的小数部分开头的连续九个数字似乎没有减少。 实际上，可以证明（√2+√3）2n的小数部分对于大n接近1。
+It looks like that the number of consecutive nines at the beginning of the fractional part of these powers is non-decreasing. In fact it can be proven that the fractional part of (√2+√3)2n approaches 1 for large n.
-考虑形式为√p+√q的所有实数，其中p和q为正整数，且p <q，使得小数部分 （√p+√q）的2n对于大n接近1。
+Consider all real numbers of the form √p+√q with p and q positive integers and p<q, such that the fractional part of (√p+√q)2n approaches 1 for large n.
-令C（p，q，n）为（√p+√q）2n的小数部分开头的连续九个数字。
+Let C(p,q,n) be the number of consecutive nines at the beginning of the fractional part of (√p+√q)2n.
-令N（p，q）为n的最小值，以使C（p，q，n）≥2011。
+Let N(p,q) be the minimal value of n such that C(p,q,n) ≥ 2011.
-求p + q≤2011的∑N（p，q）。
+Find ∑N(p,q) for p+q ≤ 2011.
# --hints--
-`euler318()`应该返回709313889。
+`euler318()` should return 709313889.
```js
assert.strictEqual(euler318(), 709313889);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-319-bounded-sequences.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-319-bounded-sequences.md
index fff5dad7bac..687eeb26cd2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-319-bounded-sequences.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-319-bounded-sequences.md
@@ -1,28 +1,30 @@
---
id: 5900f4ab1000cf542c50ffbe
-title: 问题319：有界序列
+title: 'Problem 319: Bounded Sequences'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301975
dashedName: problem-319-bounded-sequences
---
# --description--
-令x1，x2，...，xn为长度为n的序列，使得：
+Let x1, x2,..., xn be a sequence of length n such that:
x1 = 2
-对于所有1
+for all 1 < i ≤ n : xi-1 < xi
-这样的长度为2的序列只有五个，即： {2,4}，{2,5}，{2,6}，{2,7}和{2,8}。 有293个这样的长度为5的序列。 以下是三个示例： {2,5,11,25,55}，{2,6,14,36,88}，{2,8,22,64,181}。
+for all i and j with 1 ≤ i, j ≤ n : (xi) j < (xj + 1)i
-令t（n）表示长度为n的此类序列的数量。 给出t（10）= 86195和t（20）= 5227991891。
+There are only five such sequences of length 2, namely: {2,4}, {2,5}, {2,6}, {2,7} and {2,8}. There are 293 such sequences of length 5; three examples are given below: {2,5,11,25,55}, {2,6,14,36,88}, {2,8,22,64,181}.
-找出t（1010）并给出答案模块109。
+Let t(n) denote the number of such sequences of length n. You are given that t(10) = 86195 and t(20) = 5227991891.
+
+Find t(1010) and give your answer modulo 109.
# --hints--
-`euler319()`应该返回268457129。
+`euler319()` should return 268457129.
```js
assert.strictEqual(euler319(), 268457129);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-32-pandigital-products.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-32-pandigital-products.md
index 3d7af2a05c8..2559b17664f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-32-pandigital-products.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-32-pandigital-products.md
@@ -1,30 +1,30 @@
---
id: 5900f38c1000cf542c50fe9f
-title: 问题32：Pandigital产品
+title: 'Problem 32: Pandigital products'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301976
dashedName: problem-32-pandigital-products
---
# --description--
-我们可以说，如果n位数字恰好一次使用了1到n的所有数字，那么它就是pandigital。 例如，5位数字15234是1到5泛数字。
+We shall say that an `n`-digit number is pandigital if it makes use of all the digits 1 to `n` exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.
-乘积7254是不寻常的，因为其标识为39×186 = 7254，包含被乘数，乘数，乘积是1到9泛数。
+The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.
-找出所有被乘数/乘数/产品标识可以写成1到9泛数字的所有产品的总和。
+Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.
-提示：某些产品可以通过多种方式获得，因此请确保只在其总和中包括一次。
+**Hint:** Some products can be obtained in more than one way so be sure to only include it once in your sum.
# --hints--
-`pandigitalProducts()`是一个函数。
+`pandigitalProducts()` should return a number.
```js
-assert(typeof pandigitalProducts === 'function');
+assert(typeof pandigitalProducts() === 'number');
```
-`pandigitalProducts()`应该返回45228。
+`pandigitalProducts()` should return 45228.
```js
assert.strictEqual(pandigitalProducts(), 45228);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-320-factorials-divisible-by-a-huge-integer.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-320-factorials-divisible-by-a-huge-integer.md
index 10dc42944f8..5171172f984 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-320-factorials-divisible-by-a-huge-integer.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-320-factorials-divisible-by-a-huge-integer.md
@@ -1,24 +1,24 @@
---
id: 5900f4ae1000cf542c50ffbf
-title: 问题320：阶乘被整数整除
+title: 'Problem 320: Factorials divisible by a huge integer'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301977
dashedName: problem-320-factorials-divisible-by-a-huge-integer
---
# --description--
-令N（i）是最小的整数n，这样n！ 被（i！）整除1234567890
+Let N(i) be the smallest integer n such that n! is divisible by (i!)1234567890
-设S（u）= ∑N（i）且10≤i≤u。
+Let S(u)=∑N(i) for 10 ≤ i ≤ u.
-S（1000）= 614538266565663。
+S(1000)=614538266565663.
-找出S（1 000 000）mod 1018。
+Find S(1 000 000) mod 1018.
# --hints--
-`euler320()`应该返回278157919195482660。
+`euler320()` should return 278157919195482660.
```js
assert.strictEqual(euler320(), 278157919195482660);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-321-swapping-counters.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-321-swapping-counters.md
index 94ce658be02..34e52a4dea9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-321-swapping-counters.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-321-swapping-counters.md
@@ -1,26 +1,26 @@
---
id: 5900f4ae1000cf542c50ffc0
-title: 问题321：交换计数器
+title: 'Problem 321: Swapping Counters'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301978
dashedName: problem-321-swapping-counters
---
# --description--
-由2n + 1个正方形组成的水平行在一端有n个红色计数器，在另一端有n个蓝色计数器，中间由一个空的正方形隔开。 例如，当n = 3时。
+A horizontal row comprising of 2n + 1 squares has n red counters placed at one end and n blue counters at the other end, being separated by a single empty square in the centre. For example, when n = 3.
-计数器可以从一个正方形移动到下一个正方形（滑动），也可以跳过另一个计数器（跳），只要该计数器旁边的正方形未被占用即可。
+A counter can move from one square to the next (slide) or can jump over another counter (hop) as long as the square next to that counter is unoccupied.
-令M（n）表示完全反转彩色计数器位置的最小移动/动作数； 也就是说，将所有红色计数器向右移动，并将所有蓝色计数器向左移动。 可以验证M（3）= 15，它也恰好是三角形数。
+Let M(n) represent the minimum number of moves/actions to completely reverse the positions of the coloured counters; that is, move all the red counters to the right and all the blue counters to the left. It can be verified M(3) = 15, which also happens to be a triangle number.
-如果我们基于n的值创建一个序列，其中M（n）是一个三角形数字，那么前五个项将是： 1、3、10、22和63，它们的总和为99。
+If we create a sequence based on the values of n for which M(n) is a triangle number then the first five terms would be: 1, 3, 10, 22, and 63, and their sum would be 99.
-找出该序列的前四十项之和。
+Find the sum of the first forty terms of this sequence.
# --hints--
-`euler321()`应该返回2470433131948040。
+`euler321()` should return 2470433131948040.
```js
assert.strictEqual(euler321(), 2470433131948040);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-322-binomial-coefficients-divisible-by-10.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-322-binomial-coefficients-divisible-by-10.md
index 7f4343ec718..241c3a74f07 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-322-binomial-coefficients-divisible-by-10.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-322-binomial-coefficients-divisible-by-10.md
@@ -1,8 +1,8 @@
---
id: 5900f4af1000cf542c50ffc1
-title: 问题322：二项式系数可被10整除
+title: 'Problem 322: Binomial coefficients divisible by 10'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301979
dashedName: problem-322-binomial-coefficients-divisible-by-10
---
@@ -16,7 +16,7 @@ Find T(1018, 1012-10).
# --hints--
-`euler322()`应该返回999998760323314000。
+`euler322()` should return 999998760323314000.
```js
assert.strictEqual(euler322(), 999998760323314000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-323-bitwise-or-operations-on-random-integers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-323-bitwise-or-operations-on-random-integers.md
index 9eadb73896e..b3367595620 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-323-bitwise-or-operations-on-random-integers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-323-bitwise-or-operations-on-random-integers.md
@@ -1,28 +1,28 @@
---
id: 5900f4b01000cf542c50ffc2
-title: 问题323：对随机整数进行按位或运算
+title: 'Problem 323: Bitwise-OR operations on random integers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301980
dashedName: problem-323-bitwise-or-operations-on-random-integers
---
# --description--
-令y0，y1，y2，...是随机无符号32位整数的序列
+Let y0, y1, y2,... be a sequence of random unsigned 32 bit integers
-（即0≤yi <232，每个值的可能性均等）。
+(i.e. 0 ≤ yi < 232, every value equally likely).
-对于序列xi，给出以下递归：x0 = 0且
+For the sequence xi the following recursion is given:x0 = 0 and
-xi = xi-1 | yi-1，对于i>0。（|是按位或运算符）
+xi = xi-1| yi-1, for i > 0. ( | is the bitwise-OR operator)
-可以看出，对于所有i≥N，最终将存在一个索引N，使得xi = 232 -1（所有比特的位模式）。
+It can be seen that eventually there will be an index N such that xi = 232 -1 (a bit-pattern of all ones) for all i ≥ N.
-找到N的期望值。 将答案四舍五入到小数点后10位。
+Find the expected value of N. Give your answer rounded to 10 digits after the decimal point.
# --hints--
-`euler323()`应该返回6.3551758451。
+`euler323()` should return 6.3551758451.
```js
assert.strictEqual(euler323(), 6.3551758451);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-324-building-a-tower.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-324-building-a-tower.md
index ac9d7ae3399..66ce4914777 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-324-building-a-tower.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-324-building-a-tower.md
@@ -1,20 +1,22 @@
---
id: 5900f4b11000cf542c50ffc3
-title: 问题324：建造一座塔楼
+title: 'Problem 324: Building a tower'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301981
dashedName: problem-324-building-a-tower
---
# --description--
-设f（n）表示用2×1×1的块填充3×3×n塔的方式的数量。你可以以任何你喜欢的方式旋转块;然而，塔本身的旋转，反射等被视为不同的。例如（q = 100000007）：f（2）= 229，f（4）= 117805，f（10）mod q = 96149360，f（103）mod q = 24806056，f（106）mod q = 30808124。
+Let f(n) represent the number of ways one can fill a 3×3×n tower with blocks of 2×1×1. You're allowed to rotate the blocks in any way you like; however, rotations, reflections etc of the tower itself are counted as distinct.
-求f（1010000）mod 100000007。
+For example (with q = 100000007) :f(2) = 229,f(4) = 117805,f(10) mod q = 96149360,f(103) mod q = 24806056,f(106) mod q = 30808124.
+
+Find f(1010000) mod 100000007.
# --hints--
-`euler324()`应该返回96972774。
+`euler324()` should return 96972774.
```js
assert.strictEqual(euler324(), 96972774);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-325-stone-game-ii.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-325-stone-game-ii.md
index 2550e373cd2..6d0389b9fac 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-325-stone-game-ii.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-325-stone-game-ii.md
@@ -1,30 +1,30 @@
---
id: 5900f4b11000cf542c50ffc4
-title: 问题325：石头游戏II
+title: 'Problem 325: Stone Game II'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301982
dashedName: problem-325-stone-game-ii
---
# --description--
-一场比赛是用两堆石头和两个球员进行的。在轮到她时，玩家从较大的堆中移除了许多石头。她移除的石头数量必须是较小桩中石头数量的正数倍。
+A game is played with two piles of stones and two players. At her turn, a player removes a number of stones from the larger pile. The number of stones she removes must be a positive multiple of the number of stones in the smaller pile.
-例如，让有序对（6,14）描述一个配置，在较小的桩中有6个石头，在较大的桩中有14个石头，那么第一个玩家可以从较大的桩中移除6或12个石头。
+E.g., let the ordered pair(6,14) describe a configuration with 6 stones in the smaller pile and 14 stones in the larger pile, then the first player can remove 6 or 12 stones from the larger pile.
-从一堆中取出所有宝石的玩家赢得比赛。
+The player taking all the stones from a pile wins the game.
-获胜配置是第一个玩家可以强制获胜的配置。例如，（1,5），（2,6）和（3,12）是获胜配置，因为第一个玩家可以立即移除第二堆中的所有宝石。
+A winning configuration is one where the first player can force a win. For example, (1,5), (2,6) and (3,12) are winning configurations because the first player can immediately remove all stones in the second pile.
-失败的配置是第二个玩家可以强制获胜的配置，无论第一个玩家做什么。例如，（2,3）和（3,4）正在失去配置：任何合法移动都会为第二个玩家留下获胜配置。
+A losing configuration is one where the second player can force a win, no matter what the first player does. For example, (2,3) and (3,4) are losing configurations: any legal move leaves a winning configuration for the second player.
-将S（N）定义为所有丢失配置（xi，yi）的总和（xi + yi），0 <xi <yi≤N。我们可以验证S（10）= 211和S（104）= 230312207313。
+Define S(N) as the sum of (xi+yi) for all losing configurations (xi,yi), 0 < xi < yi ≤ N. We can verify that S(10) = 211 and S(104) = 230312207313.
-找到S（1016）mod 710。
+Find S(1016) mod 710.
# --hints--
-`euler325()`应该返回54672965。
+`euler325()` should return 54672965.
```js
assert.strictEqual(euler325(), 54672965);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-326-modulo-summations.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-326-modulo-summations.md
index 28501bd9cc4..1e6b648d9a0 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-326-modulo-summations.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-326-modulo-summations.md
@@ -1,28 +1,28 @@
---
id: 5900f4b21000cf542c50ffc5
-title: 问题326：模数汇总
+title: 'Problem 326: Modulo Summations'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301983
dashedName: problem-326-modulo-summations
---
# --description--
-设一个递归定义的序列：。
+Let an be a sequence recursively defined by: .
-因此，前10个元素是：1,1,0,3,0,3,5,4,1,9。
+So the first 10 elements of an are: 1,1,0,3,0,3,5,4,1,9.
-设f（N，M）表示对（p，q）的数量，使得：
+Let f(N,M) represent the number of pairs (p,q) such that:
-可以看出f（10,10）= 4，对（3,3），（5,5），（7,9）和（9,10）。
+It can be seen that f(10,10)=4 with the pairs (3,3), (5,5), (7,9) and (9,10).
-你也得到f（104,103）= 97158。
+You are also given that f(104,103)=97158.
-求f（1012,106）。
+Find f(1012,106).
# --hints--
-`euler326()`应该返回1966666166408794400。
+`euler326()` should return 1966666166408794400.
```js
assert.strictEqual(euler326(), 1966666166408794400);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-327-rooms-of-doom.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-327-rooms-of-doom.md
index ab5d26769e8..354596b1d74 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-327-rooms-of-doom.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-327-rooms-of-doom.md
@@ -1,34 +1,34 @@
---
id: 5900f4b31000cf542c50ffc6
-title: 问题327：末日的房间
+title: 'Problem 327: Rooms of Doom'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301984
dashedName: problem-327-rooms-of-doom
---
# --description--
-一系列三个房间通过自动门相互连接。
+A series of three rooms are connected to each other by automatic doors.
-每扇门都由安全卡操作。进入房间后，门会自动关闭，安全卡不能再次使用。一台机器在开始时将分配无限数量的卡，但每个房间（包括起始室）都包含扫描仪，如果他们检测到您持有三张以上的安全卡，或者如果他们在地板上检测到无人看管的安全卡，那么所有的门都将永久锁定。但是，每个房间都有一个盒子，您可以安全地存放任意数量的安全卡，以便在以后使用。
+Each door is operated by a security card. Once you enter a room the door automatically closes and that security card cannot be used again. A machine at the start will dispense an unlimited number of cards, but each room (including the starting room) contains scanners and if they detect that you are holding more than three security cards or if they detect an unattended security card on the floor, then all the doors will become permanently locked. However, each room contains a box where you may safely store any number of security cards for use at a later stage.
-如果您只是尝试一次一个地穿过房间，那么当您进入房间3时，您将使用所有三张卡片，并将永远被困在那个房间！
+If you simply tried to travel through the rooms one at a time then as you entered room 3 you would have used all three cards and would be trapped in that room forever!
-但是，如果您使用存储箱，则可以进行逃生。例如，您可以使用第一张卡进入房间1，将一张卡放入存储箱，然后使用第三张卡退回到房间。然后在从点胶机收集另外三张牌之后，您可以使用一张进入1号房间并收集刚刚放入盒子中的卡片。你现在又有三张牌，可以通过剩余的三扇门。此方法允许您使用总共六张安全卡通过所有三个房间。
+However, if you make use of the storage boxes, then escape is possible. For example, you could enter room 1 using your first card, place one card in the storage box, and use your third card to exit the room back to the start. Then after collecting three more cards from the dispensing machine you could use one to enter room 1 and collect the card you placed in the box a moment ago. You now have three cards again and will be able to travel through the remaining three doors. This method allows you to travel through all three rooms using six security cards in total.
-可以使用总共123张安全卡通过六个房间，最多可携带3张卡。
+It is possible to travel through six rooms using a total of 123 security cards while carrying a maximum of 3 cards.
-设C是任何时候可以携带的最大卡数。设R是要经过的房间数量。设M（C，R）是分配机器在任何时间通过最多携带C卡的R室所需的最小卡数。
+Let C be the maximum number of cards which can be carried at any time. Let R be the number of rooms to travel through. Let M(C,R) be the minimum number of cards required from the dispensing machine to travel through R rooms carrying up to a maximum of C cards at any time.
-例如，M（3,6）= 123并且M（4,6）= 23.并且，对于3≤C≤4，ΣM（C，6）= 146。
+For example, M(3,6)=123 and M(4,6)=23.And, ΣM(C,6)=146 for 3 ≤ C ≤ 4.
-对于3≤C≤10，给出ΣM（C，10）= 10382。
+You are given that ΣM(C,10)=10382 for 3 ≤ C ≤ 10.
-找3ΣC≤40的ΣM（C，30）。
+Find ΣM(C,30) for 3 ≤ C ≤ 40.
# --hints--
-`euler327()`应返回34315549139516。
+`euler327()` should return 34315549139516.
```js
assert.strictEqual(euler327(), 34315549139516);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-328-lowest-cost-search.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-328-lowest-cost-search.md
index 024d1f5e4ef..4a00f455cc1 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-328-lowest-cost-search.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-328-lowest-cost-search.md
@@ -1,28 +1,36 @@
---
id: 5900f4b41000cf542c50ffc7
-title: 问题328：成本最低的搜索
+title: 'Problem 328: Lowest-cost Search'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301985
dashedName: problem-328-lowest-cost-search
---
# --description--
-我们试图通过提问来找到从整数集{1,2，...，n}中选择的隐藏数字。我们要求的每个数字（问题）的成本等于所要求的数字，我们得到三个可能的答案之一：“您的猜测低于隐藏的数字”，或“是的，就是它！”或“您的猜测是高于隐藏号码“。给定n的值，最优策略最小化了最坏情况下的总成本（即所有问题的总和）。例如
+We are trying to find a hidden number selected from the set of integers {1, 2, ..., n} by asking questions.
-如果n = 3，我们能做的最好的就是问号码“2”。答案将立即引导我们找到隐藏号码（总费用= 2）。
+Each number (question) we ask, has a cost equal to the number asked and we get one of three possible answers: "Your guess is lower than the hidden number", or
-如果n = 8，我们可能决定使用“二分搜索”类型的策略：我们的第一个问题是“4”，如果隐藏数字高于4，我们将需要一个或两个额外的问题。让我们的第二个问题是“6”。如果隐藏数字仍高于6，我们将需要第三个问题以区分7和8.因此，我们的第三个问题将是“7”，这个最坏情况的总成本将是4 + 6 + 7 = 17。
+"Yes, that's it!", or
-通过将“5”作为我们的第一个问题，我们可以大大改善n = 8的最坏情况成本。如果我们被告知隐藏的数字高于5，我们的第二个问题将是“7”，那么我们将确定隐藏的数字是什么（总成本为5 + 7 = 12）。如果我们被告知隐藏号码低于5，我们的第二个问题将是“3”，如果隐藏号码低于3，我们的第三个问题将是“1”，总成本为5 + 3 + 1 = 9。自12> 9以来，该策略的最坏情况成本为12.这比我们之前使用“二元搜索”策略所取得的成本更好;它也优于或等于任何其他策略。所以，事实上，我们刚刚描述了n = 8的最优策略。
+"Your guess is higher than the hidden number".
-假设C（n）是通过n的最优策略实现的最坏情况成本，如上所述。因此，C（1）= 0，C（2）= 1，C（3）= 2，C（8）= 12.类似地，C（100）= 400，C（n）= 17575。
+Given the value of n, an optimal strategy minimizes the total cost (i.e. the sum of all the questions asked) for the worst possible case. E.g.
-找到C（n）。
+If n=3, the best we can do is obviously to ask the number "2". The answer will immediately lead us to find the hidden number (at a total cost = 2).
+
+If n=8, we might decide to use a "binary search" type of strategy: Our first question would be "4" and if the hidden number is higher than 4 we will need one or two additional questions. Let our second question be "6". If the hidden number is still higher than 6, we will need a third question in order to discriminate between 7 and 8. Thus, our third question will be "7" and the total cost for this worst-case scenario will be 4+6+7=17.
+
+We can improve considerably the worst-case cost for n=8, by asking "5" as our first question. If we are told that the hidden number is higher than 5, our second question will be "7", then we'll know for certain what the hidden number is (for a total cost of 5+7=12). If we are told that the hidden number is lower than 5, our second question will be "3" and if the hidden number is lower than 3 our third question will be "1", giving a total cost of 5+3+1=9. Since 12>9, the worst-case cost for this strategy is 12. That's better than what we achieved previously with the "binary search" strategy; it is also better than or equal to any other strategy. So, in fact, we have just described an optimal strategy for n=8.
+
+Let C(n) be the worst-case cost achieved by an optimal strategy for n, as described above. Thus C(1) = 0, C(2) = 1, C(3) = 2 and C(8) = 12. Similarly, C(100) = 400 and C(n) = 17575.
+
+Find C(n).
# --hints--
-`euler328()`应返回260511850222。
+`euler328()` should return 260511850222.
```js
assert.strictEqual(euler328(), 260511850222);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-329-prime-frog.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-329-prime-frog.md
index 643e7308e32..6b1e185daf9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-329-prime-frog.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-329-prime-frog.md
@@ -1,24 +1,28 @@
---
id: 5900f4b51000cf542c50ffc8
-title: 问题329：Prime Frog
+title: 'Problem 329: Prime Frog'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301986
dashedName: problem-329-prime-frog
---
# --description--
-苏珊有一只青蛙。她的青蛙跳过500个正方形，编号为1到500.他只能以相同的概率跳到左边或右边的一个方格，并且他不能超出范围\[1; 500]。（如果它落在任何一个结束时，它会在下一步移动时自动跳转到唯一可用的方格。）
+Susan has a prime frog.
-当他在一个带有素数的正方形上时，他在跳到下一个正方形之前以概率1/3以概率2/3或“N”（不是PRIME）弯曲'P'（PRIME）。当他在一个上面有一个不是素数的数字的正方形时，他会在跳到下一个正方形之前以概率为1/3或'N'以概率2/3歪曲'P'。
+Her frog is jumping around over 500 squares numbered 1 to 500.
-鉴于青蛙的起始位置是随机的，每个方格具有相同的概率，并且考虑到她听到他的前15个呱呱声，她听到PPPPNNPPPNPPNPN序列的概率是多少？
+He can only jump one square to the left or to the right, with equal probability, and he cannot jump outside the range \[1;500].(if it lands at either end, it automatically jumps to the only available square on the next move.)
-以简化形式给出你的答案作为p / q的分数。
+When he is on a square with a prime number on it, he croaks 'P' (PRIME) with probability 2/3 or 'N' (NOT PRIME) with probability 1/3 just before jumping to the next square. When he is on a square with a number on it that is not a prime he croaks 'P' with probability 1/3 or 'N' with probability 2/3 just before jumping to the next square.
+
+Given that the frog's starting position is random with the same probability for every square, and given that she listens to his first 15 croaks, what is the probability that she hears the sequence PPPPNNPPPNPPNPN?
+
+Give your answer as a fraction p/q in reduced form.
# --hints--
-`euler329()`应返回199740353/29386561536000。
+`euler329()` should return 199740353 / 29386561536000.
```js
assert.strictEqual(euler329(), 199740353 / 29386561536000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-33-digit-cancelling-fractions.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-33-digit-cancelling-fractions.md
index 334c673bc25..ec7756e9451 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-33-digit-cancelling-fractions.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-33-digit-cancelling-fractions.md
@@ -1,18 +1,30 @@
---
id: 5900f38d1000cf542c50fea0
-title: 问题33：数字取消分数
+title: 'Problem 33: Digit cancelling fractions'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301987
dashedName: problem-33-digit-cancelling-fractions
---
# --description--
-分数<sup><sub>九十八分之四十九</sub></sup>是好奇馏分，如在试图简化它可能会错误地认为<sup><sub>九十八分之四十九</sub></sup> = <sup><sub>4/8，</sub></sup>这是正确的，则通过取消787-9获得一个没有经验的数学家。我们应考虑馏分喜欢， <sup><sub>五十○分之三十○=</sub></sup> <sup><sub>3/5，</sub></sup>是微不足道的例子。这种类型的分数恰好有四个非平凡的例子，小于一个值，并且在分子和分母中包含两个数字。如果这四个分数的乘积以其最低公共项给出，请找到分母的值。
+The fraction <sup>49</sup>/<sub>98</sub> is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that <sup>49</sup>/<sub>98</sub> = <sup>4</sup>/<sub>8</sub>, which is correct, is obtained by cancelling the 9s.
+
+We shall consider fractions like, <sup>30</sup>/<sub>50</sub> = <sup>3</sup>/<sub>5</sub>, to be trivial examples.
+
+There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.
+
+If the product of these four fractions is given in its lowest common terms, find the value of the denominator.
# --hints--
-`digitCancellingFractions()`应该返回100。
+`digitCancellingFractions()` should return a number.
+
+```js
+assert(typeof digitCancellingFractions() === 'number');
+```
+
+`digitCancellingFractions()` should return 100.
```js
assert.strictEqual(digitCancellingFractions(), 100);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-330-eulers-number.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-330-eulers-number.md
index 9bd524fd058..c8a3feab396 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-330-eulers-number.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-330-eulers-number.md
@@ -1,36 +1,34 @@
---
id: 5900f4b71000cf542c50ffc9
-title: 问题330：欧拉数
+title: 'Problem 330: Euler''s Number'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301988
dashedName: problem-330-eulers-number
---
# --description--
-为所有整数n定义了无限的实数序列a（n），如下所示：
+An infinite sequence of real numbers a(n) is defined for all integers n as follows:
-例如，a（0）= 11！ + 12！ + 13！ + ... = e - 1 a（1）= e - 11！ + 12！ + 13！ + ... = 2e - 3 a（2）= 2e - 31！ + e - 12！ + 13！ + ... = 72 e - 6
+<!-- TODO Use MathJax and re-write from projecteuler.net -->
-e = 2.7182818 ......是欧拉常数。
+For example,a(0) = 11! + 12! + 13! + ... = e − 1 a(1) = e − 11! + 12! + 13! + ... = 2e − 3 a(2) = 2e − 31! + e − 12! + 13! + ... = 72 e − 6
-可以证明a（n）是形式
+with e = 2.7182818... being Euler's constant.
-```
- A(n) e + B(n)n! for integers A(n) and B(n).
-```
+It can be shown that a(n) is of the form
-例如a（10）=
+A(n) e + B(n)n! for integers A(n) and B(n).
-```
- 328161643 e − 65269448610! .
-```
+For example a(10) =
-求A（109）+ B（109）并给出答案mod 77 777 777。
+328161643 e − 65269448610!.
+
+Find A(109) + B(109) and give your answer mod 77 777 777.
# --hints--
-`euler330()`应该返回15955822。
+`euler330()` should return 15955822.
```js
assert.strictEqual(euler330(), 15955822);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-331-cross-flips.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-331-cross-flips.md
index 9b46b46f824..4d3d330a744 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-331-cross-flips.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-331-cross-flips.md
@@ -1,30 +1,30 @@
---
id: 5900f4b71000cf542c50ffca
-title: 问题331：交叉翻转
+title: 'Problem 331: Cross flips'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301989
dashedName: problem-331-cross-flips
---
# --description--
-N×N个盘放在方形游戏板上。每个磁盘都有黑色和白色。
+N×N disks are placed on a square game board. Each disk has a black side and white side.
-在每个回合中，您可以选择一个磁盘并翻转与该磁盘相同的行和同一列中的所有磁盘：因此翻转2×N-1个磁盘。当所有磁盘显示其白色边时，游戏结束。以下示例显示了5×5板上的游戏。
+At each turn, you may choose a disk and flip all the disks in the same row and the same column as this disk: thus 2×N-1 disks are flipped. The game ends when all disks show their white side. The following example shows a game on a 5×5 board.
-可以证明3是完成这个游戏的最小转弯次数。
+It can be proven that 3 is the minimal number of turns to finish this game.
-N×N板上的左下盘具有坐标（0,0）;右下盘具有坐标（N-1,0），左上盘具有坐标（0，N-1）。
+The bottom left disk on the N×N board has coordinates (0,0); the bottom right disk has coordinates (N-1,0) and the top left disk has coordinates (0,N-1).
-CN为具有N×N个盘的板的以下配置：（x，y）处的盘满足，表示其黑色侧;否则，它显示其白色的一面。 C5如上所示。
+Let CN be the following configuration of a board with N×N disks: A disk at (x,y) satisfying , shows its black side; otherwise, it shows its white side. C5 is shown above.
-设T（N）是从配置CN开始完成游戏的最小圈数，如果配置CN不可解，则为0。我们已经证明T（5）= 3。你还得到T（10）= 29和T（1 000）= 395253。
+Let T(N) be the minimal number of turns to finish a game starting from configuration CN or 0 if configuration CN is unsolvable. We have shown that T(5)=3. You are also given that T(10)=29 and T(1 000)=395253.
-找 。
+Find .
# --hints--
-`euler331()`应该返回467178235146843500。
+`euler331()` should return 467178235146843500.
```js
assert.strictEqual(euler331(), 467178235146843500);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-332-spherical-triangles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-332-spherical-triangles.md
index bf6aad00b40..ea3a4d0c431 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-332-spherical-triangles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-332-spherical-triangles.md
@@ -1,24 +1,24 @@
---
id: 5900f4b91000cf542c50ffcb
-title: 问题332：球形三角形
+title: 'Problem 332: Spherical triangles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301990
dashedName: problem-332-spherical-triangles
---
# --description--
-球面三角形是通过在三个顶点中成对交叉的三个大圆弧形成在球体表面上的图形。
+A spherical triangle is a figure formed on the surface of a sphere by three great circular arcs intersecting pairwise in three vertices.
-设C（r）为具有中心（0,0,0）和半径r的球体。设Z（r）为具有整数坐标的C（r）表面上的点集。令T（r）为顶点为Z（r）的球形三角形的集合。由相同大弧上的三个点形成的简并球面三角形不包括在T（r）中。设A（r）为T（r）中最小球面三角形的面积。
+Let C(r) be the sphere with the centre (0,0,0) and radius r. Let Z(r) be the set of points on the surface of C(r) with integer coordinates. Let T(r) be the set of spherical triangles with vertices in Z(r). Degenerate spherical triangles, formed by three points on the same great arc, are not included in T(r). Let A(r) be the area of the smallest spherical triangle in T(r).
-例如，A（14）是3.294040四舍五入到小数点后六位。
+For example A(14) is 3.294040 rounded to six decimal places.
-找到A（r）。将您的答案四舍五入到小数点后六位。
+Find A(r). Give your answer rounded to six decimal places.
# --hints--
-`euler332()`应该返回2717.751525。
+`euler332()` should return 2717.751525.
```js
assert.strictEqual(euler332(), 2717.751525);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-333-special-partitions.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-333-special-partitions.md
index fc2cfcb6e5b..5af490eb7ea 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-333-special-partitions.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-333-special-partitions.md
@@ -1,30 +1,30 @@
---
id: 5900f4b91000cf542c50ffcc
-title: 问题333：特殊分区
+title: 'Problem 333: Special partitions'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301991
dashedName: problem-333-special-partitions
---
# --description--
-可以以这样的方式划分所有正整数：分区的每个项可以表示为2ix3j，其中i，j≥0。
+All positive integers can be partitioned in such a way that each and every term of the partition can be expressed as 2ix3j, where i,j ≥ 0.
-我们只考虑那些没有任何术语可以划分任何其他术语的分区。例如，17 = 2 + 6 + 9 =（21x30 + 21x31 + 20x32）的分区将无效，因为2可以除以6.分区17 = 16 + 1 =（24x30 + 20x30）也不会因为1可以除16. 17的唯一有效分区是8 + 9 =（23x30 + 20x32）。
+Let's consider only those such partitions where none of the terms can divide any of the other terms. For example, the partition of 17 = 2 + 6 + 9 = (21x30 + 21x31 + 20x32) would not be valid since 2 can divide 6. Neither would the partition 17 = 16 + 1 = (24x30 + 20x30) since 1 can divide 16. The only valid partition of 17 would be 8 + 9 = (23x30 + 20x32).
-许多整数具有多个有效分区，第一个是具有以下两个分区的11。 11 = 2 + 9 =（21x30 + 20x32）11 = 8 + 3 =（23x30 + 20x31）
+Many integers have more than one valid partition, the first being 11 having the following two partitions. 11 = 2 + 9 = (21x30 + 20x32) 11 = 8 + 3 = (23x30 + 20x31)
-让我们将P（n）定义为n的有效分区数。例如，P（11）= 2。
+Let's define P(n) as the number of valid partitions of n. For example, P(11) = 2.
-让我们只考虑具有单个有效分区的素数整数q，例如P（17）。
+Let's consider only the prime integers q which would have a single valid partition such as P(17).
-素数q <100的总和使得P（q）= 1等于233。
+The sum of the primes q <100 such that P(q)=1 equals 233.
-找到质数q <1000000的总和，使得P（q）= 1。
+Find the sum of the primes q <1000000 such that P(q)=1.
# --hints--
-`euler333()`应返回3053105。
+`euler333()` should return 3053105.
```js
assert.strictEqual(euler333(), 3053105);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-334-spilling-the-beans.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-334-spilling-the-beans.md
index 6bb2bb5d5b3..fe354cec8a1 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-334-spilling-the-beans.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-334-spilling-the-beans.md
@@ -1,30 +1,28 @@
---
id: 5900f4ba1000cf542c50ffcd
-title: 问题334：溢出豆子
+title: 'Problem 334: Spilling the beans'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301992
dashedName: problem-334-spilling-the-beans
---
# --description--
-在柏拉图的天堂里，存在着无数个直线的碗。每个碗都包含一些或不包含有限数量的豆。一个孩子玩游戏，只允许一种移动：从任何碗中取出两个豆子，并在两个相邻的碗中放入一个豆子。当每个碗包含一个或没有豆时，游戏结束。
+In Plato's heaven, there exist an infinite number of bowls in a straight line. Each bowl either contains some or none of a finite number of beans. A child plays a game, which allows only one kind of move: removing two beans from any bowl, and putting one in each of the two adjacent bowls. The game ends when each bowl contains either one or no beans.
-例如，考虑两个相邻的碗分别包含2个和3个豆，所有其他碗都是空的。以下八个动作将完成游戏：
+For example, consider two adjacent bowls containing 2 and 3 beans respectively, all other bowls being empty. The following eight moves will finish the game:
-您将获得以下序列：t0 = 123456。
+<!-- TODO Use MathJax and re-write from projecteuler.net -->
-```
- ti = ti-12 , if ti-1 is even ti-12 926252, if ti-1 is odd where ⌊x⌋ is the floor function and is the bitwise XOR operator. bi = ( ti mod 211) + 1.
-```
+You are given the following sequences: t0 = 123456. ti = ti-12, if ti-1 is even ti-12 926252, if ti-1 is odd where ⌊x⌋ is the floor function and is the bitwise XOR operator. bi = ( ti mod 211) + 1.
-最后一个序列的前两个项是b1 = 289和b2 = 145.如果我们从两个相邻碗中的b1和b2豆开始，则需要3419100次移动来完成游戏。
+The first two terms of the last sequence are b1 = 289 and b2 = 145. If we start with b1 and b2 beans in two adjacent bowls, 3419100 moves would be required to finish the game.
-现在考虑1500个相邻的碗，分别包含b1，b2，...，b1500豆，所有其他碗都是空的。查看游戏结束前需要多少动作。
+Consider now 1500 adjacent bowls containing b1, b2,..., b1500 beans respectively, all other bowls being empty. Find how many moves it takes before the game ends.
# --hints--
-`euler334()`应该返回150320021261690850。
+`euler334()` should return 150320021261690850.
```js
assert.strictEqual(euler334(), 150320021261690850);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-335-gathering-the-beans.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-335-gathering-the-beans.md
index b9623d3359c..0605e0d053f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-335-gathering-the-beans.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-335-gathering-the-beans.md
@@ -1,24 +1,24 @@
---
id: 5900f4bd1000cf542c50ffce
-title: 问题335：收集豆子
+title: 'Problem 335: Gathering the beans'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301993
dashedName: problem-335-gathering-the-beans
---
# --description--
-每当彼得感到无聊时，他都会放一些碗，每个碗里都放一个豆子。在此之后，他将所有的豆子从一个碗中取出，然后顺时针将它们逐一放入碗中。他重复这一点，从碗里开始扔掉最后一根豆子，直到最初的情况再次出现。例如，有5个碗，他的行为如下：
+Whenever Peter feels bored, he places some bowls, containing one bean each, in a circle. After this, he takes all the beans out of a certain bowl and drops them one by one in the bowls going clockwise. He repeats this, starting from the bowl he dropped the last bean in, until the initial situation appears again. For example with 5 bowls he acts as follows:
-所以有5个碗需要Peter 15的动作才能恢复到初始状态。
+So with 5 bowls it takes Peter 15 moves to return to the initial situation.
-设M（x）表示从x碗开始返回初始状态所需的移动次数。因此，M（5）= 15.还可以证实M（100）= 10920。
+Let M(x) represent the number of moves required to return to the initial situation, starting with x bowls. Thus, M(5) = 15. It can also be verified that M(100) = 10920.
-找到M（2k + 1）。给出你的答案模数79。
+Find M(2k+1). Give your answer modulo 79.
# --hints--
-`euler335()`应返回5032316。
+`euler335()` should return 5032316.
```js
assert.strictEqual(euler335(), 5032316);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-336-maximix-arrangements.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-336-maximix-arrangements.md
index 604ce185dfe..3dbf18111c6 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-336-maximix-arrangements.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-336-maximix-arrangements.md
@@ -1,26 +1,30 @@
---
id: 5900f4bd1000cf542c50ffcf
-title: 问题336：Maximix安排
+title: 'Problem 336: Maximix Arrangements'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301994
dashedName: problem-336-maximix-arrangements
---
# --description--
-一列火车用于按顺序运输四辆车：ABCD。然而，有时当火车到达收集车厢时，它们的顺序不正确。为了重新安排车厢，他们都被分流到一个大型旋转转盘上。在车厢在特定点处脱开之后，列车从转盘上移开，拉动仍与其连接的托架。其余的车厢旋转180度。然后重新加入所有托架，并且根据需要重复该过程，以便获得转盘的最少使用次数。一些布置，例如ADCB，可以很容易地解决：托架在A和D之间分开，并且在DCB旋转之后，已经实现了正确的顺序。
+A train is used to transport four carriages in the order: ABCD. However, sometimes when the train arrives to collect the carriages they are not in the correct order.
-然而，火车司机简单西蒙并不知道他的效率，所以他总是通过最初将马车A放在正确的位置，然后是马车B等来解决问题。
+To rearrange the carriages they are all shunted on to a large rotating turntable. After the carriages are uncoupled at a specific point the train moves off the turntable pulling the carriages still attached with it. The remaining carriages are rotated 180 degrees. All of the carriages are then rejoined and this process is repeated as often as necessary in order to obtain the least number of uses of the turntable.
-使用四个车厢，西蒙的最坏可能安排是DACB和DBAC，我们称之为maximix安排。每个都需要他五次旋转（尽管使用最有效的方法，他们可以使用三次旋转来解决）。他用于DACB的过程如下所示。
+Some arrangements, such as ADCB, can be solved easily: the carriages are separated between A and D, and after DCB are rotated the correct order has been achieved.
-可以证实，对于六个车厢存在24个最大值布置，其中第十个词典最大化布置是DFAECB。
+However, Simple Simon, the train driver, is not known for his efficiency, so he always solves the problem by initially getting carriage A in the correct place, then carriage B, and so on.
-查找11种车厢的第11版词典格式。
+Using four carriages, the worst possible arrangements for Simon, which we shall call maximix arrangements, are DACB and DBAC; each requiring him five rotations (although, using the most efficient approach, they could be solved using just three rotations). The process he uses for DACB is shown below.
+
+It can be verified that there are 24 maximix arrangements for six carriages, of which the tenth lexicographic maximix arrangement is DFAECB.
+
+Find the 2011th lexicographic maximix arrangement for eleven carriages.
# --hints--
-`euler336()`应该返回CAGBIHEFJDK。
+`euler336()` should return CAGBIHEFJDK.
```js
assert.strictEqual(euler336(), CAGBIHEFJDK);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-337-totient-stairstep-sequences.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-337-totient-stairstep-sequences.md
index 4ab8dae78b9..677c918f053 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-337-totient-stairstep-sequences.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-337-totient-stairstep-sequences.md
@@ -1,22 +1,32 @@
---
id: 5900f4be1000cf542c50ffd0
-title: 问题337欧拉序列阶梯
+title: 'Problem 337: Totient Stairstep Sequences'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301995
dashedName: problem-337-totient-stairstep-sequences
---
# --description--
-令{a1，a2，...，an}为长度为n的整数序列，使得：a1 = 6，对于所有1≤i<n：φ（ai）<φ（ai + 1）<ai <ai + 11令S（N）为具有≤N的这种序列的数目。例如，S（10）= 4：{6}，{6,8}，{6,8,9}和{6,10}。我们可以验证S（100）= 482073668和S（10 000）mod 108 = 73808307。
+Let {a1, a2,..., an} be an integer sequence of length n such that:
-找到S（20 000 000）mod 108。
+a1 = 6
-1φ表示欧拉的函数。
+for all 1 ≤ i < n : φ(ai) < φ(ai+1) < ai < ai+11
+
+Let S(N) be the number of such sequences with an ≤ N.
+
+For example, S(10) = 4: {6}, {6, 8}, {6, 8, 9} and {6, 10}.
+
+We can verify that S(100) = 482073668 and S(10 000) mod 108 = 73808307.
+
+Find S(20 000 000) mod 108.
+
+1 φ denotes Euler's totient function.
# --hints--
-`euler337()`应该返回85068035。
+`euler337()` should return 85068035.
```js
assert.strictEqual(euler337(), 85068035);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-338-cutting-rectangular-grid-paper.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-338-cutting-rectangular-grid-paper.md
index 2381325ecb8..514dc560b01 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-338-cutting-rectangular-grid-paper.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-338-cutting-rectangular-grid-paper.md
@@ -1,26 +1,30 @@
---
id: 5900f4be1000cf542c50ffd1
-title: 问题338：切割矩形网格纸
+title: 'Problem 338: Cutting Rectangular Grid Paper'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301996
dashedName: problem-338-cutting-rectangular-grid-paper
---
# --description--
-给出了具有整数尺寸w×h的矩形网格纸。它的网格间距为1.当我们沿着网格线将纸张切割成两个部分并重新排列这些部分而没有重叠时，我们可以制作具有不同尺寸的新矩形。例如，从尺寸为9×4的纸张中，我们可以通过切割和重新排列来制作尺寸为18×2,12×3和6×6的矩形，如下所示：
+A rectangular sheet of grid paper with integer dimensions w × h is given. Its grid spacing is 1.
-同样，从尺寸为9×8的纸张中，我们可以制作尺寸为18×4和12×6的矩形。
+When we cut the sheet along the grid lines into two pieces and rearrange those pieces without overlap, we can make new rectangles with different dimensions.
-对于w和h对，让F（w，h）是可以由尺寸为w×h的薄片制成的不同矩形的数量。例如，F（2,1）= 0，F（2,2）= 1，F（9,4）= 3和F（9,8）= 2.注意，与初始一致的矩形不计算在内在F（w，h）。还要注意，尺寸为w×h且尺寸为h×w的矩形不被认为是不同的。
+For example, from a sheet with dimensions 9 × 4 , we can make rectangles with dimensions 18 × 2, 12 × 3 and 6 × 6 by cutting and rearranging as below:
-对于整数N，令G（N）为满足0 <h≤w≤N的所有w和h的F（w，h）之和。我们可以验证G（10）= 55，G（103） ）= 971745和G（105）= 9992617687。
+Similarly, from a sheet with dimensions 9 × 8 , we can make rectangles with dimensions 18 × 4 and 12 × 6 .
-找到G（1012）。给你的答案模数108。
+For a pair w and h, let F(w,h) be the number of distinct rectangles that can be made from a sheet with dimensions w × h . For example, F(2,1) = 0, F(2,2) = 1, F(9,4) = 3 and F(9,8) = 2. Note that rectangles congruent to the initial one are not counted in F(w,h). Note also that rectangles with dimensions w × h and dimensions h × w are not considered distinct.
+
+For an integer N, let G(N) be the sum of F(w,h) for all pairs w and h which satisfy 0 < h ≤ w ≤ N. We can verify that G(10) = 55, G(103) = 971745 and G(105) = 9992617687.
+
+Find G(1012). Give your answer modulo 108.
# --hints--
-`euler338()`应该返回15614292。
+`euler338()` should return 15614292.
```js
assert.strictEqual(euler338(), 15614292);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-339-peredur-fab-efrawg.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-339-peredur-fab-efrawg.md
index ef273874fe0..6844c7e65c9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-339-peredur-fab-efrawg.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-339-peredur-fab-efrawg.md
@@ -1,22 +1,22 @@
---
id: 5900f4c01000cf542c50ffd2
-title: 问题339：Peredur fab Efrawg
+title: 'Problem 339: Peredur fab Efrawg'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301997
dashedName: problem-339-peredur-fab-efrawg
---
# --description--
-“他走向一个山谷，河流穿过河谷;山谷的边界是树木繁茂的，河的两边是平坦的草地。河的一边，他看到一群白羊，另一只黑羊。每当一只白羊咩咩叫，其中一只黑羊就会越过白色;当其中一只黑羊咩咩叫的时候，其中一只白羊会越过黑羊“en.wikisource.org
+"And he came towards a valley, through which ran a river; and the borders of the valley were wooded, and on each side of the river were level meadows. And on one side of the river he saw a flock of white sheep, and on the other a flock of black sheep. And whenever one of the white sheep bleated, one of the black sheep would cross over and become white; and when one of the black sheep bleated, one of the white sheep would cross over and become black."en.wikisource.org
-最初每群由n只绵羊组成。每只绵羊（不论颜色）同样可能是下一只咩咩咩的绵羊。在一只绵羊咩咩叫，另一只羊的羊群已经越过之后，Peredur可能会移走一些白羊，以便最大化预期的最终黑羊数量。如果Peredur使用最优策略，则让E（n）成为预期的最终黑羊数量。
+Initially each flock consists of n sheep. Each sheep (regardless of colour) is equally likely to be the next sheep to bleat. After a sheep has bleated and a sheep from the other flock has crossed over, Peredur may remove a number of white sheep in order to maximize the expected final number of black sheep. Let E(n) be the expected final number of black sheep if Peredur uses an optimal strategy.
-给出E（5）= 6.871346舍入到小数点后面的6位。找到E（10 000）并将您的答案四舍五入到小数点后面的6位。
+You are given that E(5) = 6.871346 rounded to 6 places behind the decimal point. Find E(10 000) and give your answer rounded to 6 places behind the decimal point.
# --hints--
-`euler339()`应返回19823.542204。
+`euler339()` should return 19823.542204.
```js
assert.strictEqual(euler339(), 19823.542204);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-34-digit-factorials.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-34-digit-factorials.md
index ff369435582..080d068344e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-34-digit-factorials.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-34-digit-factorials.md
@@ -1,18 +1,28 @@
---
id: 5900f38e1000cf542c50fea1
-title: 问题34：数字阶乘
+title: 'Problem 34: Digit factorials'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301998
dashedName: problem-34-digit-factorials
---
# --description--
-145是一个奇怪的数字，为1！ + 4！ + 5！ = 1 + 24 + 120 = 145.找到数字和数字的总和，它们等于数字的阶乘之和。注意：为1！ = 1和2！ = 2不是它们不包括在内的总和。
+145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
+
+Find the numbers and the sum of the numbers which are equal to the sum of the factorial of their digits.
+
+**Note:** as 1! = 1 and 2! = 2 are not sums they are not included.
# --hints--
-`digitFactorial()`应该返回{sum：40730，数字：[145,40585]}。
+`digitFactorial()` should return an object.
+
+```js
+assert.typeOf(digitFactorial(), 'object');
+```
+
+`digitFactorial()` should return { sum: 40730, numbers: [145, 40585] }.
```js
assert.deepEqual(digitFactorial(), { sum: 40730, numbers: [145, 40585] });
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-340-crazy-function.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-340-crazy-function.md
index 33d7ef56940..86b29f061e2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-340-crazy-function.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-340-crazy-function.md
@@ -1,24 +1,28 @@
---
id: 5900f4c21000cf542c50ffd4
-title: 问题340：疯狂的功能
+title: 'Problem 340: Crazy Function'
challengeType: 5
-videoUrl: ''
+forumTopicId: 301999
dashedName: problem-340-crazy-function
---
# --description--
-对于固定整数a，b，c，定义疯狂函数F（n）如下：F（n）= n - c表示所有n> b F（n）= F（a + F（a + F（a +）所有n≤b的F（a + n））））。
+For fixed integers a, b, c, define the crazy function F(n) as follows:
-另外，定义S（a，b，c）=。
+F(n) = n - c for all n > b
-例如，如果a = 50，b = 2000且c = 40，则F（0）= 3240并且F（2000）= 2040.此外，S（50,2000,40）= 5204240。
+F(n) = F(a + F(a + F(a + F(a + n)))) for all n ≤ b.
-找到S（217,721,127）的最后9位数。
+Also, define S(a, b, c) = .
+
+For example, if a = 50, b = 2000 and c = 40, then F(0) = 3240 and F(2000) = 2040. Also, S(50, 2000, 40) = 5204240.
+
+Find the last 9 digits of S(217, 721, 127).
# --hints--
-`euler340()`应返回291504964。
+`euler340()` should return 291504964.
```js
assert.strictEqual(euler340(), 291504964);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-341-golombs-self-describing-sequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-341-golombs-self-describing-sequence.md
index 282bb936ad1..740ef0d7518 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-341-golombs-self-describing-sequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-341-golombs-self-describing-sequence.md
@@ -1,24 +1,24 @@
---
id: 5900f4c11000cf542c50ffd3
-title: 问题341：Golomb的自我描述序列
+title: 'Problem 341: Golomb''s self-describing sequence'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302000
dashedName: problem-341-golombs-self-describing-sequence
---
# --description--
-Golomb的自描述序列{G（n）}是唯一的非递减自然数序列，因此n在序列中恰好出现G（n）次。前几个n的G（n）值是
+The Golomb's self-describing sequence {G(n)} is the only nondecreasing sequence of natural numbers such that n appears exactly G(n) times in the sequence. The values of G(n) for the first few n are
-n123456789101112131415 ... G（n）的122334445556666 ...
+n123456789101112131415…G(n)122334445556666…
-给出G（103）= 86，G（106）= 6137.对于1≤n<103，还给出ΣG（n3）= 153506976。
+You are given that G(103) = 86, G(106) = 6137. You are also given that ΣG(n3) = 153506976 for 1 ≤ n < 103.
-找到ΣG（n3）为1≤n<106。
+Find ΣG(n3) for 1 ≤ n < 106.
# --hints--
-`euler341()`应该返回56098610614277016。
+`euler341()` should return 56098610614277016.
```js
assert.strictEqual(euler341(), 56098610614277016);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-342-the-totient-of-a-square-is-a-cube.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-342-the-totient-of-a-square-is-a-cube.md
index 508057cf3f9..2253ce390fb 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-342-the-totient-of-a-square-is-a-cube.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-342-the-totient-of-a-square-is-a-cube.md
@@ -1,22 +1,26 @@
---
id: 5900f4c31000cf542c50ffd5
-title: 问题342：正方形的总数是一个立方体
+title: 'Problem 342: The totient of a square is a cube'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302001
dashedName: problem-342-the-totient-of-a-square-is-a-cube
---
# --description--
-考虑数字50. 502 = 2500 = 22×54，因此φ（2500）= 2×4×53 = 8×53 = 23×53.1因此2500是正方形而φ（2500）是立方体。
+Consider the number 50.
-找到所有数字的总和n，1 <n <1010，使得φ（n2）是立方体。
+502 = 2500 = 22 × 54, so φ(2500) = 2 × 4 × 53 = 8 × 53 = 23 × 53. 1
-1φ表示欧拉的函数。
+So 2500 is a square and φ(2500) is a cube.
+
+Find the sum of all numbers n, 1 < n < 1010 such that φ(n2) is a cube.
+
+1 φ denotes Euler's totient function.
# --hints--
-`euler342()`应该返回5943040885644。
+`euler342()` should return 5943040885644.
```js
assert.strictEqual(euler342(), 5943040885644);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-343-fractional-sequences.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-343-fractional-sequences.md
index f93678622de..3102009e8c0 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-343-fractional-sequences.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-343-fractional-sequences.md
@@ -1,26 +1,36 @@
---
id: 5900f4c41000cf542c50ffd6
-title: 问题343：分数序列
+title: 'Problem 343: Fractional Sequences'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302002
dashedName: problem-343-fractional-sequences
---
# --description--
-对于任何正整数k，分数xi / yi的有限序列ai由以下定义：a1 = 1 / k且ai =（xi-1 + 1）/（yi-1-1）减少到最低项i> 1 。当ai达到某个整数n时，序列停止。 （即，当yi = 1时。）定义f（k）= n。例如，对于k = 20：
+For any positive integer k, a finite sequence ai of fractions xi/yi is defined by:
-1/20→2/19→3/18 = 1/6→2/5→3/4→4/3→5/2→6/1 = 6
+a1 = 1/k and
-所以f（20）= 6。
+ai = (xi-1+1)/(yi-1-1) reduced to lowest terms for i>1.
-对于1≤k≤100，f（1）= 1，f（2）= 2，f（3）= 1且Σf（k3）= 118937。
+When ai reaches some integer n, the sequence stops. (That is, when yi=1.)
-找Σf（k3）为1≤k≤2×106。
+Define f(k) = n.
+
+For example, for k = 20:
+
+1/20 → 2/19 → 3/18 = 1/6 → 2/5 → 3/4 → 4/3 → 5/2 → 6/1 = 6
+
+So f(20) = 6.
+
+Also f(1) = 1, f(2) = 2, f(3) = 1 and Σf(k3) = 118937 for 1 ≤ k ≤ 100.
+
+Find Σf(k3) for 1 ≤ k ≤ 2×106.
# --hints--
-`euler343()`应该返回269533451410884200。
+`euler343()` should return 269533451410884200.
```js
assert.strictEqual(euler343(), 269533451410884200);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-344-silver-dollar-game.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-344-silver-dollar-game.md
index c2ac58e839b..f4fb768b65b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-344-silver-dollar-game.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-344-silver-dollar-game.md
@@ -1,34 +1,34 @@
---
id: 5900f4c51000cf542c50ffd7
-title: 问题344：银元游戏
+title: 'Problem 344: Silver dollar game'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302003
dashedName: problem-344-silver-dollar-game
---
# --description--
-NG de Bruijn的银元游戏的一个变种可以描述如下：
+One variant of N.G. de Bruijn's silver dollar game can be described as follows:
-在正方形条上放置了许多硬币，每平方最多一枚硬币。只有一枚硬币，称为银元，具有任何价值。两名球员轮流做出动作。在每个回合中，玩家必须进行常规或特殊动作。
+On a strip of squares a number of coins are placed, at most one coin per square. Only one coin, called the silver dollar, has any value. Two players take turns making moves. At each turn a player must make either a regular or a special move.
-常规移动包括选择一个硬币并将其移动到左侧的一个或多个方块。硬币不能从条带中移出或跳到另一个硬币上或上方。
+A regular move consists of selecting one coin and moving it one or more squares to the left. The coin cannot move out of the strip or jump on or over another coin.
-或者，玩家可以选择将最左边的硬币扒窃而不是定期移动。如果没有可能的常规动作，则玩家被迫掏出最左边的硬币。
+Alternatively, the player can choose to make the special move of pocketing the leftmost coin rather than making a regular move. If no regular moves are possible, the player is forced to pocket the leftmost coin.
-获胜者是兜售银元的玩家。
+The winner is the player who pockets the silver dollar.
-获胜配置是在条带上的硬币排列，其中第一玩家可以强制获胜而不管第二玩家做什么。
+A winning configuration is an arrangement of coins on the strip where the first player can force a win no matter what the second player does.
-设W（n，c）为n个正方形，c个无价值硬币和1个银元的获胜配置数。
+Let W(n,c) be the number of winning configurations for a strip of n squares, c worthless coins and one silver dollar.
-给出W（10,2）= 324和W（100,10）= 1514704946113500。
+You are given that W(10,2) = 324 and W(100,10) = 1514704946113500.
-找到W（1 000 000,100）模数半数1000 036 000 099（= 1 000 003·1 000 033）。
+Find W(1 000 000, 100) modulo the semiprime 1000 036 000 099 (= 1 000 003 · 1 000 033).
# --hints--
-`euler344()`应该返回65579304332。
+`euler344()` should return 65579304332.
```js
assert.strictEqual(euler344(), 65579304332);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-345-matrix-sum.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-345-matrix-sum.md
index 91140dad1f2..9cb8157d07c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-345-matrix-sum.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-345-matrix-sum.md
@@ -1,24 +1,24 @@
---
id: 5900f4c81000cf542c50ffda
-title: 问题345：矩阵求和
+title: 'Problem 345: Matrix Sum'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302004
dashedName: problem-345-matrix-sum
---
# --description--
-我们将矩阵的矩阵和定义为矩阵元素的最大总和，每个元素是其行和列中唯一的元素。例如，下面矩阵的矩阵和等于3315（= 863 + 383 + 343 + 959 + 767）：
+We define the Matrix Sum of a matrix as the maximum sum of matrix elements with each element being the only one in his row and column. For example, the Matrix Sum of the matrix below equals 3315 ( = 863 + 383 + 343 + 959 + 767):
7 53 183 439 863 497 383 563 79 973 287 63 343 169 583 627 343 773 959 943767 473 103 699 303
-找到矩阵之和：
+Find the Matrix Sum of:
7 53 183 439 863 497 383 563 79 973 287 63 343 169 583 627 343 773 959 943 767 473 103 699 303 957 703 583 639 913 447 283 463 29 23 487 463 993 119 883 327 493 423 159 743 217 623 3 399 853 407 103 983 89 463 290 516 212 462 350 960 376 682 962 300 780 486 502 912 800 250 346 172 812 350 870 456 192 162 593 473 915 45 989 873 823 965 425 329 803 973 965 905 919 133 673 665 235 509 613 673 815 165 992 326 322 148 972 962 286 255 941 541 265 323 925 281 601 95 973 445 721 11 525 473 65 511 164 138 672 18 428 154 448 848 414 456 310 312 798 104 566 520 302 248 694 976 430 392 198 184 829 373 181 631 101 969 613 840 740 778 458 284 760 390 821 461 843 513 17 901 711 993 293 157 274 94 192 156 574 34 124 4 878 450 476 712 914 838 669 875 299 823 329 699 815 559 813 459 522 788 168 586 966 232 308 833 251 631 107 813 883 451 509 615 77 281 613 459 205 380 274 302 35 805
# --hints--
-`euler345()`应返回13938。
+`euler345()` should return 13938.
```js
assert.strictEqual(euler345(), 13938);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-346-strong-repunits.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-346-strong-repunits.md
index 4a6fe810843..49a1f6b2c25 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-346-strong-repunits.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-346-strong-repunits.md
@@ -1,22 +1,22 @@
---
id: 5900f4c71000cf542c50ffd8
-title: 问题346：强大的Repunits
+title: 'Problem 346: Strong Repunits'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302005
dashedName: problem-346-strong-repunits
---
# --description--
-数字7是特殊的，因为7是用基数2写的111，用基数6写的是11（即710 = 116 = 1112）。换句话说，7是至少两个碱基b> 1的重新排列。
+The number 7 is special, because 7 is 111 written in base 2, and 11 written in base 6 (i.e. 710 = 116 = 1112). In other words, 7 is a repunit in at least two bases b > 1.
-我们将调用一个正整数，该属性是一个强大的repunit。可以证实，在50以下有8个强大的重新命令：{1,7,13,15,21,31,40,43}。此外，1000以下的所有强大的重新命令的总和等于15864。
+We shall call a positive integer with this property a strong repunit. It can be verified that there are 8 strong repunits below 50: {1,7,13,15,21,31,40,43}. Furthermore, the sum of all strong repunits below 1000 equals 15864.
-找出1012以下所有强大的回购总和。
+Find the sum of all strong repunits below 1012.
# --hints--
-`euler346()`应该返回336108797689259260。
+`euler346()` should return 336108797689259260.
```js
assert.strictEqual(euler346(), 336108797689259260);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-347-largest-integer-divisible-by-two-primes.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-347-largest-integer-divisible-by-two-primes.md
index e4411461c0b..70845508fa3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-347-largest-integer-divisible-by-two-primes.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-347-largest-integer-divisible-by-two-primes.md
@@ -1,24 +1,28 @@
---
id: 5900f4c81000cf542c50ffd9
-title: 问题347：最大整数可被两个素数整除
+title: 'Problem 347: Largest integer divisible by two primes'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302006
dashedName: problem-347-largest-integer-divisible-by-two-primes
---
# --description--
-只能被素数2和3整除的最大整数≤100是96，因为96 = 32 \* 3 = 25 \* 3。对于两个不同的素数p和q，令M（p，q，N）是最大的正整数≤N，只能被p和q整除，并且如果不存在这样的正整数，则M（p，q，N）= 0。
+The largest integer ≤ 100 that is only divisible by both the primes 2 and 3 is 96, as 96=32\*3=25\*3.
-例如M（2,3,100）= 96。 M（3,5,100）= 75而不是90因为90可以被2,3和5整除。另外M（2,73,100）= 0因为不存在可以被2和73整除的正整数≤100。
+For two distinct primes p and q let M(p,q,N) be the largest positive integer ≤N only divisible
-设S（N）为所有不同M（p，q，N）的和。 S（100）= 2262。
+by both p and q and M(p,q,N)=0 if such a positive integer does not exist.
-找到S（10 000 000）。
+E.g. M(2,3,100)=96. M(3,5,100)=75 and not 90 because 90 is divisible by 2 ,3 and 5. Also M(2,73,100)=0 because there does not exist a positive integer ≤ 100 that is divisible by both 2 and 73.
+
+Let S(N) be the sum of all distinct M(p,q,N). S(100)=2262.
+
+Find S(10 000 000).
# --hints--
-`euler347()`应该返回11109800204052。
+`euler347()` should return 11109800204052.
```js
assert.strictEqual(euler347(), 11109800204052);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-348-sum-of-a-square-and-a-cube.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-348-sum-of-a-square-and-a-cube.md
index 8149eec422a..2fe6e2bc137 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-348-sum-of-a-square-and-a-cube.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-348-sum-of-a-square-and-a-cube.md
@@ -1,22 +1,22 @@
---
id: 5900f4c81000cf542c50ffdb
-title: 问题348：正方形和立方体的总和
+title: 'Problem 348: Sum of a square and a cube'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302007
dashedName: problem-348-sum-of-a-square-and-a-cube
---
# --description--
-许多数字可以表示为正方形和立方体的总和。其中一些不止一种。
+Many numbers can be expressed as the sum of a square and a cube. Some of them in more than one way.
-考虑可以用4种不同方式表示为正方形和立方体之和的回文数，均大于1。例如，5229225是一个回文数字，它可以用4种不同的方式表示：22852 + 203 22232 + 663 18102 + 1253 11972 + 1563
+Consider the palindromic numbers that can be expressed as the sum of a square and a cube, both greater than 1, in exactly 4 different ways. For example, 5229225 is a palindromic number and it can be expressed in exactly 4 different ways: 22852 + 203 22232 + 663 18102 + 1253 11972 + 1563
-找出五个最小的这种回文数的总和。
+Find the sum of the five smallest such palindromic numbers.
# --hints--
-`euler348()`应返回1004195061。
+`euler348()` should return 1004195061.
```js
assert.strictEqual(euler348(), 1004195061);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-349-langtons-ant.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-349-langtons-ant.md
index 3177fedac3f..e73c284fff6 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-349-langtons-ant.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-349-langtons-ant.md
@@ -1,20 +1,26 @@
---
id: 5900f4ca1000cf542c50ffdc
-title: 问题349：兰顿的蚂蚁
+title: 'Problem 349: Langton''s ant'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302008
dashedName: problem-349-langtons-ant
---
# --description--
-蚂蚁在规则的方格网格上移动，方格网格为黑色或白色。蚂蚁总是朝向一个主要方向（左，右，上或下），并根据以下规则从正方形移动到相邻正方形： - 如果它在黑色方块上，它将方形的颜色翻转为白色，逆时针旋转90度，向前移动一个方格。 - 如果它在白色方块上，它将方块的颜色翻转为黑色，顺时针旋转90度并向前移动一个方格。
+An ant moves on a regular grid of squares that are coloured either black or white.
-从一个完全是白色的网格开始，在蚂蚁1018次移动后，有多少个方块是黑色的？
+The ant is always oriented in one of the cardinal directions (left, right, up or down) and moves from square to adjacent square according to the following rules:
+
+\- if it is on a black square, it flips the color of the square to white, rotates 90 degrees counterclockwise and moves forward one square.
+
+\- if it is on a white square, it flips the color of the square to black, rotates 90 degrees clockwise and moves forward one square.
+
+Starting with a grid that is entirely white, how many squares are black after 1018 moves of the ant?
# --hints--
-`euler349()`应该返回115384615384614940。
+`euler349()` should return 115384615384614940.
```js
assert.strictEqual(euler349(), 115384615384614940);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-35-circular-primes.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-35-circular-primes.md
index 76593008c03..d3d74989d65 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-35-circular-primes.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-35-circular-primes.md
@@ -1,48 +1,62 @@
---
id: 5900f38f1000cf542c50fea2
-title: 问题35：循环素数
+title: 'Problem 35: Circular primes'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302009
dashedName: problem-35-circular-primes
---
# --description--
-这个数字197被称为循环素数，因为数字的所有旋转：197,971和719本身都是素数。在100：2,3,5,7,11,13,17,31,37,71,73,79和97之下有十三个这样的素数。在n下面有多少个圆形素数，而100 <= n < = 1000000？
+The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.
+
+There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
+
+How many circular primes are there below `n`, whereas 100 ≤ `n` ≤ 1000000?
+
+**Note:**
+
+Circular primes individual rotation can exceed `n`.
# --hints--
-`circularPrimes(100)`应该返回13。
+`circularPrimes(100)` should return a number.
+
+```js
+assert(typeof circularPrimes(100) === 'number');
+```
+
+`circularPrimes(100)` should return 13.
```js
assert(circularPrimes(100) == 13);
```
-`circularPrimes(100000)`应该返回43。
+`circularPrimes(100000)` should return 43.
```js
assert(circularPrimes(100000) == 43);
```
-`circularPrimes(250000)`应该返回45。
+`circularPrimes(250000)` should return 45.
```js
assert(circularPrimes(250000) == 45);
```
-`circularPrimes(500000)`应该返回49。
+`circularPrimes(500000)` should return 49.
```js
assert(circularPrimes(500000) == 49);
```
-`circularPrimes(750000)`应该返回49。
+`circularPrimes(750000)` should return 49.
```js
assert(circularPrimes(750000) == 49);
```
-`circularPrimes(1000000)`应该返回55。
+`circularPrimes(1000000)` should return 55.
```js
assert(circularPrimes(1000000) == 55);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-350-constraining-the-least-greatest-and-the-greatest-least.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-350-constraining-the-least-greatest-and-the-greatest-least.md
index ff4dabf6d98..ebc52998b58 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-350-constraining-the-least-greatest-and-the-greatest-least.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-350-constraining-the-least-greatest-and-the-greatest-least.md
@@ -1,28 +1,28 @@
---
id: 5900f4cb1000cf542c50ffdd
-title: 问题350：约束最小和最小
+title: 'Problem 350: Constraining the least greatest and the greatest least'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302010
dashedName: problem-350-constraining-the-least-greatest-and-the-greatest-least
---
# --description--
-大小为n的列表是n个自然数的序列。实例是（2,4,6），（2,6,4），（10,6,15,6）和（11）。
+A list of size n is a sequence of n natural numbers. Examples are (2,4,6), (2,6,4), (10,6,15,6), and (11).
-列表的最大公约数（或gcd）是划分列表中所有条目的最大自然数。例子：gcd（2,6,4）= 2，gcd（10,6,15,6）= 1，gcd（11）= 11。
+The greatest common divisor, or gcd, of a list is the largest natural number that divides all entries of the list. Examples: gcd(2,6,4) = 2, gcd(10,6,15,6) = 1 and gcd(11) = 11.
-列表的最小公倍数或lcm是列表的每个条目可分割的最小自然数。实例：1cm（2,6,4）= 12,1cm（10,6,15,6）= 30和1cm（11）= 11。
+The least common multiple, or lcm, of a list is the smallest natural number divisible by each entry of the list. Examples: lcm(2,6,4) = 12, lcm(10,6,15,6) = 30 and lcm(11) = 11.
-设f（G，L，N）为大小为N的列表数，gcd≥G且lcm≤L。例如：
+Let f(G, L, N) be the number of lists of size N with gcd ≥ G and lcm ≤ L. For example:
-f（10,100,1）= 91.f（10,100,2）= 327.f（10,100,3）= 1135.f（10,100,1000）mod 1014 = 3286053。
+f(10, 100, 1) = 91. f(10, 100, 2) = 327. f(10, 100, 3) = 1135. f(10, 100, 1000) mod 1014 = 3286053.
-找到f（106,1012,1018）mod 1014。
+Find f(106, 1012, 1018) mod 1014.
# --hints--
-`euler350()`应该返回84664213。
+`euler350()` should return 84664213.
```js
assert.strictEqual(euler350(), 84664213);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-351-hexagonal-orchards.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-351-hexagonal-orchards.md
index a3accced407..33f0089171f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-351-hexagonal-orchards.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-351-hexagonal-orchards.md
@@ -1,26 +1,26 @@
---
id: 5900f4cb1000cf542c50ffde
-title: 问题351：六角形果园
+title: 'Problem 351: Hexagonal orchards'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302011
dashedName: problem-351-hexagonal-orchards
---
# --description--
-n阶六角形果园是由正六边形中具有边n的点构成的三角形格子。以下是5阶六角形果园的示例：
+A hexagonal orchard of order n is a triangular lattice made up of points within a regular hexagon with side n. The following is an example of a hexagonal orchard of order 5:
-以绿色突出显示的是通过靠近它的点从中心隐藏的点。可以看出，对于5阶六角形果园，中心隐藏了30个点。
+Highlighted in green are the points which are hidden from the center by a point closer to it. It can be seen that for a hexagonal orchard of order 5, 30 points are hidden from the center.
-设H（n）是在n阶六角形果园中从中心隐藏的点数。
+Let H(n) be the number of points hidden from the center in a hexagonal orchard of order n.
-H（5）= 30.H（10）= 138.H（1 000）= 1177848。
+H(5) = 30. H(10) = 138. H(1 000) = 1177848.
-找H（100 000 000）。
+Find H(100 000 000).
# --hints--
-`euler351()`应该返回11762187201804552。
+`euler351()` should return 11762187201804552.
```js
assert.strictEqual(euler351(), 11762187201804552);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-352-blood-tests.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-352-blood-tests.md
index bf165c79584..76af2f075c8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-352-blood-tests.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-352-blood-tests.md
@@ -1,34 +1,36 @@
---
id: 5900f4cd1000cf542c50ffdf
-title: 问题352：验血
+title: 'Problem 352: Blood tests'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302012
dashedName: problem-352-blood-tests
---
# --description--
-必须对一群羊中的25只绵羊中的每一只进行稀有病毒检测，已知这种病毒会影响2％的绵羊群体。对血液样品存在准确且极其灵敏的PCR测试，产生明显的阳性/阴性结果，但是它非常耗时且昂贵。
+Each one of the 25 sheep in a flock must be tested for a rare virus, known to affect 2% of the sheep population.
-由于成本高，负责人建议不要进行25次单独的测试，而是可以使用以下程序：将绵羊分成5组，每组5只绵羊。对于每组，将5个样品混合在一起并进行单次测试。然后，如果结果是阴性，则该组中的所有绵羊都被认为是无病毒的。如果结果为阳性，将进行5次额外测试（对每只动物进行单独测试）以确定受影响的个体。
+An accurate and extremely sensitive PCR test exists for blood samples, producing a clear positive / negative result, but it is very time-consuming and expensive.
-由于任何特定动物的感染概率仅为0.02，因此每组的第一次测试（在合并的样本上）将为：阴性（并且不再需要测试），概率为0.985 = 0.9039207968。正（需要5次额外测试），概率为1 - 0.9039207968 = 0.0960792032。
+Because of the high cost, the vet-in-charge suggests that instead of performing 25 separate tests, the following procedure can be used instead: The sheep are split into 5 groups of 5 sheep in each group. For each group, the 5 samples are mixed together and a single test is performed. Then, If the result is negative, all the sheep in that group are deemed to be virus-free. If the result is positive, 5 additional tests will be performed (a separate test for each animal) to determine the affected individual(s).
-因此，每组的预期测试次数是1 + 0.0960792032×5 = 1.480396016。因此，所有5组都可以使用平均仅1.480396016×5 = 7.40198008的测试进行筛选，这意味着节省了超过70％的巨额费用！
+Since the probability of infection for any specific animal is only 0.02, the first test (on the pooled samples) for each group will be: Negative (and no more tests needed) with probability 0.985 = 0.9039207968. Positive (5 additional tests needed) with probability 1 - 0.9039207968 = 0.0960792032.
-尽管我们刚刚描述的方案似乎非常有效，但它仍然可以得到显着改善（总是假设测试足够敏感并且不会因混合不同样品而产生不利影响）。例如：我们可以首先对所有25个样本的混合物进行测试。可以证实，在大约60.35％的情况下，该测试将是否定的，因此不再需要进行测试。其余39.65％的案件只需要进一步测试。如果我们知道5个一组中至少有一只动物被感染并且前4个个体测试结果为阴性，则不需要对第五个动物进行测试（我们知道它必须被感染）。我们可以在每组中尝试不同数量的组/不同数量的动物，在每个级别调整这些数字，以便最小化测试的总预期数量。
+Thus, the expected number of tests for each group is 1 + 0.0960792032 × 5 = 1.480396016. Consequently, all 5 groups can be screened using an average of only 1.480396016 × 5 = 7.40198008 tests, which represents a huge saving of more than 70% !
-为了简化各种各样的可能性，我们在设计最具成本效益的测试方案时有一个限制：每当我们从混合样本开始时，必须对所有对该样本做出贡献的绵羊进行全面筛选（即对感染的判断）在我们开始检查任何其他动物之前，必须达到/无病毒。
+Although the scheme we have just described seems to be very efficient, it can still be improved considerably (always assuming that the test is sufficiently sensitive and that there are no adverse effects caused by mixing different samples). E.g.: We may start by running a test on a mixture of all the 25 samples. It can be verified that in about 60.35% of the cases this test will be negative, thus no more tests will be needed. Further testing will only be required for the remaining 39.65% of the cases. If we know that at least one animal in a group of 5 is infected and the first 4 individual tests come out negative, there is no need to run a test on the fifth animal (we know that it must be infected). We can try a different number of groups / different number of animals in each group, adjusting those numbers at each level so that the total expected number of tests will be minimised.
-对于当前的例子，事实证明，最具成本效益的测试方案（我们称之为最优策略）需要平均只需4.155452次测试！
+To simplify the very wide range of possibilities, there is one restriction we place when devising the most cost-efficient testing scheme: whenever we start with a mixed sample, all the sheep contributing to that sample must be fully screened (i.e. a verdict of infected / virus-free must be reached for all of them) before we start examining any other animals.
-使用最优策略，让T（s，p）代表筛选一群绵羊所需的平均测试次数，以获得任何个体中存在概率p的病毒。因此，四舍五入到小数点后六位，T（25,0.02）= 4.155452，T（25,0.10）= 12.702124。
+For the current example, it turns out that the most cost-efficient testing scheme (we'll call it the optimal strategy) requires an average of just 4.155452 tests!
-找到ΣT（10000，p），p = 0.01,0.02,0.03，...... 0.50。将您的答案四舍五入到小数点后六位。
+Using the optimal strategy, let T(s,p) represent the average number of tests needed to screen a flock of s sheep for a virus having probability p to be present in any individual. Thus, rounded to six decimal places, T(25, 0.02) = 4.155452 and T(25, 0.10) = 12.702124.
+
+Find ΣT(10000, p) for p=0.01, 0.02, 0.03, ... 0.50. Give your answer rounded to six decimal places.
# --hints--
-`euler352()`应该返回378563.260589。
+`euler352()` should return 378563.260589.
```js
assert.strictEqual(euler352(), 378563.260589);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-353-risky-moon.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-353-risky-moon.md
index d161381d803..74faf24c26f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-353-risky-moon.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-353-risky-moon.md
@@ -1,32 +1,32 @@
---
id: 5900f4cd1000cf542c50ffe0
-title: 问题353：危险的月亮
+title: 'Problem 353: Risky moon'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302013
dashedName: problem-353-risky-moon
---
# --description--
-月球可以由球体C（r）描述，其中心（0,0,0）和半径r。
+A moon could be described by the sphere C(r) with centre (0,0,0) and radius r.
-月球上有C（r）表面点上的整数坐标。 （0,0，r）处的站称为北极站，（0,0，-r）处的站称为南极站。
+There are stations on the moon at the points on the surface of C(r) with integer coordinates. The station at (0,0,r) is called North Pole station, the station at (0,0,-r) is called South Pole station.
-所有车站都通过车站的大弧上最短的道路相互连接。两个站之间的旅程是有风险的。如果d是两个站之间的道路长度，（d /（πr））2是旅程风险的度量（让我们称之为道路风险）。如果旅程包括两个以上的车站，则旅程的风险是使用过的道路的风险总和。
+All stations are connected with each other via the shortest road on the great arc through the stations. A journey between two stations is risky. If d is the length of the road between two stations, (d/(π r))2 is a measure for the risk of the journey (let us call it the risk of the road). If the journey includes more than two stations, the risk of the journey is the sum of risks of the used roads.
-从北极站到南极站的直接旅程长度为πr，风险为1.从北极站到南极站的路程（0，r，0）长度相同，但风险较小：（½πr/（πr））2+（1 /2πr/（πr））2 = 0.5。
+A direct journey from the North Pole station to the South Pole station has the length πr and risk 1. The journey from the North Pole station to the South Pole station via (0,r,0) has the same length, but a smaller risk: (½πr/(πr))2+(½πr/(πr))2=0.5.
-从C（r）的北极站到南极站的旅程的最小风险是M（r）。
+The minimal risk of a journey from the North Pole station to the South Pole station on C(r) is M(r).
-给出M（7）= 0.1784943998四舍五入到小数点后面的10位数。
+You are given that M(7)=0.1784943998 rounded to 10 digits behind the decimal point.
-找到ΣM（2n-1）为1≤n≤15。
+Find ∑M(2n-1) for 1≤n≤15.
-将答案四舍五入到小数点后面的10位数，形式为a.bcdefghijk。
+Give your answer rounded to 10 digits behind the decimal point in the form a.bcdefghijk.
# --hints--
-`euler353()`应返回1.2759860331。
+`euler353()` should return 1.2759860331.
```js
assert.strictEqual(euler353(), 1.2759860331);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-354-distances-in-a-bees-honeycomb.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-354-distances-in-a-bees-honeycomb.md
index 0ff44d02bfe..346a1af4515 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-354-distances-in-a-bees-honeycomb.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-354-distances-in-a-bees-honeycomb.md
@@ -1,22 +1,22 @@
---
id: 5900f4cf1000cf542c50ffe1
-title: 问题354：蜜蜂蜂窝中的距离
+title: 'Problem 354: Distances in a bee''s honeycomb'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302014
dashedName: problem-354-distances-in-a-bees-honeycomb
---
# --description--
-考虑一个蜜蜂的蜂窝，每个细胞是一个完整的正六边形，边长为1。
+Consider a honey bee's honeycomb where each cell is a perfect regular hexagon with side length 1.
-一只特定的细胞被蜂王占据。对于正实数L，让B（L）计算距蜂王细胞距离为L的细胞（所有距离从中心到中心测量）;你可以假设蜂窝大到足以容纳我们想要考虑的任何距离。例如，B（√3）= 6，B（√21）= 12，B（111 111 111）= 54。
+One particular cell is occupied by the queen bee. For a positive real number L, let B(L) count the cells with distance L from the queen bee cell (all distances are measured from centre to centre); you may assume that the honeycomb is large enough to accommodate for any distance we wish to consider. For example, B(√3) = 6, B(√21) = 12 and B(111 111 111) = 54.
-求出L≤5·1011的数，使得B（L）= 450。
+Find the number of L ≤ 5·1011 such that B(L) = 450.
# --hints--
-`euler354()`应该返回58065134。
+`euler354()` should return 58065134.
```js
assert.strictEqual(euler354(), 58065134);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-355-maximal-coprime-subset.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-355-maximal-coprime-subset.md
index ef083a2ba49..a166a1057ac 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-355-maximal-coprime-subset.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-355-maximal-coprime-subset.md
@@ -1,22 +1,22 @@
---
id: 5900f4d01000cf542c50ffe2
-title: 问题355：最大互质子集
+title: 'Problem 355: Maximal coprime subset'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302015
dashedName: problem-355-maximal-coprime-subset
---
# --description--
-将Co（n）定义为来自{1,2，...，n}的一组相互共元素的最大可能和。例如，Co（10）是30并且在子集{1,5,7,8,9}上达到最大值。
+Define Co(n) to be the maximal possible sum of a set of mutually co-prime elements from {1, 2, ..., n}. For example Co(10) is 30 and hits that maximum on the subset {1, 5, 7, 8, 9}.
-给出Co（30）= 193和Co（100）= 1356。
+You are given that Co(30) = 193 and Co(100) = 1356.
-找到Co（200000）。
+Find Co(200000).
# --hints--
-`euler355()`应返回1726545007。
+`euler355()` should return 1726545007.
```js
assert.strictEqual(euler355(), 1726545007);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-356-largest-roots-of-cubic-polynomials.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-356-largest-roots-of-cubic-polynomials.md
index 4c482f1a0de..9027dcecf96 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-356-largest-roots-of-cubic-polynomials.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-356-largest-roots-of-cubic-polynomials.md
@@ -1,22 +1,24 @@
---
id: 5900f4d01000cf542c50ffe3
-title: 问题356：三次多项式的最大根
+title: 'Problem 356: Largest roots of cubic polynomials'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302016
dashedName: problem-356-largest-roots-of-cubic-polynomials
---
# --description--
-设a是多项式的最大实根g（x）= x3 - 2n·x2 + n。例如，a2 = 3.86619826 ......
+Let an be the largest real root of a polynomial g(x) = x3 - 2n·x2 + n.
-找到最后八位数字。
+For example, a2 = 3.86619826...
-注意：代表楼层功能。
+Find the last eight digits of.
+
+Note: represents the floor function.
# --hints--
-`euler356()`应该返回28010159。
+`euler356()` should return 28010159.
```js
assert.strictEqual(euler356(), 28010159);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-357-prime-generating-integers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-357-prime-generating-integers.md
index dec2b9ae1e9..1bb3f094a21 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-357-prime-generating-integers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-357-prime-generating-integers.md
@@ -1,20 +1,22 @@
---
id: 5900f4d11000cf542c50ffe4
-title: 问题357：Prime生成整数
+title: 'Problem 357: Prime generating integers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302017
dashedName: problem-357-prime-generating-integers
---
# --description--
-考虑30：1,2,3,5,6,10,15,30的除数。可以看出，对于每个除数d为30，d + 30 / d是素数。
+Consider the divisors of 30: 1,2,3,5,6,10,15,30.
-找出所有正整数n的总和不超过100 000 000，这样对于n的每个除数d，d + n / d是素数。
+It can be seen that for every divisor d of 30, d+30/d is prime.
+
+Find the sum of all positive integers n not exceeding 100 000 000such that for every divisor d of n, d+n/d is prime.
# --hints--
-`euler357()`应该返回1739023853137。
+`euler357()` should return 1739023853137.
```js
assert.strictEqual(euler357(), 1739023853137);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-358-cyclic-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-358-cyclic-numbers.md
index ddf0e2c635e..fb08a667a8b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-358-cyclic-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-358-cyclic-numbers.md
@@ -1,28 +1,28 @@
---
id: 5900f4d21000cf542c50ffe5
-title: 问题358：循环数
+title: 'Problem 358: Cyclic numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302018
dashedName: problem-358-cyclic-numbers
---
# --description--
-具有n位数字的循环数具有一个非常有趣的属性：
+A cyclic number with n digits has a very interesting property:
-当乘以1、2、3、4，... n时，所有乘积都具有完全相同的数字，并且顺序相同，但是以循环方式旋转！
+When it is multiplied by 1, 2, 3, 4, ... n, all the products have exactly the same digits, in the same order, but rotated in a circular fashion!
-最小的循环数是6位数字142857： 142857×1 = 142857 142857×2 = 285714 142857×3 = 428571 142857×4 = 571428 142857×5 = 714285 142857×6 = 857142
+The smallest cyclic number is the 6-digit number 142857 : 142857 × 1 = 142857 142857 × 2 = 285714 142857 × 3 = 428571 142857 × 4 = 571428 142857 × 5 = 714285 142857 × 6 = 857142
-下一个循环号是16位数字0588235294117647： 0588235294117647×1 = 0588235294117647 0588235294117647×2 = 1176470588235294 0588235294117647×3 = 1764705882352941 ... 0588235294117647×16 = 9411764705882352
+The next cyclic number is 0588235294117647 with 16 digits : 0588235294117647 × 1 = 0588235294117647 0588235294117647 × 2 = 1176470588235294 0588235294117647 × 3 = 1764705882352941 ... 0588235294117647 × 16 = 9411764705882352
-请注意，对于循环数，前导零很重要。
+Note that for cyclic numbers, leading zeros are important.
-只有一个循环数，其最左边的11个数字为00000000137，最右边的5个数字为56789（即其形式为00000000137 ... 56789，中间是未知数字）。 查找所有数字的总和。
+There is only one cyclic number for which, the eleven leftmost digits are 00000000137 and the five rightmost digits are 56789 (i.e., it has the form 00000000137...56789 with an unknown number of digits in the middle). Find the sum of all its digits.
# --hints--
-`euler358()`应返回3284144505。
+`euler358()` should return 3284144505.
```js
assert.strictEqual(euler358(), 3284144505);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-359-hilberts-new-hotel.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-359-hilberts-new-hotel.md
index 1e1dc8c3951..971ae0f4012 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-359-hilberts-new-hotel.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-359-hilberts-new-hotel.md
@@ -1,28 +1,28 @@
---
id: 5900f4d31000cf542c50ffe6
-title: 问题359：希尔伯特的新酒店
+title: 'Problem 359: Hilbert''s New Hotel'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302019
dashedName: problem-359-hilberts-new-hotel
---
# --description--
-无数人（编号1,2,3等）排成一列，在希尔伯特最新的无限酒店找到一个房间。酒店包含无限数量的楼层（编号为1,2,3等），每层楼包含无限数量的房间（编号为1,2,3等）。
+An infinite number of people (numbered 1, 2, 3, etc.) are lined up to get a room at Hilbert's newest infinite hotel. The hotel contains an infinite number of floors (numbered 1, 2, 3, etc.), and each floor contains an infinite number of rooms (numbered 1, 2, 3, etc.).
-最初酒店是空的。希尔伯特宣布了关于如何为第n人分配房间的规则：人员在最低编号的楼层中获得满足以下任一项的第一个空置房间：楼层是空的，楼层不是空的，以及最新的人是否占用房间在那个楼层是人m，那么m + n是一个完美的正方形
+Initially the hotel is empty. Hilbert declares a rule on how the nth person is assigned a room: person n gets the first vacant room in the lowest numbered floor satisfying either of the following: the floor is empty the floor is not empty, and if the latest person taking a room in that floor is person m, then m + n is a perfect square
-由于1楼空了，1号人员在1楼获得1号房间。由于1 + 2 = 3不是完美的正方形，因此人2在1楼没有得到2号房间。由于2楼是空的，因此人2在2楼获得1号房间。人3在1楼获得2号房，因为1 + 3 = 4是一个完美的广场。
+Person 1 gets room 1 in floor 1 since floor 1 is empty. Person 2 does not get room 2 in floor 1 since 1 + 2 = 3 is not a perfect square. Person 2 instead gets room 1 in floor 2 since floor 2 is empty. Person 3 gets room 2 in floor 1 since 1 + 3 = 4 is a perfect square.
-最终，该线路中的每个人都在酒店获得一个房间。
+Eventually, every person in the line gets a room in the hotel.
-如果人n占据楼层f中的房间r，则将P（f，r）定义为n，如果没有人占用房间，则定义为0。以下是一些例子：P（1,1）= 1 P（1,2）= 3 P（2,1）= 2 P（10,20）= 440 P（25,75）= 4863 P（99， 100）= 19454
+Define P(f, r) to be n if person n occupies room r in floor f, and 0 if no person occupies the room. Here are a few examples: P(1, 1) = 1 P(1, 2) = 3 P(2, 1) = 2 P(10, 20) = 440 P(25, 75) = 4863 P(99, 100) = 19454
-找到所有正f和r的所有P（f，r）之和，使得f×r = 71328803586048，并将最后8位数作为答案。
+Find the sum of all P(f, r) for all positive f and r such that f × r = 71328803586048 and give the last 8 digits as your answer.
# --hints--
-`euler359()`应该返回40632119。
+`euler359()` should return 40632119.
```js
assert.strictEqual(euler359(), 40632119);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-36-double-base-palindromes.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-36-double-base-palindromes.md
index eb938f1e49c..359211b5301 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-36-double-base-palindromes.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-36-double-base-palindromes.md
@@ -1,36 +1,46 @@
---
id: 5900f3901000cf542c50fea3
-title: 问题36：双基回文
+title: 'Problem 36: Double-base palindromes'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302020
dashedName: problem-36-double-base-palindromes
---
# --description--
-十进制数，585 = 10010010012（二进制），在两个碱基中都是回文。找到所有数字的总和，小于n，而1000 <= n <= 1000000，它们在基数10和基数2中是回文的。（请注意，任一基数中的回文数可能不包括前导零。）
+The decimal number, 585 = 1001001001<sub>2</sub> (binary), is palindromic in both bases.
+
+Find the sum of all numbers, less than `n`, whereas 1000 ≤ `n` ≤ 1000000, which are palindromic in base 10 and base 2.
+
+(Please note that the palindromic number, in either base, may not include leading zeros.)
# --hints--
-`doubleBasePalindromes(1000)`应该返回1772。
+`doubleBasePalindromes(1000)` should return a number.
+
+```js
+assert(typeof doubleBasePalindromes(1000) === 'number');
+```
+
+`doubleBasePalindromes(1000)` should return 1772.
```js
assert(doubleBasePalindromes(1000) == 1772);
```
-`doubleBasePalindromes(50000)`应该返回105795。
+`doubleBasePalindromes(50000)` should return 105795.
```js
assert(doubleBasePalindromes(50000) == 105795);
```
-`doubleBasePalindromes(500000)`应该返回286602。
+`doubleBasePalindromes(500000)` should return 286602.
```js
assert(doubleBasePalindromes(500000) == 286602);
```
-`doubleBasePalindromes(1000000)`应该返回872187。
+`doubleBasePalindromes(1000000)` should return 872187.
```js
assert(doubleBasePalindromes(1000000) == 872187);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-360-scary-sphere.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-360-scary-sphere.md
index 2c345bfc1d7..f028a5d58be 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-360-scary-sphere.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-360-scary-sphere.md
@@ -1,24 +1,24 @@
---
id: 5900f4d41000cf542c50ffe7
-title: 问题360：可怕的球体
+title: 'Problem 360: Scary Sphere'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302021
dashedName: problem-360-scary-sphere
---
# --description--
-给定三维空间中的两个点（x1，y1，z1）和（x2，y2，z2），这些点之间的曼哈顿距离定义为| x1-x2 | + | y1-y2 | + | z1-z2 |。
+Given two points (x1,y1,z1) and (x2,y2,z2) in three dimensional space, the Manhattan distance between those points is defined as |x1-x2|+|y1-y2|+|z1-z2|.
-令C（r）为半径为r且在原点O（0,0,0）中心的球体。设I（r）是C（r）表面上具有整数坐标的所有点的集合。设S（r）是I（r）的所有元素与原点O的曼哈顿距离之和。
+Let C(r) be a sphere with radius r and center in the origin O(0,0,0). Let I(r) be the set of all points with integer coordinates on the surface of C(r). Let S(r) be the sum of the Manhattan distances of all elements of I(r) to the origin O.
-例如S（45）= 34518。
+E.g. S(45)=34518.
-找到S（1010）。
+Find S(1010).
# --hints--
-`euler360()`应该返回878825614395267100。
+`euler360()` should return 878825614395267100.
```js
assert.strictEqual(euler360(), 878825614395267100);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-361-subsequence-of-thue-morse-sequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-361-subsequence-of-thue-morse-sequence.md
index 863aeb63c1f..bc131597891 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-361-subsequence-of-thue-morse-sequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-361-subsequence-of-thue-morse-sequence.md
@@ -1,28 +1,34 @@
---
id: 5900f4d51000cf542c50ffe8
-title: 问题361：Thue-Morse序列的子序列
+title: 'Problem 361: Subsequence of Thue-Morse sequence'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302022
dashedName: problem-361-subsequence-of-thue-morse-sequence
---
# --description--
-Thue-Morse序列{Tn}是满足以下条件的二进制序列：T0 = 0T2n = Tn T2n + 1 = 1-Tn
+The Thue-Morse sequence {Tn} is a binary sequence satisfying:
-{Tn}的前几个术语如下：01101001100101101001011001101001 ....
+T0 = 0
-我们将{An}定义为整数的排序序列，使得每个元素的二进制表达式在{Tn}中显示为子序列。例如，十进制数18以二进制表示为10010。 10010出现在{Tn}（T8到T12）中，因此18是{An}的元素。十进制数14以二进制表示为1110。 1110永远不会出现在{Tn}中，因此14不是{An}的元素。
+T2n = Tn
-An的前几个术语如下：n0123456789101112 ... An012345691011121318 ...
+T2n+1 = 1 - Tn
-我们还可以验证A100 = 3251和A1000 = 80852364498。
+The first several terms of {Tn} are given as follows: 01101001100101101001011001101001....
-找到最后9位数。
+We define {An} as the sorted sequence of integers such that the binary expression of each element appears as a subsequence in {Tn}. For example, the decimal number 18 is expressed as 10010 in binary. 10010 appears in {Tn} (T8 to T12), so 18 is an element of {An}. The decimal number 14 is expressed as 1110 in binary. 1110 never appears in {Tn}, so 14 is not an element of {An}.
+
+The first several terms of An are given as follows: n0123456789101112…An012345691011121318…
+
+We can also verify that A100 = 3251 and A1000 = 80852364498.
+
+Find the last 9 digits of .
# --hints--
-`euler361()`应该返回178476944。
+`euler361()` should return 178476944.
```js
assert.strictEqual(euler361(), 178476944);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-362-squarefree-factors.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-362-squarefree-factors.md
index d49f1093405..4d6bc3f4a97 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-362-squarefree-factors.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-362-squarefree-factors.md
@@ -1,26 +1,32 @@
---
id: 5900f4d61000cf542c50ffe9
-title: 问题362：无广义因子
+title: 'Problem 362: Squarefree factors'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302023
dashedName: problem-362-squarefree-factors
---
# --description--
-考虑数字54. 54可以用7种不同的方式分解为大于1：54,2×27,3×18,6×9,3×3×6,2×3×9和2×的一个或多个因子3×3×3。如果我们要求所有因子都是无平方的，则只剩下两种方式：3×3×6和2×3×3×3。
+Consider the number 54.
-让我们调用Fsf（n）n可以计算为一个或多个大于1的无平方因子的方式的数量，因此Fsf（54）= 2。
+54 can be factored in 7 distinct ways into one or more factors larger than 1:
-对于k = 2到n，令S（n）为ΣFsf（k）。
+54, 2×27, 3×18, 6×9, 3×3×6, 2×3×9 and 2×3×3×3.
-S（100）= 193。
+If we require that the factors are all squarefree only two ways remain: 3×3×6 and 2×3×3×3.
-找到S（10 000 000 000）。
+Let's call Fsf(n) the number of ways n can be factored into one or more squarefree factors larger than 1, so Fsf(54)=2.
+
+Let S(n) be ∑Fsf(k) for k=2 to n.
+
+S(100)=193.
+
+Find S(10 000 000 000).
# --hints--
-`euler362()`应该返回457895958010。
+`euler362()` should return 457895958010.
```js
assert.strictEqual(euler362(), 457895958010);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-363-bzier-curves.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-363-bzier-curves.md
index 1bcd30fac49..2fde41d1eb0 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-363-bzier-curves.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-363-bzier-curves.md
@@ -1,28 +1,28 @@
---
id: 5900f4d91000cf542c50ffeb
-title: 问题363：Bézier曲线
+title: 'Problem 363: Bézier Curves'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302024
dashedName: problem-363-bzier-curves
---
# --description--
-立方贝塞尔曲线由四个点定义：P0，P1，P2和P3。
+A cubic Bézier curve is defined by four points: P0, P1, P2 and P3.
-曲线构造如下：在段P0P1，P1P2和P2P3上绘制点Q0，Q1和Q2，使得P0Q0 / P0P1 = P1Q1 / P1P2 = P2Q2 / P2P3 = t（\[0,1]中的t）。在段Q0Q1和Q1Q2上绘制点R0和R1，使得对于相同的t值，Q0R0 / Q0Q1 = Q1R1 / Q1Q2 = t。在段R0R1上绘制点B，使得对于相同的t值，R0B / R0R1 = t。由点P0，P1，P2，P3定义的贝塞尔曲线是B的轨迹，因为Q0占据了段P0P1上的所有可能位置。 （请注意，对于所有点，t的值都相同。）
+The curve is constructed as follows: On the segments P0P1, P1P2 and P2P3 the points Q0,Q1 and Q2 are drawn such that P0Q0 / P0P1 = P1Q1 / P1P2 = P2Q2 / P2P3 = t (t in \[0,1]). On the segments Q0Q1 and Q1Q2 the points R0 and R1 are drawn such that Q0R0 / Q0Q1 = Q1R1 / Q1Q2 = t for the same value of t. On the segment R0R1 the point B is drawn such that R0B / R0R1 = t for the same value of t. The Bézier curve defined by the points P0, P1, P2, P3 is the locus of B as Q0 takes all possible positions on the segment P0P1. (Please note that for all points the value of t is the same.)
-在此（外部）Web地址，您将找到一个小程序，它允许您拖动点P0，P1，P2和P3，以查看这些点定义的Bézier曲线（绿色曲线）是什么样的。您也可以沿着段P0P1拖动点Q0。
+At this (external) web address you will find an applet that allows you to drag the points P0, P1, P2 and P3 to see what the Bézier curve (green curve) defined by those points looks like. You can also drag the point Q0 along the segment P0P1.
-从构造中可以清楚地看出，Bézier曲线将与P0中的P0P1和P3中的P2P3相切。
+From the construction it is clear that the Bézier curve will be tangent to the segments P0P1 in P0 and P2P3 in P3.
-使用P0 =（1,0），P1 =（1，v），P2 =（v，1）和P3 =（0,1）的三次Bézier曲线来近似四分之一圆。选择值v> 0，使得由线OP0，OP3和曲线包围的区域等于π/ 4（四分之一圆的面积）。
+A cubic Bézier curve with P0=(1,0), P1=(1,v), P2=(v,1) and P3=(0,1) is used to approximate a quarter circle. The value v > 0 is chosen such that the area enclosed by the lines OP0, OP3 and the curve is equal to π/4 (the area of the quarter circle).
-曲线长度与四分之一圆的长度有多少百分比？也就是说，如果L是曲线的长度，则计算100×L - π/2π/ 2给你的答案四舍五入到小数点后面的10位数。
+By how many percent does the length of the curve differ from the length of the quarter circle? That is, if L is the length of the curve, calculate 100 × L − π/2π/2Give your answer rounded to 10 digits behind the decimal point.
# --hints--
-`euler363()`应返回0.0000372091。
+`euler363()` should return 0.0000372091.
```js
assert.strictEqual(euler363(), 0.0000372091);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-364-comfortable-distance.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-364-comfortable-distance.md
index 3f6d746c4f8..84637798b87 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-364-comfortable-distance.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-364-comfortable-distance.md
@@ -1,22 +1,28 @@
---
id: 5900f4d91000cf542c50ffea
-title: 问题364：舒适的距离
+title: 'Problem 364: Comfortable distance'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302025
dashedName: problem-364-comfortable-distance
---
# --description--
-连续N个座位。根据以下规则，N人互相填充座位：如果有任何座位，相邻的座位没有被占用，请使用这样的座位。如果没有这样的座位并且有任何座位仅占用一个相邻的座位，则使用这样的座位。否则请选择剩余的可用座位之一。
+There are N seats in a row. N people come after each other to fill the seats according to the following rules:
-设T（N）是具有给定规则的N个人占用N个座位的可能性的数量。下图显示T（4）= 8。
+If there is any seat whose adjacent seat(s) are not occupied take such a seat.
-我们可以验证T（10）= 61632和T（1 000）mod 100 000 007 = 47255094.求T（1 000 000）mod 100 000 007。
+If there is no such seat and there is any seat for which only one adjacent seat is occupied take such a seat.
+
+Otherwise take one of the remaining available seats.
+
+Let T(N) be the number of possibilities that N seats are occupied by N people with the given rules. The following figure shows T(4)=8.
+
+We can verify that T(10) = 61632 and T(1 000) mod 100 000 007 = 47255094. Find T(1 000 000) mod 100 000 007.
# --hints--
-`euler364()`应返回44855254。
+`euler364()` should return 44855254.
```js
assert.strictEqual(euler364(), 44855254);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-365-a-huge-binomial-coefficient.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-365-a-huge-binomial-coefficient.md
index e5b0e44248b..1647ff46ea4 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-365-a-huge-binomial-coefficient.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-365-a-huge-binomial-coefficient.md
@@ -1,22 +1,22 @@
---
id: 5900f4da1000cf542c50ffec
-title: 问题365：巨大的二项式系数
+title: 'Problem 365: A huge binomial coefficient'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302026
dashedName: problem-365-a-huge-binomial-coefficient
---
# --description--
-二项式系数C（1018,109）是具有超过90亿（9×109）个数字的数字。
+The binomial coefficient C(1018,109) is a number with more than 9 billion (9×109) digits.
-令M（n，k，m）表示二项式系数C（n，k）模m。
+Let M(n,k,m) denote the binomial coefficient C(n,k) modulo m.
-计算ΣM（1018,109，p *q* r）为1000 <p <q <r <5000和p，q，r prime。
+Calculate ∑M(1018,109,p*q*r) for 1000<p<q<r<5000 and p,q,r prime.
# --hints--
-`euler365()`应该返回162619462356610300。
+`euler365()` should return 162619462356610300.
```js
assert.strictEqual(euler365(), 162619462356610300);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-366-stone-game-iii.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-366-stone-game-iii.md
index 489f52911ae..b7f832ebf4b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-366-stone-game-iii.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-366-stone-game-iii.md
@@ -1,26 +1,34 @@
---
id: 5900f4da1000cf542c50ffed
-title: 问题366：石头游戏III
+title: 'Problem 366: Stone Game III'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302027
dashedName: problem-366-stone-game-iii
---
# --description--
-安东和伯恩哈德两位球员正在参加以下比赛。有一堆n块石头。第一个玩家可以移除任何正数的石头，但不是整个堆。此后，每个玩家可以移除最多两倍于他的对手在之前移动时所用的石头数量。移除最后一块石头的玩家获胜。
+Two players, Anton and Bernhard, are playing the following game.
-例如，n = 5如果第一个玩家获得的不仅仅是一块石头，那么下一个玩家将能够获得所有剩余的石头。如果第一个玩家拿走一块石头，留下四块石头，他的对手也将拿走一块石头，留下三块石头。第一个玩家不能拿全部三个，因为他最多可以拿2x1 = 2个宝石。所以让我们说他也拿走了一块石头，留下了2.第二名球员可以拿下剩下的两块石头并获胜。所以5对于第一个玩家来说是一个失败的位置。对于一些获胜位置，第一个玩家可能有多个可能的移动。例如，当n = 17时，第一个玩家可以移除一个或四个宝石。
+There is one pile of n stones.
-设M（n）是第一个玩家在第一个回合的胜利位置可以获得的最大结石数，M（n）= 0表示任何其他位置。
+The first player may remove any positive number of stones, but not the whole pile.
-n≤100的ΣM（n）为728。
+Thereafter, each player may remove at most twice the number of stones his opponent took on the previous move.
-找到ΣM（n）为n≤1018。给你的答案模数108。
+The player who removes the last stone wins.
+
+E.g. n=5 If the first player takes anything more than one stone the next player will be able to take all remaining stones. If the first player takes one stone, leaving four, his opponent will take also one stone, leaving three stones. The first player cannot take all three because he may take at most 2x1=2 stones. So let's say he takes also one stone, leaving 2. The second player can take the two remaining stones and wins. So 5 is a losing position for the first player. For some winning positions there is more than one possible move for the first player. E.g. when n=17 the first player can remove one or four stones.
+
+Let M(n) be the maximum number of stones the first player can take from a winning position at his first turn and M(n)=0 for any other position.
+
+∑M(n) for n≤100 is 728.
+
+Find ∑M(n) for n≤1018. Give your answer modulo 108.
# --hints--
-`euler366()`应该返回88351299。
+`euler366()` should return 88351299.
```js
assert.strictEqual(euler366(), 88351299);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-367-bozo-sort.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-367-bozo-sort.md
index f7214f6d5b3..979702592a9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-367-bozo-sort.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-367-bozo-sort.md
@@ -1,24 +1,24 @@
---
id: 5900f4db1000cf542c50ffee
-title: 问题367：Bozo排序
+title: 'Problem 367: Bozo sort'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302028
dashedName: problem-367-bozo-sort
---
# --description--
-Bozo排序，不要与稍微低效的bogo排序混淆，包括检查输入序列是否已排序，以及是否不随机交换两个元素。重复此过程，直到最终对序列进行排序。
+Bozo sort, not to be confused with the slightly less efficient bogo sort, consists out of checking if the input sequence is sorted and if not swapping randomly two elements. This is repeated until eventually the sequence is sorted.
-如果我们考虑前4个自然数的所有排列作为输入，则交换数的期望值，平均超过所有4个！输入序列是24.75。已排序的序列需要0步。
+If we consider all permutations of the first 4 natural numbers as input the expectation value of the number of swaps, averaged over all 4! input sequences is 24.75. The already sorted sequence takes 0 steps.
-在这个问题中，我们考虑以下bozo排序变体。如果序列不按顺序，我们随机选择三个元素并随机地随机抽取这三个元素。这三个元素的所有3！= 6个排列同样可能。已经排序的序列将采取0步。如果我们考虑前4个自然数的所有排列作为输入，则洗牌次数的期望值，平均超过所有4个！输入序列是27.5。将输入序列视为前11个自然数的排列。全部11个人平均！输入序列，此排序算法将执行的预期洗牌次数是多少？
+In this problem we consider the following variant on bozo sort. If the sequence is not in order we pick three elements at random and shuffle these three elements randomly. All 3!=6 permutations of those three elements are equally likely. The already sorted sequence will take 0 steps. If we consider all permutations of the first 4 natural numbers as input the expectation value of the number of shuffles, averaged over all 4! input sequences is 27.5. Consider as input sequences the permutations of the first 11 natural numbers. Averaged over all 11! input sequences, what is the expected number of shuffles this sorting algorithm will perform?
-将答案四舍五入到最接近的整数。
+Give your answer rounded to the nearest integer.
# --hints--
-`euler367()`应该返回48271207。
+`euler367()` should return 48271207.
```js
assert.strictEqual(euler367(), 48271207);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-368-a-kempner-like-series.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-368-a-kempner-like-series.md
index e34eb88d693..dc9cf2f919b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-368-a-kempner-like-series.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-368-a-kempner-like-series.md
@@ -1,26 +1,26 @@
---
id: 5900f4dd1000cf542c50ffef
-title: 问题368：类似肯珀纳的系列
+title: 'Problem 368: A Kempner-like series'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302029
dashedName: problem-368-a-kempner-like-series
---
# --description--
-谐波系列$ 1 + \\ dfrac {1} {2} + \\ dfrac {1} {3} + \\ dfrac {1} {4} + ...众所周知是不同的。
+The harmonic series $1 + \\dfrac{1}{2} + \\dfrac{1}{3} + \\dfrac{1}{4} + ...$ is well known to be divergent.
-然而，如果我们在这个系列中省略了分母中有9个的每个项，则该系列显着地收敛到大约22.9206766193。这种改进的谐波系列称为肯普纳系列。
+If we however omit from this series every term where the denominator has a 9 in it, the series remarkably enough converges to approximately 22.9206766193. This modified harmonic series is called the Kempner series.
-现在让我们通过省略分母具有3个或更多相等连续数字的每个项的谐波系列来考虑另一个修正的谐波系列。可以验证在谐波系列的前1200个项中，仅省略20个项。这20个省略的术语是：$$ \\ dfrac {1} {111}，\\ dfrac {1} {222}，\\ dfrac {1} {333}，\\ dfrac {1} {444}，\\ dfrac {1} { 555}，\\ dfrac {1} {666}，\\ dfrac {1} {777}，\\ dfrac {1} {888}，\\ dfrac {1} {999}，\\ dfrac {1} {1000}，\\ dfrac {1} {1110}，\\ \\ dfrac {1} {1111}，\\ dfrac {1} {1112}，\\ dfrac {1} {1113}，\\ dfrac {1} {1114}，\\ dfrac {1} {1115}，\\ dfrac {1} {1116}，\\ dfrac {1} {1117}，\\ dfrac {1} {1118}，\\ dfrac {1} {1119} $$
+Let us now consider another modified harmonic series by omitting from the harmonic series every term where the denominator has 3 or more equal consecutive digits. One can verify that out of the first 1200 terms of the harmonic series, only 20 terms will be omitted. These 20 omitted terms are: $$\\dfrac{1}{111}, \\dfrac{1}{222}, \\dfrac{1}{333}, \\dfrac{1}{444}, \\dfrac{1}{555}, \\dfrac{1}{666}, \\dfrac{1}{777}, \\dfrac{1}{888}, \\dfrac{1}{999}, \\dfrac{1}{1000}, \\dfrac{1}{1110}, \\\\ \\dfrac{1}{1111}, \\dfrac{1}{1112}, \\dfrac{1}{1113}, \\dfrac{1}{1114}, \\dfrac{1}{1115}, \\dfrac{1}{1116}, \\dfrac{1}{1117}, \\dfrac{1}{1118}, \\dfrac{1}{1119}$$
-这个系列也融合了。
+This series converges as well.
-找到系列收敛的值。将您的答案四舍五入到小数点后面的10位数。
+Find the value the series converges to. Give your answer rounded to 10 digits behind the decimal point.
# --hints--
-`euler368()`应该返回253.6135092068。
+`euler368()` should return 253.6135092068.
```js
assert.strictEqual(euler368(), 253.6135092068);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-369-badugi.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-369-badugi.md
index e37989e4091..abdcb4c8102 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-369-badugi.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-369-badugi.md
@@ -1,22 +1,22 @@
---
id: 5900f4de1000cf542c50fff0
-title: 问题369：Badugi
+title: 'Problem 369: Badugi'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302030
dashedName: problem-369-badugi
---
# --description--
-在标准的52张扑克牌中，一套4张牌是Badugi，如果它包含4张没有成对的牌而没有两张相同牌的牌。
+In a standard 52 card deck of playing cards, a set of 4 cards is a Badugi if it contains 4 cards with no pairs and no two cards of the same suit.
-设f（n）是选择n卡的方式的数量，其中4卡的子集是Badugi。例如，有2598960种方法可以从标准的52卡片组中选择5张卡片，其中514800包含4个卡片子集，这是Badugi，因此f（5）= 514800。
+Let f(n) be the number of ways to choose n cards with a 4 card subset that is a Badugi. For example, there are 2598960 ways to choose five cards from a standard 52 card deck, of which 514800 contain a 4 card subset that is a Badugi, so f(5) = 514800.
-找到Σf（n）为4≤n≤13。
+Find ∑f(n) for 4 ≤ n ≤ 13.
# --hints--
-`euler369()`应该返回862400558448。
+`euler369()` should return 862400558448.
```js
assert.strictEqual(euler369(), 862400558448);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-37-truncatable-primes.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-37-truncatable-primes.md
index 76064b1fb3e..7cac6e4ddb3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-37-truncatable-primes.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-37-truncatable-primes.md
@@ -1,36 +1,46 @@
---
id: 5900f3911000cf542c50fea4
-title: 问题37：可截断的素数
+title: 'Problem 37: Truncatable primes'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302031
dashedName: problem-37-truncatable-primes
---
# --description--
-3797号有一个有趣的财产。作为素数本身，可以从左到右连续删除数字，并在每个阶段保持素数：3797,797,97和7.同样，我们可以从右到左工作：3797,379,37和3。找到从左到右和从右到左都可截断的唯一n（8 <= n <= 11）个素数之和。注意：2,3,5和7不被认为是可截断的素数。
+The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.
+
+Find the sum of the only `n` (8 ≤ `n` ≤ 11) primes that are both truncatable from left to right and right to left.
+
+NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
# --hints--
-`truncatablePrimes(8)`应该返回1986年。
+`truncatablePrimes(8)` should return a number.
+
+```js
+assert(typeof truncatablePrimes(8) === 'number');
+```
+
+`truncatablePrimes(8)` should return 1986.
```js
assert(truncatablePrimes(8) == 1986);
```
-`truncatablePrimes(9)`应该返回5123。
+`truncatablePrimes(9)` should return 5123.
```js
assert(truncatablePrimes(9) == 5123);
```
-`truncatablePrimes(10)`应该返回8920。
+`truncatablePrimes(10)` should return 8920.
```js
assert(truncatablePrimes(10) == 8920);
```
-`truncatablePrimes(11)`应该返回748317。
+`truncatablePrimes(11)` should return 748317.
```js
assert(truncatablePrimes(11) == 748317);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-370-geometric-triangles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-370-geometric-triangles.md
index 7245e04dc83..469f9151e03 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-370-geometric-triangles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-370-geometric-triangles.md
@@ -1,24 +1,24 @@
---
id: 5900f4de1000cf542c50fff1
-title: 问题370：几何三角形
+title: 'Problem 370: Geometric triangles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302032
dashedName: problem-370-geometric-triangles
---
# --description--
-让我们将几何三角形定义为具有边a≤b≤c的整数边三角形，使得其边形成几何级数，即b2 = a·c。
+Let us define a geometric triangle as an integer sided triangle with sides a ≤ b ≤ c so that its sides form a geometric progression, i.e. b2 = a · c .
-这种几何三角形的一个例子是边长为a = 144，b = 156和c = 169的三角形。
+An example of such a geometric triangle is the triangle with sides a = 144, b = 156 and c = 169.
-有861805个几何三角形，周长≤106。
+There are 861805 geometric triangles with perimeter ≤ 106 .
-有多少个几何三角形，周长≤2.5·1013？
+How many geometric triangles exist with perimeter ≤ 2.5·1013 ?
# --hints--
-`euler370()`应该返回41791929448408。
+`euler370()` should return 41791929448408.
```js
assert.strictEqual(euler370(), 41791929448408);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-371-licence-plates.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-371-licence-plates.md
index 26c1117937f..005b5aaf078 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-371-licence-plates.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-371-licence-plates.md
@@ -1,24 +1,28 @@
---
id: 5900f4e01000cf542c50fff2
-title: 问题371：车牌
+title: 'Problem 371: Licence plates'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302033
dashedName: problem-371-licence-plates
---
# --description--
-俄勒冈车牌由三个字母后跟一个三位数组成（每个数字可以是\[0..9]）。开车上班时，赛斯玩下面的游戏：每当他旅行中看到的两个牌照的数量增加到1000就是胜利。
+Oregon licence plates consist of three letters followed by a three digit number (each digit can be from \[0..9]).
-例如，MIC-012和HAN-988也是胜利，RYU-500和SET-500也是如此。 （只要他在同一次旅行中看到他们）。
+While driving to work Seth plays the following game:
-找到他想要获胜的预期牌照数量。将您的答案四舍五入到小数点后面的小数点后8位。
+Whenever the numbers of two licence plates seen on his trip add to 1000 that's a win.
-注意：我们假设所看到的每个牌照同样可能有三位数字。
+E.g. MIC-012 and HAN-988 is a win and RYU-500 and SET-500 too. (as long as he sees them in the same trip).
+
+Find the expected number of plates he needs to see for a win. Give your answer rounded to 8 decimal places behind the decimal point.
+
+Note: We assume that each licence plate seen is equally likely to have any three digit number on it.
# --hints--
-`euler371()`应返回40.66368097。
+`euler371()` should return 40.66368097.
```js
assert.strictEqual(euler371(), 40.66368097);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-372-pencils-of-rays.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-372-pencils-of-rays.md
index 0b4022fdedf..1a04cacb71a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-372-pencils-of-rays.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-372-pencils-of-rays.md
@@ -1,20 +1,20 @@
---
id: 5900f4e11000cf542c50fff3
-title: 问题372：光线笔
+title: 'Problem 372: Pencils of rays'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302034
dashedName: problem-372-pencils-of-rays
---
# --description--
-令R（M，N）为满足M的格点（x，y）的数量
+Let R(M, N) be the number of lattice points (x, y) which satisfy M
-注意：代表楼层功能。
+Note: represents the floor function.
# --hints--
-`euler372()`应该返回301450082318807040。
+`euler372()` should return 301450082318807040.
```js
assert.strictEqual(euler372(), 301450082318807040);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-373-circumscribed-circles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-373-circumscribed-circles.md
index a60c0eca601..534c32ea251 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-373-circumscribed-circles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-373-circumscribed-circles.md
@@ -1,24 +1,26 @@
---
id: 5900f4e11000cf542c50fff4
-title: 问题373：外接圆圈
+title: 'Problem 373: Circumscribed Circles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302035
dashedName: problem-373-circumscribed-circles
---
# --description--
-每个三角形都有一个穿过三个顶点的外接圆。考虑所有整数边三角形，其外接圆的半径也是整数。
+Every triangle has a circumscribed circle that goes through the three vertices.
-设S（n）是半径不超过n的所有这些三角形的外接圆的半径之和。
+Consider all integer sided triangles for which the radius of the circumscribed circle is integral as well.
-S（100）= 4950并且S（1200）= 1653605。
+Let S(n) be the sum of the radii of the circumscribed circles of all such triangles for which the radius does not exceed n.
-找到S（107）。
+S(100)=4950 and S(1200)=1653605.
+
+Find S(107).
# --hints--
-`euler373()`应该返回727227472448913。
+`euler373()` should return 727227472448913.
```js
assert.strictEqual(euler373(), 727227472448913);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-374-maximum-integer-partition-product.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-374-maximum-integer-partition-product.md
index 55e9a2833c4..10fe6218919 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-374-maximum-integer-partition-product.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-374-maximum-integer-partition-product.md
@@ -1,32 +1,32 @@
---
id: 5900f4e51000cf542c50fff6
-title: 问题374：最大整数分区产品
+title: 'Problem 374: Maximum Integer Partition Product'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302036
dashedName: problem-374-maximum-integer-partition-product
---
# --description--
-数字n的整数分区是将n写为正整数之和的方法。
+An integer partition of a number n is a way of writing n as a sum of positive integers.
-仅按其加数顺序不同的分区被认为是相同的。将n分成不同部分是n的分区，其中每个部分最多出现一次。
+Partitions that differ only in the order of their summands are considered the same. A partition of n into distinct parts is a partition of n in which every part occurs at most once.
-5个不同部分的分区是：5,4 + 1和3 + 2。
+The partitions of 5 into distinct parts are: 5, 4+1 and 3+2.
-设f（n）是n的任何这种分区的部分到不同部分的最大乘积，并且令m（n）是具有该乘积的n的任何这种分区的元素的数量。
+Let f(n) be the maximum product of the parts of any such partition of n into distinct parts and let m(n) be the number of elements of any such partition of n with that product.
-所以f（5）= 6，m（5）= 2。
+So f(5)=6 and m(5)=2.
-对于n = 10，具有最大乘积的分区是10 = 2 + 3 + 5，其给出f（10）= 30和m（10）= 3。并且他们的产品，f（10）·m（10）= 30·3 = 90
+For n=10 the partition with the largest product is 10=2+3+5, which gives f(10)=30 and m(10)=3. And their product, f(10)·m(10) = 30·3 = 90
-可以证实Σf（n）·m（n）对于1≤n≤100= 1683550844462。
+It can be verified that ∑f(n)·m(n) for 1 ≤ n ≤ 100 = 1683550844462.
-找到Σf（n）·m（n）为1≤n≤1014。给出你的答案模数982451653，即第5000万个素数。
+Find ∑f(n)·m(n) for 1 ≤ n ≤ 1014. Give your answer modulo 982451653, the 50 millionth prime.
# --hints--
-`euler374()`应该返回334420941。
+`euler374()` should return 334420941.
```js
assert.strictEqual(euler374(), 334420941);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-375-minimum-of-subsequences.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-375-minimum-of-subsequences.md
index eacade3b96e..684613a2357 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-375-minimum-of-subsequences.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-375-minimum-of-subsequences.md
@@ -1,22 +1,34 @@
---
id: 5900f4e41000cf542c50fff5
-title: 问题375：最小子序列
+title: 'Problem 375: Minimum of subsequences'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302037
dashedName: problem-375-minimum-of-subsequences
---
# --description--
-设Sn是由以下伪随机数发生器产生的整数序列：S0 = 290797 Sn + 1 = Sn2 mod 50515093
+Let Sn be an integer sequence produced with the following pseudo-random number generator:
-设A（i，j）是i≤j的数字Si，Si + 1，...，Sj中的最小值。设M（N）=ΣA（i，j）为1≤i≤j≤N。我们可以验证M（10）= 432256955和M（10 000）= 3264567774119。
+S0
-找到M（2 000 000 000）。
+=
+
+290797
+
+Sn+1
+
+=
+
+Sn2 mod 50515093
+
+Let A(i, j) be the minimum of the numbers Si, Si+1, ... , Sj for i ≤ j. Let M(N) = ΣA(i, j) for 1 ≤ i ≤ j ≤ N. We can verify that M(10) = 432256955 and M(10 000) = 3264567774119.
+
+Find M(2 000 000 000).
# --hints--
-`euler375()`应该返回7435327983715286000。
+`euler375()` should return 7435327983715286000.
```js
assert.strictEqual(euler375(), 7435327983715286000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-376-nontransitive-sets-of-dice.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-376-nontransitive-sets-of-dice.md
index df376959faf..b76c4d26e8f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-376-nontransitive-sets-of-dice.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-376-nontransitive-sets-of-dice.md
@@ -1,34 +1,34 @@
---
id: 5900f4e51000cf542c50fff7
-title: 问题376：不可传递的骰子集
+title: 'Problem 376: Nontransitive sets of dice'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302038
dashedName: problem-376-nontransitive-sets-of-dice
---
# --description--
-考虑以下一组具有非标准点的骰子：
+Consider the following set of dice with nonstandard pips:
-模具A：1 4 4 4 4 4模具B：2 2 2 5 5 5模具C：3 3 3 3 3 6
+Die A: 1 4 4 4 4 4 Die B: 2 2 2 5 5 5 Die C: 3 3 3 3 3 6
-两个玩家轮流挑选骰子并滚动它来玩游戏。赢得最高价值的玩家获胜。
+A game is played by two players picking a die in turn and rolling it. The player who rolls the highest value wins.
-如果第一个玩家选择A和第二个玩家选择B，我们得到P（第二个玩家获胜）= 7/12> 1/2
+If the first player picks die A and the second player picks die B we get P(second player wins) = 7/12 > 1/2
-如果第一个玩家选择死亡B并且第二个玩家选择死亡C我们得到P（第二个玩家获胜）= 7/12> 1/2
+If the first player picks die B and the second player picks die C we get P(second player wins) = 7/12 > 1/2
-如果第一个玩家选择死C并且第二个玩家选择死A我们得到P（第二个玩家获胜）= 25/36> 1/2
+If the first player picks die C and the second player picks die A we get P(second player wins) = 25/36 > 1/2
-因此，无论第一个玩家选择什么模特，第二个玩家都可以选择另一个模具并获得超过50％的获胜机会。具有此属性的一组骰子称为非传递骰子组。
+So whatever die the first player picks, the second player can pick another die and have a larger than 50% chance of winning. A set of dice having this property is called a nontransitive set of dice.
-我们希望研究存在多少组非传递性骰子。我们将假设以下条件：有三个六面骰子，每边有1到N个点，包括1和N点。具有相同点数的骰子是相等的，无论骰子位于骰子的哪一侧。相同的点值可能出现在多个骰子上;如果两个玩家都滚动相同的值，则玩家不会获胜。骰子{A，B，C}，{B，C，A}和{C，A，B}的集合是相同的集合。
+We wish to investigate how many sets of nontransitive dice exist. We will assume the following conditions:There are three six-sided dice with each side having between 1 and N pips, inclusive. Dice with the same set of pips are equal, regardless of which side on the die the pips are located. The same pip value may appear on multiple dice; if both players roll the same value neither player wins. The sets of dice {A,B,C}, {B,C,A} and {C,A,B} are the same set.
-对于N = 7，我们发现有9780个这样的集合。 N = 30有多少？
+For N = 7 we find there are 9780 such sets. How many are there for N = 30 ?
# --hints--
-`euler376()`应该返回973059630185670。
+`euler376()` should return 973059630185670.
```js
assert.strictEqual(euler376(), 973059630185670);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-377-sum-of-digits-experience-13.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-377-sum-of-digits-experience-13.md
index 4bf024a5b73..660508c2348 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-377-sum-of-digits-experience-13.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-377-sum-of-digits-experience-13.md
@@ -1,22 +1,26 @@
---
id: 5900f4e51000cf542c50fff8
-title: 问题377：数字之和，经验13
+title: 'Problem 377: Sum of digits, experience 13'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302039
dashedName: problem-377-sum-of-digits-experience-13
---
# --description--
-有16个正整数，它们的数字没有零，并且数字和等于5，即：5,14,23,32,41,113,122,131,212,221,311,1112， 1121年，1211年，2111年和1111年。他们的总和是17891。
+There are 16 positive integers that do not have a zero in their digits and that have a digital sum equal to 5, namely:
-设f（n）为所有正整数的和，它们的数字不为零，数字和等于n。
+5, 14, 23, 32, 41, 113, 122, 131, 212, 221, 311, 1112, 1121, 1211, 2111 and 11111.
-找到$ \\ displaystyle \\ sum\_ {i = 1} ^ {17} f（13 ^ i）$。将最后9位数字作为答案。
+Their sum is 17891.
+
+Let f(n) be the sum of all positive integers that do not have a zero in their digits and have a digital sum equal to n.
+
+Find $\\displaystyle \\sum\_{i=1}^{17} f(13^i)$. Give the last 9 digits as your answer.
# --hints--
-`euler377()`应该返回732385277。
+`euler377()` should return 732385277.
```js
assert.strictEqual(euler377(), 732385277);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-378-triangle-triples.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-378-triangle-triples.md
index 6b473082a27..22589747fe8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-378-triangle-triples.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-378-triangle-triples.md
@@ -1,28 +1,28 @@
---
id: 5900f4e61000cf542c50fff9
-title: 问题378：三角形三元组
+title: 'Problem 378: Triangle Triples'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302040
dashedName: problem-378-triangle-triples
---
# --description--
-设T（n）为第n个三角形数，因此T（n）=
+Let T(n) be the nth triangle number, so T(n) =
-n（n + 1）2
+n (n+1)2
-。
+.
-设dT（n）为T（n）的除数。例如：T（7）= 28且dT（7）= 6。
+Let dT(n) be the number of divisors of T(n). E.g.: T(7) = 28 and dT(7) = 6.
-令Tr（n）为三元组数（i，j，k），使得1≤i<j <k≤n且dT（i）> dT（j）> dT（k）。 Tr（20）= 14，Tr（100）= 5772，Tr（1000）= 11174776。
+Let Tr(n) be the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n and dT(i) > dT(j) > dT(k). Tr(20) = 14, Tr(100) = 5772 and Tr(1000) = 11174776.
-找Tr（60 000 000）。给出答案的最后18位数字。
+Find Tr(60 000 000). Give the last 18 digits of your answer.
# --hints--
-`euler378()`应该返回147534623725724700。
+`euler378()` should return 147534623725724700.
```js
assert.strictEqual(euler378(), 147534623725724700);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-379-least-common-multiple-count.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-379-least-common-multiple-count.md
index 981fdbdeba8..3933d82ea2c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-379-least-common-multiple-count.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-379-least-common-multiple-count.md
@@ -1,24 +1,24 @@
---
id: 5900f4e81000cf542c50fffa
-title: 问题379：最少共同的多重计数
+title: 'Problem 379: Least common multiple count'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302041
dashedName: problem-379-least-common-multiple-count
---
# --description--
-设f（n）为具有x和y正整数的耦合数（x，y），x≤y，x和y的最小公倍数等于n。
+Let f(n) be the number of couples (x,y) with x and y positive integers, x ≤ y and the least common multiple of x and y equal to n.
-令g为f的求和函数，即：g（n）=Σf（i），1≤i≤n。
+Let g be the summatory function of f, i.e.: g(n) = ∑ f(i) for 1 ≤ i ≤ n.
-你得到g（106）= 37429395。
+You are given that g(106) = 37429395.
-找到g（1012）。
+Find g(1012).
# --hints--
-`euler379()`应该返回132314136838185。
+`euler379()` should return 132314136838185.
```js
assert.strictEqual(euler379(), 132314136838185);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-38-pandigital-multiples.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-38-pandigital-multiples.md
index 79fc723e4bf..e0ccd868bba 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-38-pandigital-multiples.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-38-pandigital-multiples.md
@@ -1,18 +1,36 @@
---
id: 5900f3931000cf542c50fea5
-title: 问题38：Pandigital倍数
+title: 'Problem 38: Pandigital multiples'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302042
dashedName: problem-38-pandigital-multiples
---
# --description--
-取数字192并乘以1,2和3中的每一个：192×1 = 192 192×2 = 384 192×3 = 576通过连接每个产品，我们得到1到9 pandigital，192384576。我们将调用192384576 192和（1,2,3）的连接产物。通过从9开始并乘以1,2,3,4和5可以实现相同的目的，得到pandigital，918273645，它是9和（1,2,3,4,5）的连接产物。什么是最大的1到9 pandigital 9位数字，可以形成一个整数与（1,2，...， `n` ）的连接乘积，其中`n` > 1？
+Take the number 192 and multiply it by each of 1, 2, and 3:
+
+<div style='margin-left: 4em;'>
+ 192 × 1 = 192<br>
+ 192 × 2 = 384<br>
+ 192 × 3 = 576<br>
+</div>
+
+By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1, 2, 3).
+
+The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1, 2, 3, 4, 5).
+
+What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1, 2, ... , `n`) where `n` > 1?
# --hints--
-pandigitalMultiples `pandigitalMultiples()`应返回932718654。
+`pandigitalMultiples()` should return a number.
+
+```js
+assert(typeof pandigitalMultiples() === 'number');
+```
+
+`pandigitalMultiples()` should return 932718654.
```js
assert.strictEqual(pandigitalMultiples(), 932718654);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-380-amazing-mazes.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-380-amazing-mazes.md
index 72c6247b28a..8596f52fa78 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-380-amazing-mazes.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-380-amazing-mazes.md
@@ -1,24 +1,24 @@
---
id: 5900f4e81000cf542c50fffb
-title: 问题380：惊人的迷宫！
+title: 'Problem 380: Amazing Mazes!'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302044
dashedName: problem-380-amazing-mazes
---
# --description--
-m×n迷宫是m×n矩形网格，其中墙壁放置在网格单元之间，使得从左上角正方形到任何其他正方形恰好有一条路径。以下是9×12迷宫和15×20迷宫的示例：
+An m×n maze is an m×n rectangular grid with walls placed between grid cells such that there is exactly one path from the top-left square to any other square. The following are examples of a 9×12 maze and a 15×20 maze:
-设C（m，n）为不同m×n个迷宫的数量。可以通过来自另一个迷宫的旋转和反射形成的迷宫被认为是不同的。
+Let C(m,n) be the number of distinct m×n mazes. Mazes which can be formed by rotation and reflection from another maze are considered distinct.
-可以证实C（1,1）= 1，C（2,2）= 4，C（3,4）= 2415，C（9,12）= 2.5720e46（科学计数法四舍五入为5显着位）。找到C（100,500）并用科学计数法将你的答案写成五位有效数字。
+It can be verified that C(1,1) = 1, C(2,2) = 4, C(3,4) = 2415, and C(9,12) = 2.5720e46 (in scientific notation rounded to 5 significant digits). Find C(100,500) and write your answer in scientific notation rounded to 5 significant digits.
-在给出答案时，使用小写e来分隔尾数和指数。例如，如果答案是1234567891011，则答案格式为1.2346e12。
+When giving your answer, use a lowercase e to separate mantissa and exponent. E.g. if the answer is 1234567891011 then the answer format would be 1.2346e12.
# --hints--
-`euler380()`应该返回Infinity。
+`euler380()` should return Infinity.
```js
assert.strictEqual(euler380(), Infinity);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-381-prime-k-factorial.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-381-prime-k-factorial.md
index 28345416927..f64768a5c7a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-381-prime-k-factorial.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-381-prime-k-factorial.md
@@ -1,24 +1,24 @@
---
id: 5900f4ea1000cf542c50fffc
-title: 问题381：（prime-k）阶乘
+title: 'Problem 381: (prime-k) factorial'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302045
dashedName: problem-381-prime-k-factorial
---
# --description--
-对于素数p，令S（p）=（Σ（pk）！）mod（p）1≤k≤5。
+For a prime p let S(p) = (∑(p-k)!) mod(p) for 1 ≤ k ≤ 5.
-例如，如果p = 7，（7-1）！ +（7-2）！ +（7-3）！ +（7-4）！ +（7-5）！ = 6！ + 5！ + 4！ + 3！ + 2！ = 720 + 120 + 24 + 6 + 2 = 872.当872 mod（7）= 4时，S（7）= 4。
+For example, if p=7, (7-1)! + (7-2)! + (7-3)! + (7-4)! + (7-5)! = 6! + 5! + 4! + 3! + 2! = 720+120+24+6+2 = 872. As 872 mod(7) = 4, S(7) = 4.
-对于5≤p<100，可以证实ΣS（p）= 480。
+It can be verified that ∑S(p) = 480 for 5 ≤ p < 100.
-求ΣS（p）为5≤p<108。
+Find ∑S(p) for 5 ≤ p < 108.
# --hints--
-`euler381()`应该返回139602943319822。
+`euler381()` should return 139602943319822.
```js
assert.strictEqual(euler381(), 139602943319822);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-382-generating-polygons.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-382-generating-polygons.md
index b49ec0ff609..ae95128f6bb 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-382-generating-polygons.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-382-generating-polygons.md
@@ -1,28 +1,28 @@
---
id: 5900f4eb1000cf542c50fffd
-title: 问题382：生成多边形
+title: 'Problem 382: Generating polygons'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302046
dashedName: problem-382-generating-polygons
---
# --description--
-多边形是由直线段组成的扁平形状，所述直线段连接以形成闭合链或电路。多边形由至少三个边组成，并且不自相交。
+A polygon is a flat shape consisting of straight line segments that are joined to form a closed chain or circuit. A polygon consists of at least three sides and does not self-intersect.
-如果：P的两边不是相同的长度，P的每一边的长度在S中，并且S不包含其他值，则称正数的集合S生成多边形P.
+A set S of positive numbers is said to generate a polygon P if: no two sides of P are the same length, the length of every side of P is in S, and S contains no other value.
-例如：集合{3,4,5}生成边3,4和5（三角形）的多边形。集合{6,9,11,24}生成具有边6,9,11和24（四边形）的多边形。集合{1,2,3}和{2,3,4,9}根本不生成任何多边形。
+For example: The set {3, 4, 5} generates a polygon with sides 3, 4, and 5 (a triangle). The set {6, 9, 11, 24} generates a polygon with sides 6, 9, 11, and 24 (a quadrilateral). The sets {1, 2, 3} and {2, 3, 4, 9} do not generate any polygon at all.
-考虑序列s，定义如下：s1 = 1，s2 = 2，s3 = 3 sn = sn-1 + sn-3，n> 3。
+Consider the sequence s, defined as follows:s1 = 1, s2 = 2, s3 = 3 sn = sn-1 + sn-3 for n > 3.
-设Un为集合{s1，s2，...，sn}。例如，U10 = {1,2,3,4,6,9,13,19,28,41}。设f（n）是产生至少一个多边形的Un子集的数量。例如，f（5）= 7，f（10）= 501，f（25）= 18635853。
+Let Un be the set {s1, s2, ..., sn}. For example, U10 = {1, 2, 3, 4, 6, 9, 13, 19, 28, 41}. Let f(n) be the number of subsets of Un which generate at least one polygon. For example, f(5) = 7, f(10) = 501 and f(25) = 18635853.
-找到f（1018）的最后9位数。
+Find the last 9 digits of f(1018).
# --hints--
-`euler382()`应该返回697003956。
+`euler382()` should return 697003956.
```js
assert.strictEqual(euler382(), 697003956);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-383-divisibility-comparison-between-factorials.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-383-divisibility-comparison-between-factorials.md
index d5b7af14279..f199cb4f946 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-383-divisibility-comparison-between-factorials.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-383-divisibility-comparison-between-factorials.md
@@ -1,22 +1,24 @@
---
id: 5900f4ed1000cf542c50ffff
-title: 问题383：阶乘之间的可比性比较
+title: 'Problem 383: Divisibility comparison between factorials'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302047
dashedName: problem-383-divisibility-comparison-between-factorials
---
# --description--
-设f5（n）是最大整数x，其中5x除以n。例如，f5（625000）= 7。
+Let f5(n) be the largest integer x for which 5x divides n.
-令T5（n）为满足f5（（2·i-1）！）<2·f5（i！）且1≤i≤n的整数i。可以证实T5（103）= 68并且T5（109）= 2408210。
+For example, f5(625000) = 7.
-找到T5（1018）。
+Let T5(n) be the number of integers i which satisfy f5((2·i-1)!) < 2·f5(i!) and 1 ≤ i ≤ n. It can be verified that T5(103) = 68 and T5(109) = 2408210.
+
+Find T5(1018).
# --hints--
-`euler383()`应该返回22173624649806。
+`euler383()` should return 22173624649806.
```js
assert.strictEqual(euler383(), 22173624649806);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-384-rudin-shapiro-sequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-384-rudin-shapiro-sequence.md
index 22bf0fef4a9..c57f0ee2f96 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-384-rudin-shapiro-sequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-384-rudin-shapiro-sequence.md
@@ -1,32 +1,34 @@
---
id: 5900f4ed1000cf542c50fffe
-title: 问题384：Rudin-Shapiro序列
+title: 'Problem 384: Rudin-Shapiro sequence'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302048
dashedName: problem-384-rudin-shapiro-sequence
---
# --description--
-将序列a（n）定义为n的二进制展开（可能重叠）中相邻的1对的数量。例如：a（5）= a（1012）= 0，a（6）= a（1102）= 1，a（7）= a（1112）= 2
+Define the sequence a(n) as the number of adjacent pairs of ones in the binary expansion of n (possibly overlapping).
-定义序列b（n）=（ - 1）a（n）。该序列称为Rudin-Shapiro序列。还要考虑b（n）的总和序列：。
+E.g.: a(5) = a(1012) = 0, a(6) = a(1102) = 1, a(7) = a(1112) = 2
-这些序列的前几个值是：n 0 1 2 3 4 5 6 7 a（n）0 0 0 1 0 0 1 2 b（n）1 1 1 -1 1 1 -1 1 s（n）1 2 3 2 3 4 3 4
+Define the sequence b(n) = (-1)a(n). This sequence is called the Rudin-Shapiro sequence. Also consider the summatory sequence of b(n): .
-序列s（n）具有显着特性，即所有元素都是正的，并且每个正整数k恰好出现k次。
+The first couple of values of these sequences are: n 0 1 2 3 4 5 6 7 a(n) 0 0 0 1 0 0 1 2 b(n) 1 1 1 -1 1 1 -1 1 s(n) 1 2 3 2 3 4 3 4
-定义g（t，c），其中1≤c≤t，作为s（n）中的索引，其中t在s（n）中出现第c次。例如：g（3,3）= 6，g（4,2）= 7，g（54321,12345）= 1220847710。
+The sequence s(n) has the remarkable property that all elements are positive and every positive integer k occurs exactly k times.
-设F（n）为由下式定义的斐波那契数：F（0）= F（1）= 1且F（n）= F（n-1）+ F（n-2），n> 1。
+Define g(t,c), with 1 ≤ c ≤ t, as the index in s(n) for which t occurs for the c'th time in s(n). E.g.: g(3,3) = 6, g(4,2) = 7 and g(54321,12345) = 1220847710.
-定义GF（t）= g（F（t），F（t-1））。
+Let F(n) be the fibonacci sequence defined by: F(0)=F(1)=1 and F(n)=F(n-1)+F(n-2) for n>1.
-找到ΣGF（t）为2≤t≤45。
+Define GF(t)=g(F(t),F(t-1)).
+
+Find ΣGF(t) for 2≤t≤45.
# --hints--
-`euler384()`应返回3354706415856333000。
+`euler384()` should return 3354706415856333000.
```js
assert.strictEqual(euler384(), 3354706415856333000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-385-ellipses-inside-triangles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-385-ellipses-inside-triangles.md
index 9747a2502b0..3cb1e307632 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-385-ellipses-inside-triangles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-385-ellipses-inside-triangles.md
@@ -1,31 +1,33 @@
---
id: 5900f4ee1000cf542c510000
-title: 问题385：三角形内的椭圆
+title: 'Problem 385: Ellipses inside triangles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302049
dashedName: problem-385-ellipses-inside-triangles
---
# --description--
-对于平面中的任何三角形T，可以显示存在唯一的椭圆，其中最大区域完全在T内。
+For any triangle T in the plane, it can be shown that there is a unique ellipse with largest area that is completely inside T.
-对于给定的n，考虑三角形T，使得：
+For a given n, consider triangles T such that:
-- T的顶点具有绝对值≤n的整数坐标，和
-- T内最大面积椭圆的焦点是（√13,0）和（-√13,0）。设A（n）为所有这些三角形的面积之和。
+- the vertices of T have integer coordinates with absolute value ≤ n, and
+- the foci1 of the largest-area ellipse inside T are (√13,0) and (-√13,0).
-例如，如果n = 8，则存在两个这样的三角形。它们的顶点是（-4，-3），（ - 4,3），（8,0）和（4,3），（4，-3），（ - 8,0），每个三角形的面积因此A（8）= 36 + 36 = 72。
+Let A(n) be the sum of the areas of all such triangles.
-可以证实A（10）= 252，A（100）= 34632和A（1000）= 3529008。
+For example, if n = 8, there are two such triangles. Their vertices are (-4,-3),(-4,3),(8,0) and (4,3),(4,-3),(-8,0), and the area of each triangle is 36. Thus A(8) = 36 + 36 = 72.
-找A（1 000 000 000）。
+It can be verified that A(10) = 252, A(100) = 34632 and A(1000) = 3529008.
-1椭圆的焦点（多个焦点）是两个点A和B，使得对于椭圆边界上的每个点P，AP + PB是恒定的。
+Find A(1 000 000 000).
+
+1The foci (plural of focus) of an ellipse are two points A and B such that for every point P on the boundary of the ellipse, AP + PB is constant.
# --hints--
-`euler385()`应该返回3776957309612154000。
+`euler385()` should return 3776957309612154000.
```js
assert.strictEqual(euler385(), 3776957309612154000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-386-maximum-length-of-an-antichain.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-386-maximum-length-of-an-antichain.md
index ac8e8ee7f7c..36a6ce965c8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-386-maximum-length-of-an-antichain.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-386-maximum-length-of-an-antichain.md
@@ -1,26 +1,26 @@
---
id: 5900f4ef1000cf542c510001
-title: 问题386：反链的最大长度
+title: 'Problem 386: Maximum length of an antichain'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302050
dashedName: problem-386-maximum-length-of-an-antichain
---
# --description--
-令n为整数，S（n）为n的因子集。
+Let n be an integer and S(n) be the set of factors of n.
-如果A仅包含一个元素，或者如果A的元素中没有一个除以A的任何其他元素，则S（n）的子集A被称为S（n）的反共。
+A subset A of S(n) is called an antichain of S(n) if A contains only one element or if none of the elements of A divides any of the other elements of A.
-例如：S（30）= {1,2,3,5,6,10,15,30} {2,5,6}不是S（30）的反链。 {2,3,5}是S（30）的反链。
+For example: S(30) = {1, 2, 3, 5, 6, 10, 15, 30} {2, 5, 6} is not an antichain of S(30). {2, 3, 5} is an antichain of S(30).
-设N（n）为S（n）的反链的最大长度。
+Let N(n) be the maximum length of an antichain of S(n).
-找ΣN（n）为1≤n≤108
+Find ΣN(n) for 1 ≤ n ≤ 108
# --hints--
-`euler386()`应该返回528755790。
+`euler386()` should return 528755790.
```js
assert.strictEqual(euler386(), 528755790);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-387-harshad-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-387-harshad-numbers.md
index b8d1e199ab5..aa3caddfe3e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-387-harshad-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-387-harshad-numbers.md
@@ -1,26 +1,34 @@
---
id: 5900f4f11000cf542c510003
-title: 问题387：Harshad数字
+title: 'Problem 387: Harshad Numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302051
dashedName: problem-387-harshad-numbers
---
# --description--
-Harshad或Niven数字是可以被其数字之和整除的数字。 201是一个Harshad数，因为它可以被3整数（它的数字之和）。当我们截断201的最后一个数字时，我们得到20，这是一个Harshad数。当我们截断20的最后一位数时，我们得到2，这也是一个Harshad数。让我们调用一个Harshad数字，在递归截断最后一个数字的同时，总是会产生一个Harshad数字，一个正可截断的Harshad数字。
+A Harshad or Niven number is a number that is divisible by the sum of its digits.
-另外：201/3 = 67这是素数。让我们称一个Harshad数，当它除以它的数字之和时，得到一个强Harshad数的素数。
+201 is a Harshad number because it is divisible by 3 (the sum of its digits.)
-现在取2011年的数字。当我们截断它的最后一位数时，我们得到201，一个强大的Harshad数也是可以截断的。让我们称这样的素数强大，正确的可截断的哈尔沙德素数。
+When we truncate the last digit from 201, we get 20, which is a Harshad number.
-你得到的是，强度小，可截断的Harshad素数小于10000的总和是90619。
+When we truncate the last digit from 20, we get 2, which is also a Harshad number.
-找到小于1014的强可，可截断的Harshad素数的总和。
+Let's call a Harshad number that, while recursively truncating the last digit, always results in a Harshad number a right truncatable Harshad number.
+
+Also: 201/3=67 which is prime. Let's call a Harshad number that, when divided by the sum of its digits, results in a prime a strong Harshad number.
+
+Now take the number 2011 which is prime. When we truncate the last digit from it we get 201, a strong Harshad number that is also right truncatable. Let's call such primes strong, right truncatable Harshad primes.
+
+You are given that the sum of the strong, right truncatable Harshad primes less than 10000 is 90619.
+
+Find the sum of the strong, right truncatable Harshad primes less than 1014.
# --hints--
-`euler387()`应该返回696067597313468。
+`euler387()` should return 696067597313468.
```js
assert.strictEqual(euler387(), 696067597313468);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-388-distinct-lines.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-388-distinct-lines.md
index bbc2175c3e8..ba13e9be3cd 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-388-distinct-lines.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-388-distinct-lines.md
@@ -1,24 +1,24 @@
---
id: 5900f4f11000cf542c510002
-title: 问题388：不同的线条
+title: 'Problem 388: Distinct Lines'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302052
dashedName: problem-388-distinct-lines
---
# --description--
-考虑所有格点（a，b，c），其中0≤a，b，c≤N。
+Consider all lattice points (a,b,c) with 0 ≤ a,b,c ≤ N.
-从原点O（0,0,0）开始，所有线都被绘制到其他格点。设D（N）是不同的这种线的数量。
+From the origin O(0,0,0) all lines are drawn to the other lattice points. Let D(N) be the number of distinct such lines.
-您被给予D（1 000 000）= 831909254469114121。
+You are given that D(1 000 000) = 831909254469114121.
-找到D（1010）。将前9位数字后跟最后9位数字作为答案。
+Find D(1010). Give as your answer the first nine digits followed by the last nine digits.
# --hints--
-`euler388()`应该返回831907372805130000。
+`euler388()` should return 831907372805130000.
```js
assert.strictEqual(euler388(), 831907372805130000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-389-platonic-dice.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-389-platonic-dice.md
index 11233238867..0fb6bb2688e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-389-platonic-dice.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-389-platonic-dice.md
@@ -1,18 +1,20 @@
---
id: 5900f4f21000cf542c510004
-title: 问题389：柏拉图式骰子
+title: 'Problem 389: Platonic Dice'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302053
dashedName: problem-389-platonic-dice
---
# --description--
-抛出一个无偏的单面4面模具，并注明其值T.T无偏见的6面骰子被抛出并且它们的分数加在一起。记录总和C.C抛出无偏见的8面骰子并将它们的分数加在一起。记录总和O.O无偏见的12面骰子被投掷并且他们的分数被加在一起。记录总和D.D抛出无偏的20面骰子并将它们的分数加在一起。总和，我，注意到。找出I的方差，并将答案四舍五入到小数点后4位。
+An unbiased single 4-sided die is thrown and its value, T, is noted.T unbiased 6-sided dice are thrown and their scores are added together. The sum, C, is noted.C unbiased 8-sided dice are thrown and their scores are added together. The sum, O, is noted.O unbiased 12-sided dice are thrown and their scores are added together. The sum, D, is noted.D unbiased 20-sided dice are thrown and their scores are added together. The sum, I, is noted.
+
+Find the variance of I, and give your answer rounded to 4 decimal places.
# --hints--
-`euler389()`应该返回2406376.3623。
+`euler389()` should return 2406376.3623.
```js
assert.strictEqual(euler389(), 2406376.3623);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-39-integer-right-triangles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-39-integer-right-triangles.md
index 46dcb7925df..d19bb816b7f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-39-integer-right-triangles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-39-integer-right-triangles.md
@@ -1,36 +1,46 @@
---
id: 5900f3931000cf542c50fea6
-title: 问题39：整数直角三角形
+title: 'Problem 39: Integer right triangles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302054
dashedName: problem-39-integer-right-triangles
---
# --description--
-如果p是具有整数长度边的直角三角形的周长{a，b，c}，则对于p = 120，恰好有三个解。{20,48,52}，{24,45,51}，{ 30,40,50}对于p≤n的值，最大化解的数量是多少？
+If `p` is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.
+
+{20,48,52}, {24,45,51}, {30,40,50}
+
+For which value of `p` ≤ `n`, is the number of solutions maximized?
# --hints--
-`intRightTriangles(500)`应该返回420。
+`intRightTriangles(500)` should return a number.
+
+```js
+assert(typeof intRightTriangles(500) === 'number');
+```
+
+`intRightTriangles(500)` should return 420.
```js
assert(intRightTriangles(500) == 420);
```
-`intRightTriangles(800)`应该返回420。
+`intRightTriangles(800)` should return 720.
```js
assert(intRightTriangles(800) == 720);
```
-`intRightTriangles(900)`应该返回840。
+`intRightTriangles(900)` should return 840.
```js
assert(intRightTriangles(900) == 840);
```
-`intRightTriangles(1000)`应该返回840。
+`intRightTriangles(1000)` should return 840.
```js
assert(intRightTriangles(1000) == 840);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-390-triangles-with-non-rational-sides-and-integral-area.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-390-triangles-with-non-rational-sides-and-integral-area.md
index 4848daa544f..e582b23f2d7 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-390-triangles-with-non-rational-sides-and-integral-area.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-390-triangles-with-non-rational-sides-and-integral-area.md
@@ -1,26 +1,28 @@
---
id: 5900f4f21000cf542c510005
-title: 问题390：具有非理性边和积分面积的三角形
+title: 'Problem 390: Triangles with non rational sides and integral area'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302055
dashedName: problem-390-triangles-with-non-rational-sides-and-integral-area
---
# --description--
-考虑边长为√5，√65和√68的三角形。可以看出，该三角形具有区域9。
+Consider the triangle with sides √5, √65 and √68.
-S（n）是所有三角形的面积之和，其边长为√（1 + b2），√（1 + c2）和√（b2 + c2）（对于正整数b和c），其积分面积不超过ñ。
+It can be shown that this triangle has area 9.
-示例三角形的b = 2且c = 8。
+S(n) is the sum of the areas of all triangles with sides √(1+b2), √(1+c2) and √(b2+c2) (for positive integers b and c ) that have an integral area not exceeding n.
-S（106）= 18018206。
+The example triangle has b=2 and c=8.
-找到S（1010）。
+S(106)=18018206.
+
+Find S(1010).
# --hints--
-`euler390()`应该返回2919133642971。
+`euler390()` should return 2919133642971.
```js
assert.strictEqual(euler390(), 2919133642971);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-391-hopping-game.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-391-hopping-game.md
index ec3811a34d0..4eda36c0a45 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-391-hopping-game.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-391-hopping-game.md
@@ -1,28 +1,32 @@
---
id: 5900f4f31000cf542c510006
-title: 问题391：跳跃游戏
+title: 'Problem 391: Hopping Game'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302056
dashedName: problem-391-hopping-game
---
# --description--
-当将数字从0写入k到二进制时，令sk为1的数。例如，以二进制形式写0到5，我们有0,1,10,11,100,101。有7个1，所以s5 = 7.序列S = {sk：k≥0}开始{0,1 ，2,4,5,7,9,12，......}。
+Let sk be the number of 1’s when writing the numbers from 0 to k in binary.
-一个游戏由两个玩家玩。在游戏开始之前，选择数字n。计数器c从0开始。每转一圈，玩家选择一个从1到n（含）的数字，然后用该数字增加c。结果值c必须是S的成员。如果没有更多有效的移动，则玩家输掉。
+For example, writing 0 to 5 in binary, we have 0, 1, 10, 11, 100, 101. There are seven 1’s, so s5 = 7.
-例如：设n = 5.c从0开始。玩家1选择4，所以c变为0 + 4 = 4.玩家2选择5，所以c变为4 + 5 = 9.玩家1选择3，所以c变为9 + 3 = 12.等等。请注意，c必须始终属于S，并且每个玩家最多可以将c增加n。
+The sequence S = {sk : k ≥ 0} starts {0, 1, 2, 4, 5, 7, 9, 12, ...}.
-设M（n）是第一个玩家在第一个回合强制获胜时可以选择的最高数字，如果没有这样的移动则M（n）= 0。例如，M（2）= 2，M（7）= 1且M（20）= 4。
+A game is played by two players. Before the game starts, a number n is chosen. A counter c starts at 0. At each turn, the player chooses a number from 1 to n (inclusive) and increases c by that number. The resulting value of c must be a member of S. If there are no more valid moves, the player loses.
-给定Σ（M（n））3 = 8150，1≤n≤20。
+For example: Let n = 5. c starts at 0. Player 1 chooses 4, so c becomes 0 + 4 = 4. Player 2 chooses 5, so c becomes 4 + 5 = 9. Player 1 chooses 3, so c becomes 9 + 3 = 12. etc. Note that c must always belong to S, and each player can increase c by at most n.
-找Σ（M（n））3表示1≤n≤1000。
+Let M(n) be the highest number the first player can choose at her first turn to force a win, and M(n) = 0 if there is no such move. For example, M(2) = 2, M(7) = 1 and M(20) = 4.
+
+Given Σ(M(n))3 = 8150 for 1 ≤ n ≤ 20.
+
+Find Σ(M(n))3 for 1 ≤ n ≤ 1000.
# --hints--
-`euler391()`应该返回61029882288。
+`euler391()` should return 61029882288.
```js
assert.strictEqual(euler391(), 61029882288);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-392-enmeshed-unit-circle.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-392-enmeshed-unit-circle.md
index b4b47b412eb..cd4d53f17da 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-392-enmeshed-unit-circle.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-392-enmeshed-unit-circle.md
@@ -1,26 +1,28 @@
---
id: 5900f4f41000cf542c510007
-title: 问题392：陷入困境的单位圆
+title: 'Problem 392: Enmeshed unit circle'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302057
dashedName: problem-392-enmeshed-unit-circle
---
# --description--
-直线网格是正交网格，其中网格线之间的间距不必是等距的。这种网格的一个例子是对数图纸。
+A rectilinear grid is an orthogonal grid where the spacing between the gridlines does not have to be equidistant.
-考虑笛卡尔坐标系中的直线网格，具有以下属性：网格线平行于笛卡尔坐标系的轴。有N + 2个垂直网格线和N + 2个水平网格线。因此存在（N + 1）x（N + 1）个矩形单元。两个外部垂直网格线的方程是x = -1且x = 1.两个外部水平网格线的方程是y = -1和y如果它们与单位圆重叠，则网格单元为红色，否则为黑色。对于这个问题，我们希望您找到剩余的N个内部水平线和N个内部垂直网格线的位置，以便红色占据的区域细胞最小化。
+An example of such grid is logarithmic graph paper.
-例如，这里是N = 10的解决方案的图片：
+Consider rectilinear grids in the Cartesian coordinate system with the following properties:The gridlines are parallel to the axes of the Cartesian coordinate system.There are N+2 vertical and N+2 horizontal gridlines. Hence there are (N+1) x (N+1) rectangular cells.The equations of the two outer vertical gridlines are x = -1 and x = 1.The equations of the two outer horizontal gridlines are y = -1 and y = 1.The grid cells are colored red if they overlap with the unit circle, black otherwise.For this problem we would like you to find the positions of the remaining N inner horizontal and N inner vertical gridlines so that the area occupied by the red cells is minimized.
-红色单元占N = 10的区域舍入到小数点后面的10位是3.3469640797。
+E.g. here is a picture of the solution for N = 10:
-找到N = 400的位置。将红色单元占用的区域四舍五入到小数点后面的10位数作为答案。
+The area occupied by the red cells for N = 10 rounded to 10 digits behind the decimal point is 3.3469640797.
+
+Find the positions for N = 400. Give as your answer the area occupied by the red cells rounded to 10 digits behind the decimal point.
# --hints--
-`euler392()`应返回3.1486734435。
+`euler392()` should return 3.1486734435.
```js
assert.strictEqual(euler392(), 3.1486734435);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-393-migrating-ants.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-393-migrating-ants.md
index b40373d7737..21f98ec10cc 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-393-migrating-ants.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-393-migrating-ants.md
@@ -1,20 +1,24 @@
---
id: 5900f4f61000cf542c510008
-title: 问题393：迁移蚂蚁
+title: 'Problem 393: Migrating ants'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302058
dashedName: problem-393-migrating-ants
---
# --description--
-n×n正方形网格包含n2个蚂蚁，每平方一个蚂蚁。所有蚂蚁决定同时移动到相邻的广场（通常有4种可能，除了网格边缘或角落处的蚂蚁）。我们将f（n）定义为在没有任何蚂蚁在同一个正方形上结束并且没有任何两个蚂蚁穿过两个正方形之间的相同边缘时可以发生的方式的数量。
+An n×n grid of squares contains n2 ants, one ant per square.
-给出f（4）= 88.求f（10）。
+All ants decide to move simultaneously to an adjacent square (usually 4 possibilities, except for ants on the edge of the grid or at the corners).
+
+We define f(n) to be the number of ways this can happen without any ants ending on the same square and without any two ants crossing the same edge between two squares.
+
+You are given that f(4) = 88. Find f(10).
# --hints--
-`euler393()`应该返回112398351350823100。
+`euler393()` should return 112398351350823100.
```js
assert.strictEqual(euler393(), 112398351350823100);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-394-eating-pie.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-394-eating-pie.md
index 791309a12fb..26c7bce3a95 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-394-eating-pie.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-394-eating-pie.md
@@ -1,22 +1,32 @@
---
id: 5900f4f71000cf542c510009
-title: 问题394：吃馅饼
+title: 'Problem 394: Eating pie'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302059
dashedName: problem-394-eating-pie
---
# --description--
-杰夫以一种不同寻常的方式吃馅饼。馅饼是圆形的。他首先沿着半径切割馅饼中的初始切口。虽然剩下至少给定的馅饼F分数，但他执行以下程序： - 他从饼图中心到馅饼边框剩余的任何一点做两个切片，剩下的馅饼边框上的任何一点都有可能。这将把剩下的馅饼分成三块。 - 从最初的切割逆时针走，他拿出前两个馅饼并吃掉它们。当剩下不到一小部分馅饼时，他不再重复这个过程。相反，他吃剩下的所有馅饼。
+Jeff eats a pie in an unusual way.
-对于x≥1，设E（x）是Jeff重复上述过程的预期次数，F = 1 / x。可以证实E（1）= 1，E（2）≈1.2676536759，和E（7.5）≈2.1215732071。
+The pie is circular. He starts with slicing an initial cut in the pie along a radius.
-找到E（40）舍入到小数点后面的10位小数。
+While there is at least a given fraction F of pie left, he performs the following procedure:
+
+\- He makes two slices from the pie centre to any point of what is remaining of the pie border, any point on the remaining pie border equally likely. This will divide the remaining pie into three pieces.
+
+\- Going counterclockwise from the initial cut, he takes the first two pie pieces and eats them.
+
+When less than a fraction F of pie remains, he does not repeat this procedure. Instead, he eats all of the remaining pie.
+
+For x ≥ 1, let E(x) be the expected number of times Jeff repeats the procedure above with F = 1/x. It can be verified that E(1) = 1, E(2) ≈ 1.2676536759, and E(7.5) ≈ 2.1215732071.
+
+Find E(40) rounded to 10 decimal places behind the decimal point.
# --hints--
-`euler394()`应返回3.2370342194。
+`euler394()` should return 3.2370342194.
```js
assert.strictEqual(euler394(), 3.2370342194);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-395-pythagorean-tree.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-395-pythagorean-tree.md
index 285bb4469ed..80903ac0ef5 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-395-pythagorean-tree.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-395-pythagorean-tree.md
@@ -1,26 +1,26 @@
---
id: 5900f4f71000cf542c51000a
-title: 问题395：毕达哥拉斯树
+title: 'Problem 395: Pythagorean tree'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302060
dashedName: problem-395-pythagorean-tree
---
# --description--
-毕达哥拉斯树是由以下程序产生的分形：
+The Pythagorean tree is a fractal generated by the following procedure:
-从单位正方形开始。然后，将其中一个侧面称为基座（在动画中，底侧是基座）：在基座对面的一侧安装一个直角三角形，斜边与该侧面重合，侧面为3-4- 5比率。请注意，三角形的较小边必须位于相对于底边的“右”侧（请参阅动画）。将一个正方形附加到右三角形的每条腿上，其一侧与该腿重合。对两个正方形重复此过程，将两侧接触三角形作为基础。
+Start with a unit square. Then, calling one of the sides its base (in the animation, the bottom side is the base): Attach a right triangle to the side opposite the base, with the hypotenuse coinciding with that side and with the sides in a 3-4-5 ratio. Note that the smaller side of the triangle must be on the 'right' side with respect to the base (see animation). Attach a square to each leg of the right triangle, with one of its sides coinciding with that leg. Repeat this procedure for both squares, considering as their bases the sides touching the triangle.
-经过无数次迭代后，得到的数字是毕达哥拉斯树。
+The resulting figure, after an infinite number of iterations, is the Pythagorean tree.
-可以看出，至少有一个矩形，其边与毕达哥拉斯树的最大正方形平行，完全包围了毕达哥拉斯树。
+It can be shown that there exists at least one rectangle, whose sides are parallel to the largest square of the Pythagorean tree, which encloses the Pythagorean tree completely.
-找到这样的边界矩形可能的最小区域，并将您的答案舍入到10个小数位。
+Find the smallest area possible for such a bounding rectangle, and give your answer rounded to 10 decimal places.
# --hints--
-`euler395()`应返回28.2453753155。
+`euler395()` should return 28.2453753155.
```js
assert.strictEqual(euler395(), 28.2453753155);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-396-weak-goodstein-sequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-396-weak-goodstein-sequence.md
index 11659a92d9d..bb98b389ba5 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-396-weak-goodstein-sequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-396-weak-goodstein-sequence.md
@@ -1,30 +1,34 @@
---
id: 5900f4f81000cf542c51000b
-title: 问题396：弱Goodstein序列
+title: 'Problem 396: Weak Goodstein sequence'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302061
dashedName: problem-396-weak-goodstein-sequence
---
# --description--
-对于任何正整数n，第n个弱Goodstein序列{g1，g2，g3，...}定义为：g1 = n，对于k> 1，gk是通过在基k中写gk-1获得的，将其解释为a基数k + 1，减去1。
+For any positive integer n, the nth weak Goodstein sequence {g1, g2, g3, ...} is defined as:
-当gk变为0时，序列终止。
+g1 = n
-例如，第6个弱Goodstein序列是{6,11,17,25，...}：g1 = 6. g2 = 11，因为6 = 1102,1103 = 12，12 - 1 = 11. g3 = 17 11 = 1023,1024 = 18,18-1 = 17.g4 = 25，因为17 = 1014,1015 = 26,26-1 = 25。
+for k > 1, gk is obtained by writing gk-1 in base k, interpreting it as a base k + 1 number, and subtracting 1.
-等等。
+The sequence terminates when gk becomes 0.
-可以证明，每个弱的Goodstein序列都会终止。
+For example, the 6th weak Goodstein sequence is {6, 11, 17, 25, ...}: g1 = 6. g2 = 11 since 6 = 1102, 1103 = 12, and 12 - 1 = 11. g3 = 17 since 11 = 1023, 1024 = 18, and 18 - 1 = 17. g4 = 25 since 17 = 1014, 1015 = 26, and 26 - 1 = 25.
-设G（n）为第n个弱Goodstein序列中的非零元素的数量。可以证实G（2）= 3，G（4）= 21和G（6）= 381.还可以证实ΣG（n）= 2517，1≤n<8。
+and so on.
-找到ΣG（n）的最后9位数，1≤n<16。
+It can be shown that every weak Goodstein sequence terminates.
+
+Let G(n) be the number of nonzero elements in the nth weak Goodstein sequence. It can be verified that G(2) = 3, G(4) = 21 and G(6) = 381. It can also be verified that ΣG(n) = 2517 for 1 ≤ n < 8.
+
+Find the last 9 digits of ΣG(n) for 1 ≤ n < 16.
# --hints--
-`euler396()`应该返回173214653。
+`euler396()` should return 173214653.
```js
assert.strictEqual(euler396(), 173214653);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-397-triangle-on-parabola.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-397-triangle-on-parabola.md
index 4f6b98a69c3..d8d95c2dd3a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-397-triangle-on-parabola.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-397-triangle-on-parabola.md
@@ -1,22 +1,22 @@
---
id: 5900f4f91000cf542c51000c
-title: 问题397：抛物线上的三角形
+title: 'Problem 397: Triangle on parabola'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302062
dashedName: problem-397-triangle-on-parabola
---
# --description--
-在抛物线y = x2 / k上，选择三个点A（a，a2 / k），B（b，b2 / k）和C（c，c2 / k）。
+On the parabola y = x2/k, three points A(a, a2/k), B(b, b2/k) and C(c, c2/k) are chosen.
-令F（K，X）为整数四元组（k，a，b，c）的数量，使得三角形ABC的至少一个角度为45度，其中1≤k≤K且-X≤a< b <c≤X。
+Let F(K, X) be the number of the integer quadruplets (k, a, b, c) such that at least one angle of the triangle ABC is 45-degree, with 1 ≤ k ≤ K and -X ≤ a < b < c ≤ X.
-例如，F（1,10）= 41并且F（10,100）= 12492.找到F（106,109）。
+For example, F(1, 10) = 41 and F(10, 100) = 12492. Find F(106, 109).
# --hints--
-`euler397()`应该返回141630459461893730。
+`euler397()` should return 141630459461893730.
```js
assert.strictEqual(euler397(), 141630459461893730);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-398-cutting-rope.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-398-cutting-rope.md
index e009b5a6a4b..4b145fdfbc2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-398-cutting-rope.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-398-cutting-rope.md
@@ -1,22 +1,22 @@
---
id: 5900f4fa1000cf542c51000d
-title: 问题398：切割绳索
+title: 'Problem 398: Cutting rope'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302063
dashedName: problem-398-cutting-rope
---
# --description--
-在长度为n的绳索内，放置n-1个点，彼此之间的距离为1，并且从端点开始。在这些点中，我们随机选择m-1个点并在这些点处切断绳索以创建m个段。
+Inside a rope of length n, n-1 points are placed with distance 1 from each other and from the endpoints. Among these points, we choose m-1 points at random and cut the rope at these points to create m segments.
-设E（n，m）为第二最短段的预期长度。例如，E（3,2）= 2且E（8,3）= 16/7。注意，如果多个段具有相同的最短长度，则将第二最短段的长度定义为与最短长度相同。
+Let E(n, m) be the expected length of the second-shortest segment. For example, E(3, 2) = 2 and E(8, 3) = 16/7. Note that if multiple segments have the same shortest length the length of the second-shortest segment is defined as the same as the shortest length.
-找到E（107,100）。将您的答案四舍五入到小数点后面的小数点后5位。
+Find E(107, 100). Give your answer rounded to 5 decimal places behind the decimal point.
# --hints--
-`euler398()`应返回2010.59096。
+`euler398()` should return 2010.59096.
```js
assert.strictEqual(euler398(), 2010.59096);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-399-squarefree-fibonacci-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-399-squarefree-fibonacci-numbers.md
index 03b126fd992..b61d13a3365 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-399-squarefree-fibonacci-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-399-squarefree-fibonacci-numbers.md
@@ -1,26 +1,34 @@
---
id: 5900f4fc1000cf542c51000e
-title: 问题399：无自由斐波纳契数
+title: 'Problem 399: Squarefree Fibonacci Numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302064
dashedName: problem-399-squarefree-fibonacci-numbers
---
# --description--
-前15个斐波纳契数是：1,1,2,3,5,8,13,21,34,55,89,144,233,377,610。可以看出8和144不是无方形的：8可以被4整除，144可以被4和9整除。所以前13个无方形的斐波纳契数是：1,1,2,3,5,13,21， 34,55,89,233,377和610。
+The first 15 fibonacci numbers are:
-第200平方免费斐波那契数是：971183874599339129547649988289594072811608739584170445。该数字的最后16位数字是：1608739584170445，并且在科学记数法中，该数字可以写为9.7e53。
+1,1,2,3,5,8,13,21,34,55,89,144,233,377,610.
-找到第100 000个squarefree fibonacci数。给出你的答案，它的最后十六位数后跟一个逗号，后跟科学记数法的数字（四舍五入到小数点后的一位数）。对于第200平方免费数字，答案应该是：1608739584170445,9.7e53
+It can be seen that 8 and 144 are not squarefree: 8 is divisible by 4 and 144 is divisible by 4 and by 9.
-注意：对于这个问题，假设对于每个素数p，可被p整除的第一个斐波纳契数不能被p2整除（这是沃尔猜想的一部分）。这已被证实适用于≤3·1015的质数，但一般尚未得到证实。
+So the first 13 squarefree fibonacci numbers are:
-如果猜测是假的，那么这个问题的接受答案不能保证是第1万个无平方的斐波纳契数，而只是它代表了该数的下限。
+1,1,2,3,5,13,21,34,55,89,233,377 and 610.
+
+The 200th squarefree fibonacci number is: 971183874599339129547649988289594072811608739584170445. The last sixteen digits of this number are: 1608739584170445 and in scientific notation this number can be written as 9.7e53.
+
+Find the 100 000 000th squarefree fibonacci number. Give as your answer its last sixteen digits followed by a comma followed by the number in scientific notation (rounded to one digit after the decimal point). For the 200th squarefree number the answer would have been: 1608739584170445,9.7e53
+
+Note: For this problem, assume that for every prime p, the first fibonacci number divisible by p is not divisible by p2 (this is part of Wall's conjecture). This has been verified for primes ≤ 3·1015, but has not been proven in general.
+
+If it happens that the conjecture is false, then the accepted answer to this problem isn't guaranteed to be the 100 000 000th squarefree fibonacci number, rather it represents only a lower bound for that number.
# --hints--
-`euler399()`应返回1508395636674243,6.5e27330467。
+`euler399()` should return 1508395636674243, 6.5e27330467.
```js
assert.strictEqual(euler399(), 1508395636674243, 6.5e27330467);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-4-largest-palindrome-product.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-4-largest-palindrome-product.md
index fea807309cb..34dbf96399d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-4-largest-palindrome-product.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-4-largest-palindrome-product.md
@@ -1,24 +1,32 @@
---
id: 5900f3701000cf542c50fe83
-title: 问题4：最大的回文产品
+title: 'Problem 4: Largest palindrome product'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302065
dashedName: problem-4-largest-palindrome-product
---
# --description--
-回文数字读取两种方式相同。由两个2位数字的乘积制成的最大回文是9009 = 91×99。找到由两个`n`数字的乘积制成的最大回文。
+A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
+
+Find the largest palindrome made from the product of two `n`-digit numbers.
# --hints--
-`largestPalindromeProduct(2)`应返回9009。
+`largestPalindromeProduct(2)` should return a number.
+
+```js
+assert(typeof largestPalindromeProduct(2) === 'number');
+```
+
+`largestPalindromeProduct(2)` should return 9009.
```js
assert.strictEqual(largestPalindromeProduct(2), 9009);
```
-`largestPalindromeProduct(3)`应返回906609。
+`largestPalindromeProduct(3)` should return 906609.
```js
assert.strictEqual(largestPalindromeProduct(3), 906609);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-40-champernownes-constant.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-40-champernownes-constant.md
index aeb1893e9c6..f7f6c8f51fb 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-40-champernownes-constant.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-40-champernownes-constant.md
@@ -1,30 +1,44 @@
---
id: 5900f3941000cf542c50fea7
-title: 问题40：Champernowne的常数
+title: 'Problem 40: Champernowne''s constant'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302066
dashedName: problem-40-champernownes-constant
---
# --description--
-通过连接正整数创建无理小数： 0.12345678910 **1** 112131415161718192021 ...可以看出小数部分的<sup>第</sup> 12位是1.如果*d <sub>n</sub>*代表小数部分的<sup>第</sup> *n*位，找到值以下表达式。 d <sub>1</sub> ×d <sub>10</sub> ×d <sub>100</sub> ×d <sub>1000</sub> ×d <sub>10000</sub> ×d <sub>100000</sub> ×d <sub>1000000</sub>
+An irrational decimal fraction is created by concatenating the positive integers:
+
+0.12345678910**1**112131415161718192021...
+
+It can be seen that the 12<sup>th</sup> digit of the fractional part is 1.
+
+If *d<sub>n</sub>* represents the *n*<sup>th</sup> digit of the fractional part, find the value of the following expression.
+
+d<sub>1</sub> × d<sub>10</sub> × d<sub>100</sub> × d<sub>1000</sub> × d<sub>10000</sub> × d<sub>100000</sub> × d<sub>1000000</sub>
# --hints--
-`champernownesConstant(100)`应该返回5。
+`champernownesConstant(100)` should return a number.
+
+```js
+assert(typeof champernownesConstant(100) === 'number');
+```
+
+`champernownesConstant(100)` should return 5.
```js
assert.strictEqual(champernownesConstant(100), 5);
```
-`champernownesConstant(1000)`应该返回15。
+`champernownesConstant(1000)` should return 15.
```js
assert.strictEqual(champernownesConstant(1000), 15);
```
-`champernownesConstant(1000000)`应该返回210。
+`champernownesConstant(1000000)` should return 210.
```js
assert.strictEqual(champernownesConstant(1000000), 210);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-400-fibonacci-tree-game.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-400-fibonacci-tree-game.md
index b89b1050c64..07e767a5ba6 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-400-fibonacci-tree-game.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-400-fibonacci-tree-game.md
@@ -1,28 +1,32 @@
---
id: 5900f4fe1000cf542c510010
-title: 问题400：斐波纳契树游戏
+title: 'Problem 400: Fibonacci tree game'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302067
dashedName: problem-400-fibonacci-tree-game
---
# --description--
-Fibonacci树是递归定义为的二叉树：T（0）是空树。 T（1）是只有一个节点的二叉树。 T（k）由具有T（k-1）和T（k-2）作为子节点的根节点组成。
+A Fibonacci tree is a binary tree recursively defined as:T(0) is the empty tree.
-在这样的树上，两个玩家玩外卖游戏。在每个回合中，玩家选择一个节点并删除该节点以及以该节点为根的子树。被迫占用整棵树根节点的玩家输了。
+T(1) is the binary tree with only one node.
-以下是T（k）从k = 1到k = 6的第一个回合的第一个玩家的获胜动作。
+T(k) consists of a root node that has T(k-1) and T(k-2) as children.
-假设f（k）是在T（k）上播放该游戏时在第一轮游戏中第一玩家的获胜移动的数量（即，第二玩家没有获胜策略的移动）。
+On such a tree two players play a take-away game. On each turn a player selects a node and removes that node along with the subtree rooted at that node. The player who is forced to take the root node of the entire tree loses.
-例如，f（5）= 1且f（10）= 17。
+Here are the winning moves of the first player on the first turn for T(k) from k=1 to k=6.
-找到f（10000）。给出答案的最后18位数字。
+Let f(k) be the number of winning moves of the first player (i.e. the moves for which the second player has no winning strategy) on the first turn of the game when this game is played on T(k).
+
+For example, f(5) = 1 and f(10) = 17.
+
+Find f(10000). Give the last 18 digits of your answer.
# --hints--
-`euler400()`应返回438505383468410600。
+`euler400()` should return 438505383468410600.
```js
assert.strictEqual(euler400(), 438505383468410600);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-401-sum-of-squares-of-divisors.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-401-sum-of-squares-of-divisors.md
index 59380877d03..d1c5ca48cc8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-401-sum-of-squares-of-divisors.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-401-sum-of-squares-of-divisors.md
@@ -1,24 +1,26 @@
---
id: 5900f4fd1000cf542c51000f
-title: 问题401：除数的平方和
+title: 'Problem 401: Sum of squares of divisors'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302069
dashedName: problem-401-sum-of-squares-of-divisors
---
# --description--
-6的除数是1,2,3和6.这些数的平方和是1 + 4 + 9 + 36 = 50。
+The divisors of 6 are 1,2,3 and 6.
-设sigma2（n）代表n的除数的平方和。因此sigma2（6）= 50。
+The sum of the squares of these numbers is 1+4+9+36=50.
-设SIGMA2表示sigma2的和函数，即对于i = 1到n，SIGMA2（n）=Σsigma2（i）。 SIGMA2的前6个值是：1,6,16,37,63和113。
+Let sigma2(n) represent the sum of the squares of the divisors of n. Thus sigma2(6)=50.
-找到SIGMA2（1015）modulo 109。
+Let SIGMA2 represent the summatory function of sigma2, that is SIGMA2(n)=∑sigma2(i) for i=1 to n. The first 6 values of SIGMA2 are: 1,6,16,37,63 and 113.
+
+Find SIGMA2(1015) modulo 109.
# --hints--
-`euler401()`应该返回281632621。
+`euler401()` should return 281632621.
```js
assert.strictEqual(euler401(), 281632621);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-402-integer-valued-polynomials.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-402-integer-valued-polynomials.md
index 4c6f8385465..b4c4a63fe8d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-402-integer-valued-polynomials.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-402-integer-valued-polynomials.md
@@ -1,28 +1,28 @@
---
id: 5900f4ff1000cf542c510011
-title: 问题402：整数值多项式
+title: 'Problem 402: Integer-valued polynomials'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302070
dashedName: problem-402-integer-valued-polynomials
---
# --description--
-可以证明，对于每个整数n，多项式n4 + 4n3 + 2n2 + 5n是6的倍数。还可以显示6是满足该属性的最大整数。
+It can be shown that the polynomial n4 + 4n3 + 2n2 + 5n is a multiple of 6 for every integer n. It can also be shown that 6 is the largest integer satisfying this property.
-将M（a，b，c）定义为最大m，使得n4 + an3 + bn2 + cn是所有整数n的m的倍数。例如，M（4,2,5）= 6。
+Define M(a, b, c) as the maximum m such that n4 + an3 + bn2 + cn is a multiple of m for all integers n. For example, M(4, 2, 5) = 6.
-此外，将S（N）定义为所有0 <a，b，c≤N的M（a，b，c）之和。
+Also, define S(N) as the sum of M(a, b, c) for all 0 < a, b, c ≤ N.
-我们可以验证S（10）= 1972和S（10000）= 2024258331114。
+We can verify that S(10) = 1972 and S(10000) = 2024258331114.
-设Fk为斐波纳契数列：对于k≥2，F0 = 0，F1 = 1且Fk = Fk-1 + Fk-2。
+Let Fk be the Fibonacci sequence: F0 = 0, F1 = 1 and Fk = Fk-1 + Fk-2 for k ≥ 2.
-求最高9位数为ΣS（Fk）为2≤k≤1234567890123。
+Find the last 9 digits of Σ S(Fk) for 2 ≤ k ≤ 1234567890123.
# --hints--
-`euler402()`应返回356019862。
+`euler402()` should return 356019862.
```js
assert.strictEqual(euler402(), 356019862);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-403-lattice-points-enclosed-by-parabola-and-line.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-403-lattice-points-enclosed-by-parabola-and-line.md
index 93d10903564..d854a56f4ac 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-403-lattice-points-enclosed-by-parabola-and-line.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-403-lattice-points-enclosed-by-parabola-and-line.md
@@ -1,24 +1,24 @@
---
id: 5900f5001000cf542c510013
-title: 问题403：由抛物线和直线包围的格点
+title: 'Problem 403: Lattice points enclosed by parabola and line'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302071
dashedName: problem-403-lattice-points-enclosed-by-parabola-and-line
---
# --description--
-对于整数a和b，我们将D（a，b）定义为由抛物线y = x2包围的域，并且线y = a·x + b：D（a，b）= {（x，y）| x2≤y≤a·x + b}。
+For integers a and b, we define D(a, b) as the domain enclosed by the parabola y = x2 and the line y = a·x + b:D(a, b) = { (x, y) | x2 ≤ y ≤ a·x + b }.
-L（a，b）定义为D（a，b）中包含的晶格点数。例如，L（1,2）= 8并且L（2，-1）= 1。
+L(a, b) is defined as the number of lattice points contained in D(a, b). For example, L(1, 2) = 8 and L(2, -1) = 1.
-我们还将S（N）定义为所有对（a，b）的L（a，b）之和，使得D（a，b）的面积是有理数并且| a |，| b | ≤N。我们可以验证S（5）= 344和S（100）= 26709528。
+We also define S(N) as the sum of L(a, b) for all the pairs (a, b) such that the area of D(a, b) is a rational number and |a|,|b| ≤ N. We can verify that S(5) = 344 and S(100) = 26709528.
-找到S（1012）。给你的答案mod 108。
+Find S(1012). Give your answer mod 108.
# --hints--
-`euler403()`应该返回18224771。
+`euler403()` should return 18224771.
```js
assert.strictEqual(euler403(), 18224771);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-404-crisscross-ellipses.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-404-crisscross-ellipses.md
index 410e876675d..6b7341cea49 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-404-crisscross-ellipses.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-404-crisscross-ellipses.md
@@ -1,24 +1,26 @@
---
id: 5900f5001000cf542c510012
-title: 问题404：交叉椭圆
+title: 'Problem 404: Crisscross Ellipses'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302072
dashedName: problem-404-crisscross-ellipses
---
# --description--
-Ea是椭圆，其形式为x2 + 4y2 = 4a2。 Ea'是Ea的旋转图像，绕原点O（0,0）逆时针旋转θ度0°<θ<90°。
+Ea is an ellipse with an equation of the form x2 + 4y2 = 4a2.
-b是距离原点最近的两个交点的原点的距离，c是另外两个交点的距离。如果a，b和c是正整数，我们称有序三元组（a，b，c）为规范椭圆体三元组。例如，（209,247,286）是规范的椭圆体三重态。
+Ea' is the rotated image of Ea by θ degrees counterclockwise around the origin O(0, 0) for 0° < θ < 90°.
-令C（N）为a≤N的不同规范椭球三胞胎（a，b，c）的数量。可以证实C（103）= 7，C（104）= 106和C（106）= 11845 。
+b is the distance to the origin of the two intersection points closest to the origin and c is the distance of the two other intersection points. We call an ordered triplet (a, b, c) a canonical ellipsoidal triplet if a, b and c are positive integers. For example, (209, 247, 286) is a canonical ellipsoidal triplet.
-找到C（1017）。
+Let C(N) be the number of distinct canonical ellipsoidal triplets (a, b, c) for a ≤ N. It can be verified that C(103) = 7, C(104) = 106 and C(106) = 11845.
+
+Find C(1017).
# --hints--
-`euler404()`应该返回1199215615081353。
+`euler404()` should return 1199215615081353.
```js
assert.strictEqual(euler404(), 1199215615081353);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-405-a-rectangular-tiling.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-405-a-rectangular-tiling.md
index ec703c6bb22..87f1f0c28f1 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-405-a-rectangular-tiling.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-405-a-rectangular-tiling.md
@@ -1,24 +1,28 @@
---
id: 5900f5021000cf542c510014
-title: 问题405：矩形平铺
+title: 'Problem 405: A rectangular tiling'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302073
dashedName: problem-405-a-rectangular-tiling
---
# --description--
-我们希望平铺一个长度是其宽度两倍的矩形。设T（0）为由单个矩形组成的平铺。对于n> 0，通过以下列方式替换所有瓦片，从T（n-1）获得T（n）：
+We wish to tile a rectangle whose length is twice its width.
-以下动画演示了从0到5的n的倾斜T（n）：
+Let T(0) be the tiling consisting of a single rectangle.
-设f（n）是四个瓦片在T（n）中相交的点数。例如，f（1）= 0，f（4）= 82并且f（109）mod 177 = 126897180。
+For n > 0, let T(n) be obtained from T(n-1) by replacing all tiles in the following manner:
-找到f（10k）为k = 1018，给出你的答案模数为177。
+The following animation demonstrates the tilings T(n) for n from 0 to 5:
+
+Let f(n) be the number of points where four tiles meet in T(n). For example, f(1) = 0, f(4) = 82 and f(109) mod 177 = 126897180.
+
+Find f(10k) for k = 1018, give your answer modulo 177.
# --hints--
-`euler405()`应返回237696125。
+`euler405()` should return 237696125.
```js
assert.strictEqual(euler405(), 237696125);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-406-guessing-game.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-406-guessing-game.md
index cf530579ae3..fd6535eb66d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-406-guessing-game.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-406-guessing-game.md
@@ -1,28 +1,36 @@
---
id: 5900f5021000cf542c510015
-title: 问题406：猜猜游戏
+title: 'Problem 406: Guessing Game'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302074
dashedName: problem-406-guessing-game
---
# --description--
-我们试图通过提问来找到从整数集{1,2，...，n}中选择的隐藏数字。我们问的每个数字（问题），我们得到三个可能的答案之一：“你的猜测低于隐藏的数字”（并且你需要花费一个成本），或者“你的猜测高于隐藏的数字”（和你承担b）的费用，或“是的，就是这样！” （游戏结束）。给定n，a和b的值，最优策略最小化最坏情况下的总成本。
+We are trying to find a hidden number selected from the set of integers {1, 2, ..., n} by asking questions.
-例如，如果n = 5，a = 2，b = 3，那么我们可以先问“2”作为我们的第一个问题。
+Each number (question) we ask, we get one of three possible answers: "Your guess is lower than the hidden number" (and you incur a cost of a), or
-如果我们被告知2高于隐藏号码（b = 3的成本），那么我们确定“1”是隐藏号码（总成本为3）。如果我们被告知2低于隐藏号码（a = 2的成本），那么我们的下一个问题将是“4”。如果我们被告知4高于隐藏号码（b = 3的成本），那么我们确定“3”是隐藏号码（总成本为2 + 3 = 5）。如果我们被告知4低于隐藏号码（a = 2的成本），那么我们确定“5”是隐藏号码（总成本为2 + 2 = 4）。因此，该策略实现的最坏情况成本为5.还可以证明这是可以实现的最低的最坏情况成本。所以，事实上，我们刚刚描述了给定n，a和b值的最优策略。
+"Your guess is higher than the hidden number" (and you incur a cost of b), or
-设C（n，a，b）是针对给定n，a和b值的最优策略实现的最坏情况成本。
+"Yes, that's it!" (and the game ends).
-以下是几个例子：C（5,2,3）= 5 C（500，√2，√3）= 13.22073197 ... C（20000,5,7）= 82 C（2000000，√5，√7 ）= 49.63755955 ......
+Given the value of n, a, and b, an optimal strategy minimizes the total cost for the worst possible case.
-设Fk为斐波纳契数：Fk = Fk-1 + Fk-2，基本情况F1 = F2 =1.FindΣ1≤k≤30C（1012，√k，√Fk），并将答案四舍五入为8小数点后面的小数位数。
+For example, if n = 5, a = 2, and b = 3, then we may begin by asking "2" as our first question.
+
+If we are told that 2 is higher than the hidden number (for a cost of b=3), then we are sure that "1" is the hidden number (for a total cost of 3). If we are told that 2 is lower than the hidden number (for a cost of a=2), then our next question will be "4". If we are told that 4 is higher than the hidden number (for a cost of b=3), then we are sure that "3" is the hidden number (for a total cost of 2+3=5). If we are told that 4 is lower than the hidden number (for a cost of a=2), then we are sure that "5" is the hidden number (for a total cost of 2+2=4). Thus, the worst-case cost achieved by this strategy is 5. It can also be shown that this is the lowest worst-case cost that can be achieved. So, in fact, we have just described an optimal strategy for the given values of n, a, and b.
+
+Let C(n, a, b) be the worst-case cost achieved by an optimal strategy for the given values of n, a, and b.
+
+Here are a few examples: C(5, 2, 3) = 5 C(500, √2, √3) = 13.22073197... C(20000, 5, 7) = 82 C(2000000, √5, √7) = 49.63755955...
+
+Let Fk be the Fibonacci numbers: Fk = Fk-1 + Fk-2 with base cases F1 = F2 = 1.Find ∑1≤k≤30 C(1012, √k, √Fk), and give your answer rounded to 8 decimal places behind the decimal point.
# --hints--
-`euler406()`应返回36813.12757207。
+`euler406()` should return 36813.12757207.
```js
assert.strictEqual(euler406(), 36813.12757207);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-407-idempotents.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-407-idempotents.md
index 5e1a942cbf3..eb473f7ba0b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-407-idempotents.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-407-idempotents.md
@@ -1,22 +1,22 @@
---
id: 5900f5041000cf542c510016
-title: 问题407：幂等元素
+title: 'Problem 407: Idempotents'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302075
dashedName: problem-407-idempotents
---
# --description--
-如果我们计算a2 mod 6为0≤a≤5，我们得到：0,1,4,3,4,1。
+If we calculate a2 mod 6 for 0 ≤ a ≤ 5 we get: 0,1,4,3,4,1.
-a2≡amod 6的最大值为4.让我们将M（n）称为a <n的最大值，使得a2≡a（mod n）。所以M（6）= 4。
+The largest value of a such that a2 ≡ a mod 6 is 4. Let's call M(n) the largest value of a < n such that a2 ≡ a (mod n). So M(6) = 4.
-找ΣM（n）为1≤n≤107。
+Find ∑M(n) for 1 ≤ n ≤ 107.
# --hints--
-`euler407()`应该返回39782849136421。
+`euler407()` should return 39782849136421.
```js
assert.strictEqual(euler407(), 39782849136421);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-408-admissible-paths-through-a-grid.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-408-admissible-paths-through-a-grid.md
index 5ac4f9a6ae9..0c3decbbe7c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-408-admissible-paths-through-a-grid.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-408-admissible-paths-through-a-grid.md
@@ -1,24 +1,26 @@
---
id: 5900f5091000cf542c51001b
-title: 问题408：通过网格的可允许路径
+title: 'Problem 408: Admissible paths through a grid'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302076
dashedName: problem-408-admissible-paths-through-a-grid
---
# --description--
-如果x，y和x + y都是正的正方形，那么我们称格子点（x，y）是不允许的。例如，（9,16）是不允许的，而（0,4），（3,1）和（9,4）则不允许。
+Let's call a lattice point (x, y) inadmissible if x, y and x + y are all positive perfect squares.
-考虑从点（x1，y1）到点（x2，y2）的路径，仅使用北或东的单位步长。如果其中间点都不允许，我们可以称这样的路径是可以接受的。
+For example, (9, 16) is inadmissible, while (0, 4), (3, 1) and (9, 4) are not.
-令P（n）是从（0,0）到（n，n）的可允许路径的数量。可以证实P（5）= 252，P（16）= 596994440和P（1000）mod 1 000 000 007 = 341920854。
+Consider a path from point (x1, y1) to point (x2, y2) using only unit steps north or east. Let's call such a path admissible if none of its intermediate points are inadmissible.
-求P（10 000 000）mod 1 000 000 007。
+Let P(n) be the number of admissible paths from (0, 0) to (n, n). It can be verified that P(5) = 252, P(16) = 596994440 and P(1000) mod 1 000 000 007 = 341920854.
+
+Find P(10 000 000) mod 1 000 000 007.
# --hints--
-`euler408()`应该返回299742733。
+`euler408()` should return 299742733.
```js
assert.strictEqual(euler408(), 299742733);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-409-nim-extreme.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-409-nim-extreme.md
index fc2fb3bc80d..d6dbeffce98 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-409-nim-extreme.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-409-nim-extreme.md
@@ -1,20 +1,28 @@
---
id: 5900f5061000cf542c510017
-title: 问题409：Nim Extreme
+title: 'Problem 409: Nim Extreme'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302077
dashedName: problem-409-nim-extreme
---
# --description--
-设n是正整数。考虑nim位置：有n个非空桩。每堆的尺寸小于2n。没有两个桩具有相同的尺寸。设W（n）是满足上述条件的获胜nim位置的数量（如果第一个玩家具有获胜策略，则获胜的位置）。例如，W（1）= 1，W（2）= 6，W（3）= 168，W（5）= 19764360，W（100）mod 1 000 000 007 = 384777056。
+Let n be a positive integer. Consider nim positions where:There are n non-empty piles.
-求W（10 000 000）mod 1 000 000 007。
+Each pile has size less than 2n.
+
+No two piles have the same size.
+
+Let W(n) be the number of winning nim positions satisfying the above
+
+conditions (a position is winning if the first player has a winning strategy). For example, W(1) = 1, W(2) = 6, W(3) = 168, W(5) = 19764360 and W(100) mod 1 000 000 007 = 384777056.
+
+Find W(10 000 000) mod 1 000 000 007.
# --hints--
-`euler409()`应该返回253223948。
+`euler409()` should return 253223948.
```js
assert.strictEqual(euler409(), 253223948);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-41-pandigital-prime.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-41-pandigital-prime.md
index aa3d5877fc4..41465478fc3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-41-pandigital-prime.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-41-pandigital-prime.md
@@ -1,24 +1,32 @@
---
id: 5900f3951000cf542c50fea8
-title: 问题41：Pandigital prime
+title: 'Problem 41: Pandigital prime'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302078
dashedName: problem-41-pandigital-prime
---
# --description--
-我们将说*n*数字是pandigital，如果它使用所有数字1到*n*恰好一次。例如，2143是一个4位数的pandigital，也是素数。什么是最大的*n长*数字pandigital素数？
+We shall say that an `n`-digit number is pandigital if it makes use of all the digits 1 to `n` exactly once. For example, 2143 is a 4-digit pandigital and is also prime.
+
+What is the largest `n`-length digit pandigital prime that exists?
# --hints--
-`pandigitalPrime(4)`应该返回4231。
+`pandigitalPrime(4)` should return a number.
+
+```js
+assert(typeof pandigitalPrime(4) === 'number');
+```
+
+`pandigitalPrime(4)` should return 4231.
```js
assert(pandigitalPrime(4) == 4231);
```
-`pandigitalPrime(7)`应该返回7652413。
+`pandigitalPrime(7)` should return 7652413.
```js
assert(pandigitalPrime(7) == 7652413);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-410-circle-and-tangent-line.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-410-circle-and-tangent-line.md
index e23de43437f..cd55b71cbc0 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-410-circle-and-tangent-line.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-410-circle-and-tangent-line.md
@@ -1,24 +1,24 @@
---
id: 5900f5071000cf542c510018
-title: 问题410：圆和切线
+title: 'Problem 410: Circle and tangent line'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302079
dashedName: problem-410-circle-and-tangent-line
---
# --description--
-设C为半径为r的圆，x2 + y2 = r2。我们选择两个点P（a，b）和Q（-a，c），使得穿过P和Q的线与C相切。
+Let C be the circle with radius r, x2 + y2 = r2. We choose two points P(a, b) and Q(-a, c) so that the line passing through P and Q is tangent to C.
-例如，四元组（r，a，b，c）=（2,6,2，-7）满足该属性。
+For example, the quadruplet (r, a, b, c) = (2, 6, 2, -7) satisfies this property.
-令F（R，X）为具有该性质的整数四元组（r，a，b，c）的数量，并且0 <r≤R且0 <a≤X。
+Let F(R, X) be the number of the integer quadruplets (r, a, b, c) with this property, and with 0 < r ≤ R and 0 < a ≤ X.
-我们可以验证F（1,5）= 10，F（2,10）= 52和F（10,100）= 3384.求F（108,109）+ F（109,108）。
+We can verify that F(1, 5) = 10, F(2, 10) = 52 and F(10, 100) = 3384. Find F(108, 109) + F(109, 108).
# --hints--
-`euler410()`应该返回799999783589946600。
+`euler410()` should return 799999783589946600.
```js
assert.strictEqual(euler410(), 799999783589946600);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-411-uphill-paths.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-411-uphill-paths.md
index b535dbddc6b..db9dd2cb866 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-411-uphill-paths.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-411-uphill-paths.md
@@ -1,26 +1,26 @@
---
id: 5900f5081000cf542c510019
-title: 问题411：上坡路径
+title: 'Problem 411: Uphill paths'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302080
dashedName: problem-411-uphill-paths
---
# --description--
-设n是正整数。假设坐标（x，y）=（2i mod n，3i mod n）的站点为0≤i≤2n。我们将考虑与同一站点具有相同坐标的站点。
+Let n be a positive integer. Suppose there are stations at the coordinates (x, y) = (2i mod n, 3i mod n) for 0 ≤ i ≤ 2n. We will consider stations with the same coordinates as the same station.
-我们希望形成从（0,0）到（n，n）的路径，使得x和y坐标永不减少。设S（n）是路径可以通过的最大站数。
+We wish to form a path from (0, 0) to (n, n) such that the x and y coordinates never decrease. Let S(n) be the maximum number of stations such a path can pass through.
-例如，如果n = 22，则有11个不同的站，并且有效路径最多可以通过5个站。因此，S（22）= 5.下面举例说明了一个最佳路径的例子：
+For example, if n = 22, there are 11 distinct stations, and a valid path can pass through at most 5 stations. Therefore, S(22) = 5. The case is illustrated below, with an example of an optimal path:
-还可以证实S（123）= 14并且S（10000）= 48。
+It can also be verified that S(123) = 14 and S(10000) = 48.
-求1Σk≤30，ΣS（k5）。
+Find ∑ S(k5) for 1 ≤ k ≤ 30.
# --hints--
-`euler411()`应返回9936352。
+`euler411()` should return 9936352.
```js
assert.strictEqual(euler411(), 9936352);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-412-gnomon-numbering.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-412-gnomon-numbering.md
index 53d91bce348..f0dfab1bf60 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-412-gnomon-numbering.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-412-gnomon-numbering.md
@@ -1,28 +1,28 @@
---
id: 5900f5081000cf542c51001a
-title: 问题412：Gnomon编号
+title: 'Problem 412: Gnomon numbering'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302081
dashedName: problem-412-gnomon-numbering
---
# --description--
-对于整数m，n（0≤n<m），令L（m，n）为m×m网格，其中右上n×n网格被移除。
+For integers m, n (0 ≤ n < m), let L(m, n) be an m×m grid with the top-right n×n grid removed.
-例如，L（5,3）看起来像这样：
+For example, L(5, 3) looks like this:
-我们希望用连续的整数1,2,3 ......对L（m，n）的每个单元进行编号，使得每个单元格中的数字小于它下面和左边的数字。
+We want to number each cell of L(m, n) with consecutive integers 1, 2, 3, ... such that the number in every cell is smaller than the number below it and to the left of it.
-例如，这里有两个有效的L（5,3）编号：
+For example, here are two valid numberings of L(5, 3):
-设LC（m，n）为L（m，n）的有效编号。可以证实LC（3,0）= 42，LC（5,3）= 250250，LC（6,3）= 406029023400和LC（10,5）mod 76543217 = 61251715。
+Let LC(m, n) be the number of valid numberings of L(m, n). It can be verified that LC(3, 0) = 42, LC(5, 3) = 250250, LC(6, 3) = 406029023400 and LC(10, 5) mod 76543217 = 61251715.
-找到LC（10000,5000）mod 76543217。
+Find LC(10000, 5000) mod 76543217.
# --hints--
-`euler412()`应该返回38788800。
+`euler412()` should return 38788800.
```js
assert.strictEqual(euler412(), 38788800);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-413-one-child-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-413-one-child-numbers.md
index 282f693a44b..a3a7ac23c57 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-413-one-child-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-413-one-child-numbers.md
@@ -1,24 +1,24 @@
---
id: 5900f50a1000cf542c51001c
-title: 问题413：独生子女号码
+title: 'Problem 413: One-child Numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302082
dashedName: problem-413-one-child-numbers
---
# --description--
-我们说如果其子串中的一个子串可被d整除，则d位正数（无前导零）是一个子数。
+We say that a d-digit positive number (no leading zeros) is a one-child number if exactly one of its sub-strings is divisible by d.
-例如，5671是一个4位数的单子号码。在其所有子串5,6,7,1,56,67,71,567,671和5671中，只有56可被4整除。类似地，104是3位单子数，因为只有0可被整除3. 1132451是一个7位数的单子号码，因为只有245可被7整除。
+For example, 5671 is a 4-digit one-child number. Among all its sub-strings 5, 6, 7, 1, 56, 67, 71, 567, 671 and 5671, only 56 is divisible by 4. Similarly, 104 is a 3-digit one-child number because only 0 is divisible by 3. 1132451 is a 7-digit one-child number because only 245 is divisible by 7.
-设F（N）是小于N的一子数的数。我们可以验证F（10）= 9，F（103）= 389和F（107）= 277674。
+Let F(N) be the number of the one-child numbers less than N. We can verify that F(10) = 9, F(103) = 389 and F(107) = 277674.
-找到F（1019）。
+Find F(1019).
# --hints--
-`euler413()`应该返回3079418648040719。
+`euler413()` should return 3079418648040719.
```js
assert.strictEqual(euler413(), 3079418648040719);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-414-kaprekar-constant.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-414-kaprekar-constant.md
index 9e2aa4bf335..9cf45b79ab3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-414-kaprekar-constant.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-414-kaprekar-constant.md
@@ -1,30 +1,40 @@
---
id: 5900f50b1000cf542c51001d
-title: 问题414：标题常量
+title: 'Problem 414: Kaprekar constant'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302083
dashedName: problem-414-kaprekar-constant
---
# --description--
-6174是一个了不起的数字;如果我们按递增顺序对其数字进行排序，并从按降序排序数字时得到的数字中减去该数字，我们得到7641-1467 = 6174。更值得注意的是，如果我们从任何4位数字开始并重复这个排序和减法过程，我们最终将以6174结束或者如果所有数字相等则立即结束0。如果我们用前导零填充数字直到我们有4位数，这也适用于少于4位的数字。例如，让我们从数字0837开始：8730-0378 = 8352 8532-2358 = 6174
+6174 is a remarkable number; if we sort its digits in increasing order and subtract that number from the number you get when you sort the digits in decreasing order, we get 7641-1467=6174.
-6174被称为Kaprekar常数。排序和减去并重复这一过程直到0或达到Kaprekar常数的过程称为Kaprekar例程。
+Even more remarkable is that if we start from any 4 digit number and repeat this process of sorting and subtracting, we'll eventually end up with 6174 or immediately with 0 if all digits are equal.
-我们可以考虑其他基数和位数的Kaprekar例程。不幸的是，并不能保证在所有情况下都存在Kaprekar常数;例程可以在某个输入数字的循环中结束，或者例程到达的常数对于不同的输入数字可以是不同的。然而，可以证明，对于5位数并且基数b = 6t + 3≠9，存在Kaprekar常数。例如，基数15：（10,4,14,9,5）15基数21：（14,6,20,13,7）21
+This also works with numbers that have less than 4 digits if we pad the number with leading zeroes until we have 4 digits.
-将Cb定义为基数b中的Kaprekar常数，为5位数。如果i = Cb，则将函数sb（i）定义为0，或者如果i写入基数b，则由5个相同的数字组成，基数b中的Kaprekar例程到达Cb所需的迭代次数，否则
+E.g. let's start with the number 0837:
-注意，我们可以为所有整数i <b5定义sb（i）。如果我在基数b中写入少于5位数，则在应用Kaprekar例程之前，我们有5位数字，前面加零数字。
+8730-0378=8352
-将S（b）定义为0 <i <b5的sb（i）之和。例如S（15）= 5274369 S（111）= 400668930299
+8532-2358=6174
-找到S（6k + 3）的总和2≤k≤300。给出最后18位数字作为答案。
+6174 is called the Kaprekar constant. The process of sorting and subtracting and repeating this until either 0 or the Kaprekar constant is reached is called the Kaprekar routine.
+
+We can consider the Kaprekar routine for other bases and number of digits. Unfortunately, it is not guaranteed a Kaprekar constant exists in all cases; either the routine can end up in a cycle for some input numbers or the constant the routine arrives at can be different for different input numbers. However, it can be shown that for 5 digits and a base b = 6t+3≠9, a Kaprekar constant exists. E.g. base 15: (10,4,14,9,5)15 base 21: (14,6,20,13,7)21
+
+Define Cb to be the Kaprekar constant in base b for 5 digits. Define the function sb(i) to be 0 if i = Cb or if i written in base b consists of 5 identical digits the number of iterations it takes the Kaprekar routine in base b to arrive at Cb, otherwise
+
+Note that we can define sb(i) for all integers i < b5. If i written in base b takes less than 5 digits, the number is padded with leading zero digits until we have 5 digits before applying the Kaprekar routine.
+
+Define S(b) as the sum of sb(i) for 0 < i < b5. E.g. S(15) = 5274369 S(111) = 400668930299
+
+Find the sum of S(6k+3) for 2 ≤ k ≤ 300. Give the last 18 digits as your answer.
# --hints--
-`euler414()`应该返回552506775824935500。
+`euler414()` should return 552506775824935500.
```js
assert.strictEqual(euler414(), 552506775824935500);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-415-titanic-sets.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-415-titanic-sets.md
index 16748f22d88..951684acc05 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-415-titanic-sets.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-415-titanic-sets.md
@@ -1,26 +1,26 @@
---
id: 5900f50c1000cf542c51001e
-title: 问题415：泰坦尼克号集
+title: 'Problem 415: Titanic sets'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302084
dashedName: problem-415-titanic-sets
---
# --description--
-如果存在一条直线恰好经过S中的两个点，则一组晶格点S称为钛酸组。
+A set of lattice points S is called a titanic set if there exists a line passing through exactly two points in S.
-泰坦尼克集的示例是S = {（0，0），（0，1），（0，2），（1，1），（2，0），（1，0）}，其中 通过（0，1）和（2，0）的通道不会通过S中的任何其他点。
+An example of a titanic set is S = {(0, 0), (0, 1), (0, 2), (1, 1), (2, 0), (1, 0)}, where the line passing through (0, 1) and (2, 0) does not pass through any other point in S.
-另一方面，集合{（0，0），（1，1），（2，2），（4，4）}不是泰坦尼克号集合，因为穿过集合中任意两点的线也经过 通过其他两个。
+On the other hand, the set {(0, 0), (1, 1), (2, 2), (4, 4)} is not a titanic set since the line passing through any two points in the set also passes through the other two.
-对于任何正整数N，令T（N）为每个点（x，y）都满足0≤x，y≤N的钛酸集合S的数量。 可以验证T（1）= 11，T（2）= 494，T（4）= 33554178，T（111）mod 108 = 13500401和T（105）mod 108 = 63259062。
+For any positive integer N, let T(N) be the number of titanic sets S whose every point (x, y) satisfies 0 ≤ x, y ≤ N. It can be verified that T(1) = 11, T(2) = 494, T(4) = 33554178, T(111) mod 108 = 13500401 and T(105) mod 108 = 63259062.
-找出T（1011）mod 108。
+Find T(1011) mod 108.
# --hints--
-`euler415()`应该返回55859742。
+`euler415()` should return 55859742.
```js
assert.strictEqual(euler415(), 55859742);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-416-a-frogs-trip.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-416-a-frogs-trip.md
index b6f291ff5f4..96e9fe1bb0f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-416-a-frogs-trip.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-416-a-frogs-trip.md
@@ -1,22 +1,22 @@
---
id: 5900f50e1000cf542c510020
-title: 问题416：青蛙的旅行
+title: 'Problem 416: A frog''s trip'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302085
dashedName: problem-416-a-frogs-trip
---
# --description--
-一排n个正方形在最左边的正方形中包含一个青蛙。 通过连续跳跃，青蛙到达最右边的方块，然后回到最左边的方块。 在向外旅行中，他向右跳一个，两个或三个方块，在向家旅行中，他以类似的方式向左跳。 他不能跳出广场。 我已经重复了两次往返旅行。
+A row of n squares contains a frog in the leftmost square. By successive jumps the frog goes to the rightmost square and then back to the leftmost square. On the outward trip he jumps one, two or three squares to the right, and on the homeward trip he jumps to the left in a similar manner. He cannot jump outside the squares. He repeats the round-trip travel m times.
-令F（m，n）为青蛙可以通过的方式的数目，这样最多一个正方形就不会出现。 例如，F（1，3）= 4，F（1，4）= 15，F（1，5）= 46，F（2，3）= 16和F（2，100）mod 109 = 429619151。
+Let F(m, n) be the number of the ways the frog can travel so that at most one square remains unvisited. For example, F(1, 3) = 4, F(1, 4) = 15, F(1, 5) = 46, F(2, 3) = 16 and F(2, 100) mod 109 = 429619151.
-查找F的最后9位数字（10，1012）。
+Find the last 9 digits of F(10, 1012).
# --hints--
-`euler416()`应该返回898082747。
+`euler416()` should return 898082747.
```js
assert.strictEqual(euler416(), 898082747);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-417-reciprocal-cycles-ii.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-417-reciprocal-cycles-ii.md
index 7cc8a238330..e73dd646e83 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-417-reciprocal-cycles-ii.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-417-reciprocal-cycles-ii.md
@@ -1,28 +1,28 @@
---
id: 5900f50d1000cf542c51001f
-title: 问题417：倒数周期II
+title: 'Problem 417: Reciprocal cycles II'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302086
dashedName: problem-417-reciprocal-cycles-ii
---
# --description--
-单位分数在分子中包含1。给出分母2到10的单位分数的十进制表示：
+A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
-1/2 = 0.5 1/3 = 0.（3）1/4 = 0.25 1/5 = 0.2 1/6 = 0.1（6）1/7 = 0.（142857）1/8 = 0.125 1/9 = 0.（1）1/10 = 0.1
+1/2= 0.5 1/3= 0.(3) 1/4= 0.25 1/5= 0.2 1/6= 0.1(6) 1/7= 0.(142857) 1/8= 0.125 1/9= 0.(1) 1/10= 0.1
-其中0.1（6）表示0.166666 ...，并具有1位循环周期。可以看出1/7具有6位循环周期。
+Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
-分母没有其他素数因子而不是2和/或5的单位分数不被认为具有重复周期。我们将这些单位分数的重复周期的长度定义为0。
+Unit fractions whose denominator has no other prime factors than 2 and/or 5 are not considered to have a recurring cycle. We define the length of the recurring cycle of those unit fractions as 0.
-令L（n）表示1 / n的循环周期的长度。给出ΣL（n）3≤n≤1000 000等于55535191115。
+Let L(n) denote the length of the recurring cycle of 1/n. You are given that ∑L(n) for 3 ≤ n ≤ 1 000 000 equals 55535191115.
-找ΣL（n）为3≤n≤100000 000
+Find ∑L(n) for 3 ≤ n ≤ 100 000 000
# --hints--
-`euler417()`应该返回446572970925740。
+`euler417()` should return 446572970925740.
```js
assert.strictEqual(euler417(), 446572970925740);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-418-factorisation-triples.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-418-factorisation-triples.md
index e1d635f2abc..ac11b1083f2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-418-factorisation-triples.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-418-factorisation-triples.md
@@ -1,26 +1,26 @@
---
id: 5900f50f1000cf542c510021
-title: 问题418：三重分解
+title: 'Problem 418: Factorisation triples'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302087
dashedName: problem-418-factorisation-triples
---
# --description--
-令n为正整数。 在以下情况下，整数三元组（a，b，c）称为n的分解三元组：
+Let n be a positive integer. An integer triple (a, b, c) is called a factorisation triple of n if: 1 ≤ a ≤ b ≤ c
-a·b·c = n。
+a·b·c = n.
-将f（n）定义为n的因式分解三元组（a，b，c）的a + b + c，使c / a最小。 可以证明这一三元组是独特的。
+Define f(n) to be a + b + c for the factorisation triple (a, b, c) of n which minimises c / a. One can show that this triple is unique.
-例如，f（165）= 19，f（100100）= 142，f（20！）= 4034872。
+For example, f(165) = 19, f(100100) = 142 and f(20!) = 4034872.
-查找f（43！）。
+Find f(43!).
# --hints--
-`euler418()`应该返回1177163565297340400。
+`euler418()` should return 1177163565297340400.
```js
assert.strictEqual(euler418(), 1177163565297340400);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-419-look-and-say-sequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-419-look-and-say-sequence.md
index d577cec1e51..2d9f3650907 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-419-look-and-say-sequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-419-look-and-say-sequence.md
@@ -1,38 +1,38 @@
---
id: 5900f5101000cf542c510022
-title: 问题419：先说顺序
+title: 'Problem 419: Look and say sequence'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302088
dashedName: problem-419-look-and-say-sequence
---
# --description--
-外观序列依次为1，11，21，1211，111221，312211，13112221，1113213211，...
+The look and say sequence goes 1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, ...
-该序列以1开头，并且所有其他成员都是通过用连续数字描述前一个成员而获得的。
+The sequence starts with 1 and all other members are obtained by describing the previous member in terms of consecutive digits.
-大声地这样做有助于：
+It helps to do this out loud:
-1是“一个”→11
+1 is 'one one' → 11
-11是“两个”→21
+11 is 'two ones' → 21
-21是“一二一一”→1211
+21 is 'one two and one one' → 1211
-1211是\`\`一一，一二和二一''→111221
+1211 is 'one one, one two and two ones' → 111221
-111221是“三，二，二，一”→312211
+111221 is 'three ones, two twos and one one' → 312211
...
-将A（n），B（n）和C（n）分别定义为序列的第n个元素中的1、2和3的数目。 可以验证A（40）= 31254，B（40）= 20259和C（40）= 11625。
+Define A(n), B(n) and C(n) as the number of ones, twos and threes in the n'th element of the sequence respectively. One can verify that A(40) = 31254, B(40) = 20259 and C(40) = 11625.
-找到n（1012）的A（n），B（n）和C（n）。 给您的答案模块230，并用逗号分隔A，B和C的值。 例如 对于n = 40，答案将是31254,20259,11625
+Find A(n), B(n) and C(n) for n = 1012. Give your answer modulo 230 and separate your values for A, B and C by a comma. E.g. for n = 40 the answer would be 31254,20259,11625
# --hints--
-`euler419()`应该返回998567458, 1046245404, 43363922。
+`euler419()` should return 998567458, 1046245404, 43363922.
```js
assert.strictEqual(euler419(), 998567458, 1046245404, 43363922);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-42-coded-triangle-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-42-coded-triangle-numbers.md
index b899d0990fb..bc3de086231 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-42-coded-triangle-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-42-coded-triangle-numbers.md
@@ -1,42 +1,48 @@
---
id: 5900f3961000cf542c50fea9
-title: 问题42：编码的三角形数字
+title: 'Problem 42: Coded triangle numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302089
dashedName: problem-42-coded-triangle-numbers
---
# --description--
-三角形编号序列的第n个项由下式给出：tn =½n（n +1）; 所以前十个三角形数字是：
+The `n`<sup>th</sup> term of the sequence of triangle numbers is given by, `tn` = ½`n`(`n`+1); so the first ten triangle numbers are:
-1，3，6，10，15，21，28，36，45，55，...
+<div style='margin-left: 4em;'>1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...</div>
-通过将单词中的每个字母转换为与其字母位置相对应的数字并添加这些值，我们形成了单词值。 例如，SKY的单词值为19 + 11 + 25 = 55 = t10。 如果单词值是一个三角形数字，那么我们将其称为三角形单词。
+By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value. For example, the word value for SKY is 19 + 11 + 25 = 55 = `t`<sub>10</sub>. If the word value is a triangle number then we shall call the word a triangle word.
-使用n长度的单词数组，三角形单词有几个？
+Using words array of `n`-length, how many are triangle words?
# --hints--
-`codedTriangleNumbers(1400)`应该返回129。
+`codedTriangleNumbers(1400)` should return a number.
+
+```js
+assert(typeof codedTriangleNumbers(1400) === 'number');
+```
+
+`codedTriangleNumbers(1400)` should return 129.
```js
assert(codedTriangleNumbers(1400) == 129);
```
-`codedTriangleNumbers(1500)`应该返回137。
+`codedTriangleNumbers(1500)` should return 137.
```js
assert(codedTriangleNumbers(1500) == 137);
```
-`codedTriangleNumbers(1600)`应该返回141。
+`codedTriangleNumbers(1600)` should return 141.
```js
assert(codedTriangleNumbers(1600) == 141);
```
-`codedTriangleNumbers(1786)`应该返回162。
+`codedTriangleNumbers(1786)` should return 162.
```js
assert(codedTriangleNumbers(1786) == 162);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-420-2x2-positive-integer-matrix.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-420-2x2-positive-integer-matrix.md
index 36d882071a6..e65a335aada 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-420-2x2-positive-integer-matrix.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-420-2x2-positive-integer-matrix.md
@@ -1,24 +1,24 @@
---
id: 5900f5111000cf542c510023
-title: 问题420：2x2正整数矩阵
+title: 'Problem 420: 2x2 positive integer matrix'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302090
dashedName: problem-420-2x2-positive-integer-matrix
---
# --description--
-正整数矩阵是其元素均为正整数的矩阵。
+A positive integer matrix is a matrix whose elements are all positive integers.
-一些正整数矩阵可以两种不同方式表示为正整数矩阵的平方。 这是一个例子：
+Some positive integer matrices can be expressed as a square of a positive integer matrix in two different ways. Here is an example:
-我们将F（N）定义为迹线小于N的2x2正整数矩阵的数目，并且可以用两种不同的方式将其表示为正整数矩阵的平方。 我们可以验证F（50）= 7和F（1000）= 1019。
+We define F(N) as the number of the 2x2 positive integer matrices which have a trace less than N and which can be expressed as a square of a positive integer matrix in two different ways. We can verify that F(50) = 7 and F(1000) = 1019.
-找出F（107）。
+Find F(107).
# --hints--
-`euler420()`应该返回145159332。
+`euler420()` should return 145159332.
```js
assert.strictEqual(euler420(), 145159332);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-421-prime-factors-of-n151.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-421-prime-factors-of-n151.md
index c6281b69e08..61e0b86a92d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-421-prime-factors-of-n151.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-421-prime-factors-of-n151.md
@@ -1,24 +1,24 @@
---
id: 5900f5131000cf542c510024
-title: 问题421：n15 +1的素因子
+title: 'Problem 421: Prime factors of n15+1'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302091
dashedName: problem-421-prime-factors-of-n151
---
# --description--
-对于n> 1的每个整数，n15 +1形式的数字是合成的。
+Numbers of the form n15+1 are composite for every integer n > 1.
-对于正整数n和m，将s（n，m）定义为不超过m的n15 +1个不同素数因子的总和。
+For positive integers n and m let s(n,m) be defined as the sum of the distinct prime factors of n15+1 not exceeding m.
-例如 215 + 1 = 3×3×11×331。 因此s（2.10）= 3且s（2,1000）= 3 + 11 + 331 = 345。
+E.g. 215+1 = 3×3×11×331. So s(2,10) = 3 and s(2,1000) = 3+11+331 = 345.
-同样是1015 +1 = 7×11×13×211×241×2161×9091。 因此s（10,100）= 31，而s（10,1000）= 483。 求出∑ s（n，108）为1≤n≤1011。
+Also 1015+1 = 7×11×13×211×241×2161×9091. So s(10,100) = 31 and s(10,1000) = 483. Find ∑ s(n,108) for 1 ≤ n ≤ 1011.
# --hints--
-`euler421()`应该返回2304215802083466200。
+`euler421()` should return 2304215802083466200.
```js
assert.strictEqual(euler421(), 2304215802083466200);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-422-sequence-of-points-on-a-hyperbola.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-422-sequence-of-points-on-a-hyperbola.md
index cd3be28acd6..cf7455ac9d6 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-422-sequence-of-points-on-a-hyperbola.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-422-sequence-of-points-on-a-hyperbola.md
@@ -1,26 +1,26 @@
---
id: 5900f5131000cf542c510025
-title: 问题422：双曲线上的点序列
+title: 'Problem 422: Sequence of points on a hyperbola'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302092
dashedName: problem-422-sequence-of-points-on-a-hyperbola
---
# --description--
-假设H是由等式12x2 + 7xy-12y2 = 625定义的双曲线。
+Let H be the hyperbola defined by the equation 12x2 + 7xy - 12y2 = 625.
-接下来，将X定义为点（7，1）。 可以看出X在H中。
+Next, define X as the point (7, 1). It can be seen that X is in H.
-现在，我们将H中的点序列{Pi：i≥1}定义为： P1 =（13，61/4）。 P2 =（-43/6，-4）。 对于i> 2，Pi是H中与Pi-1不同的唯一点，因此线PiPi-1与线Pi-2X平行。 可以证明Pi是定义明确的，并且其坐标始终是有理的。 您得到P3 =（-19/2，-229/24），P4 =（1267/144，-37/12）和P7 =（17194218091/143327232，274748766781/1719926784）。
+Now we define a sequence of points in H, {Pi : i ≥ 1}, as: P1 = (13, 61/4). P2 = (-43/6, -4). For i > 2, Pi is the unique point in H that is different from Pi-1 and such that line PiPi-1 is parallel to line Pi-2X. It can be shown that Pi is well-defined, and that its coordinates are always rational. You are given that P3 = (-19/2, -229/24), P4 = (1267/144, -37/12) and P7 = (17194218091/143327232, 274748766781/1719926784).
-用以下格式找到n = 1114的Pn：如果Pn =（a / b，c / d），其中分数是最低项，而分母是正数，则答案是（a + b + c + d）mod 1 000 007。
+Find Pn for n = 1114 in the following format:If Pn = (a/b, c/d) where the fractions are in lowest terms and the denominators are positive, then the answer is (a + b + c + d) mod 1 000 000 007.
-对于n = 7，答案应该是：806236837。
+For n = 7, the answer would have been: 806236837.
# --hints--
-`euler422()`应该返回92060460。
+`euler422()` should return 92060460.
```js
assert.strictEqual(euler422(), 92060460);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-423-consecutive-die-throws.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-423-consecutive-die-throws.md
index f42aa478054..a19a5ce19d3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-423-consecutive-die-throws.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-423-consecutive-die-throws.md
@@ -1,30 +1,30 @@
---
id: 5900f5141000cf542c510027
-title: 问题423：连续死球
+title: 'Problem 423: Consecutive die throws'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302093
dashedName: problem-423-consecutive-die-throws
---
# --description--
-令n为正整数。
+Let n be a positive integer.
-一个6面的骰子被抛出n次。 令c为给出相同值的连续抛出的对数。
+A 6-sided die is thrown n times. Let c be the number of pairs of consecutive throws that give the same value.
-例如，如果n = 7并且掷骰的值为（1,1,5,6,6,6,3），那么以下连续投掷对将给出相同的值： （1,1,5,6,6,6,3） （1,1,5,6,6,6,3） （1,1,5,6,6,6,3） 因此，对于（1,1,5,6,6,6,3），c = 3。
+For example, if n = 7 and the values of the die throws are (1,1,5,6,6,6,3), then the following pairs of consecutive throws give the same value: (1,1,5,6,6,6,3) (1,1,5,6,6,6,3) (1,1,5,6,6,6,3) Therefore, c = 3 for (1,1,5,6,6,6,3).
-将C（n）定义为n次抛出6面骰子的结果数，以使c不超过π（n）.1 例如，C（3）= 216，C（4）= 1290，C（11）= 361912500和C（24）= 4727547363281250000。
+Define C(n) as the number of outcomes of throwing a 6-sided die n times such that c does not exceed π(n).1 For example, C(3) = 216, C(4) = 1290, C(11) = 361912500 and C(24) = 4727547363281250000.
-对于1≤n≤L，将S（L）定义为∑ C（n）。 例如，S（50）mod 1 000 000 007 = 832833871。
+Define S(L) as ∑ C(n) for 1 ≤ n ≤ L. For example, S(50) mod 1 000 000 007 = 832833871.
-求S（50000000）mod 1000000007。
+Find S(50 000 000) mod 1 000 000 007.
-1π表示素数计数函数，即 π（n）是质数≤n的素数。
+1 π denotes the prime-counting function, i.e. π(n) is the number of primes ≤ n.
# --hints--
-`euler423()`应该返回653972374。
+`euler423()` should return 653972374.
```js
assert.strictEqual(euler423(), 653972374);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-424-kakuro.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-424-kakuro.md
index f7ba5e1fd08..827b969ce69 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-424-kakuro.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-424-kakuro.md
@@ -1,34 +1,34 @@
---
id: 5900f5141000cf542c510026
-title: 问题424：九郎
+title: 'Problem 424: Kakuro'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302094
dashedName: problem-424-kakuro
---
# --description--
-上面是一个神秘的kakuro（也称为交叉和，甚至是交叉和）难题的示例，其最终解决方案在右侧。 （在许多互联网站点上都可以轻松找到kakuro谜题的通用规则。当前还可以在krazydad.com上找到其他相关信息，其作者已提供了此挑战的谜题数据。）
+The above is an example of a cryptic kakuro (also known as cross sums, or even sums cross) puzzle, with its final solution on the right. (The common rules of kakuro puzzles can be found easily on numerous internet sites. Other related information can also be currently found at krazydad.com whose author has provided the puzzle data for this challenge.)
-可下载的文本文件（kakuro200.txt）包含200个此类难题的描述，混合了5x5和6x6类型。文件中的第一个难题是上面的示例，其编码如下：
+The downloadable text file (kakuro200.txt) contains the description of 200 such puzzles, a mix of 5x5 and 6x6 types. The first puzzle in the file is the above example which is coded as follows:
-6，X，X，（vCC），（vI），X，X，X，（hH），B，O，（vCA），（vJE），X，（hFE，vD），O，O，O， O，（hA），O，I，（hJC，vB），O，O，（hJC），H，O，O，O，X，X，X，（hJE），O，O，X
+6,X,X,(vCC),(vI),X,X,X,(hH),B,O,(vCA),(vJE),X,(hFE,vD),O,O,O,O,(hA),O,I,(hJC,vB),O,O,(hJC),H,O,O,O,X,X,X,(hJE),O,O,X
-第一个字符是指示信息网格大小的数字。它可能是6（用于5x5的kakuro拼图）或7（用于6x6的拼图），后跟逗号（，）。需要额外的顶行和左列来插入信息。
+The first character is a numerical digit indicating the size of the information grid. It would be either a 6 (for a 5x5 kakuro puzzle) or a 7 (for a 6x6 puzzle) followed by a comma (,). The extra top line and left column are needed to insert information.
-然后描述每个单元格的内容，并在其后跟一个逗号，从左到右并从顶行开始。 X =灰色单元格，不需要用数字填充。 O（大写字母）=要用数字填充的空白单元格。 A =或从A到J的大写字母中的任何一个都由已解决难题中的等效数字替换。 （）=加密总和的位置。水平总和前面加上小写的“ v”。后面跟着一两个大写字母，具体取决于总和是一位还是两位数。对于两位数的总和，第一个字母表示“十”，第二个字母表示“单位”。当单元格必须同时包含水平和垂直信息时，第一个始终是水平和信息，并且两者在同一组括号内用逗号分隔，例如：（.hFE，vD）。每组括号后还紧跟一个逗号。
+The content of each cell is then described and followed by a comma, going left to right and starting with the top line. X = Gray cell, not required to be filled by a digit. O (upper case letter)= White empty cell to be filled by a digit. A = Or any one of the upper case letters from A to J to be replaced by its equivalent digit in the solved puzzle. ( ) = Location of the encrypted sums. Horizontal sums are preceded by a lower case "h" and vertical sums are preceded by a lower case "v". Those are followed by one or two upper case letters depending if the sum is a single digit or double digit one. For double digit sums, the first letter would be for the "tens" and the second one for the "units". When the cell must contain information for both a horizontal and a vertical sum, the first one is always for the horizontal sum and the two are separated by a comma within the same set of brackets, ex.: (hFE,vD). Each set of brackets is also immediately followed by a comma.
-最后一个单元格的描述后跟回车符/换行符（CRLF），而不是逗号。
+The description of the last cell is followed by a Carriage Return/Line Feed (CRLF) instead of a comma.
-每个谜题所需的答案均基于得出解决方案所需的每个字母的值，并根据字母顺序。如示例难题所示，其答案将为8426039571。10个加密字母中至少有9个始终是问题描述的一部分。如果仅给出9，则必须为丢失的数字分配剩余的数字。
+The required answer to each puzzle is based on the value of each letter necessary to arrive at the solution and according to the alphabetical order. As indicated under the example puzzle, its answer would be 8426039571. At least 9 out of the 10 encrypting letters are always part of the problem description. When only 9 are given, the missing one must be assigned the remaining digit.
-您将获得文件中前10个谜题的答案总和为64414157580。
+You are given that the sum of the answers for the first 10 puzzles in the file is 64414157580.
-找到200个难题的答案总和。
+Find the sum of the answers for the 200 puzzles.
# --hints--
-`euler424()`应该返回1059760019628。
+`euler424()` should return 1059760019628.
```js
assert.strictEqual(euler424(), 1059760019628);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-425-prime-connection.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-425-prime-connection.md
index 56b58c2c47a..d46740b0d58 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-425-prime-connection.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-425-prime-connection.md
@@ -1,30 +1,30 @@
---
id: 5900f5151000cf542c510028
-title: 问题425：主要连接
+title: 'Problem 425: Prime connection'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302095
dashedName: problem-425-prime-connection
---
# --description--
-如果满足以下条件之一，则称两个正数A和B被连接（用“ A↔B”表示）：
+Two positive numbers A and B are said to be connected (denoted by "A ↔ B") if one of these conditions holds:
-（1）A和B的长度相同，且仅相差一位数； 例如123↔173。
+(1) A and B have the same length and differ in exactly one digit; for example, 123 ↔ 173.
-（2）在A（或B）的左边加一位使B（或A）成为数字； 例如23↔223和123↔23。
+(2) Adding one digit to the left of A (or B) makes B (or A); for example, 23 ↔ 223 and 123 ↔ 23.
-如果在2和P之间存在连接的素数链，并且链中没有素数超过P，则我们称素数为P a 2的相对数。
+We call a prime P a 2's relative if there exists a chain of connected primes between 2 and P and no prime in the chain exceeds P.
-例如，127是2的亲戚。 可能的链之一如下所示： 2分3分13分113分103分107分127 但是，11和103不是2的亲戚。
+For example, 127 is a 2's relative. One of the possible chains is shown below: 2 ↔ 3 ↔ 13 ↔ 113 ↔ 103 ↔ 107 ↔ 127 However, 11 and 103 are not 2's relatives.
-令F（N）为素数≤N的和，它们不是2的相对数。 我们可以验证F（103）= 431和F（104）= 78728。
+Let F(N) be the sum of the primes ≤ N which are not 2's relatives. We can verify that F(103) = 431 and F(104) = 78728.
-找出F（107）。
+Find F(107).
# --hints--
-`euler425()`应该返回46479497324。
+`euler425()` should return 46479497324.
```js
assert.strictEqual(euler425(), 46479497324);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-426-box-ball-system.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-426-box-ball-system.md
index 50aa1248696..5e2c48c34e0 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-426-box-ball-system.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-426-box-ball-system.md
@@ -1,8 +1,8 @@
---
id: 5900f5171000cf542c510029
-title: 问题426：箱式球系统
+title: 'Problem 426: Box-ball system'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302096
dashedName: problem-426-box-ball-system
---
@@ -24,7 +24,7 @@ Starting from the initial configuration (t0, t1, …, t10), the final state beco
# --hints--
-`euler426()`应该返回31591886008。
+`euler426()` should return 31591886008.
```js
assert.strictEqual(euler426(), 31591886008);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-427-n-sequences.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-427-n-sequences.md
index 0ae14574b55..8f399eeaf0a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-427-n-sequences.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-427-n-sequences.md
@@ -1,28 +1,28 @@
---
id: 5900f5181000cf542c51002a
-title: 问题427：n序列
+title: 'Problem 427: n-sequences'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302097
dashedName: problem-427-n-sequences
---
# --description--
-整数序列S = {si}如果具有n个元素，则每个元素满足1≤si≤n，则称为n序列。 因此，总共有nn个不同的n序列。
+A sequence of integers S = {si} is called an n-sequence if it has n elements and each element si satisfies 1 ≤ si ≤ n. Thus there are nn distinct n-sequences in total.
-例如，序列S = {1、5、5、10、7、7、7、2、3、7}是10个序列。
+For example, the sequence S = {1, 5, 5, 10, 7, 7, 7, 2, 3, 7} is a 10-sequence.
-对于任何序列S，令L（S）为具有相同值的S的最长连续存在的长度。 例如，对于上面给定的序列S，由于三个连续的7，L（S）= 3。
+For any sequence S, let L(S) be the length of the longest contiguous subsequence of S with the same value. For example, for the given sequence S above, L(S) = 3, because of the three consecutive 7's.
-对于所有n序列S，令f（n）= ∑ L（S）。
+Let f(n) = ∑ L(S) for all n-sequences S.
-例如，f（3）= 45，f（7）= 1403689和f（11）= 481496895121。
+For example, f(3) = 45, f(7) = 1403689 and f(11) = 481496895121.
-找出f（7,500,000）mod 1 000 009。
+Find f(7 500 000) mod 1 000 000 009.
# --hints--
-`euler427()`应该返回97138867。
+`euler427()` should return 97138867.
```js
assert.strictEqual(euler427(), 97138867);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-428-necklace-of-circles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-428-necklace-of-circles.md
index 86fda9d6e52..5919151606c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-428-necklace-of-circles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-428-necklace-of-circles.md
@@ -1,30 +1,34 @@
---
id: 5900f5191000cf542c51002b
-title: 问题428：圆圈项链
+title: 'Problem 428: Necklace of Circles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302098
dashedName: problem-428-necklace-of-circles
---
# --description--
-令`a`，`b`和`c`为正数。
+Let `a`, `b` and `c` be positive numbers.
-令W，X，Y，Z为四个共线点，其中| WX | = `到`，| XY | = `b`，| YZ | = `c`和| WZ | = `a` + `b` + `c`。
+Let W, X, Y, Z be four collinear points where |WX| = `a`, |XY| = `b`, |YZ| = `c` and |WZ| = `a` + `b` + `c`.
-令中的C <sub>为直径XY的圆。 令C <sub>out</sub>为直径WZ的圆。</sub>
+Let C<sub>in</sub> be the circle having the diameter XY.
-如果可以放置`k，则三元组（a，b，c）被称为项链三元组</ em>。`*≥3个不同的圆C <sub>1</sub>，C <sub>2</sub>，...，C <sub><var>k</var></sub> ：*
+Let C<sub>out</sub> be the circle having the diameter WZ.
-<ul> <li> C <sub> <var> i </var> </sub>与任何C <sub> <var> j </var> </sub>都没有公共内点，且1≤<var > i </var>，<var> j </var>≤<var> k </var>和<var> i </var>≠<var> j </var>，</ li> <li> C <sub> <var> i </var> </sub>与</sub>中的C <sub>和 C <sub> out </sub>表示1≤<var> i </var>≤<var> k </var>，</ li> <li> C <sub> <var> i </var> < / sub>与C <sub> <var> i </var> +1 </sub>相切1≤<var> i </var> < <var> k </var>和</ li> <li> C <sub> <var> k </var> </sub>与C <sub> 1 </sub>相切。</ li> </ ul>
-例如，（5,5,5）和（4,3,21）是项链三胞胎，而可以证明（2,2,5）不是。
-<img src="https://projecteuler.net/project/images/p428_necklace.png" alt="a visual representation of a necklace triplet">
+The triplet (`a`, `b`, `c`) is called a *necklace triplet* if you can place `k` ≥ 3 distinct circles C<sub>1</sub>, C<sub>2</sub>, ..., C<sub><var>k</var></sub> such that:
-令T（`n`）为项链三联体的数量（`a`，`b`，`c`），使得`a`，`b`和`c`是正整数，并且`b`≤`n`。 例如，T（1）= 9，T（20）= 732和T（3000）= 438106。 求T（1000000000000）。
+<ul><li>C<sub><var>i</var></sub> has no common interior points with any C<sub><var>j</var></sub> for 1 ≤ <var>i</var>, <var>j</var> ≤ <var>k</var> and <var>i</var> ≠ <var>j</var>,</li><li>C<sub><var>i</var></sub> is tangent to both C<sub>in</sub> and C<sub>out</sub> for 1 ≤ <var>i</var> ≤ <var>k</var>,</li><li>C<sub><var>i</var></sub> is tangent to C<sub><var>i</var>+1</sub> for 1 ≤ <var>i</var> < <var>k</var>, and</li><li>C<sub><var>k</var></sub> is tangent to C<sub>1</sub>.</li></ul>
+For example, (5, 5, 5) and (4, 3, 21) are necklace triplets, while it can be shown that (2, 2, 5) is not.
+<img src="https://projecteuler.net/project/images/p428_necklace.png" alt="a visual representation of a necklace triplet" />
+
+Let T(`n`) be the number of necklace triplets (`a`, `b`, `c`) such that `a`, `b` and `c` are positive integers, and `b` ≤ `n`. For example, T(1) = 9, T(20) = 732 and T(3000) = 438106.
+
+Find T(1 000 000 000).
# --hints--
-`necklace(1000000000)`应该返回747215561862。
+`necklace(1000000000)` should return 747215561862.
```js
assert.strictEqual(necklace(1000000000), 747215561862);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-429-sum-of-squares-of-unitary-divisors.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-429-sum-of-squares-of-unitary-divisors.md
index 5f14cceedc4..8b9c108829c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-429-sum-of-squares-of-unitary-divisors.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-429-sum-of-squares-of-unitary-divisors.md
@@ -1,26 +1,26 @@
---
id: 5900f5191000cf542c51002c
-title: 问题429：单一除数的平方和
+title: 'Problem 429: Sum of squares of unitary divisors'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302099
dashedName: problem-429-sum-of-squares-of-unitary-divisors
---
# --description--
-数为n的unit除数d是n的除数，其性质为gcd（d，n / d）= 1。
+A unitary divisor d of a number n is a divisor of n that has the property gcd(d, n/d) = 1.
-4的除数！ = 24是1、3、8和24。
+The unitary divisors of 4! = 24 are 1, 3, 8 and 24.
-它们的平方和是12 + 32 + 82 + 242 = 650。
+The sum of their squares is 12 + 32 + 82 + 242 = 650.
-令S（n）表示n的一元除数的平方和。 因此，S（4！）＝ 650。
+Let S(n) represent the sum of the squares of the unitary divisors of n. Thus S(4!)=650.
-求S（100000000！）取模1000000009。
+Find S(100 000 000!) modulo 1 000 000 009.
# --hints--
-`euler429()`应该返回98792821。
+`euler429()` should return 98792821.
```js
assert.strictEqual(euler429(), 98792821);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-43-sub-string-divisibility.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-43-sub-string-divisibility.md
index d80300b9c70..6a0ed4ec73c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-43-sub-string-divisibility.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-43-sub-string-divisibility.md
@@ -1,39 +1,41 @@
---
id: 5900f3971000cf542c50feaa
-title: 问题43：子串可分性
+title: 'Problem 43: Sub-string divisibility'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302100
dashedName: problem-43-sub-string-divisibility
---
# --description--
-数字1406357289是0到9的全数字，因为它是由0到9的每个数字以某种顺序组成的，但是它也具有相当有趣的子字符串可除性。
+The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.
-令d1为第一位数，d2为第二位数，依此类推。 这样，我们注意以下几点：
+Let d<sub>1</sub> be the 1<sup>st</sup> digit, d<sub>2</sub> be the 2<sup>nd</sup> digit, and so on. In this way, we note the following:
-d2d3d4 = 406可被2整除
+<ul>
+ <li>d<sub>2</sub>d<sub>3</sub>d<sub>4</sub> = 406 is divisible by 2</li>
+ <li>d<sub>3</sub>d<sub>4</sub>d<sub>5</sub> = 063 is divisible by 3</li>
+ <li>d<sub>4</sub>d<sub>5</sub>d<sub>6</sub> = 635 is divisible by 5</li>
+ <li>d<sub>5</sub>d<sub>6</sub>d<sub>7</sub> = 357 is divisible by 7</li>
+ <li>d<sub>6</sub>d<sub>7</sub>d<sub>8</sub> = 572 is divisible by 11</li>
+ <li>d<sub>7</sub>d<sub>8</sub>d<sub>9</sub> = 728 is divisible by 13</li>
+ <li>d<sub>8</sub>d<sub>9</sub>d<sub>10</sub> = 289 is divisible by 17</li>
+</ul>
-d3d4d5 = 063被3整除
-
-d4d5d6 = 635可被5整除
-
-d5d6d7 = 357可被7整除
-
-d6d7d8 = 572被11整除
-
-d7d8d9 = 728被13整除
-
-d8d9d10 = 289被17整除
-
-使用此属性查找所有0到9个泛数字的数字。
+Find the numbers of all 0 to 9 pandigital numbers with this property.
# --hints--
-`substringDivisibility()`应该返回[1430952867, 1460357289, 1406357289, 4130952867, 4160357289, 4106357289]。
+`substringDivisibility()` should return an array.
```js
-assert.deepEqual(substringDivisibility(), [
+assert(Array.isArray(substringDivisibility()));
+```
+
+`substringDivisibility()` should return [ 1430952867, 1460357289, 1406357289, 4130952867, 4160357289, 4106357289 ].
+
+```js
+assert.sameMembers(substringDivisibility(), [
1430952867,
1460357289,
1406357289,
@@ -43,6 +45,18 @@ assert.deepEqual(substringDivisibility(), [
]);
```
+You should not copy and return the array.
+
+```js
+assert(
+ !__helpers
+ .removeJSComments(code)
+ .match(
+ /(1430952867)|(1460357289)|(1406357289)|(4130952867)|(4160357289)|(4106357289)/
+ )
+);
+```
+
# --seed--
## --seed-contents--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-430-range-flips.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-430-range-flips.md
index 2369d6a4143..ba66983c7b3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-430-range-flips.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-430-range-flips.md
@@ -1,28 +1,28 @@
---
id: 5900f51a1000cf542c51002d
-title: 问题430：范围翻转
+title: 'Problem 430: Range flips'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302101
dashedName: problem-430-range-flips
---
# --description--
-N个磁盘连续放置，从左到右从1到N索引。
+N disks are placed in a row, indexed 1 to N from left to right.
-每个磁盘都有黑色和白色的一面。 最初，所有磁盘都显示白色的一面。
+Each disk has a black side and white side. Initially all disks show their white side.
-在每一回合，随机地均匀地选择介于1和N（含）之间的两个整数A和B（不一定是不同的）。 索引从A到B（包括B）的所有磁盘均被翻转。
+At each turn, two, not necessarily distinct, integers A and B between 1 and N (inclusive) are chosen uniformly at random. All disks with an index from A to B (inclusive) are flipped.
-以下示例显示了N = 8的情况。在第一圈A = 5且B = 2，在第二圈A = 4且B = 6。
+The following example shows the case N = 8. At the first turn A = 5 and B = 2, and at the second turn A = 4 and B = 6.
-令E（N，M）为M圈后显示白色面的预期磁盘数量。 我们可以验证E（3，1）= 10/9，E（3，2）= 5/3，E（10，4）≈5,157和E（100，10）≈51,893。
+Let E(N, M) be the expected number of disks that show their white side after M turns. We can verify that E(3, 1) = 10/9, E(3, 2) = 5/3, E(10, 4) ≈ 5.157 and E(100, 10) ≈ 51.893.
-找出E（1010，4000）。 将答案四舍五入到小数点后两位。
+Find E(1010, 4000). Give your answer rounded to 2 decimal places behind the decimal point.
# --hints--
-`euler430()`应该返回5000624921.38。
+`euler430()` should return 5000624921.38.
```js
assert.strictEqual(euler430(), 5000624921.38);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-431-square-space-silo.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-431-square-space-silo.md
index 9bda780d49f..51a104b47a0 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-431-square-space-silo.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-431-square-space-silo.md
@@ -1,26 +1,26 @@
---
id: 5900f51b1000cf542c51002e
-title: 问题431：方形筒仓
+title: 'Problem 431: Square Space Silo'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302102
dashedName: problem-431-square-space-silo
---
# --description--
-弗雷德农夫安排在他的农场安装一个新的储存仓，并对所有方面进行痴迷，当他发现它是圆形时，他绝对是毁灭性的。安装筒仓的公司代表Quentin解释说，他们只制造圆柱形筒仓，但他指出它是放在方形底座上。弗雷德并不觉得有趣并且坚持将其从他的财产中删除。
+Fred the farmer arranges to have a new storage silo installed on his farm and having an obsession for all things square he is absolutely devastated when he discovers that it is circular. Quentin, the representative from the company that installed the silo, explains that they only manufacture cylindrical silos, but he points out that it is resting on a square base. Fred is not amused and insists that it is removed from his property.
-快速思考Quentin解释说，当从上方输送粒状材料时，形成锥形斜面，与水平面形成的自然角度称为休止角。例如，如果休息角度为$ \\ alpha = 30 $度，并且在筒仓的中心处传送颗粒，则将朝向圆柱体的顶部形成完美的锥体。在这个直径为6米的筒仓的情况下，浪费的空间量约为32.648388556立方米。然而，如果谷物在顶部的一个点处被输送，该点与中心的水平距离为$ x $ m，则形成具有奇怪弯曲和倾斜的基部的锥体。他向弗雷德展示了一张照片
+Quick thinking Quentin explains that when granular materials are delivered from above a conical slope is formed and the natural angle made with the horizontal is called the angle of repose. For example if the angle of repose, $\\alpha = 30$ degrees, and grain is delivered at the centre of the silo then a perfect cone will form towards the top of the cylinder. In the case of this silo, which has a diameter of 6m, the amount of space wasted would be approximately 32.648388556 m3. However, if grain is delivered at a point on the top which has a horizontal distance of $x$ metres from the centre then a cone with a strangely curved and sloping base is formed. He shows Fred a picture.
-我们将以$ V（x）$给出以立方米浪费的空间量。如果$ x = 1.114785284 $，恰好有三个小数位，那么浪费的空间金额为$ V（1.114785284）\\约36 $。鉴于此问题的可能解决方案的范围，还有另一种选择：$ V（2.511167869）\\约49 $。这就像知道广场是筒仓之王，坐在你的粮食上的灿烂荣耀。
+We shall let the amount of space wasted in cubic metres be given by $V(x)$. If $x = 1.114785284$, which happens to have three squared decimal places, then the amount of space wasted, $V(1.114785284) \\approx 36$. Given the range of possible solutions to this problem there is exactly one other option: $V(2.511167869) \\approx 49$. It would be like knowing that the square is king of the silo, sitting in splendid glory on top of your grain.
-弗雷德的眼睛高兴地看着这个优雅的分辨率，但仔细观察昆汀的绘画和计算后，他的幸福再次变成了沮丧。弗雷德向昆汀指出，这是筒仓的半径是6米，而不是直径，他的谷物的休止角是40度。但是，如果Quentin可以为这个特定的孤岛找到一套解决方案，那么他将非常乐意保留它。
+Fred's eyes light up with delight at this elegant resolution, but on closer inspection of Quentin's drawings and calculations his happiness turns to despondency once more. Fred points out to Quentin that it's the radius of the silo that is 6 metres, not the diameter, and the angle of repose for his grain is 40 degrees. However, if Quentin can find a set of solutions for this particular silo then he will be more than happy to keep it.
-如果快速思考昆汀要满足沮丧地挑剔弗雷德农民对所有方格的兴趣，然后确定所有可能的方形空间浪费选项的$ x $的值并计算$ \\ sum x $正确到9位小数。
+If Quick thinking Quentin is to satisfy frustratingly fussy Fred the farmer's appetite for all things square then determine the values of $x$ for all possible square space wastage options and calculate $\\sum x$ correct to 9 decimal places.
# --hints--
-`euler431()`应该返回23.386029052。
+`euler431()` should return 23.386029052.
```js
assert.strictEqual(euler431(), 23.386029052);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-432-totient-sum.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-432-totient-sum.md
index dce4784ece5..4f0f5928ef5 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-432-totient-sum.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-432-totient-sum.md
@@ -1,22 +1,22 @@
---
id: 5900f51e1000cf542c510030
-title: 问题432欧拉
+title: 'Problem 432: Totient sum'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302103
dashedName: problem-432-totient-sum
---
# --description--
-令S（n，m）= ∑φ（n×i）且1≤i≤m。 （φ是欧拉的上位函数）
+Let S(n,m) = ∑φ(n × i) for 1 ≤ i ≤ m. (φ is Euler's totient function)
-给出S（510510,106）= 45480596821125120。
+You are given that S(510510,106 )= 45480596821125120.
-找出S（510510,1011）。 输入答案的最后9位数字。
+Find S(510510,1011). Give the last 9 digits of your answer.
# --hints--
-`euler432()`应该返回754862080。
+`euler432()` should return 754862080.
```js
assert.strictEqual(euler432(), 754862080);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-433-steps-in-euclids-algorithm.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-433-steps-in-euclids-algorithm.md
index 74582500264..5423c8691c9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-433-steps-in-euclids-algorithm.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-433-steps-in-euclids-algorithm.md
@@ -1,26 +1,26 @@
---
id: 5900f51d1000cf542c51002f
-title: 问题433：欧几里得算法的步骤
+title: 'Problem 433: Steps in Euclid''s algorithm'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302104
dashedName: problem-433-steps-in-euclids-algorithm
---
# --description--
-设E（x0，y0）为用Euclid算法确定x0和y0的最大公约数所需要的步数。 更正式地说：x1 = y0，y1 = x0 mod y0xn = yn-1，yn = xn-1 mod yn-1
+Let E(x0, y0) be the number of steps it takes to determine the greatest common divisor of x0 and y0 with Euclid's algorithm. More formally:x1 = y0, y1 = x0 mod y0xn = yn-1, yn = xn-1 mod yn-1
-E（x0，y0）是最小的n，因此yn = 0。
+E(x0, y0) is the smallest n such that yn = 0.
-我们有E（1,1）= 1，E（10,6）= 3和E（6,10）= 4。
+We have E(1,1) = 1, E(10,6) = 3 and E(6,10) = 4.
-将S（N）定义为1≤x，y≤N的E（x，y）之和。 我们有S（1）= 1，S（10）= 221和S（100）= 39826。
+Define S(N) as the sum of E(x,y) for 1 ≤ x,y ≤ N. We have S(1) = 1, S(10) = 221 and S(100) = 39826.
-求S（5·106）。
+Find S(5·106).
# --hints--
-`euler433()`应该返回326624372659664。
+`euler433()` should return 326624372659664.
```js
assert.strictEqual(euler433(), 326624372659664);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-434-rigid-graphs.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-434-rigid-graphs.md
index c757a3e56bc..48a01b7a3ba 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-434-rigid-graphs.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-434-rigid-graphs.md
@@ -1,32 +1,32 @@
---
id: 5900f51f1000cf542c510031
-title: 问题434：刚性图
+title: 'Problem 434: Rigid graphs'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302105
dashedName: problem-434-rigid-graphs
---
# --description--
-回想一下，图是顶点和连接顶点的边的集合，并且通过边连接的两个顶点称为相邻。
+Recall that a graph is a collection of vertices and edges connecting the vertices, and that two vertices connected by an edge are called adjacent.
-通过将每个顶点与欧氏空间中的一个点相关联，可以将图嵌入到欧氏空间中。
+Graphs can be embedded in Euclidean space by associating each vertex with a point in the Euclidean space.
-柔性图是图的嵌入，其中可以连续移动一个或多个顶点，以便至少两个不相邻顶点之间的距离发生变化，而每对相邻顶点之间的距离保持恒定。
+A flexible graph is an embedding of a graph where it is possible to move one or more vertices continuously so that the distance between at least two nonadjacent vertices is altered while the distances between each pair of adjacent vertices is kept constant.
-刚性图是不灵活的图的嵌入。
+A rigid graph is an embedding of a graph which is not flexible.
-非正式地，如果通过用完全旋转的铰链替换顶点，并用不弯曲且无弹性的杆代替边，则图形是刚性的，则图形的任何部分都不能独立于图形的其余部分移动。
+Informally, a graph is rigid if by replacing the vertices with fully rotating hinges and the edges with rods that are unbending and inelastic, no parts of the graph can be moved independently from the rest of the graph.
-嵌入在欧几里得平面中的网格图不是刚性的，如以下动画所示： 但是，可以通过向单元格添加对角线边缘来使它们变硬。例如，对于2x3网格图，有19种方法可以使图变刚性： 请注意，出于解决此问题的目的，我们不考虑更改对角线边缘的方向或将两个对角线边缘添加到单元格，这是使网格图变硬的另一种方法。
+The grid graphs embedded in the Euclidean plane are not rigid, as the following animation demonstrates: However, one can make them rigid by adding diagonal edges to the cells. For example, for the 2x3 grid graph, there are 19 ways to make the graph rigid: Note that for the purposes of this problem, we do not consider changing the orientation of a diagonal edge or adding both diagonal edges to a cell as a different way of making a grid graph rigid.
-令R（m，n）为使m×n网格图变硬的方法的数量。 例如。 R（2,3）= 19和R（5,5）= 23679901
+Let R(m,n) be the number of ways to make the m × n grid graph rigid. E.g. R(2,3) = 19 and R(5,5) = 23679901
-对于1≤i，j≤N，将S（N）定义为∑R（i，j）。 例如。 S（5）= 25021721。 求S（100），以1000000033为模给出答案
+Define S(N) as ∑R(i,j) for 1 ≤ i, j ≤ N. E.g. S(5) = 25021721. Find S(100), give your answer modulo 1000000033
# --hints--
-`euler434()`应该返回863253606。
+`euler434()` should return 863253606.
```js
assert.strictEqual(euler434(), 863253606);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-435-polynomials-of-fibonacci-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-435-polynomials-of-fibonacci-numbers.md
index 8d5f9f45529..ee078a73824 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-435-polynomials-of-fibonacci-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-435-polynomials-of-fibonacci-numbers.md
@@ -1,24 +1,24 @@
---
id: 5900f5201000cf542c510032
-title: 问题435：斐波那契数的多项式
+title: 'Problem 435: Polynomials of Fibonacci numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302106
dashedName: problem-435-polynomials-of-fibonacci-numbers
---
# --description--
-斐波纳契数{fn，n≥0}被递归定义为fn = fn-1 + fn-2，基本情况为f0 = 0和f1 = 1。
+The Fibonacci numbers {fn, n ≥ 0} are defined recursively as fn = fn-1 + fn-2 with base cases f0 = 0 and f1 = 1.
-对于0≤i≤n，将多项式{Fn，n≥0}定义为Fn（x）= ∑fixi。
+Define the polynomials {Fn, n ≥ 0} as Fn(x) = ∑fixi for 0 ≤ i ≤ n.
-例如，F7（x）= x + x2 + 2x3 + 3x4 + 5x5 + 8x6 + 13x7，而F7（11）= 268357683。
+For example, F7(x) = x + x2 + 2x3 + 3x4 + 5x5 + 8x6 + 13x7, and F7(11) = 268357683.
-令n =1015。求和\[∑0≤x≤100 Fn（x）] mod 1307674368000（= 15！）。
+Let n = 1015. Find the sum \[∑0≤x≤100 Fn(x)] mod 1307674368000 (= 15!).
# --hints--
-`euler435()`252541322550。
+`euler435()` should return 252541322550.
```js
assert.strictEqual(euler435(), 252541322550);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-436-unfair-wager.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-436-unfair-wager.md
index abd6208ce23..43b111b868f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-436-unfair-wager.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-436-unfair-wager.md
@@ -1,34 +1,34 @@
---
id: 5900f5221000cf542c510033
-title: 问题436：不公平的赌注
+title: 'Problem 436: Unfair wager'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302107
dashedName: problem-436-unfair-wager
---
# --description--
-朱莉向妹妹路易丝提出以下赌注。
+Julie proposes the following wager to her sister Louise.
-她建议他们玩一场机会游戏，以确定谁来洗碗。
+She suggests they play a game of chance to determine who will wash the dishes.
-对于此游戏，他们应使用独立随机数的生成器，该生成器均匀分布在0和1之间。
+For this game, they shall use a generator of independent random numbers uniformly distributed between 0 and 1.
-游戏从S = 0开始。
+The game starts with S = 0.
-第一个玩家路易丝（Louise）向生成器添加了S个不同的随机数，直到S> 1并记录了她的最后一个随机数'x'。
+The first player, Louise, adds to S different random numbers from the generator until S > 1 and records her last random number 'x'.
-第二个玩家朱莉继续从生成器中添加S个不同的随机数，直到S> 2并记录她的最后一个随机数“ y”。
+The second player, Julie, continues adding to S different random numbers from the generator until S > 2 and records her last random number 'y'.
-拥有最高数字的玩家获胜，而输家则洗碗，即，如果y> x，则第二名玩家获胜。
+The player with the highest number wins and the loser washes the dishes, i.e. if y > x the second player wins.
-例如，如果第一个玩家抽出0.62和0.44，则第一个玩家回合结束，因为0.62 + 0.44> 1且x = 0.44。 如果第二名玩家抽出0.1、0.27和0.91，则第二名玩家回合结束，因为0.62 + 0.44 + 0.1 + 0.27 + 0.91> 2且y = 0.91。 由于y> x，第二名玩家获胜。
+For example, if the first player draws 0.62 and 0.44, the first player turn ends since 0.62+0.44 > 1 and x = 0.44. If the second players draws 0.1, 0.27 and 0.91, the second player turn ends since 0.62+0.44+0.1+0.27+0.91 > 2 and y = 0.91. Since y > x, the second player wins.
-路易丝考虑了一秒钟，然后反对：“那不公平”。 第二名玩家获胜的概率是多少？ 以0.abcdefghij的形式将答案四舍五入到小数点后10位。
+Louise thinks about it for a second, and objects: "That's not fair". What is the probability that the second player wins? Give your answer rounded to 10 places behind the decimal point in the form 0.abcdefghij
# --hints--
-`euler436()`应该返回0.5276662759。
+`euler436()` should return 0.5276662759.
```js
assert.strictEqual(euler436(), 0.5276662759);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-437-fibonacci-primitive-roots.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-437-fibonacci-primitive-roots.md
index 08eda4e8a29..0072d0f9a9b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-437-fibonacci-primitive-roots.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-437-fibonacci-primitive-roots.md
@@ -1,20 +1,44 @@
---
id: 5900f5241000cf542c510036
-title: 问题437：斐波那契原始根
+title: 'Problem 437: Fibonacci primitive roots'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302108
dashedName: problem-437-fibonacci-primitive-roots
---
# --description--
-当我们计算8n模11为n = 0到9时，我们得到：1,8,9,6,4,10,3,2,5,7。我们看到所有可能的值从1到10出现。所以8是11的原始根。但还有更多：如果我们仔细看看，我们看到：1 + 8 = 9 8 + 9 =17≡6mod11 9 + 6 =15≡4mod11 6 + 4 = 10 4 + 10 =14≡3mod11 10 + 3 =13≡2mod11 3 + 2 = 5 2 + 5 = 7 5 + 7 =12≡1mod11。
+When we calculate 8n modulo 11 for n=0 to 9 we get: 1, 8, 9, 6, 4, 10, 3, 2, 5, 7.
-因此，8 mod 11的幂是循环的，具有周期10，并且8n + 8n +1≡8n+ 2（mod 11）。 8被称为11的斐波那契原始根。不是每个素数都有斐波那契原始根。有一个或多个Fibonacci原始根有323个小于10000的素数，这些素数的总和是1480491.用至少一个Fibonacci原始根找到小于100,000,000的素数之和。
+As we see all possible values from 1 to 10 occur. So 8 is a primitive root of 11.
+
+But there is more:
+
+If we take a closer look we see:
+
+1+8=9
+
+8+9=17≡6 mod 11
+
+9+6=15≡4 mod 11
+
+6+4=10
+
+4+10=14≡3 mod 11
+
+10+3=13≡2 mod 11
+
+3+2=5
+
+2+5=7
+
+5+7=12≡1 mod 11.
+
+So the powers of 8 mod 11 are cyclic with period 10, and 8n + 8n+1 ≡ 8n+2 (mod 11). 8 is called a Fibonacci primitive root of 11. Not every prime has a Fibonacci primitive root. There are 323 primes less than 10000 with one or more Fibonacci primitive roots and the sum of these primes is 1480491. Find the sum of the primes less than 100,000,000 with at least one Fibonacci primitive root.
# --hints--
-`euler437()`应该返回74204709657207。
+`euler437()` should return 74204709657207.
```js
assert.strictEqual(euler437(), 74204709657207);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-438-integer-part-of-polynomial-equations-solutions.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-438-integer-part-of-polynomial-equations-solutions.md
index 2f62782b75a..112326a2fa3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-438-integer-part-of-polynomial-equations-solutions.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-438-integer-part-of-polynomial-equations-solutions.md
@@ -1,24 +1,24 @@
---
id: 5900f5231000cf542c510034
-title: 问题438：多项式方程解的整数部分
+title: 'Problem 438: Integer part of polynomial equation''s solutions'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302109
dashedName: problem-438-integer-part-of-polynomial-equations-solutions
---
# --description--
-对于整数的n元组t =（a1，...，an），let（x1，...，xn）是多项式方程xn + a1xn-1 + a2xn-2 + ... +的解。 an-1x + an = 0。
+For an n-tuple of integers t = (a1, ..., an), let (x1, ..., xn) be the solutions of the polynomial equation xn + a1xn-1 + a2xn-2 + ... + an-1x + an = 0.
-考虑以下两个条件：x1，...，xn都是真实的。如果x1，...，xn被排序，则⌊xi⌋= i，1≤i≤n。 （⌊·⌋：地板功能。）
+Consider the following two conditions: x1, ..., xn are all real. If x1, ..., xn are sorted, ⌊xi⌋ = i for 1 ≤ i ≤ n. (⌊·⌋: floor function.)
-在n = 4的情况下，有12个n元组的整数满足两个条件。我们将S（t）定义为t中整数绝对值的总和。对于n = 4，我们可以验证满足两个条件的所有n元组t的ΣS（t）= 2087。
+In the case of n = 4, there are 12 n-tuples of integers which satisfy both conditions. We define S(t) as the sum of the absolute values of the integers in t. For n = 4 we can verify that ∑S(t) = 2087 for all n-tuples t which satisfy both conditions.
-找到ΣS（t）为n = 7。
+Find ∑S(t) for n = 7.
# --hints--
-`euler438()`应该返回2046409616809。
+`euler438()` should return 2046409616809.
```js
assert.strictEqual(euler438(), 2046409616809);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-439-sum-of-sum-of-divisors.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-439-sum-of-sum-of-divisors.md
index f3b303ffc6e..4fb1871a853 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-439-sum-of-sum-of-divisors.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-439-sum-of-sum-of-divisors.md
@@ -1,24 +1,24 @@
---
id: 5900f5231000cf542c510035
-title: 问题439：除数之和
+title: 'Problem 439: Sum of sum of divisors'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302110
dashedName: problem-439-sum-of-sum-of-divisors
---
# --description--
-令d（k）是k的所有除数之和。
+Let d(k) be the sum of all divisors of k.
-我们定义函数S（N）= ∑1≤i≤N ∑1≤j≤Nd（i·j）。
+We define the function S(N) = ∑1≤i≤N ∑1≤j≤Nd(i·j).
-例如，S（3）= d（1）+ d（2）+ d（3）+ d（2）+ d（4）+ d（6）+ d（3）+ d（6）+ d（ 9）= 59。
+For example, S(3) = d(1) + d(2) + d(3) + d(2) + d(4) + d(6) + d(3) + d(6) + d(9) = 59.
-给出S（103）= 563576517282和S（105）mod 109 = 215766508。 找出S（1011）mod 109。
+You are given that S(103) = 563576517282 and S(105) mod 109 = 215766508. Find S(1011) mod 109.
# --hints--
-`euler439()`应该返回968697378。
+`euler439()` should return 968697378.
```js
assert.strictEqual(euler439(), 968697378);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-44-pentagon-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-44-pentagon-numbers.md
index a0e3e9aec4c..3b89176ca1a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-44-pentagon-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-44-pentagon-numbers.md
@@ -1,24 +1,30 @@
---
id: 5900f3981000cf542c50feab
-title: 问题44：五角大楼数字
+title: 'Problem 44: Pentagon numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302111
dashedName: problem-44-pentagon-numbers
---
# --description--
-五边形数由公式P <sub>n</sub> = `n`（3 `n` -1）/ 2生成。 前十个五边形数字是：
+Pentagonal numbers are generated by the formula, P<sub>n</sub>=`n`(3`n`−1)/2. The first ten pentagonal numbers are:
-1，5，12，22，35，51，70，92，117，145，...
+1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
-可以看出，P <sub>4</sub> + P <sub>7</sub> = 22 + 70 = 92 = P <sub>8</sub>。 但是，它们的差70-22 = 48不是五边形。
+It can be seen that P<sub>4</sub> + P<sub>7</sub> = 22 + 70 = 92 = P<sub>8</sub>. However, their difference, 70 − 22 = 48, is not pentagonal.
-找出一对五边形数P <sub>j</sub>和P <sub>k</sub>，它们的和与差为五角形，并且D = | P <sub>k</sub>-P <sub>j</sub> | 最小化； D的值是多少？
+Find the pair of pentagonal numbers, P<sub>j</sub> and P<sub>k</sub>, for which their sum and difference are pentagonal and D = |P<sub>k</sub> − P<sub>j</sub>| is minimized; what is the value of D?
# --hints--
-`pentagonNumbers()`应该返回5482660。
+`pentagonNumbers()` should return a number.
+
+```js
+assert(typeof pentagonNumbers() === 'number');
+```
+
+`pentagonNumbers()` should return 5482660.
```js
assert.strictEqual(pentagonNumbers(), 5482660);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-440-gcd-and-tiling.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-440-gcd-and-tiling.md
index 499a5eae169..a68f99a00ad 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-440-gcd-and-tiling.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-440-gcd-and-tiling.md
@@ -1,28 +1,28 @@
---
id: 5900f5241000cf542c510037
-title: 问题440：GCD和平铺
+title: 'Problem 440: GCD and Tiling'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302112
dashedName: problem-440-gcd-and-tiling
---
# --description--
-我们要完全平铺一块长度为n且高度为1的板，上面有1×2块或1×1块，上面有一个十进制数字：
+We want to tile a board of length n and height 1 completely, with either 1 × 2 blocks or 1 × 1 blocks with a single decimal digit on top:
-例如，以下是铺砌长度为n = 8的板的一些方法：
+For example, here are some of the ways to tile a board of length n = 8:
-令T（n）是如上所述的平铺长度为n的板的方式的数量。
+Let T(n) be the number of ways to tile a board of length n as described above.
-例如，T（1）= 10且T（2）= 101。
+For example, T(1) = 10 and T(2) = 101.
-令S（L）为1≤a，b，c≤L的三次和∑a，b，c gcd（T（ca），T（cb））。 例如： S（2）= 10444 S（3）= 1292115238446807016106539989 S（4）模数987898789 = 670616280。
+Let S(L) be the triple sum ∑a,b,c gcd(T(ca), T(cb)) for 1 ≤ a, b, c ≤ L. For example: S(2) = 10444 S(3) = 1292115238446807016106539989 S(4) mod 987 898 789 = 670616280.
-找出S（2000）mod 987898898。
+Find S(2000) mod 987 898 789.
# --hints--
-`euler440()`应该返回970746056。
+`euler440()` should return 970746056.
```js
assert.strictEqual(euler440(), 970746056);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-441-the-inverse-summation-of-coprime-couples.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-441-the-inverse-summation-of-coprime-couples.md
index bf2bf02fc0e..6cc5fd7e2eb 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-441-the-inverse-summation-of-coprime-couples.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-441-the-inverse-summation-of-coprime-couples.md
@@ -1,24 +1,24 @@
---
id: 5900f5261000cf542c510038
-title: 问题441：互质夫妇的反向总和
+title: 'Problem 441: The inverse summation of coprime couples'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302113
dashedName: problem-441-the-inverse-summation-of-coprime-couples
---
# --description--
-对于整数M，我们将R（M）定义为满足所有这些条件的所有整数对p和q的1 /（p·q）之和：
+For an integer M, we define R(M) as the sum of 1/(p·q) for all the integer pairs p and q which satisfy all of these conditions:
-1≤p<q≤Mp +q≥Mp和q是互质的。
+1 ≤ p < q ≤ M p + q ≥ M p and q are coprime.
-我们还将S（N）定义为R（i）的总和为2≤i≤N。我们可以验证S（2）= R（2）= 1/2，S（10）≈6.9147和S（100） ）≈58.2962。
+We also define S(N) as the sum of R(i) for 2 ≤ i ≤ N. We can verify that S(2) = R(2) = 1/2, S(10) ≈ 6.9147 and S(100) ≈ 58.2962.
-找到S（107）。将您的答案四舍五入到小数点后四位。
+Find S(107). Give your answer rounded to four decimal places.
# --hints--
-`euler441()`应返回5000088.8395。
+`euler441()` should return 5000088.8395.
```js
assert.strictEqual(euler441(), 5000088.8395);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-442-eleven-free-integers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-442-eleven-free-integers.md
index bbe35c7af27..e5b1e35319a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-442-eleven-free-integers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-442-eleven-free-integers.md
@@ -1,24 +1,24 @@
---
id: 5900f5271000cf542c510039
-title: 问题442：十一个整数
+title: 'Problem 442: Eleven-free integers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302114
dashedName: problem-442-eleven-free-integers
---
# --description--
-如果整数的十进制扩展不包含任何表示11的幂的子串（除1之外），则称其为十一无码。
+An integer is called eleven-free if its decimal expansion does not contain any substring representing a power of 11 except 1.
-例如，2404和13431不包含11，而911和4121331不是。
+For example, 2404 and 13431 are eleven-free, while 911 and 4121331 are not.
-令E（n）为第n个正十一无整数。 例如，E（3）= 3，E（200）= 213和E（500’000）= 531563。
+Let E(n) be the nth positive eleven-free integer. For example, E(3) = 3, E(200) = 213 and E(500 000) = 531563.
-找出E（1018）。
+Find E(1018).
# --hints--
-`euler442()`应该返回1295552661530920200。
+`euler442()` should return 1295552661530920200.
```js
assert.strictEqual(euler442(), 1295552661530920200);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-443-gcd-sequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-443-gcd-sequence.md
index 07856e75b42..9afda46059a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-443-gcd-sequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-443-gcd-sequence.md
@@ -1,28 +1,28 @@
---
id: 5900f5271000cf542c51003a
-title: 问题443：GCD序列
+title: 'Problem 443: GCD sequence'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302115
dashedName: problem-443-gcd-sequence
---
# --description--
-设g（n）为如下定义的序列：g（4）= 13，g（n）= g（n-1）+ gcd（n，g（n-1）），n> 4。
+Let g(n) be a sequence defined as follows: g(4) = 13, g(n) = g(n-1) + gcd(n, g(n-1)) for n > 4.
-前几个值是：
+The first few values are:
-```
- n4567891011121314151617181920... g(n)1314161718272829303132333451545560...
-```
+n 4567891011121314151617181920... g(n) 1314161718272829303132333451545560...
-给出g（1 000）= 2524和g（1 000 000）= 2624152。
+<!-- TODO Use MathJax -->
-找到g（1015）。
+You are given that g(1 000) = 2524 and g(1 000 000) = 2624152.
+
+Find g(1015).
# --hints--
-`euler443()`应该返回2744233049300770。
+`euler443()` should return 2744233049300770.
```js
assert.strictEqual(euler443(), 2744233049300770);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-444-the-roundtable-lottery.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-444-the-roundtable-lottery.md
index b4943db3186..04d71de4854 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-444-the-roundtable-lottery.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-444-the-roundtable-lottery.md
@@ -1,33 +1,33 @@
---
id: 5900f52a1000cf542c51003b
-title: 问题444：圆桌会议彩票
+title: 'Problem 444: The Roundtable Lottery'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302116
dashedName: problem-444-the-roundtable-lottery
---
# --description--
-一群人决定坐在圆桌旁玩彩票交易游戏。每个人都是从随机分配的未经批准的彩票开始的。每张票都被划伤，显示全票奖金从1英镑到1英镑不等，没有两张票。游戏的目标是让每个人在离开游戏时最大化他的奖金。
+A group of p people decide to sit down at a round table and play a lottery-ticket trading game. Each person starts off with a randomly-assigned, unscratched lottery ticket. Each ticket, when scratched, reveals a whole-pound prize ranging anywhere from £1 to £p, with no two tickets alike. The goal of the game is for each person to maximize his ticket winnings upon leaving the game.
-选择任意人作为第一个玩家。围绕桌子，每个玩家只有两个选项之一：
+An arbitrary person is chosen to be the first player. Going around the table, each player has only one of two options:
-1. 玩家可以抓住他的票，并向桌上的每个人透露其价值。
-2. 玩家可以将他未击中的门票换成先前玩家划伤的门票，然后将该门票留给游戏。之前的玩家随后抓住了他新买的机票，并向桌上的每个人展示了它的价值。
+1. The player can scratch his ticket and reveal its worth to everyone at the table.
+2. The player can trade his unscratched ticket for a previous player's scratched ticket, and then leave the game with that ticket. The previous player then scratches his newly-acquired ticket and reveals its worth to everyone at the table.
-所有门票都被划伤后游戏结束。仍然留在桌上的所有玩家必须带着他们当前持有的门票。
+The game ends once all tickets have been scratched. All players still remaining at the table must leave with their currently-held tickets.
-假设每个玩家使用最优策略来最大化他的票奖金的期望值。
+Assume that each player uses the optimal strategy for maximizing the expected value of his ticket winnings.
-设E（p）表示当游戏在由p个玩家组成的游戏中结束时留在桌旁的玩家的预期数量（例如当舍入到5个有效数字时E（111）= 5.2912）。
+Let E(p) represent the expected number of players left at the table when the game ends in a game consisting of p players (e.g. E(111) = 5.2912 when rounded to 5 significant digits).
-设S1（N）= E（p）令Sk（N）= Sk-1（p），其中k> 1
+Let S1(N) = E(p) Let Sk(N) = Sk-1(p) for k > 1
-找到S20（1014）并用科学记数法将答案写成10位有效数字。使用小写e来分隔尾数和指数（例如S3（100）= 5.983679014e5）。
+Find S20(1014) and write the answer in scientific notation rounded to 10 significant digits. Use a lowercase e to separate mantissa and exponent (e.g. S3(100) = 5.983679014e5).
# --hints--
-`euler444()`应该返回1.200856722e+263。
+`euler444()` should return 1.200856722e+263.
```js
assert.strictEqual(euler444(), 1.200856722e263);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-445-retractions-a.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-445-retractions-a.md
index 5fec157d821..39c80f706d7 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-445-retractions-a.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-445-retractions-a.md
@@ -1,22 +1,24 @@
---
id: 5900f52a1000cf542c51003c
-title: 问题445：撤回A.
+title: 'Problem 445: Retractions A'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302117
dashedName: problem-445-retractions-a
---
# --description--
-对于每个整数n> 1，函数族fn，a，b由fn，a，b（x）≡ax+ b mod n定义为a，b，x整数和0
+For every integer n>1, the family of functions fn,a,b is defined
-对于c = C（100 000，k），1≤k≤99999≡628701600（mod 1 000 000 007），给出ΣR（c）。 （C（n，k）是二项式系数）。
+by fn,a,b(x)≡ax+b mod n for a,b,x integer and 0
-找到ΣR（c）得c = C（10 000 000，k），1≤k≤9999 999.给出你的答案模1 000 000 007。
+You are given that ∑ R(c) for c=C(100 000,k), and 1 ≤ k ≤99 999 ≡628701600 (mod 1 000 000 007). (C(n,k) is the binomial coefficient).
+
+Find ∑ R(c) for c=C(10 000 000,k), and 1 ≤k≤ 9 999 999. Give your answer modulo 1 000 000 007.
# --hints--
-`euler445()`应该返回659104042。
+`euler445()` should return 659104042.
```js
assert.strictEqual(euler445(), 659104042);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-446-retractions-b.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-446-retractions-b.md
index 0762c88a754..a46677999e8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-446-retractions-b.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-446-retractions-b.md
@@ -1,22 +1,24 @@
---
id: 5900f52c1000cf542c51003d
-title: 问题446：撤回B.
+title: 'Problem 446: Retractions B'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302118
dashedName: problem-446-retractions-b
---
# --description--
-对于每个整数n> 1，函数族fn，a，b由fn，a，b（x）≡ax+ b mod n定义为a，b，x整数和0
+For every integer n>1, the family of functions fn,a,b is defined
-对于1≤n≤N，F（N）=ΣR（n4 + 4）。 F（1024）= 77532377300600。
+by fn,a,b(x)≡ax+b mod n for a,b,x integer and 0
-求F（107）（mod 1 000 000 007）
+F(N)=∑R(n4+4) for 1≤n≤N. F(1024)=77532377300600.
+
+Find F(107) (mod 1 000 000 007)
# --hints--
-`euler446()`应该返回907803852。
+`euler446()` should return 907803852.
```js
assert.strictEqual(euler446(), 907803852);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-447-retractions-c.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-447-retractions-c.md
index 8b0b3628985..c8cb09065ca 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-447-retractions-c.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-447-retractions-c.md
@@ -1,22 +1,24 @@
---
id: 5900f52c1000cf542c51003e
-title: 问题447：撤回C.
+title: 'Problem 447: Retractions C'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302119
dashedName: problem-447-retractions-c
---
# --description--
-对于每个整数n> 1，函数族fn，a，b由fn，a，b（x）≡ax+ b mod n定义为a，b，x整数和0
+For every integer n>1, the family of functions fn,a,b is defined
-对于2≤n≤N，F（N）=ΣR（n）。 F（107）≡638042271（mod 1 000 000 007）。
+by fn,a,b(x)≡ax+b mod n for a,b,x integer and 0
-求F（1014）（mod 1 000 000 007）。
+F(N)=∑R(n) for 2≤n≤N. F(107)≡638042271 (mod 1 000 000 007).
+
+Find F(1014) (mod 1 000 000 007).
# --hints--
-`euler447()`应返回530553372。
+`euler447()` should return 530553372.
```js
assert.strictEqual(euler447(), 530553372);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-448-average-least-common-multiple.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-448-average-least-common-multiple.md
index 609b3de8681..6bf716c1d34 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-448-average-least-common-multiple.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-448-average-least-common-multiple.md
@@ -1,22 +1,26 @@
---
id: 5900f52c1000cf542c51003f
-title: 问题448：平均最小公倍数
+title: 'Problem 448: Average least common multiple'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302120
dashedName: problem-448-average-least-common-multiple
---
# --description--
-函数lcm（a，b）表示a和b的最小公倍数。设A（n）为1≤i≤n的lcm（n，i）的平均值。例如：A（2）=（2 + 2）/ 2 = 2且A（10）=（10 + 10 + 30 + 20 + 10 + 30 + 70 + 40 + 90 + 10）/ 10 = 32。
+The function lcm(a,b) denotes the least common multiple of a and b.
-令S（n）=ΣA（k）为1≤k≤n。 S（100）= 122726。
+Let A(n) be the average of the values of lcm(n,i) for 1≤i≤n.
-找到S（99999999019）mod 999999017。
+E.g: A(2)=(2+2)/2=2 and A(10)=(10+10+30+20+10+30+70+40+90+10)/10=32.
+
+Let S(n)=∑A(k) for 1≤k≤n. S(100)=122726.
+
+Find S(99999999019) mod 999999017.
# --hints--
-`euler448()`应该返回106467648。
+`euler448()` should return 106467648.
```js
assert.strictEqual(euler448(), 106467648);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-449-chocolate-covered-candy.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-449-chocolate-covered-candy.md
index 07f2e675a4d..18c19230958 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-449-chocolate-covered-candy.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-449-chocolate-covered-candy.md
@@ -1,28 +1,28 @@
---
id: 5900f52d1000cf542c510040
-title: 问题449：巧克力糖果
+title: 'Problem 449: Chocolate covered candy'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302121
dashedName: problem-449-chocolate-covered-candy
---
# --description--
-菲尔糖果制造商正在制作一批新的巧克力糖果。每个糖果中心的形状类似于由以下等式定义的旋转椭圆体：b2x2 + b2y2 + a2z2 = a2b2。
+Phil the confectioner is making a new batch of chocolate covered candy. Each candy centre is shaped like an ellipsoid of revolution defined by the equation: b2x2 + b2y2 + a2z2 = a2b2.
-菲尔想知道需要多少巧克力来覆盖一个糖果中心，用一层厚厚的巧克力涂层。如果a = 1 mm且b = 1 mm，则所需的巧克力量为
+Phil wants to know how much chocolate is needed to cover one candy centre with a uniform coat of chocolate one millimeter thick. If a=1 mm and b=1 mm, the amount of chocolate required is
-```
- 283 π mm3
-```
+283 π mm3
-如果a = 2 mm且b = 1 mm，则所需的巧克力量约为60.35475635 mm3。
+<!-- TODO Use MathJax -->
-如果a = 3 mm且b = 1 mm，则找出所需的mm3巧克力量。将数字舍入到小数点后面的小数点后8位。
+If a=2 mm and b=1 mm, the amount of chocolate required is approximately 60.35475635 mm3.
+
+Find the amount of chocolate in mm3 required if a=3 mm and b=1 mm. Give your answer as the number rounded to 8 decimal places behind the decimal point.
# --hints--
-`euler449()`应返回103.37870096。
+`euler449()` should return 103.37870096.
```js
assert.strictEqual(euler449(), 103.37870096);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-45-triangular-pentagonal-and-hexagonal.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-45-triangular-pentagonal-and-hexagonal.md
index a0518d4a81c..5c3b224b4a1 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-45-triangular-pentagonal-and-hexagonal.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-45-triangular-pentagonal-and-hexagonal.md
@@ -1,28 +1,32 @@
---
id: 5900f3991000cf542c50feac
-title: 问题45：三角形，五边形和六边形
+title: 'Problem 45: Triangular, pentagonal, and hexagonal'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302122
dashedName: problem-45-triangular-pentagonal-and-hexagonal
---
# --description--
-三角形，五边形和六边形数由以下公式生成：
+Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
-<div style='display：inline-grid; 文本对齐：居中； 网格模板列：repeat（3，minmax（117px，12％））; grid-template-rows：auto;'> <div>三角形</div> <div> T <sub> n </sub> = <var> n </var>（<var> n </var> +1 ）/ 2 </div> <div> 1、3、6、10、15 ... </div> </div>
+<div style='display: inline-grid; text-align: center; grid-template-columns: 135px 135px 260px; grid-template-rows: auto;'><div>Triangle</div><div>T<sub>n</sub>=<var>n</var>(<var>n</var>+1)/2</div><div>1, 3, 6, 10, 15, ...</div></div>
+<div style='display: inline-grid; text-align: center; grid-template-columns: 135px 135px 260px; grid-template-rows: auto;'><div>Pentagonal</div><div>P<sub>n</sub>=<var>n</var>(3<var>n</var>−1)/2</div><div>1, 5, 12, 22, 35, ...</div></div>
+<div style='display: inline-grid; text-align: center; grid-template-columns: 135px 135px 260px; grid-template-rows: auto;'><div>Hexagonal</div><div>H<sub>n</sub>=<var>n</var>(2<var>n</var>−1)</div><div>1, 6, 15, 28, 45, ...</div></div>
-<div style='display：inline-grid; 文本对齐：居中； 网格模板列：repeat（3，minmax（117px，12％））; grid-template-rows：auto;'> <div>五角形</div> <div> P <sub> n </sub> = <var> n </var>（3 <var> n </var>- 1）/ 2 </div> <div> 1、5、12、22、35，... </div> </div>
+It can be verified that T<sub>285</sub> = P<sub>165</sub> = H<sub>143</sub> = 40755.
-<div style='display：inline-grid; 文本对齐：居中； 网格模板列：repeat（3，minmax（117px，12％））; grid-template-rows：auto;'> <div>六角形</div> <div> H <sub> n </sub> = <var> n </var>（2 <var> n </var>- 1）</div> <div> 1、6、15、28、45，... </div> </div>
-
-可以证明T <sub>285</sub> = P <sub>165</sub> = H <sub>143</sub> = 40755。
-
-查找下一个也是五边形和六边形的三角形。
+Find the next triangle number that is also pentagonal and hexagonal.
# --hints--
-`triPentaHexa(40756)`应该返回1533776805。
+`triPentaHexa(40756)` should return a number.
+
+```js
+assert(typeof triPentaHexa(40756) === 'number');
+```
+
+`triPentaHexa(40756)` should return 1533776805.
```js
assert.strictEqual(triPentaHexa(40756), 1533776805);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-450-hypocycloid-and-lattice-points.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-450-hypocycloid-and-lattice-points.md
index faec320af7a..17cf3af082d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-450-hypocycloid-and-lattice-points.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-450-hypocycloid-and-lattice-points.md
@@ -1,34 +1,40 @@
---
id: 5900f52e1000cf542c510041
-title: 问题450：Hypocycloid和Lattice点
+title: 'Problem 450: Hypocycloid and Lattice points'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302123
dashedName: problem-450-hypocycloid-and-lattice-points
---
# --description--
-内摆线是由在较大圆内滚动的小圆上的点绘制的曲线。以原点为中心，从最右边开始的内摆线的参数方程由下式给出：$ x（t）=（R - r）\\ cos（t）+ r \\ cos（\\ frac {R - r} rt）$ $ y（t）=（R - r）\\ sin（t） - r \\ sin（\\ frac {R - r} rt）$其中R是大圆的半径，r是小圆的半径圈。
+A hypocycloid is the curve drawn by a point on a small circle rolling inside a larger circle. The parametric equations of a hypocycloid centered at the origin, and starting at the right most point is given by:
-设$ C（R，r）$是具有半径为R和r的内摆线上的整数坐标的不同点的集合，并且对应的值为t，使得$ \\ sin（t）$和$ \\ cos（ t）$是有理数。
+$x(t) = (R - r) \\cos(t) + r \\cos(\\frac {R - r} r t)$
-设$ S（R，r）= \\ sum \_ {（x，y）\\ in C（R，r）} | x | + | y | $是$ C（R，r）$中点的x和y坐标的绝对值之和。
+$y(t) = (R - r) \\sin(t) - r \\sin(\\frac {R - r} r t)$
-设$ T（N）= \\ sum *{R = 3} ^ N \\ sum* {r = 1} ^ {\\ lfloor \\ frac {R - 1} 2 \\ rfloor} S（R，r）$是$的总和S（R，r）$表示R和r正整数，$ R \\ leq N $和$ 2r <R $。
+Where R is the radius of the large circle and r the radius of the small circle.
-给出：C（3,1）= {（3,0），（-1,2），（ - 1,0），（ - 1，-2）} C（2500,1000）= {（2500 ，0），（772,2376），（772，-2376），（516,1792），（516，-1792），（500,0），（68,504），（68，-504），（ -1356,1088），（ - 1356，-1088），（ - 1500,1000），（ - 1500，-1000）}
+Let $C(R, r)$ be the set of distinct points with integer coordinates on the hypocycloid with radius R and r and for which there is a corresponding value of t such that $\\sin(t)$ and $\\cos(t)$ are rational numbers.
-注意：（ - 625,0）不是C（2500,1000）的元素，因为$ \\ sin（t）$不是t的相应值的有理数。
+Let $S(R, r) = \\sum\_{(x,y) \\in C(R, r)} |x| + |y|$ be the sum of the absolute values of the x and y coordinates of the points in $C(R, r)$.
-S（3,1）=（| 3 | + | 0 |）+（| -1 | + | 2 |）+（| -1 | + | 0 |）+（| -1 | + | -2 |） = 10
+Let $T(N) = \\sum*{R = 3}^N \\sum*{r=1}^{\\lfloor \\frac {R - 1} 2 \\rfloor} S(R, r)$ be the sum of $S(R, r)$ for R and r positive integers, $R\\leq N$ and $2r < R$.
-T（3）= 10; T（10）= 524; T（100）= 580442; T（103）= 583108600。
+You are given:C(3, 1) = {(3, 0), (-1, 2), (-1,0), (-1,-2)} C(2500, 1000) = {(2500, 0), (772, 2376), (772, -2376), (516, 1792), (516, -1792), (500, 0), (68, 504), (68, -504),(-1356, 1088), (-1356, -1088), (-1500, 1000), (-1500, -1000)}
-求T（106）。
+Note: (-625, 0) is not an element of C(2500, 1000) because $\\sin(t)$ is not a rational number for the corresponding values of t.
+
+S(3, 1) = (|3| + |0|) + (|-1| + |2|) + (|-1| + |0|) + (|-1| + |-2|) = 10
+
+T(3) = 10; T(10) = 524 ;T(100) = 580442; T(103) = 583108600.
+
+Find T(106).
# --hints--
-`euler450()`应该返回583333163984220900。
+`euler450()` should return 583333163984220900.
```js
assert.strictEqual(euler450(), 583333163984220900);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-451-modular-inverses.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-451-modular-inverses.md
index c8a5dbcdb19..1b12ec1c227 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-451-modular-inverses.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-451-modular-inverses.md
@@ -1,22 +1,40 @@
---
id: 5900f5311000cf542c510042
-title: 问题451：模逆
+title: 'Problem 451: Modular inverses'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302124
dashedName: problem-451-modular-inverses
---
# --description--
-考虑数字15.有八个正数小于15，它们与15：1,2,4,7,8,11,13,14相互作用。这些数模15的模数逆是：1,8,4 ，13,2,11,7,14因为1 \* 1 mod 15 = 1 2 \* 8 = 16 mod 15 = 1 4 \* 4 = 16 mod 15 = 1 7 \* 13 = 91 mod 15 = 1 11 \* 11 = 121 mod 15 = 1 14 \* 14 = 196 mod 15 = 1
+Consider the number 15.
-设I（n）是小于n-1的最大正数m，使得m modulo n的模逆与m本身相等。所以我（15）= 11。我（100）= 51和I（7）= 1。
+There are eight positive numbers less than 15 which are coprime to 15: 1, 2, 4, 7, 8, 11, 13, 14.
-求3Σn≤2·107的ΣI（n）
+The modular inverses of these numbers modulo 15 are: 1, 8, 4, 13, 2, 11, 7, 14
+
+because
+
+1\*1 mod 15=1
+
+2\*8=16 mod 15=1
+
+4\*4=16 mod 15=1
+
+7\*13=91 mod 15=1
+
+11\*11=121 mod 15=1
+
+14\*14=196 mod 15=1
+
+Let I(n) be the largest positive number m smaller than n-1 such that the modular inverse of m modulo n equals m itself. So I(15)=11. Also I(100)=51 and I(7)=1.
+
+Find ∑I(n) for 3≤n≤2·107
# --hints--
-`euler451()`应该返回153651073760956。
+`euler451()` should return 153651073760956.
```js
assert.strictEqual(euler451(), 153651073760956);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-452-long-products.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-452-long-products.md
index eb791b4ffe8..b52ca22d506 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-452-long-products.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-452-long-products.md
@@ -1,24 +1,24 @@
---
id: 5900f5311000cf542c510043
-title: 问题452：长期产品
+title: 'Problem 452: Long Products'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302125
dashedName: problem-452-long-products
---
# --description--
-将F（m，n）定义为元素的乘积不超过m的n个正整数的元组数。
+Define F(m,n) as the number of n-tuples of positive integers for which the product of the elements doesn't exceed m.
-F（10，10）= 571。
+F(10, 10) = 571.
-F（106，106）mod 1234567891 = 252903833。
+F(106, 106) mod 1 234 567 891 = 252903833.
-找出F（109，109）mod 1234567891。
+Find F(109, 109) mod 1 234 567 891.
# --hints--
-`euler452()`应该返回345558983。
+`euler452()` should return 345558983.
```js
assert.strictEqual(euler452(), 345558983);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-453-lattice-quadrilaterals.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-453-lattice-quadrilaterals.md
index 5191216aef7..69b668a626e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-453-lattice-quadrilaterals.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-453-lattice-quadrilaterals.md
@@ -1,26 +1,26 @@
---
id: 5900f5311000cf542c510044
-title: 问题453：格子四边形
+title: 'Problem 453: Lattice Quadrilaterals'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302126
dashedName: problem-453-lattice-quadrilaterals
---
# --description--
-简单的四边形是具有四个不同顶点的多边形，没有直角并且不会自相交。
+A simple quadrilateral is a polygon that has four distinct vertices, has no straight angles and does not self-intersect.
-设Q（m，n）为简单四边形的数量，其顶点为格点，坐标（x，y）满足0≤x≤m且0≤y≤n。
+Let Q(m, n) be the number of simple quadrilaterals whose vertices are lattice points with coordinates (x,y) satisfying 0 ≤ x ≤ m and 0 ≤ y ≤ n.
-例如，Q（2,2）= 94，如下所示：
+For example, Q(2, 2) = 94 as can be seen below:
-还可以证实Q（3,7）= 39590，Q（12,3）= 309000和Q（123,45）= 70542215894646。
+It can also be verified that Q(3, 7) = 39590, Q(12, 3) = 309000 and Q(123, 45) = 70542215894646.
-找到Q（12345,6789）mod 135707531。
+Find Q(12345, 6789) mod 135707531.
# --hints--
-`euler453()`应该返回104354107。
+`euler453()` should return 104354107.
```js
assert.strictEqual(euler453(), 104354107);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-454-diophantine-reciprocals-iii.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-454-diophantine-reciprocals-iii.md
index c92f61e4d33..414cfab965d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-454-diophantine-reciprocals-iii.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-454-diophantine-reciprocals-iii.md
@@ -1,24 +1,24 @@
---
id: 5900f5331000cf542c510045
-title: 问题454：丢番图倒数III
+title: 'Problem 454: Diophantine reciprocals III'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302127
dashedName: problem-454-diophantine-reciprocals-iii
---
# --description--
-在下面的等式中，x，y和n是正整数。
+In the following equation x, y, and n are positive integers. 1/x + 1/y= 1/n
-1x + 1y = 1n
+<!-- TODO Use MathJax -->
-对于极限L，我们将F（L）定义为满足x <y≤L的解的数量。
+For a limit L we define F(L) as the number of solutions which satisfy x < y ≤ L.
-我们可以验证F（15）= 4和F（1000）= 1069.找到F（1012）。
+We can verify that F(15) = 4 and F(1000) = 1069. Find F(1012).
# --hints--
-`euler454()`应该返回5435004633092。
+`euler454()` should return 5435004633092.
```js
assert.strictEqual(euler454(), 5435004633092);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-455-powers-with-trailing-digits.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-455-powers-with-trailing-digits.md
index 70a3874d3de..71202a5f3f2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-455-powers-with-trailing-digits.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-455-powers-with-trailing-digits.md
@@ -1,22 +1,22 @@
---
id: 5900f5331000cf542c510046
-title: 问题455：使用尾随数字的权力
+title: 'Problem 455: Powers With Trailing Digits'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302129
dashedName: problem-455-powers-with-trailing-digits
---
# --description--
-设f（n）是小于109的最大正整数x，使得nx的最后9位数字形成数字x（包括前导零），或者如果不存在这样的整数则为零。
+Let f(n) be the largest positive integer x less than 109 such that the last 9 digits of nx form the number x (including leading zeros), or zero if no such integer exists.
-例如：
+For example:
-f（4）= 411728896（4411728896 = ... 490411728896）f（10）= 0 f（157）= 743757（157743757 = ... 567000743757）Σf（n），2≤n≤103= 442530011399查找Σf（n ），2≤n≤106。
+f(4) = 411728896 (4411728896 = ...490411728896) f(10) = 0 f(157) = 743757 (157743757 = ...567000743757) Σf(n), 2 ≤ n ≤ 103 = 442530011399 Find Σf(n), 2 ≤ n ≤ 106.
# --hints--
-`euler455()`应该返回450186511399999。
+`euler455()` should return 450186511399999.
```js
assert.strictEqual(euler455(), 450186511399999);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-456-triangles-containing-the-origin-ii.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-456-triangles-containing-the-origin-ii.md
index e3b24ba01d6..fba2743a699 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-456-triangles-containing-the-origin-ii.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-456-triangles-containing-the-origin-ii.md
@@ -1,26 +1,28 @@
---
id: 5900f5351000cf542c510047
-title: 问题456：包含原点II的三角形
+title: 'Problem 456: Triangles containing the origin II'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302130
dashedName: problem-456-triangles-containing-the-origin-ii
---
# --description--
-定义：xn =（1248n mod 32323） - 16161yn =（8421n mod 30103） - 15051 Pn = {（x1，y1），（x2，y2），...，（xn，yn）}
+Define:xn = (1248n mod 32323) - 16161yn = (8421n mod 30103) - 15051
-例如，P8 = {（ - 14913，-6630），（ - 10161,5625），（5226,11889），（8340，-10778），（15852，-5203），（ - 15165,11295），（ - 1427，-14495），（12407,1060）}。
+Pn = {(x1, y1), (x2, y2), ..., (xn, yn)}
-设C（n）是顶点在Pn中的三角形的数量，其中包含内部的原点。
+For example, P8 = {(-14913, -6630), (-10161, 5625), (5226, 11896), (8340, -10778), (15852, -5203), (-15165, 11295), (-1427, -14495), (12407, 1060)}.
-示例：C（8）= 20 C（600）= 8950634 C（40 000）= 2666610948988
+Let C(n) be the number of triangles whose vertices are in Pn which contain the origin in the interior.
-找到C（2 000 000）。
+Examples: C(8) = 20 C(600) = 8950634 C(40 000) = 2666610948988
+
+Find C(2 000 000).
# --hints--
-`euler456()`应返回333333208685971500。
+`euler456()` should return 333333208685971500.
```js
assert.strictEqual(euler456(), 333333208685971500);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-457-a-polynomial-modulo-the-square-of-a-prime.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-457-a-polynomial-modulo-the-square-of-a-prime.md
index 2b8686a5afa..77c3da532c9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-457-a-polynomial-modulo-the-square-of-a-prime.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-457-a-polynomial-modulo-the-square-of-a-prime.md
@@ -1,22 +1,26 @@
---
id: 5900f5361000cf542c510048
-title: 问题457：多项式以素数的平方为模
+title: 'Problem 457: A polynomial modulo the square of a prime'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302131
dashedName: problem-457-a-polynomial-modulo-the-square-of-a-prime
---
# --description--
-设f（n）= n2 - 3n - 1.设p为素数。令R（p）是最小的正整数n，使得如果存在这样的整数n则f（n）mod p2 = 0，否则R（p）= 0。
+Let f(n) = n2 - 3n - 1.
-对于不超过L的所有素数，令SR（L）为ΣR（p）。
+Let p be a prime.
-找到SR（107）。
+Let R(p) be the smallest positive integer n such that f(n) mod p2 = 0 if such an integer n exists, otherwise R(p) = 0.
+
+Let SR(L) be ∑R(p) for all primes not exceeding L.
+
+Find SR(107).
# --hints--
-`euler457()`应该返回2647787126797397000。
+`euler457()` should return 2647787126797397000.
```js
assert.strictEqual(euler457(), 2647787126797397000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-458-permutations-of-project.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-458-permutations-of-project.md
index 8149afb95d7..de844d0fe53 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-458-permutations-of-project.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-458-permutations-of-project.md
@@ -1,22 +1,24 @@
---
id: 5900f5361000cf542c510049
-title: 问题458：项目的排列
+title: 'Problem 458: Permutations of Project'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302132
dashedName: problem-458-permutations-of-project
---
# --description--
-考虑由“项目”一词的字母组成的字母A：A = {c，e，j，o，p，r，t}。令T（n）为长度为n的字符串数，该字符串由来自A的字母组成，其中没有子字符串，该子字符串是“项目”的5040个排列之一。
+Consider the alphabet A made out of the letters of the word "project": A={c,e,j,o,p,r,t}.
-T（7）= 77-7！= 818503。
+Let T(n) be the number of strings of length n consisting of letters from A that do not have a substring that is one of the 5040 permutations of "project".
-找到T（1012）。给出答案的最后9位数字。
+T(7)=77-7!=818503.
+
+Find T(1012). Give the last 9 digits of your answer.
# --hints--
-`euler458()`应该返回423341841。
+`euler458()` should return 423341841.
```js
assert.strictEqual(euler458(), 423341841);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-459-flipping-game.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-459-flipping-game.md
index bd51eeb3016..72134eec524 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-459-flipping-game.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-459-flipping-game.md
@@ -1,28 +1,32 @@
---
id: 5900f5371000cf542c51004a
-title: 问题459：翻转游戏
+title: 'Problem 459: Flipping game'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302133
dashedName: problem-459-flipping-game
---
# --description--
-翻转游戏是在N×N方板上玩的双人游戏。每个方块包含一面白色和一面黑色的圆盘。游戏开始时所有磁盘都显示白色。
+The flipping game is a two player game played on a N by N square board.
-转弯包括翻转具有以下属性的矩形中的所有磁盘：矩形的右上角包含一个白色圆盘，矩形宽度是一个完美的正方形（1,4,9,16，...），矩形高度为三角形数字（1,3,6,10 ......）
+Each square contains a disk with one side white and one side black.
-球员轮流转换。玩家通过将网格全黑变为胜利。
+The game starts with all disks showing their white side.
-假设完美游戏，让W（N）为N×N板上第一个玩家的获胜动作数，所有盘都是白色的。 W（1）= 1，W（2）= 0，W（5）= 8，W（102）= 31395。
+A turn consists of flipping all disks in a rectangle with the following properties: the upper right corner of the rectangle contains a white disk the rectangle width is a perfect square (1, 4, 9, 16, ...) the rectangle height is a triangular number (1, 3, 6, 10, ...)
-对于N = 5，第一个玩家的八个获胜第一步是：
+Players alternate turns. A player wins by turning the grid all black.
-找到W（106）。
+Let W(N) be the number of winning moves for the first player on a N by N board with all disks white, assuming perfect play. W(1) = 1, W(2) = 0, W(5) = 8 and W(102) = 31395.
+
+For N=5, the first player's eight winning first moves are:
+
+Find W(106).
# --hints--
-`euler459()`应该返回3996390106631。
+`euler459()` should return 3996390106631.
```js
assert.strictEqual(euler459(), 3996390106631);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-46-goldbachs-other-conjecture.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-46-goldbachs-other-conjecture.md
index c95bd903f71..9cd82e5bdf2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-46-goldbachs-other-conjecture.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-46-goldbachs-other-conjecture.md
@@ -1,18 +1,37 @@
---
id: 5900f39a1000cf542c50fead
-title: 问题46：哥德巴赫的另一个猜想
+title: 'Problem 46: Goldbach''s other conjecture'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302134
dashedName: problem-46-goldbachs-other-conjecture
---
# --description--
-由Christian Goldbach提出，每个奇数的复合数可以写成素数和两个平方的总和。 9 = 7 + 2×1 <sup>2</sup> 15 = 7 + 2×2 <sup>2</sup> 21 = 3 + 2×3 <sup>2</sup> 25 = 7 + 2×3 <sup>2</sup> 27 = 19 + 2×2 <sup>2</sup> 33 = 31 + 2×1 <sup>2</sup>转这个猜想是假的。什么是最小的奇数复合，不能写为素数和两倍平方的总和？
+It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square.
+
+<div style='margin-left: 2em;'>
+ 9 = 7 + 2×1<sup>2</sup><br>
+ 15 = 7 + 2×2<sup>2</sup><br>
+ 21 = 3 + 2×3<sup>2</sup><br>
+ 25 = 7 + 2×3<sup>2</sup><br>
+ 27 = 19 + 2×2<sup>2</sup><br>
+ 33 = 31 + 2×1<sup>2</sup>
+</div>
+
+It turns out that the conjecture was false.
+
+What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?
# --hints--
-`goldbachsOtherConjecture()`应返回5777。
+`goldbachsOtherConjecture()` should return a number.
+
+```js
+assert(typeof goldbachsOtherConjecture() === 'number');
+```
+
+`goldbachsOtherConjecture()` should return 5777.
```js
assert.strictEqual(goldbachsOtherConjecture(), 5777);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-460-an-ant-on-the-move.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-460-an-ant-on-the-move.md
index 09cd6d984cf..467a9693c3a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-460-an-ant-on-the-move.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-460-an-ant-on-the-move.md
@@ -1,28 +1,28 @@
---
id: 5900f5381000cf542c51004b
-title: 问题460：移动中的蚂蚁
+title: 'Problem 460: An ant on the move'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302135
dashedName: problem-460-an-ant-on-the-move
---
# --description--
-在欧几里得平面上，蚂蚁从点A（0,1）行进到点B（d，1）得到整数d。
+On the Euclidean plane, an ant travels from point A(0, 1) to point B(d, 1) for an integer d.
-在每个步骤中，点（x0，y0）处的蚂蚁选择满足x1≥0且y1≥1的格点（x1，y1）之一，并以恒定速度v直接到（x1，y1）。 v取决于y0和y1如下：如果y0 = y1，则v的值等于y0。如果y0≠y1，则v的值等于（y1-y0）/（ln（y1）-ln（y0））。
+In each step, the ant at point (x0, y0) chooses one of the lattice points (x1, y1) which satisfy x1 ≥ 0 and y1 ≥ 1 and goes straight to (x1, y1) at a constant velocity v. The value of v depends on y0 and y1 as follows: If y0 = y1, the value of v equals y0. If y0 ≠ y1, the value of v equals (y1 - y0) / (ln(y1) - ln(y0)).
-左图是d = 4的可能路径之一。首先，蚂蚁以速度（3 - 1）/（ln（3） - ln（1）从A（0,1）到达P1（1,3） ）≈1.8205。然后所需的时间是sqrt（5）/1.8205≈1.2283。从P1（1,3）到P2（3,3），蚂蚁以速度3行进，因此所需时间为2 /3≈0.6667。从P2（3,3）到B（4,1），蚂蚁以速度（1 - 3）/（ln（1） - ln（3））≈1.8205，所以需要的时间是sqrt（5）/ 1.8205 ...“抓鸟”英语词典1.2283。因此总的所需时间是1.2283 + 0.6667 + 1.2283 = 3.1233。
+The left image is one of the possible paths for d = 4. First the ant goes from A(0, 1) to P1(1, 3) at velocity (3 - 1) / (ln(3) - ln(1)) ≈ 1.8205. Then the required time is sqrt(5) / 1.8205 ≈ 1.2283. From P1(1, 3) to P2(3, 3) the ant travels at velocity 3 so the required time is 2 / 3 ≈ 0.6667. From P2(3, 3) to B(4, 1) the ant travels at velocity (1 - 3) / (ln(1) - ln(3)) ≈ 1.8205 so the required time is sqrt(5) / 1.8205 ≈ 1.2283. Thus the total required time is 1.2283 + 0.6667 + 1.2283 = 3.1233.
-正确的形象是另一条道路。总的所需时间计算为0.98026 + 1 + 0.98026 = 2.96052。可以证明这是d = 4的最快路径。
+The right image is another path. The total required time is calculated as 0.98026 + 1 + 0.98026 = 2.96052. It can be shown that this is the quickest path for d = 4.
-如果蚂蚁选择最快的路径，则让F（d）成为所需的总时间。例如，F（4）≈2.960516287。我们可以验证F（10）≈4.6668787834和F（100）≈9.217221972。
+Let F(d) be the total required time if the ant chooses the quickest path. For example, F(4) ≈ 2.960516287. We can verify that F(10) ≈ 4.668187834 and F(100) ≈ 9.217221972.
-找到F（10000）。将您的答案四舍五入到小数点后九位。
+Find F(10000). Give your answer rounded to nine decimal places.
# --hints--
-`euler460()`应该返回18.420738199。
+`euler460()` should return 18.420738199.
```js
assert.strictEqual(euler460(), 18.420738199);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-461-almost-pi.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-461-almost-pi.md
index d4894df611e..50a12c051c0 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-461-almost-pi.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-461-almost-pi.md
@@ -1,18 +1,30 @@
---
id: 5900f53a1000cf542c51004c
-title: 问题461：几乎是Pi
+title: 'Problem 461: Almost Pi'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302136
dashedName: problem-461-almost-pi
---
# --description--
-对于所有非负整数k，设fn（k）= ek / n-1。值得注意的是，f200（6）+ f200（75）+ f200（89）+ f200（226）= 3.141592644529 ...≈π。事实上，对于n = 200，它是fn（a）+ fn（b）+ fn（c）+ fn（d）形式的π的最佳近似值。设g（n）= a2 + b2 + c2 + d 2为a，b，c，d最小化错误：| fn（a）+ fn（b）+ fn（c）+ fn（d） - π| （其中| x |表示x的绝对值）。给出g（200）= 62 + 752 + 892 + 2262 = 64658.求g（10000）。
+Let fn(k) = ek/n - 1, for all non-negative integers k.
+
+Remarkably, f200(6) + f200(75) + f200(89) + f200(226) = 3.141592644529… ≈ π.
+
+In fact, it is the best approximation of π of the form fn(a) + fn(b) + fn(c) + fn(d) for n = 200.
+
+Let g(n) = a2 + b2 + c2 + d 2 for a, b, c, d that minimize the error: | fn(a) + fn(b) + fn(c) + fn(d) - π|
+
+(where |x| denotes the absolute value of x).
+
+You are given g(200) = 62 + 752 + 892 + 2262 = 64658.
+
+Find g(10000).
# --hints--
-`euler461()`应该返回159820276。
+`euler461()` should return 159820276.
```js
assert.strictEqual(euler461(), 159820276);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-462-permutation-of-3-smooth-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-462-permutation-of-3-smooth-numbers.md
index 281b919e575..6eb9550e454 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-462-permutation-of-3-smooth-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-462-permutation-of-3-smooth-numbers.md
@@ -1,27 +1,32 @@
---
id: 5900f53b1000cf542c51004d
-title: 问题462：3个平滑数的排列
+title: 'Problem 462: Permutation of 3-smooth numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302137
dashedName: problem-462-permutation-of-3-smooth-numbers
---
# --description--
-3平滑数是没有素数因子大于3的整数。对于整数N，我们将S（N）定义为小于或等于N的3平滑数的集合。例如，S（20）= {1,2,3,4,6,8,9,12,16,18}。
+<!-- TODO use MathJax -->
-我们将F（N）定义为S（N）的排列数，其中每个元素都在其所有适当的除数之后。
+A 3-smooth number is an integer which has no prime factor larger than 3. For an integer N, we define S(N) as the set of 3-smooth numbers less than or equal to N . For example, S(20) = { 1, 2, 3, 4, 6, 8, 9, 12, 16, 18 }.
-这是N = 20的可能排列之一。
+We define F(N) as the number of permutations of S(N) in which each element comes after all of its proper divisors.
-- 1,2,4,3,9,8,16,6,18,12。这不是有效的排列，因为12在它的除数6之前出现。
-- 1,2,4,3,9,8,12,16,6,18。
+This is one of the possible permutations for N = 20.
-我们可以验证F（6）= 5，F（8）= 9，F（20）= 450和F（1000）≈8.8521816557e21。找到F（1018）。给出你的答案，它的科学记数在小数点后四舍五入到十位数。在给出答案时，使用小写e来分隔尾数和指数。例如，如果答案是112,233,445,566,778,899，则答案格式为1.1223344557e17。
+- 1, 2, 4, 3, 9, 8, 16, 6, 18, 12.
+
+This is not a valid permutation because 12 comes before its divisor 6.
+
+- 1, 2, 4, 3, 9, 8, 12, 16, 6, 18.
+
+We can verify that F(6) = 5, F(8) = 9, F(20) = 450 and F(1000) ≈ 8.8521816557e21. Find F(1018). Give as your answer its scientific notation rounded to ten digits after the decimal point. When giving your answer, use a lowercase e to separate mantissa and exponent. E.g. if the answer is 112,233,445,566,778,899 then the answer format would be 1.1223344557e17.
# --hints--
-`euler462()`应返回Infinity。
+`euler462()` should return Infinity.
```js
assert.strictEqual(euler462(), Infinity);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-463-a-weird-recurrence-relation.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-463-a-weird-recurrence-relation.md
index 09210929664..014935108f6 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-463-a-weird-recurrence-relation.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-463-a-weird-recurrence-relation.md
@@ -1,20 +1,30 @@
---
id: 5900f53c1000cf542c51004e
-title: 问题463：一种奇怪的复发关系
+title: 'Problem 463: A weird recurrence relation'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302138
dashedName: problem-463-a-weird-recurrence-relation
---
# --description--
-函数$ f $定义为所有正整数，如下所示：$ f（1）= 1 $ $ f（3）= 3 $ $ f（2n）= f（n）$ $ f（4n + 1）= 2f （2n + 1） - f（n）$ $ f（4n + 3）= 3f（2n + 1） - 2f（n）$
+The function $f$ is defined for all positive integers as follows:
-函数$ S（n）$定义为$ \\ sum\_ {i = 1} ^ {n} f（i）$。 $ S（8）= 22 $和$ S（100）= 3604 $。找到$ S（3 ^ {37}）$。给出答案的最后9位数字。
+$f(1)=1$
+
+$f(3)=3$
+
+$f(2n)=f(n)$
+
+$f(4n + 1)=2f(2n + 1) - f(n)$
+
+$f(4n + 3)=3f(2n + 1) - 2f(n)$
+
+The function $S(n)$ is defined as $\\sum\_{i=1}^{n}f(i)$. $S(8)=22$ and $S(100)=3604$. Find $S(3^{37})$. Give the last 9 digits of your answer.
# --hints--
-`euler463()`应该返回808981553。
+`euler463()` should return 808981553.
```js
assert.strictEqual(euler463(), 808981553);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-464-mbius-function-and-intervals.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-464-mbius-function-and-intervals.md
index a0907c4b0b5..5b209dd1d75 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-464-mbius-function-and-intervals.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-464-mbius-function-and-intervals.md
@@ -1,26 +1,30 @@
---
id: 5900f53d1000cf542c51004f
-title: 问题464：莫比乌斯函数和区间
+title: 'Problem 464: Möbius function and intervals'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302139
dashedName: problem-464-mbius-function-and-intervals
---
# --description--
-Möbius函数，表示为μ（n），定义为：μ（n）=（ - 1）ω（n），如果n是无平方（其中ω（n）是n的不同素因子的数量）μ（n） ）如果n不是无平方，则= 0。
+The Möbius function, denoted μ(n), is defined as:
-令P（a，b）为区间\[a，b]中的整数n，使得μ（n）= 1.设N（a，b）为区间\[a，b]中的整数n这样μ（n）= -1。例如，P（2,10）= 2且N（2,10）= 4。
+μ(n) = (-1)ω(n) if n is squarefree (where ω(n) is the number of distinct prime factors of n)
-令C（n）为整数对（a，b）的数目，使得：1≤a≤b≤n，99·N（a，b）≤100·P（a，b），和99·P（ a，b）≤100·N（a，b）。
+μ(n) = 0 if n is not squarefree.
-例如，C（10）= 13，C（500）= 16676和C（10000）= 20155319。
+Let P(a,b) be the number of integers n in the interval \[a,b] such that μ(n) = 1. Let N(a,b) be the number of integers n in the interval \[a,b] such that μ(n) = -1. For example, P(2,10) = 2 and N(2,10) = 4.
-找C（2 000 000）。
+Let C(n) be the number of integer pairs (a,b) such that: 1 ≤ a ≤ b ≤ n, 99·N(a,b) ≤ 100·P(a,b), and 99·P(a,b) ≤ 100·N(a,b).
+
+For example, C(10) = 13, C(500) = 16676 and C(10 000) = 20155319.
+
+Find C(20 000 000).
# --hints--
-`euler464()`应返回198775297232878。
+`euler464()` should return 198775297232878.
```js
assert.strictEqual(euler464(), 198775297232878);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-465-polar-polygons.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-465-polar-polygons.md
index f62349bd5aa..2baaf0d7751 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-465-polar-polygons.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-465-polar-polygons.md
@@ -1,32 +1,32 @@
---
id: 5900f53d1000cf542c510050
-title: 问题465：极地多边形
+title: 'Problem 465: Polar polygons'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302140
dashedName: problem-465-polar-polygons
---
# --description--
-多边形的内核由一组点定义，整个多边形的边界是可见的。我们将极坐标多边形定义为多边形，其原点严格包含在其内核中。
+The kernel of a polygon is defined by the set of points from which the entire polygon's boundary is visible. We define a polar polygon as a polygon for which the origin is strictly contained inside its kernel.
-对于此问题，多边形可以具有共线的连续顶点。但是，多边形仍然不能具有自相交，并且不能具有零面积。
+For this problem, a polygon can have collinear consecutive vertices. However, a polygon still cannot have self-intersection and cannot have zero area.
-例如，只有下面的第一个是极多边形（第二个，第三个和第四个的内核不严格包含原点，第五个根本没有内核）：
+For example, only the first of the following is a polar polygon (the kernels of the second, third, and fourth do not strictly contain the origin, and the fifth does not have a kernel at all):
-请注意，第一个多边形有三个连续的共线顶点。
+Notice that the first polygon has three consecutive collinear vertices.
-令P（n）为极坐标多边形的数量，使得顶点（x，y）具有绝对值不大于n的整数坐标。
+Let P(n) be the number of polar polygons such that the vertices (x, y) have integer coordinates whose absolute values are not greater than n.
-请注意，如果多边形具有不同的边集，即使它们包含相同的区域，也应该计为不同的多边形。例如，具有顶点\[（0,0），（0,3），（1,1），（3,0）]的多边形与具有顶点\[（0,0），（0,3）的多边形不同），（1,1），（3,0），（1,0）]。
+Note that polygons should be counted as different if they have different set of edges, even if they enclose the same area. For example, the polygon with vertices \[(0,0),(0,3),(1,1),(3,0)] is distinct from the polygon with vertices \[(0,0),(0,3),(1,1),(3,0),(1,0)].
-例如，P（1）= 131，P（2）= 1648531，P（3）= 1099461296175，P（343）mod 1 000 000 007 = 937293740。
+For example, P(1) = 131, P(2) = 1648531, P(3) = 1099461296175 and P(343) mod 1 000 000 007 = 937293740.
-求P（713）mod 1 000 000 007。
+Find P(713) mod 1 000 000 007.
# --hints--
-`euler465()`应该返回585965659。
+`euler465()` should return 585965659.
```js
assert.strictEqual(euler465(), 585965659);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-466-distinct-terms-in-a-multiplication-table.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-466-distinct-terms-in-a-multiplication-table.md
index e8a69b373ba..3050ea410b1 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-466-distinct-terms-in-a-multiplication-table.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-466-distinct-terms-in-a-multiplication-table.md
@@ -1,28 +1,28 @@
---
id: 5900f53e1000cf542c510051
-title: 问题466：乘法表中的不同术语
+title: 'Problem 466: Distinct terms in a multiplication table'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302141
dashedName: problem-466-distinct-terms-in-a-multiplication-table
---
# --description--
-令P（m，n）为m×n乘法表中的不同项的数量。
+Let P(m,n) be the number of distinct terms in an m×n multiplication table.
-例如，3×4乘法表如下所示：
+For example, a 3×4 multiplication table looks like this:
-×12341 12342 24683 36912
+× 12341 12342 24683 36912
-有8个不同的术语{1,2,3,4,6,8,9,12}，因此P（3,4）= 8。
+There are 8 distinct terms {1,2,3,4,6,8,9,12}, therefore P(3,4) = 8.
-给出：P（64,64）= 1263，P（12,345）= 1998，P（32,1015）= 13826382602124302。
+You are given that: P(64,64) = 1263, P(12,345) = 1998, and P(32,1015) = 13826382602124302.
-求P（64,1016）。
+Find P(64,1016).
# --hints--
-`euler466()`应该返回258381958195474750。
+`euler466()` should return 258381958195474750.
```js
assert.strictEqual(euler466(), 258381958195474750);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-467-superinteger.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-467-superinteger.md
index 15e2e3d38b7..8cfd9dc16ab 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-467-superinteger.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-467-superinteger.md
@@ -1,28 +1,30 @@
---
id: 5900f5411000cf542c510052
-title: 问题467：超级整数
+title: 'Problem 467: Superinteger'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302142
dashedName: problem-467-superinteger
---
# --description--
-如果n的数字形成s的数字的子序列，则整数s被称为另一整数n的超级整数。例如，2718281828是18828的超级整数，而314159不是151的超级整数。
+An integer s is called a superinteger of another integer n if the digits of n form a subsequence of the digits of s.
-令p（n）为第n个素数，并且令c（n）为第n个复合数。例如，p（1）= 2，p（10）= 29，c（1）= 4且c（10）= 18. {p（i）：i≥1} = {2,3,5,7 ，11,13,17,19,23,29，...} {c（i）：i≥1} = {4,6,8,9,10,12,14,15,16,18，.... ..}
+For example, 2718281828 is a superinteger of 18828, while 314159 is not a superinteger of 151.
-设PD为{p（i）}的数字根的序列（CD对{c（i）}的定义类似）：PD = {2,3,5,7,2,4,8,1,5， 2，...} CD = {4,6,8,9,1,3,5,6,7,9 ......}
+Let p(n) be the nth prime number, and let c(n) be the nth composite number. For example, p(1) = 2, p(10) = 29, c(1) = 4 and c(10) = 18. {p(i) : i ≥ 1} = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ...} {c(i) : i ≥ 1} = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18, ...}
-令Pn为通过连接PD的前n个元素形成的整数（Cn类似地定义为CD）。 P10 = 2357248152 C10 = 4689135679
+Let PD the sequence of the digital roots of {p(i)} (CD is defined similarly for {c(i)}): PD = {2, 3, 5, 7, 2, 4, 8, 1, 5, 2, ...} CD = {4, 6, 8, 9, 1, 3, 5, 6, 7, 9, ...}
-设f（n）是最小的正整数，它是Pn和Cn的共同超整数。例如，f（10）= 2357246891352679，并且f（100）mod 1 000 000 007 = 771661825。
+Let Pn be the integer formed by concatenating the first n elements of PD (Cn is defined similarly for CD). P10 = 2357248152 C10 = 4689135679
-求f（10 000）mod 1 000 000 007。
+Let f(n) be the smallest positive integer that is a common superinteger of Pn and Cn. For example, f(10) = 2357246891352679, and f(100) mod 1 000 000 007 = 771661825.
+
+Find f(10 000) mod 1 000 000 007.
# --hints--
-`euler467()`应该返回775181359。
+`euler467()` should return 775181359.
```js
assert.strictEqual(euler467(), 775181359);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-468-smooth-divisors-of-binomial-coefficients.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-468-smooth-divisors-of-binomial-coefficients.md
index 1979a5673c1..239d2cf56d4 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-468-smooth-divisors-of-binomial-coefficients.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-468-smooth-divisors-of-binomial-coefficients.md
@@ -1,24 +1,24 @@
---
id: 5900f5411000cf542c510054
-title: 问题468：二项式系数的平滑除数
+title: 'Problem 468: Smooth divisors of binomial coefficients'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302143
dashedName: problem-468-smooth-divisors-of-binomial-coefficients
---
# --description--
-如果没有一个整数因子大于B，则整数称为B-smooth。
+An integer is called B-smooth if none of its prime factors is greater than B.
-设SB（n）是n的最大B-平滑除数。示例：S1（10）= 1 S4（2100）= 12 S17（2496144）= 5712
+Let SB(n) be the largest B-smooth divisor of n. Examples: S1(10) = 1 S4(2100) = 12 S17(2496144) = 5712
-定义F（n）=Σ1≤B≤nΣ0≤r≤nSB（C（n，r））。这里，C（n，r）表示二项式系数。示例：F（11）= 3132 F（1 111）mod 1 000 000 993 = 706036312 F（111 111）mod 1 000 000 993 = 22156169
+Define F(n) = ∑1≤B≤n ∑0≤r≤n SB(C(n,r)). Here, C(n,r) denotes the binomial coefficient. Examples: F(11) = 3132 F(1 111) mod 1 000 000 993 = 706036312 F(111 111) mod 1 000 000 993 = 22156169
-求F（11 111 111）mod 1 000 000 993。
+Find F(11 111 111) mod 1 000 000 993.
# --hints--
-`euler468()`应该返回852950321。
+`euler468()` should return 852950321.
```js
assert.strictEqual(euler468(), 852950321);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-469-empty-chairs.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-469-empty-chairs.md
index d10b3d7986a..4e8be025a2f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-469-empty-chairs.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-469-empty-chairs.md
@@ -1,22 +1,26 @@
---
id: 5900f5411000cf542c510053
-title: 问题469：空椅子
+title: 'Problem 469: Empty chairs'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302144
dashedName: problem-469-empty-chairs
---
# --description--
-在一个房间里，N个椅子摆放在圆桌旁。骑士一个接一个地进入房间，随意选择一把可用的空椅子。为了拥有足够的肘部空间，骑士总是在彼此之间留下至少一把空座椅。
+In a room N chairs are placed around a round table.
-当没有任何合适的椅子时，确定空椅子的分数C.我们还将E（N）定义为C的期望值。我们可以验证E（4）= 1/2和E（6）= 5/9。
+Knights enter the room one by one and choose at random an available empty chair.
-找到E（1018）。将答案四舍五入到小数点后十四位，形式为0.abcdefghijklmn。
+To have enough elbow room the knights always leave at least one empty chair between each other.
+
+When there aren't any suitable chairs left, the fraction C of empty chairs is determined. We also define E(N) as the expected value of C. We can verify that E(4) = 1/2 and E(6) = 5/9.
+
+Find E(1018). Give your answer rounded to fourteen decimal places in the form 0.abcdefghijklmn.
# --hints--
-`euler469()`应该返回0.56766764161831。
+`euler469()` should return 0.56766764161831.
```js
assert.strictEqual(euler469(), 0.56766764161831);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-47-distinct-primes-factors.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-47-distinct-primes-factors.md
index 4ef488a0076..998b035f79d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-47-distinct-primes-factors.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-47-distinct-primes-factors.md
@@ -1,44 +1,51 @@
---
id: 5900f39c1000cf542c50feae
-title: 问题47：不同的素数因素
+title: 'Problem 47: Distinct primes factors'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302145
dashedName: problem-47-distinct-primes-factors
---
# --description--
-前两个连续数字有两个不同的素数因子是：
+The first two consecutive numbers to have two distinct prime factors are:
-14 = 2×7
+<div style='padding-left: 4em;'>
+ 14 = 2 × 7<br>
+ 15 = 3 × 5
+</div>
-15 = 3×5
+The first three consecutive numbers to have three distinct prime factors are:
-前三个连续数字有三个不同的素因子：
+<div style='padding-left: 4em;'>
+ 644 = 2<sup>2</sup> × 7 × 23<br>
+ 645 = 3 × 5 × 43<br>
+ 646 = 2 × 17 × 19
+</div>
-644 =2²×7×23
-
-645 = 3×5×43
-
-646 = 2×17×19
-
-找到前四个连续的整数，每个整数有四个不同的素数因子。这些数字中的第一个是什么？
+Find the first four consecutive integers to have four distinct prime factors each. What is the first of these numbers?
# --hints--
-`distinctPrimeFactors(2, 2)`应该返回14。
+`distinctPrimeFactors(2, 2)` should return a number.
+
+```js
+assert(typeof distinctPrimeFactors(2, 2) === 'number');
+```
+
+`distinctPrimeFactors(2, 2)` should return 14.
```js
assert.strictEqual(distinctPrimeFactors(2, 2), 14);
```
-`distinctPrimeFactors(3, 3)`应该返回644。
+`distinctPrimeFactors(3, 3)` should return 644.
```js
assert.strictEqual(distinctPrimeFactors(3, 3), 644);
```
-`distinctPrimeFactors(4, 4)`应该返回134043。
+`distinctPrimeFactors(4, 4)` should return 134043.
```js
assert.strictEqual(distinctPrimeFactors(4, 4), 134043);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-470-super-ramvok.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-470-super-ramvok.md
index 6c72e7d562c..10ee6d95d8b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-470-super-ramvok.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-470-super-ramvok.md
@@ -1,30 +1,30 @@
---
id: 5900f5431000cf542c510055
-title: 问题470：超级拉姆沃克
+title: 'Problem 470: Super Ramvok'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302146
dashedName: problem-470-super-ramvok
---
# --description--
-考虑一下Ramvok的单一游戏：
+Consider a single game of Ramvok:
-设t表示游戏持续的最大转弯次数。如果t = 0，则游戏立即结束。否则，在每个回合i，玩家掷骰子。在滚动之后，如果玩家可以停止游戏并获得等于当前滚动值的奖励，或者丢弃滚动并在下一轮再试一次。如果i = t，则不能丢弃卷筒并且必须接受奖品。在比赛开始之前，t由玩家选择，然后玩家必须为某些常数c支付前期成本ct。对于c = 0，可以选择t为无穷大（前期成本为0）。设R（d，c）是玩家从单次最佳玩法Ramvok游戏中获得的预期利润（即净收益），给定公平的d-die模具和成本常数c。例如，R（4,0.2）= 2.65。假设玩家有足够的资金支付任何/所有前期费用。
+Let t represent the maximum number of turns the game lasts. If t = 0, then the game ends immediately. Otherwise, on each turn i, the player rolls a die. After rolling, if i < t the player can either stop the game and receive a prize equal to the value of the current roll, or discard the roll and try again next turn. If i = t, then the roll cannot be discarded and the prize must be accepted. Before the game begins, t is chosen by the player, who must then pay an up-front cost ct for some constant c. For c = 0, t can be chosen to be infinite (with an up-front cost of 0). Let R(d, c) be the expected profit (i.e. net gain) that the player receives from a single game of optimally-played Ramvok, given a fair d-sided die and cost constant c. For example, R(4, 0.2) = 2.65. Assume that the player has sufficient funds for paying any/all up-front costs.
-现在考虑一个超级拉姆沃克游戏：
+Now consider a game of Super Ramvok:
-在超级拉姆沃克，拉姆沃克的比赛是反复播放，但略有修改。每场比赛后，骰子都会被改变。改变过程如下：模具滚动一次，如果得到的面部有可见的点，则该面被改为空白。如果面部已经是空白，则会将其更改回原始值。在进行改变之后，可以开始另一场Ramvok游戏（并且在这样的游戏中，在每个回合中，掷骰子被滚动直到出现具有其值的面部）。玩家知道哪些面是空白的，哪些面不是。一旦模具的所有面都空白，超级拉姆沃克的游戏结束。
+In Super Ramvok, the game of Ramvok is played repeatedly, but with a slight modification. After each game, the die is altered. The alteration process is as follows: The die is rolled once, and if the resulting face has its pips visible, then that face is altered to be blank instead. If the face is already blank, then it is changed back to its original value. After the alteration is made, another game of Ramvok can begin (and during such a game, at each turn, the die is rolled until a face with a value on it appears). The player knows which faces are blank and which are not at all times. The game of Super Ramvok ends once all faces of the die are blank.
-设S（d，c）是玩家从Super Ramvok最佳玩游戏中获得的预期利润，给出一个公平的d-side模具开始（所有边都可见），并且成本常数c。例如，S（6,1）= 208.3。
+Let S(d, c) be the expected profit that the player receives from an optimally-played game of Super Ramvok, given a fair d-sided die to start (with all sides visible), and cost constant c. For example, S(6, 1) = 208.3.
-设F（n）=Σ4≤d≤nΣ0≤c≤nS（d，c）。
+Let F(n) = ∑4≤d≤n ∑0≤c≤n S(d, c).
-计算F（20），四舍五入到最接近的整数。
+Calculate F(20), rounded to the nearest integer.
# --hints--
-`euler470()`应该返回147668794。
+`euler470()` should return 147668794.
```js
assert.strictEqual(euler470(), 147668794);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-471-triangle-inscribed-in-ellipse.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-471-triangle-inscribed-in-ellipse.md
index fd00149d31b..95385ddec9c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-471-triangle-inscribed-in-ellipse.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-471-triangle-inscribed-in-ellipse.md
@@ -1,20 +1,24 @@
---
id: 5900f5431000cf542c510056
-title: 问题471：以椭圆形刻的三角形
+title: 'Problem 471: Triangle inscribed in ellipse'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302148
dashedName: problem-471-triangle-inscribed-in-ellipse
---
# --description--
-三角形ΔABC刻在椭圆中，方程为$ \\ frac {x ^ 2} {a ^ 2} + \\ frac {y ^ 2} {b ^ 2} = 1 $，0 <2b <a，a和b整数。设r（a，b）为内圆（2b，0）时A的圆周半径，A有坐标$ \\ left（\\ frac a 2，\\ frac {\\ sqrt 3} 2 b \\ right） $。例如，r（3,1）= 1/2，r（6,2）= 1，r（12,3）= 2。
+The triangle ΔABC is inscribed in an ellipse with equation $\\frac {x^2} {a^2} + \\frac {y^2} {b^2} = 1$, 0 < 2b < a, a and b integers.
-设$ G（n）= \\ sum *{a = 3} ^ n \\ sum* {b = 1} ^ {\\ lfloor \\ frac {a - 1} 2 \\ rfloor} r（a，b）$给你G（ 10）= 20.59722222，G（100）= 19223.60980（舍入为10位有效数字）。找到G（1011）。以科学计数法给出答案四舍五入到10位有效数字。使用小写e分隔尾数和指数。对于G（10），答案应该是2.059722222e1。
+Let r(a,b) be the radius of the incircle of ΔABC when the incircle has center (2b, 0) and A has coordinates $\\left( \\frac a 2, \\frac {\\sqrt 3} 2 b\\right)$.
+
+For example, r(3,1) = ½, r(6,2) = 1, r(12,3) = 2.
+
+Let $G(n) = \\sum*{a=3}^n \\sum*{b=1}^{\\lfloor \\frac {a - 1} 2 \\rfloor} r(a, b)$ You are given G(10) = 20.59722222, G(100) = 19223.60980 (rounded to 10 significant digits). Find G(1011). Give your answer in scientific notation rounded to 10 significant digits. Use a lowercase e to separate mantissa and exponent. For G(10) the answer would have been 2.059722222e1.
# --hints--
-`euler471()`应返回1.895093981e + 31。
+`euler471()` should return 1.895093981e+31.
```js
assert.strictEqual(euler471(), 1.895093981e31);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-472-comfortable-distance-ii.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-472-comfortable-distance-ii.md
index bb2691a508a..f4e90155496 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-472-comfortable-distance-ii.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-472-comfortable-distance-ii.md
@@ -1,28 +1,36 @@
---
id: 5900f5451000cf542c510057
-title: 问题472：舒适的距离II
+title: 'Problem 472: Comfortable Distance II'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302149
dashedName: problem-472-comfortable-distance-ii
---
# --description--
-连续N个座位。 N人按照以下规则一个接一个地来到座位：没有人坐在另一个人旁边。第一个人选择任何座位。只要不违反规则1，每个后续人员选择离已经就座的人最远的座位。如果满足该条件的选择不止一个，则该人选择最左边的选择。请注意，由于规则1，某些座位肯定会未被占用，并且可以坐着的最大人数小于N（对于N> 1）。
+There are N seats in a row. N people come one after another to fill the seats according to the following rules:
-以下是N = 15的可能座位安排：
+No person sits beside another.
-我们看到，如果第一个人选择正确，15个座位最多可容纳7人。我们还可以看到第一个人有9个选择来最大化可能坐着的人数。
+The first person chooses any seat.
-设f（N）是第一个人必须选择的数量，以最大化连续N个座位的占用者数量。因此，f（1）= 1，f（15）= 9，f（20）= 6，并且f（500）= 16。
+Each subsequent person chooses the seat furthest from anyone else already seated, as long as it does not violate rule 1. If there is more than one choice satisfying this condition, then the person chooses the leftmost choice.
-此外，对于1≤N≤20，Σf（N）= 83，对于1≤N≤500，Σf（N）= 13343。
+Note that due to rule 1, some seats will surely be left unoccupied, and the maximum number of people that can be seated is less than N (for N > 1).
-找到Σf（N）为1≤N≤1012。给出答案的最后8位数字。
+Here are the possible seating arrangements for N = 15:
+
+We see that if the first person chooses correctly, the 15 seats can seat up to 7 people. We can also see that the first person has 9 choices to maximize the number of people that may be seated.
+
+Let f(N) be the number of choices the first person has to maximize the number of occupants for N seats in a row. Thus, f(1) = 1, f(15) = 9, f(20) = 6, and f(500) = 16.
+
+Also, ∑f(N) = 83 for 1 ≤ N ≤ 20 and ∑f(N) = 13343 for 1 ≤ N ≤ 500.
+
+Find ∑f(N) for 1 ≤ N ≤ 1012. Give the last 8 digits of your answer.
# --hints--
-`euler472()`应该返回73811586。
+`euler472()` should return 73811586.
```js
assert.strictEqual(euler472(), 73811586);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-473-phigital-number-base.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-473-phigital-number-base.md
index 88f3314cc2e..505c8a757ba 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-473-phigital-number-base.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-473-phigital-number-base.md
@@ -1,24 +1,34 @@
---
id: 5900f5461000cf542c510058
-title: 问题473：数字基数
+title: 'Problem 473: Phigital number base'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302150
dashedName: problem-473-phigital-number-base
---
# --description--
-让$ \\ varphi $成为黄金比例：$ \\ varphi = \\ frac {1+ \\ sqrt {5}} {2}。$值得注意的是，有可能将每个正整数写为$ \\ varphi $的幂的总和如果我们要求$ \\ varphi $的每个幂在此总和中最多使用一次。即使这样，这种表现也不是唯一的。我们可以通过要求不使用具有连续指数的幂并且表示是有限的来使其独特。例如：$ 2 = \\ varphi + \\ varphi ^ { - 2} $和$ 3 = \\ varphi ^ {2} + \\ varphi ^ { - 2} $
+Let $\\varphi$ be the golden ratio: $\\varphi=\\frac{1+\\sqrt{5}}{2}.$
-为了表示$ \\ varphi $的这个幂的总和，我们使用一个0和1的字符串以及一个指示负指数开始位置的点。我们将其称为数字数字库中的表示。所以$ 1 = 1 *{\\ varphi} $，$ 2 = 10.01* {\\ varphi} $，$ 3 = 100.01 *{\\ varphi} $和$ 14 = 100100.001001* {\\ varphi} $。数字基数中代表1,2和14的字符串是回文，而表示3的字符串则不是。 （数字点不是中间字符）。
+Remarkably it is possible to write every positive integer as a sum of powers of $\\varphi$ even if we require that every power of $\\varphi$ is used at most once in this sum.
-正数整数不超过1000，其数字表示为回文的总和为4345。
+Even then this representation is not unique.
-找出不超过$ 10 ^ {10} $的正整数之和，其数字表示为回文。
+We can make it unique by requiring that no powers with consecutive exponents are used and that the representation is finite.
+
+E.g:
+
+$2=\\varphi+\\varphi^{-2}$ and $3=\\varphi^{2}+\\varphi^{-2}$
+
+To represent this sum of powers of $\\varphi$ we use a string of 0's and 1's with a point to indicate where the negative exponents start. We call this the representation in the phigital numberbase. So $1=1*{\\varphi}$, $2=10.01*{\\varphi}$, $3=100.01*{\\varphi}$ and $14=100100.001001*{\\varphi}$. The strings representing 1, 2 and 14 in the phigital number base are palindromic, while the string representing 3 is not. (the phigital point is not the middle character).
+
+The sum of the positive integers not exceeding 1000 whose phigital representation is palindromic is 4345.
+
+Find the sum of the positive integers not exceeding $10^{10}$ whose phigital representation is palindromic.
# --hints--
-`euler473()`应该返回35856681704365。
+`euler473()` should return 35856681704365.
```js
assert.strictEqual(euler473(), 35856681704365);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-474-last-digits-of-divisors.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-474-last-digits-of-divisors.md
index 331327a316f..5c77726bd25 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-474-last-digits-of-divisors.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-474-last-digits-of-divisors.md
@@ -1,22 +1,24 @@
---
id: 5900f5471000cf542c510059
-title: 问题474：除数的最后数字
+title: 'Problem 474: Last digits of divisors'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302151
dashedName: problem-474-last-digits-of-divisors
---
# --description--
-对于正整数n和数字d，我们将F（n，d）定义为n的除数的数，其最后的数字等于d。例如，F（84,4）= 3.在84的除数中（1,2,3,4,6,7,12,14,21,28,42,84），其中三个（4,14） ，84）有最后一位数字4。
+For a positive integer n and digits d, we define F(n, d) as the number of the divisors of n whose last digits equal d.
-我们还可以验证F（12！，12）= 11和F（50！，123）= 17888。
+For example, F(84, 4) = 3. Among the divisors of 84 (1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84), three of them (4, 14, 84) have the last digit 4.
-找到F（106！，65432）modulo（1016 + 61）。
+We can also verify that F(12!, 12) = 11 and F(50!, 123) = 17888.
+
+Find F(106!, 65432) modulo (1016 + 61).
# --hints--
-`euler474()`应该返回9690646731515010。
+`euler474()` should return 9690646731515010.
```js
assert.strictEqual(euler474(), 9690646731515010);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-475-music-festival.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-475-music-festival.md
index 2993bd48f23..dd8cf3538b7 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-475-music-festival.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-475-music-festival.md
@@ -1,22 +1,26 @@
---
id: 5900f5481000cf542c51005a
-title: 问题475：音乐节
+title: 'Problem 475: Music festival'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302152
dashedName: problem-475-music-festival
---
# --description--
-12n音乐家参加音乐节。在第一天，他们形成3n四重奏并整日练习。这是一场灾难。在一天结束时，所有音乐家都决定再也不会同意与他们四重奏的任何成员一起玩。在第二天，他们形成4n三人组，每个音乐家都避开他以前的四重奏伙伴。
+12n musicians participate at a music festival. On the first day, they form 3n quartets and practice all day.
-设f（12n）是在12n音乐家中组织三人组合的方式。给出f（12）= 576和f（24）mod 1 000 000 007 = 509089824。
+It is a disaster. At the end of the day, all musicians decide they will never again agree to play with any member of their quartet.
-求f（600）mod 1 000 000 007。
+On the second day, they form 4n trios, each musician avoiding his previous quartet partners.
+
+Let f(12n) be the number of ways to organize the trios amongst the 12n musicians. You are given f(12) = 576 and f(24) mod 1 000 000 007 = 509089824.
+
+Find f(600) mod 1 000 000 007.
# --hints--
-`euler475()`应返回75780067。
+`euler475()` should return 75780067.
```js
assert.strictEqual(euler475(), 75780067);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-476-circle-packing-ii.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-476-circle-packing-ii.md
index 651654ddf99..a5a8bf868b0 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-476-circle-packing-ii.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-476-circle-packing-ii.md
@@ -1,18 +1,24 @@
---
id: 5900f5481000cf542c51005b
-title: 问题476：圆形包装II
+title: 'Problem 476: Circle Packing II'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302153
dashedName: problem-476-circle-packing-ii
---
# --description--
-令R（a，b，c）为具有边长a，b和c的三角形内的三个非重叠圆所覆盖的最大面积。令S（n）为R（a，b，c）在所有整数三元组（a，b，c）上的平均值，使得1≤a≤b≤c<a +b≤n给出S（2） ）= R（1,1,1）≈0.31998，S（5）≈1.25899。查找S（1803）四舍五入到小数点后面的小数点后5位。
+Let R(a, b, c) be the maximum area covered by three non-overlapping circles inside a triangle with edge lengths a, b and c.
+
+Let S(n) be the average value of R(a, b, c) over all integer triplets (a, b, c) such that 1 ≤ a ≤ b ≤ c < a + b ≤ n
+
+You are given S(2) = R(1, 1, 1) ≈ 0.31998, S(5) ≈ 1.25899.
+
+Find S(1803) rounded to 5 decimal places behind the decimal point.
# --hints--
-`euler476()`应返回110242.87794。
+`euler476()` should return 110242.87794.
```js
assert.strictEqual(euler476(), 110242.87794);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-477-number-sequence-game.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-477-number-sequence-game.md
index e67ac58479f..f4c95faba72 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-477-number-sequence-game.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-477-number-sequence-game.md
@@ -1,18 +1,46 @@
---
id: 5900f54a1000cf542c51005c
-title: 问题477：数字序列游戏
+title: 'Problem 477: Number Sequence Game'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302154
dashedName: problem-477-number-sequence-game
---
# --description--
-数字序列游戏以写在一行上的N个数字的序列S开始。两名球员交替轮流。在轮到他时，玩家必须选择并删除序列中剩余的第一个或最后一个数字。球员得分是他所取得的所有数字的总和。每个玩家都试图最大化自己的总和。如果N = 4并且S = {1,2,10,3}，则每个玩家最大化他的得分如下：玩家1：移除第一个数字（1）玩家2：从剩余序列移除最后一个数字（3）玩家1：从剩余序列中移除最后一个号码（10）玩家2：移除剩余号码（2）玩家1得分为1 + 10 = 11.如果两个玩家都遵循，则F（N）为玩家1的得分序列的最优策略S = {s1，s2，...，sN}定义为：s1 = 0 si + 1 =（si2 + 45）modulo 1 000 000 007序列以S = {0,45,2070开头，4284945,753524550,478107844,894218625，...}。给出F（2）= 45，F（4）= 4284990，F（100）= 26365463243，F（104）= 2495838522951。求F（108）。
+The number sequence game starts with a sequence S of N numbers written on a line.
+
+Two players alternate turns. At his turn, a player must select and remove either the first or the last number remaining in the sequence.
+
+The player score is the sum of all the numbers he has taken. Each player attempts to maximize his own sum.
+
+If N = 4 and S = {1, 2, 10, 3}, then each player maximizes his score as follows:
+
+Player 1: removes the first number (1)
+
+Player 2: removes the last number from the remaining sequence (3)
+
+Player 1: removes the last number from the remaining sequence (10)
+
+Player 2: removes the remaining number (2)
+
+Player 1 score is 1 + 10 = 11.
+
+Let F(N) be the score of player 1 if both players follow the optimal strategy for the sequence S = {s1, s2, ..., sN} defined as:
+
+s1 = 0
+
+si+1 = (si2 + 45) modulo 1 000 000 007
+
+The sequence begins with S = {0, 45, 2070, 4284945, 753524550, 478107844, 894218625, ...}.
+
+You are given F(2) = 45, F(4) = 4284990, F(100) = 26365463243, F(104) = 2495838522951.
+
+Find F(108).
# --hints--
-`euler477()`应该返回25044905874565164。
+`euler477()` should return 25044905874565164.
```js
assert.strictEqual(euler477(), 25044905874565164);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-478-mixtures.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-478-mixtures.md
index b7e289dd614..37e7681ecff 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-478-mixtures.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-478-mixtures.md
@@ -1,30 +1,30 @@
---
id: 5900f54c1000cf542c51005e
-title: 问题478：混合物
+title: 'Problem 478: Mixtures'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302155
dashedName: problem-478-mixtures
---
# --description--
-让我们考虑三种物质的混合物：A，B和C.混合物可以用其中A，B和C的量的比例来描述，即（a：b：c）。例如，由比例（2：3：5）描述的混合物含有20％A，30％B和50％C。
+Let us consider mixtures of three substances: A, B and C. A mixture can be described by a ratio of the amounts of A, B, and C in it, i.e., (a : b : c). For example, a mixture described by the ratio (2 : 3 : 5) contains 20% A, 30% B and 50% C.
-出于这个问题的目的，我们不能将各个组分与混合物分开。但是，我们可以将不同量的不同混合物组合以形成具有新比例的混合物。
+For the purposes of this problem, we cannot separate the individual components from a mixture. However, we can combine different amounts of different mixtures to form mixtures with new ratios.
-例如，假设我们有三种比例（3：0：2），（3：6：11）和（3：3：4）的混合物。通过混合10个单位的第一个，20个单位的第二个和30个单位的第三个，我们得到一个比例（6：5：9）的新混合物，因为：（10·3/5 + 20·3/20 + 30·3/10：10·0/5 + 20·6/20 + 30·3/10：10·2/5 + 20·11/20 + 30·4/10）=（18：15：27） =（6：5：9）
+For example, say we have three mixtures with ratios (3 : 0 : 2), (3 : 6 : 11) and (3 : 3 : 4). By mixing 10 units of the first, 20 units of the second and 30 units of the third, we get a new mixture with ratio (6 : 5 : 9), since: (10·3/5 + 20·3/20 + 30·3/10 : 10·0/5 + 20·6/20 + 30·3/10 : 10·2/5 + 20·11/20 + 30·4/10) = (18 : 15 : 27) = (6 : 5 : 9)
-然而，使用相同的三种混合物，不可能形成比例（3：2：1），因为B的量总是小于C的量。
+However, with the same three mixtures, it is impossible to form the ratio (3 : 2 : 1), since the amount of B is always less than the amount of C.
-设n是正整数。假设对于0≤a，b，c≤n和gcd（a，b，c）= 1的整数（a，b，c）的每三个，我们得到具有比率（a：b：c）的混合。设M（n）为所有这些混合物的集合。
+Let n be a positive integer. Suppose that for every triple of integers (a, b, c) with 0 ≤ a, b, c ≤ n and gcd(a, b, c) = 1, we have a mixture with ratio (a : b : c). Let M(n) be the set of all such mixtures.
-例如，M（2）包含具有以下比率的19种混合物：{（0：0：1），（0：1：0），（0：1：1），（0：1：2），（ 0：2：1），（1：0：0），（1：0：1），（1：0：2），（1：1：0），（1：1：1），（1： 1：2），（1：2：0），（1：2：1），（1：2：2），（2：0：1），（2：1：0），（2：1： 1），（2：1：2），（2：2：1）}。
+For example, M(2) contains the 19 mixtures with the following ratios: {(0 : 0 : 1), (0 : 1 : 0), (0 : 1 : 1), (0 : 1 : 2), (0 : 2 : 1), (1 : 0 : 0), (1 : 0 : 1), (1 : 0 : 2), (1 : 1 : 0), (1 : 1 : 1), (1 : 1 : 2), (1 : 2 : 0), (1 : 2 : 1), (1 : 2 : 2), (2 : 0 : 1), (2 : 1 : 0), (2 : 1 : 1), (2 : 1 : 2), (2 : 2 : 1)}.
-令E（n）为M（n）的子集数，其可以产生具有比率（1：1：1）的混合物，即具有相等部分A，B和C的混合物。我们可以验证E（1） ）= 103，E（2）= 520447，E（10）mod 118 = 82608406和E（500）mod 118 = 13801403.求E（10 000 000）mod 118。
+Let E(n) be the number of subsets of M(n) which can produce the mixture with ratio (1 : 1 : 1), i.e., the mixture with equal parts A, B and C. We can verify that E(1) = 103, E(2) = 520447, E(10) mod 118 = 82608406 and E(500) mod 118 = 13801403. Find E(10 000 000) mod 118.
# --hints--
-`euler478()`应该返回59510340。
+`euler478()` should return 59510340.
```js
assert.strictEqual(euler478(), 59510340);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-479-roots-on-the-rise.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-479-roots-on-the-rise.md
index 50c428c24a6..202e7e04be7 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-479-roots-on-the-rise.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-479-roots-on-the-rise.md
@@ -1,26 +1,26 @@
---
id: 5900f54b1000cf542c51005d
-title: 问题479：崛起的根源
+title: 'Problem 479: Roots on the Rise'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302156
dashedName: problem-479-roots-on-the-rise
---
# --description--
-设ak，bk和ck表示表达式1 / x =（k / x）2（k + x2）-kx的三个解（实数或复数）。
+Let ak, bk, and ck represent the three solutions (real or complex numbers) to the expression 1/x = (k/x)2(k+x2) - kx.
-例如，对于k = 5，我们看到{a5，b5，c5}约为{5.727244，-0.363622 + 2.057397i，-0.363622-2.057397i}。
+For instance, for k = 5, we see that {a5, b5, c5} is approximately {5.727244, -0.363622+2.057397i, -0.363622-2.057397i}.
-令所有整数p，k的S（n）=Σ（ak + bk）p（bk + ck）p（ck + ak）p，使得1≤p，k≤n。
+Let S(n) = Σ (ak+bk)p(bk+ck)p(ck+ak)p for all integers p, k such that 1 ≤ p, k ≤ n.
-有趣的是，S（n）总是一个整数。例如，S（4）= 51160。
+Interestingly, S(n) is always an integer. For example, S(4) = 51160.
-求S（106）modulo 1 000 000 007。
+Find S(106) modulo 1 000 000 007.
# --hints--
-`euler479()`应该返回191541795。
+`euler479()` should return 191541795.
```js
assert.strictEqual(euler479(), 191541795);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-48-self-powers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-48-self-powers.md
index 16205f066be..58e654f8938 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-48-self-powers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-48-self-powers.md
@@ -1,36 +1,44 @@
---
id: 5900f39c1000cf542c50feaf
-title: 问题48：自我权力
+title: 'Problem 48: Self powers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302157
dashedName: problem-48-self-powers
---
# --description--
-系列，1 <sup>1</sup> + 2 <sup>2</sup> + 3 <sup>3</sup> + ... + 10 <sup>10</sup> = 10405071317.查找该系列的最后十位数字，1 <sup>1</sup> + 2 <sup>2</sup> + 3 <sup>3</sup> + ... + 1000 <sup>1000</sup> 。
+The series, 1<sup>1</sup> + 2<sup>2</sup> + 3<sup>3</sup> + ... + 10<sup>10</sup> = 10405071317.
+
+Find the last ten digits of the series, 1<sup>1</sup> + 2<sup>2</sup> + 3<sup>3</sup> + ... + 1000<sup>1000</sup>.
# --hints--
-`selfPowers(10, 3)`应该返回317。
+`selfPowers(10, 3)` should return a number.
+
+```js
+assert(typeof selfPowers(10, 3) === 'number');
+```
+
+`selfPowers(10, 3)` should return 317.
```js
assert.strictEqual(selfPowers(10, 3), 317);
```
-`selfPowers(150, 6)`应返回29045。
+`selfPowers(150, 6)` should return 29045.
```js
assert.strictEqual(selfPowers(150, 6), 29045);
```
-`selfPowers(673, 7)`应该返回2473989。
+`selfPowers(673, 7)` should return 2473989.
```js
assert.strictEqual(selfPowers(673, 7), 2473989);
```
-`selfPowers(1000, 10)`应该返回9110846700。
+`selfPowers(1000, 10)` should return 9110846700.
```js
assert.strictEqual(selfPowers(1000, 10), 9110846700);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-480-the-last-question.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-480-the-last-question.md
index d62f342c8d8..9e2ac8d96d8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-480-the-last-question.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-480-the-last-question.md
@@ -1,18 +1,94 @@
---
id: 5900f54c1000cf542c51005f
-title: 问题480：最后一个问题
+title: 'Problem 480: The Last Question'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302158
dashedName: problem-480-the-last-question
---
# --description--
-考虑所有可以通过从短语中选择任何顺序的字母形成的单词：thereisasyetinsufficientdataforameaningfulanswer假设15个字母或更少的字母按字母顺序列出并从1开始按顺序编号。列表包括：1：a 2： aa 3：aaa 4：aaaa 5：aaaaa 6：aaaaaa 7：aaaaaac 8：aaaaaacd 9：aaaaaacde 10：aaaaaacdee 11：aaaaaacdeee 12：aaaaaacdeeee 13：aaaaaacdeeeee 14：aaaaaacdeeeeee 15：aaaaaacdeeeeeef 16：aaaaaacdeeeeeeee 17：aaaaaacdeeeeeee 17：aaaaaacdeeeeeeh ... 28 ：aaaaaacdeeeeeey 29：aaaaaacdeeeeef 30：aaaaaacdeeeeefe ... 115246685191495242：euleoywuttttsss 115246685191495243：euler 115246685191495244：eulera ... 525069350231428029：ywuuttttssssrrrDefine P（w）作为单词w的位置。将W（p）定义为位置p中的单词。我们可以看到P（w）和W（p）是逆的：P（W（p））= p和W（P（w））= w。示例：W（10）= aaaaaacdee P（aaaaaacdee）= 10 W（115246685191495243）= euler P（euler）= 115246685191495243找到W（P（军团）+ P（量热计） - P（歼灭）+ P（精心策划） - P（飘飘））。使用小写字符（没有标点符号或空格）给出答案。
+Consider all the words which can be formed by selecting letters, in any order, from the phrase:
+
+thereisasyetinsufficientdataforameaningfulanswer
+
+Suppose those with 15 letters or less are listed in alphabetical order and numbered sequentially starting at 1.
+
+The list would include:
+
+1 : a
+
+2 : aa
+
+3 : aaa
+
+4 : aaaa
+
+5 : aaaaa
+
+6 : aaaaaa
+
+7 : aaaaaac
+
+8 : aaaaaacd
+
+9 : aaaaaacde
+
+10 : aaaaaacdee
+
+11 : aaaaaacdeee
+
+12 : aaaaaacdeeee
+
+13 : aaaaaacdeeeee
+
+14 : aaaaaacdeeeeee
+
+15 : aaaaaacdeeeeeef
+
+16 : aaaaaacdeeeeeeg
+
+17 : aaaaaacdeeeeeeh
+
+...
+
+28 : aaaaaacdeeeeeey
+
+29 : aaaaaacdeeeeef
+
+30 : aaaaaacdeeeeefe
+
+...
+
+115246685191495242: euleoywuttttsss
+
+115246685191495243: euler
+
+115246685191495244: eulera
+
+...
+
+525069350231428029: ywuuttttssssrrrDefine P(w) as the position of the word w.
+
+Define W(p) as the word in position p.
+
+We can see that P(w) and W(p) are inverses: P(W(p)) = p and W(P(w)) = w.
+
+Examples:
+
+W(10) = aaaaaacdee
+
+P(aaaaaacdee) = 10
+
+W(115246685191495243) = euler
+
+P(euler) = 115246685191495243Find W(P(legionary) + P(calorimeters) - P(annihilate) + P(orchestrated) - P(fluttering)).
+
+Give your answer using lowercase characters (no punctuation or space).
# --hints--
-`euler480()`应该返回turnthestarson。
+`euler480()` should return turnthestarson.
```js
assert.strictEqual(euler480(), turnthestarson);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-49-prime-permutations.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-49-prime-permutations.md
index a8fb8f99701..0dc7ae0ddcf 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-49-prime-permutations.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-49-prime-permutations.md
@@ -1,18 +1,28 @@
---
id: 5900f39d1000cf542c50feb0
-title: 问题49：主要排列
+title: 'Problem 49: Prime permutations'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302159
dashedName: problem-49-prime-permutations
---
# --description--
-算术序列1487,4817,8147，其中每个项增加3330，在两个方面是不寻常的：（i）三个项中的每一个都是素数，并且，（ii）每个4位数字是彼此的排列。没有由三个1位，2位或3位数的素数组成的算术序列，表现出这种性质，但还有另外一个4位数的增加序列。你通过连接这个序列中的三个术语来形成12位数字？
+The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit numbers are permutations of one another.
+
+There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit increasing sequence.
+
+What 12-digit number do you form by concatenating the three terms in this sequence?
# --hints--
-`primePermutations()`应该返回296962999629。
+`primePermutations()` should return a number.
+
+```js
+assert(typeof primePermutations() === 'number');
+```
+
+`primePermutations()` should return 296962999629.
```js
assert.strictEqual(primePermutations(), 296962999629);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-5-smallest-multiple.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-5-smallest-multiple.md
index b631e65df3e..bdece251d77 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-5-smallest-multiple.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-5-smallest-multiple.md
@@ -1,42 +1,50 @@
---
id: 5900f3711000cf542c50fe84
-title: 问题5：最小的倍数
+title: 'Problem 5: Smallest multiple'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302160
dashedName: problem-5-smallest-multiple
---
# --description--
-2520是可以除以1到10中的每个数字而没有任何余数的最小数字。从1到`n`所有数字均可被整除的最小正数是多少？
+2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
+
+What is the smallest positive number that is evenly divisible by all of the numbers from 1 to `n`?
# --hints--
-`smallestMult(5)`应该返回60。
+`smallestMult(5)` should return a number.
+
+```js
+assert(typeof smallestMult(5) === 'number');
+```
+
+`smallestMult(5)` should return 60.
```js
assert.strictEqual(smallestMult(5), 60);
```
-`smallestMult(7)`应该返回420。
+`smallestMult(7)` should return 420.
```js
assert.strictEqual(smallestMult(7), 420);
```
-`smallestMult(10)`应返回2520。
+`smallestMult(10)` should return 2520.
```js
assert.strictEqual(smallestMult(10), 2520);
```
-`smallestMult(13)`应返回360360。
+`smallestMult(13)` should return 360360.
```js
assert.strictEqual(smallestMult(13), 360360);
```
-`smallestMult(20)`应该返回232792560。
+`smallestMult(20)` should return 232792560.
```js
assert.strictEqual(smallestMult(20), 232792560);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-50-consecutive-prime-sum.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-50-consecutive-prime-sum.md
index d3eba492668..92924bf85e1 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-50-consecutive-prime-sum.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-50-consecutive-prime-sum.md
@@ -1,24 +1,38 @@
---
id: 5900f39e1000cf542c50feb1
-title: 问题50：连续的总和
+title: 'Problem 50: Consecutive prime sum'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302161
dashedName: problem-50-consecutive-prime-sum
---
# --description--
-素数41可以写成六个连续素数的总和：41 = 2 + 3 + 5 + 7 + 11 + 13这是连续素数的最长和，它加到低于一百的素数。连续素数低于1000的连续素数加上一个素数，包含21个项，等于953.哪个素数低于一百万，可以写成最连续素数的总和？
+The prime 41, can be written as the sum of six consecutive primes:
+
+<div style='text-align: center;'>41 = 2 + 3 + 5 + 7 + 11 + 13</div>
+
+This is the longest sum of consecutive primes that adds to a prime below one-hundred.
+
+The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
+
+Which prime, below one-million, can be written as the sum of the most consecutive primes?
# --hints--
-`consecutivePrimeSum(1000)`应该返回953。
+`consecutivePrimeSum(1000)` should return a number.
+
+```js
+assert(typeof consecutivePrimeSum(1000) === 'number');
+```
+
+`consecutivePrimeSum(1000)` should return 953.
```js
assert.strictEqual(consecutivePrimeSum(1000), 953);
```
-`consecutivePrimeSum(1000000)`应该返回997651。
+`consecutivePrimeSum(1000000)` should return 997651.
```js
assert.strictEqual(consecutivePrimeSum(1000000), 997651);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-51-prime-digit-replacements.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-51-prime-digit-replacements.md
index 6a8b3f079e6..7190e580ae4 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-51-prime-digit-replacements.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-51-prime-digit-replacements.md
@@ -1,21 +1,31 @@
---
id: 5900f39f1000cf542c50feb2
-title: 问题51：主要数字替换
+title: 'Problem 51: Prime digit replacements'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302162
dashedName: problem-51-prime-digit-replacements
---
# --description--
-通过替换2位数\* 3的第1位，可以看出9个可能值中的6个：13,23,43,53,73和83都是素数。通过用相同的数字替换56 \*\* 3的第3和第4位数字，这个5位数字是在10个生成的数字中具有7个素数的第一个例子，产生该族：56003,56113,56333,56443,56663,56773因此56003，作为这个家族的第一个成员，是拥有这个属性的最小的素数。找到最小的素数，通过用相同的数字替换部分数字（不一定是相邻的数字），它是8个素数值族的一部分。
+By replacing the 1st digit of the 2-digit number \*3, it turns out that six of the nine possible values: 13, 23, 43, 53, 73, and 83, are all prime.
+
+By replacing the 3rd and 4th digits of 56\*\*3 with the same digit, this 5-digit number is the first example having seven primes among the ten generated numbers, yielding the family: 56003, 56113, 56333, 56443, 56663, 56773, and 56993. Consequently 56003, being the first member of this family, is the smallest prime with this property.
+
+Find the smallest prime which, by replacing part of the number (not necessarily adjacent digits) with the same digit, is part of an eight prime value family.
# --hints--
-`euler51()`应返回121313。
+`primeDigitReplacements()` should return a number.
```js
-assert.strictEqual(euler51(), 121313);
+assert(typeof primeDigitReplacements() === 'number');
+```
+
+`primeDigitReplacements()` should return 121313.
+
+```js
+assert.strictEqual(primeDigitReplacements(), 121313);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-52-permuted-multiples.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-52-permuted-multiples.md
index b0f821002cd..080123c5aa8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-52-permuted-multiples.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-52-permuted-multiples.md
@@ -1,18 +1,26 @@
---
id: 5900f3a01000cf542c50feb3
-title: 问题52：置换倍数
+title: 'Problem 52: Permuted multiples'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302163
dashedName: problem-52-permuted-multiples
---
# --description--
-可以看出，数字125874及其双精度数251748包含完全相同的数字，但顺序不同。找到最小的正整数x，使得2x，3x，4x，5x和6x包含相同的数字。
+It can be seen that the number, 125874, and its double, 251748, contain exactly the same digits, but in a different order.
+
+Find the smallest positive integer, `x`, such that `2x`, `3x`, `4x`, `5x`, and `6x`, contain the same digits.
# --hints--
-`permutedMultiples()`应该返回142857。
+`permutedMultiples()` should return a number.
+
+```js
+assert(typeof permutedMultiples() === 'number');
+```
+
+`permutedMultiples()` should return 142857.
```js
assert.strictEqual(permutedMultiples(), 142857);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-53-combinatoric-selections.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-53-combinatoric-selections.md
index 335bc013b39..dbd39ef9755 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-53-combinatoric-selections.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-53-combinatoric-selections.md
@@ -1,40 +1,52 @@
---
id: 5900f3a11000cf542c50feb4
-title: 问题53：组合选择
+title: 'Problem 53: Combinatoric selections'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302164
dashedName: problem-53-combinatoric-selections
---
# --description--
-有十种方法从五种中选择三种，12345：123,124,125,134,135,145,234,235,245和345在组合学中，我们使用符号，5C3 = 10.一般来说，
+There are exactly ten ways of selecting three from five, 12345:
-nCr = n！r！（n-r）！ ，其中r≤n，n！ = n×（n-1）×...×3×2×1和0！ = 1。
+<div style='text-align: center;'>123, 124, 125, 134, 135, 145, 234, 235, 245, and 345</div>
-直到n = 23，一个值超过一百万：23C10 = 1144066.对于1≤n≤100，nCr的多少，不一定是不同的值大于一百万？
+In combinatorics, we use the notation, $\\displaystyle \\binom 5 3 = 10$
+
+In general, $\\displaystyle \\binom n r = \\dfrac{n!}{r!(n-r)!}$, where $r \\le n$, $n! = n \\times (n-1) \\times ... \\times 3 \\times 2 \\times 1$, and $0! = 1$.
+
+It is not until $n = 23$, that a value exceeds one-million: $\\displaystyle \\binom {23} {10} = 1144066$.
+
+How many, not necessarily distinct, values of $\\displaystyle \\binom n r$ for $1 \\le n \\le 100$, are greater than one-million?
# --hints--
-`combinatoricSelections(1000)`应返回4626。
+`combinatoricSelections(1000)` should return a number.
+
+```js
+assert(typeof combinatoricSelections(1000) === 'number');
+```
+
+`combinatoricSelections(1000)` should return 4626.
```js
assert.strictEqual(combinatoricSelections(1000), 4626);
```
-`combinatoricSelections(10000)`应该返回4431。
+`combinatoricSelections(10000)` should return 4431.
```js
assert.strictEqual(combinatoricSelections(10000), 4431);
```
-`combinatoricSelections(100000)`应返回4255。
+`combinatoricSelections(100000)` should return 4255.
```js
assert.strictEqual(combinatoricSelections(100000), 4255);
```
-`combinatoricSelections(1000000)`应该返回4075。
+`combinatoricSelections(1000000)` should return 4075.
```js
assert.strictEqual(combinatoricSelections(1000000), 4075);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-54-poker-hands.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-54-poker-hands.md
index 32ac8870159..fd260cddf92 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-54-poker-hands.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-54-poker-hands.md
@@ -1,25 +1,64 @@
---
id: 5900f3a21000cf542c50feb5
-title: 问题54：扑克手
+title: 'Problem 54: Poker hands'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302165
dashedName: problem-54-poker-hands
---
# --description--
-在纸牌游戏扑克牌中，一手牌由五张牌组成，并按以下方式从最低到最高排名：高牌：最高价值牌。一对：两张价值相同的牌。两对：两对不同。三种类型：三张相同价值的牌。直：所有卡都是连续值。同花顺：同一套牌的所有牌。满屋：三种和一对。四种：四张相同价值的牌。同花顺：所有牌都是同一套牌的连续值。皇家同花顺：十，杰克，女王，国王，王牌，同样的诉讼。卡的价值依次为：2,3,4,5,6,7,8,9,10，Jack，Queen，King，Ace。如果两个玩家拥有相同的排名牌，则排名由最高值赢得;例如，一对八个击打一对五（见下面的例子1）。但是，如果两个级别相关，例如，两个玩家都有一对皇后，则比较每手牌中的最高牌数（参见下面的例子4）;如果最高牌结合，则比较下一张最高牌，依此类推。考虑以下五手交给两名球员：
+In the card game poker, a hand consists of five cards and are ranked, from lowest to highest, in the following way:
-手牌1球员2冠军1 5H 5C 6S 7S KDPair of Fives 2C 3S 8S 8D TDPair of Eights Player 2 2 5D 8C 9S JS ACHighest card Ace 2C 5C 7D 8S QHHighest card Queen Player 1 3 2D 9C AS AH ACThree Aces 3D 6D 7D TD QDFlush与钻石玩家2 4 4D 6S 9H QH QCPair of Queens最高卡九3D 6D 7H QD QSPair皇后最高卡七人玩家1 5 2H 2D 4C 4D 4SFull HouseWith Three Fours 3C 3D 3S 9S 9DFull Housewith Three Threes Player 1
+<ul>
+ <li>High Card: Highest value card.</li>
+ <li>One Pair: Two cards of the same value.</li>
+ <li>Two Pairs: Two different pairs.</li>
+ <li>Three of a Kind: Three cards of the same value.</li>
+ <li>Straight: All cards are consecutive values.</li>
+ <li>Flush: All cards of the same suit.</li>
+ <li>Full House: Three of a kind and a pair.</li>
+ <li>Four of a Kind: Four cards of the same value.</li>
+ <li>Straight Flush: All cards are consecutive values of same suit.</li>
+ <li>Royal Flush: Ten, Jack, Queen, King, Ace, in same suit.</li>
+</ul>
-文件poker.txt包含一千个随机发给两个玩家的牌。该文件的每一行包含十张牌（由一个空格分隔）：前五张是Player 1的牌，后五张是Player 2的牌。你可以假设所有的牌都是有效的（没有无效的角色或重复的牌），每个牌手的牌都没有特定的顺序，而且每手牌都有明显的赢家。玩家1赢了多少手牌？
+The cards are valued in the order: 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace.
+
+If two players have the same ranked hands then the rank made up of the highest value wins; for example, a pair of eights beats a pair of fives (see example 1 below). But if two ranks tie, for example, both players have a pair of queens, then highest cards in each hand are compared (see example 4 below); if the highest cards tie then the next highest cards are compared, and so on.
+
+Consider the following five hands dealt to two players:
+
+| Hand | Player 1 | Player 2 | Winner |
+| ------------------------- | --------------------------------------------------------------------- | ---------------------------------------------------------------------- | -------- |
+| <strong>1</strong> | 5H 5C 6S 7S KD <br> Pair of Fives | 2C 3S 8S 8D TD <br> Pair of Eights | Player 2 |
+| <strong>2</strong> | 5D 8C 9S JS AC <br> Highest card Ace | 2C 5C 7D 8S QH <br> Highest card Queen | Player 1 |
+| <strong>3</strong> | 2D 9C AS AH AC <br> Three Aces | 3D 6D 7D TD QD <br> Flush with Diamonds | Player 2 |
+| <strong>4</strong> | 4D 6S 9H QH QC <br> Pair of Queens <br> Highest card Nine | 3D 6D 7H QD QS <br> Pair of Queens <br> Highest card Seven | Player 1 |
+| <strong>5</strong> | 2H 2D 4C 4D 4S <br> Full House <br> with Three Fours | 3C 3D 3S 9S 9D <br> Full House <br> with Three Threes | Player 1 |
+
+The global array (`handsArr`) passed to the function, contains one-thousand random hands dealt to two players. Each line of the file contains ten cards (separated by a single space): the first five are Player 1's cards and the last five are Player 2's cards. You can assume that all hands are valid (no invalid characters or repeated cards), each player's hand is in no specific order, and in each hand there is a clear winner.
+
+How many hands does Player 1 win?
# --hints--
-`euler54()`应该返回376。
+`pokerHands(testArr)` should return a number.
```js
-assert.strictEqual(euler54(), 376);
+assert(typeof pokerHands(testArr) === 'number');
+```
+
+`pokerHands(testArr)` should return 2.
+
+```js
+assert.strictEqual(pokerHands(testArr), 2);
+```
+
+`pokerHands(handsArr)` should return 376.
+
+```js
+assert.strictEqual(pokerHands(handsArr), 376);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-55-lychrel-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-55-lychrel-numbers.md
index 1d96c63859c..8b074ba2594 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-55-lychrel-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-55-lychrel-numbers.md
@@ -1,47 +1,71 @@
---
id: 5900f3a31000cf542c50feb6
-title: 问题55：Lychrel数字
+title: 'Problem 55: Lychrel numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302166
dashedName: problem-55-lychrel-numbers
---
# --description--
-如果我们采取47，反向并添加，47 + 74 = 121，这是回文。并非所有数字都如此迅速地产生回文。例如，349 + 943 = 1292,1292 + 2921 = 4213 4213 + 3124 = 7337也就是说，349进行了三次迭代以到达回文。虽然还没有人证明这一点，但据认为有些数字，如196，从未产生回文。通过反向和添加过程从不形成回文的数字称为Lychrel数。由于这些数字的理论性质，并且出于这个问题的目的，我们将假设一个数字是Lychrel，直到证明不是这样。另外，对于每万个低于一万的数字，你将得到（i）在不到五十次迭代中成为回文，或者（ii）没有一个，具有所有存在的计算能力，到目前为止已经管理到将它映射到回文结构。事实上，10677是第一个在产生回文之前需要超过50次迭代的数字：4668731596684224866951378664（53次迭代，28位数）。令人惊讶的是，有一些回文数字本身就是Lychrel数字;第一个例子是4994.有多少Lychrel数字在`num`以下？注：2007年4月24日略微修改了措辞，以强调Lychrel数的理论性质。
+If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.
+
+Not all numbers produce palindromes so quickly. For example,
+
+<div style="margin-left: 4em;">
+ 349 + 943 = 1292,<br>
+ 1292 + 2921 = 4213<br>
+ 4213 + 3124 = 7337<br>
+</div>
+
+That is, 349 took three iterations to arrive at a palindrome.
+
+Although no one has proved it yet, it is thought that some numbers, like 196, never produce a palindrome. A number that never forms a palindrome through the reverse and add process is called a Lychrel number. Due to the theoretical nature of these numbers, and for the purpose of this problem, we shall assume that a number is Lychrel until proven otherwise. In addition you are given that for every number below ten-thousand, it will either (i) become a palindrome in less than fifty iterations, or, (ii) no one, with all the computing power that exists, has managed so far to map it to a palindrome. In fact, 10677 is the first number to be shown to require over fifty iterations before producing a palindrome: 4668731596684224866951378664 (53 iterations, 28-digits).
+
+Surprisingly, there are palindromic numbers that are themselves Lychrel numbers; the first example is 4994.
+
+How many Lychrel numbers are there below `num`?
+
+**Note:** Wording was modified slightly on 24 April 2007 to emphasize the theoretical nature of Lychrel numbers.
# --hints--
-`countLychrelNumbers(1000)`应该返回13。
+`countLychrelNumbers(1000)` should return a number.
+
+```js
+assert(typeof countLychrelNumbers(1000) === 'number');
+```
+
+`countLychrelNumbers(1000)` should return 13.
```js
assert.strictEqual(countLychrelNumbers(1000), 13);
```
-`countLychrelNumbers(5000)`应该返回76。
-
-```js
-assert.strictEqual(countLychrelNumbers(5000), 76);
-```
-
-`countLychrelNumbers(10000)`应该返回249。
-
-```js
-assert.strictEqual(countLychrelNumbers(10000), 249);
-```
-
-你的函数应该计算所有Lychrel数。
+`countLychrelNumbers(3243)` should return 39.
```js
assert.strictEqual(countLychrelNumbers(3243), 39);
```
-您的函数应该通过所有测试用例。
+`countLychrelNumbers(5000)` should return 76.
+
+```js
+assert.strictEqual(countLychrelNumbers(5000), 76);
+```
+
+`countLychrelNumbers(7654)` should return 140.
```js
assert.strictEqual(countLychrelNumbers(7654), 140);
```
+`countLychrelNumbers(10000)` should return 249.
+
+```js
+assert.strictEqual(countLychrelNumbers(10000), 249);
+```
+
# --seed--
## --seed-contents--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-56-powerful-digit-sum.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-56-powerful-digit-sum.md
index 6f3d2b64e4b..fbe0693bb2b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-56-powerful-digit-sum.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-56-powerful-digit-sum.md
@@ -1,21 +1,29 @@
---
id: 5900f3a41000cf542c50feb7
-title: 问题56：强大的数字总和
+title: 'Problem 56: Powerful digit sum'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302167
dashedName: problem-56-powerful-digit-sum
---
# --description--
-googol（10100）是一个庞大的数字：一个跟着一百个零; 100100几乎是难以想象的大：一个跟着两百个零。尽管它们的大小，每个数字的数字之和仅为1.考虑到形式的自然数，ab，其中a，b <100，最大数字和是多少？
+A googol (10<sup>100</sup>) is a massive number: one followed by one-hundred zeros; 100<sup>100</sup> is almost unimaginably large: one followed by two-hundred zeros. Despite their size, the sum of the digits in each number is only 1.
+
+Considering natural numbers of the form, `ab`, where `a`, `b` < 100, what is the maximum digital sum?
# --hints--
-`euler56()`应返回972。
+`powerfulDigitSum()` should return a number.
```js
-assert.strictEqual(euler56(), 972);
+assert(typeof powerfulDigitSum() === 'number');
+```
+
+`powerfulDigitSum()` should return 972.
+
+```js
+assert.strictEqual(powerfulDigitSum(), 972);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-57-square-root-convergents.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-57-square-root-convergents.md
index 8d81bc905fa..e57107565f7 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-57-square-root-convergents.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-57-square-root-convergents.md
@@ -1,21 +1,43 @@
---
id: 5900f3a51000cf542c50feb8
-title: 问题57：平方根收敛
+title: 'Problem 57: Square root convergents'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302168
dashedName: problem-57-square-root-convergents
---
# --description--
-可以证明两个的平方根可以表示为无限连续分数。 √2= 1 + 1 /（2 + 1 /（2 + 1 /（2 + ...）））= 1.414213 ...通过扩展前四次迭代，得到：1 + 1/2 = 3 / 2 = 1.5 1 + 1 /（2 + 1/2）= 7/5 = 1.4 1 + 1 /（2 + 1 /（2 + 1/2））= 17/12 = 1.41666 ... 1 + 1 /（2 + 1 /（2 + 1 /（2 + 1/2）））= 41/29 = 1.41379 ...接下来的三次扩展是99 / 70,239 / 169和577/408，但是第八次扩展，1393/985，是分子中位数超过分母中位数的第一个例子。在前一千次扩展中，有多少分数包含一个数字比分母更多的分子？
+It is possible to show that the square root of two can be expressed as an infinite continued fraction.
+
+<div style='text-align: center;'>$\sqrt 2 =1+ \frac 1 {2+ \frac 1 {2 +\frac 1 {2+ \dots}}}$</div>
+
+By expanding this for the first four iterations, we get:
+
+$1 + \\frac 1 2 = \\frac 32 = 1.5$
+
+$1 + \\frac 1 {2 + \\frac 1 2} = \\frac 7 5 = 1.4$
+
+$1 + \\frac 1 {2 + \\frac 1 {2+\\frac 1 2}} = \\frac {17}{12} = 1.41666 \\dots$
+
+$1 + \\frac 1 {2 + \\frac 1 {2+\\frac 1 {2+\\frac 1 2}}} = \\frac {41}{29} = 1.41379 \\dots$
+
+The next three expansions are $\\frac {99}{70}$, $\\frac {239}{169}$, and $\\frac {577}{408}$, but the eighth expansion, $\\frac {1393}{985}$, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.
+
+In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?
# --hints--
-`euler57()`应返回153。
+`squareRootConvergents()` should return a number.
```js
-assert.strictEqual(euler57(), 153);
+assert(typeof squareRootConvergents() === 'number');
+```
+
+`squareRootConvergents()` should return 153.
+
+```js
+assert.strictEqual(squareRootConvergents(), 153);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-58-spiral-primes.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-58-spiral-primes.md
index c0ba66e8052..d9fb67db87e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-58-spiral-primes.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-58-spiral-primes.md
@@ -1,21 +1,41 @@
---
id: 5900f3a61000cf542c50feb9
-title: 问题58：螺旋素数
+title: 'Problem 58: Spiral primes'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302169
dashedName: problem-58-spiral-primes
---
# --description--
-从1开始并以下列方式逆时针旋转，形成边长为7的方形螺旋。 37 36 35 34 33 32 31 38 17 16 15 14 13 30 39 18 5 4 3 12 29 40 19 6 1 2 11 28 41 20 7 8 9 10 27 42 21 22 23 24 25 2643 44 45 46 47 48 49这是有趣的是，奇数方块位于右下角，但更有意思的是沿着两条对角线的13个数字中有8个是素数;也就是说，比例为8 /13≈62％。如果在上面的螺旋周围缠绕一个完整的新层，则将形成具有边长9的方形螺旋。如果继续这个过程，那么沿着两条对角线的素数比首先低于10％的方形螺旋的边长是多少？
+Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.
+
+<div style='text-align: center;'>
+ <strong><span style='color: red;'>37</span></strong> 36 35 34 33 32 <strong><span style='color: red;'>31</span></strong><br>
+ 38 <strong><span style='color: red;'>17</span></strong> 16 15 14 <strong><span style='color: red;'>13</span></strong> 30<br>
+ 39 18 <strong><span style='color: red;'>5</span></strong> 4 <strong><span style='color: red;'>3</span></strong> 12 29<br>
+ 40 19 6 1 2 11 28<br>
+ 41 20 <strong><span style='color: red;'>7</span></strong> 8 9 10 27<br>
+ 42 21 22 23 24 25 26<br>
+ <strong><span style='color: red;'>43</span></strong> 44 45 46 47 48 49<br>
+</div>
+
+It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.
+
+If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?
# --hints--
-`euler58()`应返回26241。
+`spiralPrimes()` should return a number.
```js
-assert.strictEqual(euler58(), 26241);
+assert(typeof spiralPrimes() === 'number');
+```
+
+`spiralPrimes()` should return 26241.
+
+```js
+assert.strictEqual(spiralPrimes(), 26241);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-59-xor-decryption.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-59-xor-decryption.md
index e50c455d96c..683db15249a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-59-xor-decryption.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-59-xor-decryption.md
@@ -1,21 +1,35 @@
---
id: 5900f3a81000cf542c50feba
-title: 问题59：XOR解密
+title: 'Problem 59: XOR decryption'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302170
dashedName: problem-59-xor-decryption
---
# --description--
-计算机上的每个字符都分配有唯一的代码，首选标准是ASCII（美国信息交换标准代码）。例如，大写A = 65，星号（\*）= 42，小写k = 107.现代加密方法是获取文本文件，将字节转换为ASCII，然后使用给定值对每个字节进行异或，取自密钥。 XOR功能的优点是在密文上使用相同的加密密钥，恢复纯文本;例如，65 XOR 42 = 107，然后107 XOR 42 = 65.对于不可破解的加密，密钥与纯文本消息的长度相同，密钥由随机字节组成。用户将加密的消息和加密密钥保持在不同的位置，并且没有两个“一半”，就不可能解密该消息。不幸的是，这种方法对大多数用户来说是不切实际的，因此修改后的方法是使用密码作为密钥。如果密码短于可能的消息，则密钥在整个消息中循环重复。此方法的余额是使用足够长的密码密钥来保证安全性，但又要足够短，以便令人难忘。您的任务变得简单，因为加密密钥由三个小写字符组成。使用cipher.txt（右键单击并“保存链接/目标为...”），包含加密的ASCII代码的文件，以及纯文本必须包含常用英语单词的知识，解密消息并找到总和原始文本中的ASCII值。
+Each character on a computer is assigned a unique code and the preferred standard is ASCII (American Standard Code for Information Interchange). For example, uppercase A = 65, asterisk (\*) = 42, and lowercase k = 107.
+
+A modern encryption method is to take a text file, convert the bytes to ASCII, then XOR each byte with a given value, taken from a secret key. The advantage with the XOR function is that using the same encryption key on the cipher text, restores the plain text; for example, 65 XOR 42 = 107, then 107 XOR 42 = 65.
+
+For unbreakable encryption, the key is the same length as the plain text message, and the key is made up of random bytes. The user would keep the encrypted message and the encryption key in different locations, and without both "halves", it is impossible to decrypt the message.
+
+Unfortunately, this method is impractical for most users, so the modified method is to use a password as a key. If the password is shorter than the message, which is likely, the key is repeated cyclically throughout the message. The balance for this method is using a sufficiently long password key for security, but short enough to be memorable.
+
+Your task has been made easy, as the encryption key consists of three lower case characters. Using `cipher`, an array containing the encrypted ASCII codes, and the knowledge that the plain text must contain common English words, decrypt the message and find the sum of the ASCII values in the original text.
# --hints--
-`euler59()`应返回107359。
+`XORDecryption(cipher)` should return a number.
```js
-assert.strictEqual(euler59(), 107359);
+assert(typeof XORDecryption(cipher) === 'number');
+```
+
+`XORDecryption(cipher)` should return 129448.
+
+```js
+assert.strictEqual(XORDecryption(cipher), 129448);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-6-sum-square-difference.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-6-sum-square-difference.md
index 8495aef8d14..ec60601f106 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-6-sum-square-difference.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-6-sum-square-difference.md
@@ -1,58 +1,46 @@
---
id: 5900f3721000cf542c50fe85
-title: 问题6：求和方差
+title: 'Problem 6: Sum square difference'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302171
dashedName: problem-6-sum-square-difference
---
# --description--
-前十个自然数的平方和是，
+The sum of the squares of the first ten natural numbers is,
-1
+<div style='text-align: center;'>1<sup>2</sup> + 2<sup>2</sup> + ... + 10<sup>2</sup> = 385</div>
-<sup>2</sup>
+The square of the sum of the first ten natural numbers is,
-- 2
+<div style='text-align: center;'>(1 + 2 + ... + 10)<sup>2</sup> = 55<sup>2</sup> = 3025</div>
-<sup>2</sup>
+Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.
-- ... + 10
-
-<sup>2</sup>
-
- = 385
-
-前十个自然数之和的平方是，
-
-（1 + 2 + ... + 10）
-
-<sup>2</sup>
-
- = 55
-
-<sup>2</sup>
-
- = 3025
-
-因此，前十个自然数的平方和与和的平方之间的差值为3025 - 385 = 2640.求出前`n`自然数的平方和与总和的平方之间的差值。
+Find the difference between the sum of the squares of the first `n` natural numbers and the square of the sum.
# --hints--
-`sumSquareDifference(10)`应该返回2640。
+`sumSquareDifference(10)` should return a number.
+
+```js
+assert(typeof sumSquareDifference(10) === 'number');
+```
+
+`sumSquareDifference(10)` should return 2640.
```js
assert.strictEqual(sumSquareDifference(10), 2640);
```
-`sumSquareDifference(20)`应该返回41230。
+`sumSquareDifference(20)` should return 41230.
```js
assert.strictEqual(sumSquareDifference(20), 41230);
```
-`sumSquareDifference(100)`应该返回25164150。
+`sumSquareDifference(100)` should return 25164150.
```js
assert.strictEqual(sumSquareDifference(100), 25164150);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-60-prime-pair-sets.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-60-prime-pair-sets.md
index 64988b85b94..5acc6c21769 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-60-prime-pair-sets.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-60-prime-pair-sets.md
@@ -1,21 +1,29 @@
---
id: 5900f3a81000cf542c50febb
-title: 问题60：素数对设置
+title: 'Problem 60: Prime pair sets'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302172
dashedName: problem-60-prime-pair-sets
---
# --description--
-素数3,7,109和673非常值得注意。通过取任何两个素数并以任何顺序连接它们，结果将始终是素数。例如，取7和109，7109和1097都是素数。这四个素数的总和，792，代表具有此属性的一组四个素数的最低和。找到一组五个素数的最低和，其中任何两个素数连接以产生另一个素数。
+The primes 3, 7, 109, and 673, are quite remarkable. By taking any two primes and concatenating them in any order the result will always be prime. For example, taking 7 and 109, both 7109 and 1097 are prime. The sum of these four primes, 792, represents the lowest sum for a set of four primes with this property.
+
+Find the lowest sum for a set of five primes for which any two primes concatenate to produce another prime.
# --hints--
-`euler60()`应返回26033。
+`primePairSets()` should return a number.
```js
-assert.strictEqual(euler60(), 26033);
+assert(typeof primePairSets() === 'number');
+```
+
+`primePairSets()` should return 26033.
+
+```js
+assert.strictEqual(primePairSets(), 26033);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-61-cyclical-figurate-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-61-cyclical-figurate-numbers.md
index bd7889c471b..04ace60de8e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-61-cyclical-figurate-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-61-cyclical-figurate-numbers.md
@@ -1,45 +1,46 @@
---
id: 5900f3a91000cf542c50febc
-title: 问题61：循环图号
+title: 'Problem 61: Cyclical figurate numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302173
dashedName: problem-61-cyclical-figurate-numbers
---
# --description--
-三角形，正方形，五边形，六边形，七边形和八边形数字都是图形（多边形）数字，由以下公式生成：三角形
+Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae:
-P3中，n = N（N + 1）/ 2
+| Type of Number | Formula | Sequence |
+| -------------- | --------------------------------------------------------------------- | --------------------- |
+| Triangle | P<sub>3</sub>,<var><sub>n</sub></var>=<var>n</var>(<var>n</var>+1)/2 | 1, 3, 6, 10, 15, ... |
+| Square | P<sub>4</sub>,<var><sub>n</sub></var>=<var>n</var><sup>2</sup> | 1, 4, 9, 16, 25, ... |
+| Pentagonal | P<sub>5</sub>,<var><sub>n</sub></var>=<var>n</var>(3<var>n</var>−1)/2 | 1, 5, 12, 22, 35, ... |
+| Hexagonal | P<sub>6</sub>,<var><sub>n</sub></var>=<var>n</var>(2<var>n</var>−1) | 1, 6, 15, 28, 45, ... |
+| Heptagonal | P<sub>7</sub>,<var><sub>n</sub></var>=<var>n</var>(5<var>n</var>−3)/2 | 1, 7, 18, 34, 55, ... |
+| Octagonal | P<sub>8</sub>,<var><sub>n</sub></var>=<var>n</var>(3<var>n</var>−2) | 1, 8, 21, 40, 65, ... |
-1,3,6,10,15 ......方形
+The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three interesting properties.
-P4 N = N2
+<ol>
+ <li>The set is cyclic, in that the last two digits of each number is the first two digits of the next number (including the last number with the first).</li>
+ <li>Each polygonal type: triangle (P<sub>3,127</sub> = 8128), square (P<sub>4,91</sub> = 8281), and pentagonal (P<sub>5,44</sub> = 2882), is represented by a different number in the set.</li>
+ <li>This is the only set of 4-digit numbers with this property.</li>
+</ol>
-1,4,9,16,25 ......五角形
-
-P5，N = N（3N-1）/ 2
-
-1,5,12,22,35 ......六角形
-
-P6，N = N（2N-1）
-
-1,6,15,28,45 ...... Heptagonal
-
-P7，N = N（5N-3）/ 2
-
-1,7,18,34,55，......八角形
-
-P8，N = N（3N-2）
-
-1,8,21,40,65，......三个4位数字的有序集合：8128,2882,8281，有三个有趣的属性。该集合是循环的，因为每个数字的最后两位数字是下一个数字的前两位数字（包括与第一个数字相关的最后一个数字）。每个多边形类型：三角形（P3,127 = 8128），方形（P4,91 = 8281）和五边形（P5,44 = 2882），由集合中的不同数字表示。这是具有此属性的唯一一组4位数字。求出六个循环4位数字的唯一有序集合的总和，其中每个多边形类型：三角形，正方形，五边形，六边形，七边形和八边形，由集合中的不同数字表示。
+Find the sum of the only ordered set of six cyclic 4-digit numbers for which each polygonal type: triangle, square, pentagonal, hexagonal, heptagonal, and octagonal, is represented by a different number in the set.
# --hints--
-`euler61()`应返回28684。
+`cyclicalFigurateNums()` should return a number.
```js
-assert.strictEqual(euler61(), 28684);
+assert(typeof cyclicalFigurateNums() === 'number');
+```
+
+`cyclicalFigurateNums()` should return 28684.
+
+```js
+assert.strictEqual(cyclicalFigurateNums(), 28684);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-62-cubic-permutations.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-62-cubic-permutations.md
index bcd5dc4b6f6..fb98324efec 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-62-cubic-permutations.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-62-cubic-permutations.md
@@ -1,21 +1,29 @@
---
id: 5900f3aa1000cf542c50febd
-title: 问题62：立方排列
+title: 'Problem 62: Cubic permutations'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302174
dashedName: problem-62-cubic-permutations
---
# --description--
-可以置换立方体41063625（3453）以产生另外两个立方体：56623104（3843）和66430125（4053）。实际上，41063625是最小的立方体，其正好具有三个排列的数字也是立方体。找到最小的立方体，其数字的五个排列正好是立方体。
+The cube, 41063625 (345<sup>3</sup>), can be permuted to produce two other cubes: 56623104 (384<sup>3</sup>) and 66430125 (405<sup>3</sup>). In fact, 41063625 is the smallest cube which has exactly three permutations of its digits which are also cube.
+
+Find the smallest cube for which exactly five permutations of its digits are cube.
# --hints--
-`euler62()`应该返回127035954683。
+`cubicPermutations()` should return a number.
```js
-assert.strictEqual(euler62(), 127035954683);
+assert(typeof cubicPermutations() === 'number');
+```
+
+`cubicPermutations()` should return 127035954683.
+
+```js
+assert.strictEqual(cubicPermutations(), 127035954683);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-63-powerful-digit-counts.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-63-powerful-digit-counts.md
index 4838f0a2ba4..661240b19cf 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-63-powerful-digit-counts.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-63-powerful-digit-counts.md
@@ -1,21 +1,29 @@
---
id: 5900f3ab1000cf542c50febe
-title: 问题63：强大的数字计数
+title: 'Problem 63: Powerful digit counts'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302175
dashedName: problem-63-powerful-digit-counts
---
# --description--
-5位数字，16807 = 75，也是第五种力量。类似地，9位数字134217728 = 89是第九次幂。存在多少个n位正整数，它们也是n次幂？
+The 5-digit number, 16807 = 7<sup>5</sup>, is also a fifth power. Similarly, the 9-digit number, 134217728 = 8<sup>9</sup>, is a ninth power.
+
+How many `n`-digit positive integers exist which are also an `n`th power?
# --hints--
-`euler63()`应该返回49。
+`powerfulDigitCounts()` should return a number.
```js
-assert.strictEqual(euler63(), 49);
+assert(typeof powerfulDigitCounts() === 'number');
+```
+
+`powerfulDigitCounts()` should return 49.
+
+```js
+assert.strictEqual(powerfulDigitCounts(), 49);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-64-odd-period-square-roots.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-64-odd-period-square-roots.md
index 549ed778d81..dedc83de20d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-64-odd-period-square-roots.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-64-odd-period-square-roots.md
@@ -1,73 +1,83 @@
---
id: 5900f3ac1000cf542c50febf
-title: 问题64：奇数期平方根
+title: 'Problem 64: Odd period square roots'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302176
dashedName: problem-64-odd-period-square-roots
---
# --description--
-所有平方根都是周期性的，当写为连续分数时，可以写成以下形式：
+All square roots are periodic when written as continued fractions and can be written in the form:
-√N= a0 + 1
+$\\displaystyle \\quad \\quad \\sqrt{N}=a_0+\\frac 1 {a_1+\\frac 1 {a_2+ \\frac 1 {a3+ \\dots}}}$
-a1 + 1
+For example, let us consider $\\sqrt{23}:$:
-a2 + 1
+$\\quad \\quad \\sqrt{23}=4+\\sqrt{23}-4=4+\\frac 1 {\\frac 1 {\\sqrt{23}-4}}=4+\\frac 1 {1+\\frac{\\sqrt{23}-3}7}$
-a3 + ......
+If we continue we would get the following expansion:
-例如，让我们考虑√23：
+$\\displaystyle \\quad \\quad \\sqrt{23}=4+\\frac 1 {1+\\frac 1 {3+ \\frac 1 {1+\\frac 1 {8+ \\dots}}}}$
-√23= 4 +√23 - 4 = 4 + 1 = 4 + 1
+The process can be summarized as follows:
-1√23-4
+$\\quad \\quad a_0=4, \\frac 1 {\\sqrt{23}-4}=\\frac {\\sqrt{23}+4} 7=1+\\frac {\\sqrt{23}-3} 7$
-1 +√23 - 37
+$\\quad \\quad a_1=1, \\frac 7 {\\sqrt{23}-3}=\\frac {7(\\sqrt{23}+3)} {14}=3+\\frac {\\sqrt{23}-3} 2$
-如果我们继续，我们将得到以下扩展：
+$\\quad \\quad a_2=3, \\frac 2 {\\sqrt{23}-3}=\\frac {2(\\sqrt{23}+3)} {14}=1+\\frac {\\sqrt{23}-4} 7$
-√23= 4 + 1
+$\\quad \\quad a_3=1, \\frac 7 {\\sqrt{23}-4}=\\frac {7(\\sqrt{23}+4)} 7=8+\\sqrt{23}-4$
-1 + 1
+$\\quad \\quad a_4=8, \\frac 1 {\\sqrt{23}-4}=\\frac {\\sqrt{23}+4} 7=1+\\frac {\\sqrt{23}-3} 7$
-3 + 1
+$\\quad \\quad a_5=1, \\frac 7 {\\sqrt{23}-3}=\\frac {7 (\\sqrt{23}+3)} {14}=3+\\frac {\\sqrt{23}-3} 2$
-1 + 1
+$\\quad \\quad a_6=3, \\frac 2 {\\sqrt{23}-3}=\\frac {2(\\sqrt{23}+3)} {14}=1+\\frac {\\sqrt{23}-4} 7$
-8 + ......
+$\\quad \\quad a_7=1, \\frac 7 {\\sqrt{23}-4}=\\frac {7(\\sqrt{23}+4)} {7}=8+\\sqrt{23}-4$
-该过程可归纳如下：
+It can be seen that the sequence is repeating. For conciseness, we use the notation $\\sqrt{23}=\[4;(1,3,1,8)]$, to indicate that the block (1,3,1,8) repeats indefinitely.
-a0 = 4，
+The first ten continued fraction representations of (irrational) square roots are:
-1√23-4=√23+ 47 = 1 +√23-37a1 = 1，
+$\\quad \\quad \\sqrt{2}=\[1;(2)]$, period = 1
-7√23-3= 7（√23+ 3）14 = 3 +√23-32a2= 3，
+$\\quad \\quad \\sqrt{3}=\[1;(1,2)]$, period = 2
-2√23-3= 2（√23+ 3）14 = 1 +√23-47a3 = 1，
+$\\quad \\quad \\sqrt{5}=\[2;(4)]$, period = 1
-7√23-4= 7（√23+ 4）7 = 8 +√23-4a4= 8，
+$\\quad \\quad \\sqrt{6}=\[2;(2,4)]$, period = 2
-1√23-4=√23+ 47 = 1 +√23-37a5 = 1，
+$\\quad \\quad \\sqrt{7}=\[2;(1,1,1,4)]$, period = 4
-7√23-3= 7（√23+ 3）14 = 3 +√23-32a6= 3，
+$\\quad \\quad \\sqrt{8}=\[2;(1,4)]$, period = 2
-2√23-3= 2（√23+ 3）14 = 1 +√23-47a7 = 1，
+$\\quad \\quad \\sqrt{10}=\[3;(6)]$, period = 1
-7√23-4= 7（√23+ 4）7 = 8 +√23-4
+$\\quad \\quad \\sqrt{11}=\[3;(3,6)]$, period = 2
-可以看出序列是重复的。为简明起见，我们使用符号√23= \[4;（1,3,1,8）]来表示块（1,3,1,8）无限重复。
+$\\quad \\quad \\sqrt{12}=\[3;(2,6)]$, period = 2
-（无理）平方根的前十个连续分数表示为：√2= \[1;（2）]，周期=1√3= \[1;（1,2）]，周期=2√5= \[2; （4）]，期间=1√6= \[2;（2,4）]，期间=2√7= \[2;（1,1,1,4）]，期间=4√8= \[2; （1,4）]，期间=2√10= \[3;（6）]，期间=1√11= \[3;（3,6）]，期间=2√12= \[3;（2,6 ）]，period =2√13= \[3;（1,1,1,1,6）]，period = 5对于N≤13，恰好四个连续分数具有奇数周期。 N≤10000的连续分数有多少个奇数周期？
+$\\quad \\quad \\sqrt{13}=\[3;(1,1,1,1,6)]$, period = 5
+
+Exactly four continued fractions, for $N \\le 13$, have an odd period.
+
+How many continued fractions for $N \\le 10\\,000$ have an odd period?
# --hints--
-`euler64()`应返回1322。
+`oddPeriodSqrts()` should return a number.
```js
-assert.strictEqual(euler64(), 1322);
+assert(typeof oddPeriodSqrts() === 'number');
+```
+
+`oddPeriodSqrts()` should return 1322.
+
+```js
+assert.strictEqual(oddPeriodSqrts(), 1322);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-65-convergents-of-e.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-65-convergents-of-e.md
index 5fe2acc975f..ca61b8fa362 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-65-convergents-of-e.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-65-convergents-of-e.md
@@ -1,63 +1,45 @@
---
id: 5900f3ad1000cf542c50fec0
-title: 问题65：e的收敛
+title: 'Problem 65: Convergents of e'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302177
dashedName: problem-65-convergents-of-e
---
# --description--
-2的平方根可以写成无限连续分数。
+The square root of 2 can be written as an infinite continued fraction.
-√2= 1 + 1
+$\\sqrt{2} = 1 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2 + ...}}}}$
-2 + 1
+The infinite continued fraction can be written, $\\sqrt{2} = \[1; (2)]$ indicates that 2 repeats *ad infinitum*. In a similar way, $\\sqrt{23} = \[4; (1, 3, 1, 8)]$. It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for $\\sqrt{2}$.
-2 + 1
+$1 + \\dfrac{1}{2} = \\dfrac{3}{2}\\\\ 1 + \\dfrac{1}{2 + \\dfrac{1}{2}} = \\dfrac{7}{5}\\\\ 1 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2}}} = \\dfrac{17}{12}\\\\ 1 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2}}}} = \\dfrac{41}{29}$
-2 + 1
+Hence the sequence of the first ten convergents for $\\sqrt{2}$ are:
-2 + ......
+$1, \\dfrac{3}{2}, \\dfrac{7}{5}, \\dfrac{17}{12}, \\dfrac{41}{29}, \\dfrac{99}{70}, \\dfrac{239}{169}, \\dfrac{577}{408}, \\dfrac{1393}{985}, \\dfrac{3363}{2378}, ...$
-可以写出无限连续分数，√2= \[1;（2）]，（2）表示2无限重复。以类似的方式，√23= \[4;（1,3,1,8）]。事实证明，平方根的连续分数的部分值序列提供了最佳的有理近似。让我们考虑√2的收敛。
+What is most surprising is that the important mathematical constant, $e = \[2; 1, 2, 1, 1, 4, 1, 1, 6, 1, ... , 1, 2k, 1, ...]$. The first ten terms in the sequence of convergents for `e` are:
-1 + 1 = 3/2
+$2, 3, \\dfrac{8}{3}, \\dfrac{11}{4}, \\dfrac{19}{7}, \\dfrac{87}{32}, \\dfrac{106}{39}, \\dfrac{193}{71}, \\dfrac{1264}{465}, \\dfrac{1457}{536}, ...$
-2
+The sum of digits in the numerator of the 10<sup>th</sup> convergent is $1 + 4 + 5 + 7 = 17$.
-1 + 1 = 7/5
-
-2 + 1
-
-2
-
-1 + 1 = 17/12
-
-2 + 1
-
-2 + 1
-
-2
-
-1 + 1 = 41/29
-
-2 + 1
-
-2 + 1
-
-2 + 1
-
-2
-
-因此，√2的前十个收敛的序列是：1,3 / 2,7 / 5,17 / 12,41 / 29,99 / 70,239 / 169,577 / 408,1333 / 985,3333 / 2378 ，...最令人惊讶的是重要的数学常数，e = \[2; 1,2,1,1,4,1,1,6,1，......，1,2k，1，......]。 e的会聚序列中的前十个项是：2,3,8 / 3,11 / 4,19 / 7,87 / 32,106 / 39,193 / 71,1264 / 465,1457 / 536，.... ..第10个收敛的分子中的数字之和为1 + 4 + 5 + 7 = 17。求e的连续分数的第100个收敛的分子中的位数之和。
+Find the sum of digits in the numerator of the 100<sup>th</sup> convergent of the continued fraction for `e`.
# --hints--
-`euler65()`应该返回272。
+`convergentsOfE()` should return a number.
```js
-assert.strictEqual(euler65(), 272);
+assert(typeof convergentsOfE() === 'number');
+```
+
+`convergentsOfE()` should return 272.
+
+```js
+assert.strictEqual(convergentsOfE(), 272);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-66-diophantine-equation.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-66-diophantine-equation.md
index 9f9a62ccfa2..956b98cbefd 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-66-diophantine-equation.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-66-diophantine-equation.md
@@ -1,21 +1,47 @@
---
id: 5900f3ae1000cf542c50fec1
-title: 问题66：丢番图方程
+title: 'Problem 66: Diophantine equation'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302178
dashedName: problem-66-diophantine-equation
---
# --description--
-考虑形式的二次丢番图方程：x2 - Dy2 = 1例如，当D = 13时，x中的最小解是6492 - 13×1802 = 1.可以假设当D是正整数时没有解广场。通过在D中找到D = {2,3,5,6,7}的最小解，我们得到以下结果：32 - 2×22 = 1 22 - 3×12 = 192 - 5×42 = 1 52 - 6× 22 = 1 82 - 7×32 = 1因此，通过考虑D中对于D≤7的最小解，当D = 5时获得最大的x。在x的最小解中找到D≤1000的值，其中获得x的最大值。
+Consider quadratic Diophantine equations of the form:
+
+<div style='text-align: center;'>x<sup>2</sup> – Dy<sup>2</sup> = 1</div>
+
+For example, when D=13, the minimal solution in x is 649<sup>2</sup> – 13×180<sup>2</sup> = 1.
+
+It can be assumed that there are no solutions in positive integers when D is square.
+
+By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following:
+
+<div style='margin-left: 2em;'>
+ 3<sup>2</sup> – 2×2<sup>2</sup> = 1<br>
+ 2<sup>2</sup> – 3×1<sup>2</sup> = 1<br>
+ <strong><span style='color: red;'>9</span></strong><sup>2</sup> – 5×4<sup>2</sup> = 1<br>
+ 5<sup>2</sup> – 6×2<sup>2</sup> = 1<br>
+ 8<sup>2</sup> – 7×3<sup>2</sup> = 1<br>
+</div>
+
+Hence, by considering minimal solutions in `x` for D ≤ 7, the largest `x` is obtained when D=5.
+
+Find the value of D ≤ 1000 in minimal solutions of `x` for which the largest value of `x` is obtained.
# --hints--
-`euler66()`应返回661。
+`diophantineEquation()` should return a number.
```js
-assert.strictEqual(euler66(), 661);
+assert(typeof diophantineEquation() === 'number');
+```
+
+`diophantineEquation()` should return 661.
+
+```js
+assert.strictEqual(diophantineEquation(), 661);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-67-maximum-path-sum-ii.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-67-maximum-path-sum-ii.md
index 6fad0a17999..da2c7676e0d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-67-maximum-path-sum-ii.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-67-maximum-path-sum-ii.md
@@ -1,21 +1,46 @@
---
id: 5900f3b01000cf542c50fec2
-title: 问题67：最大路径总和II
+title: 'Problem 67: Maximum path sum II'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302179
dashedName: problem-67-maximum-path-sum-ii
---
# --description--
-从下面三角形的顶部开始并移动到下面一行的相邻数字，从上到下的最大总数为23. 37 4 2 4 6 8 5 9 3即3 + 7 + 4 + 9 = 23 。在triangle.txt（右键单击和'Save Link / Target As ...'）中找到从上到下的最大总数，这是一个包含一百行三角形的15K文本文件。注意：这是问题18的一个更难的版本。不可能尝试每条路线来解决这个问题，因为共有299个！如果你每秒可以检查一万亿（1012）条路线，则需要花费超过二百亿年才能检查它们。有一种有效的算法可以解决它。 ; O）
+By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
+
+<div style='text-align: center;'>
+ <strong style='color: red;'>3</strong><br>
+ <strong style='color: red;'>7</strong> 4<br>
+ 2 <strong style='color: red;'>4</strong> 6<br>
+ 8 5 <strong style='color: red;'>9</strong> 3
+</div>
+
+That is, 3 + 7 + 4 + 9 = 23.
+
+Find the maximum total from top to bottom in `numTriangle`, a 2D array defined in the background containing a triangle with one-hundred rows.
+
+**Note:** This is a much more difficult version of Problem 18. It is not possible to try every route to solve this problem, as there are 2<sup>99</sup> altogether! If you could check one trillion (10<sup>12</sup>) routes every second it would take over twenty billion years to check them all. There is an efficient algorithm to solve it. ;o)
# --hints--
-`euler67()`应该返回7273。
+`maximumPathSumII(testTriangle)` should return a number.
```js
-assert.strictEqual(euler67(), 7273);
+assert(typeof maximumPathSumII(testTriangle) === 'number');
+```
+
+`maximumPathSumII(testTriangle)` should return 23.
+
+```js
+assert.strictEqual(maximumPathSumII(testTriangle), 23);
+```
+
+`maximumPathSumII(numTriangle)` should return 7273.
+
+```js
+assert.strictEqual(maximumPathSumII(numTriangle), 7273);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-68-magic-5-gon-ring.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-68-magic-5-gon-ring.md
index e799c20d2c9..1ae580513f2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-68-magic-5-gon-ring.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-68-magic-5-gon-ring.md
@@ -1,27 +1,54 @@
---
id: 5900f3b01000cf542c50fec3
-title: 问题68：魔法5-gon戒指
+title: 'Problem 68: Magic 5-gon ring'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302180
dashedName: problem-68-magic-5-gon-ring
---
# --description--
-考虑以下“魔术”3-gon环，填充数字1到6，每行加9。
+Consider the following "magic" 3-gon ring, filled with the numbers 1 to 6, and each line adding to nine.
-顺时针工作，并从具有数字最低外部节点（在该示例中为4,3,2）的三个一组开始，可以唯一地描述每个解决方案。例如，上述解决方案可以通过集合来描述：4,3,2; 6,2,1; 5,1,3。可以用四种不同的总数来完成环：9,10,11和12.总共有八种解决方案。
+<img class="img-responsive center-block" alt="a completed example of a 3-gon ring" src="https://cdn-media-1.freecodecamp.org/project-euler/3-gon-ring.png" style="background-color: white; padding: 10px;" />
-TotalSolution Set 94,2,3; 5,3,1; 6,1,2 94,3,2; 6,2,1; 5,1,3 102,3,5; 4,5,1; 6,1,3 102,5,3; 6,3,1; 4,1,5 111,4,6; 3,6,2; 5,2,4 111,6,4; 5,4,2; 3,2,6 121,5,6; 2,6,4; 3,4,5 121,6,5; 3,5,4; 2,4,6
+Working **clockwise**, and starting from the group of three with the numerically lowest external node (4,3,2 in this example), each solution can be described uniquely. For example, the above solution can be described by the set: 4,3,2; 6,2,1; 5,1,3.
-通过连接每个组，可以形成9位数的字符串; 3-gon环的最大字符串是432621513.使用数字1到10，并根据排列，可以形成16位和17位字符串。 “魔法”5-gon戒指的最大16位数字串是多少？
+It is possible to complete the ring with four different totals: 9, 10, 11, and 12. There are eight solutions in total.
+
+<div style='text-align: center;'>
+
+| !!crwdBlockTags_6_sgaTkcolBdwrc!! |!!crwdBlockTags_7_sgaTkcolBdwrc!! |
+| -------------------------------------- | --------------------------------------------- |
+| 9 | 4,2,3; 5,3,1; 6,1,2 |
+| 9 | 4,3,2; 6,2,1; 5,1,3 |
+| 10 | 2,3,5; 4,5,1; 6,1,3 |
+| 10 | 2,5,3; 6,3,1; 4,1,5 |
+| 11 | 1,4,6; 3,6,2; 5,2,4 |
+| 11 | 1,6,4; 5,4,2; 3,2,6 |
+| 12 | 1,5,6; 2,6,4; 3,4,5 |
+| 12 | 1,6,5; 3,5,4; 2,4,6 |
+
+</div>
+
+By concatenating each group it is possible to form 9-digit strings; the maximum string for a 3-gon ring is 432621513.
+
+Using the numbers 1 to 10, and depending on arrangements, it is possible to form 16- and 17-digit strings. What is the maximum **16-digit** string for a "magic" 5-gon ring?
+
+<img class="img-responsive center-block" alt="a blank diagram of a 5-gon ring" src="https://cdn-media-1.freecodecamp.org/project-euler/5-gon-ring.png" style="background-color: white; padding: 10px;" />
# --hints--
-`euler68()`应该返回6531031914842725。
+`magic5GonRing()` should return a number.
```js
-assert.strictEqual(euler68(), 6531031914842725);
+assert(typeof magic5GonRing() === 'number');
+```
+
+`magic5GonRing()` should return 6531031914842725.
+
+```js
+assert.strictEqual(magic5GonRing(), 6531031914842725);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-69-totient-maximum.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-69-totient-maximum.md
index 35c0772a732..06e4e360f91 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-69-totient-maximum.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-69-totient-maximum.md
@@ -1,25 +1,47 @@
---
id: 5900f3b11000cf542c50fec4
-title: 问题69大欧拉
+title: 'Problem 69: Totient maximum'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302181
dashedName: problem-69-totient-maximum
---
# --description--
-Euler的Totient函数φ（n）\[有时称为phi函数]用于确定小于n的数字的数量，这些数字是n的相对素数。例如，如1,2,4,5,7和8都小于9且相对于9，φ（9）= 6。
+Euler's Totient function, φ(`n`) \[sometimes called the phi function], is used to determine the number of numbers less than `n` which are relatively prime to `n`. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6.
-n相对素数φ（n）n /φ（n）2 1 1 2 3 1,2 2 1.5 4 1,3 2 2 5 1,2,3,4 4 1.25 6 1,5 2 3 7 1,2， 3,4,5,6 6 1.1666 ... 8 1,3,5,7 4 2 9 1,2,4,5,7,8 6 1.5 10 1,3,7,9 4 2.5
+<div style='margin-left: 4em;'>
-可以看出，对于n≤10，n = 6产生最大n /φ（n）。找到n≤1,000,000的值，其中n /φ（n）是最大值。
+| !!crwdBlockTags_15_sgaTkcolBdwrc!! | Relatively Prime | φ(!!crwdBlockTags_16_sgaTkcolBdwrc!!) |!!crwdBlockTags_17_sgaTkcolBdwrc!!/φ(!!crwdBlockTags_18_sgaTkcolBdwrc!!) |
+| ------------ | ---------------- | --------------- | ---------------------------- |
+| 2 | 1 | 1 | 2 |
+| 3 | 1,2 | 2 | 1.5 |
+| 4 | 1,3 | 2 | 2 |
+| 5 | 1,2,3,4 | 4 | 1.25 |
+| 6 | 1,5 | 2 | 3 |
+| 7 | 1,2,3,4,5,6 | 6 | 1.1666... |
+| 8 | 1,3,5,7 | 4 | 2 |
+| 9 | 1,2,4,5,7,8 | 6 | 1.5 |
+| 10 | 1,3,7,9 | 4 | 2.5 |
+
+</div>
+
+It can be seen that `n`=6 produces a maximum `n`/φ(`n`) for `n` ≤ 10.
+
+Find the value of `n` ≤ 1,000,000 for which n/φ(`n`) is a maximum.
# --hints--
-`euler69()`应该返回510510。
+`totientMaximum()` should return a number.
```js
-assert.strictEqual(euler69(), 510510);
+assert(typeof totientMaximum() === 'number');
+```
+
+`totientMaximum()` should return 510510.
+
+```js
+assert.strictEqual(totientMaximum(), 510510);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-7-10001st-prime.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-7-10001st-prime.md
index 4031e9f31d9..18b18b2bf1a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-7-10001st-prime.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-7-10001st-prime.md
@@ -1,42 +1,50 @@
---
id: 5900f3731000cf542c50fe86
-title: 问题7：10001个素数
+title: 'Problem 7: 10001st prime'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302182
dashedName: problem-7-10001st-prime
---
# --description--
-通过列出前六个素数：2,3,5,7,11和13，我们可以看到第6个素数是13.第`n`个素数是多少？
+By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
+
+What is the `n`th prime number?
# --hints--
-`nthPrime(6)`应该返回13。
+`nthPrime(6)` should return a number.
+
+```js
+assert(typeof nthPrime(6) === 'number');
+```
+
+`nthPrime(6)` should return 13.
```js
assert.strictEqual(nthPrime(6), 13);
```
-`nthPrime(10)`应该返回29。
+`nthPrime(10)` should return 29.
```js
assert.strictEqual(nthPrime(10), 29);
```
-`nthPrime(100)`应该返回541。
+`nthPrime(100)` should return 541.
```js
assert.strictEqual(nthPrime(100), 541);
```
-`nthPrime(1000)`应该返回7919。
+`nthPrime(1000)` should return 7919.
```js
assert.strictEqual(nthPrime(1000), 7919);
```
-`nthPrime(10001)`应该返回104743。
+`nthPrime(10001)` should return 104743.
```js
assert.strictEqual(nthPrime(10001), 104743);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-70-totient-permutation.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-70-totient-permutation.md
index db6a20edb6b..5f3d2bb3619 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-70-totient-permutation.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-70-totient-permutation.md
@@ -1,21 +1,31 @@
---
id: 5900f3b21000cf542c50fec5
-title: 问题70欧拉置换
+title: 'Problem 70: Totient permutation'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302183
dashedName: problem-70-totient-permutation
---
# --description--
-Euler的Totient函数φ（n）\[有时称为phi函数]用于确定小于或等于n的正数的数量，它们是n的相对素数。例如，如1,2,4,5,7和8都小于9且相对素数为9，φ（9）= 6.数字1被认为是每个正数的相对素数，所以φ（1）= 1。有趣的是，φ（87109）= 79180，可以看出87109是79180的置换。找到n的值，1 <n <107，其中φ（n）是n的置换和比率n / φ（n）产生最小值。
+Euler's Totient function, φ(`n`) \[sometimes called the phi function], is used to determine the number of positive numbers less than or equal to `n` which are relatively prime to `n`. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6. The number 1 is considered to be relatively prime to every positive number, so φ(1)=1.
+
+Interestingly, φ(87109)=79180, and it can be seen that 87109 is a permutation of 79180.
+
+Find the value of `n`, 1 < `n` < 10<sup>7</sup>, for which φ(`n`) is a permutation of `n` and the ratio `n`/φ(`n`) produces a minimum.
# --hints--
-`euler70()`应该返回8319823。
+`totientPermutation()` should return a number.
```js
-assert.strictEqual(euler70(), 8319823);
+assert(typeof totientPermutation() === 'number');
+```
+
+`totientPermutation()` should return 8319823.
+
+```js
+assert.strictEqual(totientPermutation(), 8319823);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-71-ordered-fractions.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-71-ordered-fractions.md
index 4ab4cf19e9d..3e0d91f5cf1 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-71-ordered-fractions.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-71-ordered-fractions.md
@@ -1,6 +1,6 @@
---
id: 5900f3b31000cf542c50fec6
-title: 关卡 71：有序分数
+title: 'Problem 71: Ordered fractions'
challengeType: 5
forumTopicId: 302184
dashedName: problem-71-ordered-fractions
@@ -8,25 +8,25 @@ dashedName: problem-71-ordered-fractions
# --description--
-考虑形如 `n`/`d` 的分数，其中 `n` 和 `d` 均为正整数。如果 `n`<`d`，且最大公约数 HCF(`n`,`d`)=1，则该分数被称为最简真分数。
+Consider the fraction, `n`/`d`, where `n` and `d` are positive integers. If `n`<`d` and HCF(`n`,`d`)=1, it is called a reduced proper fraction.
-如果我们将 `d` ≤ 8 的最简真分数构成的集合按大小升序列出，将得到：
+If we list the set of reduced proper fractions for `d` ≤ 8 in ascending order of size, we get:
<div style='text-align: center;'>1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, <strong>2/5</strong>, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8</div>
-可以看出 2/5 是 3/7 的直接左邻的分数。
+It can be seen that 2/5 is the fraction immediately to the left of 3/7.
-将所有 `d` ≤ 1,000,000 的最简真分数按大小升序排列，求此时 3/7 直接左邻的分数的分子。
+By listing the set of reduced proper fractions for `d` ≤ 1,000,000 in ascending order of size, find the numerator of the fraction immediately to the left of 3/7.
# --hints--
-`orderedFractions()` 应该返回一个数字。
+`orderedFractions()` should return a number.
```js
assert(typeof orderedFractions() === 'number');
```
-`orderedFractions()` 应该返回 428570。
+`orderedFractions()` should return 428570.
```js
assert.strictEqual(orderedFractions(), 428570);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-72-counting-fractions.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-72-counting-fractions.md
index 498071169f2..ac66afa5a42 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-72-counting-fractions.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-72-counting-fractions.md
@@ -1,6 +1,6 @@
---
id: 5900f3b41000cf542c50fec7
-title: 关卡 72：分数计数
+title: 'Problem 72: Counting fractions'
challengeType: 5
forumTopicId: 302185
dashedName: problem-72-counting-fractions
@@ -8,25 +8,25 @@ dashedName: problem-72-counting-fractions
# --description--
-考虑形如 `n`/`d` 的分数，其中 n 和 d 均为正整数。如果 `n`<`d`，且最大公约数 HCF(`n`,`d`)=1，则该分数被称为最简真分数。
+Consider the fraction, `n`/`d`, where n and d are positive integers. If `n`<`d` and HCF(`n`,`d`)=1, it is called a reduced proper fraction.
-如果我们将 `d` ≤ 8 的最简真分数构成的集合按大小升序列出，将得到：
+If we list the set of reduced proper fractions for `d` ≤ 8 in ascending order of size, we get:
<div style='text-align: center;'>1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8</div>
-可以看出该集合中共有 21 个元素。
+It can be seen that there are 21 elements in this set.
-求 `d` ≤ 1,000,000 的最简真分数构成的集合中共有多少个元素。
+How many elements would be contained in the set of reduced proper fractions for `d` ≤ 1,000,000?
# --hints--
-`countingFractions()` 应该返回一个数字。
+`countingFractions()` should return a number.
```js
assert(typeof countingFractions() === 'number');
```
-`countingFractions()` 应该返回 303963552391。
+`countingFractions()` should return 303963552391.
```js
assert.strictEqual(countingFractions(), 303963552391);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-73-counting-fractions-in-a-range.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-73-counting-fractions-in-a-range.md
index be906aafebf..68b3e0d7024 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-73-counting-fractions-in-a-range.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-73-counting-fractions-in-a-range.md
@@ -1,6 +1,6 @@
---
id: 5900f3b61000cf542c50fec8
-title: 关卡 73：区间内的分数个数
+title: 'Problem 73: Counting fractions in a range'
challengeType: 5
forumTopicId: 302186
dashedName: problem-73-counting-fractions-in-a-range
@@ -8,25 +8,25 @@ dashedName: problem-73-counting-fractions-in-a-range
# --description--
-考虑形如 `n`/`d` 的分数，其中 n 和 d 均为正整数。如果 `n`<`d`，且其最大公约数 HCF(`n`,`d`)=1，则该分数被称为最简真分数。
+Consider the fraction, `n`/`d`, where n and d are positive integers. If `n`<`d` and HCF(`n`,`d`)=1, it is called a reduced proper fraction.
-如果我们将 `d` ≤ 8 的最简真分数构成的集合按大小升序列出，将得到：
+If we list the set of reduced proper fractions for `d` ≤ 8 in ascending order of size, we get:
<div style='text-align: center;'>1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, <strong>3/8</strong>, <strong>2/5</strong>, <strong>3/7</strong>, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8</div>
-可以看出在 1/3 和 1/2 之间有3个分数。
+It can be seen that there are 3 fractions between 1/3 and 1/2.
-将 `d` ≤ 12,000 的最简真分数构成的集合排序后，在 1/3 和 1/2 之间有多少个分数？
+How many fractions lie between 1/3 and 1/2 in the sorted set of reduced proper fractions for `d` ≤ 12,000?
# --hints--
-`countingFractionsInARange()` 应该返回一个数字。
+`countingFractionsInARange()` should return a number.
```js
assert(typeof countingFractionsInARange() === 'number');
```
-`countingFractionsInARange()` 应该返回 7295372。
+`countingFractionsInARange()` should return 7295372.
```js
assert.strictEqual(countingFractionsInARange(), 7295372);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-74-digit-factorial-chains.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-74-digit-factorial-chains.md
index 56f85825074..c1eeeda4d75 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-74-digit-factorial-chains.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-74-digit-factorial-chains.md
@@ -1,21 +1,48 @@
---
id: 5900f3b61000cf542c50fec9
-title: 问题74：数字因子链
+title: 'Problem 74: Digit factorial chains'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302187
dashedName: problem-74-digit-factorial-chains
---
# --description--
-数字145是众所周知的，其数字的阶乘之和等于145：1！ + 4！ + 5！ = 1 + 24 + 120 = 145也许知名度较低的是169，因为它产生了最长的数字链，可以链接到169;事实证明，只存在三个这样的循环：169→363601→1454→169 871→45361→871 872→45362→872不难证明每个起始编号最终都会陷入循环。例如，69→363600→1454→169→363601（→1454）78→45360→871→45361（→871）540→145（→145）从69开始产生五个非重复项链，但最长起始数低于一百万的非重复链是60个项。起始数低于一百万的链中有多少个包含正好六十个非重复项？
+The number 145 is well known for the property that the sum of the factorial of its digits is equal to 145:
+
+<div style='margin-left: 4em;'>1! + 4! + 5! = 1 + 24 + 120 = 145</div>
+
+Perhaps less well known is 169, in that it produces the longest chain of numbers that link back to 169; it turns out that there are only three such loops that exist:
+
+<div style='margin-left: 4em;'>
+ 169 → 363601 → 1454 → 169<br>
+ 871 → 45361 → 871<br>
+ 872 → 45362 → 872<br>
+</div>
+It is not difficult to prove that EVERY starting number will eventually get stuck in a loop. For example,
+
+<div style='margin-left: 4em;'>
+ 69 → 363600 → 1454 → 169 → 363601 (→ 1454)<br>
+ 78 → 45360 → 871 → 45361 (→ 871)<br>
+ 540 → 145 (→ 145)<br>
+</div>
+
+Starting with 69 produces a chain of five non-repeating terms, but the longest non-repeating chain with a starting number below one million is sixty terms.
+
+How many chains, with a starting number below one million, contain exactly sixty non-repeating terms?
# --hints--
-`euler74()`应返回402。
+`digitFactorialChains()` should return a number.
```js
-assert.strictEqual(euler74(), 402);
+assert(typeof digitFactorialChains() === 'number');
+```
+
+`digitFactorialChains()` should return 402.
+
+```js
+assert.strictEqual(digitFactorialChains(), 402);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-75-singular-integer-right-triangles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-75-singular-integer-right-triangles.md
index 6b133dcc5d9..4a71f87c728 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-75-singular-integer-right-triangles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-75-singular-integer-right-triangles.md
@@ -1,21 +1,44 @@
---
id: 5900f3b71000cf542c50feca
-title: 问题75：奇异整数直角三角形
+title: 'Problem 75: Singular integer right triangles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302188
dashedName: problem-75-singular-integer-right-triangles
---
# --description--
-事实证明，12厘米是最小的导线长度，可以弯曲形成一个完整的直角三角形，只有一种方式，但还有更多的例子。 12 cm：（3,4,5）24 cm：（6,8,10）30 cm：（5,12,13）36 cm：（9,12,15）40 cm：（8,15,17） 48厘米：（12,16,20）相比之下，一些长度的线，如20厘米，不能弯曲形成整数个直角三角形，其他长度允许找到多个解决方案;例如，使用120厘米，可以精确地形成三个不同的整数侧直角三角形。 120厘米：（30,40,50），（20,48,52），（24,45,51）假设L是线的长度，L≤1,500,000的多少个数值恰好可以是一个整数的右边角三角形成？
+It turns out that 12 cm is the smallest length of wire that can be bent to form an integer sided right angle triangle in exactly one way, but there are many more examples.
+
+<div style='margin-left: 4em;'>
+ <strong>12 cm:</strong> (3,4,5)<br>
+ <strong>24 cm:</strong> (6,8,10)<br>
+ <strong>30 cm:</strong> (5,12,13)<br>
+ <strong>36 cm:</strong> (9,12,15)<br>
+ <strong>40 cm:</strong> (8,15,17)<br>
+ <strong>48 cm:</strong> (12,16,20)<br>
+</div>
+
+In contrast, some lengths of wire, like 20 cm, cannot be bent to form an integer sided right angle triangle, and other lengths allow more than one solution to be found; for example, using 120 cm it is possible to form exactly three different integer sided right angle triangles.
+
+<div style='margin-left: 4em;'>
+ <strong>120 cm:</strong> (30,40,50), (20,48,52), (24,45,51)
+</div>
+
+Given that L is the length of the wire, for how many values of L ≤ 1,500,000 can exactly one integer sided right angle triangle be formed?
# --hints--
-`euler75()`应返回161667。
+`singularIntRightTriangles()` should return a number.
```js
-assert.strictEqual(euler75(), 161667);
+assert(typeof singularIntRightTriangles() === 'number');
+```
+
+`singularIntRightTriangles()` should return 161667.
+
+```js
+assert.strictEqual(singularIntRightTriangles(), 161667);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-76-counting-summations.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-76-counting-summations.md
index 03cedae2131..dded902e520 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-76-counting-summations.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-76-counting-summations.md
@@ -1,21 +1,38 @@
---
id: 5900f3b81000cf542c50fecb
-title: 问题76：计算总结
+title: 'Problem 76: Counting summations'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302189
dashedName: problem-76-counting-summations
---
# --description--
-可以用六种不同的方式将五个作为总和写成：4 + 1 3 + 2 3 + 1 + 1 2 + 2 + 1 2 + 1 + 1 + 1 1 + 1 + 1 + 1 + 1多少个不同一百种方式可以写成至少两个正整数的总和？
+It is possible to write five as a sum in exactly six different ways:
+
+<div style='margin-left: 4em;'>
+ 4 + 1<br>
+ 3 + 2<br>
+ 3 + 1 + 1<br>
+ 2 + 2 + 1<br>
+ 2 + 1 + 1 + 1<br>
+ 1 + 1 + 1 + 1 + 1<br>
+</div>
+
+How many different ways can one hundred be written as a sum of at least two positive integers?
# --hints--
-`euler76()`应该返回190569291。
+`countingSummations()` should return a number.
```js
-assert.strictEqual(euler76(), 190569291);
+assert(typeof countingSummations() === 'number');
+```
+
+`countingSummations()` should return 190569291.
+
+```js
+assert.strictEqual(countingSummations(), 190569291);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-77-prime-summations.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-77-prime-summations.md
index 2b0ea054762..fade056d037 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-77-prime-summations.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-77-prime-summations.md
@@ -1,21 +1,37 @@
---
id: 5900f3b91000cf542c50fecc
-title: 问题77：总结
+title: 'Problem 77: Prime summations'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302190
dashedName: problem-77-prime-summations
---
# --description--
-可以用五种不同的方式将十个作为素数之和：7 + 3 5 + 5 5 + 3 + 2 3 + 3 + 2 + 2 2 + 2 + 2 + 2 + 2第一个值是多少可以用五千多种不同的方式写成素数之和？
+It is possible to write ten as the sum of primes in exactly five different ways:
+
+<div style='margin-left: 4em;'>
+ 7 + 3<br>
+ 5 + 5<br>
+ 5 + 3 + 2<br>
+ 3 + 3 + 2 + 2<br>
+ 2 + 2 + 2 + 2 + 2<br>
+</div>
+
+What is the first value which can be written as the sum of primes in over five thousand different ways?
# --hints--
-`euler77()`应该返回71。
+`primeSummations()` should return a number.
```js
-assert.strictEqual(euler77(), 71);
+assert(typeof primeSummations() === 'number');
+```
+
+`primeSummations()` should return 71.
+
+```js
+assert.strictEqual(primeSummations(), 71);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-78-coin-partitions.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-78-coin-partitions.md
index 20dc5eb8484..c5d96870e5a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-78-coin-partitions.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-78-coin-partitions.md
@@ -1,25 +1,43 @@
---
id: 5900f3ba1000cf542c50fecd
-title: 问题78：硬币分区
+title: 'Problem 78: Coin partitions'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302191
dashedName: problem-78-coin-partitions
---
# --description--
-设p（n）表示n个硬币可以分成桩的不同方式的数量。例如，五个硬币可以用七种不同的方式分成几堆，所以p（5）= 7。
+Let p(n) represent the number of different ways in which n coins can be separated into piles. For example, five coins can be separated into piles in exactly seven different ways, so p(5)=7.
-OOOOO OOOO O OOO OO OOO O O OO OO O O O O O O O O O O O O O O O O O O O O O O O O O
+<div style='text-align: center;'>
-找到n的最小值，p（n）可以被一百万整除。
+| Coin piles |
+| ----------------- |
+| OOOOO |
+| OOOO O |
+| OOO OO |
+| OOO O O |
+| OO OO O |
+| OO O O O |
+| O O O O O |
+
+</div>
+
+Find the least value of `n` for which p(`n`) is divisible by one million.
# --hints--
-`euler78()`应返回55374。
+`coinPartitions()` should return a number.
```js
-assert.strictEqual(euler78(), 55374);
+assert(typeof coinPartitions() === 'number');
+```
+
+`coinPartitions()` should return 55374.
+
+```js
+assert.strictEqual(coinPartitions(), 55374);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-79-passcode-derivation.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-79-passcode-derivation.md
index 8d09da70f85..a88a40e6aed 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-79-passcode-derivation.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-79-passcode-derivation.md
@@ -1,21 +1,31 @@
---
id: 5900f3bb1000cf542c50fece
-title: 问题79：密码派生
+title: 'Problem 79: Passcode derivation'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302192
dashedName: problem-79-passcode-derivation
---
# --description--
-用于在线银行业务的常见安全方法是向用户询问密码中的三个随机字符。例如，如果密码是531278，他们可能会要求输入第2，第3和第5个字符;预期的回复是：317。文本文件keylog.txt包含50次成功的登录尝试。鉴于总是按顺序询问三个字符，分析该文件以确定未知长度的最短可能秘密密码。
+A common security method used for online banking is to ask the user for three random characters from a passcode. For example, if the passcode was 531278, they may ask for the 2nd, 3rd, and 5th characters; the expected reply would be: 317.
+
+The array, `keylog`, contains fifty successful login attempts.
+
+Given that the three characters are always asked for in order, analyze the array so as to determine the shortest possible secret passcode of unknown length.
# --hints--
-`euler79()`应该返回73162890。
+`passcodeDerivation(keylog)` should return a number.
```js
-assert.strictEqual(euler79(), 73162890);
+assert(typeof passcodeDerivation(keylog) === 'number');
+```
+
+`passcodeDerivation(keylog)` should return 73162890.
+
+```js
+assert.strictEqual(passcodeDerivation(keylog), 73162890);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-8-largest-product-in-a-series.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-8-largest-product-in-a-series.md
index dd1b8c4dc66..0fd4703af75 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-8-largest-product-in-a-series.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-8-largest-product-in-a-series.md
@@ -1,66 +1,53 @@
---
id: 5900f3741000cf542c50fe87
-title: 问题8：系列中最大的产品
+title: 'Problem 8: Largest product in a series'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302193
dashedName: problem-8-largest-product-in-a-series
---
# --description--
-具有最大乘积的1000位数字中的四个相邻数字是9×9×8×9 = 5832。
+The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.
-73167176531330624919225119674426574742355349194934
+<div style='text-align: center;'>73167176531330624919225119674426574742355349194934</div>
+<div style='text-align: center;'>96983520312774506326239578318016984801869478851843</div>
+<div style='text-align: center;'>85861560789112949495459501737958331952853208805511</div>
+<div style='text-align: center;'>12540698747158523863050715693290963295227443043557</div>
+<div style='text-align: center;'>66896648950445244523161731856403098711121722383113</div>
+<div style='text-align: center;'>62229893423380308135336276614282806444486645238749</div>
+<div style='text-align: center;'>30358907296290491560440772390713810515859307960866</div>
+<div style='text-align: center;'>70172427121883998797908792274921901699720888093776</div>
+<div style='text-align: center;'>65727333001053367881220235421809751254540594752243</div>
+<div style='text-align: center;'>52584907711670556013604839586446706324415722155397</div>
+<div style='text-align: center;'>53697817977846174064955149290862569321978468622482</div>
+<div style='text-align: center;'>83972241375657056057490261407972968652414535100474</div>
+<div style='text-align: center;'>82166370484403199890008895243450658541227588666881</div>
+<div style='text-align: center;'>16427171479924442928230863465674813919123162824586</div>
+<div style='text-align: center;'>17866458359124566529476545682848912883142607690042</div>
+<div style='text-align: center;'>24219022671055626321111109370544217506941658960408</div>
+<div style='text-align: center;'>07198403850962455444362981230987879927244284909188</div>
+<div style='text-align: center;'>84580156166097919133875499200524063689912560717606</div>
+<div style='text-align: center;'>05886116467109405077541002256983155200055935729725</div>
+<div style='text-align: center;'>71636269561882670428252483600823257530420752963450</div>
-96983520312774506326239578318016984801869478851843
-
-85861560789112949495459501737958331952853208805511
-
-12540698747158523863050715693290963295227443043557
-
-66896648950445244523161731856403098711121722383113
-
-62229893423380308135336276614282806444486645238749
-
-30358907296290491560440772390713810515859307960866
-
-70172427121883998797908792274921901699720888093776
-
-65727333001053367881220235421809751254540594752243
-
-52584907711670556013604839586446706324415722155397
-
-53697817977846174064955149290862569321978468622482
-
-83972241375657056057490261407972968652414535100474
-
-82166370484403199890008895243450658541227588666881
-
-16427171479924442928230863465674813919123162824586
-
-17866458359124566529476545682848912883142607690042
-
-24219022671055626321111109370544217506941658960408
-
-07198403850962455444362981230987879927244284909188
-
-84580156166097919133875499200524063689912560717606
-
-05886116467109405077541002256983155200055935729725
-
-71636269561882670428252483600823257530420752963450
-
-找到具有最大乘积的1000位数字中的`n`相邻数字。这个产品有什么价值？
+Find the `n` adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?
# --hints--
-`largestProductinaSeries(4)`应该返回5832。
+`largestProductinaSeries(4)` should return a number.
+
+```js
+assert(typeof largestProductinaSeries(4) === 'number');
+```
+
+`largestProductinaSeries(4)` should return 5832.
```js
assert.strictEqual(largestProductinaSeries(4), 5832);
```
-`largestProductinaSeries(13)`应返回23514624000。
+`largestProductinaSeries(13)` should return 23514624000.
```js
assert.strictEqual(largestProductinaSeries(13), 23514624000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-80-square-root-digital-expansion.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-80-square-root-digital-expansion.md
index 37e6ce66d06..2f8be71b438 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-80-square-root-digital-expansion.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-80-square-root-digital-expansion.md
@@ -1,21 +1,31 @@
---
id: 5900f3bc1000cf542c50fecf
-title: 问题80：平方根数字扩展
+title: 'Problem 80: Square root digital expansion'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302194
dashedName: problem-80-square-root-digital-expansion
---
# --description--
-众所周知，如果自然数的平方根不是整数，那么它是不合理的。这种平方根的十进制扩展是无限的，没有任何重复模式。二的平方根是1.41421356237309504880 ...，前一百个十进制数字的数字和是475.对于前一百个自然数，找到所有的前一百个十进制数字的数字总和的总和。不合理的平方根。
+It is well known that if the square root of a natural number is not an integer, then it is irrational. The decimal expansion of such square roots is infinite without any repeating pattern at all.
+
+The square root of two is 1.41421356237309504880..., and the digital sum of the first one hundred decimal digits is 475.
+
+For the first one hundred natural numbers, find the total of the digital sums of the first one hundred decimal digits for all the irrational square roots.
# --hints--
-`euler80()`应返回40886。
+`sqrtDigitalExpansion()` should return a number.
```js
-assert.strictEqual(euler80(), 40886);
+assert(typeof sqrtDigitalExpansion() === 'number');
+```
+
+`sqrtDigitalExpansion()` should return 40886.
+
+```js
+assert.strictEqual(sqrtDigitalExpansion(), 40886);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-81-path-sum-two-ways.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-81-path-sum-two-ways.md
index 31d8fe78eca..630903b7ccb 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-81-path-sum-two-ways.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-81-path-sum-two-ways.md
@@ -1,25 +1,45 @@
---
id: 5900f3bd1000cf542c50fed0
-title: 问题81：路径总和：两种方式
+title: 'Problem 81: Path sum: two ways'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302195
dashedName: problem-81-path-sum-two-ways
---
# --description--
-在下面的5乘5矩阵中，从左上角到右下角的最小路径总和，仅向右和向下移动，以粗体红色表示，等于2427。
+In the 5 by 5 matrix below, the minimal path sum from the top left to the bottom right, by **only moving to the right and down**, is indicated in bold red and is equal to 2427.
-$$ \\ begin {pmatrix} \\ color {red} {131}＆673＆234＆103＆18 \\ \\ color {red} {201}＆\\ color {red} {96}＆\\ color {red} {342} ＆965＆150 \\ 630＆803＆\\ color {red} {746}＆\\ color {red} {422}＆111 \\ 537＆699＆497＆\\ color {red} {121}＆956 \\ 805＆732＆ 524＆\\ color {red} {37}＆\\ color {red} {331} \\ end {pmatrix} $$
+<div style='text-align: center;'>
+ $\begin{pmatrix}
+ \color{red}{131} & 673 & 234 & 103 & 18\\\\
+ \color{red}{201} & \color{red}{96} & \color{red}{342} & 965 & 150\\\\
+ 630 & 803 & \color{red}{746} & \color{red}{422} & 111\\\\
+ 537 & 699 & 497 & \color{red}{121} & 956\\\\
+ 805 & 732 & 524 & \color{red}{37} & \color{red}{331}
+ \end{pmatrix}$
+</div>
-在matrix.txt中找到最小路径总和（右键单击并“保存链接/目标为...”），一个包含80 x 80矩阵的31K文本文件，从左上角到右下角仅向右移动下。
+Find the minimal path sum from the top left to the bottom right by only moving right and down in `matrix`, a 2D array containing an 80 by 80 matrix.
# --hints--
-`euler81()`应该返回427337。
+`pathSumTwoWays(testMatrix)` should return a number.
```js
-assert.strictEqual(euler81(), 427337);
+assert(typeof pathSumTwoWays(testMatrix) === 'number');
+```
+
+`pathSumTwoWays(testMatrix)` should return 2427.
+
+```js
+assert.strictEqual(pathSumTwoWays(testMatrix), 2427);
+```
+
+`pathSumTwoWays(matrix)` should return 427337.
+
+```js
+assert.strictEqual(pathSumTwoWays(matrix), 427337);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-82-path-sum-three-ways.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-82-path-sum-three-ways.md
index fe0f9184f0e..f2947d34a64 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-82-path-sum-three-ways.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-82-path-sum-three-ways.md
@@ -1,25 +1,47 @@
---
id: 5900f3be1000cf542c50fed1
-title: 问题82：路径总和：三种方式
+title: 'Problem 82: Path sum: three ways'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302196
dashedName: problem-82-path-sum-three-ways
---
# --description--
-注意：这个问题是问题81的一个更具挑战性的版本。在下面的5乘5矩阵中的最小路径总和，从左列中的任何单元格开始并在右列中的任何单元格中完成，并且仅向上，向下移动，右边用红色和粗体表示;总和等于994。
+**Note:** This problem is a more challenging version of Problem 81.
-$$ \\ begin {pmatrix} 131＆673＆\\ color {red} {234}＆\\ color {red} {103}＆\\ color {red} {18} \\ \\ color {red} {201}＆\\ color {红色} {96}＆\\ color {red} {342}＆965＆150 \\ 630＆803＆746＆422＆111 \\ 537＆699＆497＆121＆956 \\ 805＆732＆524＆37＆331 \\ end {pmatrix} $$
+The minimal path sum in the 5 by 5 matrix below, by starting in any cell in the left column and finishing in any cell in the right column, and only moving up, down, and right, is indicated in red and bold; the sum is equal to 994.
-在matrix.txt（右键单击和“Save Link / Target As ...”）中查找最小路径总和，这是一个包含80 x 80矩阵的31K文本文件，从左列到右列。
+<div style='text-align: center;'>
+ $\begin{pmatrix}
+ 131 & 673 & \color{red}{234} & \color{red}{103} & \color{red}{18}\\\\
+ \color{red}{201} & \color{red}{96} & \color{red}{342} & 965 & 150\\\\
+ 630 & 803 & 746 & 422 & 111\\\\
+ 537 & 699 & 497 & 121 & 956\\\\
+ 805 & 732 & 524 & 37 & 331
+ \end{pmatrix}$
+</div>
+
+Find the minimal path sum from the left column to the right column in `matrix`, a 2D array containing an 80 by 80 matrix.
# --hints--
-`euler82()`应返回260324。
+`pathSumThreeWays(testMatrix)` should return a number.
```js
-assert.strictEqual(euler82(), 260324);
+assert(typeof pathSumThreeWays(testMatrix) === 'number');
+```
+
+`pathSumThreeWays(testMatrix)` should return 994.
+
+```js
+assert.strictEqual(pathSumThreeWays(testMatrix), 994);
+```
+
+`pathSumThreeWays(matrix)` should return 260324.
+
+```js
+assert.strictEqual(pathSumThreeWays(matrix), 260324);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-83-path-sum-four-ways.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-83-path-sum-four-ways.md
index f6f66386a59..40a008132a1 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-83-path-sum-four-ways.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-83-path-sum-four-ways.md
@@ -1,25 +1,47 @@
---
id: 5900f3bf1000cf542c50fed2
-title: 问题83：路径总和：四种方式
+title: 'Problem 83: Path sum: four ways'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302197
dashedName: problem-83-path-sum-four-ways
---
# --description--
-注意：此问题是问题81的一个更具挑战性的版本。在下面的5乘5矩阵中，通过向左，向右，向上和向下移动，从左上角到右下角的最小路径总和以粗体表示红色等于2297。
+**Note:** This problem is a significantly more challenging version of Problem 81.
-$$ \\ begin {pmatrix} \\ color {red} {131}＆673＆\\ color {red} {234}＆\\ color {red} {103}＆\\ color {red} {18} \\ \\ color {red} {201}＆\\ color {red} {96}＆\\ color {red} {342}＆965＆\\ color {red} {150} \\ 630＆803＆746＆\\ color {red} {422}＆\\ color {red} {111} \\ 537＆699＆497＆\\ color {red} {121}＆956 \\ 805＆732＆524＆\\ color {red} {37}＆\\ color {red} {331} \\ end { pmatrix} $$
+In the 5 by 5 matrix below, the minimal path sum from the top left to the bottom right, by moving left, right, up, and down, is indicated in bold red and is equal to 2297.
-在matrix.txt中找到最小路径总和（右键单击并“保存链接/目标为...”），一个包含80 x 80矩阵的31K文本文件，通过向左，向右移动从左上角到右下角， 上和下。
+<div style='text-align: center;'>
+ $\begin{pmatrix}
+ \color{red}{131} & 673 & \color{red}{234} & \color{red}{103} & \color{red}{18}\\\\
+ \color{red}{201} & \color{red}{96} & \color{red}{342} & 965 & \color{red}{150}\\\\
+ 630 & 803 & 746 & \color{red}{422} & \color{red}{111}\\\\
+ 537 & 699 & 497 & \color{red}{121} & 956\\\\
+ 805 & 732 & 524 & \color{red}{37} & \color{red}{331}
+ \end{pmatrix}$
+</div>
+
+Find the minimal path sum from the top left to the bottom right by moving left, right, up, and down in `matrix`, a 2D array defined in the background, containing an 80 by 80 matrix.
# --hints--
-`euler83()`应返回425185。
+`pathSumFourWays(testMatrix)` should return a number.
```js
-assert.strictEqual(euler83(), 425185);
+assert(typeof pathSumFourWays(testMatrix) === 'number');
+```
+
+`pathSumFourWays(testMatrix)` should return 2297.
+
+```js
+assert.strictEqual(pathSumFourWays(testMatrix), 2297);
+```
+
+`pathSumFourWays(matrix)` should return 425185.
+
+```js
+assert.strictEqual(pathSumFourWays(matrix), 425185);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-84-monopoly-odds.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-84-monopoly-odds.md
index 3c87cc08ae8..f9429edeafa 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-84-monopoly-odds.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-84-monopoly-odds.md
@@ -1,47 +1,141 @@
---
id: 5900f3c11000cf542c50fed3
-title: 问题84：垄断赔率
+title: 'Problem 84: Monopoly odds'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302198
dashedName: problem-84-monopoly-odds
---
# --description--
-在游戏“大富翁”中，标准板的设置方式如下：
+In the game, *Monopoly*, the standard board is set up in the following way:
-GO A1 CC1 A2 T1 R1 B1 CH1 B2 B3 JAIL H2
+<div style="text-align: center;">
+ <table cellspacing="1" cellpadding="5" border="0" style="background-color: black; color: black;" align="center">
+ <tbody>
+ <tr>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">GO</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">A1</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">CC1</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">A2</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">T1</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">R1</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">B1</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">CH1</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">B2</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">B3</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">JAIL</td>
+ </tr>
+ <tr>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">H2</td>
+ <td colspan="9"> </td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">C1</td>
+ </tr>
+ <tr>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">T2</td>
+ <td colspan="9"> </td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">U1</td>
+ </tr>
+ <tr>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">H1</td>
+ <td colspan="9"> </td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">C2</td>
+ </tr>
+ <tr>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">CH3</td>
+ <td colspan="9"> </td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">C3</td>
+ </tr>
+ <tr>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">R4</td>
+ <td colspan="9"> </td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">R2</td>
+ </tr>
+ <tr>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">G3</td>
+ <td colspan="9"> </td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">D1</td>
+ </tr>
+ <tr>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">CC3</td>
+ <td colspan="9"> </td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">CC2</td>
+ </tr>
+ <tr>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">G2</td>
+ <td colspan="9"> </td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">D2</td>
+ </tr>
+ <tr>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">G1</td>
+ <td colspan="9"> </td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">D3</td>
+ </tr>
+ <tr>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">G2J</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">F3</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">U2</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">F2</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">F1</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">R3</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">E3</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">E2</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">CH2</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">E1</td>
+ <td style="background-color: #ffffff; color: black; padding: 5px; border: 1px solid black;">FP</td>
+ </tr>
+ </tbody>
+ </table>
+</div>
-C1 T2
+A player starts on the GO square and adds the scores on two 6-sided dice to determine the number of squares they advance in a clockwise direction. Without any further rules we would expect to visit each square with equal probability: 2.5%. However, landing on G2J (Go To Jail), CC (community chest), and CH (chance) changes this distribution.
-U1 H1
+In addition to G2J, and one card from each of CC and CH, that orders the player to go directly to jail, if a player rolls three consecutive doubles, they do not advance the result of their 3rd roll. Instead they proceed directly to jail.
-C2 CH3
+At the beginning of the game, the CC and CH cards are shuffled. When a player lands on CC or CH they take a card from the top of the respective pile and, after following the instructions, it is returned to the bottom of the pile. There are sixteen cards in each pile, but for the purpose of this problem we are only concerned with cards that order a movement; any instruction not concerned with movement will be ignored and the player will remain on the CC/CH square.
-C3 R4
+<ul>
+ <li>Community Chest (2/16 cards):</li>
+ <ol>
+ <li>Advance to GO</li>
+ <li>Go to JAIL</li>
+ </ol>
-R2 G3
+ <li>Chance (10/16 cards):</li>
+ <ol>
+ <li>Advance to GO</li>
+ <li>Go to JAIL</li>
+ <li>Go to C1</li>
+ <li>Go to E3</li>
+ <li>Go to H2</li>
+ <li>Go to R1</li>
+ <li>Go to next R (railway company)</li>
+ <li>Go to next R</li>
+ <li>Go to next U (utility company)</li>
+ <li>Go back 3 squares.</li>
+ </ol>
+</ul>
-D1 CC3
+The heart of this problem concerns the likelihood of visiting a particular square. That is, the probability of finishing at that square after a roll. For this reason it should be clear that, with the exception of G2J for which the probability of finishing on it is zero, the CH squares will have the lowest probabilities, as 5/8 request a movement to another square, and it is the final square that the player finishes at on each roll that we are interested in. We shall make no distinction between "Just Visiting" and being sent to JAIL, and we shall also ignore the rule about requiring a double to "get out of jail", assuming that they pay to get out on their next turn.
-CC2 G2
+By starting at GO and numbering the squares sequentially from 00 to 39 we can concatenate these two-digit numbers to produce strings that correspond with sets of squares.
-D2 G1
+Statistically it can be shown that the three most popular squares, in order, are JAIL (6.24%) = Square 10, E3 (3.18%) = Square 24, and GO (3.09%) = Square 00. So these three most popular squares can be listed with the six-digit modal string: 102400.
-D3 G2J F3 U2 F2 F1 R3 E3 E2 CH2 E1 FP
-
-玩家在GO广场上开始并在两个6面骰子上添加分数以确定它们以顺时针方向前进的方格数。如果没有任何进一步的规则，我们期望以相同的概率访问每个广场：2.5％。但是，登陆G2J（Go To Jail），CC（社区胸部）和CH（机会）会改变这种分布。除了G2J，还有来自CC和CH的一张牌，命令玩家直接进入监狱，如果玩家连续三次打败，他们不会推进他们的第三次掷骰结果。相反，他们直接进入监狱。在游戏开始时，CC和CH卡被洗牌。当玩家登陆CC或CH时，他们从相应堆的顶部取出一张牌，并按照说明后将其返回到堆的底部。每堆中有16张牌，但出于这个问题的目的，我们只关心命令运动的牌;任何与运动无关的指令都将被忽略，玩家将保留在CC / CH广场上。公益金（2/16卡）：前往GO去JAIL
-
-机会（10/16卡）：前往GO转到JAIL转到C1转到E3转到H2转到R1转到下一个R（铁路公司）转到下一个R转到下一个U（公用事业公司）返回3个方块。
-
-这个问题的核心是访问特定广场的可能性。也就是说，在滚动之后在该正方形处完成的概率。由于这个原因，应该清楚的是，除了完成它的概率为零的G2J之外，CH方块将具有最低概率，因为5/8请求移动到另一个方格，并且它是最终的玩家在我们感兴趣的每一卷上完成的方格。我们不会区分“Just Visiting”和被送到JAIL，我们也会忽略关于要求双重“离开监狱”的规则，假设他们支付下一轮出去的费用。通过从GO开始并按顺序从00到39对方块进行编号，我们可以连接这些两位数字以产生与方块组对应的字符串。统计上可以看出，按顺序排列的三个最受欢迎的正方形是JAIL（6.24％）= Square 10，E3（3.18％）= Square 24和GO（3.09％）= Square 00.所以这三个最受欢迎的正方形可以使用六位数模态字符串列出：102400。如果使用两个4面骰子而不是使用两个6面骰子，则找到六位数模态字符串。
+If, instead of using two 6-sided dice, two 4-sided dice are used, find the six-digit modal string.
# --hints--
-`euler84()`应返回101524。
+`monopolyOdds()` should return a number.
```js
-assert.strictEqual(euler84(), 101524);
+assert(typeof monopolyOdds() === 'number');
+```
+
+`monopolyOdds()` should return 101524.
+
+```js
+assert.strictEqual(monopolyOdds(), 101524);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-85-counting-rectangles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-85-counting-rectangles.md
index 55142112255..50935108e62 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-85-counting-rectangles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-85-counting-rectangles.md
@@ -1,23 +1,31 @@
---
id: 5900f3c11000cf542c50fed4
-title: 问题85：计算矩形
+title: 'Problem 85: Counting rectangles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302199
dashedName: problem-85-counting-rectangles
---
# --description--
-通过仔细计数可以看出，3×2的矩形网格包含十八个矩形：
+By counting carefully it can be seen that a rectangular grid measuring 3 by 2 contains eighteen rectangles:
-虽然不存在包含正好两百万个矩形的矩形网格，但是找到最近解的网格区域。
+<img class="img-responsive center-block" alt="a diagram of the different rectangles found within a 3 by 2 rectangular grid" src="https://cdn-media-1.freecodecamp.org/project-euler/counting-rectangles.png" style="background-color: white; padding: 10px;" />
+
+Although there exists no rectangular grid that contains exactly two million rectangles, find the area of the grid with the nearest solution.
# --hints--
-`euler85()`应该返回2772。
+`countingRectangles()` should return a number.
```js
-assert.strictEqual(euler85(), 2772);
+assert(typeof countingRectangles() === 'number');
+```
+
+`countingRectangles()` should return 2772.
+
+```js
+assert.strictEqual(countingRectangles(), 2772);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-86-cuboid-route.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-86-cuboid-route.md
index c042972dd98..7019da3ff29 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-86-cuboid-route.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-86-cuboid-route.md
@@ -1,23 +1,35 @@
---
id: 5900f3c31000cf542c50fed5
-title: 问题86：长方体路线
+title: 'Problem 86: Cuboid route'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302200
dashedName: problem-86-cuboid-route
---
# --description--
-一只蜘蛛S坐在一个长方形房间的一个角落里，尺寸为6乘5乘3，一只苍蝇F坐在对面的角落里。通过在房间的表面上行进，从S到F的最短“直线”距离是10并且路径在图上示出。
+A spider, S, sits in one corner of a cuboid room, measuring 6 by 5 by 3, and a fly, F, sits in the opposite corner. By travelling on the surfaces of the room the shortest "straight line" distance from S to F is 10 and the path is shown on the diagram.
-但是，对于任何给定的长方体，最多有三个“最短”路径候选，并且最短路径并不总是具有整数长度。可以证明，正好有2060个不同的长方体，忽略旋转，具有整数尺寸，最大尺寸为M×M×M，当M = 100时，最短路径具有整数长度。这是最小值M的解决方案数首先超过两千;当M = 99时的解的数量是1975.找到M的最小值，使得解的数量首先超过一百万。
+<img class="img-responsive center-block" alt="a diagram of a spider and fly's path from one corner of a cuboid room to the opposite corner" src="https://cdn-media-1.freecodecamp.org/project-euler/cuboid-route.png" style="background-color: white; padding: 10px;" />
+
+However, there are up to three "shortest" path candidates for any given cuboid and the shortest route doesn't always have integer length.
+
+It can be shown that there are exactly 2060 distinct cuboids, ignoring rotations, with integer dimensions, up to a maximum size of M by M by M, for which the shortest route has integer length when M = 100. This is the least value of M for which the number of solutions first exceeds two thousand; the number of solutions when M = 99 is 1975.
+
+Find the least value of M such that the number of solutions first exceeds one million.
# --hints--
-`euler86()`应该返回1818年。
+`cuboidRoute()` should return a number.
```js
-assert.strictEqual(euler86(), 1818);
+assert(typeof cuboidRoute() === 'number');
+```
+
+`cuboidRoute()` should return 1818.
+
+```js
+assert.strictEqual(cuboidRoute(), 1818);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-87-prime-power-triples.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-87-prime-power-triples.md
index 7736bce0fa6..fcf84bac7b0 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-87-prime-power-triples.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-87-prime-power-triples.md
@@ -1,21 +1,36 @@
---
id: 5900f3c51000cf542c50fed8
-title: 问题87：Prime power三倍
+title: 'Problem 87: Prime power triples'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302201
dashedName: problem-87-prime-power-triples
---
# --description--
-可表示为素数平方，素数立方和素数四次幂之和的最小数是28.事实上，正好有四个数字低于五十，可以用这样的方式表示：28 = 22 + 23 + 24 33 = 32 + 23 + 24 49 = 52 + 23 + 24 47 = 22 + 33 + 24五千万以下的数字可以表示为素数，素数和素数四次幂之和？
+The smallest number expressible as the sum of a prime square, prime cube, and prime fourth power is 28. In fact, there are exactly four numbers below fifty that can be expressed in such a way:
+
+<div style='margin-left: 4em;'>
+ 28 = 2<sup>2</sup> + 2<sup>3</sup> + 2<sup>4</sup><br>
+ 33 = 3<sup>2</sup> + 2<sup>3</sup> + 2<sup>4</sup><br>
+ 49 = 5<sup>2</sup> + 2<sup>3</sup> + 2<sup>4</sup><br>
+ 47 = 2<sup>2</sup> + 3<sup>3</sup> + 2<sup>4</sup>
+</div>
+
+How many numbers below fifty million can be expressed as the sum of a prime square, prime cube, and prime fourth power?
# --hints--
-`euler87()`应该返回1097343。
+`primePowerTriples()` should return a number.
```js
-assert.strictEqual(euler87(), 1097343);
+assert(typeof primePowerTriples() === 'number');
+```
+
+`primePowerTriples()` should return 1097343.
+
+```js
+assert.strictEqual(primePowerTriples(), 1097343);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-88-product-sum-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-88-product-sum-numbers.md
index 19413b7d27a..96efafdf5d2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-88-product-sum-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-88-product-sum-numbers.md
@@ -1,21 +1,45 @@
---
id: 5900f3c51000cf542c50fed6
-title: 问题88：产品总和数
+title: 'Problem 88: Product-sum numbers'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302203
dashedName: problem-88-product-sum-numbers
---
# --description--
-可以写为至少两个自然数{a1，a2，...，ak}的给定集合的和和乘积的自然数N称为乘积和数：N = a1 + a2 + ... + ak = a1×a2×...×ak。例如，6 = 1 + 2 + 3 = 1×2×3。对于给定的大小集合k，我们将使用此属性调用最小的N作为最小乘积和数。尺寸组k = 2,3,4,5和6的最小乘积和数如下。 k = 2：4 = 2×2 = 2 + 2k = 3：6 = 1×2×3 = 1 + 2 + 3k = 4：8 = 1×1×2×4 = 1 + 1 + 2 + 4k = 5：8 = 1×1×2×2×2 = 1 + 1 + 2 + 2 + 2k = 6：12 = 1×1×1×1×2×6 = 1 + 1 + 1 + 1 + 2 +因此，对于2≤k≤6，所有最小乘积和数之和为4 + 6 + 8 + 12 = 30;请注意，8只计算总和一次。实际上，由于2≤k≤12的完整最小乘积和数是{4,6,8,12,15,16}，因此总和为61.所有最小乘积和的总和是多少数字为2≤k≤12000？
+A natural number, N, that can be written as the sum and product of a given set of at least two natural numbers, {`a`<sub>1</sub>, `a`<sub>2</sub>, ... , `a`<sub>k</sub>} is called a product-sum number: N = `a`<sub>1</sub> + `a`<sub>2</sub> + ... + `a`<sub>k</sub> = `a`<sub>1</sub> × `a`<sub>2</sub> × ... × `a`<sub>k</sub>.
+
+For example, 6 = 1 + 2 + 3 = 1 × 2 × 3.
+
+For a given set of size, `k`, we shall call the smallest N with this property a minimal product-sum number. The minimal product-sum numbers for sets of size, `k` = 2, 3, 4, 5, and 6 are as follows.
+
+<div style='margin-left: 4em;'>
+ <var>k</var>=2: 4 = 2 × 2 = 2 + 2<br>
+ <var>k</var>=3: 6 = 1 × 2 × 3 = 1 + 2 + 3<br>
+ <var>k</var>=4: 8 = 1 × 1 × 2 × 4 = 1 + 1 + 2 + 4<br>
+ <var>k</var>=5: 8 = 1 × 1 × 2 × 2 × 2 = 1 + 1 + 2 + 2 + 2<br>
+ <var>k</var>=6: 12 = 1 × 1 × 1 × 1 × 2 × 6 = 1 + 1 + 1 + 1 + 2 + 6
+</div>
+
+Hence for 2≤`k`≤6, the sum of all the minimal product-sum numbers is 4+6+8+12 = 30; note that 8 is only counted once in the sum.
+
+In fact, as the complete set of minimal product-sum numbers for 2≤`k`≤12 is {4, 6, 8, 12, 15, 16}, the sum is 61.
+
+What is the sum of all the minimal product-sum numbers for 2≤`k`≤12000?
# --hints--
-`euler88()`应该返回7587457。
+`productSumNumbers()` should return a number.
```js
-assert.strictEqual(euler88(), 7587457);
+assert(typeof productSumNumbers() === 'number');
+```
+
+`productSumNumbers()` should return 7587457.
+
+```js
+assert.strictEqual(productSumNumbers(), 7587457);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-89-roman-numerals.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-89-roman-numerals.md
index ea452052a0e..259dea83cd4 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-89-roman-numerals.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-89-roman-numerals.md
@@ -1,21 +1,46 @@
---
id: 5900f3c51000cf542c50fed7
-title: 问题89：罗马数字
+title: 'Problem 89: Roman numerals'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302204
dashedName: problem-89-roman-numerals
---
# --description--
-对于以罗马数字书写的数字被认为有效，必须遵循基本规则。尽管规则允许以不止一种方式表达某些数字，但始终存在编写特定数字的“最佳”方式。例如，似乎至少有六种方式编写十六号：IIIIIIIIIIIIIII VIIIIIIIIII VVIIIIII XIIIIII VVVI XVI但是，根据规则，只有XIIIIII和XVI是有效的，最后一个例子被认为是最有效的，因为它使用最少数量的数字。 11K文本文件roman.txt（右键单击和'Save Link / Target As ...'）包含一千个用有效但不一定是最小的罗马数字写的数字;请参阅关于...罗马数字，了解此问题的明确规则。通过以最小的形式编写每个字符来查找保存的字符数。注意：您可以假设文件中的所有罗马数字包含不超过四个连续的相同单位。
+For a number written in Roman numerals to be considered valid there are basic rules which must be followed. Even though the rules allow some numbers to be expressed in more than one way there is always a "best" way of writing a particular number.
+
+For example, it would appear that there are at least six ways of writing the number sixteen:
+
+<div style="margin-left: 4em; font-family: 'courier new', monospace;">
+ IIIIIIIIIIIIIIII<br>
+ VIIIIIIIIIII<br>
+ VVIIIIII<br>
+ XIIIIII<br>
+ VVVI<br>
+ XVI<br>
+</div>
+
+However, according to the rules only XIIIIII and XVI are valid, and the last example is considered to be the most efficient, as it uses the least number of numerals.
+
+The array, `roman`, contains one thousand numbers written in valid, but not necessarily minimal, Roman numerals; see [About... Roman Numerals](https://projecteuler.net/about=roman_numerals) for the definitive rules for this problem.
+
+Find the number of characters saved by writing each of these in their minimal form.
+
+**Note:** You can assume that all the Roman numerals in the array contain no more than four consecutive identical units.
# --hints--
-`euler89()`应该返回743。
+`romanNumerals(roman)` should return a number.
```js
-assert.strictEqual(euler89(), 743);
+assert(typeof romanNumerals(roman) === 'number');
+```
+
+`romanNumerals(roman)` should return 743.
+
+```js
+assert.strictEqual(romanNumerals(roman), 743);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-9-special-pythagorean-triplet.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-9-special-pythagorean-triplet.md
index 7425d5f1ee3..3ce9715e401 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-9-special-pythagorean-triplet.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-9-special-pythagorean-triplet.md
@@ -1,61 +1,47 @@
---
id: 5900f3761000cf542c50fe88
-title: 问题9：特殊的毕达哥拉斯三重奏
+title: 'Problem 9: Special Pythagorean triplet'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302205
dashedName: problem-9-special-pythagorean-triplet
---
# --description--
-毕达哥拉斯三元组是一组三个自然数， `a` < `b` < `c` ，其中，
+A Pythagorean triplet is a set of three natural numbers, `a` < `b` < `c`, for which,
-`a`
+<div style='text-align: center;'><var>a</var><sup>2</sup> + <var>b</var><sup>2</sup> = <var>c</var><sup>2</sup></div>
-<sup>2</sup>
+For example, 3<sup>2</sup> + 4<sup>2</sup> = 9 + 16 = 25 = 5<sup>2</sup>.
-- `b`
-
-<sup>2</sup>
-
- = `c`
-
-<sup>2</sup>
-
-例如，3
-
-<sup>2</sup>
-
-- 4
-
-<sup>2</sup>
-
- = 9 + 16 = 25 = 5
-
-<sup>2</sup>
-
- 。恰好存在一个毕达哥拉斯三元组，其中`a` + `b` + `c` = 1000.求产品`abc`使得`a` + `b` + `c` = `n` 。
+There exists exactly one Pythagorean triplet for which `a` + `b` + `c` = 1000. Find the product `abc` such that `a` + `b` + `c` = `n`.
# --hints--
-`specialPythagoreanTriplet(1000)`应返回31875000。
+`specialPythagoreanTriplet(24)` should return a number.
```js
-assert.strictEqual(specialPythagoreanTriplet(1000), 31875000);
+assert(typeof specialPythagoreanTriplet(24) === 'number');
```
-`specialPythagoreanTriplet(24)`应该返回480。
+`specialPythagoreanTriplet(24)` should return 480.
```js
assert.strictEqual(specialPythagoreanTriplet(24), 480);
```
-`specialPythagoreanTriplet(120)`应该返回49920。
+`specialPythagoreanTriplet(120)` should return 49920, 55080 or 60000
```js
assert([49920, 55080, 60000].includes(specialPythagoreanTriplet(120)));
```
+`specialPythagoreanTriplet(1000)` should return 31875000.
+
+```js
+assert.strictEqual(specialPythagoreanTriplet(1000), 31875000);
+```
+
# --seed--
## --seed-contents--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-90-cube-digit-pairs.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-90-cube-digit-pairs.md
index fdfddb6a30d..ef2182b215d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-90-cube-digit-pairs.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-90-cube-digit-pairs.md
@@ -1,37 +1,48 @@
---
id: 5900f3c61000cf542c50fed9
-title: 问题90：立方体数字对
+title: 'Problem 90: Cube digit pairs'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302207
dashedName: problem-90-cube-digit-pairs
---
# --description--
-立方体上的六个面中的每一个都有一个写在其上的不同数字（0到9）;第二个立方体也是如此。通过将两个立方体并排放置在不同的位置，我们可以形成各种2位数字。
+Each of the six faces on a cube has a different digit (0 to 9) written on it; the same is done to a second cube. By placing the two cubes side-by-side in different positions we can form a variety of 2-digit numbers.
-例如，可以形成平方数64：
+For example, the square number 64 could be formed:
-实际上，通过仔细选择两个立方体上的数字，可以显示低于100的所有正方形数字：01,04,09,16,25,36,49,64和81。
+<img class="img-responsive center-block" alt="two cubes, one with the number 6 and the other with number 4" src="https://cdn-media-1.freecodecamp.org/project-euler/cube-digit-pairs.png" style="background-color: white; padding: 10px;" />
-例如，可以实现的一种方法是将{0,5,6,7,8,9}放在一个立方体上，将{1,2,3,4,8,9}放在另一个立方体上。
+In fact, by carefully choosing the digits on both cubes it is possible to display all of the square numbers below one-hundred: 01, 04, 09, 16, 25, 36, 49, 64, and 81.
-但是，对于这个问题，我们应该允许将6或9颠倒，以便安排像{0,5,6,7,8,9}和{1,2,3,4,6,7}允许显示所有九个方形数字;否则就不可能获得09。
+For example, one way this can be achieved is by placing {0, 5, 6, 7, 8, 9} on one cube and {1, 2, 3, 4, 8, 9} on the other cube.
-在确定不同的排列时，我们对每个立方体上的数字感兴趣，而不是订单。
+However, for this problem we shall allow the 6 or 9 to be turned upside-down so that an arrangement like {0, 5, 6, 7, 8, 9} and {1, 2, 3, 4, 6, 7} allows for all nine square numbers to be displayed; otherwise it would be impossible to obtain 09.
-{1,2,3,4,5,6}相当于{3,6,4,1,2,5} {1,2,3,4,5,6}不同于{1,2， 3,4,5,9}
+In determining a distinct arrangement we are interested in the digits on each cube, not the order.
-但是因为我们允许反转6和9，所以最后一个例子中的两个不同的集合都代表扩展集{1,2,3,4,5,6,9}以形成2位数字。
+<div style="margin-left: 4em;">
+ {1, 2, 3, 4, 5, 6} is equivalent to {3, 6, 4, 1, 2, 5}<br>
+ {1, 2, 3, 4, 5, 6} is distinct from {1, 2, 3, 4, 5, 9}
+</div>
-两个立方体有多少不同的排列可以显示所有的方形数字？
+But because we are allowing 6 and 9 to be reversed, the two distinct sets in the last example both represent the extended set {1, 2, 3, 4, 5, 6, 9} for the purpose of forming 2-digit numbers.
+
+How many distinct arrangements of the two cubes allow for all of the square numbers to be displayed?
# --hints--
-`euler90()`应该返回1217。
+`cubeDigitPairs()` should return a number.
```js
-assert.strictEqual(euler90(), 1217);
+assert(typeof cubeDigitPairs() === 'number');
+```
+
+`cubeDigitPairs()` should return 1217.
+
+```js
+assert.strictEqual(cubeDigitPairs(), 1217);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-91-right-triangles-with-integer-coordinates.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-91-right-triangles-with-integer-coordinates.md
index dac4423e390..b309ce36f9e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-91-right-triangles-with-integer-coordinates.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-91-right-triangles-with-integer-coordinates.md
@@ -1,25 +1,35 @@
---
id: 5900f3c71000cf542c50feda
-title: 问题91：带有整数坐标的直角三角形
+title: 'Problem 91: Right triangles with integer coordinates'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302208
dashedName: problem-91-right-triangles-with-integer-coordinates
---
# --description--
-点P（x1，y1）和Q（x2，y2）绘制在整数坐标处并连接到原点O（0,0），以形成ΔOPQ。
+The points P (`x`<sub>1</sub>, `y`<sub>1</sub>) and Q (`x`<sub>2</sub>, `y`<sub>2</sub>) are plotted at integer co-ordinates and are joined to the origin, O(0,0), to form ΔOPQ.
-正好有十四个三角形包含一个直角，当每个坐标位于0和2之间时，可以形成直角;即，0≤x1，y1，x2，y2≤2。
+<img class="img-responsive center-block" alt="a graph plotting points P (x_1, y_1) and Q(x_2, y_2) at integer coordinates that are joined to the origin O (0, 0)" src="https://cdn-media-1.freecodecamp.org/project-euler/right-triangles-integer-coordinates-1.png" style="background-color: white; padding: 10px;" />
-假设0≤x1，y1，x2，y2≤50，可以形成多少个直角三角形？
+There are exactly fourteen triangles containing a right angle that can be formed when each co-ordinate lies between 0 and 2 inclusive; that is, 0 ≤ `x`<sub>1</sub>, `y`<sub>1</sub>, `x`<sub>2</sub>, `y`<sub>2</sub> ≤ 2.
+
+<img class="img-responsive center-block" alt="a diagram showing the 14 triangles containing a right angle that can be formed when each coordinate is between 0 and 2" src="https://cdn-media-1.freecodecamp.org/project-euler/right-triangles-integer-coordinates-2.png" style="background-color: white; padding: 10px;" />
+
+Given that 0 ≤ `x`<sub>1</sub>, `y`<sub>1</sub>, `x`<sub>2</sub>, `y`<sub>2</sub> ≤ 50, how many right triangles can be formed?
# --hints--
-`euler91()`应返回`euler91()` 。
+`rightTrianglesIntCoords()` should return a number.
```js
-assert.strictEqual(euler91(), 14234);
+assert(typeof rightTrianglesIntCoords() === 'number');
+```
+
+`rightTrianglesIntCoords()` should return 14234.
+
+```js
+assert.strictEqual(rightTrianglesIntCoords(), 14234);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-92-square-digit-chains.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-92-square-digit-chains.md
index 07a2bff221f..7c0ecc19d05 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-92-square-digit-chains.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-92-square-digit-chains.md
@@ -1,21 +1,38 @@
---
id: 5900f3c81000cf542c50fedb
-title: 问题92：方形数字链
+title: 'Problem 92: Square digit chains'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302209
dashedName: problem-92-square-digit-chains
---
# --description--
-通过在数字中连续添加数字的平方来创建数字链，以形成新的数字，直到之前看到它为止。例如，44→32→13→10→1→1 85→89→145→42→20→4→16→37→58→89因此，任何到达1或89的链条都会卡在无限循环中。最令人惊讶的是，每个起始号码最终都会达到1或89.低于一千万的起始号码将达到89个？
+A number chain is created by continuously adding the square of the digits in a number to form a new number until it has been seen before.
+
+For example,
+
+<div style='margin-left: 4em;'>
+ 44 → 32 → 13 → 10 → <strong>1</strong> → <strong>1</strong><br>
+ 85 → <strong>89</strong> → 145 → 42 → 20 → 4 → 16 → 37 → 58 → <strong>89</strong>
+</div>
+
+Therefore any chain that arrives at 1 or 89 will become stuck in an endless loop. What is most amazing is that EVERY starting number will eventually arrive at 1 or 89.
+
+How many starting numbers below ten million will arrive at 89?
# --hints--
-`euler92()`应返回8581146。
+`squareDigitChains()` should return a number.
```js
-assert.strictEqual(euler92(), 8581146);
+assert(typeof squareDigitChains() === 'number');
+```
+
+`squareDigitChains()` should return 8581146.
+
+```js
+assert.strictEqual(squareDigitChains(), 8581146);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-93-arithmetic-expressions.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-93-arithmetic-expressions.md
index 8749d2a08a1..81b5dce63a3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-93-arithmetic-expressions.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-93-arithmetic-expressions.md
@@ -1,21 +1,42 @@
---
id: 5900f3ca1000cf542c50fedc
-title: 问题93：算术表达式
+title: 'Problem 93: Arithmetic expressions'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302210
dashedName: problem-93-arithmetic-expressions
---
# --description--
-通过使用集合中的每个数字{1,2,3,4}，恰好一次，并利用四个算术运算（+， - ，\*，/）和括号/括号，可以形成不同的正整数目标。例如，8 =（4 \*（1 + 3））/ 2 14 = 4 \*（3 + 1/2）19 = 4 \*（2 + 3） - 1 36 = 3 \* 4 \*（2 + 1）注数字的连接，如12 + 34，是不允许的。使用集合{1,2,3,4}，可以获得31个不同的目标数，其中36个是最大的，并且在遇到第一个不可表达之前可以获得数字1到28中的每一个。数。找到四个不同数字的集合，a <b <c <d，可以获得最长的一组连续正整数，1到n，给出你的答案字符串：abcd。
+By using each of the digits from the set, {1, 2, 3, 4}, exactly once, and making use of the four arithmetic operations (+, −, \*, /) and brackets/parentheses, it is possible to form different positive integer targets.
+
+For example,
+
+<div style='margin-left: 4em;'>
+ 8 = (4 * (1 + 3)) / 2<br>
+ 14 = 4 * (3 + 1 / 2)<br>
+ 19 = 4 * (2 + 3) − 1<br>
+ 36 = 3 * 4 * (2 + 1)
+</div>
+
+Note that concatenations of the digits, like 12 + 34, are not allowed.
+
+Using the set, {1, 2, 3, 4}, it is possible to obtain thirty-one different target numbers of which 36 is the maximum, and each of the numbers 1 to 28 can be obtained before encountering the first non-expressible number.
+
+Find the set of four distinct digits, `a` < `b` < `c` < `d`, for which the longest set of consecutive positive integers, 1 to `n`, can be obtained, giving your answer as a string: `abcd`.
# --hints--
-`euler93()`应返回1258。
+`arithmeticExpressions()` should return a number.
```js
-assert.strictEqual(euler93(), 1258);
+assert(typeof arithmeticExpressions() === 'number');
+```
+
+`arithmeticExpressions()` should return 1258.
+
+```js
+assert.strictEqual(arithmeticExpressions(), 1258);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-94-almost-equilateral-triangles.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-94-almost-equilateral-triangles.md
index 056dbc7d482..fbd3f04a8b3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-94-almost-equilateral-triangles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-94-almost-equilateral-triangles.md
@@ -1,21 +1,31 @@
---
id: 5900f3ca1000cf542c50fedd
-title: 问题94：几乎等边三角形
+title: 'Problem 94: Almost equilateral triangles'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302211
dashedName: problem-94-almost-equilateral-triangles
---
# --description--
-很容易证明，不存在具有整体长边和整体面积的等边三角形。然而，几乎等边三角形5-5-6的面积为12平方单位。我们将几乎等边三角形定义为三边形，其中两边相等，第三边相差不超过一个单位。求出所有几乎等边三角形的周长之和，其中边长和面积不超过十亿（1,000,000,000）。
+It is easily proved that no equilateral triangle exists with integral length sides and integral area. However, the <dfn>almost equilateral triangle</dfn> 5-5-6 has an area of 12 square units.
+
+We shall define an <dfn>almost equilateral triangle</dfn> to be a triangle for which two sides are equal and the third differs by no more than one unit.
+
+Find the sum of the perimeters of all <dfn>almost equilateral triangle</dfn> with integral side lengths and area and whose perimeters do not exceed one billion (1,000,000,000).
# --hints--
-`euler94()`应该返回518408346。
+`almostEquilateralTriangles()` should return a number.
```js
-assert.strictEqual(euler94(), 518408346);
+assert(typeof almostEquilateralTriangles() === 'number');
+```
+
+`almostEquilateralTriangles()` should return 518408346.
+
+```js
+assert.strictEqual(almostEquilateralTriangles(), 518408346);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-95-amicable-chains.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-95-amicable-chains.md
index 90e9390e6b0..037330b6608 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-95-amicable-chains.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-95-amicable-chains.md
@@ -1,21 +1,39 @@
---
id: 5900f3cc1000cf542c50fede
-title: 问题95：友好的链条
+title: 'Problem 95: Amicable chains'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302212
dashedName: problem-95-amicable-chains
---
# --description--
-一个数的适当除数是除数字本身之外的所有除数。例如，28的正确除数是1,2,4,7和14.由于这些除数的总和等于28，我们称之为完全数。有趣的是，220的适当除数之和为284，284的适当除数之和为220，形成两个数的链。出于这个原因，220和284被称为友好对。也许鲜为人知的是较长的链条。例如，从12496开始，我们形成一个五个数字的链：12496→14288→15472→14536→14264（→12496→...）由于该链返回其起始点，因此称为友好链。找到最长友好链中最小的成员，没有超过一百万的元素。
+The proper divisors of a number are all the divisors excluding the number itself. For example, the proper divisors of 28 are 1, 2, 4, 7, and 14. As the sum of these divisors is equal to 28, we call it a perfect number.
+
+Interestingly the sum of the proper divisors of 220 is 284 and the sum of the proper divisors of 284 is 220, forming a chain of two numbers. For this reason, 220 and 284 are called an amicable pair.
+
+Perhaps less well known are longer chains. For example, starting with 12496, we form a chain of five numbers:
+
+<div style="text-align: center;">
+ 12496 → 14288 → 15472 → 14536 → 14264 (→ 12496 → ...)
+</div>
+
+Since this chain returns to its starting point, it is called an amicable chain.
+
+Find the smallest member of the longest amicable chain with no element exceeding one million.
# --hints--
-`euler95()`应返回14316。
+`amicableChains()` should return a number.
```js
-assert.strictEqual(euler95(), 14316);
+assert(typeof amicableChains() === 'number');
+```
+
+`amicableChains()` should return 14316.
+
+```js
+assert.strictEqual(amicableChains(), 14316);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-96-su-doku.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-96-su-doku.md
index 12dfcc613f3..22442b67c2d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-96-su-doku.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-96-su-doku.md
@@ -1,27 +1,129 @@
---
id: 5900f3cc1000cf542c50fedf
-title: 问题96：苏杜库
+title: 'Problem 96: Su Doku'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302213
dashedName: problem-96-su-doku
---
# --description--
-Su Doku（日语含义数字位置）是流行拼图概念的名称。它的起源尚不清楚，但必须归功于莱昂哈德·欧拉（Leonhard Euler），他发明了一种类似的，更加困难的拼图游戏，称为拉丁广场（Latin Squares）。然而，Su Doku谜题的目标是替换9乘9网格中的空白（或零），使得每行，每列和3乘3包含每个数字1到9.下面是一个示例一个典型的起始拼图网格及其解决方案网格。
+Su Doku (Japanese meaning *number place*) is the name given to a popular puzzle concept. Its origin is unclear, but credit must be attributed to Leonhard Euler who invented a similar, and much more difficult, puzzle idea called Latin Squares. The objective of Su Doku puzzles, however, is to replace the blanks (or zeros) in a 9 by 9 grid in such that each row, column, and 3 by 3 box contains each of the digits 1 to 9. Below is an example of a typical starting puzzle grid and its solution grid.
-0 0 39 0 00 0 1 0 2 03 0 58 0 6 6 0 00 0 14 0 0 0 0 87 0 00 0 6 1 0 20 0 07 0 8 9 0 00 0 82 0 0 0 0 28 0 00 0 5 6 0 92 0 30 1 0 5 0 00 0 93 0 0
+<div style="margin: auto; background-color: white; padding: 10px; width: 80%; text-align: center;">
+ <table border="0" cellpadding="0" cellspacing="0" align="center">
+ <tbody>
+ <tr>
+ <td>
+ <table cellpadding="5" cellspacing="0" border="1">
+ <tbody>
+ <tr>
+ <td style="font-family:'courier new';font-size:14pt; color: black; padding: 5px; border: 2px solid black;">
+ 0 0 3<br />9 0 0<br />0 0 1
+ </td>
+ <td style="font-family:'courier new';font-size:14pt; color: black; padding: 5px; border: 2px solid black;">
+ 0 2 0<br />3 0 5<br />8 0 6
+ </td>
+ <td style="font-family:'courier new';font-size:14pt; color: black; padding: 5px; border: 2px solid black;">
+ 6 0 0<br />0 0 1<br />4 0 0
+ </td>
+ </tr>
+ <tr>
+ <td style="font-family:'courier new';font-size:14pt; color: black; padding: 5px; border: 2px solid black;">
+ 0 0 8<br />7 0 0<br />0 0 6
+ </td>
+ <td style="font-family:'courier new';font-size:14pt; color: black; padding: 5px; border: 2px solid black;">
+ 1 0 2<br />0 0 0<br />7 0 8
+ </td>
+ <td style="font-family:'courier new';font-size:14pt; color: black; padding: 5px; border: 2px solid black;">
+ 9 0 0<br />0 0 8<br />2 0 0
+ </td>
+ </tr>
+ <tr>
+ <td style="font-family:'courier new';font-size:14pt; color: black; padding: 5px; border: 2px solid black;">
+ 0 0 2<br />8 0 0<br />0 0 5
+ </td>
+ <td style="font-family:'courier new';font-size:14pt; color: black; padding: 5px; border: 2px solid black;">
+ 6 0 9<br />2 0 3<br />0 1 0
+ </td>
+ <td style="font-family:'courier new';font-size:14pt; color: black; padding: 5px; border: 2px solid black;">
+ 5 0 0<br />0 0 9<br />3 0 0
+ </td>
+ </tr>
+ </tbody>
+ </table>
+ </td>
+ <td width="50">
+ <img src="images/spacer.gif" width="50" height="1" alt="" /><br />
+ </td>
+ <td>
+ <table cellpadding="5" cellspacing="0" border="1">
+ <tbody>
+ <tr>
+ <td style="font-family:'courier new';font-size:14pt; color: black; padding: 5px; border: 2px solid black;">
+ 4 8 3<br />9 6 7<br />2 5 1
+ </td>
+ <td style="font-family:'courier new';font-size:14pt; color: black; padding: 5px; border: 2px solid black;">
+ 9 2 1<br />3 4 5<br />8 7 6
+ </td>
+ <td style="font-family:'courier new';font-size:14pt; color: black; padding: 5px; border: 2px solid black;">
+ 6 5 7<br />8 2 1<br />4 9 3
+ </td>
+ </tr>
+ <tr>
+ <td style="font-family:'courier new';font-size:14pt; color: black; padding: 5px; border: 2px solid black;">
+ 5 4 8<br />7 2 9<br />1 3 6
+ </td>
+ <td style="font-family:'courier new';font-size:14pt; color: black; padding: 5px; border: 2px solid black;">
+ 1 3 2<br />5 6 4<br />7 9 8
+ </td>
+ <td style="font-family:'courier new';font-size:14pt; color: black; padding: 5px; border: 2px solid black;">
+ 9 7 6<br />1 3 8<br />2 4 5
+ </td>
+ </tr>
+ <tr>
+ <td style="font-family:'courier new';font-size:14pt; color: black; padding: 5px; border: 2px solid black;">
+ 3 7 2<br />8 1 4<br />6 9 5
+ </td>
+ <td style="font-family:'courier new';font-size:14pt; color: black; padding: 5px; border: 2px solid black;">
+ 6 8 9<br />2 5 3<br />4 1 7
+ </td>
+ <td style="font-family:'courier new';font-size:14pt; color: black; padding: 5px; border: 2px solid black;">
+ 5 1 4<br />7 6 9<br />3 8 2
+ </td>
+ </tr>
+ </tbody>
+ </table>
+ </td>
+ </tr>
+ </tbody>
+ </table>
+</div>
-4 8 39 6 72 5 1 9 2 13 4 58 7 6 6 5 78 2 14 9 3 5 4 87 2 91 3 6 1 3 25 6 47 9 8 9 7 61 3 82 4 5 3 7 28 1 46 9 5 6 8 92 5 34 1 7 5 1 47 6 93 8 2
+A well constructed Su Doku puzzle has a unique solution and can be solved by logic, although it may be necessary to employ "guess and test" methods in order to eliminate options (there is much contested opinion over this). The complexity of the search determines the difficulty of the puzzle; the example above is considered easy because it can be solved by straight forward direct deduction.
-一个构造良好的Su Doku拼图有一个独特的解决方案，可以通过逻辑解决，虽然可能需要采用“猜测和测试”方法来消除选项（对此有很多争议的意见）。搜索的复杂性决定了拼图的难度;上面的例子被认为是简单的，因为它可以通过直接直接扣除来解决。 6K文本文件，sudoku.txt（右键单击和'保存链接/目标为...'），包含五十个不同的Su Doku难题，但都有独特的解决方案（文件中的第一个谜题是上面的例子） ）。通过解决所有五十个谜题，找到每个解决方案网格左上角找到的3位数字的总和;例如，483是上面解决方案网格左上角的3位数字。
+The `puzzlesArr` array contains fifty different Su Doku puzzle strings ranging in difficulty, but all with unique solutions (the first puzzle in the array is the example in the challenge description).
+
+By solving all fifty puzzles find the sum of the 3-digit numbers found in the top left corner of each solution grid; for example, 483 is the 3-digit number found in the top left corner of the solution grid above.
# --hints--
-`euler96()`应返回24702。
+`suDoku(testPuzzles)` should return a number.
```js
-assert.strictEqual(euler96(), 24702);
+assert(typeof suDoku(testPuzzles) === 'number');
+```
+
+`suDoku(testPuzzles)` should return 1190.
+
+```js
+assert.strictEqual(suDoku(testPuzzles), 1190);
+```
+
+`suDoku(puzzlesArr)` should return 24702.
+
+```js
+assert.strictEqual(suDoku(puzzlesArr), 24702);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-97-large-non-mersenne-prime.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-97-large-non-mersenne-prime.md
index e869da35abb..b2b651f2678 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-97-large-non-mersenne-prime.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-97-large-non-mersenne-prime.md
@@ -1,21 +1,31 @@
---
id: 5900f3ce1000cf542c50fee0
-title: 问题97：大的非梅森素数
+title: 'Problem 97: Large non-Mersenne prime'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302214
dashedName: problem-97-large-non-mersenne-prime
---
# --description--
-1999年发现了第一个发现超过一百万个数字的已知素数，它是形式为26972593-1的梅森素数;它包含正好2,098,960位数字。随后发现其他形式为2p-1的梅森素数包含更多数字。然而，在2004年发现了一个巨大的非梅森素数，其中包含2,357,207个数字：28433×27830457 + 1。找到此素数的最后十位数字。
+The first known prime found to exceed one million digits was discovered in 1999, and is a Mersenne prime of the form 2<sup>6972593</sup>−1; it contains exactly 2,098,960 digits. Subsequently other Mersenne primes, of the form 2<sup><var>p</var></sup>−1, have been found which contain more digits.
+
+However, in 2004 there was found a massive non-Mersenne prime which contains 2,357,207 digits: 28433×2<sup>7830457</sup>+1.
+
+Find the last ten digits of this prime number.
# --hints--
-`euler97()`应该返回8739992577。
+`lrgNonMersennePrime()` should return a number.
```js
-assert.strictEqual(euler97(), 8739992577);
+assert(typeof lrgNonMersennePrime() === 'number');
+```
+
+`lrgNonMersennePrime()` should return 8739992577.
+
+```js
+assert.strictEqual(lrgNonMersennePrime(), 8739992577);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-98-anagramic-squares.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-98-anagramic-squares.md
index 5fb5b36fe76..7397b3f7b59 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-98-anagramic-squares.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-98-anagramic-squares.md
@@ -1,21 +1,33 @@
---
id: 5900f3cf1000cf542c50fee1
-title: 问题98：Anagramic正方形
+title: 'Problem 98: Anagramic squares'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302215
dashedName: problem-98-anagramic-squares
---
# --description--
-通过分别用1,2,9和6替换单词CARE中的每个字母，我们形成一个正方形数字：1296 = 362.值得注意的是，通过使用相同的数字替换，anagram，RACE，形成一个正方形数字：9216 = 962.我们将CARE（和RACE）称为方形字谜字对，并进一步指定不允许前导零，不同的字母也不能与另一个字母具有相同的数字值。使用words.txt（右键单击和“保存链接/目标为...”），一个包含近两千个常用英语单词的16K文本文件，找到所有方形字谜词对（一个回文词不被认为是一个自己的字谜）。这种货币对的任何成员形成的最大平方数是多少？注意：形成的所有字谜必须包含在给定的文本文件中。
+By replacing each of the letters in the word CARE with 1, 2, 9, and 6 respectively, we form a square number: 1296 = 36<sup>2</sup>. What is remarkable is that, by using the same digital substitutions, the anagram, RACE, also forms a square number: 9216 = 96<sup>2</sup>. We shall call CARE (and RACE) a square anagram word pair and specify further that leading zeroes are not permitted, neither may a different letter have the same digital value as another letter.
+
+Using the included array, find all the square anagram word pairs (a palindromic word is NOT considered to be an anagram of itself).
+
+What is the largest square number formed by any member of such a pair?
+
+**Note:** All anagrams formed must be contained in the given `words` array.
# --hints--
-`euler98()`应该返回18769。
+`anagramicSquares()` should return a number.
```js
-assert.strictEqual(euler98(), 18769);
+assert(typeof anagramicSquares() === 'number');
+```
+
+`anagramicSquares()` should return 18769.
+
+```js
+assert.strictEqual(anagramicSquares(), 18769);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-99-largest-exponential.md b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-99-largest-exponential.md
index 58d43bfe9b1..959ce076576 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-99-largest-exponential.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-99-largest-exponential.md
@@ -1,21 +1,33 @@
---
id: 5900f3d01000cf542c50fee2
-title: 问题99：最大的指数
+title: 'Problem 99: Largest exponential'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302216
dashedName: problem-99-largest-exponential
---
# --description--
-比较以索引形式（如211和37）编写的两个数字并不困难，因为任何计算器都会确认211 = 2048 <37 = 2187.但是，确认632382518061> 519432525806会更加困难，因为这两个数字都包含超过三百万个数字。使用base_exp.txt（右键单击并“将链接/目标另存为...”），一个22K文本文件包含每行上带有基数/指数对的一千行，确定哪个行号具有最大数值。注意：文件中的前两行代表上面给出的示例中的数字。
+Comparing two numbers written in index form like 2<sup>11</sup> and 3<sup>7</sup> is not difficult, as any calculator would confirm that 2<sup>11</sup> = 2048 < 3<sup>7</sup> = 2187.
+
+However, confirming that 632382<sup>518061</sup> > 519432<sup>525806</sup> would be much more difficult, as both numbers contain over three million digits.
+
+Using the included 2D array of base/exponent pairs, determine which pair has the greatest numerical value.
+
+**Note:** The first array in `baseExp` represent the numbers in the example given above.
# --hints--
-`euler99()`应该返回709。
+`largestExponential()` should return a number.
```js
-assert.strictEqual(euler99(), 709);
+assert(typeof largestExponential() === 'number');
+```
+
+`largestExponential()` should return 709.
+
+```js
+assert.strictEqual(largestExponential(), 709);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/100-doors.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/100-doors.md
index df78411a98c..e791c065581 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/100-doors.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/100-doors.md
@@ -1,30 +1,34 @@
---
id: 594810f028c0303b75339acb
-title: 100门
+title: 100 doors
challengeType: 5
-videoUrl: ''
+forumTopicId: 302217
dashedName: 100-doors
---
# --description--
-<p>连续100个门都是最初关闭的。你可以在门口进行100次通行证。第一次通过，访问每扇门并“切换”门（如果门关闭，打开它;如果它打开，关闭它）。第二次，只访问每个第二个门（即门＃2，＃4，＃6，......）并切换它。第三次，访问每个第3门（即3号门，＃6号，＃9号，......）等，直到您只访问第100个门。 </p><p>实现一个功能，以确定最后一次通过后门的状态。将最终结果返回到数组中，如果数组打开，则只包含数字中包含的门号。 </p>
+There are 100 doors in a row that are all initially closed. You make 100 passes by the doors. The first time through, visit every door and 'toggle' the door (if the door is closed, open it; if it is open, close it). The second time, only visit every 2nd door (i.e., door #2, #4, #6, ...) and toggle it. The third time, visit every 3rd door (i.e., door #3, #6, #9, ...), etc., until you only visit the 100th door.
+
+# --instructions--
+
+Implement a function to determine the state of the doors after the last pass. Return the final result in an array, with only the door number included in the array if it is open.
# --hints--
-`getFinalOpenedDoors`是一个函数。
+`getFinalOpenedDoors` should be a function.
```js
assert(typeof getFinalOpenedDoors === 'function');
```
-`getFinalOpenedDoors`应该返回一个数组。
+`getFinalOpenedDoors` should return an array.
```js
assert(Array.isArray(getFinalOpenedDoors(100)));
```
-`getFinalOpenedDoors`没有产生正确的结果。
+`getFinalOpenedDoors` should produce the correct result.
```js
assert.deepEqual(getFinalOpenedDoors(100), solution);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/24-game.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/24-game.md
index 6bcc9002a5f..7bf28fe4df4 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/24-game.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/24-game.md
@@ -1,45 +1,66 @@
---
id: 5951e88f64ebf159166a1176
-title: 24场比赛
+title: 24 game
challengeType: 5
-videoUrl: ''
+forumTopicId: 302218
dashedName: 24-game
---
# --description--
-<p>实现一个以四位数字串为参数的函数，每个数字从1──►9（含），允许重复，并返回一个算术表达式，其值为24。如果不存在这样的解，则返回“没有解决方案。“ </p><p>规则： </p>只允许以下运算符/函数：乘法，除法，加法，减法除法应使用浮点或有理算术等来保留余数。不允许从提供的数字中形成多位数字。 （所以当给出1,2,2和1时，12 + 12的答案是错误的）。给定的数字顺序不必保留。 <p>示例输入： </p> <code>solve24("4878");</code> <code>solve24("1234");</code> <code>solve24("6789");</code> <code>solve24("1127");</code> <p>示例输出（字符串）： </p> <code>(7-8/8)\*4</code> <code>3\*1\*4\*2</code> <code>(6\*8)/(9-7)</code> <code>(1+7)\*(2+1)</code>
+The [24 Game](https://en.wikipedia.org/wiki/24_Game) tests a person's mental arithmetic.
+
+The aim of the game is to arrange four numbers in a way that when evaluated, the result is 24
+
+# --instructions--
+
+Implement a function that takes a string of four digits as its argument, with each digit from 1 to 9 (inclusive) with repetitions allowed, and returns an arithmetic expression that evaluates to the number 24. If no such solution exists, return "no solution exists".
+
+**Rules:**
+<ul>
+ <li> Only the following operators/functions are allowed: multiplication, division, addition, subtraction. </li>
+ <li> Division should use floating point or rational arithmetic, etc, to preserve remainders. </li>
+ <li> Forming multiple digit numbers from the supplied digits is disallowed. (So an answer of 12+12 when given 1, 2, 2, and 1 is wrong). </li>
+ <li> The order of the digits when given does not have to be preserved. </li>
+</ul>
+
+| Example input | Example output |
+| ------------------------- | ------------------------- |
+| <code>solve24("4878");</code> | <code>(7-8/8)\*4</code> |
+| <code>solve24("1234");</code> | <code>3\*1\*4\*2</code> |
+| <code>solve24("6789");</code> | <code>(6\*8)/(9-7)</code> |
+| <code>solve24("1127");</code> | <code>(1+7)\*(2+1)</code> |
# --hints--
-`solve24`是一个函数。
+`solve24` should be a function.
```js
assert(typeof solve24 === 'function');
```
-`solve24("4878")`应返回`(7-8/8)*4`或`4*(7-8/8)`
+`solve24("4878")` should return `(7-8/8)*4` or `4*(7-8/8)`
```js
-assert(include(answers[0], solve24(testCases[0])));
+assert(include(answers[0], removeParentheses(solve24(testCases[0]))));
```
-`solve24("1234")`应返回`1*2*3*4`任何排列
+`solve24("1234")` should return any arrangement of `1*2*3*4`
```js
-assert(include(answers[1], solve24(testCases[1])));
+assert(include(answers[1], removeParentheses(solve24(testCases[1]))));
```
-`solve24("6789")`应返回`(6*8)/(9-7)`或`(8*6)/(9-7)`
+`solve24("6789")` should return `(6*8)/(9-7)` or `(8*6)/(9-7)`
```js
-assert(include(answers[2], solve24(testCases[2])));
+assert(include(answers[2], removeParentheses(solve24(testCases[2]))));
```
-`solve24("1127")`应该返回`(1+7)*(1*2)`的排列
+`solve24("1127")` should return a permutation of `(1+7)*(1+2)`
```js
-assert(include(answers[3], solve24(testCases[3])));
+assert(include(answers[3], removeParentheses(solve24(testCases[3]))));
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/9-billion-names-of-god-the-integer.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/9-billion-names-of-god-the-integer.md
index 0805daf3efe..125236ca75f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/9-billion-names-of-god-the-integer.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/9-billion-names-of-god-the-integer.md
@@ -1,60 +1,84 @@
---
id: 5949b579404977fbaefcd736
-title: 90亿上帝的名字整数
+title: 9 billion names of God the integer
challengeType: 5
-videoUrl: ''
+forumTopicId: 302219
dashedName: 9-billion-names-of-god-the-integer
---
# --description--
-<p>这项任务是<a href='https://en.wikipedia.org/wiki/The Nine Billion Names of God#Plot_summary' title='wp：上帝的九十亿名字#Plot_summary'>Arthur C. Clarke</a>的<a href='https://en.wikipedia.org/wiki/The Nine Billion Names of God#Plot_summary' title='wp：上帝的九十亿名字#Plot_summary'>短篇小说改编</a> 。 </p><p> （求解者应该意识到完成这项任务的后果。） </p><p>详细说明，指定“名称”的含义： </p><p>整数1有1个名称“1”。 </p><p>整数2有2个名称“1 + 1”和“2”。 </p><p>整数3具有3个名称“1 + 1 + 1”，“2 + 1”和“3”。 </p><p>整数4具有5个名称“1 + 1 + 1 + 1”，“2 + 1 + 1”，“2 + 2”，“3 + 1”，“4”。 </p><p>整数5有7个名称“1 + 1 + 1 + 1 + 1”，“2 + 1 + 1 + 1”，“2 + 2 + 1”，“3 + 1 + 1”，“3 + 2”， “4 + 1”，“5”。 </p><p>这可以通过以下形式显示： </p><pre> 1
- 1 1
- 1 1 1
- 1 2 1 1
- 1 2 2 1 1
-1 3 3 2 1 1
-</pre><p>其中row $ n $对应于整数$ n $，而行$ m $中从左到右的每列$ C $对应于以$ C $开头的名称数。 </p><p> （可选）请注意$ n $ -th行$ P（n）$的总和是<a href='http://mathworld.wolfram.com/PartitionFunctionP.html' title='链接：http：//mathworld.wolfram.com/PartitionFunctionP.html'>整数分区函数</a> 。 </p>任务<p>实现一个返回$ n $ -th行之和的函数。 </p>
+This task is a variation of the [short story by Arthur C. Clarke](https://en.wikipedia.org/wiki/The Nine Billion Names of God#Plot_summary "wp: The Nine Billion Names of God#Plot_summary").
+
+(Solvers should be aware of the consequences of completing this task.)
+
+In detail, to specify what is meant by a "name":
+
+<ul>
+ <li>The integer 1 has 1 name "1".</li>
+ <li>The integer 2 has 2 names "1+1" and "2".</li>
+ <li>The integer 3 has 3 names "1+1+1", "2+1", and "3".</li>
+ <li>The integer 4 has 5 names "1+1+1+1", "2+1+1", "2+2", "3+1", "4".</li>
+ <li>The integer 5 has 7 names "1+1+1+1+1", "2+1+1+1", "2+2+1", "3+1+1", "3+2", "4+1", "5".</li>
+</ul>
+
+This can be visualized in the following form:
+
+<pre> 1
+ 1 1
+ 1 1 1
+ 1 2 1 1
+ 1 2 2 1 1
+1 3 3 2 1 1
+</pre>
+
+Where row $n$ corresponds to integer $n$, and each column $C$ in row $m$ from left to right corresponds to the number of names beginning with $C$.
+
+Optionally note that the sum of the $n$-th row $P(n)$ is the integer partition function.
+
+# --instructions--
+
+Implement a function that returns the sum of the $n$-th row.
# --hints--
-`numberOfNames`是一个函数。
+`numberOfNames` should be function.
```js
assert(typeof numberOfNames === 'function');
```
-`numberOfNames(5)`应该等于7。
+`numberOfNames(5)` should equal 7.
```js
assert.equal(numberOfNames(5), 7);
```
-`numberOfNames(12)`应该等于77。
+`numberOfNames(12)` should equal 77.
```js
assert.equal(numberOfNames(12), 77);
```
-`numberOfNames(18)`应该等于385。
+`numberOfNames(18)` should equal 385.
```js
assert.equal(numberOfNames(18), 385);
```
-`numberOfNames(23)`应该等于1255。
+`numberOfNames(23)` should equal 1255.
```js
assert.equal(numberOfNames(23), 1255);
```
-`numberOfNames(42)`应该等于53174。
+`numberOfNames(42)` should equal 53174.
```js
assert.equal(numberOfNames(42), 53174);
```
-`numberOfNames(123)`应该等于2552338241。
+`numberOfNames(123)` should equal 2552338241.
```js
assert.equal(numberOfNames(123), 2552338241);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/abc-problem.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/abc-problem.md
index b9b5af522e2..b42b619749d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/abc-problem.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/abc-problem.md
@@ -1,60 +1,93 @@
---
id: 594810f028c0303b75339acc
-title: ABC问题
+title: ABC Problem
challengeType: 5
-videoUrl: ''
+forumTopicId: 302220
dashedName: abc-problem
---
# --description--
-<p>您将获得ABC块的集合（例如，童年字母块）。每个街区有20个街区，两个字母。块的所有侧面都保证有完整的字母表。块的样本集合： </p><p> （BO） </p><p> （XK） </p><p> （DQ） </p><p> （CP） </p><p> （NA） </p><p> （GT） </p><p> （回覆） </p><p> （TG） </p><p> （QD） </p><p> （FS） </p><p> （JW） </p><p> （HU） </p><p> （VI） </p><p> （一个） </p><p> （OB） </p><p> （ER） </p><p> （FS） </p><p> （LY） </p><p> （PC） </p><p> （ZM） </p><p>要记住一些规则： </p>一旦使用了块上的字母，就不能再使用该块。该函数应该不区分大小写。 <p>实现一个带字符串（单词）的函数，并确定该单词是否可以与给定的块集合拼写。 </p>
+You are given a collection of ABC blocks (e.g., childhood alphabet blocks). There are 20 blocks with two letters on each block. A complete alphabet is guaranteed amongst all sides of the blocks. The sample collection of blocks:
+
+<pre>(B O)
+(X K)
+(D Q)
+(C P)
+(N A)
+(G T)
+(R E)
+(T G)
+(Q D)
+(F S)
+(J W)
+(H U)
+(V I)
+(A N)
+(O B)
+(E R)
+(F S)
+(L Y)
+(P C)
+(Z M)
+</pre>
+
+# --instructions--
+
+Implement a function that takes a string (word) and determines whether the word can be spelled with the given collection of blocks.
+
+Some rules to keep in mind:
+
+<ul>
+ <li>Once a letter on a block is used, that block cannot be used again.</li>
+ <li>The function should be case-insensitive.</li>
+</ul>
# --hints--
-`canMakeWord`是一个功能。
+`canMakeWord` should be a function.
```js
assert(typeof canMakeWord === 'function');
```
-`canMakeWord`应该返回一个布尔值。
+`canMakeWord` should return a boolean.
```js
assert(typeof canMakeWord('hi') === 'boolean');
```
-`canMakeWord("bark")`应该返回true。
+`canMakeWord("bark")` should return true.
```js
assert(canMakeWord(words[0]));
```
-`canMakeWord("BooK")`应该返回false。
+`canMakeWord("BooK")` should return false.
```js
assert(!canMakeWord(words[1]));
```
-`canMakeWord("TReAT")`应该返回true。
+`canMakeWord("TReAT")` should return true.
```js
assert(canMakeWord(words[2]));
```
-`canMakeWord("COMMON")`应返回false。
+`canMakeWord("COMMON")` should return false.
```js
assert(!canMakeWord(words[3]));
```
-`canMakeWord("squAD")`应该返回true。
+`canMakeWord("squAD")` should return true.
```js
assert(canMakeWord(words[4]));
```
-`canMakeWord("conFUSE")`应该返回true。
+`canMakeWord("conFUSE")` should return true.
```js
assert(canMakeWord(words[5]));
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/abundant-deficient-and-perfect-number-classifications.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/abundant-deficient-and-perfect-number-classifications.md
index 4d601e5bc9a..ca8465f0baa 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/abundant-deficient-and-perfect-number-classifications.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/abundant-deficient-and-perfect-number-classifications.md
@@ -1,36 +1,50 @@
---
id: 594810f028c0303b75339acd
-title: 丰富，不足和完善的数字分类
+title: 'Abundant, deficient and perfect number classifications'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302221
dashedName: abundant-deficient-and-perfect-number-classifications
---
# --description--
-<p>它们根据<a href='http://rosettacode.org/wiki/Proper divisors' title='适当的除数'>适当的除数</a>定义了三个正整数分类。 </p><p>设$ P（n）$是n的适当除数的总和，其中适当的除数都是n本身以外的正整数。 </p><p>如果<code>P(n) < n</code>那么n被归类为“缺陷” </p><p>如果<code>P(n) === n</code>那么n被归类为“完美” </p><p>如果<code>P(n) > n</code>则n被归类为“丰富” </p><p>例： </p><p> 6具有1,2和3的适当除数。 </p><p> 1 + 2 + 3 = 6，因此6被归类为完美数字。 </p><p>实现一个函数，计算三个类中每个类中1到20,000（包括）的整数。以下列格式将结果输出为数组<code>[deficient, perfect, abundant]</code> 。 </p>
+These define three classifications of positive integers based on their [proper divisors](https://rosettacode.org/wiki/Proper divisors "Proper divisors").
+
+Let $P(n)$ be the sum of the proper divisors of `n` where proper divisors are all positive integers `n` other than `n` itself.
+
+If `P(n) < n` then `n` is classed as `deficient`
+
+If `P(n) === n` then `n` is classed as `perfect`
+
+If `P(n) > n` then `n` is classed as `abundant`
+
+**Example**: `6` has proper divisors of `1`, `2`, and `3`. `1 + 2 + 3 = 6`, so `6` is classed as a perfect number.
+
+# --instructions--
+
+Implement a function that calculates how many of the integers from `1` to `20,000` (inclusive) are in each of the three classes. Output the result as an array in the following format `[deficient, perfect, abundant]`.
# --hints--
-`getDPA`是一个功能。
+`getDPA` should be a function.
```js
assert(typeof getDPA === 'function');
```
-`getDPA`应该返回一个数组。
+`getDPA` should return an array.
```js
assert(Array.isArray(getDPA(100)));
```
-`getDPA`返回值的长度应为3。
+`getDPA` return value should have a length of 3.
```js
assert(getDPA(100).length === 3);
```
-`getDPA(20000)`应该等于[15043,4,4953]
+`getDPA(20000)` should equal [15043, 4, 4953]
```js
assert.deepEqual(getDPA(20000), solution);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/accumulator-factory.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/accumulator-factory.md
index 2f3401b343c..fd90df9f347 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/accumulator-factory.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/accumulator-factory.md
@@ -1,36 +1,48 @@
---
id: 594810f028c0303b75339ace
-title: 蓄能器工厂
+title: Accumulator factory
challengeType: 5
-videoUrl: ''
+forumTopicId: 302222
dashedName: accumulator-factory
---
# --description--
-<p>创建一个带有单个（数字）参数的函数，并返回另一个作为累加器的函数。返回的累加器函数又接受一个数字参数，并返回到目前为止传递给该累加器的所有数值的总和（包括创建累加器时传递的初始值）。 </p><p>规则： </p><p>不要使用全局变量。 </p><p>暗示： </p><p>闭包可以保存外部状态。 </p>
+A problem posed by [Paul Graham](https://en.wikipedia.org/wiki/Paul_Graham_(programmer)) is that of creating a function that takes a single (numeric) argument and which returns another function that is an accumulator. The returned accumulator function in turn also takes a single numeric argument, and returns the sum of all the numeric values passed in so far to that accumulator (including the initial value passed when the accumulator was created).
+
+# --instructions--
+
+Create a function that takes a number $n$ and generates accumulator functions that return the sum of every number ever passed to them.
+
+**Rules:**
+
+Do not use global variables.
+
+**Hint:**
+
+Closures save outer state.
# --hints--
-`accumulator`是一个功能。
+`accumulator` should be a function.
```js
assert(typeof accumulator === 'function');
```
-`accumulator(0)`应该返回一个函数。
+`accumulator(0)` should return a function.
```js
assert(typeof accumulator(0) === 'function');
```
-`accumulator(0)(2)`应该返回一个数字。
+`accumulator(0)(2)` should return a number.
```js
assert(typeof accumulator(0)(2) === 'number');
```
-传递值3，-4,1.5和5应返回5.5。
+Passing in the values 3, -4, 1.5, and 5 should return 5.5.
```js
assert(testFn(5) === 5.5);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/ackermann-function.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/ackermann-function.md
index d4a17310f4b..f80262dc40d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/ackermann-function.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/ackermann-function.md
@@ -1,42 +1,52 @@
---
id: 594810f028c0303b75339acf
-title: 阿克曼功能
+title: Ackermann function
challengeType: 5
-videoUrl: ''
+forumTopicId: 302223
dashedName: ackermann-function
---
# --description--
-<p> Ackermann函数是递归函数的典型示例，尤其值得注意的是它不是原始递归函数。它的值增长非常快，其调用树的大小也是如此。 </p><p> Ackermann函数通常定义如下： </p> $$ A（m，n）= \\ begin {cases} n + 1＆\\ mbox {if} m = 0 \\\\ A（m-1,1）＆\\ mbox {if} m> 0 \\ mbox {和} n = 0 \\\\ A（m-1，A（m，n-1））＆\\ mbox {if} m> 0 \\ mbox {和} n> 0. \\ end {cases} $$ <p>它的论点永远不会消极，它总是终止。编写一个返回$ A（m，n）$的值的函数。任意精度是首选（因为函数增长如此之快），但不是必需的。 </p>
+The Ackermann function is a classic example of a recursive function, notable especially because it is not a primitive recursive function. It grows very quickly in value, as does the size of its call tree.
+
+The Ackermann function is usually defined as follows:
+
+$A(m, n) = \\begin{cases} n+1 & \\mbox{if } m = 0 \\\\ A(m-1, 1) & \\mbox{if } m > 0 \\mbox{ and } n = 0 \\\\ A(m-1, A(m, n-1)) & \\mbox{if } m > 0 \\mbox{ and } n > 0. \\end{cases}$
+
+Its arguments are never negative and it always terminates.
+
+# --instructions--
+
+Write a function which returns the value of $A(m, n)$. Arbitrary precision is preferred (since the function grows so quickly), but not required.
# --hints--
-`ack`是一个功能。
+`ack` should be a function.
```js
assert(typeof ack === 'function');
```
-`ack(0, 0)`应该返回1。
+`ack(0, 0)` should return 1.
```js
assert(ack(0, 0) === 1);
```
-`ack(1, 1)`应该返回3。
+`ack(1, 1)` should return 3.
```js
assert(ack(1, 1) === 3);
```
-`ack(2, 5)`应该返回13。
+`ack(2, 5)` should return 13.
```js
assert(ack(2, 5) === 13);
```
-`ack(3, 3)`应该返回61。
+`ack(3, 3)` should return 61.
```js
assert(ack(3, 3) === 61);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/align-columns.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/align-columns.md
index cee98417082..45ff5fa2f35 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/align-columns.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/align-columns.md
@@ -1,46 +1,64 @@
---
id: 594810f028c0303b75339ad0
-title: 对齐列
+title: Align columns
challengeType: 5
-videoUrl: ''
+forumTopicId: 302224
dashedName: align-columns
---
# --description--
-<p>给定一个包含许多行的文本文件，其中一行中的字段由单个<code>$</code>字符描述，编写一个程序，通过确保每列中的单词至少由一个空格分隔来对齐每列字段。此外，允许列中的每个单词在其列中左对齐，右对齐或居中对齐。 </p><p>使用以下文本测试您的程序： </p><pre>的$ $多行给出$ A $文本$文件$
-其中$领域内$ A $线$ $
-是$由$ A $单个$ '美元' $字符划定$
-写$ A $程序
-即$对齐$ $场的每个$列$
-由$ $确保该$字美元$ $每
-列$ $是在$最少$一张价值$空间被分隔$ $。
-此外，$允许为$ $ $中的每个字$ $ A $ $列到$为$要么离开$ $
-对齐，右$ $有道理
-或$ $中心内$有道理$ $其列。
-</pre><p>注意： </p>示例输入文本行可以或可以不具有尾随美元字符。所有列应共享相同的对齐方式。为了完成任务，在行尾附近产生的连续空格字符是无关紧要的。输出文本将以纯文本编辑器或基本终端上的单行间隔字体查看。列之间的最小间距应根据文本计算，而不是硬编码。不需要在列之间或列周围添加分隔字符。
+Given a text file of many lines, where fields within a line are delineated by a single `$` character, write a program that aligns each column of fields by ensuring that words in each column are separated by at least one space. Further, allow for each word in a column to be either left justified, right justified, or center justified within its column.
+
+# --instructions--
+
+Use the following text to test your programs:
+
+<pre>
+Given$a$text$file$of$many$lines
+where$fields$within$a$line$
+are$delineated$by$a$single$'dollar'$character
+write$a$program
+that$aligns$each$column$of$fields
+by$ensuring$that$words$in$each$
+column$are$separated$by$at$least$one$space.
+Further,$allow$for$each$word$in$a$column$to$be$either$left$
+justified,$right$justified
+or$center$justified$within$its$column.
+</pre>
+
+**Note that:**
+
+<ul>
+ <li>The example input texts lines may, or may not, have trailing dollar characters.</li>
+ <li>All columns should share the same alignment.</li>
+ <li>Consecutive space characters produced adjacent to the end of lines are insignificant for the purposes of the task.</li>
+ <li>Output text will be viewed in a mono-spaced font on a plain text editor or basic terminal.</li>
+ <li>The minimum space between columns should be computed from the text and not hard-coded.</li>
+ <li>It is not a requirement to add separating characters between or around columns.</li>
+</ul>
# --hints--
-`formatText`是一个函数。
+`formatText` should be a function.
```js
assert(typeof formatText === 'function');
```
-具有上述输入和“右”对齐的`formatText`应产生以下内容：
+`formatText` with the above input and "right" justification should produce the following:
```js
assert.strictEqual(formatText(testInput, 'right'), rightAligned);
```
-具有上述输入和“左”对齐的`formatText`应产生以下内容：
+`formatText` with the above input and "left" justification should produce the following:
```js
assert.strictEqual(formatText(testInput, 'left'), leftAligned);
```
-具有上述输入和“居中”对齐的`formatText`应产生以下内容：
+`formatText` with the above input and "center" justification should produce the following:
```js
assert.strictEqual(formatText(testInput, 'center'), centerAligned);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/amicable-pairs.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/amicable-pairs.md
index 4fdd3fa5516..207ab439f3f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/amicable-pairs.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/amicable-pairs.md
@@ -1,36 +1,49 @@
---
id: 5949b579404977fbaefcd737
-title: 友好的对
+title: Amicable pairs
challengeType: 5
-videoUrl: ''
+forumTopicId: 302225
dashedName: amicable-pairs
---
# --description--
-如果$ N \\ neq M $和N $ $的[适当除数之](<http://rosettacode.org/wiki/Proper divisors> "适当的除数")和（$ \\ mathrm {sum}（\\ mathrm {propDivs}（N）），两个整数$ N $和$ M $被认为是[友好对](<https://en.wikipedia.org/wiki/Amicable numbers> "wp：友善号码") $）$ = M $以及$ \\ mathrm {sum}（\\ mathrm {propDivs}（M））= N $。示例：1184和1210是友好对，具有适当的除数：1,2,4,8,16,32,37,74,148,296,592和1,2,5,10,11,25,55，分别为110,121,242,605。任务：计算并显示低于20,000的Amicable对（有八个）。相关任务[适当的除数](<http://rosettacode.org/wiki/Proper divisors> "适当的除数") [丰富，缺陷和完善的数字分类](<http://rosettacode.org/wiki/Abundant, deficient and perfect number classifications> "丰富，不足和完善的数字分类") [等分序列分类](<http://rosettacode.org/wiki/Aliquot sequence classifications> "等分序列分类")及其友好分类。
+Two integers $N$ and $M$ are said to be [amicable pairs](https://en.wikipedia.org/wiki/Amicable numbers "wp: Amicable numbers") if $N \\neq M$ and the sum of the [proper divisors](https://rosettacode.org/wiki/Proper divisors "Proper divisors") of $N$ ($\\mathrm{sum}(\\mathrm{propDivs}(N))$) $= M$ as well as $\\mathrm{sum}(\\mathrm{propDivs}(M)) = N$.
+
+**Example:**
+
+**1184** and **1210** are an amicable pair, with proper divisors:
+
+<ul>
+ <li>1, 2, 4, 8, 16, 32, 37, 74, 148, 296, 592 and</li>
+ <li>1, 2, 5, 10, 11, 22, 55, 110, 121, 242, 605 respectively.</li>
+</ul>
+
+# --instructions--
+
+Calculate and show here the Amicable pairs below 20,000 (there are eight).
# --hints--
-`amicablePairsUpTo`是一个函数。
+`amicablePairsUpTo` should be a function.
```js
assert(typeof amicablePairsUpTo === 'function');
```
-`[[220,284]]` `amicablePairsUpTo(300)`应返回`[[220,284]]` 。
+`amicablePairsUpTo(300)` should return `[[220,284]]`.
```js
assert.deepEqual(amicablePairsUpTo(300), answer300);
```
-`[[220,284],[1184,1210],[2620,2924]]` `amicablePairsUpTo(3000)`应返回`[[220,284],[1184,1210],[2620,2924]]` 。
+`amicablePairsUpTo(3000)` should return `[[220,284],[1184,1210],[2620,2924]]`.
```js
assert.deepEqual(amicablePairsUpTo(3000), answer3000);
```
-`[[220,284],[1184,1210],[2620,2924],[5020,5564],[6232,6368],[10744,10856],[12285,14595],[17296,18416]]` `amicablePairsUpTo(20000)`应返回`[[220,284],[1184,1210],[2620,2924],[5020,5564],[6232,6368],[10744,10856],[12285,14595],[17296,18416]]` 。
+`amicablePairsUpTo(20000)` should return `[[220,284],[1184,1210],[2620,2924],[5020,5564],[6232,6368],[10744,10856],[12285,14595],[17296,18416]]`.
```js
assert.deepEqual(amicablePairsUpTo(20000), answer20000);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/averages-mode.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/averages-mode.md
index 8e836f1eb32..166050bd516 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/averages-mode.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/averages-mode.md
@@ -1,30 +1,34 @@
---
id: 594d8d0ab97724821379b1e6
-title: 平均值模式
+title: Averages/Mode
challengeType: 5
-videoUrl: ''
+forumTopicId: 302226
dashedName: averagesmode
---
# --description--
-<p>编写程序以查找集合的<a href='https://en.wikipedia.org/wiki/Mode (statistics)' title='wp：模式（统计）'>模式</a>值。 </p><p>可以忽略集合为空的情况。必须小心处理模式不唯一的情况。 </p><p>如果不适合或不可能支持常规集合，请尽可能使用向量（数组）。如果不适合或不可能支持未指定的值类型，请使用整数。 </p>
+Write a program to find the [mode](https://en.wikipedia.org/wiki/Mode (statistics) "wp: Mode (statistics)") value of a collection.
+
+The case where the collection is empty may be ignored. Care must be taken to handle the case where the mode is non-unique.
+
+If it is not appropriate or possible to support a general collection, use a vector (array), if possible. If it is not appropriate or possible to support an unspecified value type, use integers.
# --hints--
-`mode`是一种功能。
+`mode` should be a function.
```js
assert(typeof mode === 'function');
```
-`mode([1, 3, 6, 6, 6, 6, 7, 7, 12, 12, 17])`应该相等`[6]`
+`mode([1, 3, 6, 6, 6, 6, 7, 7, 12, 12, 17])` should equal `[6]`
```js
assert.deepEqual(mode(arr1), [6]);
```
-`mode([1, 2, 4, 4, 1])`应该等于`[1, 4]` 。
+`mode([1, 2, 4, 4, 1])` should equal `[1, 4]`.
```js
assert.deepEqual(mode(arr2).sort(), [1, 4]);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/averages-pythagorean-means.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/averages-pythagorean-means.md
index 588affe26fd..45425141e3b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/averages-pythagorean-means.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/averages-pythagorean-means.md
@@ -1,32 +1,52 @@
---
id: 594d966a1467eb84194f0086
-title: 平均值 - 毕达哥拉斯指的是
+title: Averages/Pythagorean means
challengeType: 5
-videoUrl: ''
+forumTopicId: 302227
dashedName: averagespythagorean-means
---
# --description--
-<p class='rosetta__paragraph'>计算整数<big>1</big>到<big>10</big> （包括）的所有三个<a class='rosetta__link--wiki' href='https://en.wikipedia.org/wiki/Pythagorean means' title='wp：毕达哥拉斯意味着'>毕达哥拉斯方法</a> 。 </p><p class='rosetta__paragraph'>为这组正整数显示<big>$ A（x_1，\ ldots，x_n）\ geq G（x_1，\ ldots，x_n）\ geq H（x_1，\ ldots，x_n）$</big> 。 </p>这三种方法中最常见的<a class='rosetta__link--rosetta' href='http://rosettacode.org/wiki/Averages/Arithmetic mean' title='平均值/算术平均值'>算术平均值</a>是列表的总和除以其长度： <big>$ A（x_1，\\ ldots，x_n）= \\ frac {x_1 + \\ cdots + x_n} {n} $</big> <a class='rosetta__link--wiki' href='https://en.wikipedia.org/wiki/Geometric mean' title='wp：几何平均值'>几何mean</a>是列表产品的$ n $ th根： <big>$ G（x_1，\\ ldots，x_n）= \\ sqrt \[n] {x_1 \\ cdots x_n} $</big> <a class='rosetta__link--wiki' href='https://en.wikipedia.org/wiki/Harmonic mean' title='wp：谐波意味着'>调和平均值</a>是$ n $除以总和列表中每个项目的倒数： <big>$ H（x_1，\\ ldots，x_n）= \\ frac {n} {\\ frac {1} {x_1} + \\ cdots + \\ frac {1} {x_n}} $</big> <p class='rosetta__paragraph'>假设输入是包含所有数字的有序数组。 </p><p class='rosetta__paragraph'>要获得答案，请按以下格式输出对象： </p><pre class='rosetta__pre'> {
- 值：{
- 算术：5.5，
- 几何：4.528728688116765，
- 谐波：3.414171521474055
- }，
- 测试：'是A> = G> = H？是'
+Compute all three of the [Pythagorean means](https://en.wikipedia.org/wiki/Pythagorean means "wp: Pythagorean means") of the set of integers $1$ through $10$ (inclusive).
+
+Show that $A(x_1,\\ldots,x_n) \\geq G(x_1,\\ldots,x_n) \\geq H(x_1,\\ldots,x_n)$ for this set of positive integers.
+
+<ul>
+ <li>The most common of the three means, the <a class='rosetta__link--rosetta' href='https://rosettacode.org/wiki/Averages/Arithmetic mean' title='Averages/Arithmetic mean' target='_blank'>arithmetic mean</a>, is the sum of the list divided by its length:<br>
+ <big>$ A(x_1, \ldots, x_n) = \frac{x_1 + \cdots + x_n}{n}$</big></li>
+ <li>The <a class='rosetta__link--wiki' href='https://en.wikipedia.org/wiki/Geometric mean' title='wp: Geometric mean' target='_blank'>geometric mean</a> is the $n$th root of the product of the list:<br>
+ <big>$ G(x_1, \ldots, x_n) = \sqrt[n]{x_1 \cdots x_n} $</big></li>
+ <li>The <a class='rosetta__link--wiki' href='https://en.wikipedia.org/wiki/Harmonic mean' title='wp: Harmonic mean' target='_blank'>harmonic mean</a> is $n$ divided by the sum of the reciprocal of each item in the list:<br>
+ <big>$ H(x_1, \ldots, x_n) = \frac{n}{\frac{1}{x_1} + \cdots + \frac{1}{x_n}} $</big></li>
+</ul>
+
+# --instructions--
+
+When writing your function, assume the input is an ordered array of all inclusive numbers.
+
+For the answer, please output an object in the following format:
+
+```js
+{
+ values: {
+ Arithmetic: 5.5,
+ Geometric: 4.528728688116765,
+ Harmonic: 3.414171521474055
+ },
+ test: 'is A >= G >= H ? yes'
}
-</pre>
+```
# --hints--
-`pythagoreanMeans`是一种功能。
+`pythagoreanMeans` should be a function.
```js
assert(typeof pythagoreanMeans === 'function');
```
-`pythagoreanMeans([1, 2, ..., 10])`应该等于上面相同的输出。
+`pythagoreanMeans([1, 2, ..., 10])` should equal the same output above.
```js
assert.deepEqual(pythagoreanMeans(range1), answer1);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/averages-root-mean-square.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/averages-root-mean-square.md
index 22267fa5e25..d7bb3ef969c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/averages-root-mean-square.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/averages-root-mean-square.md
@@ -1,24 +1,30 @@
---
id: 594da033de4190850b893874
-title: 平均值 - 均方根
+title: Averages/Root mean square
challengeType: 5
-videoUrl: ''
+forumTopicId: 302228
dashedName: averagesroot-mean-square
---
# --description--
-<p>计算数字1到10（包括1和10）的<a href='https://en.wikipedia.org/wiki/Root mean square' title='wp：均方根'>均方根</a> 。 </p><p>均方根也以其首字母RMS（或rms）和二次均值来表示。 </p><p> RMS计算为数字平方的平均值，平方根： </p><p> <big>$$ x _ {\ mathrm {rms}} = \ sqrt {{{x_1} ^ 2 + {x_2} ^ 2 + \ cdots + {x_n} ^ 2} \ over n}。 $$</big> </p>
+Compute the [Root mean square](https://en.wikipedia.org/wiki/Root mean square "wp: Root mean square") of the numbers 1 through 10 inclusive.
+
+The *root mean square* is also known by its initials RMS (or rms), and as the **quadratic mean**.
+
+The RMS is calculated as the mean of the squares of the numbers, square-rooted:
+
+$$x\_{\\mathrm{rms}} = \\sqrt {{{x_1}^2 + {x_2}^2 + \\cdots + {x_n}^2} \\over n}. $$
# --hints--
-`rms`是一个功能。
+`rms` should be a function.
```js
assert(typeof rms === 'function');
```
-`rms([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])`应等于`6.2048368229954285` 。
+`rms([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])` should equal `6.2048368229954285`.
```js
assert.equal(rms(arr1), answer1);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/babbage-problem.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/babbage-problem.md
index 30e2b661f93..27054cccc54 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/babbage-problem.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/babbage-problem.md
@@ -1,24 +1,37 @@
---
id: 594db4d0dedb4c06a2a4cefd
-title: 巴贝奇问题
+title: Babbage problem
challengeType: 5
-videoUrl: ''
+forumTopicId: 302229
dashedName: babbage-problem
---
# --description--
-<p> <a href='https://en.wikipedia.org/wiki/Charles_Babbage' title='wp：Charles_Babbage'>Charles Babbage</a> ，展望他的分析引擎能解决的各种问题，给出了这个例子： </p><blockquote>什么是正方形以数字269,696结尾的最小正整数？ </blockquote><p> - 巴贝奇，致鲍登勋爵的信，1837年;见Hollingdale和Tootill， <i>电子计算机</i> ，第二版，1970年，p。 125。 </p><p>他认为答案可能是99,736，其平方是9,947,269,696;但他无法确定。 </p><p>任务是找出巴贝奇是否有正确的答案。 </p><p>实现一个函数来返回满足Babbage问题的最小整数。如果巴贝奇是对的，返回巴贝奇的号码。 </p>
+[Charles Babbage](https://en.wikipedia.org/wiki/Charles_Babbage "wp: Charles_Babbage"), looking ahead to the sorts of problems his Analytical Engine would be able to solve, gave this example:
+
+<blockquote>
+ What is the smallest positive integer whose square ends in the digits 269,696?
+ <footer style='margin-left: 2em;'>Babbage, letter to Lord Bowden, 1837; see Hollingdale and Tootill, <i>Electronic Computers</i>, second edition, 1970, p. 125.</footer>
+</blockquote>
+
+He thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.
+
+The task is to find out if Babbage had the right answer.
+
+# --instructions--
+
+Implement a function to return the lowest integer that satisfies the Babbage problem. If Babbage was right, return Babbage's number.
# --hints--
-`babbage`是一种功能。
+`babbage` should be a function.
```js
assert(typeof babbage === 'function');
```
-`babbage(99736, 269696)`不应该返回99736（答案较小）。
+`babbage(99736, 269696)` should not return 99736 (there is a smaller answer).
```js
assert.equal(babbage(babbageAns, endDigits), answer);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/balanced-brackets.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/balanced-brackets.md
index f2155e78e27..5045fbd2cad 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/balanced-brackets.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/balanced-brackets.md
@@ -1,126 +1,136 @@
---
id: 594dc6c729e5700999302b45
-title: 平衡括号
+title: Balanced brackets
challengeType: 5
-videoUrl: ''
+forumTopicId: 302230
dashedName: balanced-brackets
---
# --description--
-<p>确定生成的括号字符串是否平衡;也就是说，它是否完全由成对的开/关括号（按此顺序）组成，其中没有一个错误嵌套。 </p>例子： <p class='rosetta__paragraph'> （空）是的</p><p class='rosetta__paragraph'> <code>[]</code>是的</p><p class='rosetta__paragraph'> <code>][</code>假</p><p class='rosetta__paragraph'> <code>[][]</code>是的</p><p class='rosetta__paragraph'> <code>][][</code>假</p><p class='rosetta__paragraph'> <code>[]][[]</code> false </p><p class='rosetta__paragraph'> <code>[[[[]]]]</code>是的</p>
+Determine whether a generated string of brackets is balanced; that is, whether it consists entirely of pairs of opening/closing brackets (in that order), none of which mis-nest.
+
+**Examples:**
+| Input | Output |
+| ------------------------- | ------ |
+| <code>\[]</code> | true |
+| <code>]\[</code> | false |
+| <code>[][]</code> | true |
+| <code>]\[]</code> | false |
+| <code>\[]]\[\[]</code> | false |
+| <code>\[\[\[\[]]]]</code> | true |
# --hints--
-`isBalanced`是一个函数。
+`isBalanced` should be a function.
```js
assert(typeof isBalanced === 'function');
```
-`isBalanced("[]")`应该返回true。
+`isBalanced("[]")` should return true.
```js
assert(isBalanced(testCases[0]));
```
-`isBalanced("]][[[][][][]][")`应该返回false。
+`isBalanced("]][[[][][][]][")` should return false.
```js
assert(!isBalanced(testCases[1]));
```
-`isBalanced("[][[[[][][[[]]]]]]")`应该返回true。
+`isBalanced("[][[[[][][[[]]]]]]")` should return true.
```js
assert(isBalanced(testCases[2]));
```
-`isBalanced("][")`应该返回true。
+`isBalanced("][")` should return false.
```js
assert(!isBalanced(testCases[3]));
```
-`isBalanced("[[[]]]][[]")`应该返回true。
+`isBalanced("[[[]]]][[]")` should return false.
```js
assert(!isBalanced(testCases[4]));
```
-`isBalanced("][[]")`应该返回true。
+`isBalanced("][[]")` should return false.
```js
assert(!isBalanced(testCases[5]));
```
-`isBalanced("][[][]][[[]]")`应该返回true。
+`isBalanced("][[][]][[[]]")` should return false.
```js
assert(!isBalanced(testCases[6]));
```
-`isBalanced("[[][]]][")`应该返回true。
+`isBalanced("[[][]]][")` should return false.
```js
assert(!isBalanced(testCases[7]));
```
-`isBalanced("[[[]]][[]]]][][[")`应该返回true。
+`isBalanced("[[[]]][[]]]][][[")` should return false.
```js
assert(!isBalanced(testCases[8]));
```
-`isBalanced("[]][[]]][[[[][]]")`应该返回true。
+`isBalanced("[]][[]]][[[[][]]")` should return false.
```js
assert(!isBalanced(testCases[9]));
```
-`isBalanced("][]][[][")`应该返回true。
+`isBalanced("][]][[][")` should return false.
```js
assert(!isBalanced(testCases[10]));
```
-`isBalanced("[[]][[][]]")`应该返回true。
+`isBalanced("[[]][[][]]")` should return true.
```js
assert(isBalanced(testCases[11]));
```
-`isBalanced("[[]]")`应该返回true。
+`isBalanced("[[]]")` should return true.
```js
assert(isBalanced(testCases[12]));
```
-`isBalanced("]][]][[]][[[")`应该返回true。
+`isBalanced("]][]][[]][[[")` should return false.
```js
assert(!isBalanced(testCases[13]));
```
-`isBalanced("][]][][[")`应该返回true。
+`isBalanced("][]][][[")` should return false.
```js
assert(!isBalanced(testCases[14]));
```
-`isBalanced("][][")`应该返回true。
+`isBalanced("][][")` should return false.
```js
assert(!isBalanced(testCases[15]));
```
-`isBalanced("[[]]][][][[]][")`应该返回true。
+`isBalanced("[]]]")` should return false.
```js
assert(!isBalanced(testCases[16]));
```
-`isBalanced("")`应该返回true。
+`isBalanced("")` should return true.
```js
assert(isBalanced(testCases[17]));
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/circles-of-given-radius-through-two-points.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/circles-of-given-radius-through-two-points.md
index 589b93f6fee..bbd910fb6c5 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/circles-of-given-radius-through-two-points.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/circles-of-given-radius-through-two-points.md
@@ -1,54 +1,79 @@
---
id: 5951815dd895584b06884620
-title: 给定半径的圆圈通过两个点
+title: Circles of given radius through two points
challengeType: 5
-videoUrl: ''
+forumTopicId: 302231
dashedName: circles-of-given-radius-through-two-points
---
# --description--
-<p>给定平面上的两个点和半径，通常可以通过这些点绘制给定半径的两个圆。 </p>例外：零半径应视为从不描述圆（除非点是重合的）。如果这些点是重合的，则可以绘制无限数量的圆，其圆周上的点可以被绘制，除非半径也等于零，然后将圆圈折叠到一个点。如果点形成直径，则返回单个圆。如果这些点相距太远则无法绘制圆圈。任务：实现一个取两个点和一个半径的函数，并通过这些点返回两个圆。对于每个结果圆，提供每个圆的中心的坐标，四舍五入到四个十进制数字。将每个坐标作为数组返回，并作为数组数组进行坐标。对于边缘情况，请返回以下内容：如果点在直径上，则返回一个点。如果半径也为零，则返回<code>"Radius Zero"</code> 。如果点重合，则返回<code>"Coincident point. Infinite solutions"</code> 。如果点与直径相距更远，则返回<code>"No intersection. Points further apart than circle diameter"</code>更远的<code>"No intersection. Points further apart than circle diameter"</code> 。样本输入： <pre> p1 p2 r
-0.1234,0.9876 0.8765,0.2345 2.0
-0.0000,2.000000 0.0000,0.0000 1.0
-0.1234,0.9876 0.1234,0.9876 2.0
-0.1234,0.9876 0.8765,0.2345 0.5
-0.1234,0.9876 0.1234,0.9876 0.0
-</pre>参考：从数学论坛@Drexel的<a href='http://mathforum.org/library/drmath/view/53027.html' title='链接：http：//mathforum.org/library/drmath/view/53027.html'>2点和半径中找到一个</a>圆心
+Given two points on a plane and a radius, usually two circles of given radius can be drawn through the points.
+
+**Exceptions:**
+
+<ul>
+ <li>A radius of zero should be treated as never describing circles (except in the case where the points are coincident).</li>
+ <li>If the points are coincident then an infinite number of circles with the point on their circumference can be drawn, unless the radius is equal to zero as well which then collapses the circles to a point.</li>
+ <li>If the points form a diameter then return a single circle.</li>
+ <li>If the points are too far apart then no circles can be drawn.</li>
+</ul>
+
+# --instructions--
+
+Implement a function that takes two points and a radius and returns the two circles through those points. For each resulting circle, provide the coordinates for the center of each circle rounded to four decimal digits. Return each coordinate as an array, and coordinates as an array of arrays.
+
+**For edge cases, return the following:**
+
+<ul>
+ <li>If points are on the diameter, return one point. If the radius is also zero however, return <code>"Radius Zero"</code>.</li>
+ <li>If points are coincident, return <code>"Coincident point. Infinite solutions"</code>.</li>
+ <li>If points are farther apart than the diameter, return <code>"No intersection. Points further apart than circle diameter"</code>.</li>
+</ul>
+
+**Sample inputs:**
+
+<pre> p1 p2 r
+0.1234, 0.9876 0.8765, 0.2345 2.0
+0.0000, 2.0000 0.0000, 0.0000 1.0
+0.1234, 0.9876 0.1234, 0.9876 2.0
+0.1234, 0.9876 0.8765, 0.2345 0.5
+0.1234, 0.9876 0.1234, 0.9876 0.0
+</pre>
# --hints--
-`getCircles`是一个函数。
+`getCircles` should be a function.
```js
assert(typeof getCircles === 'function');
```
-`getCircles([0.1234, 0.9876], [0.8765, 0.2345], 2.0)`应该返回`[[1.8631, 1.9742], [-0.8632, -0.7521]]` 。
+`getCircles([0.1234, 0.9876], [0.8765, 0.2345], 2.0)` should return `[[1.8631, 1.9742], [-0.8632, -0.7521]]`.
```js
assert.deepEqual(getCircles(...testCases[0]), answers[0]);
```
-`getCircles([0.0000, 2.0000], [0.0000, 0.0000], 1.0)`应该返回`[0, 1]`
+`getCircles([0.0000, 2.0000], [0.0000, 0.0000], 1.0)` should return `[0, 1]`
```js
assert.deepEqual(getCircles(...testCases[1]), answers[1]);
```
-`getCircles([0.1234, 0.9876], [0.1234, 0.9876], 2.0)`应返回`Coincident point. Infinite solutions`
+`getCircles([0.1234, 0.9876], [0.1234, 0.9876], 2.0)` should return `Coincident point. Infinite solutions`
```js
assert.deepEqual(getCircles(...testCases[2]), answers[2]);
```
-`getCircles([0.1234, 0.9876], [0.8765, 0.2345], 0.5)`应返回`No intersection. Points further apart than circle diameter`
+`getCircles([0.1234, 0.9876], [0.8765, 0.2345], 0.5)` should return `No intersection. Points further apart than circle diameter`
```js
assert.deepEqual(getCircles(...testCases[3]), answers[3]);
```
-`getCircles([0.1234, 0.9876], [0.1234, 0.9876], 0.0)`应返回`Radius Zero`
+`getCircles([0.1234, 0.9876], [0.1234, 0.9876], 0.0)` should return `Radius Zero`
```js
assert.deepEqual(getCircles(...testCases[4]), answers[4]);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/closest-pair-problem.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/closest-pair-problem.md
index 048a3df4077..3ca20812317 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/closest-pair-problem.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/closest-pair-problem.md
@@ -1,86 +1,96 @@
---
id: 5951a53863c8a34f02bf1bdc
-title: 最近对的问题
+title: Closest-pair problem
challengeType: 5
-videoUrl: ''
+forumTopicId: 302232
dashedName: closest-pair-problem
---
# --description--
-任务：
+Provide a function to find the closest two points among a set of given points in two dimensions, i.e. to solve the [Closest pair of points problem](https://en.wikipedia.org/wiki/Closest pair of points problem "wp: Closest pair of points problem") in the *planar* case.
-提供一个函数来在二维中找到一组给定点中最接近的两个点，即求解平面情况下的[最近点对问题](<https://en.wikipedia.org/wiki/Closest pair of points problem> "wp：最近点的问题") 。
+The straightforward solution is a O(n<sup>2</sup>) algorithm (which we can call *brute-force algorithm*); the pseudo-code (using indexes) could be simply:
-直接的解决方案是O（n <sup>2</sup> ）算法（我们可以称之为强力算法）;伪代码（使用索引）可以简单地：
+<pre><strong>bruteForceClosestPair</strong> of P(1), P(2), ... P(N)
+<strong>if</strong> N < 2 <strong>then</strong>
+ <strong>return</strong> ∞
+<strong>else</strong>
+ minDistance ← |P(1) - P(2)|
+ minPoints ← { P(1), P(2) }
+ <strong>foreach</strong> i ∈ [1, N-1]
+ <strong>foreach</strong> j ∈ [i+1, N]
+ <strong>if</strong> |P(i) - P(j)| < minDistance <strong>then</strong>
+ minDistance ← |P(i) - P(j)|
+ minPoints ← { P(i), P(j) }
+ <strong>endif</strong>
+ <strong>endfor</strong>
+ <strong>endfor</strong>
+ <strong>return</strong> minDistance, minPoints
+<strong>endif</strong>
+</pre>
-```
- bruteForceClosestPair of P（1），P（2），... P（N）
-如果N <2那么
-返回∞
-其他
-minDistance←| P（1） - P（2）|
-minPoints←{P（1），P（2）}
-foreachi∈[1，N-1]
-foreachj∈[i + 1，N]
-if | P（i） - P（j）| <minDistance然后
-minDistance←| P（i） - P（j）|
-minPoints←{P（i），P（j）}
-万一
-ENDFOR
-ENDFOR
-return minDistance，minPoints
-万一
-```
+A better algorithm is based on the recursive divide and conquer approach, as explained also at [Wikipedia's Closest pair of points problem](https://en.wikipedia.org/wiki/Closest pair of points problem#Planar_case "wp: Closest pair of points problem#Planar_case"), which is `O(nlog(n))` a pseudo-code could be:
-</pre><p>一个更好的算法是基于递归分而治之的方法，正如<a href='https://en.wikipedia.org/wiki/Closest pair of points problem#Planar_case' title='wp：最近点的问题#Planar_case'>维基百科最近的一对点问题</a>所解释的那样，即O（n log n）;伪代码可以是： </p><pre>最近的（xP，yP）
- 其中xP是P（1）.. P（N）按x坐标排序，和
- yP是P（1）.. P（N）按y坐标排序（升序）
-如果N≤3那么
- 使用强力算法返回xP的最近点
-其他
- xL←xP点从1到⌈N/2⌉
- xR←xP点从⌈N/2⌉+ 1到N.
- xm←xP（⌈N/2⌉） <sub>x</sub>
- 基←{P∈YP：P <sub>X≤XM}</sub>
- yR←{p∈yP：p <sub>x</sub> > xm}
- （dL，pairL）←nearestPair（xL，yL）
- （dR，pairR）←nearestRair（xR，yR）
- （dmin，pairMin）←（dR，pairR）
- 如果dL <dR则
- （dmin，pairMin）←（dL，pairL）
- 万一
- yS←{p∈yP：| xm - p <sub>x</sub> | <dmin}
- nS←yS中的点数
- （最近，最近的公里）←（dmin，pairMin）
- 我从1到nS - 1
- k←i + 1
- 而k≤nS和yS（k） <sub>y</sub> -yS（i） <sub>y</sub> <dmin
- if | yS（k） - yS（i）| <最接近的
- （最近，最近的公里）←（| yS（k） - yS（i）|，{yS（k），yS（i）}）
- 万一
- k←k + 1
- ENDWHILE
- ENDFOR
- 返回最近，最近的
-万一
-</pre>参考和进一步阅读： <a href='https://en.wikipedia.org/wiki/Closest pair of points problem' title='wp：最近点的问题'>最近的一对点问题</a> <a href='http://www.cs.mcgill.ca/~cs251/ClosestPair/ClosestPairDQ.html' title='链接：http：//www.cs.mcgill.ca/~cs251/ClosestPair/ClosestPairDQ.html'>最近的一对（麦吉尔）</a> <a href='http://www.cs.ucsb.edu/~suri/cs235/ClosestPair.pdf' title='链接：http：//www.cs.ucsb.edu/~suri/cs235/ClosestPair.pdf'>最近的一对（UCSB）</a> <a href='http://classes.cec.wustl.edu/~cse241/handouts/closestpair.pdf' title='链接：http：//classes.cec.wustl.edu/~cse241/handouts/closestpair.pdf'>最近的一对（WUStL）</a> <a href='http://www.cs.iupui.edu/~xkzou/teaching/CS580/Divide-and-conquer-closestPair.ppt' title='链接：http：//www.cs.iupui.edu/~xkzou/teaching/CS580/Divide-and-conquer-closestPair.ppt'>最近的一对（IUPUI）</a> <p>对于输入，期望参数是一个对象（点）数组，其中<code>x</code>和<code>y</code>成员设置为数字。对于输出，返回一个包含键的对象： <code>distance</code>和<code>pair</code>值对（即两对最近点的对）。 </p>
+<pre><strong>closestPair</strong> of (xP, yP)
+ where xP is P(1) .. P(N) sorted by x coordinate, and
+ yP is P(1) .. P(N) sorted by y coordinate (ascending order)
+<strong>if</strong> N ≤ 3 <strong>then</strong>
+ <strong>return</strong> closest points of xP using brute-force algorithm
+<strong>else</strong>
+ xL ← points of xP from 1 to ⌈N/2⌉
+ xR ← points of xP from ⌈N/2⌉+1 to N
+ xm ← xP(⌈N/2⌉)<sub>x</sub>
+ yL ← { p ∈ yP : p<sub>x</sub> ≤ xm }
+ yR ← { p ∈ yP : p<sub>x</sub> > xm }
+ (dL, pairL) ← closestPair of (xL, yL)
+ (dR, pairR) ← closestPair of (xR, yR)
+ (dmin, pairMin) ← (dR, pairR)
+ <strong>if</strong> dL < dR <strong>then</strong>
+ (dmin, pairMin) ← (dL, pairL)
+ <strong>endif</strong>
+ yS ← { p ∈ yP : |xm - p<sub>x</sub>| < dmin }
+ nS ← number of points in yS
+ (closest, closestPair) ← (dmin, pairMin)
+ <strong>for</strong> i <strong>from</strong> 1 <strong>to</strong> nS - 1
+ k ← i + 1
+ <strong>while</strong> k ≤ nS <strong>and</strong> yS(k)<sub>y</sub> - yS(i)<sub>y</sub> < dmin
+ <strong>if</strong> |yS(k) - yS(i)| < closest <strong>then</strong>
+ (closest, closestPair) ← (|yS(k) - yS(i)|, {yS(k), yS(i)})
+ <strong>endif</strong>
+ k ← k + 1
+ <strong>endwhile</strong>
+ <strong>endfor</strong>
+ <strong>return</strong> closest, closestPair
+<strong>endif</strong>
+</pre>
+
+For the input, expect the argument to be an array of objects (points) with `x` and `y` members set to numbers. For the output, return an object containing the key:value pairs for `distance` and `pair` (the pair of two closest points).
+
+**References and further readings:**
+
+<ul>
+ <li><a href='https://en.wikipedia.org/wiki/Closest pair of points problem' title='wp: Closest pair of points problem' target='_blank'>Closest pair of points problem</a></li>
+ <li><a href='https://www.cs.mcgill.ca/~cs251/ClosestPair/ClosestPairDQ.html' target='_blank'>Closest Pair (McGill)</a></li>
+ <li><a href='https://www.cs.ucsb.edu/~suri/cs235/ClosestPair.pdf' target='_blank'>Closest Pair (UCSB)</a></li>
+ <li><a href='https://classes.cec.wustl.edu/~cse241/handouts/closestpair.pdf' target='_blank'>Closest pair (WUStL)</a></li>
+ </ul>
# --hints--
-`getClosestPair`是一个函数。
+`getClosestPair` should be a function.
```js
assert(typeof getClosestPair === 'function');
```
-距离应如下。
+Distance should be the following.
```js
assert.equal(getClosestPair(points1).distance, answer1.distance);
```
-要点应如下。
+Points should be the following.
```js
assert.deepEqual(
@@ -89,13 +99,13 @@ assert.deepEqual(
);
```
-距离应如下。
+Distance should be the following.
```js
assert.equal(getClosestPair(points2).distance, answer2.distance);
```
-要点应如下。
+Points should be the following.
```js
assert.deepEqual(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/combinations.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/combinations.md
index b542e83ed4d..27e9a3fa6d4 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/combinations.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/combinations.md
@@ -1,23 +1,20 @@
---
id: 5958469238c0d8d2632f46db
-title: 组合
+title: Combinations
challengeType: 5
-videoUrl: ''
+forumTopicId: 302233
dashedName: combinations
---
# --description--
-任务：
+Given non-negative integers `m` and `n`, generate all size `m` combinations of the integers from `0` (zero) to `n-1` in sorted order (each combination is sorted and the entire table is sorted).
-给定非负整数m和n ，以排序顺序生成从0 （零）到n-1的整数的所有大小m个 [组合](http://mathworld.wolfram.com/Combination.html "链接：http：//mathworld.wolfram.com/Combination.html") （每个组合被排序并且整个表被排序）。
+**Example:**
-例：
+`3` comb `5` is:
-3梳子5是：
-
-```
- 0 1 2
+<pre>0 1 2
0 1 3
0 1 4
0 2 3
@@ -27,25 +24,23 @@ dashedName: combinations
1 2 4
1 3 4
2 3 4
-```
-
</pre>
# --hints--
-`combinations`是一种功能。
+`combinations` should be a function.
```js
assert(typeof combinations === 'function');
```
-`combinations(3, 5)`应返回`[[0, 1, 2], [0, 1, 3], [0, 1, 4], [0, 2, 3], [0, 2, 4], [0, 3, 4], [1, 2, 3], [1, 2, 4], [1, 3, 4], [2, 3, 4]]` 。
+`combinations(3, 5)` should return `[[0, 1, 2], [0, 1, 3], [0, 1, 4], [0, 2, 3], [0, 2, 4], [0, 3, 4], [1, 2, 3], [1, 2, 4], [1, 3, 4], [2, 3, 4]]`.
```js
assert.deepEqual(combinations(testInput1[0], testInput1[1]), testOutput1);
```
-`combinations(4, 6)`应返回`[[0,1,2,3], [0,1,2,4], [0,1,2,5], [0,1,3,4], [0,1,3,5], [0,1,4,5], [0,2,3,4], [0,2,3,5], [0,2,4,5], [0,3,4,5], [1,2,3,4], [1,2,3,5], [1,2,4,5], [1,3,4,5], [2,3,4,5]]`
+`combinations(4, 6)` should return `[[0,1,2,3], [0,1,2,4], [0,1,2,5], [0,1,3,4], [0,1,3,5], [0,1,4,5], [0,2,3,4], [0,2,3,5], [0,2,4,5], [0,3,4,5], [1,2,3,4], [1,2,3,5], [1,2,4,5], [1,3,4,5], [2,3,4,5]]`
```js
assert.deepEqual(combinations(testInput2[0], testInput2[1]), testOutput2);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/comma-quibbling.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/comma-quibbling.md
index a02023d1392..6223a523cee 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/comma-quibbling.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/comma-quibbling.md
@@ -1,48 +1,70 @@
---
id: 596e414344c3b2872167f0fe
-title: 逗号狡猾
+title: Comma quibbling
challengeType: 5
-videoUrl: ''
+forumTopicId: 302234
dashedName: comma-quibbling
---
# --description--
-<p> Comma quibbling是Eric Lippert在他的<a href='http://blogs.msdn.com/b/ericlippert/archive/2009/04/15/comma-quibbling.aspx' title='链接：http：//blogs.msdn.com/b/ericlippert/archive/2009/04/15/comma-quibbling.aspx'>博客中</a>最初设定的任务。 </p>任务： <p>编写一个函数来生成一个字符串输出，它是列表/序列中输入字的串联，其中： </p>没有单词的输入产生仅两个大括号字符“{}”的输出字符串。只有一个单词的输入，例如\[“ABC”]，会在两个大括号内产生单词的输出字符串，例如“{ABC}”。两个单词的输入，例如\[“ABC”，“DEF”]，产生两个大括号内的两个单词的输出字符串，其中单词由字符串“和”分隔，例如“{ABC和DEF}”。三个或更多单词的输入，例如\[“ABC”，“DEF”，“G”，“H”]，产生除了最后一个单词之外的所有输出字符串，用“，”分隔，最后一个单词用“和”分隔。 “并且都在括号内;例如“{ABC，DEF，G和H}”。 <p>在此页面上显示输出的以下一系列输入测试您的功能： </p> \[]＃（无输入字）。 \[“ABC”] \[“ABC”，“DEF”] \[“ABC”，“DEF”，“G”，“H”] <p>注意：假设此单词是此任务的非空字符串大写字符。 </p>
+Comma quibbling is a task originally set by Eric Lippert in his [blog](https://blogs.msdn.com/b/ericlippert/archive/2009/04/15/comma-quibbling.aspx).
+
+# --instructions--
+
+Write a function to generate a string output which is the concatenation of input words from a list/sequence where:
+
+<ol>
+ <li>An input of no words produces the output string of just the two brace characters (<code>"{}"</code>)</li>
+ <li>An input of just one word, e.g. <code>["ABC"]</code>, produces the output string of the word inside the two braces, e.g. <code>"{ABC}"</code></li>
+ <li>An input of two words, e.g. <code>["ABC", "DEF"]</code>, produces the output string of the two words inside the two braces with the words separated by the string <code>" and "</code>, e.g. <code>"{ABC and DEF}"</code></li>
+ <li>An input of three or more words, e.g. <code>["ABC", "DEF", "G", "H"]</code>, produces the output string of all but the last word separated by <code>", "</code> with the last word separated by <code>" and "</code> and all within braces; e.g. <code>"{ABC, DEF, G and H}"</code></li>
+</ol>
+
+Test your function with the following series of inputs showing your output here on this page:
+
+<ul>
+ <li>[] # (No input words).</li>
+ <li>["ABC"]</li>
+ <li>["ABC", "DEF"]</li>
+ <li>["ABC", "DEF", "G", "H"]</li>
+</ul>
+
+**Note:** Assume words are non-empty strings of uppercase characters for this task.
# --hints--
-`quibble`是一种功能。
+`quibble` should be a function.
```js
assert(typeof quibble === 'function');
```
-`quibble(["ABC"])`应该返回一个字符串。
+`quibble(["ABC"])` should return a string.
```js
assert(typeof quibble(['ABC']) === 'string');
```
-`quibble([])`应返回“{}”。
+`quibble([])` should return "{}".
```js
assert.equal(quibble(testCases[0]), results[0]);
```
-`quibble(["ABC"])`应该返回“{ABC}”。
+`quibble(["ABC"])` should return "{ABC}".
```js
assert.equal(quibble(testCases[1]), results[1]);
```
-`quibble(["ABC", "DEF"])`应返回“{ABC和DEF}”。
+`quibble(["ABC", "DEF"])` should return "{ABC and DEF}".
```js
assert.equal(quibble(testCases[2]), results[2]);
```
-`quibble(["ABC", "DEF", "G", "H"])`应返回“{ABC，DEF，G和H}”。
+`quibble(["ABC", "DEF", "G", "H"])` should return "{ABC,DEF,G and H}".
```js
assert.equal(quibble(testCases[3]), results[3]);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/compare-a-list-of-strings.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/compare-a-list-of-strings.md
index 01132febbba..313cda7dcc3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/compare-a-list-of-strings.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/compare-a-list-of-strings.md
@@ -1,84 +1,89 @@
---
id: 596e457071c35c882915b3e4
-title: 比较字符串列表
+title: Compare a list of strings
challengeType: 5
-videoUrl: ''
+forumTopicId: 302235
dashedName: compare-a-list-of-strings
---
# --description--
-<p>给定一个任意多个字符串的<a href='https://en.wikipedia.org/wiki/List_(abstract_data_type)' title='wp：List_（abstract_data_type）'>列表</a> ，为以下每个条件实现一个函数： </p>测试它们是否都是词法上相等的测试，如果每个字符串在词法上小于它之后的字符串（即列表是否按严格的升序排列）
+Given a [list](https://en.wikipedia.org/wiki/List_(abstract_data_type) "wp: List\_(abstract_data_type)") of arbitrarily many strings, implement a function for each of the following conditions:
+
+<ul>
+ <li>test if they are all lexically equal</li>
+ <li>test if every string is lexically less than the one after it (i.e. whether the list is in strict ascending order)</li>
+</ul>
# --hints--
-`allEqual`是一个函数。
+`allEqual` should be a function.
```js
assert(typeof allEqual === 'function');
```
-`azSorted`是一个函数。
+`azSorted` should be a function.
```js
assert(typeof azSorted === 'function');
```
-`allEqual(["AA", "AA", "AA", "AA"])`返回true。
+`allEqual(["AA", "AA", "AA", "AA"])` should return true.
```js
assert(allEqual(testCases[0]));
```
-`azSorted(["AA", "AA", "AA", "AA"])`返回false。
+`azSorted(["AA", "AA", "AA", "AA"])` should return false.
```js
assert(!azSorted(testCases[0]));
```
-`allEqual(["AA", "ACB", "BB", "CC"])`返回false。
+`allEqual(["AA", "ACB", "BB", "CC"])` should return false.
```js
assert(!allEqual(testCases[1]));
```
-`azSorted(["AA", "ACB", "BB", "CC"])`返回true。
+`azSorted(["AA", "ACB", "BB", "CC"])` should return true.
```js
assert(azSorted(testCases[1]));
```
-`allEqual([])`返回true。
+`allEqual([])` should return true.
```js
assert(allEqual(testCases[2]));
```
-`azSorted([])`返回true。
+`azSorted([])` should return true.
```js
assert(azSorted(testCases[2]));
```
-`allEqual(["AA"])`返回true。
+`allEqual(["AA"])` should return true.
```js
assert(allEqual(testCases[3]));
```
-`azSorted(["AA"])`返回true。
+`azSorted(["AA"])` should return true.
```js
assert(azSorted(testCases[3]));
```
-`allEqual(["BB", "AA"])`返回false。
+`allEqual(["BB", "AA"])` should return false.
```js
assert(!allEqual(testCases[4]));
```
-`azSorted(["BB", "AA"])`返回false。
+`azSorted(["BB", "AA"])` should return false.
```js
assert(!azSorted(testCases[4]));
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/convert-seconds-to-compound-duration.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/convert-seconds-to-compound-duration.md
index 221db573d18..a0a594031ca 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/convert-seconds-to-compound-duration.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/convert-seconds-to-compound-duration.md
@@ -1,66 +1,76 @@
---
id: 596fd036dc1ab896c5db98b1
-title: 将秒转换为复合持续时间
+title: Convert seconds to compound duration
challengeType: 5
-videoUrl: ''
+forumTopicId: 302236
dashedName: convert-seconds-to-compound-duration
---
# --description--
-任务：
+Implement a function which:
-实现一个功能：
+<ul>
+ <li>takes a positive integer representing a duration in seconds as input (e.g., <code>100</code>), and</li>
+ <li>returns a string which shows the same duration decomposed into weeks, days, hours, minutes, and seconds as detailed below (e.g., <code>1 min, 40 sec</code>).</li>
+</ul>
-取一个表示以秒为单位的持续时间的正整数作为输入（例如， `100` ），并返回一个字符串，该字符串显示分解为周，日，小时，分钟和秒的相同持续时间，如下所述（例如，“ `1 min, 40 sec` “）。
+Demonstrate that it passes the following three test-cases:
-证明它通过以下三个测试用例：
+<div style='font-size:115%; font-weight: bold;'>Test Cases</div>
-测试用例
+| Input number | Output number |
+| ------------ | ------------------------- |
+| 7259 | <code>2 hr, 59 sec</code> |
+| 728640059 | <code>1 d</code> |
+| 6000000 | <code>9 wk, 6 d, 10 hr, 40 min</code> |
-| 输入号码 | 输出数量 |
-| ------- | -------------------------- |
-| 7259 | `2 hr, 59 sec` |
-| 86400 | `1 d` |
-| 6000000 | `9 wk, 6 d, 10 hr, 40 min` |
+<div style="font-size:115%; font-weight: bold;">Details</div>
+<ul>
+ <li>
+ The following five units should be used:
-细节
+| Unit | Suffix used in Output | Conversion |
+| ------ | --------------------- | --------------------- |
+| week |!!crwdBlockTags_18_sgaTkcolBdwrc!! | 1 week = 7 days |
+| day |!!crwdBlockTags_19_sgaTkcolBdwrc!! | 1 day = 24 hours |
+| hour |!!crwdBlockTags_20_sgaTkcolBdwrc!! | 1 hour = 60 minutes |
+| minute |!!crwdBlockTags_21_sgaTkcolBdwrc!! | 1 minute = 60 seconds |
+| second |!!crwdBlockTags_22_sgaTkcolBdwrc!! | --- |
-应使用以下五个单位：
-
-| 单元 | 输出中使用的后缀 | 转变 |
-| -- | -------- | --------- |
-| 周 | `wk` | 1周= 7天 |
-| 天 | `d` | 1天= 24小时 |
-| 小时 | `hr` | 1小时= 60分钟 |
-| 分钟 | `min` | 1分钟= 60秒 |
-| 第二 | `sec` | |
-
-但是，仅包括输出中具有非零值的数量（例如，返回“ `1 d` ”而不是“ `0 wk, 1 d, 0 hr, 0 min, 0 sec` ”）。更大的单位优先于较小的单位尽可能（例如，返回`2 min, 10 sec`而不是`1 min, 70 sec`或`130 sec` ）模拟测试用例中显示的格式（从最大单位到最小单位的数量，以逗号+空格分隔;数值和单位每个数量用空格分隔）。
-
-* * *
+ </li>
+ <li>
+ However, <strong>only</strong> include quantities with non-zero values in the output (e.g., return <code>1 d</code> and not <code>0 wk, 1 d, 0 hr, 0 min, 0 sec</code>).
+ </li>
+ <li>
+ Give larger units precedence over smaller ones as much as possible (e.g., return <code>2 min, 10 sec</code> and not <code>1 min, 70 sec</code> or <code>130 sec</code>).
+ </li>
+ <li>
+ Mimic the formatting shown in the test-cases (quantities sorted from largest unit to smallest and separated by comma+space; value and unit of each quantity separated by space).
+ </li>
+</ul>
# --hints--
-`convertSeconds`是一个函数。
+`convertSeconds` should be a function.
```js
assert(typeof convertSeconds === 'function');
```
-`convertSeconds(7259)`应该返回`2 hr, 59 sec` 。
+`convertSeconds(7259)` should return `2 hr, 59 sec`.
```js
assert.equal(convertSeconds(testCases[0]), results[0]);
```
-`convertSeconds(86400)`应返回`1 d` 。
+`convertSeconds(86400)` should return `1 d`.
```js
assert.equal(convertSeconds(testCases[1]), results[1]);
```
-`convertSeconds(6000000)`应该返回`9 wk, 6 d, 10 hr, 40 min` 。
+`convertSeconds(6000000)` should return `9 wk, 6 d, 10 hr, 40 min`.
```js
assert.equal(convertSeconds(testCases[2]), results[2]);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/count-occurrences-of-a-substring.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/count-occurrences-of-a-substring.md
index 7a465a5e961..50d6c16f8dd 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/count-occurrences-of-a-substring.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/count-occurrences-of-a-substring.md
@@ -1,48 +1,49 @@
---
id: 596fda99c69f779975a1b67d
-title: 计算子字符串的出现次数
+title: Count occurrences of a substring
challengeType: 5
-videoUrl: ''
+forumTopicId: 302237
dashedName: count-occurrences-of-a-substring
---
# --description--
-任务：
+Create a function, or show a built-in function, to count the number of non-overlapping occurrences of a substring inside a string.
-创建函数或显示内置函数，以计算字符串中子字符串的非重叠出现次数。
+The function should take two arguments:
-该函数应该有两个参数：
+<ul>
+ <li>the first argument being the string to search, and</li>
+ <li>the second a substring to be searched for.</li>
+</ul>
-第一个参数是要搜索的字符串，第二个参数是要搜索的子字符串。
+It should return an integer count.
-它应该返回一个整数计数。
+The matching should yield the highest number of non-overlapping matches.
-匹配应产生最多数量的非重叠匹配。
-
-通常，这实质上意味着从左到右或从右到左匹配。
+In general, this essentially means matching from left-to-right or right-to-left.
# --hints--
-`countSubstring`是一个函数。
+`countSubstring` should be a function.
```js
assert(typeof countSubstring === 'function');
```
-`countSubstring("the three truths", "th")`应该返回`3` 。
+`countSubstring("the three truths", "th")` should return `3`.
```js
assert.equal(countSubstring(testCases[0], searchString[0]), results[0]);
```
-`countSubstring("ababababab", "abab")`应返回`2` 。
+`countSubstring("ababababab", "abab")` should return `2`.
```js
assert.equal(countSubstring(testCases[1], searchString[1]), results[1]);
```
-`countSubstring("abaabba*bbaba*bbab", "a*b")`应返回`2` 。
+`countSubstring("abaabba*bbaba*bbab", "a*b")` should return `2`.
```js
assert.equal(countSubstring(testCases[2], searchString[2]), results[2]);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/count-the-coins.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/count-the-coins.md
index 1e468e64933..5d900eb107f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/count-the-coins.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/count-the-coins.md
@@ -1,24 +1,46 @@
---
id: 59713bd26bdeb8a594fb9413
-title: 计算硬币
+title: Count the coins
challengeType: 5
-videoUrl: ''
+forumTopicId: 302238
dashedName: count-the-coins
---
# --description--
-<p> <a href='https://en.wikipedia.org/wiki/United_States' title='链接：https：//en.wikipedia.org/wiki/United_States'>美国</a>货币有四种常见硬币： </p>季度（25美分）硬币（10美分）镍（5美分）和便士（1美分） <p>有六种方法可以换15美分： </p>一角钱和一角钱一角钱和5便士3镍2镍和5便士一镍和10便士15便士任务： <p>实现一个功能，以确定使用这些普通硬币改变一美元的方式有多少？ （1美元= 100美分）。 </p>参考： <a href='http://mitpress.mit.edu/sicp/full-text/book/book-Z-H-11.html#%_sec_Temp_52' title='链接：http：//mitpress.mit.edu/sicp/full-text/book/book-Z-H-11.html#%_sec_Temp_52'>麻省理工学院出版社的算法</a> 。
+There are four types of common coins in [US](https://en.wikipedia.org/wiki/United_States) currency:
+
+<ul>
+ <li>quarters (25 cents)</li>
+ <li>dimes (10 cents)</li>
+ <li>nickels (5 cents), and</li>
+ <li>pennies (1 cent)</li>
+</ul>
+
+<p>There are six ways to make change for 15 cents:</p>
+
+<ul>
+ <li>A dime and a nickel</li>
+ <li>A dime and 5 pennies</li>
+ <li>3 nickels</li>
+ <li>2 nickels and 5 pennies</li>
+ <li>A nickel and 10 pennies</li>
+ <li>15 pennies</li>
+</ul>
+
+# --instructions--
+
+Implement a function to determine how many ways there are to make change for a dollar using these common coins (1 dollar = 100 cents)
# --hints--
-`countCoins`是一个函数。
+`countCoins` should be a function.
```js
assert(typeof countCoins === 'function');
```
-`countCoints()`应该返回242。
+`countCoins()` should return 242.
```js
assert.equal(countCoins(), 242);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/cramers-rule.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/cramers-rule.md
index 9d79110f9f9..7580e28c50e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/cramers-rule.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/cramers-rule.md
@@ -1,30 +1,50 @@
---
id: 59713da0a428c1a62d7db430
-title: 克莱默的统治
+title: Cramer's rule
challengeType: 5
-videoUrl: ''
+forumTopicId: 302239
dashedName: cramers-rule
---
# --description--
-<p> <a href='https://en.wikipedia.org/wiki/linear algebra' title='wp：线性代数'>在线性代数中</a> ， <a href='https://en.wikipedia.org/wiki/Cramer's rule' title='wp：Cramer的规则'>Cramer规则</a>是一个<a href='https://en.wikipedia.org/wiki/system of linear equations' title='wp：线性方程组'>线性方程组</a>解的显式公式，其中包含与未知数一样多的方程，只要系统具有唯一解，就有效。它通过用方程右边的矢量替换一列来表示（方形）系数矩阵的决定因素和从它获得的矩阵的解决方案。 </p><p>特定</p><p><big></big></p><p> <big>$ \ left \ {\ begin {matrix} a_1x + b_1y + c_1z＆= {\ color {red} d_1} \\ a_2x + b_2y + c_2z＆= {\ color {red} d_2} \\ a_3x + b_3y + c_3z＆= {\颜色{红} D_3} \ {结束矩阵} \权。$</big> </p><p>以矩阵格式表示</p><p><big></big></p><p> <big>$ \ begin {bmatrix} a_1＆b_1＆c_1 \\ a_2＆b_2＆c_2 \\ a_3＆b_3＆c_3 \ end {bmatrix} \ begin {bmatrix} x \\ y \\ z \ end {bmatrix} = \ begin {bmatrix} {\ color {red} d_1} \\ {\ color {red} d_2} \\ {\ color {red} d_3} \ end {bmatrix}。$</big> </p><p>然后可以找到$ x，y $和$ z $的值，如下所示： </p><p><big></big></p><p> <big>$ x = \ frac {\ begin {vmatrix} {\ color {red} d_1}＆b_1＆c_1 \\ {\ color {red} d_2}＆b_2＆c_2 \\ {\ color {red} d_3}＆b_3＆ c_3 \ end {vmatrix}} {\ begin {vmatrix} a_1＆b_1＆c_1 \\ a_2＆b_2＆c_2 \\ a_3＆b_3＆c_3 \ end {vmatrix}}，\ quad y = \ frac {\ begin {vmatrix } a_1＆{\ color {red} d_1}＆c_1 \\ a_2＆{\ color {red} d_2}＆c_2 \\ a_3＆{\ color {red} d_3}＆c_3 \ end {vmatrix}} {\ begin {vmatrix} a_1＆b_1＆c_1 \\ a_2＆b_2＆c_2 \\ a_3＆b_3＆c_3 \ end {vmatrix}}，\ text {和} z = \ frac {\ begin {vmatrix} a_1＆b_1＆{\ color {red} d_1} \\ a_2＆b_2＆{\ color {red} d_2} \\ a_3＆b_3＆{\ color {red} d_3} \ end {vmatrix}} {\ begin {vmatrix} a_1＆b_1＆ c_1 \\ a_2＆b_2＆c_2 \\ a_3＆b_3＆c_3 \ end {vmatrix}}。$</big> </p>任务<p>给定以下方程组： </p><p> <big>$ \ begin {例} 2w-x + 5y + z = -3 \\ 3w + 2x + 2y-6z = -32 \\ w + 3x + 3y-z = -47 \\ 5w-2x-3y + 3z = 49 \\ \ end {cases} $</big> </p><p>使用Cramer的规则解决<big>$ w $，$ x $，$ y $</big>和<big>$ z $</big> 。 </p>
+In [linear algebra](https://en.wikipedia.org/wiki/linear algebra "wp: linear algebra"), [Cramer's rule](https://en.wikipedia.org/wiki/Cramer's rule "wp: Cramer's rule") is an explicit formula for the solution of a [system of linear equations](https://en.wikipedia.org/wiki/system of linear equations "wp: system of linear equations") with as many equations as unknowns, valid whenever the system has a unique solution. It expresses the solution in terms of the determinants of the (square) coefficient matrix and of matrices obtained from it by replacing one column by the vector of right hand sides of the equations.
+
+Given
+
+$\\left\\{\\begin{matrix}a_1x + b_1y + c_1z&= {\\color{red}d_1}\\\\a_2x + b_2y + c_2z&= {\\color{red}d_2}\\\\a_3x + b_3y + c_3z&= {\\color{red}d_3}\\end{matrix}\\right.$
+
+which in matrix format is
+
+$\\begin{bmatrix} a_1 & b_1 & c_1 \\\\ a_2 & b_2 & c_2 \\\\ a_3 & b_3 & c_3 \\end{bmatrix}\\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix}=\\begin{bmatrix} {\\color{red}d_1} \\\\ {\\color{red}d_2} \\\\ {\\color{red}d_3} \\end{bmatrix}.$
+
+Then the values of $x, y$ and $z$ can be found as follows:
+
+$x = \\frac{\\begin{vmatrix} {\\color{red}d_1} & b_1 & c_1 \\\\ {\\color{red}d_2} & b_2 & c_2 \\\\ {\\color{red}d_3} & b_3 & c_3 \\end{vmatrix} } { \\begin{vmatrix} a_1 & b_1 & c_1 \\\\ a_2 & b_2 & c_2 \\\\ a_3 & b_3 & c_3 \\end{vmatrix}}, \\quad y = \\frac {\\begin{vmatrix} a_1 & {\\color{red}d_1} & c_1 \\\\ a_2 & {\\color{red}d_2} & c_2 \\\\ a_3 & {\\color{red}d_3} & c_3 \\end{vmatrix}} {\\begin{vmatrix} a_1 & b_1 & c_1 \\\\ a_2 & b_2 & c_2 \\\\ a_3 & b_3 & c_3 \\end{vmatrix}}, \\text{ and }z = \\frac { \\begin{vmatrix} a_1 & b_1 & {\\color{red}d_1} \\\\ a_2 & b_2 & {\\color{red}d_2} \\\\ a_3 & b_3 & {\\color{red}d_3} \\end{vmatrix}} {\\begin{vmatrix} a_1 & b_1 & c_1 \\\\ a_2 & b_2 & c_2 \\\\ a_3 & b_3 & c_3 \\end{vmatrix} }.$
+
+# --instructions--
+
+Given the following system of equations:
+
+$\\begin{cases} 2w-x+5y+z=-3 \\\\ 3w+2x+2y-6z=-32 \\\\ w+3x+3y-z=-47 \\\\ 5w-2x-3y+3z=49 \\\\ \\end{cases}$
+
+solve for $w$, $x$, $y$ and $z$, using Cramer's rule.
# --hints--
-`cramersRule`是一个函数。
+`cramersRule` should be a function.
```js
assert(typeof cramersRule === 'function');
```
-`cramersRule([[2, -1, 5, 1], [3, 2, 2, -6], [1, 3, 3, -1], [5, -2, -3, 3]], [-3, -32, -47, 49])`应返回`[2, -12, -4, 1]` 。
+`cramersRule([[2, -1, 5, 1], [3, 2, 2, -6], [1, 3, 3, -1], [5, -2, -3, 3]], [-3, -32, -47, 49])` should return `[2, -12, -4, 1]`.
```js
assert.deepEqual(cramersRule(matrices[0], freeTerms[0]), answers[0]);
```
-`cramersRule([[3, 1, 1], [2, 2, 5], [1, -3, -4]], [3, -1, 2])`应返回`[1, 1, -1]` 。
+`cramersRule([[3, 1, 1], [2, 2, 5], [1, -3, -4]], [3, -1, 2])` should return `[1, 1, -1]`.
```js
assert.deepEqual(cramersRule(matrices[1], freeTerms[1]), answers[1]);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/cumulative-standard-deviation.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/cumulative-standard-deviation.md
index 4ca5496cb74..28aa14710d4 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/cumulative-standard-deviation.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/cumulative-standard-deviation.md
@@ -8,7 +8,7 @@ dashedName: cumulative-standard-deviation
# --description--
-Write a function that takes an array of numbers as parameter and returns the [standard deviation](<https://en.wikipedia.org/wiki/Standard Deviation>) of the series.
+Write a function that takes an array of numbers as parameter and returns the [standard deviation](https://en.wikipedia.org/wiki/Standard Deviation) of the series.
# --hints--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/cut-a-rectangle.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/cut-a-rectangle.md
index 9e37a3146af..60d72baa9d2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/cut-a-rectangle.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/cut-a-rectangle.md
@@ -18,7 +18,7 @@ A given rectangle is made from *m* × *n* squares. If *m* and *n* are not both o
.s2 { fill: #fbf }
.d { stroke:black; fill:none}
</style>
- <defs> <g id="m">
+ <defs> <g id="m">
<g id="h4"><g id="h2">
<path id="h" d="m0 10h 640" class="g"/>
<use xlink:href="#h" transform="translate(0,20)"/></g>
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/date-format.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/date-format.md
index 3cd15732205..b3d9c7c5903 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/date-format.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/date-format.md
@@ -1,44 +1,43 @@
---
id: 59669d08d75b60482359409f
-title: 日期格式
+title: Date format
challengeType: 5
-videoUrl: ''
+forumTopicId: 302243
dashedName: date-format
---
# --description--
-任务：
+Return an array with the current date in the formats:
-返回包含以下格式的当前日期的数组：
+<ul>
+ <li>2007-11-23</li>
+ <li>Friday, November 23, 2007</li>
+</ul>
-\- 2007-11-23和
-
-\- 2007年11月23日星期日
-
-示例输出： `['2007-11-23', 'Sunday, November 23, 2007']`
+Example output: `['2007-11-23', 'Friday, November 23, 2007']`
# --hints--
-`getDateFormats`是一个函数。
+`getDateFormats` should be a function.
```js
assert(typeof getDateFormats === 'function');
```
-应该返回一个对象。
+`getDateFormats` should return an object.
```js
assert(typeof getDateFormats() === 'object');
```
-应该返回一个包含2个元素的数组。
+`getDateFormats` should return an array with 2 elements.
```js
assert(getDateFormats().length === 2);
```
-应以正确的格式返回正确的日期
+`getDateFormats` should return the correct date in the right format
```js
assert.deepEqual(getDateFormats(), dates, equalsMessage);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/date-manipulation.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/date-manipulation.md
index bfc494c6560..3412cd12fda 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/date-manipulation.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/date-manipulation.md
@@ -1,42 +1,34 @@
---
id: 5966c21cf732a95f1b67dd28
-title: 日期操纵
+title: Date manipulation
challengeType: 5
-videoUrl: ''
+forumTopicId: 302244
dashedName: date-manipulation
---
# --description--
-任务：
+Given a date string in EST, output the given date as a string with 12 hours added to the time. Time zone should be preserved.
-给定EST中的日期字符串，将给定日期输出为字符串，并添加12小时。
+Example input: `"March 6 2009 7:30pm EST"`
-时区应该保留。
-
-示例输入：
-
-`"March 7 2009 7:30pm EST"`
-
-输出示例：
-
-`"March 8 2009 7:30am EST"`
+Example output: `"March 7 2009 7:30am EST"`
# --hints--
-`add12Hours`是一个功能。
+`add12Hours` should be a function.
```js
assert(typeof add12Hours === 'function');
```
-`add12Hours(dateString)`应该返回一个字符串。
+`add12Hours(dateString)` should return a string.
```js
assert(typeof add12Hours('January 17 2017 11:43am EST') === 'string');
```
-`add12Hours("" + tests[0] + "")`应该返回`"" + answers[0] + ""`
+`add12Hours("January 17 2017 11:43am EST")` should return `"January 17 2017 11:43pm EST"`
```js
assert(
@@ -44,25 +36,25 @@ assert(
);
```
-汉德尔应该改变一天。 `add12Hours("" + tests[1] + "")`应返回`"" + answers[1] + ""`
+Should handle day change. `add12Hours("March 6 2009 7:30pm EST")` should return `"March 7 2009 7:30am EST"`
```js
-assert(add12Hours('March 7 2009 7:30pm EST') === 'March 8 2009 7:30am EST');
+assert(add12Hours('March 6 2009 7:30pm EST') === 'March 7 2009 7:30am EST');
```
-汉德尔月份应该在闰年中发生变化。 `add12Hours("" + tests[2] + "")`应返回`"" + answers[2] + ""`
+Should handle month change in a leap years. `add12Hours("February 29 2004 9:15pm EST")` should return `"March 1 2004 9:15am EST"`
```js
assert(add12Hours('February 29 2004 9:15pm EST') === 'March 1 2004 9:15am EST');
```
-应该在一个共同的年份改变汉德尔月份。 `add12Hours("" + tests[3] + "")`应该返回`"" + answers[3] + ""`
+Should handle month change in a common years. `add12Hours("February 28 1999 3:15pm EST")` should return `"March 1 1999 3:15am EST"`
```js
assert(add12Hours('February 28 1999 3:15pm EST') === 'March 1 1999 3:15am EST');
```
-汉德尔应该改变一年。 `add12Hours("" + tests[4] + "")`应该返回`"" + answers[4] + ""`
+Should handle year change. `add12Hours("December 31 2020 1:45pm EST")` should return `"January 1 2021 1:45am EST"`
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/day-of-the-week.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/day-of-the-week.md
index 148a02251d9..0260694251c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/day-of-the-week.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/day-of-the-week.md
@@ -1,36 +1,40 @@
---
id: 5966f99c45e8976909a85575
-title: 一周中的天
+title: Day of the week
challengeType: 5
-videoUrl: ''
+forumTopicId: 302245
dashedName: day-of-the-week
---
# --description--
-<p>一家公司决定，每当圣诞节落在星期天，他们将给予他们的工人所有额外带薪假期，这样，在任何公共假期，工人将不必在下一周（12月25日到1月1日之间）工作。 </p><p>任务： </p><p>编写一个开始年份和结束年份的函数，并返回12月25日为星期日的所有年份的数组。 </p>
+A company decides that whenever Xmas falls on a Sunday they will give their workers all extra paid holidays so that, together with any public holidays, workers will not have to work the following week (between the 25th of December and the first of January).
+
+# --instructions--
+
+Write a function that takes a start year and an end year and return an array of all the years where the 25th of December will be a Sunday.
# --hints--
-`findXmasSunday`是一个函数。
+`findXmasSunday` should be a function.
```js
assert(typeof findXmasSunday === 'function');
```
-`findChristmasSunday(2000, 2100)`应该返回一个数组。
+`findXmasSunday(2000, 2100)` should return an array.
```js
assert(typeof findXmasSunday(2000, 2100) === 'object');
```
-`findChristmasSunday(2008, 2121`应该回归[1977,1983,1988,1994,2005,2011,2016]
+`findXmasSunday(1970, 2017)` should return `[1977, 1983, 1988, 1994, 2005, 2011, 2016]`
```js
assert.deepEqual(findXmasSunday(1970, 2017), firstSolution);
```
-`findChristmasSunday(2008, 2121`应该返回[2011,2016,2022,2033,2039,2044,2050,2061,2067,2072,2078,2089,2095,2101,2107,2112,2118]
+`findXmasSunday(2008, 2121)` should return `[2011, 2016, 2022, 2033, 2039, 2044, 2050, 2061, 2067, 2072, 2078, 2089, 2095, 2101, 2107, 2112, 2118]`
```js
assert.deepEqual(findXmasSunday(2008, 2121), secondSolution);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/deal-cards-for-freecell.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/deal-cards-for-freecell.md
index 8fe4273eb8e..f9989155956 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/deal-cards-for-freecell.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/deal-cards-for-freecell.md
@@ -1,64 +1,113 @@
---
id: 59694356a6e7011f7f1c5f4e
-title: FreeCell的交易卡
+title: Deal cards for FreeCell
challengeType: 5
-videoUrl: ''
+forumTopicId: 302246
dashedName: deal-cards-for-freecell
---
# --description--
-<p> Free Cell是Paul Alfille在1978年向PLATO系统推出的单人纸牌游戏。微软的Jim Horne将名称更改为FreeCell并重新实现了<a href='http://rosettacode.org/wiki/DOS' title='DOS'>DOS</a>游戏，然后是<a href='http://rosettacode.org/wiki/Windows' title='视窗'>Windows</a>游戏。 </p><p>这个版本引入了32000个编号的交易。 （ <a href='http://www.solitairelaboratory.com/fcfaq.html' title='链接：http：//www.solitairelaboratory.com/fcfaq.html'>FreeCell FAQ</a>讲述了这段历史。） </p><p>随着游戏的流行，Jim Horne披露<a href='http://www.solitairelaboratory.com/mshuffle.txt' title='链接：http：//www.solitairelaboratory.com/mshuffle.txt'>了这种算法</a> ，FreeCell的其他实现开始重现微软的交易。 </p><p>这些交易编号从1到32000。 </p><p>微软的新版本有100万笔交易，编号从1到1000000;某些实现允许该范围之外的数字。 </p><p>该算法使用Microsoft C的这种<a href='http://rosettacode.org/wiki/linear congruential generator' title='线性同余发生器'>线性同余生成器</a> ： </p> $ state\_ {n + 1} \\ equiv 214013 \\ times state_n + 2531011 \\ pmod {2 ^ {31}} $ $ rand_n = state_n \\ div 2 ^ {16} $ $ rand_n $的范围是0到32767。 <p>该算法如下： </p>使用交易编号为RNG播种。创建一个由52张牌组成的<a href='http://rosettacode.org/wiki/array' title='排列'>阵列</a> ：俱乐部的王牌，钻石王牌，心中的王牌，黑桃王牌，2个俱乐部，2个钻石等等等等：Ace，2,3,4,5,6， 7,8,9,10，杰克，女王，国王。数组索引为0到51，其中俱乐部的Ace为0，黑桃之王为51.直到数组为空：在索引≡下一个随机数（mod数组长度）中选择随机卡。将此随机卡与阵列的最后一张卡交换。从阵列中删除此随机卡。 （数组长度减少1.）处理此随机卡。交易所有52张牌，面朝上，横跨8列。前8张牌分为8列，后8张牌分为8张，依此类推。例： <p>订购交易卡</p><p></p><pre> 1 2 3 4 5 6 7 8
+*FreeCell* is the solitaire card game that Paul Alfille introduced to the PLATO system in 1978. Jim Horne, at Microsoft, changed the name to FreeCell and reimplemented the game for [DOS](https://rosettacode.org/wiki/DOS "DOS"), then [Windows](https://rosettacode.org/wiki/Windows "Windows"). This version introduced 32000 numbered deals.
+
+As the game became popular, Jim Horne disclosed the algorithm, and other implementations of FreeCell began to reproduce the Microsoft deals. These deals are numbered from 1 to 32000. Newer versions from Microsoft have 1 million deals, numbered from 1 to 1000000; some implementations allow numbers outside that range.
+
+The algorithm uses this [linear congruential generator](https://rosettacode.org/wiki/linear congruential generator "linear congruential generator") from Microsoft C:
+
+<ul>
+ <li>$state_{n + 1} \equiv 214013 \times state_n + 2531011 \pmod{2^{31}}$</li>
+ <li>$rand_n = state_n \div 2^{16}$</li>
+ <li>$rand_n$ is in range 0 to 32767.</li>
+</ul>
+
+The algorithm follows:
+
+<ol>
+ <li>Seed the RNG with the number of the deal.
+ </li><li>Create an <a href='https://rosettacode.org/wiki/array' title='array' target='_blank'>array</a> of 52 cards: Ace of Clubs, Ace of Diamonds, Ace of Hearts, Ace of Spades, 2 of Clubs, 2 of Diamonds, and so on through the ranks: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King. The array indexes are 0 to 51, with Ace of Clubs at 0, and King of Spades at 51.</li>
+ <li>Until the array is empty:</li>
+ <li>Choose a random card at index ≡ next random number (mod array length).</li>
+ <ul>
+ <li>Swap this random card with the last card of the array.</li>
+ <li>Remove this random card from the array. (Array length goes down by 1.)</li>
+ <li>Deal this random card.</li>
+ </ul>
+ <li>Deal all 52 cards, face up, across 8 columns. The first 8 cards go in 8 columns, the next 8 cards go on the first 8 cards, and so on.</li>
+</ol>
+
+**Example:**
+
+**Order to deal cards**
+
+<pre> 1 2 3 4 5 6 7 8
9 10 11 12 13 14 15 16
17 18 19 20 21 22 23 24
25 26 27 28 29 30 31 32
33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48
-49 50 51 52 </pre><p></p><p>游戏＃1 </p><p></p><pre> [
-['JD'，'2D'，'9H'，'JC'，'5D'，'7H'，'7C'，'5H']，
-['KD'，'KC'，'9S'，'5S'，'AD'，'QC'，'KH'，'3H']，
-['2S'，'KS'，'9D'，'QD'，'JS'，'AS'，'AH'，'3C']，
-['4C'，'5C'，'TS'，'QH'，'4H'，'AC'，'4D'，'7S']，
-['3S'，'TD'，'4S'，'TH'，'8H'，'2C'，'JH'，'7D']，
-['6D'，'8S'，'8D'，'QS'，'6C'，'3D'，'8C'，'TC']，
-['6S'，'9C'，'2H'，'6H']
-] </pre><p></p><p>游戏＃617 </p><p></p><pre> [
-['7D'，'AD'，'5C'，'3S'，'5S'，'8C'，'2D'，'AH']，
-['TD'，'7S'，'QD'，'AC'，'6D'，'8H'，'AS'，'KH']，
-['TH'，'QC'，'3H'，'9D'，'6S'，'8D'，'3D'，'TC']，
-['KD'，'5H'，'9S'，'3C'，'8S'，'7H'，'4D'，'JS']，
-['4C'，'QS'，'9C'，'9H'，'7C'，'6H'，'2C'，'2S']，
-['4S'，'TS'，'2H'，'5D'，'JC'，'6C'，'JH'，'QH']，
-['JD'，'KS'，'KC'，'4H']
-] </pre><p></p>任务： <p>编写一个函数来获取交易编号并按照与此算法相同的顺序处理卡。 </p><p>该函数必须返回表示FreeCell板的二维数组。 </p><p>还可以针对<a href='http://freecellgamesolutions.com/' title='链接：http：//freecellgamesolutions.com/'>1000000游戏的FreeCell解决方案</a>检查交易。 </p><p> （召唤一个视频解决方案，它会显示初始交易。） </p>
+49 50 51 52</pre>
+
+**Game #1**
+
+```js
+[
+ ['JD', '2D', '9H', 'JC', '5D', '7H', '7C', '5H'],
+ ['KD', 'KC', '9S', '5S', 'AD', 'QC', 'KH', '3H'],
+ ['2S', 'KS', '9D', 'QD', 'JS', 'AS', 'AH', '3C'],
+ ['4C', '5C', 'TS', 'QH', '4H', 'AC', '4D', '7S'],
+ ['3S', 'TD', '4S', 'TH', '8H', '2C', 'JH', '7D'],
+ ['6D', '8S', '8D', 'QS', '6C', '3D', '8C', 'TC'],
+ ['6S', '9C', '2H', '6H']
+]
+```
+
+**Game #617**
+
+```js
+[
+ ['7D', 'AD', '5C', '3S', '5S', '8C', '2D', 'AH'],
+ ['TD', '7S', 'QD', 'AC', '6D', '8H', 'AS', 'KH'],
+ ['TH', 'QC', '3H', '9D', '6S', '8D', '3D', 'TC'],
+ ['KD', '5H', '9S', '3C', '8S', '7H', '4D', 'JS'],
+ ['4C', 'QS', '9C', '9H', '7C', '6H', '2C', '2S'],
+ ['4S', 'TS', '2H', '5D', 'JC', '6C', 'JH', 'QH'],
+ ['JD', 'KS', 'KC', '4H']
+]
+```
+
+# --instructions--
+
+Write a function to take a deal number and deal cards in the same order as this algorithm. The function must return a two dimensional array representing the FreeCell board.
+
+Deals can also be checked against [FreeCell solutions to 1000000 games](https://freecellgamesolutions.com/). (Summon a video solution, and it displays the initial deal.)
# --hints--
-`dealFreeCell`是一个功能。
+`dealFreeCell` should be a function.
```js
assert(typeof dealFreeCell === 'function');
```
-`dealFreeCell(seed)`应该返回一个对象。
+`dealFreeCell(seed)` should return an object.
```js
assert(typeof dealFreeCell(1) === 'object');
```
-`dealFreeCell(seed)`应该返回一个长度为7的数组。
+`dealFreeCell(seed)` should return an array of length 7.
```js
assert(dealFreeCell(1).length === 7);
```
-`dealFreeCell(1)`应该返回一个与示例“Game＃1”相同的数组
+`dealFreeCell(1)` should return an array identical to example "Game #1"
```js
assert.deepEqual(dealFreeCell(1), game1);
```
-`dealFreeCell(617)`应该返回一个与示例“Game＃617”相同的数组
+`dealFreeCell(617)` should return an array identical to example "Game #617"
```js
assert.deepEqual(dealFreeCell(617), game617);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/deepcopy.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/deepcopy.md
index 3d55181970f..24b950d8d55 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/deepcopy.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/deepcopy.md
@@ -1,50 +1,51 @@
---
id: 596a8888ab7c01048de257d5
-title: deepcopy的
+title: Deepcopy
challengeType: 5
-videoUrl: ''
+forumTopicId: 302247
dashedName: deepcopy
---
# --description--
-任务：
+Write a function that returns a deep copy of a given object. The copy must not be the same object that was given.
-编写一个返回给定对象的深层副本的函数。
+This task will not test for:
-副本不得与给定的对象相同。
-
-此任务不会测试：
-
-具有属性属性的对象Date对象或具有Date对象属性的对象RegEx或具有RegEx对象属性的对象原型复制
+<ul>
+ <li>Objects with properties that are functions</li>
+ <li>Date objects or object with properties that are Date objects</li>
+ <li>RegEx or object with properties that are RegEx objects</li>
+ <li>Prototype copying</li>
+</ul>
# --hints--
-`deepcopy`应该是一个功能。
+`deepcopy` should be a function.
```js
assert(typeof deepcopy === 'function');
```
-`deepcopy({test: "test"})`应返回一个对象。
+`deepcopy({test: "test"})` should return an object.
```js
assert(typeof deepcopy(obj1) === 'object');
```
-不应该返回提供的相同对象。
+`deepcopy` should not return the same object that was provided.
```js
assert(deepcopy(obj2) != obj2);
```
-传递包含数组的对象时，应返回该对象的深层副本。
+When passed an object containing an array, `deepcopy` should return a deep copy of the object.
```js
assert.deepEqual(deepcopy(obj2), obj2);
```
-传递包含另一个对象的对象时，应返回该对象的深层副本。
+When passed an object containing another object, `deepcopy` should return a deep copy of the object.
```js
assert.deepEqual(deepcopy(obj3), obj3);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/define-a-primitive-data-type.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/define-a-primitive-data-type.md
index ef026dd478b..2bc3f2dd1b7 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/define-a-primitive-data-type.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/define-a-primitive-data-type.md
@@ -1,100 +1,103 @@
---
id: 597089c87eec450c68aa1643
-title: 定义原始数据类型
+title: Define a primitive data type
challengeType: 5
-videoUrl: ''
+forumTopicId: 302248
dashedName: define-a-primitive-data-type
---
# --description--
-任务：
+Define a type that behaves like an integer but has a lowest valid value of 1 and a highest valid value of 10.
-定义一个行为类似于整数但最低有效值为1且最高有效值为10的类型。
+Error handling:
-错误：如果您尝试实例化一个值超出1 - 10的`Num` ，它应该抛出一个错误消息为`'Out of range'`的`TypeError` 。如果您尝试使用不是`Num`的值实例化`Num` ，则应抛出`TypeError`并显示错误消息`'Not a Number'` 。
+<ul>
+ <li>If you try to instantiate a <code>Num</code> with a value outside of 1 - 10, it should throw a <code>TypeError</code> with an error message of <code>'Out of range'</code>.</li>
+ <li>If you try to instantiate a <code>Num</code> with a value that is not a number, it should throw a <code>TypeError</code> with an error message of <code>'Not a Number'</code>.</li>
+</ul>
# --hints--
-`Num`应该是一个功能。
+`Num` should be a function.
```js
assert(typeof Num === 'function');
```
-`new Num(4)`应返回一个对象。
+`new Num(4)` should return an object.
```js
assert(typeof new Num(4) === 'object');
```
-`new Num(\'test\')`应抛出一个带有消息“非数字”的TypeError。
+`new Num('test')` should throw a TypeError with message 'Not a Number'.
```js
assert.throws(() => new Num('test'), TypeError);
```
-`new Num(0)`应该抛出一个带有消息“超出范围”的TypeError。
+`new Num(0)` should throw a TypeError with message 'Out of range'.
```js
assert.throws(() => new Num(0), TypeError);
```
-`new Num(-5)`应该抛出一个带有消息“超出范围”的TypeError。
+`new Num(-5)` should throw a TypeError with message 'Out of range'.
```js
assert.throws(() => new Num(-5), TypeError);
```
-`new Num(10)`应抛出一个带有消息“超出范围”的TypeError。
+`new Num(10)` should throw a TypeError with message 'Out of range'.
```js
assert.throws(() => new Num(11), TypeError);
```
-`new Num(20)`应抛出一个带有消息“超出范围”的TypeError。
+`new Num(20)` should throw a TypeError with message 'Out of range'.
```js
assert.throws(() => new Num(20), TypeError);
```
-`new Num(3) + new Num(4)`应该等于7。
+`new Num(3) + new Num(4)` should equal 7.
```js
assert.equal(new Num(3) + new Num(4), 7);
```
-`new Num(3) - new Num(4)`应该等于-1。
+`new Num(3) - new Num(4)` should equal -1.
```js
assert.equal(new Num(3) - new Num(4), -1);
```
-`new Num(3) * new Num(4)`应该等于12。
+`new Num(3) * new Num(4)` should equal 12.
```js
assert.equal(new Num(3) * new Num(4), 12);
```
-`new Num(3) / new Num(4)`应该等于0.75。
+`new Num(3) / new Num(4)` should equal 0.75.
```js
assert.equal(new Num(3) / new Num(4), 0.75);
```
-`new Num(3) < new Num(4)`应该是真的。
+`new Num(3) < new Num(4)` should be true.
```js
assert(new Num(3) < new Num(4));
```
-`new Num(3) > new Num(4)`应该是假的。
+`new Num(3) > new Num(4)` should be false.
```js
assert(!(new Num(3) > new Num(4)));
```
-`(new Num(5)).toString()`应返回\\'5 \\'
+`(new Num(5)).toString()` should return '5'
```js
assert.equal(new Num(5).toString(), '5');
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/department-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/department-numbers.md
index 1b00670710e..0eba1bec54a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/department-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/department-numbers.md
@@ -1,36 +1,60 @@
---
id: 59f40b17e79dbf1ab720ed7a
-title: 部门编号
+title: Department Numbers
challengeType: 5
-videoUrl: ''
+forumTopicId: 302249
dashedName: department-numbers
---
# --description--
-<p>有一个高度组织化的城市决定为每个部门分配一个号码： </p>警察局环卫部门消防部门<p>每个部门的数字可以在1到7之间（含）。 </p><p>这三个部门编号应该是唯一的（彼此不同），并且必须加起来为12。 </p><p>警察局长不喜欢奇怪的号码，并希望他的部门有一个偶数。 </p>任务： <p>编写一个输出所有有效组合的程序： </p><p> [2,3,7] </p><p> [2,4,6] </p><p> [2,6,4] </p><p> [2,7,3] </p><p> [4,1,7] </p><p> [4,2,6] </p><p> [4,3,5] </p><p> [4,5,3] </p><p> [4,6,2] </p><p> [4,7,1] </p><p> [6,1,5] </p><p> [6,2,4] </p><p> [6,4,2] </p><p> [6,5,1] </p>
+There is a highly organized city that has decided to assign a number to each of their departments:
+
+<ul>
+ <li>Police department</li>
+ <li>Sanitation department</li>
+ <li>Fire department</li>
+</ul>
+
+Each department can have a number between 1 and 7 (inclusive).
+
+The three department numbers are to be unique (different from each other) and must add up to the number 12.
+
+The Chief of the Police doesn't like odd numbers and wants to have an even number for his department.
+
+# --instructions--
+
+Write a program which outputs all valid combinations as an array.
+
+```js
+[2, 3, 7] [2, 4, 6] [2, 6, 4]
+[2, 7, 3] [4, 1, 7] [4, 2, 6]
+[4, 3, 5] [4, 5, 3] [4, 6, 2]
+[4, 7, 1] [6, 1, 5] [6, 2, 4]
+[6, 4, 2] [6, 5, 1]
+```
# --hints--
-`combinations`应该是一个功能。
+`combinations` should be a function.
```js
assert(typeof combinations === 'function');
```
-`combinations([1, 2, 3], 6)`应该返回一个数组。
+`combinations([1, 2, 3], 6)` should return an Array.
```js
assert(Array.isArray(combinations([1, 2, 3], 6)));
```
-`combinations([1, 2, 3, 4, 5, 6, 7], 12)`应返回长度为14的数组。
+`combinations([1, 2, 3, 4, 5, 6, 7], 12)` should return an array of length 14.
```js
assert(combinations(nums, total).length === len);
```
-`combinations([1, 2, 3, 4, 5, 6, 7], 12)`应返回所有有效组合。
+`combinations([1, 2, 3, 4, 5, 6, 7], 12)` should return all valid combinations.
```js
assert.deepEqual(combinations(nums, total), result);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/discordian-date.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/discordian-date.md
index 840cb87533b..811a3b867ec 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/discordian-date.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/discordian-date.md
@@ -1,26 +1,24 @@
---
id: 59f4eafba0343628bb682785
-title: Discordian日期
+title: Discordian date
challengeType: 5
-videoUrl: ''
+forumTopicId: 302250
dashedName: discordian-date
---
# --description--
-任务：
-
-将给定日期从[公历](<https://en.wikipedia.org/wiki/Gregorian calendar> "wp：阳历")转换为[Discordian日历](<https://en.wikipedia.org/wiki/Discordian calendar> "wp：Discordian日历") 。
+Convert a given date from the [Gregorian calendar](https://en.wikipedia.org/wiki/Gregorian calendar "wp: Gregorian calendar") to the [Discordian calendar](https://en.wikipedia.org/wiki/Discordian calendar "wp: Discordian calendar").
# --hints--
-`discordianDate`是一个函数。
+`discordianDate` should be a function.
```js
assert(typeof discordianDate === 'function');
```
-`discordianDate(new Date(2010, 6, 22))` `"Pungenday, the 57th day of Confusion in the YOLD 3176"` `discordianDate(new Date(2010, 6, 22))`应该返回`"Pungenday, the 57th day of Confusion in the YOLD 3176"` 。
+`discordianDate(new Date(2010, 6, 22))` should return `"Pungenday, the 57th day of Confusion in the YOLD 3176"`.
```js
assert(
@@ -29,7 +27,7 @@ assert(
);
```
-`discordianDate(new Date(2012, 1, 28))`应该返回`"Prickle-Prickle, the 59th day of Chaos in the YOLD 3178"` 。
+`discordianDate(new Date(2012, 1, 28))` should return `"Prickle-Prickle, the 59th day of Chaos in the YOLD 3178"`.
```js
assert(
@@ -38,7 +36,7 @@ assert(
);
```
-`discordianDate(new Date(2012, 1, 29))` `"Setting Orange, the 60th day of Chaos in the YOLD 3178. Celebrate St. Tib's Day!"` `discordianDate(new Date(2012, 1, 29))`应该返回`"Setting Orange, the 60th day of Chaos in the YOLD 3178. Celebrate St. Tib's Day!"` 。
+`discordianDate(new Date(2012, 1, 29))` should return `"Setting Orange, the 60th day of Chaos in the YOLD 3178. Celebrate St. Tib\'s Day!"`.
```js
assert(
@@ -47,7 +45,7 @@ assert(
);
```
-`discordianDate(new Date(2012, 2, 1))`应该返回`"Setting Orange, the 60th day of Chaos in the YOLD 3178"` 。
+`discordianDate(new Date(2012, 2, 1))` should return `"Setting Orange, the 60th day of Chaos in the YOLD 3178"`.
```js
assert(
@@ -56,7 +54,7 @@ assert(
);
```
-`discordianDate(new Date(2010, 0, 5))` `"Setting Orange, the 5th day of Chaos in the YOLD 3176. Celebrate Mungday!"` `discordianDate(new Date(2010, 0, 5))`应该返回`"Setting Orange, the 5th day of Chaos in the YOLD 3176. Celebrate Mungday!"` 。
+`discordianDate(new Date(2010, 0, 5))` should return `"Setting Orange, the 5th day of Chaos in the YOLD 3176. Celebrate Mungday!"`.
```js
assert(
@@ -65,7 +63,7 @@ assert(
);
```
-`discordianDate(new Date(2011, 4, 3))`应该返回`"Pungenday, the 50th day of Discord in the YOLD 3177. Celebrate Discoflux!"` 。
+`discordianDate(new Date(2011, 4, 3))` should return `"Pungenday, the 50th day of Discord in the YOLD 3177. Celebrate Discoflux!"`.
```js
assert(
@@ -74,7 +72,7 @@ assert(
);
```
-`discordianDate(new Date(2015, 9, 19))` `"Boomtime, the 73rd day of Bureaucracy in the YOLD 3181"` `discordianDate(new Date(2015, 9, 19))`应该返回`"Boomtime, the 73rd day of Bureaucracy in the YOLD 3181"` 。
+`discordianDate(new Date(2015, 9, 19))` should return `"Boomtime, the 73rd day of Bureaucracy in the YOLD 3181"`.
```js
assert(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/dot-product.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/dot-product.md
index 5584a7e26a7..c1c61d1b124 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/dot-product.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/dot-product.md
@@ -8,7 +8,7 @@ dashedName: dot-product
# --description--
-Create a function, to compute the **[dot product](<https://en.wikipedia.org/wiki/Dot product>)**, also known as the **scalar product** of two vectors.
+Create a function, to compute the **[dot product](https://en.wikipedia.org/wiki/Dot product)**, also known as the **scalar product** of two vectors.
# --hints--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/element-wise-operations.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/element-wise-operations.md
index ff8bf3e2c91..52a15c8d11d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/element-wise-operations.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/element-wise-operations.md
@@ -1,24 +1,36 @@
---
id: 599c333915e0ea32d04d4bec
-title: 元素操作
+title: Element-wise operations
challengeType: 5
-videoUrl: ''
+forumTopicId: 302252
dashedName: element-wise-operations
---
# --description--
-<p>实现基本的元素矩阵 - 矩阵和标量矩阵运算。 </p><p>实行： </p><p> :: *另外</p><p> :: *减法</p><p> :: *乘法</p><p> :: *分裂</p><p> :: *取幂</p><p>第一个参数是要执行的操作，例如：用于矩阵加法的“m_add”和用于标量加法的“s_add”。第二和第三参数将是要在其上执行操作的矩阵。 </p>
+Implement basic element-wise matrix-matrix and scalar-matrix operations.
+
+**Implement:**
+
+<ul>
+ <li>addition</li>
+ <li>subtraction</li>
+ <li>multiplication</li>
+ <li>division</li>
+ <li>exponentiation</li>
+</ul>
+
+The first parameter will be the operation to be performed, for example, "m_add" for matrix addition and "s_add" for scalar addition. The second and third parameters will be the matrices on which the operations are to be performed.
# --hints--
-`operation`是一种功能。
+`operation` should be a function.
```js
assert(typeof operation === 'function');
```
-`operation("m_add",[[1,2],[3,4]],[[1,2],[3,4]])`应返回`[[2,4],[6,8]]` 。
+`operation("m_add",[[1,2],[3,4]],[[1,2],[3,4]])` should return `[[2,4],[6,8]]`.
```js
assert.deepEqual(
@@ -40,7 +52,7 @@ assert.deepEqual(
);
```
-`operation("s_add",[[1,2],[3,4]],[[1,2],[3,4]])`应返回`[[3,4],[5,6]]` 。
+`operation("s_add",[[1,2],[3,4]],2)` should return `[[3,4],[5,6]]`.
```js
assert.deepEqual(
@@ -59,7 +71,7 @@ assert.deepEqual(
);
```
-`operation("m_sub",[[1,2],[3,4]],[[1,2],[3,4]])`应返回`[[0,0],[0,0]]` 。
+`operation("m_sub",[[1,2],[3,4]],[[1,2],[3,4]])` should return `[[0,0],[0,0]]`.
```js
assert.deepEqual(
@@ -81,7 +93,7 @@ assert.deepEqual(
);
```
-`operation("m_mult",[[1,2],[3,4]],[[1,2],[3,4]])`应该返回`[[1,4],[9,16]]` 。
+`operation("m_mult",[[1,2],[3,4]],[[1,2],[3,4]])` should return `[[1,4],[9,16]]`.
```js
assert.deepEqual(
@@ -103,7 +115,7 @@ assert.deepEqual(
);
```
-`operation("m_div",[[1,2],[3,4]],[[1,2],[3,4]])`应返回`[[1,1],[1,1]]` 。
+`operation("m_div",[[1,2],[3,4]],[[1,2],[3,4]])` should return `[[1,1],[1,1]]`.
```js
assert.deepEqual(
@@ -125,7 +137,7 @@ assert.deepEqual(
);
```
-`operation("m_exp",[[1,2],[3,4]],[[1,2],[3,4]])`应返回`[[1,4],[27,256]]` 。
+`operation("m_exp",[[1,2],[3,4]],[[1,2],[3,4]])` should return `[[1,4],[27,256]]`.
```js
assert.deepEqual(
@@ -147,7 +159,7 @@ assert.deepEqual(
);
```
-`operation("m_add",[[1,2,3,4],[5,6,7,8]],[[9,10,11,12],[13,14,15,16]])`应该返回`[[10,12,14,16],[18,20,22,24]]` 。
+`operation("m_add",[[1,2,3,4],[5,6,7,8]],[[9,10,11,12],[13,14,15,16]])` should return `[[10,12,14,16],[18,20,22,24]]`.
```js
assert.deepEqual(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/emirp-primes.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/emirp-primes.md
index 33683431c11..f8a6d221058 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/emirp-primes.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/emirp-primes.md
@@ -1,24 +1,37 @@
---
id: 599d0ba974141b0f508b37d5
-title: Emirp奖金
+title: Emirp primes
challengeType: 5
-videoUrl: ''
+forumTopicId: 302253
dashedName: emirp-primes
---
# --description--
-<p>一个emirp（向后拼写的主要拼写）是素数，当它们被反转时（在它们的十进制表示中）是一个不同的素数。 </p><p>编写应能功能：显示前<b>n</b> eprimes numbers.Show在range.Show的eprimes数字eprimes的在range.Show <b><sup>第</sup> n <sup>个</sup></b> eprimes数目。 </p><p>该函数应该有两个参数。第一个将接收<b>n</b>或作为数组的范围。第二个将接收一个布尔值，它指定函数是否将eprimes作为数组或单个数字（范围中的素数或第<b>n <sup>个</sup></b>素数）返回。根据参数，函数应返回数组或数字。 </p>
+An emirp (**prime** spelled backwards) are primes that when reversed (in their decimal representation) are a different prime.
+
+# --instructions--
+
+Write a function that:
+
+<ul>
+ <li>Shows the first <code>n</code> emirp numbers.</li>
+ <li>Shows the emirp numbers in a range.</li>
+ <li>Shows the number of emirps in a range.</li>
+ <li>Shows the <code>n<sup>th</sup></code> emirp number.</li>
+</ul>
+
+The function should accept two parameters. The first will receive `n` or the range as an array. The second will receive a boolean, that specifies if the function returns the emirps as an array or a single number (the number of primes in the range or the <code>n<sup>th</sup></code> prime). According to the parameters the function should return an array or a number.
# --hints--
-`emirps`是一个功能。
+`emirps` should be a function.
```js
assert(typeof emirps === 'function');
```
-`emirps(20,true)`应该返回`[13,17,31,37,71,73,79,97,107,113,149,157,167,179,199,311,337,347,359,389]`
+`emirps(20,true)` should return `[13,17,31,37,71,73,79,97,107,113,149,157,167,179,199,311,337,347,359,389]`
```js
assert.deepEqual(emirps(20, true), [
@@ -45,13 +58,13 @@ assert.deepEqual(emirps(20, true), [
]);
```
-`emirps(10000)`应该返回`948349`
+`emirps(1000)` should return `70529`
```js
assert.deepEqual(emirps(1000), 70529);
```
-`emirps([7700,8000],true)`应该返回`[7717,7757,7817,7841,7867,7879,7901,7927,7949,7951,7963]`
+`emirps([7700,8000],true)` should return `[7717,7757,7817,7841,7867,7879,7901,7927,7949,7951,7963]`
```js
assert.deepEqual(emirps([7700, 8000], true), [
@@ -69,7 +82,7 @@ assert.deepEqual(emirps([7700, 8000], true), [
]);
```
-`emirps([7700,8000],true)`应该返回`11`
+`emirps([7700,8000],true)` should return `11`
```js
assert.deepEqual(emirps([7700, 8000], false), 11);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/entropy.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/entropy.md
index 33a0a51eac9..75d61ccc5d9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/entropy.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/entropy.md
@@ -1,62 +1,60 @@
---
id: 599d15309e88c813a40baf58
-title: 熵
+title: Entropy
challengeType: 5
-videoUrl: ''
+forumTopicId: 302254
dashedName: entropy
---
# --description--
-任务：
+Calculate the Shannon entropy H of a given input string.
-计算给定输入字符串的香农熵H.
+Given the discreet random variable $X$ that is a string of $N$ "symbols" (total characters) consisting of $n$ different characters (n=2 for binary), the Shannon entropy of X in bits/symbol is:
-给定谨慎的随机变量$ X $，它是$ N $“符号”（总字符）的字符串，由$ n $个不同的字符组成（对于二进制，n = 2），位/符号中X的香农熵是：
+$H_2(X) = -\\sum\_{i=1}^n \\frac{count_i}{N} \\log_2 \\left(\\frac{count_i}{N}\\right)$
-$ H*2（X）= - \\ sum* {i = 1} ^ n \\ frac {count_i} {N} \\ log_2 \\ left（\\ frac {count_i} {N} \\ right）$
-
-其中$ count_i $是字符$ n_i $的计数。
+where $count_i$ is the count of character $n_i$.
# --hints--
-`entropy`是一种功能。
+`entropy` should be a function.
```js
assert(typeof entropy === 'function');
```
-`entropy("0")`应该返回`0`
+`entropy("0")` should return `0`
```js
assert.equal(entropy('0'), 0);
```
-`entropy("01")`应该返回`1`
+`entropy("01")` should return `1`
```js
assert.equal(entropy('01'), 1);
```
-`entropy("0123")`应该返回`2`
+`entropy("0123")` should return `2`
```js
assert.equal(entropy('0123'), 2);
```
-`entropy("01234567")`应该返回`3`
+`entropy("01234567")` should return `3`
```js
assert.equal(entropy('01234567'), 3);
```
-`entropy("0123456789abcdef")`应返回`4`
+`entropy("0123456789abcdef")` should return `4`
```js
assert.equal(entropy('0123456789abcdef'), 4);
```
-`entropy("1223334444")`应返回`1.8464393446710154`
+`entropy("1223334444")` should return `1.8464393446710154`
```js
assert.equal(entropy('1223334444'), 1.8464393446710154);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/equilibrium-index.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/equilibrium-index.md
index fd1f2670e5d..a7a33f9d37a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/equilibrium-index.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/equilibrium-index.md
@@ -1,54 +1,88 @@
---
id: 5987fd532b954e0f21b5d3f6
-title: 均衡指数
+title: Equilibrium index
challengeType: 5
-videoUrl: ''
+forumTopicId: 302255
dashedName: equilibrium-index
---
# --description--
-<p>序列的平衡指数是序列的索引，使得较低指数处的元素之和等于较高指数处的元素之和。 </p><p>例如，在序列<big>$ A $中</big> ： </p><p> :::: <big>$ A_0 = -7 $</big> </p><p> :::: <big>$ A_1 = 1 $</big> </p><p> :::: <big>$ A_2 = 5 $</big> </p><p> :::: <big>$ A_3 = 2 $</big> </p><p> :::: <big>$ A_4 = -4 $</big> </p><p> :::: <big>$ A_5 = 3 $</big> </p><p> :::: <big>$ A_6 = 0 $</big> </p><p> 3是均衡指数，因为： </p><p> :::: <big>$ A_0 + A_1 + A_2 = A_4 + A_5 + A_6 $</big> </p><p> 6也是均衡指数，因为： </p><p> :::: <big>$ A_0 + A_1 + A_2 + A_3 + A_4 + A_5 = 0 $</big> </p><p> （零元素之和为零） </p><p> 7不是均衡指数，因为它不是序列<big>$ A $</big>的有效索引。 </p><p>编写一个函数，给定一个序列，返回其平衡指数（如果有的话）。 </p><p>假设序列可能很长。 </p>
+An equilibrium index of a sequence is an index into the sequence such that the sum of elements at lower indices is equal to the sum of elements at higher indices.
+
+For example, in a sequence $A$:
+
+<ul style='list-style: none;'>
+ <li><big>$A_0 = -7$</big></li>
+ <li><big>$A_1 = 1$</big></li>
+ <li><big>$A_2 = 5$</big></li>
+ <li><big>$A_3 = 2$</big></li>
+ <li><big>$A_4 = -4$</big></li>
+ <li><big>$A_5 = 3$</big></li>
+ <li><big>$A_6 = 0$</big></li>
+</ul>
+
+`3` is an equilibrium index, because:
+
+<ul style='list-style: none;'>
+ <li><big>$A_0 + A_1 + A_2 = A_4 + A_5 + A_6$</big></li>
+</ul>
+
+`6` is also an equilibrium index, because:
+
+<ul style='list-style: none;'>
+ <li><big>$A_0 + A_1 + A_2 + A_3 + A_4 + A_5 = 0$</big></li>
+</ul>
+
+(sum of zero elements is zero)
+
+`7` is not an equilibrium index, because it is not a valid index of sequence $A$.
+
+# --instructions--
+
+Write a function that, given a sequence, returns its equilibrium indices (if any).
+
+Assume that the sequence may be very long.
# --hints--
-`equilibrium`是一种功能。
+`equilibrium` should be a function.
```js
assert(typeof equilibrium === 'function');
```
-`equilibrium([-7, 1, 5, 2, -4, 3, 0])` `[3,6]` `equilibrium([-7, 1, 5, 2, -4, 3, 0])`应该返回`[3,6]` 。
+`equilibrium([-7, 1, 5, 2, -4, 3, 0])` should return `[3,6]`.
```js
assert.deepEqual(equilibrium(equilibriumTests[0]), ans[0]);
```
-`equilibrium([2, 4, 6])`应该返回`[]` 。
+`equilibrium([2, 4, 6])` should return `[]`.
```js
assert.deepEqual(equilibrium(equilibriumTests[1]), ans[1]);
```
-`equilibrium([2, 9, 2])`应该返回`[1]` 。
+`equilibrium([2, 9, 2])` should return `[1]`.
```js
assert.deepEqual(equilibrium(equilibriumTests[2]), ans[2]);
```
-`equilibrium([1, -1, 1, -1, 1, -1, 1])`应该返回`[0,1,2,3,4,5,6]` 。
+`equilibrium([1, -1, 1, -1, 1, -1, 1])` should return `[0,1,2,3,4,5,6]`.
```js
assert.deepEqual(equilibrium(equilibriumTests[3]), ans[3]);
```
-`equilibrium([1])`应该返回`[0]` 。
+`equilibrium([1])` should return `[0]`.
```js
assert.deepEqual(equilibrium(equilibriumTests[4]), ans[4]);
```
-`equilibrium([])`应该返回`[]` 。
+`equilibrium([])` should return `[]`.
```js
assert.deepEqual(equilibrium(equilibriumTests[5]), ans[5]);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/ethiopian-multiplication.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/ethiopian-multiplication.md
index 15d0634740b..a6d1c4cb349 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/ethiopian-multiplication.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/ethiopian-multiplication.md
@@ -1,48 +1,121 @@
---
id: 599d1566a02b571412643b84
-title: 埃塞俄比亚的乘法
+title: Ethiopian multiplication
challengeType: 5
-videoUrl: ''
+forumTopicId: 302257
dashedName: ethiopian-multiplication
---
# --description--
-<p>埃塞俄比亚乘法是一种仅使用加法，加倍和减半来乘以整数的方法。 </p><p>方法： </p>取两个数字相乘，然后将它们写在两列的顶部。在左侧列中反复将最后一个数字减半，丢弃任何余数，并将结果写入同一列中的最后一个，直到您写入值1.在右侧列中重复加倍最后一个数字并写入结果如下。在左侧列显示的同一行中添加结果时停止1.检查生成的表并丢弃左列中的值为偶数的任何行。将右侧列中的值相加，以产生将原始两个数相乘的结果<p>例如：17×34 </p><p> 17 34 </p><p>将第一列减半： </p><p> 17 34 </p><p> 8 </p><p> 4 </p><p> 2 </p><p> 1 </p><p>加倍第二列： </p><p> 17 34 </p><p> 8 68 </p><p> 4 136 </p><p> 2 272 </p><p> 1 544 </p><p>第一个单元格为偶数的删除行： </p><p> 17 34 </p><p> 8 <strike>68</strike> </p><p> 4 <strike>136</strike> </p><p> 2 <strike>272</strike> </p><p> 1 544 </p><p>将右侧列中的剩余数字相加： </p><p> 17 34 </p><p> 8 - </p><p> 4 --- </p><p> 2 --- </p><p> 1 544 </p><p> ==== </p><p> 578 </p><p>所以17乘以34，埃塞俄比亚方法是578。 </p>任务： <p>任务是定义三个命名函数/方法/过程/子例程： </p>一个将一个整数减半，一个减半整数，一个整数是偶数。 <p>使用这些函数创建一个执行埃塞俄比亚乘法的函数。 </p>
+Ethiopian multiplication is a method of multiplying integers using only addition, doubling, and halving.
+
+**Method:**
+
+<ol>
+ <li>Take two numbers to be multiplied and write them down at the top of two columns</li>
+ <li>In the left-hand column repeatedly halve the last number, discarding any remainders, and write the result below the last in the same column, until you write a value of <code>1</code></li>
+ <li>In the right-hand column repeatedly double the last number and write the result below. stop when you add a result in the same row as where the left hand column shows <code>1</code></li>
+ <li>Examine the table produced and discard any row where the value in the left column is even</li>
+ <li>Sum the values in the right-hand column that remain to produce the result of multiplying the original two numbers together</li>
+</ol>
+
+**For example:** `17 × 34`
+
+<pre>17 34
+</pre>
+
+Halving the first column:
+
+<pre>17 34
+8
+4
+2
+1
+</pre>
+
+Doubling the second column:
+
+<pre>17 34
+8 68
+4 136
+2 272
+1 544
+</pre>
+
+Strike-out rows whose first cell is even:
+
+<pre>17 34
+8 <strike>68</strike>
+4 <strike>136</strike>
+2 <strike>272</strike>
+1 544
+</pre>
+
+Sum the remaining numbers in the right-hand column:
+
+<!-- markdownlint-disable MD003 -->
+
+<pre>17 34
+8 --
+4 ---
+2 ---
+1 544
+ ====
+ 578
+</pre>
+
+<!-- markdownlint-enable MD003 -->
+
+So `17` multiplied by `34`, by the Ethiopian method is `578`.
+
+# --instructions--
+
+The task is to define three named functions/methods/procedures/subroutines:
+
+<ol>
+ <li>one to halve an integer,</li>
+ <li>one to double an integer, and</li>
+ <li>one to state if an integer is even</li>
+</ol>
+
+Use these functions to create a function that does Ethiopian multiplication.
+
+<!-- markdownlint-disable MD046-->
# --hints--
-`eth_mult`是一个功能。
+`eth_mult` should be a function.
```js
assert(typeof eth_mult === 'function');
```
-`eth_mult(17,34)`应该返回`578` 。
+`eth_mult(17,34)` should return `578`.
```js
assert.equal(eth_mult(17, 34), 578);
```
-`eth_mult(23,46)`应该返回`1058` 。
+`eth_mult(23,46)` should return `1058`.
```js
assert.equal(eth_mult(23, 46), 1058);
```
-`eth_mult(12,27)`应该返回`324` 。
+`eth_mult(12,27)` should return `324`.
```js
assert.equal(eth_mult(12, 27), 324);
```
-`eth_mult(56,98)`应该返回`5488` 。
+`eth_mult(56,98)` should return `5488`.
```js
assert.equal(eth_mult(56, 98), 5488);
```
-`eth_mult(63,74)`应该返回`4662` 。
+`eth_mult(63,74)` should return `4662`.
```js
assert.equal(eth_mult(63, 74), 4662);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/euler-method.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/euler-method.md
index b8d8579d5fc..4a17c09a25d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/euler-method.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/euler-method.md
@@ -1,42 +1,125 @@
---
id: 59880443fb36441083c6c20e
-title: 欧拉方法
+title: Euler method
challengeType: 5
-videoUrl: ''
+forumTopicId: 302258
dashedName: euler-method
---
# --description--
-<p>欧拉方法在数值上近似具有给定初始值的一阶常微分方程（ODE）的解。它是解决初始值问题（IVP）的一种显式方法，如<a href='https://en.wikipedia.org/wiki/Euler method' title='wp：欧拉方法'>维基百科页面中所述</a> 。 </p><p> ODE必须以下列形式提供： </p><p> :: <big>$ \ frac {dy（t）} {dt} = f（t，y（t））$</big> </p><p>具有初始值</p><p> :: <big>$ y（t_0）= y_0 $</big> </p><p>为了得到数值解，我们用有限差分近似替换LHS上的导数： </p><p> :: <big>$ \ frac {dy（t）} {dt} \ approx \ frac {y（t + h）-y（t）} {h} $</big> </p><p>然后解决$ y（t + h）$： </p><p> :: <big>$ y（t + h）\ about y（t）+ h \，\ frac {dy（t）} {dt} $</big> </p><p>这是一样的</p><p> :: <big>$ y（t + h）\ about y（t）+ h \，f（t，y（t））$</big> </p><p>然后迭代解决方案规则是： </p><p> :: <big>$ y_ {n + 1} = y_n + h \，f（t_n，y_n）$</big> </p><p>其中<big>$ h $</big>是步长，是解决方案准确性最相关的参数。较小的步长会提高精度，但也会增加计算成本，因此必须根据手头的问题手工挑选。 </p><p>示例：牛顿冷却法</p><p> Newton的冷却定律描述了在温度<big>$ T_R $</big>的环境中初始温度<big>$ T（t_0）= T_0 $</big>的对象如何冷却： </p><p> :: <big>$ \ frac {dT（t）} {dt} = -k \，\ Delta T $</big> </p><p>要么</p><p> :: <big>$ \ frac {dT（t）} {dt} = -k \，（T（t） - T_R）$</big> </p><p>它表示物体的冷却速率<big>$ \ frac {dT（t）} {dt} $</big>与周围环境的当前温差<big>$ \ Delta T =（T（t） - T_R）$成正比</big> 。 </p><p>我们将与数值近似进行比较的解析解是</p><p> :: <big>$ T（t）= T_R +（T_0 - T_R）\; Ë^ { -克拉} $</big> </p>任务： <p>实现欧拉方法的一个例程，然后用它来解决牛顿冷却定律的给定例子，它有三种不同的步长： </p><p> :: * 2秒</p><p> :: * 5秒和</p><p> :: * 10秒</p><p>并与分析解决方案进行比较。 </p>初始值： <p> :: *初始温度<big>$ T_0 $</big>应为100°C </p><p> :: *室温<big>$ T_R $</big>应为20°C </p><p> :: *冷却常数<big>$ k $</big>应为0.07 </p><p> :: *计算的时间间隔应为0s──►100s </p>
+Euler's method numerically approximates solutions of first-order ordinary differential equations (ODEs) with a given initial value. It is an explicit method for solving initial value problems (IVPs), as described in [the wikipedia page](https://en.wikipedia.org/wiki/Euler method "wp: Euler method").
+
+The ODE has to be provided in the following form:
+
+<ul style='list-style: none;'>
+ <li><big>$\frac{dy(t)}{dt} = f(t,y(t))$</big></li>
+</ul>
+
+with an initial value
+
+<ul style='list-style: none;'>
+ <li><big>$y(t_0) = y_0$</big></li>
+</ul>
+
+To get a numeric solution, we replace the derivative on the LHS with a finite difference approximation:
+
+<ul style='list-style: none;'>
+ <li><big>$\frac{dy(t)}{dt} \approx \frac{y(t+h)-y(t)}{h}$</big></li>
+</ul>
+
+then solve for $y(t+h)$:
+
+<ul style='list-style: none;'>
+ <li><big>$y(t+h) \approx y(t) + h \, \frac{dy(t)}{dt}$</big></li>
+</ul>
+
+which is the same as
+
+<ul style='list-style: none;'>
+ <li><big>$y(t+h) \approx y(t) + h \, f(t,y(t))$</big></li>
+</ul>
+
+The iterative solution rule is then:
+
+<ul style='list-style: none;'>
+ <li><big>$y_{n+1} = y_n + h \, f(t_n, y_n)$</big></li>
+</ul>
+
+where $h$ is the step size, the most relevant parameter for accuracy of the solution. A smaller step size increases accuracy but also the computation cost, so it has always has to be hand-picked according to the problem at hand.
+
+**Example: Newton's Cooling Law**
+
+Newton's cooling law describes how an object of initial temperature $T(t_0) = T_0$ cools down in an environment of temperature $T_R$:
+
+<ul style='list-style: none;'>
+ <li><big>$\frac{dT(t)}{dt} = -k \, \Delta T$</big></li>
+</ul>
+
+or
+
+<ul style='list-style: none;'>
+ <li><big>$\frac{dT(t)}{dt} = -k \, (T(t) - T_R)$</big></li>
+</ul>
+
+It says that the cooling rate $\\frac{dT(t)}{dt}$ of the object is proportional to the current temperature difference $\\Delta T = (T(t) - T_R)$ to the surrounding environment.
+
+The analytical solution, which we will compare to the numerical approximation, is
+
+<ul style='list-style: none;'>
+ <li><big>$T(t) = T_R + (T_0 - T_R) \; e^{-k t}$</big></li>
+</ul>
+
+# --instructions--
+
+Implement a routine of Euler's method and then use it to solve the given example of Newton's cooling law for three different step sizes of:
+
+<ul>
+ <li><code>2 s</code></li>
+ <li><code>5 s</code> and</li>
+ <li><code>10 s</code></li>
+</ul>
+
+and compare with the analytical solution.
+
+**Initial values:**
+
+<ul>
+ <li>initial temperature <big>$T_0$</big> shall be <code>100 °C</code></li>
+ <li>room temperature <big>$T_R$</big> shall be <code>20 °C</code></li>
+ <li>cooling constant <big>$k$</big> shall be <code>0.07</code></li>
+ <li>time interval to calculate shall be from <code>0 s</code> to <code>100 s</code></li>
+</ul>
+
+First parameter to the function is initial time, second parameter is initial temperature, third parameter is elapsed time and fourth parameter is step size.
# --hints--
-`eulersMethod`是一个函数。
+`eulersMethod` should be a function.
```js
assert(typeof eulersMethod === 'function');
```
-`eulersMethod(0, 100, 100, 10)`应该返回一个数字。
+`eulersMethod(0, 100, 100, 2)` should return a number.
```js
-assert(typeof eulersMethod(0, 100, 100, 10) === 'number');
+assert(typeof eulersMethod(0, 100, 100, 2) === 'number');
```
-`eulersMethod(0, 100, 100, 10)`应返回20.0424631833732。
+`eulersMethod(0, 100, 100, 2)` should return 20.0424631833732.
```js
assert.equal(eulersMethod(0, 100, 100, 2), 20.0424631833732);
```
-`eulersMethod(0, 100, 100, 10)`应返回20.01449963666907。
+`eulersMethod(0, 100, 100, 5)` should return 20.01449963666907.
```js
assert.equal(eulersMethod(0, 100, 100, 5), 20.01449963666907);
```
-`eulersMethod(0, 100, 100, 10)`应返回20.000472392。
+`eulersMethod(0, 100, 100, 10)` should return 20.000472392.
```js
assert.equal(eulersMethod(0, 100, 100, 10), 20.000472392);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/evaluate-binomial-coefficients.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/evaluate-binomial-coefficients.md
index efe5d045ffa..5b48751996c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/evaluate-binomial-coefficients.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/evaluate-binomial-coefficients.md
@@ -1,48 +1,52 @@
---
id: 598de241872ef8353c58a7a2
-title: 评估二项式系数
+title: Evaluate binomial coefficients
challengeType: 5
-videoUrl: ''
+forumTopicId: 302259
dashedName: evaluate-binomial-coefficients
---
# --description--
-<p>写一个函数来计算给定n和k值的二项式系数。 </p><p>推荐这个公式： </p> $ \\ binom {n} {k} = \\ frac {n！} {（nk）！k！} = \\ frac {n（n-1）（n-2）\\ ldots（n-k + 1）} { k（k-1）（k-2）\\ ldots 1} $
+Write a function to calculate the binomial coefficient for the given value of n and k.
+
+This formula is recommended:
+
+$\\binom{n}{k} = \\frac{n!}{(n-k)!k!} = \\frac{n(n-1)(n-2)\\ldots(n-k+1)}{k(k-1)(k-2)\\ldots 1}$
# --hints--
-`binom`是一个功能。
+`binom` should be a function.
```js
assert(typeof binom === 'function');
```
-`binom(5,3)`应该返回10。
+`binom(5,3)` should return 10.
```js
assert.equal(binom(5, 3), 10);
```
-`binom(7,2)`应该返回21。
+`binom(7,2)` should return 21.
```js
assert.equal(binom(7, 2), 21);
```
-`binom(10,4)`应该返回210。
+`binom(10,4)` should return 210.
```js
assert.equal(binom(10, 4), 210);
```
-`binom(6,1)`应该返回6。
+`binom(6,1)` should return 6.
```js
assert.equal(binom(6, 1), 6);
```
-`binom(12,8)`应该返回495。
+`binom(12,8)` should return 495.
```js
assert.equal(binom(12, 8), 495);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/execute-a-markov-algorithm.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/execute-a-markov-algorithm.md
index 3f8ccc9895e..499a48f0a69 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/execute-a-markov-algorithm.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/execute-a-markov-algorithm.md
@@ -1,18 +1,16 @@
---
id: 59e09e6d412c5939baa02d16
-title: 执行马尔可夫算法
+title: Execute a Markov algorithm
challengeType: 5
-videoUrl: ''
+forumTopicId: 302260
dashedName: execute-a-markov-algorithm
---
# --description--
-任务：
+Create an interpreter for a [Markov Algorithm](https://en.wikipedia.org/wiki/Markov algorithm "wp: Markov algorithm").
-为[马尔可夫算法](<https://en.wikipedia.org/wiki/Markov algorithm> "wp：马尔可夫算法")创建解释器。
-
-规则的语法如下：
+Rules have the syntax:
<pre>[ruleset] ::= (([comment] | [rule]) [newline]+)*
[comment] ::= # {[any character]}
@@ -20,60 +18,60 @@ dashedName: execute-a-markov-algorithm
[whitespace] ::= ([tab] | [space]) [[whitespace]]
</pre>
-每行有一条规则。
+There is one rule per line.
-如果在 \[replacement] 之前有一个 `.`（句点），那么这就是一个终止规则。在这种情况下，解释器必须停止执行。
+If there is a `.` (period) present before the \[replacement], then this is a terminating rule in which case the interpreter must halt execution.
-规则集由一系列规则组成，可能包含一些注释。
+A ruleset consists of a sequence of rules, with optional comments.
-规则集
+Rulesets
-我们会对你提交的代码进行如下测试：
+Use the following tests on entries:
-**规则集 1：**
+**Ruleset 1:**
-<pre># 此条规则来自 Wikipedia：
+<pre># This rules file is extracted from Wikipedia:
# <code>http://en.wikipedia.org/wiki/Markov_Algorithm</code>
A -> apple
B -> bag
S -> shop
T -> the
the shop -> my brother
-(终止规则) -> .
+a never used -> .terminating rule
</pre>
-对于这段文本：
+Sample text of:
`I bought a B of As from T S.`
-应该输出：
+Should generate the output:
`I bought a bag of apples from my brother.`
-**规则集 2：**
+**Ruleset 2:**
-终止规则的测试
+A test of the terminating rule
-<pre># 基于 Wikipedia 的规则稍做修改
+<pre># Slightly modified from the rules on Wikipedia
A -> apple
B -> bag
S -> .shop
T -> the
the shop -> my brother
-(终止规则) -> .
+a never used -> .terminating rule
</pre>
-对于这段文本：
+Sample text of:
`I bought a B of As from T S.`
-应该输出：
+Should generate:
`I bought a bag of apples from T shop.`
-**规则集 3：**
+**Ruleset 3:**
-这条不仅可以用来测试替换顺序是否正确，还可以测试你的代码中对正则表达式的处理是否完备。如果你的代码没有对正则表达式进行正确的转义处理，那在替换的时候就会出现问题。
+This tests for correct substitution order and may trap simple regexp based replacement routines if special regexp characters are not escaped.
<pre># BNF Syntax testing rules
A -> apple
@@ -85,31 +83,32 @@ W -> WW
S -> .shop
T -> the
the shop -> my brother
-(终止规则) -> .
+a never used -> .terminating rule
</pre>
-对于这段文本：
+Sample text of:
`I bought a B of As W my Bgage from T S.`
-应该输出：
+Should generate:
-`I bought a bag of apples with my money from T shop.`
+`I bought a bag of apples with my money from T shop.`
-**规则集 4：**
+**Ruleset 4:**
-这条是用来测试规则扫描的顺序是否正确，并可能捕获以错误顺序扫描的替换例程。这里我们选取了通用的一元乘法引擎（请注意，在此实现中，输入的表达式必须放在两个下划线之间）。
+This tests for correct order of scanning of rules, and may trap replacement routines that scan in the wrong order. It implements a general unary multiplication engine. (Note that the input expression must be placed within underscores in this implementation.)
-<pre> ##一元乘法引擎，用于测试马尔可夫算法实现
-### by Donal Fellows
-# 一元加法引擎
+<pre>### Unary Multiplication Engine, for testing Markov Algorithm implementations
+### By Donal Fellows.
+# Unary addition engine
_+1 -> _1+
1+1 -> 11+
-# 将乘法转换为普通加法
+# Pass for converting from the splitting of multiplication into ordinary
+# addition
1! -> !1
,! -> !+
_! -> _
-# 一元乘法，左侧为被乘数，右侧为乘数
+# Unary multiplication by duplicating left side, right side times
1*1 -> x,@y
1x -> xX
X, -> 1,1
@@ -118,91 +117,94 @@ _x -> _X
,x -> ,X
y1 -> 1y
y_ -> _
-# 下一阶段
+# Next phase of applying
1@1 -> x,@y
1@_ -> @_
,@_ -> !_
++ -> +
-# 加法的终止条件
+# Termination cleanup for addition
_1 -> 1
1+_ -> 1
_+_ ->
</pre>
-对于这段文本：
+Sample text of:
`_1111*11111_`
-应该输出：
+should generate the output:
`11111111111111111111`
-**规则集5：**
+**Ruleset 5:**
-一台简单的[图灵机](http://en.wikipedia.org/wiki/Turing_machine)包含三个状态的["忙碌海狸"](http://en.wikipedia.org/wiki/Busy_beaver)。纸带由 0 和 1 组成，状态为 A、B、C 和代表终止（Halt）的 H。通过在字符前写状态字母来的方式来指示读写头的位置。机器运行时需要的初始纸带必须通过输入在一开始全部给出。
+A simple [Turing machine](http://en.wikipedia.org/wiki/Turing_machine "link: http://en.wikipedia.org/wiki/Turing_machine"), implementing a three-state [busy beaver](http://en.wikipedia.org/wiki/Busy_beaver "link: http://en.wikipedia.org/wiki/Busy_beaver").
-这一规则集除了可以证明 Markov 算法是图灵完备的，它还帮我找出了使用 C++ 完成此题中的一个错误，而且这个错误没有被前四个规则集抓到。
+The tape consists of `0`s and `1`s, the states are `A`, `B`, `C` and `H` (for `H`alt), and the head position is indicated by writing the state letter before the character where the head is. All parts of the initial tape the machine operates on have to be given in the input.
-<pre># 图灵机：三个状态的"忙碌海狸"
-# 状态 A，符号 0 => 写入 1，向右移动，新状态 B
+Besides demonstrating that the Markov algorithm is Turing-complete, it also made me catch a bug in the C++ implementation which wasn't caught by the first four rulesets.
+
+<pre># Turing machine: three-state busy beaver
+#
+# state A, symbol 0 => write 1, move right, new state B
A0 -> 1B
-# 状态 A，符号 1 => 写入 1，向左移动，新状态 C
+# state A, symbol 1 => write 1, move left, new state C
0A1 -> C01
1A1 -> C11
-# 状态 B，符号 0 => 写入 1，向左移动，新状态 A
+# state B, symbol 0 => write 1, move left, new state A
0B0 -> A01
1B0 -> A11
-# 状态 B，符号 1 => 写入 1，向右移动，新状态 B
-B1 - > 1B
-# 状态 C，符号 0 => 写入 1，向左移动，新状态 B
-0C0 - > B01
-1C0 - > B11
-# 状态 C，符号 1 => 写入 1，向左移动，停止
-0C1 - > H01
-1C1 - > H11
+# state B, symbol 1 => write 1, move right, new state B
+B1 -> 1B
+# state C, symbol 0 => write 1, move left, new state B
+0C0 -> B01
+1C0 -> B11
+# state C, symbol 1 => write 1, move left, halt
+0C1 -> H01
+1C1 -> H11
</pre>
-这个规则集应将这段输入：
+This ruleset should turn
-`000000A000000`
+`000000A000000`
-转换成：
+into
`00011H1111000`
# --hints--
-`markov` 应是一个函数。
+`markov` should be a function.
```js
assert(typeof markov === 'function');
```
-`markov(["A -> apple","B -> bag","S -> shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],"I bought a B of As from T S.")` 应返回 "I bought a bag of apples from my brother."。
+`markov(["A -> apple","B -> bag","S -> shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],"I bought a B of As from T S.")` should return "I bought a bag of apples from my brother.".
```js
assert.deepEqual(markov(rules[0], tests[0]), outputs[0]);
```
-`markov(["A -> apple","B -> bag","S -> .shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],"I bought a B of As from T S.")` 应返回 "I bought a bag of apples from T shop."。
+`markov(["A -> apple","B -> bag","S -> .shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],"I bought a B of As from T S.")` should return "I bought a bag of apples from T shop.".
```js
assert.deepEqual(markov(rules[1], tests[1]), outputs[1]);
```
-`markov(["A -> apple","WWWW -> with","Bgage -> ->.*","B -> bag","->.* -> money","W -> WW","S -> .shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],"I bought a B of As W my Bgage from T S.")` 应返回 "I bought a bag of apples with my money from T shop."。
+`markov(["A -> apple","WWWW -> with","Bgage -> ->.*","B -> bag","->.* -> money","W -> WW","S -> .shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],"I bought a B of As W my Bgage from T S.")` should return "I bought a bag of apples with my money from T shop.".
```js
assert.deepEqual(markov(rules[2], tests[2]), outputs[2]);
```
-`markov(["_+1 -> _1+","1+1 -> 11+","1! -> !1",",! -> !+","_! -> _","1*1 -> x,@y","1x -> xX","X, -> 1,1","X1 -> 1X","_x -> _X",",x -> ,X","y1 -> 1y","y_ -> _","1@1 -> x,@y","1@_ -> @_",",@_ -> !_","++ -> +","_1 -> 1","1+_ -> 1","_+_ -> "],"_1111*11111_")` 应返回 "11111111111111111111"。
+`markov(["_+1 -> _1+","1+1 -> 11+","1! -> !1",",! -> !+","_! -> _","1*1 -> x,@y","1x -> xX","X, -> 1,1","X1 -> 1X","_x -> _X",",x -> ,X","y1 -> 1y","y_ -> _","1@1 -> x,@y","1@_ -> @_",",@_ -> !_","++ -> +","_1 -> 1","1+_ -> 1","_+_ -> "],"_1111*11111_")` should return "11111111111111111111".
```js
assert.deepEqual(markov(rules[3], tests[3]), outputs[3]);
```
-`markov(["A0 -> 1B","0A1 -> C01","1A1 -> C11","0B0 -> A01","1B0 -> A11","B1 -> 1B","0C0 -> B01","1C0 -> B11","0C1 -> H01","1C1 -> H11"],"")` 应返回 "00011H1111000"。
+`markov(["A0 -> 1B","0A1 -> C01","1A1 -> C11","0B0 -> A01","1B0 -> A11","B1 -> 1B","0C0 -> B01","1C0 -> B11","0C1 -> H01","1C1 -> H11"],"")` should return "00011H1111000".
```js
assert.deepEqual(markov(rules[4], tests[4]), outputs[4]);
@@ -210,6 +212,29 @@ assert.deepEqual(markov(rules[4], tests[4]), outputs[4]);
# --seed--
+## --after-user-code--
+
+```js
+
+let rules=[["A -> apple","B -> bag","S -> shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],
+ ["A -> apple","B -> bag","S -> .shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],
+ ["A -> apple","WWWW -> with","Bgage -> ->.*","B -> bag","->.* -> money","W -> WW","S -> .shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],
+ ["_+1 -> _1+","1+1 -> 11+","1! -> !1",",! -> !+","_! -> _","1*1 -> x,@y","1x -> xX","X, -> 1,1","X1 -> 1X","_x -> _X",",x -> ,X","y1 -> 1y","y_ -> _","1@1 -> x,@y","1@_ -> @_",",@_ -> !_","++ -> +","_1 -> 1","1+_ -> 1","_+_ -> "],
+ ["A0 -> 1B","0A1 -> C01","1A1 -> C11","0B0 -> A01","1B0 -> A11","B1 -> 1B","0C0 -> B01","1C0 -> B11","0C1 -> H01","1C1 -> H11"]];
+let tests=["I bought a B of As from T S.",
+ "I bought a B of As from T S.",
+ "I bought a B of As W my Bgage from T S.",
+ "_1111*11111_",
+ "000000A000000"];
+let outputs=["I bought a bag of apples from my brother.",
+ "I bought a bag of apples from T shop.",
+ "I bought a bag of apples with my money from T shop.",
+ "11111111111111111111",
+ "00011H1111000"]
+
+```
+
+
## --seed-contents--
```js
@@ -248,21 +273,4 @@ function markov(rules,test) {
}
return test;
}
-
-// tail:
-let rules=[["A -> apple","B -> bag","S -> shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],
- ["A -> apple","B -> bag","S -> .shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],
- ["A -> apple","WWWW -> with","Bgage -> ->.*","B -> bag","->.* -> money","W -> WW","S -> .shop","T -> the","the shop -> my brother","a never used -> .terminating rule"],
- ["_+1 -> _1+","1+1 -> 11+","1! -> !1",",! -> !+","_! -> _","1*1 -> x,@y","1x -> xX","X, -> 1,1","X1 -> 1X","_x -> _X",",x -> ,X","y1 -> 1y","y_ -> _","1@1 -> x,@y","1@_ -> @_",",@_ -> !_","++ -> +","_1 -> 1","1+_ -> 1","_+_ -> "],
- ["A0 -> 1B","0A1 -> C01","1A1 -> C11","0B0 -> A01","1B0 -> A11","B1 -> 1B","0C0 -> B01","1C0 -> B11","0C1 -> H01","1C1 -> H11"]];
-let tests=["I bought a B of As from T S.",
- "I bought a B of As from T S.",
- "I bought a B of As W my Bgage from T S.",
- "_1111*11111_",
- "000000A000000"];
-let outputs=["I bought a bag of apples from my brother.",
- "I bought a bag of apples from T shop.",
- "I bought a bag of apples with my money from T shop.",
- "11111111111111111111",
- "00011H1111000"];
```
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/execute-brain.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/execute-brain.md
index 802f929b944..7504589e582 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/execute-brain.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/execute-brain.md
@@ -1,42 +1,61 @@
---
id: 59e0a8df964e4540d5abe599
-title: 执行大脑****
+title: Execute Brain****
challengeType: 5
-videoUrl: ''
+forumTopicId: 302261
dashedName: execute-brain
---
# --description--
-<p>编写一个函数来实现Brain ****解释器。该函数将字符串作为参数，并应返回一个字符串作为输出。更多细节如下： </p><p> <a href='http://rosettacode.org/wiki/Brainf***' title='Brainf ***'>RCBF</a>是一套为各种语言的Rosetta Code编写的<a href='http://rosettacode.org/wiki/Brainf***' title='Brainf ***'>Brainf ***</a>编译器和解释器。 </p><p>以下是RCBF每个版本的链接。 </p><p>实现只需要正确实现以下指令： </p><p> {| </p><p> ！命令</p><p> ！描述</p><p> | - </p><p> |风格=“文本对齐：中心” | <code>></code> ||将指针向右移动</p><p> | - </p><p> |风格=“文本对齐：中心” | <code><</code> ||将指针移到左侧</p><p> | - </p><p> |风格=“文本对齐：中心” | <code>+</code> ||增加指针下的内存单元格</p><p> | - </p><p> |风格=“文本对齐：中心” | <code>-</code> ||减少指针下的内存单元格</p><p> | - </p><p> |风格=“文本对齐：中心” | <code>.</code> ||输出指针处单元格表示的字符</p><p> | - </p><p> |风格=“文本对齐：中心” | <code>,</code> ||输入一个字符并将其存储在指针的单元格中</p><p> | - </p><p> |风格=“文本对齐：中心” | <code>[</code> ||如果指针下的单元格为0，则跳过匹配<code>]</code> </p><p> | - </p><p> |风格=“文本对齐：中心” | <code>]</code> ||跳回匹配<code>[</code>如果指针下的单元格非零</p><p> |} </p><p>允许任何单元格大小，EOF（ <u>E</u> nd- <u>O</u> - <u>F</u> ile）支持是可选的，无论您是否有有界或无界内存。 </p>
+Write a function to implement a Brain\*\*\*\* interpreter. The function will take a string as a parameter and should return a string as the output. More details are given below:
+
+RCBF is a set of [Brainf\*\*\*](https://rosettacode.org/wiki/Brainf*** "Brainf\*\*\*") compilers and interpreters written for Rosetta Code in a variety of languages.
+
+Below are links to each of the versions of RCBF.
+
+An implementation need only properly implement the following instructions:
+
+| Command | Description |
+| ------------------------- | ------------------------------------------------------------------------------------------ |
+| <code>></code> | Move the pointer to the right |
+| <code><</code> | Move the pointer to the left |
+| <code>+</code> | Increment the memory cell under the pointer |
+| <code>-</code> | Decrement the memory cell under the pointer |
+| <code>.</code> | Output the character signified by the cell at the pointer |
+| <code>,</code> | Input a character and store it in the cell at the pointer |
+| <code>\[</code> | Jump past the matching <code>]</code> if the cell under the pointer is 0 |
+| <code>]</code> | Jump back to the matching <code>\[</code> if the cell under the pointer is nonzero |
+
+Any cell size is allowed, EOF (*E*nd-*O*-*F*ile) support is optional, as is whether you have bounded or unbounded memory.
# --hints--
-`brain(bye)`应该重新调整一个字符串
+`brain(bye)` should return a string
```js
assert(typeof brain(bye) === 'string');
```
-<code>brain("++++++[>++++++++++<-]>+++++.")</code should return "A"
+`brain("++++++[>++++++++++<-]>+++++.")` should return "A"
```js
assert.equal(brain('++++++[>++++++++++<-]>+++++.'), 'A');
```
-`brain(bye)`应该回归`Goodbye, World!\\r\\n`
+`brain(bye)` should return `Goodbye, World!\r\n`
```js
assert.equal(brain(bye), 'Goodbye, World!\r\n');
```
-`brain(hello)`应该回归`Hello World!\\n`
+`brain(hello)` should return `Hello World!\n`
```js
assert.equal(brain(hello), 'Hello World!\n');
```
-`brain(fib)`应该返回`1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89`
+`brain(fib)` should return `1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89`
```js
assert.equal(brain(fib), '1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89');
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/extensible-prime-generator.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/extensible-prime-generator.md
index d37956690b5..620da453ffe 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/extensible-prime-generator.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/extensible-prime-generator.md
@@ -1,24 +1,35 @@
---
id: 598ee8b91b410510ae82efef
-title: 可扩展的素发生器
+title: Extensible prime generator
challengeType: 5
-videoUrl: ''
+forumTopicId: 302262
dashedName: extensible-prime-generator
---
# --description--
-<p>按顺序编写素数生成器，它将自动调整以适应任何合理高素数的生成。 </p>发电机应能：显示前<b>N个</b>黄金numbers.Show在range.Show的素数的素数在range.Show数<b><sup>第</sup> n <sup>个</sup></b>素数。 <p>该函数应该有两个参数。第一个将接收<b>n</b>或作为数组的范围。第二个将接收一个布尔值，它指定函数是否将素数作为数组或单个数字（范围中的素数或第<b>n <sup>个</sup></b>素数）返回。根据参数，函数应该返回一个数组。 </p>
+Write a generator of prime numbers, in order, that will automatically adjust to accommodate the generation of any reasonably high prime.
+
+The generator should be able to:
+
+<ul>
+ <li>Show the first <code>n</code> prime numbers</li>
+ <li>Show the prime numbers in a range</li>
+ <li>Show the number of primes in a range</li>
+ <li>Show the <code>n<sup>th</sup></code> prime number</li>
+</ul>
+
+The function should have two parameters. The first will receive `n` or the range as an array. The second will receive a boolean, that specifies if the function returns the prime numbers as an array or a single number(the number of primes in the range or the <code>n<sup>th</sup></code> prime). According to the parameters the function should return an array.
# --hints--
-`primeGenerator`是一个函数。
+`primeGenerator` should be a function.
```js
assert(typeof primeGenerator === 'function');
```
-`primeGenerator`是一个函数。
+`primeGenerator(20, true)` should return `[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71]`.
```js
assert.deepEqual(primeGenerator(20, true), [
@@ -45,7 +56,7 @@ assert.deepEqual(primeGenerator(20, true), [
]);
```
-`primeGenerator`是一个函数。
+`primeGenerator([100, 150], true)` should return `[101, 103, 107, 109, 113, 127, 131, 137, 139, 149]`.
```js
assert.deepEqual(primeGenerator([100, 150], true), [
@@ -62,13 +73,13 @@ assert.deepEqual(primeGenerator([100, 150], true), [
]);
```
-`primeGenerator`是一个函数。
+`primeGenerator([7700, 8000], false)` should return `30`.
```js
assert.equal(primeGenerator([7700, 8000], false), 30);
```
-`primeGenerator`是一个函数。
+`primeGenerator(10000, false)` should return `104729`.
```js
assert.equal(primeGenerator(10000, false), 104729);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/factorial.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/factorial.md
index 84b89ffce25..f58bdffe25e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/factorial.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/factorial.md
@@ -1,42 +1,56 @@
---
id: 597b2b2a2702b44414742771
-title: 阶乘
+title: Factorial
challengeType: 5
-videoUrl: ''
+forumTopicId: 302263
dashedName: factorial
---
# --description--
-<p>编写一个函数来返回一个数字的阶乘。 </p><p>一个数字的因子由下式给出： </p> N！ = n \*（n-1）\*（n-2）\* ..... \* 1 <p>例如：3！ = 3 * 2 * 1 = 6 4！ = 4 * 3 * 2 * 1 = 24 </p><p>注意：0！ = 1 </p>
+Write a function to return the factorial of a number.
+
+Factorial of a number is given by:
+
+<pre><big>n! = n * (n-1) * (n-2) * ..... * 1</big>
+</pre>
+
+For example:
+
+<ul>
+ <li><code>3! = 3 * 2 * 1 = 6</code></li>
+ <li><code>4! = 4 * 3 * 2 * 1 = 24</code></li>
+</ul>
+
+**Note:** `0! = 1`
# --hints--
-`factorial`是一种功能。
+`factorial` should be a function.
```js
assert(typeof factorial === 'function');
```
-`factorial(2)`应该返回一个数字。
+`factorial(2)` should return a number.
```js
assert(typeof factorial(2) === 'number');
```
-`factorial(3)`应该返回6.“）
+`factorial(3)` should return 6.
```js
assert.equal(factorial(3), 6);
```
-`factorial(3)`应返回120.“）
+`factorial(5)` should return 120.
```js
assert.equal(factorial(5), 120);
```
-`factorial(3)`应返回3,628,800。“）
+`factorial(10)` should return 3,628,800.
```js
assert.equal(factorial(10), 3628800);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/factors-of-a-mersenne-number.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/factors-of-a-mersenne-number.md
index 9f72beb2b3f..f450b13ddbe 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/factors-of-a-mersenne-number.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/factors-of-a-mersenne-number.md
@@ -1,42 +1,88 @@
---
id: 598eea87e5cf4b116c3ff81a
-title: 梅森数的因素
+title: Factors of a Mersenne number
challengeType: 5
-videoUrl: ''
+forumTopicId: 302264
dashedName: factors-of-a-mersenne-number
---
# --description--
-<p>梅森数是2 <sup>P</sup> -1形式的数字。 </p><p>如果P是素数，那么梅森数可能是梅森素数</p><p> （如果P不是素数，则梅森数也不是素数）。 </p><p>在搜索梅森素数时，通过在开始可能冗长的<a href='http://rosettacode.org/wiki/Lucas-Lehmer test' title='Lucas-Lehmer测试'>Lucas-Lehmer检验</a>之前找到一个小因子来消除指数是有利的。 </p><p>有非常有效的算法来确定数字是否除以2 <sup>P</sup> -1（或等效地，如果2 <sup>P</sup> mod（数字）= 1）。 </p><p>有些语言已经有了这个exponent-and-mod操作的内置实现（称为modPow或类似的）。 </p><p>以下是如何自己实现这个modPow： </p><p>例如，让我们计算2 <sup>23</sup> mod 47。 </p><p>将指数23转换为二进制，得到10111.从<tt>square</tt> = 1开始，重复平方。 </p><p>卸下指数的最高位，并且如果它是1 <tt>平方</tt>乘以由所述幂（2）的基础上，然后计算<tt>平方</tt>模47。 </p><p>在下一步中使用最后一步的模数结果作为<tt>square</tt>的初始值： </p><p>删除可选</p><p>方形顶部位乘以2 mod 47 </p><p> ------------ ------- ------------- ------ </p><p> 1 * 1 = 1 1 0111 1 * 2 = 2 2 </p><p> 2 * 2 = 4 0 111否4 </p><p> 4 * 4 = 16 1 11 16 * 2 = 32 32 </p><p> 32 * 32 = 1024 1 1 1024 * 2 = 2048 27 </p><p> 27 * 27 = 729 1 729 * 2 = 1458 1 </p><p>由于2 <sup>23</sup> mod 47 = 1,47是2 <sup>P</sup> -1的因子。 </p><p> （要看到这一点，从两边减去1：2 <sup>23</sup> -1 = 0 mod 47.） </p><p>由于我们已经证明47是一个因子，因此2 <sup>23</sup> -1不是素数。 </p><p> Mersenne数字的其他属性使我们能够进一步完善这一过程。 </p><p>任何因子q为2 <sup>P</sup> -1必须是2kP + 1的形式，k是正整数或零。此外，q必须是1或7 mod 8。 </p><p>最后任何潜在因素q必须是<a href='http://rosettacode.org/wiki/Primality by Trial Division' title='审判分庭的原始性'>素数</a> 。 </p><p>与其他试验划分算法一样，算法在2kP + 1> sqrt（N）时停止。 </p><p>这些素性测试仅适用于P为素数的Mersenne数。例如，M <sub>4</sub> = 15不使用这些技术产生因子，而是产生3和5的因子，两者都不符合2kP + 1。 </p>任务： <p>使用上述方法找到因子2 <sup>929</sup> -1（又名M929） </p>相关任务： <a href='http://rosettacode.org/wiki/count in factors' title='算上因素'>计数因素</a> <a href='http://rosettacode.org/wiki/prime decomposition' title='主要分解'>素数分解</a> <a href='http://rosettacode.org/wiki/factors of an integer' title='整数的因子'>的整数的因素</a> <a href='http://rosettacode.org/wiki/Sieve of Eratosthenes' title='Eratosthenes的筛子'>埃拉托塞尼的筛</a> <a href='http://rosettacode.org/wiki/primality by trial division' title='审判分裂的素性'>通过试验除法素性</a> <a href='http://rosettacode.org/wiki/trial factoring of a Mersenne number' title='试用Mensenne数的因式'>梅森数的试验理</a> <a href='http://rosettacode.org/wiki/partition an integer X into N primes' title='将整数X划分为N个素数'>分区的整数X为N个素数</a> <a href='http://rosettacode.org/wiki/sequence of primes by Trial Division' title='审判分庭的素数序列'>由审判庭素数的序列</a> <a href='https://www.youtube.com/watch?v=SNwvJ7psoow' title='链接：https：//www.youtube.com/watch？v = SNwvJ7psoow'>在1948年计算机：2¹²⁷-1</a>
+A Mersenne number is a number in the form of <code>2<sup>P</sup>-1</code>.
+
+If `P` is prime, the Mersenne number may be a Mersenne prime. (If `P` is not prime, the Mersenne number is also not prime.)
+
+In the search for Mersenne prime numbers it is advantageous to eliminate exponents by finding a small factor before starting a, potentially lengthy, [Lucas-Lehmer test](https://rosettacode.org/wiki/Lucas-Lehmer test "Lucas-Lehmer test").
+
+There are very efficient algorithms for determining if a number divides <code>2<sup>P</sup>-1</code> (or equivalently, if <code>2<sup>P</sup> mod (the number) = 1</code>).
+
+Some languages already have built-in implementations of this exponent-and-mod operation (called modPow or similar).
+
+The following is how to implement this modPow yourself:
+
+For example, let's compute <code>2<sup>23</sup> mod 47</code>.
+
+Convert the exponent 23 to binary, you get 10111. Starting with <code><tt>square</tt> = 1</code>, repeatedly square it.
+
+Remove the top bit of the exponent, and if it's 1 multiply `square` by the base of the exponentiation (2), then compute <code><tt>square</tt> modulo 47</code>.
+
+Use the result of the modulo from the last step as the initial value of `square` in the next step:
+
+<pre>Remove Optional
+square top bit multiply by 2 mod 47
+------------ ------- ------------- ------
+1*1 = 1 1 0111 1*2 = 2 2
+2*2 = 4 0 111 no 4
+4*4 = 16 1 11 16*2 = 32 32
+32*32 = 1024 1 1 1024*2 = 2048 27
+27*27 = 729 1 729*2 = 1458 1
+</pre>
+
+Since <code>2<sup>23</sup> mod 47 = 1</code>, 47 is a factor of <code>2<sup>P</sup>-1</code>.
+
+(To see this, subtract 1 from both sides: <code>2<sup>23</sup>-1 = 0 mod 47</code>.)
+
+Since we've shown that 47 is a factor, <code>2<sup>23</sup>-1</code> is not prime.
+
+Further properties of Mersenne numbers allow us to refine the process even more.
+
+Any factor `q` of <code>2<sup>P</sup>-1</code> must be of the form `2kP+1`, `k` being a positive integer or zero. Furthermore, `q` must be `1` or `7 mod 8`.
+
+Finally any potential factor `q` must be [prime](https://rosettacode.org/wiki/Primality by Trial Division "Primality by Trial Division").
+
+As in other trial division algorithms, the algorithm stops when `2kP+1 > sqrt(N)`.These primarily tests only work on Mersenne numbers where `P` is prime. For example, <code>M<sub>4</sub>=15</code> yields no factors using these techniques, but factors into 3 and 5, neither of which fit `2kP+1`.
+
+# --instructions--
+
+Using the above method find a factor of <code>2<sup>929</sup>-1</code> (aka M929)
# --hints--
-`check_mersenne`是一个函数。
+`check_mersenne` should be a function.
```js
assert(typeof check_mersenne === 'function');
```
-`check_mersenne(3)`应该返回一个字符串。
+`check_mersenne(3)` should return a string.
```js
assert(typeof check_mersenne(3) == 'string');
```
-`check_mersenne(3)`应该返回“M3 = 2 ^ 3-1是素数”。
+`check_mersenne(3)` should return "M3 = 2^3-1 is prime".
```js
assert.equal(check_mersenne(3), 'M3 = 2^3-1 is prime');
```
-`check_mersenne(23)`应返回“M23 = 2 ^ 23-1与因子47复合”。
+`check_mersenne(23)` should return "M23 = 2^23-1 is composite with factor 47".
```js
assert.equal(check_mersenne(23), 'M23 = 2^23-1 is composite with factor 47');
```
-`check_mersenne(929)`应返回“M929 = 2 ^ 929-1与因子13007复合
+`check_mersenne(929)` should return "M929 = 2^929-1 is composite with factor 13007
```js
assert.equal(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/factors-of-an-integer.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/factors-of-an-integer.md
index 12dcf06644b..31d9b722b7f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/factors-of-an-integer.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/factors-of-an-integer.md
@@ -1,36 +1,38 @@
---
id: 597f1e7fbc206f0e9ba95dc4
-title: 整数因子
+title: Factors of an integer
challengeType: 5
-videoUrl: ''
+forumTopicId: 302265
dashedName: factors-of-an-integer
---
# --description--
-<p>编写一个返回正整数因子的函数。 </p><p>这些因子是正整数，通过该正整数可以将被分解的数量除以产生正整数结果。 </p> ///
+Write a function that returns the factors of a positive integer as an array.
+
+These factors are the positive integers by which the number being factored can be divided to yield a positive integer result.
# --hints--
-`factors`是一种功能。
+`factors` should be a function.
```js
assert(typeof factors === 'function');
```
-`factors(45)`应该返回`[1,3,5,9,15,45]` 。
+`factors(45)` should return `[1,3,5,9,15,45]`.
```js
assert.deepEqual(factors(45), ans[0]);
```
-`factors(53)`应该返回`[1,53]` 。
+`factors(53)` should return `[1,53]`.
```js
assert.deepEqual(factors(53), ans[1]);
```
-`factors(64)`应该返回`[1,2,4,8,16,32,64]` 。
+`factors(64)` should return `[1,2,4,8,16,32,64]`.
```js
assert.deepEqual(factors(64), ans[2]);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/farey-sequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/farey-sequence.md
index 2641088dd2d..59b4edc9a54 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/farey-sequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/farey-sequence.md
@@ -1,42 +1,65 @@
---
id: 59c3ec9f15068017c96eb8a3
-title: Farey序列
+title: Farey sequence
challengeType: 5
-videoUrl: ''
+forumTopicId: 302266
dashedName: farey-sequence
---
# --description--
-<p>编写一个返回n阶Farey序列的函数。该函数应该有一个参数n。它应该将序列作为数组返回。阅读以下内容了解更多详情： </p><p>阶数n的<a href='https://en.wikipedia.org/wiki/Farey sequence' title='wp：Farey序列'>Farey序列</a> F <sub>n</sub>是在0和1之间的完全减少的分数的序列，当在最低阶段时，具有小于或等于n的分母，按照增大的大小排列。 </p><p> Farey序列有时被错误地称为Farey系列。 </p><p>每个Farey序列： </p><p> :: *以值0开头，由分数$ \ frac {0} {1} $表示</p><p> :: *以值1结尾，由$ \ frac {1} {1} $分数表示。 </p><p>订单1到5的Farey序列是： </p><p> $ {\ bf \ it {F}} _ 1 = \ frac {0} {1}，\ frac {1} {1} $ </p><p></p><p> $ {\ bf \ it {F}} _ 2 = \ frac {0} {1}，\ frac {1} {2}，\ frac {1} {1} $ </p><p></p><p> $ {\ bf \ it {F}} _ 3 = \ frac {0} {1}，\ frac {1} {3}，\ frac {1} {2}，\ frac {2} {3}，\ frac {1} {1} $ </p><p></p><p> $ {\ bf \ it {F}} _ 4 = \ frac {0} {1}，\ frac {1} {4}，\ frac {1} {3}，\ frac {1} {2}，\ frac {2} {3}，\ frac {3} {4}，\ frac {1} {1} $ </p><p></p><p> $ {\ bf \ it {F}} _ 5 = \ frac {0} {1}，\ frac {1} {5}，\ frac {1} {4}，\ frac {1} {3}，\ frac {2} {5}，\ frac {1} {2}，\ frac {3} {5}，\ frac {2} {3}，\ frac {3} {4}，\ frac {4} {5 }，\ frac {1} {1} $ </p>
+The [Farey sequence](https://en.wikipedia.org/wiki/Farey sequence "wp: Farey sequence") <code>F<sub>n</sub></code> of order `n` is the sequence of completely reduced fractions between `0` and `1` which, when in lowest terms, have denominators less than or equal to `n`, arranged in order of increasing size.
+
+The *Farey sequence* is sometimes incorrectly called a *Farey series*.
+
+Each Farey sequence:
+
+<ul>
+ <li>starts with the value 0, denoted by the fraction $ \frac{0}{1} $</li>
+ <li>ends with the value 1, denoted by the fraction $ \frac{1}{1}$.</li>
+</ul>
+
+The Farey sequences of orders `1` to `5` are:
+
+<ul>
+ <li style='list-style: none;'>${\bf\it{F}}_1 = \frac{0}{1}, \frac{1}{1}$</li>
+ <li style='list-style: none;'>${\bf\it{F}}_2 = \frac{0}{1}, \frac{1}{2}, \frac{1}{1}$</li>
+ <li style='list-style: none;'>${\bf\it{F}}_3 = \frac{0}{1}, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, \frac{1}{1}$</li>
+ <li style='list-style: none;'>${\bf\it{F}}_4 = \frac{0}{1}, \frac{1}{4}, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{1}{1}$</li>
+ <li style='list-style: none;'>${\bf\it{F}}_5 = \frac{0}{1}, \frac{1}{5}, \frac{1}{4}, \frac{1}{3}, \frac{2}{5}, \frac{1}{2}, \frac{3}{5}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{1}{1}$</li>
+</ul>
+
+# --instructions--
+
+Write a function that returns the Farey sequence of order `n`. The function should have one parameter that is `n`. It should return the sequence as an array.
# --hints--
-`farey`是一种功能。
+`farey` should be a function.
```js
assert(typeof farey === 'function');
```
-`farey(3)`应该返回一个数组
+`farey(3)` should return an array
```js
assert(Array.isArray(farey(3)));
```
-`farey(3)`应该返回`["1/3","1/2","2/3"]`
+`farey(3)` should return `["1/3","1/2","2/3"]`
```js
assert.deepEqual(farey(3), ['1/3', '1/2', '2/3']);
```
-`farey(4)`应该返回`["1/4","1/3","1/2","2/4","2/3","3/4"]`
+`farey(4)` should return `["1/4","1/3","1/2","2/4","2/3","3/4"]`
```js
assert.deepEqual(farey(4), ['1/4', '1/3', '1/2', '2/4', '2/3', '3/4']);
```
-`farey(5)`应返回`["1/5","1/4","1/3","2/5","1/2","2/4","3/5","2/3","3/4","4/5"]`
+`farey(5)` should return `["1/5","1/4","1/3","2/5","1/2","2/4","3/5","2/3","3/4","4/5"]`
```js
assert.deepEqual(farey(5), [
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/fibonacci-n-step-number-sequences.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/fibonacci-n-step-number-sequences.md
index 03c846f96fe..456b87de792 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/fibonacci-n-step-number-sequences.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/fibonacci-n-step-number-sequences.md
@@ -1,60 +1,89 @@
---
id: 598eef80ba501f1268170e1e
-title: 斐波那契n步数序列
+title: Fibonacci n-step number sequences
challengeType: 5
-videoUrl: ''
+forumTopicId: 302267
dashedName: fibonacci-n-step-number-sequences
---
# --description--
-<p>编写一个函数来生成Fibonacci n步数序列和Lucas序列。第一个参数是n。第二个参数是要返回的元素数。第三个参数将指定是输出Fibonacci序列还是Lucas序列。如果参数为“f”，则返回Fibonacci序列，如果为“l”，则返回Lucas序列。序列必须作为数组返回。更多细节如下： </p><p>这些数字序列是普通<a href='http://rosettacode.org/wiki/Fibonacci sequence' title='斐波那契序列'>斐波纳契数列</a>的扩展，其中： </p>对于$ n = 2 $，我们有Fibonacci序列;初始值$ \[1,1] $和$ F_k ^ 2 = F\_ {k-1} ^ 2 + F\_ {k-2} ^ 2 $对于$ n = 3 $，我们有tribonacci序列;初始值$ \[1,1,2] $和$ F_k ^ 3 = F\_ {k-1} ^ 3 + F\_ {k-2} ^ 3 + F\_ {k-3} ^ 3 $ $ $ = 4 $我们有tetranacci序列;初始值$ \[1,1,2,4] $和$ F_k ^ 4 = F\_ {k-1} ^ 4 + F\_ {k-2} ^ 4 + F\_ {k-3} ^ 4 + F\_ {k -4} ^ 4 $ ...对于一般的$ n> 2 $，我们有斐波那契$ n $ -step序列 - $ F_k ^ n $; $（n-1）$'斐波那契$ n $ -step序列$ F_k ^ {n-1} $的前$ n $值的初始值;和$ k $'这个$ n $'序列的值是$ F_k ^ n = \\ sum\_ {i = 1} ^ {（n）} {F\_ {ki} ^ {（n）}} $ <p>对于$ n $的小值， <a href='https://en.wikipedia.org/wiki/Number prefix#Greek_series' title='wp：数字前缀＃Greek_series'>希腊数字前缀</a>有时用于单独命名每个系列。 </p><p> {| style =“text-align：left;” border =“4”cellpadding =“2”cellspacing =“2” </p><p> | + Fibonacci $ n $ -step序列</p><p> | - style =“background-color：rgb（255,204,255）;” </p><p> ！ $ n $ !!系列名称!!值</p><p> | - </p><p> | 2 ||斐波那契|| 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 ...... </p><p> | - </p><p> | 3 || tribonacci || 1 1 2 4 7 13 24 44 81 149 274 504 927 1705 3136 ...... </p><p> | - </p><p> | 4 || tetranacci || 1 1 2 4 8 15 29 56 108 208 401 773 1490 2872 5536 ...... </p><p> | - </p><p> | 5 || pentanacci || 1 1 2 4 8 16 31 61 120 236 464 912 1793 3525 6930 ...... </p><p> | - </p><p> | 6 || hexanacci || 1 1 2 4 8 16 32 63 125 248 492 976 1936 3840 7617 ...... </p><p> | - </p><p> | 7 || heptanacci || 1 1 2 4 8 16 32 64 127 253 504 1004 2000 3984 7936 ...... </p><p> | - </p><p> | 8 || octonacci || 1 1 2 4 8 16 32 64 128 255 509 1016 2028 4048 8080 ... </p><p> | - </p><p> | 9 || nonanacci || 1 1 2 4 8 16 32 64 128 256 511 1021 2040 4076 8144 ...... </p><p> | - </p><p> | 10 || decanacci || 1 1 2 4 8 16 32 64 128 256 512 1023 2045 4088 8172 ... </p><p> |} </p><p>可以在更改初始值的位置生成联合序列： </p><p> <a href='https://en.wikipedia.org/wiki/Lucas number' title='wp：卢卡斯号码'>Lucas系列</a>将两个前面的值相加，例如$ n = 2 $的斐波那契数列，但使用$ [2,1] $作为其初始值。 </p><p><!-- Lucas numbers, Lucas number, Lucas series [added to make searches easier.] --></p>
+These number series are an expansion of the ordinary [Fibonacci sequence](https://rosettacode.org/wiki/Fibonacci sequence "Fibonacci sequence") where:
+
+<ol>
+ <li>For $n = 2$ we have the Fibonacci sequence; with initial values $[1, 1]$ and $F_k^2 = F_{k-1}^2 + F_{k-2}^2$</li>
+ <li>For $n = 3$ we have the tribonacci sequence; with initial values $[1, 1, 2]$ and $F_k^3 = F_{k-1}^3 + F_{k-2}^3 + F_{k-3}^3$</li>
+ <li>For $n = 4$ we have the tetranacci sequence; with initial values $[1, 1, 2, 4]$ and $F_k^4 = F_{k-1}^4 + F_{k-2}^4 + F_{k-3}^4 + F_{k-4}^4$...</li>
+ <li>For general $n>2$ we have the Fibonacci $n$-step sequence - $F_k^n$; with initial values of the first $n$ values of the $(n-1)$'th Fibonacci $n$-step sequence $F_k^{n-1}$; and $k$'th value of this $n$'th sequence being $F_k^n = \sum_{i=1}^{(n)} {F_{k-i}^{(n)}}$</li>
+</ol>
+
+For small values of $n$, [Greek numeric prefixes](https://en.wikipedia.org/wiki/Number prefix#Greek_series "wp: Number prefix#Greek_series") are sometimes used to individually name each series.
+
+Fibonacci $n$-step sequences:
+
+| $n$ | Series name | Values |
+| --- | ----------- | ------------------------------------------------------ |
+| 2 | fibonacci | 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 ... |
+| 3 | tribonacci | 1 1 2 4 7 13 24 44 81 149 274 504 927 1705 3136 ... |
+| 4 | tetranacci | 1 1 2 4 8 15 29 56 108 208 401 773 1490 2872 5536 ... |
+| 5 | pentanacci | 1 1 2 4 8 16 31 61 120 236 464 912 1793 3525 6930 ... |
+| 6 | hexanacci | 1 1 2 4 8 16 32 63 125 248 492 976 1936 3840 7617 ... |
+| 7 | heptanacci | 1 1 2 4 8 16 32 64 127 253 504 1004 2000 3984 7936 ... |
+| 8 | octonacci | 1 1 2 4 8 16 32 64 128 255 509 1016 2028 4048 8080 ... |
+| 9 | nonanacci | 1 1 2 4 8 16 32 64 128 256 511 1021 2040 4076 8144 ... |
+| 10 | decanacci | 1 1 2 4 8 16 32 64 128 256 512 1023 2045 4088 8172 ... |
+
+Allied sequences can be generated where the initial values are changed: The [Lucas series](https://en.wikipedia.org/wiki/Lucas number "wp: Lucas number") sums the two preceding values like the fibonacci series for $n=2$ but uses $\[2, 1]$ as its initial values.
+
+# --instructions--
+
+Write a function to generate Fibonacci $n$-step number sequences and Lucas sequences. The first parameter will be $n$. The second parameter will be the number of elements to be returned. The third parameter will specify whether to output the Fibonacci sequence or the Lucas sequence. If the parameter is `"f"` then return the Fibonacci sequence and if it is `"l"`, then return the Lucas sequence. The sequences must be returned as an array.
# --hints--
-`fib_luc`是一个功能。
+`fib_luc` should be a function.
```js
assert(typeof fib_luc === 'function');
```
-`fib_luc(2,10,"f")`应返回`[1,1,2,3,5,8,13,21,34,55]` 。
+`fib_luc(2,10,"f")` should return `[1,1,2,3,5,8,13,21,34,55]`.
```js
assert.deepEqual(fib_luc(2, 10, 'f'), ans[0]);
```
-`fib_luc(3,15,"f")`应返回`[1,1,2,4,7,13,24,44,81,149,274,504,927,1705,3136]` 。
+`fib_luc(3,15,"f")` should return `[1,1,2,4,7,13,24,44,81,149,274,504,927,1705,3136]`.
```js
assert.deepEqual(fib_luc(3, 15, 'f'), ans[1]);
```
-`fib_luc(4,15,"f")`应返回`[1,1,2,4,8,15,29,56,108,208,401,773,1490,2872,5536]` 。
+`fib_luc(4,15,"f")` should return `[1,1,2,4,8,15,29,56,108,208,401,773,1490,2872,5536]`.
```js
assert.deepEqual(fib_luc(4, 15, 'f'), ans[2]);
```
-`fib_luc(2,10,"l")`应返回`[ 2, 1, 3, 4, 7, 11, 18, 29, 47, 76]` `fib_luc(2,10,"l")` `[ 2, 1, 3, 4, 7, 11, 18, 29, 47, 76]` 。
+`fib_luc(2,10,"l")` should return `[ 2, 1, 3, 4, 7, 11, 18, 29, 47, 76]`.
```js
assert.deepEqual(fib_luc(2, 10, 'l'), ans[3]);
```
-`fib_luc(3,15,"l")`应返回`[ 2, 1, 3, 6, 10, 19, 35, 64, 118, 217, 399, 734, 1350, 2483, 4567 ]` `fib_luc(3,15,"l")` `[ 2, 1, 3, 6, 10, 19, 35, 64, 118, 217, 399, 734, 1350, 2483, 4567 ]` 。
+`fib_luc(3,15,"l")` should return `[ 2, 1, 3, 6, 10, 19, 35, 64, 118, 217, 399, 734, 1350, 2483, 4567 ]`.
```js
assert.deepEqual(fib_luc(3, 15, 'l'), ans[4]);
```
-`fib_luc(4,15,"l")`应该返回`[ 2, 1, 3, 6, 12, 22, 43, 83, 160, 308, 594, 1145, 2207, 4254, 8200 ]` `fib_luc(4,15,"l")` `[ 2, 1, 3, 6, 12, 22, 43, 83, 160, 308, 594, 1145, 2207, 4254, 8200 ]` 。
+`fib_luc(4,15,"l")` should return `[ 2, 1, 3, 6, 12, 22, 43, 83, 160, 308, 594, 1145, 2207, 4254, 8200 ]`.
```js
assert.deepEqual(fib_luc(4, 15, 'l'), ans[5]);
```
-`fib_luc(5,15,"l")`应该返回`[ 2, 1, 3, 6, 12, 24, 46, 91, 179, 352, 692, 1360, 2674, 5257, 10335 ]` 。
+`fib_luc(5,15,"l")` should return `[ 2, 1, 3, 6, 12, 24, 46, 91, 179, 352, 692, 1360, 2674, 5257, 10335 ]`.
```js
assert.deepEqual(fib_luc(5, 15, 'l'), ans[6]);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/fibonacci-sequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/fibonacci-sequence.md
index 5fa1a5e1466..2cce69c1856 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/fibonacci-sequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/fibonacci-sequence.md
@@ -1,45 +1,53 @@
---
id: 597f24c1dda4e70f53c79c81
-title: 斐波那契序列
+title: Fibonacci sequence
challengeType: 5
-videoUrl: ''
+forumTopicId: 302268
dashedName: fibonacci-sequence
---
# --description--
-<p>编写一个函数来生成第<big>n <sup>个</sup></big> Fibonacci数。 </p> /// <p>第<big>n <sup>个</sup></big> Fibonacci数由下式给出：/// </p><p> F <sub>n</sub> = F <sub>n-1</sub> + F <sub>n-2</sub> </p> /// <p>该系列的前两个术语是0,1。 </p> /// <p>因此，该系列是：0,1,1,2,3,5,8,13 ...... </p> ///
+Write a function to generate the <code>n<sup>th</sup></code> Fibonacci number.
+
+The <code>n<sup>th</sup></code> Fibonacci number is given by:
+
+<code>F<sub>n</sub> = F<sub>n-1</sub> + F<sub>n-2</sub></code>
+
+The first two terms of the series are 0 and 1.
+
+Hence, the series is: 0, 1, 1, 2, 3, 5, 8, 13...
# --hints--
-`fibonacci`是一种功能。
+`fibonacci` should be a function.
```js
assert(typeof fibonacci === 'function');
```
-`fibonacci(2)`应该返回一个数字。
+`fibonacci(2)` should return a number.
```js
assert(typeof fibonacci(2) == 'number');
```
-`fibonacci(3)`应该返回1.“）
+`fibonacci(3)` should return 2.
```js
-assert.equal(fibonacci(3), 1);
+assert.equal(fibonacci(3), 2);
```
-`fibonacci(5)`应该返回3.“）
+`fibonacci(5)` should return 5.
```js
-assert.equal(fibonacci(5), 3);
+assert.equal(fibonacci(5), 5);
```
-`fibonacci(10)`应该返回34.“）
+`fibonacci(10)` should return 55.
```js
-assert.equal(fibonacci(10), 34);
+assert.equal(fibonacci(10), 55);
```
# --seed--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/fibonacci-word.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/fibonacci-word.md
index 12a36bc499f..708f743add3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/fibonacci-word.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/fibonacci-word.md
@@ -1,30 +1,40 @@
---
id: 5992e222d397f00d21122931
-title: 斐波那契字
+title: Fibonacci word
challengeType: 5
-videoUrl: ''
+forumTopicId: 302269
dashedName: fibonacci-word
---
# --description--
-<p>编写一个函数将Fibonacci字返回到N.N将作为参数提供给函数。该函数应返回一个对象数组。对象的形式应为：{N：1，长度：1，熵：0，单词：'1'}。更多细节如下： </p><p> Fibonacci Word可以类似于Fibonacci Sequence的方式创建， <a href='http://hal.archives-ouvertes.fr/docs/00/36/79/72/PDF/The_Fibonacci_word_fractal.pdf' title='链接：http：//hal.archives-ouvertes.fr/docs/00/36/79/72/PDF/The_Fibonacci_word_fractal.pdf'>如下所述</a> ： </p><p>将F_Word <sub>1</sub>定义为1 </p><p>将F_Word <sub>2</sub>定义为0 </p><p>将F_Word <sub>3表示</sub>为F_Word <sub>2</sub>与F_Word <sub>1</sub>连接，即：01 </p><p>将F_Word <sub>n表示</sub>为F_Word <sub>n-1</sub>与F_word <sub>n-2连接</sub> </p>
+The Fibonacci Word may be created in a manner analogous to the Fibonacci Sequence [as described here](https://hal.archives-ouvertes.fr/docs/00/36/79/72/PDF/The_Fibonacci_word_fractal.pdf):
+
+<pre>Define F_Word<sub>1</sub> as <strong>1</strong>
+Define F_Word<sub>2</sub> as <strong>0</strong>
+Form F_Word<sub>3</sub> as F_Word<sub>2</sub> concatenated with F_Word<sub>1</sub> i.e.: <strong>01</strong>
+Form F_Word<sub>n</sub> as F_Word<sub>n-1</sub> concatenated with F_word<sub>n-2</sub>
+</pre>
+
+# --instructions--
+
+Write a function to return the Fibonacci Words up to `n`. `n` will be provided as a parameter to the function. The function should return an array of objects. The objects should be of the form: `{ N: 1, Length: 1, Entropy: 0, Word: '1' }`.
# --hints--
-`fibWord`是一个功能。
+`fibWord` should be a function.
```js
assert(typeof fibWord === 'function');
```
-`fibWord(5)`应该返回一个数组。
+`fibWord(5)` should return an array.
```js
assert(Array.isArray(fibWord(5)));
```
-`fibWord(5)`应该返回`'+JSON.stringify(ans)+'` 。
+`fibWord(5)` should return `[{ N:1, Length:1, Entropy:0, Word:"1" },{ N:2, Length:1, Entropy:0, Word:"0" },{ N:3, Length:2, Entropy:1, Word:"01" },{ N:4, Length:3, Entropy:0.9182958340544896, Word:"010" },{ N:5, Length:5, Entropy:0.9709505944546688, Word:"01001" }]`.
```js
assert.deepEqual(fibWord(5), ans);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/fractran.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/fractran.md
index 79f8f9f7cf7..0bcda09bdec 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/fractran.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/fractran.md
@@ -1,54 +1,81 @@
---
id: 5a7dad05be01840e1778a0d1
title: Fractran
-challengeType: 3
-videoUrl: ''
+challengeType: 5
+forumTopicId: 302270
dashedName: fractran
---
# --description--
-<div class='rosetta'><p class='rosetta__paragraph'> <span class='rosetta__text--bold'><a class='rosetta__link--wiki' href='https://en.wikipedia.org/wiki/FRACTRAN' title='wp：FRACTRAN'>FRACTRAN</a></span>是由数学家<a class='rosetta__link--wiki' href='https://en.wikipedia.org/wiki/John Horton Conway' title='wp：John Horton Conway'>John Horton Conway</a>发明的图灵完备的深奥编程语言。 </p><br><p class='rosetta__paragraph'> FRACTRAN程序是正分数$ P =（f_1，f_2，\ ldots，f_m）$的有序列表，以及初始正整数输入$ n $。 </p><br><p class='rosetta__paragraph'>该程序通过更新整数$ n $来运行，如下所示： </p><br><ul class='rosetta__unordered-list'><li class='rosetta__list-item--unordered'>对于第一个分数$ f_i $，在$ nf_i $为整数的列表中，用$ nf_i $替换$ n $; </li><li class='rosetta__list-item--unordered'>重复此规则，直到列表中没有分数乘以$ n $时产生整数，然后停止。 </li></ul><br><p class='rosetta__paragraph'>康威为FRACTRAN提供了素数计划： </p><br><p class='rosetta__paragraph'> <span class='rosetta__text--indented'>$ 17/91 $，$ 78/85 $，$ 19/51 $，$ 23/38 $，$ 29/33 $，$ 77/29 $，$ 95/23 $，$ 77/19 $，$ 1/17 $，$ 11/13 $， $ 13/11 $ $，$ 15/14 $，$ 15/2 $，$ 55/1 $</span> </p><br><p class='rosetta__paragraph'>从$ n = 2 $开始，此FRACTRAN程序将$ n $更改为$ 15 = 2 \ times（15/2）$，然后$ 825 = 15 \ times（55/1）$，生成以下整数序列： </p><br><p class='rosetta__paragraph'> <span class='rosetta__text--indented'>$ 2 $，$ 15 $，$ 825 $，$ 725 $，$ 1925 $，$ 2275 $，$ 425 $，$ 390 $，$ 330 $，$ 290 $，$ 770 $，$ \ ldots $</span> </p><br><p class='rosetta__paragraph'> 2之后，此序列包含以下2的幂： </p><br><p class='rosetta__paragraph'> <span class='rosetta__text--indented'>$ 2 ^ 2 = 4 $，$ 2 ^ 3 = 8 $，$ 2 ^ 5 = 32 $，$ 2 ^ 7 = 128 $，$ 2 ^ {11} = 2048 $，$ 2 ^ {13} = 8192 $，$ 2 ^ {17 } = 131072 $，$ 2 ^ {19} = 524288 $，$ \ ldots $</span> </p><br><p class='rosetta__paragraph'>这是2的主要权力。 </p><br><dl class='rosetta__description-list'><dt class='rosetta__description-title'>任务： </dt></dl><p class='rosetta__paragraph'>编写一个函数，将fractran程序作为字符串参数，并将程序的前10个数字作为数组返回。如果结果没有10个数字，则按原样返回数字。 </p></div>
+[FRACTRAN](https://en.wikipedia.org/wiki/FRACTRAN "wp: FRACTRAN") is a Turing-complete esoteric programming language invented by the mathematician [John Horton Conway](https://en.wikipedia.org/wiki/John Horton Conway "wp: John Horton Conway").
+
+A FRACTRAN program is an ordered list of positive fractions $P = (f_1, f_2, \\ldots, f_m)$, together with an initial positive integer input $n$.
+
+The program is run by updating the integer $n$ as follows:
+
+<ul>
+ <li>for the first fraction, $f_i$, in the list for which $nf_i$ is an integer, replace $n$ with $nf_i$ ;</li>
+ <li>repeat this rule until no fraction in the list produces an integer when multiplied by $n$, then halt.</li>
+</ul>
+
+Conway gave a program for primes in FRACTRAN:
+
+$\\dfrac{17}{91}$, $\\dfrac{78}{85}$, $\\dfrac{19}{51}$, $\\dfrac{23}{38}$, $\\dfrac{29}{33}$, $\\dfrac{77}{29}$, $\\dfrac{95}{23}$, $\\dfrac{77}{19}$, $\\dfrac{1}{17}$, $\\dfrac{11}{13}$, $\\dfrac{13}{11}$, $\\dfrac{15}{14}$, $\\dfrac{15}{2}$, $\\dfrac{55}{1}$
+
+Starting with $n=2$, this FRACTRAN program will change $n$ to $15=2\\times (\\frac{15}{2})$, then $825=15\\times (\\frac{55}{1})$, generating the following sequence of integers:
+
+$2$, $15$, $825$, $725$, $1925$, $2275$, $425$, $390$, $330$, $290$, $770$, $\\ldots$
+
+After 2, this sequence contains the following powers of 2:
+
+$2^2=4$, $2^3=8$, $2^5=32$, $2^7=128$, $2^{11}=2048$, $2^{13}=8192$, $2^{17}=131072$, $2^{19}=524288$, $\\ldots$
+
+which are the prime powers of 2.
+
+# --instructions--
+
+Write a function that takes a fractran program as a string parameter and returns the first 10 numbers of the program as an array. If the result does not have 10 numbers then return the numbers as is.
# --hints--
-`fractran`应该是一个功能。
+`fractran` should be a function.
```js
assert(typeof fractran == 'function');
```
-`fractran(""+tests[0]+"")`应该返回一个数组。
+`fractran("3/2, 1/3")` should return an array.
```js
assert(Array.isArray(fractran('3/2, 1/3')));
```
-`fractran(""+tests[0]+"")`应返回`"+JSON.stringify(results[0])+"` 。
+`fractran("3/2, 1/3")` should return `[ 2, 3, 1 ]`.
```js
assert.deepEqual(fractran('3/2, 1/3'), [2, 3, 1]);
```
-`fractran(""+tests[1]+"")`应返回`"+JSON.stringify(results[1])+"` 。
+`fractran("3/2, 5/3, 1/5")` should return `[ 2, 3, 5, 1 ]`.
```js
assert.deepEqual(fractran('3/2, 5/3, 1/5'), [2, 3, 5, 1]);
```
-`fractran(""+tests[2]+"")`应返回`"+JSON.stringify(results[2])+"` 。
+`fractran("3/2, 6/3")` should return `[ 2, 3, 6, 9, 18, 27, 54, 81, 162, 243 ]`.
```js
assert.deepEqual(fractran('3/2, 6/3'), [2, 3, 6, 9, 18, 27, 54, 81, 162, 243]);
```
-`fractran(""+tests[3]+"")`应返回`"+JSON.stringify(results[3])+"` 。
+`fractran("2/7, 7/2")` should return `[ 2, 7, 2, 7, 2, 7, 2, 7, 2, 7 ]`.
```js
assert.deepEqual(fractran('2/7, 7/2'), [2, 7, 2, 7, 2, 7, 2, 7, 2, 7]);
```
-`fractran(""+tests[4]+"")`应返回`"+JSON.stringify(results[4])+"` 。
+`fractran("17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1")` should return `[ 2, 15, 825, 725, 1925, 2275, 425, 390, 330, 290 ]`.
```js
assert.deepEqual(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/gamma-function.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/gamma-function.md
index 152f2916264..f105a0a1bd5 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/gamma-function.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/gamma-function.md
@@ -1,56 +1,58 @@
---
id: 5a23c84252665b21eecc7e76
-title: 伽玛功能
+title: Gamma function
challengeType: 5
-videoUrl: ''
+forumTopicId: 302271
dashedName: gamma-function
---
# --description--
-实现一个算法（或更多）来计算[Gamma](<https://en.wikipedia.org/wiki/Gamma function>) （$ \\ Gamma $）函数（仅在实际字段中）。 Gamma功能可以定义为：
+Implement one algorithm (or more) to compute the [Gamma](https://en.wikipedia.org/wiki/Gamma function) ($\\Gamma$) function (in the real field only).
-$ \\ Gamma（x）= \\ displaystyle \\ int_0 ^ \\ infty t ^ {x-1} e ^ { - t} dt $
+The Gamma function can be defined as:
+
+<div style='padding-left: 4em;'><big><big>$\Gamma(x) = \displaystyle\int_0^\infty t^{x-1}e^{-t} dt$</big></big></div>
# --hints--
-`gamma`应该是一个功能。
+`gamma` should be a function.
```js
assert(typeof gamma == 'function');
```
-`gamma("+tests[0]+")`应该返回一个数字。
+`gamma(.1)` should return a number.
```js
assert(typeof gamma(0.1) == 'number');
```
-`gamma("+tests[0]+")`应该返回`"+results[0]+"` 。
+`gamma(.1)` should return `9.513507698668736`.
```js
assert.equal(round(gamma(0.1)), round(9.513507698668736));
```
-`gamma("+tests[1]+")`应该返回`"+results[1]+"` 。
+`gamma(.2)` should return `4.590843711998803`.
```js
assert.equal(round(gamma(0.2)), round(4.590843711998803));
```
-`gamma("+tests[2]+")`应该返回`"+results[2]+"` 。
+`gamma(.3)` should return `2.9915689876875904`.
```js
assert.equal(round(gamma(0.3)), round(2.9915689876875904));
```
-`gamma("+tests[3]+")`应该返回`"+results[3]+"` 。
+`gamma(.4)` should return `2.218159543757687`.
```js
assert.equal(round(gamma(0.4)), round(2.218159543757687));
```
-`gamma("+tests[4]+")`应返回`"+results[4]+"` 。
+`gamma(.5)` should return `1.7724538509055159`.
```js
assert.equal(round(gamma(0.5)), round(1.7724538509055159));
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/gaussian-elimination.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/gaussian-elimination.md
index bfef3fa4012..2e5adf4bd9e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/gaussian-elimination.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/gaussian-elimination.md
@@ -1,24 +1,28 @@
---
id: 5a23c84252665b21eecc7e77
-title: 高斯消除
+title: Gaussian elimination
challengeType: 5
-videoUrl: ''
+forumTopicId: 302272
dashedName: gaussian-elimination
---
# --description--
-编写一个函数来解决\\（Ax = b \\）使用高斯消除然后向后替换。 \\（A \\）是\\（n \\次n \\）矩阵。此外，\\（x \\）和\\（b \\）是\\（n \\）乘以1个向量。要提高准确性，请使用部分旋转和缩放。
+Write a function to solve \\(Ax = b\\) using Gaussian elimination then backwards substitution.
+
+\\(A\\) being an \\(n \\times n\\) matrix. Also, \\(x\\) and \\(b\\) are \\(n\\) by 1 vectors.
+
+To improve accuracy, please use partial pivoting and scaling.
# --hints--
-`gaussianElimination`应该是一个函数。
+`gaussianElimination` should be a function.
```js
assert(typeof gaussianElimination == 'function');
```
-`gaussianElimination("+JSON.stringify(tests[0][0])+","+JSON.stringify(tests[0][1])+")`应该返回一个数组。
+`gaussianElimination([[1,1],[1,-1]], [5,1])` should return an array.
```js
assert(
@@ -34,7 +38,7 @@ assert(
);
```
-`gaussianElimination("+JSON.stringify(tests[0][0])+","+JSON.stringify(tests[0][1])+")`应返回`"+JSON.stringify(results[0])+"` 。
+`gaussianElimination([[1,1],[1,-1]], [5,1])` should return `[ 3, 2 ]`.
```js
assert.deepEqual(
@@ -49,7 +53,7 @@ assert.deepEqual(
);
```
-`gaussianElimination("+JSON.stringify(tests[1][0])+","+JSON.stringify(tests[1][1])+")`应返回`"+JSON.stringify(results[1])+"` 。
+`gaussianElimination([[2,3],[2,1]] , [8,4])` should return `[ 1, 2 ]`.
```js
assert.deepEqual(
@@ -64,7 +68,7 @@ assert.deepEqual(
);
```
-`gaussianElimination("+JSON.stringify(tests[2][0])+","+JSON.stringify(tests[2][1])+")`应返回`"+JSON.stringify(results[2])+"` 。
+`gaussianElimination([[1,3],[5,-2]], [14,19])` should return `[ 5, 3 ]`.
```js
assert.deepEqual(
@@ -79,7 +83,7 @@ assert.deepEqual(
);
```
-`gaussianElimination("+JSON.stringify(tests[3][0])+","+JSON.stringify(tests[3][1])+")`应返回`"+JSON.stringify(results[3])+"` 。
+`gaussianElimination([[1,1],[5,-1]] , [10,14])` should return `[ 4, 6 ]`.
```js
assert.deepEqual(
@@ -94,7 +98,7 @@ assert.deepEqual(
);
```
-`gaussianElimination("+JSON.stringify(tests[4][0])+","+JSON.stringify(tests[4][1])+")`应返回`"+JSON.stringify(results[4])+"` 。
+`gaussianElimination([[1,2,3],[4,5,6],[7,8,8]] , [6,15,23])` should return `[ 1, 1, 1 ]`.
```js
assert.deepEqual(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/general-fizzbuzz.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/general-fizzbuzz.md
index c1a0c3b59ec..4af8e94f0e3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/general-fizzbuzz.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/general-fizzbuzz.md
@@ -1,24 +1,32 @@
---
id: 5a23c84252665b21eecc7e78
-title: 一般的FizzBuzz
+title: General FizzBuzz
challengeType: 5
-videoUrl: ''
+forumTopicId: 302273
dashedName: general-fizzbuzz
---
# --description--
-编写[FizzBuzz](http://rosettacode.org/wiki/FizzBuzz)的通用版本，适用于任何因子列表及其单词。这基本上是一种“fizzbuzz”实现，其中游戏规则被提供给用户。创建一个实现此功能的函数。该函数应该有两个参数。第一个是带有FizzBuzz规则的数组。例如： `[ [3,"Fizz"] , [5,"Buzz"] ]` 。此indcates该`Fizz` ，如果数量是3的倍数，并应被打印`Buzz`如果是5的倍数。如果它是两则字符串应该在阵列中指定的顺序被连结的倍数。在这种情况下，如果数字是3和5的倍数，则为`FizzBuzz` 。第二个参数是函数应返回如上所述的字符串的数字。
+Write a generalized version of [FizzBuzz](https://rosettacode.org/wiki/FizzBuzz) that works for any list of factors, along with their words.
+
+This is basically a "fizzbuzz" implementation where the rules of the game are supplied to the user. Create a function to implement this. The function should take two parameters.
+
+The first will be an array with the FizzBuzz rules. For example: `[ [3, "Fizz"] , [5, "Buzz"] ]`.
+
+This indicates that `Fizz` should be printed if the number is a multiple of 3 and `Buzz` if it is a multiple of 5. If it is a multiple of both then the strings should be concatenated in the order specified in the array. In this case, `FizzBuzz` if the number is a multiple of 3 and 5.
+
+The second parameter is the number for which the function should return a string as stated above.
# --hints--
-`genFizzBuzz`应该是一个功能。
+`genFizzBuzz` should be a function.
```js
assert(typeof genFizzBuzz == 'function');
```
-`genFizzBuzz("+JSON.stringify(tests[0][0])+","+tests[0][1]+")`应该返回一个类型。
+`genFizzBuzz([[3, "Fizz"],[5, "Buzz"]], 6)` should return a string.
```js
assert(
@@ -32,7 +40,7 @@ assert(
);
```
-`genFizzBuzz("+JSON.stringify(tests[0][0])+","+tests[0][1]+")`应返回`""+results[0]+""` 。
+`genFizzBuzz([[3, "Fizz"],[5, "Buzz"]], 6)` should return `"Fizz"`.
```js
assert.equal(
@@ -47,7 +55,7 @@ assert.equal(
);
```
-`genFizzBuzz("+JSON.stringify(tests[1][0])+","+tests[1][1]+")`应返回`""+results[1]+""` 。
+`genFizzBuzz([[3, "Fizz"],[5, "Buzz"]], 10)` should return `"Buzz"`.
```js
assert.equal(
@@ -62,7 +70,7 @@ assert.equal(
);
```
-`genFizzBuzz("+JSON.stringify(tests[2][0])+","+tests[2][1]+")`应返回`""+results[2]+""` 。
+`genFizzBuzz([[3, "Buzz"],[5, "Fizz"]], 12)` should return `"Buzz"`.
```js
assert.equal(
@@ -77,7 +85,7 @@ assert.equal(
);
```
-`genFizzBuzz("+JSON.stringify(tests[3][0])+","+tests[3][1]+")`应返回`""+results[3]+""` 。
+`genFizzBuzz([[3, "Buzz"],[5, "Fizz"]], 13)` should return `"13"`.
```js
assert.equal(
@@ -92,7 +100,7 @@ assert.equal(
);
```
-`genFizzBuzz("+JSON.stringify(tests[4][0])+","+tests[4][1]+")`应该返回`""+results[4]+""` 。
+`genFizzBuzz([[3, "Buzz"],[5, "Fizz"]], 15)` should return `"BuzzFizz"`.
```js
assert.equal(
@@ -107,7 +115,7 @@ assert.equal(
);
```
-`genFizzBuzz("+JSON.stringify(tests[5][0])+","+tests[5][1]+")`应返回`""+results[5]+""` 。
+`genFizzBuzz([[3, "Fizz"],[5, "Buzz"]], 15)` should return `"FizzBuzz"`.
```js
assert.equal(
@@ -122,7 +130,7 @@ assert.equal(
);
```
-`genFizzBuzz("+JSON.stringify(tests[6][0])+","+tests[6][1]+")`应该返回`""+results[6]+""` 。
+`genFizzBuzz([[3, "Fizz"],[5, "Buzz"],[7, "Baxx"]], 105)` should return `"FizzBuzzBaxx"`.
```js
assert.equal(
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/generate-lower-case-ascii-alphabet.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/generate-lower-case-ascii-alphabet.md
index d6aad561f10..5f643c3a38b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/generate-lower-case-ascii-alphabet.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/generate-lower-case-ascii-alphabet.md
@@ -1,54 +1,54 @@
---
id: 5a23c84252665b21eecc7e7a
-title: 生成小写ASCII字母表
+title: Generate lower case ASCII alphabet
challengeType: 5
-videoUrl: ''
+forumTopicId: 302274
dashedName: generate-lower-case-ascii-alphabet
---
# --description--
-编写一个函数以生成给定范围的小写ASCII字符数组。例如：对于范围1到4，函数应返回`['a','b','c','d']` 。
+Write a function to generate an array of lower case ASCII characters for a given range. For example, given the range `['a', 'd']`, the function should return `['a', 'b', 'c', 'd']`.
# --hints--
-`lascii`应该是一个功能。
+`lascii` should be a function.
```js
assert(typeof lascii == 'function');
```
-`lascii("a","d")`应该返回一个数组。
+`lascii("a","d")` should return an array.
```js
assert(Array.isArray(lascii('a', 'd')));
```
-`lascii("a","d")`应该返回`[ "a", "b", "c", "d" ]` 。
+`lascii('a','d')` should return `[ 'a', 'b', 'c', 'd' ]`.
```js
assert.deepEqual(lascii('a', 'd'), results[0]);
```
-`lascii("c","i")`应该返回`[ "c", "d", "e", "f", "g", "h", "i" ]` 。
+`lascii('c','i')` should return `[ 'c', 'd', 'e', 'f', 'g', 'h', 'i' ]`.
```js
assert.deepEqual(lascii('c', 'i'), results[1]);
```
-`lascii("m","q")`应该返回`[ "m", "n", "o", "p", "q" ]` 。
+`lascii('m','q')` should return `[ 'm', 'n', 'o', 'p', 'q' ]`.
```js
assert.deepEqual(lascii('m', 'q'), results[2]);
```
-`lascii("k","n")`应返回`[ "k", "l", "m", "n" ]` 。
+`lascii('k','n')` should return `[ 'k', 'l', 'm', 'n' ]`.
```js
assert.deepEqual(lascii('k', 'n'), results[3]);
```
-`lascii("t","z")`应该返回`[ "t", "u", "v", "w", "x", "y", "z" ]` 。
+`lascii('t','z')` should return `[ 't', 'u', 'v', 'w', 'x', 'y', 'z' ]`.
```js
assert.deepEqual(lascii('t', 'z'), results[4]);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/gray-code.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/gray-code.md
index 2515bf0d142..f46350f0be6 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/gray-code.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/gray-code.md
@@ -1,64 +1,94 @@
---
id: 5a23c84252665b21eecc7e80
-title: 格雷码
+title: Gray code
challengeType: 5
-videoUrl: ''
+forumTopicId: 302276
dashedName: gray-code
---
# --description--
-[格雷码](<https://en.wikipedia.org/wiki/Gray code>)是二进制编码的一种形式，其中连续数字之间的转换仅相差一位。这是一种有用的编码，用于减少硬件数据危险，其值快速变化和/或连接到较慢的硬件作为输入。从左到右或从上到下依次为[卡诺图](<https://en.wikipedia.org/wiki/Karnaugh map>)生成输入也很有用。创建一个函数来编码数字并解码格雷码中的数字。该函数应该有2个参数。第一个是布尔值。该函数应编码为true，解码为false。第二个参数是要编码/解码的数字。显示所有5位二进制数的正常二进制表示，格雷码表示和解码格雷码值（0-31包括0，不需要前导0）。有许多可能的格雷码。以下编码所谓的“二进制反射格雷码”。
-编码（MSB为0位，b为二进制，g为格雷码）： <code><br>if b[i-1] = 1<br><span style='padding-left:1em'>g[i] = not b[i]</span><br>else<br><span style='padding-left:1em'>g[i] = b[i]</span><br></code>要么：
-`g = b xor (b logically right shifted 1 time)`
-解码（MSB为0位，b为二进制，g为格雷码）：
-<code>b[0] = g[0]<br>for other bits:<br>b[i] = g[i] xor b[i-1]<br></code>
+[Gray code](https://en.wikipedia.org/wiki/Gray code) is a form of binary encoding where transitions between consecutive numbers differ by only one bit.
+
+This is a useful encoding for reducing hardware data hazards with values that change rapidly and/or connect to slower hardware as inputs.
+
+It is also useful for generating inputs for [Karnaugh maps](https://en.wikipedia.org/wiki/Karnaugh map) in order from left to right or top to bottom.
+
+# --instructions--
+
+Create a function to encode a number to and decode a number from Gray code. The function should will have 2 parameters.
+
+The first would be a boolean. The function should encode for true and decode for false. The second parameter would be the number to be encoded/decoded.
+
+Display the normal binary representations, Gray code representations, and decoded Gray code values for all 5-bit binary numbers (0-31 inclusive, leading 0's not necessary).
+
+There are many possible Gray codes. The following encodes what is called "binary reflected Gray code."
+
+Encoding (MSB is bit 0, b is binary, g is Gray code):
+
+<pre>if b[i-1] = 1
+ g[i] = not b[i]
+else
+ g[i] = b[i]
+</pre>
+
+Or:
+
+<pre>g = b xor (b logically right shifted 1 time)
+</pre>
+
+Decoding (MSB is bit 0, b is binary, g is Gray code):
+
+<pre>b[0] = g[0]<br>
+for other bits:
+b[i] = g[i] xor b[i-1]
+</pre>
# --hints--
-`gray`应该是一个功能。
+`gray` should be a function.
```js
assert(typeof gray == 'function');
```
-`gray(true,177)`应该返回一个数字。
+`gray(true,177)` should return a number.
```js
assert(typeof gray(true, 177) == 'number');
```
-`gray(true,177)`应该返回`233` 。
+`gray(true,177)` should return `233`.
```js
assert.equal(gray(true, 177), 233);
```
-`gray(true,425)`应该返回`381` 。
+`gray(true,425)` should return `381`.
```js
assert.equal(gray(true, 425), 381);
```
-`gray(true,870)`应该返回`725` 。
+`gray(true,870)` should return `725`.
```js
assert.equal(gray(true, 870), 725);
```
-`gray(false,233)`应该返回`177` 。
+`gray(false,233)` should return `177`.
```js
assert.equal(gray(false, 233), 177);
```
-`gray(false,381)`应该返回`425` 。
+`gray(false,381)` should return `425`.
```js
assert.equal(gray(false, 381), 425);
```
-`gray(false,725)`应该返回`870` 。
+`gray(false,725)` should return `870`.
```js
assert.equal(gray(false, 725), 870);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/greatest-common-divisor.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/greatest-common-divisor.md
index e70a483fe3b..60b96750188 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/greatest-common-divisor.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/greatest-common-divisor.md
@@ -1,60 +1,60 @@
---
id: 5a23c84252665b21eecc7e82
-title: 最大公约数
+title: Greatest common divisor
challengeType: 5
-videoUrl: ''
+forumTopicId: 302277
dashedName: greatest-common-divisor
---
# --description--
-编写一个函数，返回两个整数的最大公约数。
+Write a function that returns the greatest common divisor of two integers.
# --hints--
-`gcd`应该是一个功能。
+`gcd` should be a function.
```js
assert(typeof gcd == 'function');
```
-`gcd(24,36)`应该返回一个数字。
+`gcd(24,36)` should return a number.
```js
assert(typeof gcd(24, 36) == 'number');
```
-`gcd(24,36)`应该返回`12` 。
+`gcd(24,36)` should return `12`.
```js
assert.equal(gcd(24, 36), 12);
```
-`gcd(30,48)`应该返回`6` 。
+`gcd(30,48)` should return `6`.
```js
assert.equal(gcd(30, 48), 6);
```
-`gcd(10,15)`应该返回`5` 。
+`gcd(10,15)` should return `5`.
```js
assert.equal(gcd(10, 15), 5);
```
-`gcd(100,25)`应该返回`25` 。
+`gcd(100,25)` should return `25`.
```js
assert.equal(gcd(100, 25), 25);
```
-`gcd(13,250)`应该返回`1` 。
+`gcd(13,250)` should return `1`.
```js
assert.equal(gcd(13, 250), 1);
```
-`gcd(1300,250)`应该返回`50` 。
+`gcd(1300,250)` should return `50`.
```js
assert.equal(gcd(1300, 250), 50);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/greatest-subsequential-sum.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/greatest-subsequential-sum.md
index 03deb01b433..4a90c111fe2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/greatest-subsequential-sum.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/greatest-subsequential-sum.md
@@ -1,36 +1,38 @@
---
id: 5a23c84252665b21eecc7e84
-title: 最重要的后续总和
+title: Greatest subsequential sum
challengeType: 5
-videoUrl: ''
+forumTopicId: 302278
dashedName: greatest-subsequential-sum
---
# --description--
-给定一个整数序列，找到一个连续的子序列，它最大化其元素的总和，也就是说，没有其他单个子序列的元素加起来大于这一个的值。空子序列被认为具有\\（0 \\）的总和;因此，如果所有元素都是负数，则结果必须是空序列。
+Given a sequence of integers, find a continuous subsequence which maximizes the sum of its elements, that is, the elements of no other single subsequence add up to a value larger than this one.
+
+An empty subsequence is considered to have the sum of \\( 0 \\); thus if all elements are negative, the result must be the empty sequence.
# --hints--
-`maximumSubsequence`应该是一个函数。
+`maximumSubsequence` should be a function.
```js
assert(typeof maximumSubsequence == 'function');
```
-`maximumSubsequence("+JSON.stringify(tests[0])+")`应该返回一个数组。
+`maximumSubsequence([ 1, 2, -1, 3, 10, -10 ])` should return an array.
```js
assert(Array.isArray(maximumSubsequence([1, 2, -1, 3, 10, -10])));
```
-`maximumSubsequence("+JSON.stringify(tests[0])+")`应返回`"+JSON.stringify(results[0])+"` 。
+`maximumSubsequence([ 1, 2, -1, 3, 10, -10 ])` should return `[ 1, 2, -1, 3, 10 ]`.
```js
assert.deepEqual(maximumSubsequence([1, 2, -1, 3, 10, -10]), [1, 2, -1, 3, 10]);
```
-`maximumSubsequence("+JSON.stringify(tests[1])+")`应返回`"+JSON.stringify(results[1])+"` 。
+`maximumSubsequence([ 0, 8, 10, -2, -4, -1, -5, -3 ])` should return `[ 0, 8, 10 ]`.
```js
assert.deepEqual(maximumSubsequence([0, 8, 10, -2, -4, -1, -5, -3]), [
@@ -40,25 +42,25 @@ assert.deepEqual(maximumSubsequence([0, 8, 10, -2, -4, -1, -5, -3]), [
]);
```
-`maximumSubsequence("+JSON.stringify(tests[2])+")`应返回`"+JSON.stringify(results[2])+"` 。
+`maximumSubsequence([ 9, 9, -10, 1 ])` should return `[ 9, 9 ]`.
```js
assert.deepEqual(maximumSubsequence([9, 9, -10, 1]), [9, 9]);
```
-`maximumSubsequence("+JSON.stringify(tests[3])+")`应返回`"+JSON.stringify(results[3])+"` 。
+`maximumSubsequence([ 7, 1, -5, -3, -8, 1 ])` should return `[ 7, 1 ]`.
```js
assert.deepEqual(maximumSubsequence([7, 1, -5, -3, -8, 1]), [7, 1]);
```
-`maximumSubsequence("+JSON.stringify(tests[4])+")`应返回`"+JSON.stringify(results[4])+"` 。
+`maximumSubsequence([ -3, 6, -1, 4, -4, -6 ])` should return `[ 6, -1, 4 ]`.
```js
assert.deepEqual(maximumSubsequence([-3, 6, -1, 4, -4, -6]), [6, -1, 4]);
```
-`maximumSubsequence("+JSON.stringify(tests[5])+")`应返回`"+JSON.stringify(results[5])+"` 。
+`maximumSubsequence([ -1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1 ])` should return `[ 3, 5, 6, -2, -1, 4 ]`.
```js
assert.deepEqual(maximumSubsequence([-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1]), [
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hailstone-sequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hailstone-sequence.md
index 9d0e4ca1e3f..e7587e2f999 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hailstone-sequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hailstone-sequence.md
@@ -1,24 +1,48 @@
---
id: 595608ff8bcd7a50bd490181
-title: 冰雹序列
+title: Hailstone sequence
challengeType: 5
-videoUrl: ''
+forumTopicId: 302279
dashedName: hailstone-sequence
---
# --description--
-<p> Hailstone数字序列可以从起始正整数n生成： </p>如果n为1，则序列结束。如果n是偶数，那么序列的下一个n <code>= n/2</code>如果n是奇数，那么序列的下一个n <code>= (3 \* n) + 1</code> <p> （未经证实的） <a href='https://en.wikipedia.org/wiki/Collatz conjecture' title='wp：Collatz猜想'>Collatz猜想</a>是任何起始编号的冰雹序列总是终止。 </p><p>冰雹序列也称为冰雹数（因为这些值通常受到多个下降和上升，如云中的冰雹），或者作为Collatz序列。 </p>任务：创建例程以生成数字的hailstone序列。使用例程表明，对于27号的冰雹序列具有开始与112个元件<code>27, 82, 41, 124</code> ，结束时用<code>8, 4, 2, 1</code>与显示具有最长冰雹序列的数目少于100,000一起序列的长度。 （但不要显示实际的序列！）参见： <a href='http://xkcd.com/710' title='链接：http：//xkcd.com/710'>xkcd</a> （幽默）。
+The Hailstone sequence of numbers can be generated from a starting positive integer, `n` by:
+
+<ul>
+ <li>If <code>n</code> is <code>1</code> then the sequence ends</li>
+ <li>If <code>n</code> is <code>even</code> then the next <code>n</code> of the sequence <code>= n/2</code></li>
+ <li>If <code>n</code> is <code>odd</code> then the next <code>n</code> of the sequence <code>= (3 * n) + 1</code></li>
+</ul>
+
+The (unproven) [Collatz conjecture](https://en.wikipedia.org/wiki/Collatz conjecture "wp: Collatz conjecture") is that the hailstone sequence for any starting number always terminates.
+
+The hailstone sequence is also known as hailstone numbers (because the values are usually subject to multiple descents and ascents like hailstones in a cloud), or as the Collatz sequence.
+
+# --instructions--
+
+<ol>
+ <li>Create a routine to generate the hailstone sequence for a number</li>
+ <li>Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with <code>27, 82, 41, 124</code> and ending with <code>8, 4, 2, 1</code></li>
+ <li>Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length. (But don't show the actual sequence!)</li>
+</ol>
+
+**See also:**
+
+<ul>
+ <li><a href='https://xkcd.com/710' target='_blank'>xkcd</a> (humourous).</li>
+</ul>
# --hints--
-`hailstoneSequence`是一个函数。
+`hailstoneSequence` should be a function.
```js
assert(typeof hailstoneSequence === 'function');
```
-`[[27,82,41,124,8,4,2,1], [351, 77031]]` `hailstoneSequence()`应返回`[[27,82,41,124,8,4,2,1], [351, 77031]]`
+`hailstoneSequence()` should return `[[27,82,41,124,8,4,2,1], [351, 77031]]`
```js
assert.deepEqual(hailstoneSequence(), res);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/happy-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/happy-numbers.md
index 5aba3634788..6eb49885391 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/happy-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/happy-numbers.md
@@ -1,90 +1,96 @@
---
id: 594810f028c0303b75339ad1
-title: 快乐的数字
+title: Happy numbers
challengeType: 5
-videoUrl: ''
+forumTopicId: 302280
dashedName: happy-numbers
---
# --description--
-<p>幸福的数字由以下过程定义： </p><p>从任何正整数开始，将数字替换为其数字的平方和，并重复该过程直到数字等于1（它将保持不变），或者它在一个不包括1的循环中无休止地循环。这些数字这个过程在1结束的是幸福的数字，而那些不以1结尾的数字是不愉快的数字。 </p><p>实现一个函数，如果数字是满意的，则返回true，否则返回false。 </p>
+A [happy number](https://en.wikipedia.org/wiki/Happy_number) is defined by the following process:
+
+Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals `1` (where it will stay), or it loops endlessly in a cycle which does not include `1`. Those numbers for which this process ends in `1` are happy numbers, while those that do not end in `1` are unhappy numbers.
+
+# --instructions--
+
+Implement a function that returns true if the number is happy, or false if not.
# --hints--
-`happy`是一种功能。
+`happy` should be a function.
```js
assert(typeof happy === 'function');
```
-`happy(1)`应该返回一个布尔值。
+`happy(1)` should return a boolean.
```js
assert(typeof happy(1) === 'boolean');
```
-`happy(1)`应该回归真实。
+`happy(1)` should return true.
```js
assert(happy(1));
```
-`happy(2)`应该返回虚假。
+`happy(2)` should return false.
```js
assert(!happy(2));
```
-`happy(7)`应该回归真实。
+`happy(7)` should return true.
```js
assert(happy(7));
```
-`happy(10)`应该回归真实。
+`happy(10)` should return true.
```js
assert(happy(10));
```
-`happy(13)`应该回归真实。
+`happy(13)` should return true.
```js
assert(happy(13));
```
-`happy(19)`应该回归真实。
+`happy(19)` should return true.
```js
assert(happy(19));
```
-`happy(23)`应该回归真实。
+`happy(23)` should return true.
```js
assert(happy(23));
```
-`happy(28)`应该回归真实。
+`happy(28)` should return true.
```js
assert(happy(28));
```
-`happy(31)`应该回归真实。
+`happy(31)` should return true.
```js
assert(happy(31));
```
-`happy(32)`应该回归真实：
+`happy(32)` should return true:.
```js
assert(happy(32));
```
-`happy(33)`应该返回虚假。
+`happy(33)` should return false.
```js
assert(!happy(33));
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/harshad-or-niven-series.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/harshad-or-niven-series.md
index fecd4a27bfb..383989816ae 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/harshad-or-niven-series.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/harshad-or-niven-series.md
@@ -1,24 +1,34 @@
---
id: 595668ca4cfe1af2fb9818d4
-title: Harshad或Niven系列
+title: Harshad or Niven series
challengeType: 5
-videoUrl: ''
+forumTopicId: 302281
dashedName: harshad-or-niven-series
---
# --description--
-<p> <a href='http://mathworld.wolfram.com/HarshadNumber.html' title='链接：http：//mathworld.wolfram.com/HarshadNumber.html'>Harshad</a>或Niven数是正整数≥1，可以被它们的数字之和整除。 </p><p>例如，42是<a href='http://rosettacode.org/wiki/oeis:A005349' title='OEIS：A005349'>Harshad数，</a>因为42可以被（4 + 2）整除而没有余数。 </p>假设系列被定义为按递增顺序排列的数字。任务： <p>实现一个函数来生成Harshad序列的连续成员。 </p><p>使用它列出序列的前20个成员并列出第一个大于1000的Harshad数。 </p>
+The Harshad or Niven numbers are positive integers ≥ 1 that are divisible by the sum of their digits.
+
+For example, `42` is a [Harshad number](https://rosettacode.org/wiki/Harshad_or_Niven_series "Harshad or Niven series") as `42` is divisible by `(4 + 2)` without remainder.
+
+Assume that the series is defined as the numbers in increasing order.
+
+# --instructions--
+
+Implement a function to generate successive members of the Harshad sequence.
+
+Use it to list the first twenty members of the sequence and list the first Harshad number greater than 1000.
# --hints--
-`isHarshadOrNiven`是一个函数。
+`isHarshadOrNiven` should be a function.
```js
assert(typeof isHarshadOrNiven === 'function');
```
-`isHarshadOrNiven()`应该返回`{"firstTwenty": [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42],"firstOver1000": 1002}`
+`isHarshadOrNiven()` should return `{"firstTwenty": [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42],"firstOver1000": 1002}`
```js
assert.deepEqual(isHarshadOrNiven(), res);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hash-from-two-arrays.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hash-from-two-arrays.md
index ee9e64ec850..d5a9a396bb9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hash-from-two-arrays.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hash-from-two-arrays.md
@@ -1,58 +1,60 @@
---
id: 595671d4d2cdc305f0d5b36f
-title: 来自两个数组的哈希
+title: Hash from two arrays
challengeType: 5
-videoUrl: ''
+forumTopicId: 302283
dashedName: hash-from-two-arrays
---
# --description--
-任务：
+Using two Arrays of equal length, create a Hash object where the elements from one array (the keys) are linked to the elements of the other (the values).
-使用两个相等长度的数组，创建一个Hash对象，其中一个数组中的元素（键）链接到另一个数组（值）
+**Related task:**
-相关任务： [关联数组/创建](<http://rosettacode.org/wiki/Associative arrays/Creation> "关联数组/创建")
+<ul>
+ <li><a href='https://rosettacode.org/wiki/Associative arrays/Creation' title='Associative arrays/Creation' target='_blank'>Associative arrays/Creation</a></li>
+</ul>
# --hints--
-`arrToObj`是一个功能。
+`arrToObj` should be a function.
```js
assert(typeof arrToObj === 'function');
```
-`arrToObj([1, 2, 3, 4, 5], ["a", "b", "c", "d", "e"])`应返回`{ 1: "a", 2: "b", 3: "c", 4: "d", 5: "e" }`
+`arrToObj([1, 2, 3, 4, 5], ["a", "b", "c", "d", "e"])` should return `{ 1: "a", 2: "b", 3: "c", 4: "d", 5: "e" }`
```js
assert.deepEqual(arrToObj(...testCases[0]), res[0]);
```
-`arrToObj([1, 2, 3, 4, 5], ["a", "b", "c", "d"])`应返回`{ 1: "a", 2: "b", 3: "c", 4: "d", 5: undefined }`
+`arrToObj([1, 2, 3, 4, 5], ["a", "b", "c", "d"])` should return `{ 1: "a", 2: "b", 3: "c", 4: "d", 5: undefined }`
```js
assert.deepEqual(arrToObj(...testCases[1]), res[1]);
```
-`arrToObj([1, 2, 3], ["a", "b", "c", "d", "e"])`应返回`{ 1: "a", 2: "b", 3: "c" }`
+`arrToObj([1, 2, 3], ["a", "b", "c", "d", "e"])` should return `{ 1: "a", 2: "b", 3: "c" }`
```js
assert.deepEqual(arrToObj(...testCases[2]), res[2]);
```
-`arrToObj(["a", "b", "c", "d", "e"], [1, 2, 3, 4, 5])`应返回`{ "a": 1, "b": 2, "c": 3 , "d": 4, "e": 5 }`
+`arrToObj(["a", "b", "c", "d", "e"], [1, 2, 3, 4, 5])` should return `{ "a": 1, "b": 2, "c": 3 , "d": 4, "e": 5 }`
```js
assert.deepEqual(arrToObj(...testCases[3]), res[3]);
```
-`arrToObj(["a", "b", "c", "d", "e"], [1, 2, 3, 4])`应返回`{ "a": 1, "b": 2, "c": 3 , "d": 4, "e": undefined }`
+`arrToObj(["a", "b", "c", "d", "e"], [1, 2, 3, 4])` should return `{ "a": 1, "b": 2, "c": 3 , "d": 4, "e": undefined }`
```js
assert.deepEqual(arrToObj(...testCases[4]), res[4]);
```
-`arrToObj(["a", "b", "c"], [1, 2, 3, 4, 5])`应返回`{ "a": 1, "b": 2, "c": 3 }`
+`arrToObj(["a", "b", "c"], [1, 2, 3, 4, 5])` should return `{ "a": 1, "b": 2, "c": 3 }`
```js
assert.deepEqual(arrToObj(...testCases[5]), res[5]);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hash-join.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hash-join.md
index efdad81ee39..7dc6b729cb6 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hash-join.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hash-join.md
@@ -1,36 +1,155 @@
---
id: 5956795bc9e2c415eb244de1
-title: 哈希加入
+title: Hash join
challengeType: 5
-videoUrl: ''
+forumTopicId: 302284
dashedName: hash-join
---
# --description--
-<p> <a href='https://en.wikipedia.org/wiki/Join_(SQL)#Inner_join' title='wp：Join_（SQL）#Inner_join'>内连接</a>是一种操作，它根据匹配的列值将两个数据表组合到一个表中。实现此操作的最简单方法是<a href='https://en.wikipedia.org/wiki/Nested loop join' title='wp：嵌套循环连接'>嵌套循环连接</a>算法，但更可扩展的替代方法是<a href='https://en.wikipedia.org/wiki/hash join' title='wp：哈希联接'>散列连接</a>算法。 </p><p>实现“散列连接”算法，并演示它通过下面列出的测试用例。 </p><p>您应该将表表示为在编程语言中感觉自然的数据结构。 </p><p> “散列连接”算法包含两个步骤： </p>哈希阶段：从两个表中的一个表创建一个<a href='https://en.wikipedia.org/wiki/Multimap' title='wp：Multimap'>多</a>图，从每个连接列值映射到包含它的所有行。多图必须支持基于散列的查找，它比简单的线性搜索更好地扩展，因为这是该算法的重点。理想情况下，我们应该为较小的表创建多图，从而最小化其创建时间和内存大小。加入阶段：扫描另一个表，通过查看之前创建的多图来查找匹配的行。 <p>在伪代码中，算法可以表示如下： </p><pre>让A =第一个输入表（或理想情况下，更大的输入表）
-设B =第二个输入表（或理想情况下，较小的输入表）
-令j <sub>A</sub> =表A的连接列ID
-令j <sub>B</sub> =表B的连接列ID
-让M <sub>B</sub> =一个多图，用于从单个值映射到表B的多行（从空白开始）
-让C =输出表（从空开始）
-对于表B中的每一行b：
- 将b放在密钥b（j <sub>B</sub> ）下的多映射M <sub>B中</sub>
-对于表A中的每一行a：
- 对于a（j <sub>A</sub> ）项下多图M <sub>B中的</sub>每一行b：
- 设c =第a行和第b行的串联
- 将行c放在表C中<p></p>
-</pre>测试用例<p>输入</p><table><tbody><tr><td style='padding: 4px; margin: 5px;'><table style='border:none; border-collapse:collapse;'><tbody><tr><td style='border:none'> <i>A =</i> </td><td style='border:none'><table><tbody><tr><th style='padding: 4px; margin: 5px;'>年龄</th><th style='padding: 4px; margin: 5px;'>名称</th></tr><tr><td style='padding: 4px; margin: 5px;'> 27 </td><td style='padding: 4px; margin: 5px;'>约拿</td></tr><tr><td style='padding: 4px; margin: 5px;'> 18 </td><td style='padding: 4px; margin: 5px;'>艾伦</td></tr><tr><td style='padding: 4px; margin: 5px;'> 28 </td><td style='padding: 4px; margin: 5px;'>荣耀</td></tr><tr><td style='padding: 4px; margin: 5px;'> 18 </td><td style='padding: 4px; margin: 5px;'>大力水手</td></tr><tr><td style='padding: 4px; margin: 5px;'> 28 </td><td style='padding: 4px; margin: 5px;'>艾伦</td></tr></tbody></table></td><td style='border:none; padding-left:1.5em;' rowspan='2'></td><td style='border:none'> <i>B =</i> </td><td style='border:none'><table><tbody><tr><th style='padding: 4px; margin: 5px;'>字符</th><th style='padding: 4px; margin: 5px;'>复仇者</th></tr><tr><td style='padding: 4px; margin: 5px;'>约拿</td><td style='padding: 4px; margin: 5px;'>鲸鱼</td></tr><tr><td style='padding: 4px; margin: 5px;'>约拿</td><td style='padding: 4px; margin: 5px;'>蜘蛛</td></tr><tr><td style='padding: 4px; margin: 5px;'>艾伦</td><td style='padding: 4px; margin: 5px;'>鬼</td></tr><tr><td style='padding: 4px; margin: 5px;'>艾伦</td><td style='padding: 4px; margin: 5px;'>植物大战僵尸</td></tr><tr><td style='padding: 4px; margin: 5px;'>荣耀</td><td style='padding: 4px; margin: 5px;'>巴菲</td></tr></tbody></table></td></tr><tr><td style='border:none'> <i>j <sub>A</sub> =</i> </td><td style='border:none'> <i><code>Name</code> （即第1栏）</i> </td><td style='border:none'> <i>j <sub>B</sub> =</i> </td><td style='border:none'> <i><code>Character</code> （即第0列）</i> </td></tr></tbody></table></td><td style='padding: 4px; margin: 5px;'></td></tr></tbody></table><p>产量</p><table><tbody><tr><th style='padding: 4px; margin: 5px;'> A.Age </th><th style='padding: 4px; margin: 5px;'>一个名字</th><th style='padding: 4px; margin: 5px;'> B.Character </th><th style='padding: 4px; margin: 5px;'> B.Nemesis </th></tr><tr><td style='padding: 4px; margin: 5px;'> 27 </td><td style='padding: 4px; margin: 5px;'>约拿</td><td style='padding: 4px; margin: 5px;'>约拿</td><td style='padding: 4px; margin: 5px;'>鲸鱼</td></tr><tr><td style='padding: 4px; margin: 5px;'> 27 </td><td style='padding: 4px; margin: 5px;'>约拿</td><td style='padding: 4px; margin: 5px;'>约拿</td><td style='padding: 4px; margin: 5px;'>蜘蛛</td></tr><tr><td style='padding: 4px; margin: 5px;'> 18 </td><td style='padding: 4px; margin: 5px;'>艾伦</td><td style='padding: 4px; margin: 5px;'>艾伦</td><td style='padding: 4px; margin: 5px;'>鬼</td></tr><tr><td style='padding: 4px; margin: 5px;'> 18 </td><td style='padding: 4px; margin: 5px;'>艾伦</td><td style='padding: 4px; margin: 5px;'>艾伦</td><td style='padding: 4px; margin: 5px;'>植物大战僵尸</td></tr><tr><td style='padding: 4px; margin: 5px;'> 28 </td><td style='padding: 4px; margin: 5px;'>荣耀</td><td style='padding: 4px; margin: 5px;'>荣耀</td><td style='padding: 4px; margin: 5px;'>巴菲</td></tr><tr><td style='padding: 4px; margin: 5px;'> 28 </td><td style='padding: 4px; margin: 5px;'>艾伦</td><td style='padding: 4px; margin: 5px;'>艾伦</td><td style='padding: 4px; margin: 5px;'>鬼</td></tr><tr><td style='padding: 4px; margin: 5px;'> 28 </td><td style='padding: 4px; margin: 5px;'>艾伦</td><td style='padding: 4px; margin: 5px;'>艾伦</td><td style='padding: 4px; margin: 5px;'>植物大战僵尸</td></tr></tbody></table><p></p><p></p><p>输出表中行的顺序并不重要。 </p><p>如果你使用数字索引数组来表示表行（而不是按名称引用列），你可以用<code style='white-space:nowrap'>[[27, "Jonah"], ["Jonah", "Whales"]]</code>的形式表示输出行。 。 </p><hr>
+An [inner join](https://en.wikipedia.org/wiki/Join_(SQL)#Inner_join "wp: Join\_(SQL)#Inner_join") is an operation that combines two data tables into one table, based on matching column values. The simplest way of implementing this operation is the [nested loop join](https://en.wikipedia.org/wiki/Nested loop join "wp: Nested loop join") algorithm, but a more scalable alternative is the [hash join](https://en.wikipedia.org/wiki/hash join "wp: hash join") algorithm.
+
+The "hash join" algorithm consists of two steps:
+
+<ol>
+ <li><strong>Hash phase:</strong> Create a <a href='https://en.wikipedia.org/wiki/Multimap' title='wp: Multimap' target='_blank'>multimap</a> from one of the two tables, mapping from each join column value to all the rows that contain it.</li>
+ <ul>
+ <li>The multimap must support hash-based lookup which scales better than a simple linear search, because that's the whole point of this algorithm.</li>
+ <li>Ideally we should create the multimap for the smaller table, thus minimizing its creation time and memory size.</li>
+ </ul>
+ <li><strong>Join phase:</strong> Scan the other table, and find matching rows by looking in the multimap created before.</li>
+</ol>
+
+In pseudo-code, the algorithm could be expressed as follows:
+
+<pre><strong>let</strong> <i>A</i> = the first input table (or ideally, the larger one)
+<strong>let</strong> <i>B</i> = the second input table (or ideally, the smaller one)
+<strong>let</strong> <i>j<sub>A</sub></i> = the join column ID of table <i>A</i>
+<strong>let</strong> <i>j<sub>B</sub></i> = the join column ID of table <i>B</i>
+<strong>let</strong> <i>M<sub>B</sub></i> = a multimap for mapping from single values to multiple rows of table <i>B</i> (starts out empty)
+<strong>let</strong> <i>C</i> = the output table (starts out empty)
+<strong>for each</strong> row <i>b</i> in table <i>B</i>:
+ <strong>place</strong> <i>b</i> in multimap <i>M<sub>B</sub></i> under key <i>b(j<sub>B</sub>)</i>
+<strong>for each</strong> row <i>a</i> in table <i>A</i>:
+ <strong>for each</strong> row <i>b</i> in multimap <i>M<sub>B</sub></i> under key <i>a(j<sub>A</sub>)</i>:
+ <strong>let</strong> <i>c</i> = the concatenation of row <i>a</i> and row <i>b</i>
+ <strong>place</strong> row <i>c</i> in table <i>C</i>
+</pre>
+
+# --instructions--
+
+Implement the "hash join" algorithm as a function and demonstrate that it passes the test-case listed below. The function should accept two arrays of objects and return an array of combined objects.
+
+**Input**
+
+<table>
+ <tr>
+ <td style="padding: 4px; margin: 5px;">
+ <table style="border:none; border-collapse:collapse;">
+ <tr>
+ <td style="border:none"><i>A =</i></td>
+ <td style="border:none">
+ <table>
+ <tr>
+ <th style="padding: 4px; margin: 5px;">Age</th>
+ <th style="padding: 4px; margin: 5px;">Name</th>
+ </tr>
+ <tr>
+ <td style="padding: 4px; margin: 5px;">27</td>
+ <td style="padding: 4px; margin: 5px;">Jonah</td>
+ </tr>
+ <tr>
+ <td style="padding: 4px; margin: 5px;">18</td>
+ <td style="padding: 4px; margin: 5px;">Alan</td>
+ </tr>
+ <tr>
+ <td style="padding: 4px; margin: 5px;">28</td>
+ <td style="padding: 4px; margin: 5px;">Glory</td>
+ </tr>
+ <tr>
+ <td style="padding: 4px; margin: 5px;">18</td>
+ <td style="padding: 4px; margin: 5px;">Popeye</td>
+ </tr>
+ <tr>
+ <td style="padding: 4px; margin: 5px;">28</td>
+ <td style="padding: 4px; margin: 5px;">Alan</td>
+ </tr>
+ </table>
+ </td>
+ <td style="border:none; padding-left:1.5em;" rowspan="2"></td>
+ <td style="border:none"><i>B =</i></td>
+ <td style="border:none">
+ <table>
+ <tr>
+ <th style="padding: 4px; margin: 5px;">Character</th>
+ <th style="padding: 4px; margin: 5px;">Nemesis</th>
+ </tr>
+ <tr>
+ <td style="padding: 4px; margin: 5px;">Jonah</td>
+ <td style="padding: 4px; margin: 5px;">Whales</td>
+ </tr>
+ <tr>
+ <td style="padding: 4px; margin: 5px;">Jonah</td>
+ <td style="padding: 4px; margin: 5px;">Spiders</td>
+ </tr>
+ <tr>
+ <td style="padding: 4px; margin: 5px;">Alan</td>
+ <td style="padding: 4px; margin: 5px;">Ghosts</td>
+ </tr>
+ <tr>
+ <td style="padding: 4px; margin: 5px;">Alan</td>
+ <td style="padding: 4px; margin: 5px;">Zombies</td>
+ </tr>
+ <tr>
+ <td style="padding: 4px; margin: 5px;">Glory</td>
+ <td style="padding: 4px; margin: 5px;">Buffy</td>
+ </tr>
+ </table>
+ </td>
+ </tr>
+ <tr>
+ <td style="border:none">
+ <i>j<sub>A</sub> =</i>
+ </td>
+ <td style="border:none">
+ <i><code>Name</code> (i.e. column 1)</i>
+ </td>
+ <td style="border:none">
+ <i>j<sub>B</sub> =</i>
+ </td>
+ <td style="border:none">
+ <i><code>Character</code> (i.e. column 0)</i>
+ </td>
+ </tr>
+ </table>
+ </td>
+ </tr>
+</table>
+
+**Output**
+
+| A_age | A_name | B_character | B_nemesis |
+| ----- | ------ | ----------- | --------- |
+| 27 | Jonah | Jonah | Whales |
+| 27 | Jonah | Jonah | Spiders |
+| 18 | Alan | Alan | Ghosts |
+| 18 | Alan | Alan | Zombies |
+| 28 | Glory | Glory | Buffy |
+| 28 | Alan | Alan | Ghosts |
+| 28 | Alan | Alan | Zombies |
+
+The order of the rows in the output table is not significant.
# --hints--
-`hashJoin`是一个函数。
+`hashJoin` should be a function.
```js
assert(typeof hashJoin === 'function');
```
-`hashJoin([{ age: 27, name: "Jonah" }, { age: 18, name: "Alan" }, { age: 28, name: "Glory" }, { age: 18, name: "Popeye" }, { age: 28, name: "Alan" }], [{ character: "Jonah", nemesis: "Whales" }, { character: "Jonah", nemesis: "Spiders" }, { character: "Alan", nemesis: "Ghosts" }, { character:"Alan", nemesis: "Zombies" }, { character: "Glory", nemesis: "Buffy" }, { character: "Bob", nemesis: "foo" }])`应该返回`[{"A_age": 27,"A_name": "Jonah", "B_character": "Jonah", "B_nemesis": "Whales"}, {"A_age": 27,"A_name": "Jonah", "B_character": "Jonah", "B_nemesis": "Spiders"}, {"A_age": 18,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Ghosts"}, {"A_age": 18,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Zombies"}, {"A_age": 28,"A_name": "Glory", "B_character": "Glory", "B_nemesis": "Buffy"}, {"A_age": 28,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Ghosts"}, {"A_age": 28,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Zombies"}]`
+`hashJoin([{ age: 27, name: "Jonah" }, { age: 18, name: "Alan" }, { age: 28, name: "Glory" }, { age: 18, name: "Popeye" }, { age: 28, name: "Alan" }], [{ character: "Jonah", nemesis: "Whales" }, { character: "Jonah", nemesis: "Spiders" }, { character: "Alan", nemesis: "Ghosts" }, { character:"Alan", nemesis: "Zombies" }, { character: "Glory", nemesis: "Buffy" }, { character: "Bob", nemesis: "foo" }])` should return `[{"A_age": 27,"A_name": "Jonah", "B_character": "Jonah", "B_nemesis": "Whales"}, {"A_age": 27,"A_name": "Jonah", "B_character": "Jonah", "B_nemesis": "Spiders"}, {"A_age": 18,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Ghosts"}, {"A_age": 18,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Zombies"}, {"A_age": 28,"A_name": "Glory", "B_character": "Glory", "B_nemesis": "Buffy"}, {"A_age": 28,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Ghosts"}, {"A_age": 28,"A_name": "Alan", "B_character": "Alan", "B_nemesis": "Zombies"}]`
```js
assert.deepEqual(hashJoin(hash1, hash2), res);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/heronian-triangles.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/heronian-triangles.md
index 347befa8dd7..2162dc57eb4 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/heronian-triangles.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/heronian-triangles.md
@@ -1,42 +1,64 @@
---
id: 595b98f8b5a2245e243aa831
-title: 苍鹭三角形
+title: Heronian triangles
challengeType: 5
-videoUrl: ''
+forumTopicId: 302285
dashedName: heronian-triangles
---
# --description--
-<p>给出三边<big>a，</big> <big>b</big>和<big>c</big>长度的三角形区域的<a href='https://en.wikipedia.org/wiki/Heron's formula' title='wp：苍鹭的配方'>英雄公式</a>由下<a href='https://en.wikipedia.org/wiki/Heron's formula' title='wp：苍鹭的配方'>式</a>给出： </p><p> <big>$$ A = \ sqrt {s（sa）（sb）（sc）}，$$</big> </p><p>其中<big>s</big>是三角形的一半周长;那是， </p><p> <big>$$ S = \压裂{A + B + C} {2} $$。</big> </p><p> <a href='http://www.had2know.com/academics/heronian-triangles-generator-calculator.html' title='链接：http：//www.had2know.com/academics/heronian-triangles-generator-calculator.html'>Heronian三角形</a>是三角形，其边和面都是整数。 </p><p>一个例子是边长为3,4,5的三角形，其面积为6（其周长为12）。 </p><p>注意任何三角形的边都是3,4,5的整数倍;如6,8,10，也将是一个苍鹭三角形。 </p><p>将原始的Heronian三角形定义为最大公约数的Heronian三角形</p><p>三方都是1（统一）。 </p><p>这将排除例如三角形6,8,10。 </p>任务： <p>实现一个基于Hero公式的函数，该函数返回数组数组中的前<code>n <sub>th</sub></code>有序三角形。 </p>
+[Hero's formula](https://en.wikipedia.org/wiki/Heron's formula "wp: Heron's formula") for the area of a triangle given the length of its three sides `a`, `b`, and `c` is given by:
+
+$A = \\sqrt{s(s-a)(s-b)(s-c)},$
+
+where `s` is half the perimeter of the triangle; that is,
+
+$s=\\frac{a+b+c}{2}.$
+
+Heronian triangles are triangles whose sides and area are all integers.
+
+An example is the triangle with sides `3, 4, 5` whose area is `6` (and whose perimeter is `12`).
+
+Note that any triangle whose sides are all an integer multiple of `3, 4, 5`; such as `6, 8, 10,` will also be a Heronian triangle.
+
+Define a Primitive Heronian triangle as a Heronian triangle where the greatest common divisor
+
+of all three sides is `1` (unity).
+
+This will exclude, for example, triangle `6, 8, 10.`
+
+# --instructions--
+
+Implement a function based on Hero's formula that returns the first <code>n<sub>th</sub></code> ordered triangles in an array of arrays.
# --hints--
-`heronianTriangle`是一个函数。
+`heronianTriangle` should be a function.
```js
assert(typeof heronianTriangle === 'function');
```
-`heronianTriangle()`应返回`[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17]]` `heronianTriangle()` `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17]]` `heronianTriangle()` `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17]]` `heronianTriangle()` `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17]]` `heronianTriangle()` `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17]]`
+`heronianTriangle(10)` should return `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17]]`
```js
assert.deepEqual(heronianTriangle(testCases[0]), res[0]);
```
-`heronianTriangle()`应返回`[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17], [13, 13, 24], [6, 25, 29], [11, 13, 20], [5, 29, 30], [13, 14, 15]],` `heronianTriangle()` `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17], [13, 13, 24], [6, 25, 29], [11, 13, 20], [5, 29, 30], [13, 14, 15]],` `heronianTriangle()` `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17], [13, 13, 24], [6, 25, 29], [11, 13, 20], [5, 29, 30], [13, 14, 15]],` `heronianTriangle()` `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17], [13, 13, 24], [6, 25, 29], [11, 13, 20], [5, 29, 30], [13, 14, 15]],` `heronianTriangle()` `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17], [13, 13, 24], [6, 25, 29], [11, 13, 20], [5, 29, 30], [13, 14, 15]],`
+`heronianTriangle(15)` should return `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17], [13, 13, 24], [6, 25, 29], [11, 13, 20], [5, 29, 30], [13, 14, 15]],`
```js
assert.deepEqual(heronianTriangle(testCases[1]), res[1]);
```
-`heronianTriangle()`应返回`[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17], [13, 13, 24], [6, 25, 29], [11, 13, 20], [5, 29, 30], [13, 14, 15], [10, 17, 21], [7, 24, 25], [8, 29, 35], [12, 17, 25], [4, 51, 53]],` `heronianTriangle()` `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17], [13, 13, 24], [6, 25, 29], [11, 13, 20], [5, 29, 30], [13, 14, 15], [10, 17, 21], [7, 24, 25], [8, 29, 35], [12, 17, 25], [4, 51, 53]],` `heronianTriangle()` `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17], [13, 13, 24], [6, 25, 29], [11, 13, 20], [5, 29, 30], [13, 14, 15], [10, 17, 21], [7, 24, 25], [8, 29, 35], [12, 17, 25], [4, 51, 53]],` `heronianTriangle()` `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17], [13, 13, 24], [6, 25, 29], [11, 13, 20], [5, 29, 30], [13, 14, 15], [10, 17, 21], [7, 24, 25], [8, 29, 35], [12, 17, 25], [4, 51, 53]],` `heronianTriangle()` `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17], [13, 13, 24], [6, 25, 29], [11, 13, 20], [5, 29, 30], [13, 14, 15], [10, 17, 21], [7, 24, 25], [8, 29, 35], [12, 17, 25], [4, 51, 53]],`
+`heronianTriangle(20)` should return `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17], [13, 13, 24], [6, 25, 29], [11, 13, 20], [5, 29, 30], [13, 14, 15], [10, 17, 21], [7, 24, 25], [8, 29, 35], [12, 17, 25], [4, 51, 53]],`
```js
assert.deepEqual(heronianTriangle(testCases[2]), res[2]);
```
-`heronianTriangle()`应返回`[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17], [13, 13, 24], [6, 25, 29], [11, 13, 20], [5, 29, 30], [13, 14, 15], [10, 17, 21], [7, 24, 25], [8, 29, 35], [12, 17, 25], [4, 51, 53], [19, 20, 37],[16, 17, 17], [17, 17, 30], [16, 25, 39], [13, 20, 21]]` `heronianTriangle()` `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17], [13, 13, 24], [6, 25, 29], [11, 13, 20], [5, 29, 30], [13, 14, 15], [10, 17, 21], [7, 24, 25], [8, 29, 35], [12, 17, 25], [4, 51, 53], [19, 20, 37],[16, 17, 17], [17, 17, 30], [16, 25, 39], [13, 20, 21]]` `heronianTriangle()` `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17], [13, 13, 24], [6, 25, 29], [11, 13, 20], [5, 29, 30], [13, 14, 15], [10, 17, 21], [7, 24, 25], [8, 29, 35], [12, 17, 25], [4, 51, 53], [19, 20, 37],[16, 17, 17], [17, 17, 30], [16, 25, 39], [13, 20, 21]]` `heronianTriangle()` `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17], [13, 13, 24], [6, 25, 29], [11, 13, 20], [5, 29, 30], [13, 14, 15], [10, 17, 21], [7, 24, 25], [8, 29, 35], [12, 17, 25], [4, 51, 53], [19, 20, 37],[16, 17, 17], [17, 17, 30], [16, 25, 39], [13, 20, 21]]` `heronianTriangle()` `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17], [13, 13, 24], [6, 25, 29], [11, 13, 20], [5, 29, 30], [13, 14, 15], [10, 17, 21], [7, 24, 25], [8, 29, 35], [12, 17, 25], [4, 51, 53], [19, 20, 37],[16, 17, 17], [17, 17, 30], [16, 25, 39], [13, 20, 21]]`
+`heronianTriangle(25)` should return `[[3, 4, 5], [5, 5, 6], [5, 5, 8], [4, 13, 15], [5, 12, 13], [9, 10, 17], [3, 25, 26], [7, 15, 20], [10, 13, 13], [8, 15, 17], [13, 13, 24], [6, 25, 29], [11, 13, 20], [5, 29, 30], [13, 14, 15], [10, 17, 21], [7, 24, 25], [8, 29, 35], [12, 17, 25], [4, 51, 53], [19, 20, 37],[16, 17, 17], [17, 17, 30], [16, 25, 39], [13, 20, 21]]`
```js
assert.deepEqual(heronianTriangle(testCases[3]), res[3]);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hofstadter-figure-figure-sequences.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hofstadter-figure-figure-sequences.md
index b68fcd4ede1..1b5b864eda0 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hofstadter-figure-figure-sequences.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hofstadter-figure-figure-sequences.md
@@ -1,84 +1,113 @@
---
id: 59622f89e4e137560018a40e
-title: Hofstadter图 - 图序列
+title: Hofstadter Figure-Figure sequences
challengeType: 5
-videoUrl: ''
+forumTopicId: 302286
dashedName: hofstadter-figure-figure-sequences
---
# --description--
-<p>这两个正整数序列定义为： </p><p> <big>$$ R（1）= 1 \; \ S（1）= 2 \\ R（n）= R（n-1）+ S（n-1），\ quad n> 1. $$</big> </p><p>序列<big>$ S（n）$</big>进一步定义为<big>$ R（n）$中</big>不存在的正整数序列。 </p><p>序列<big>$ R $</big>开始： </p><p> 1,3,7,12,18 ...... </p><p>序列<big>$ S $</big>开始： </p><p> 2,4,5,6,8 ...... </p>任务：创建两个名为ffr和ffs的函数，当给定n分别返回R（n）或S（n）时（注意R（1）= 1且S（1）= 2以避免逐个错误） 。不应假设n的最大值。 Sloane的<a href='http://oeis.org/A005228' title='链接：http：//oeis.org/A005228'>A005228</a>和<a href='http://oeis.org/A030124' title='链接：http：//oeis.org/A030124'>A030124</a> 。 <a href='http://mathworld.wolfram.com/HofstadterFigure-FigureSequence.html' title='链接：http：//mathworld.wolfram.com/HofstadterFigure-FigureSequence.html'>Wolfram MathWorld</a>维基百科： <a href='https://en.wikipedia.org/wiki/Hofstadter_sequence#Hofstadter_Figure-Figure_sequences' title='wp：Hofstadter_sequence＃Hofstadter_Figure-Figure_sequences'>Hofstadter图 - 图序列</a> 。
+These two sequences of positive integers are defined as:
+
+$R(1)=1\\ ;\\ S(1)=2 \\\\R(n)=R(n-1)+S(n-1), \\quad n>1.$
+
+The sequence $S(n)$ is further defined as the sequence of positive integers not present in $R(n)$.
+
+Sequence $R$ starts:
+
+<pre>1, 3, 7, 12, 18, ...</pre>
+
+Sequence $S$ starts:
+
+<pre>2, 4, 5, 6, 8, ...</pre>
+
+# --instructions--
+
+Create two functions named `ffr` and `ffs` that when given `n` return `R(n)` or `S(n)` respectively. (Note that R(1) = 1 and S(1) = 2 to avoid off-by-one errors).
+
+No maximum value for `n` should be assumed.
+
+**References**
+
+<ul>
+ <li>
+ Sloane's <a href='https://oeis.org/A005228' target='_blank'>A005228</a> and <a href='https://oeis.org/A030124' target='_blank'>A030124</a>.
+ </li>
+ <li>
+ Wikipedia: <a href='https://en.wikipedia.org/wiki/Hofstadter_sequence#Hofstadter_Figure-Figure_sequences' title='wp: Hofstadter_sequence#Hofstadter_Figure-Figure_sequences' target='_blank'>Hofstadter Figure-Figure sequences</a>.
+ </li>
+</ul>
# --hints--
-`ffr`是一个功能。
+`ffr` should be a function.
```js
assert(typeof ffr === 'function');
```
-`ffs`是一个函数。
+`ffs` should be a function.
```js
assert(typeof ffs === 'function');
```
-`ffr`应该返回整数。
+`ffr` should return integer.
```js
assert(Number.isInteger(ffr(1)));
```
-`ffs`应该返回整数。
+`ffs` should return integer.
```js
assert(Number.isInteger(ffs(1)));
```
-`ffr()`应该返回`69`
+`ffr(10)` should return `69`
```js
assert.equal(ffr(ffrParamRes[0][0]), ffrParamRes[0][1]);
```
-`ffr()`应返回`1509`
+`ffr(50)` should return `1509`
```js
assert.equal(ffr(ffrParamRes[1][0]), ffrParamRes[1][1]);
```
-`ffr()`应返回`5764`
+`ffr(100)` should return `5764`
```js
assert.equal(ffr(ffrParamRes[2][0]), ffrParamRes[2][1]);
```
-`ffr()`应返回`526334`
+`ffr(1000)` should return `526334`
```js
assert.equal(ffr(ffrParamRes[3][0]), ffrParamRes[3][1]);
```
-`ffs()`应该返回`14`
+`ffs(10)` should return `14`
```js
assert.equal(ffs(ffsParamRes[0][0]), ffsParamRes[0][1]);
```
-`ffs()`应该返回`59`
+`ffs(50)` should return `59`
```js
assert.equal(ffs(ffsParamRes[1][0]), ffsParamRes[1][1]);
```
-`ffs()`应该返回`112`
+`ffs(100)` should return `112`
```js
assert.equal(ffs(ffsParamRes[2][0]), ffsParamRes[2][1]);
```
-`ffs()`应该返回`1041`
+`ffs(1000)` should return `1041`
```js
assert.equal(ffs(ffsParamRes[3][0]), ffsParamRes[3][1]);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hofstadter-q-sequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hofstadter-q-sequence.md
index 49b9557ca89..c8ab7d438ce 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hofstadter-q-sequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hofstadter-q-sequence.md
@@ -1,48 +1,56 @@
---
id: 59637c4d89f6786115efd814
-title: Hofstadter Q序列
+title: Hofstadter Q sequence
challengeType: 5
-videoUrl: ''
+forumTopicId: 302287
dashedName: hofstadter-q-sequence
---
# --description--
-<p> <a href='https://en.wikipedia.org/wiki/Hofstadter_sequence#Hofstadter_Q_sequence' title='wp：Hofstadter_sequence＃Hofstadter_Q_sequence'>Hofstadter Q序列</a>定义为： </p><p> $ Q（1）= Q（2）= 1，\\ Q（n）= Q \ big（nQ（n-1）\ big）+ Q \ big（nQ（n-2）），\ quad n> 2. $ </p><p>它定义为<a href='http://rosettacode.org/wiki/Fibonacci sequence' title='斐波那契序列'>Fibonacci序列</a> ，但<a href='http://rosettacode.org/wiki/Fibonacci sequence' title='斐波那契序列'>Fibonacci序列中</a>的下一个术语是前两个术语的总和，在Q序列中，前两个术语告诉您在Q序列中返回多远以找到两个数字总结以制作序列的下一个术语。 </p>任务：将Hofstadter Q Sequence方程实现为JavaScript
+The [Hofstadter Q sequence](https://en.wikipedia.org/wiki/Hofstadter_sequence#Hofstadter_Q_sequence "wp: Hofstadter_sequence#Hofstadter_Q_sequence") is defined as:
+
+$Q(1)=Q(2)=1, \\\\ Q(n)=Q\\big(n-Q(n-1)\\big)+Q\\big(n-Q(n-2)), \\quad n>2.$
+
+It is defined like the [Fibonacci sequence](https://rosettacode.org/wiki/Fibonacci sequence "Fibonacci sequence"), but whereas the next term in the Fibonacci sequence is the sum of the previous two terms, in the Q sequence the previous two terms tell you how far to go back in the Q sequence to find the two numbers to sum to make the next term of the sequence.
+
+# --instructions--
+
+Implement the Hofstadter Q Sequence equation as a function. The function should accept number, `n`, and return an integer.
# --hints--
-`hofstadterQ`是一个函数。
+`hofstadterQ` should be a function.
```js
assert(typeof hofstadterQ === 'function');
```
-`hofstadterQ()`应该返回`integer`
+`hofstadterQ()` should return `integer`
```js
assert(Number.isInteger(hofstadterQ(1000)));
```
-`hofstadterQ(1000)`应该返回`502`
+`hofstadterQ(1000)` should return `502`
```js
assert.equal(hofstadterQ(testCase[0]), res[0]);
```
-`hofstadterQ(1500)`应该返回`755`
+`hofstadterQ(1500)` should return `755`
```js
assert.equal(hofstadterQ(testCase[1]), res[1]);
```
-`hofstadterQ(2000)`应该返回`1005`
+`hofstadterQ(2000)` should return `1005`
```js
assert.equal(hofstadterQ(testCase[2]), res[2]);
```
-`hofstadterQ(2500)`应该返回`1261`
+`hofstadterQ(2500)` should return `1261`
```js
assert.equal(hofstadterQ(testCase[3]), res[3]);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/i-before-e-except-after-c.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/i-before-e-except-after-c.md
index c82bf3244ee..afac6dc5121 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/i-before-e-except-after-c.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/i-before-e-except-after-c.md
@@ -1,65 +1,77 @@
---
id: 5a23c84252665b21eecc7eb0
-title: 我在E之前除了C之后
+title: I before E except after C
challengeType: 5
-videoUrl: ''
+forumTopicId: 302288
dashedName: i-before-e-except-after-c
---
# --description--
-短语[“我在E之前，除了C之后”](<https://en.wikipedia.org/wiki/I before E except after C>)是一个广为人知的助记符，它应该有助于拼写英语单词。使用提供的单词，检查短语的两个子句是否合理：
+The phrase ["I before E, except after C"](https://en.wikipedia.org/wiki/I before E except after C) is a widely known mnemonic which is supposed to help when spelling English words.
-1. *“我在E之前没有前面的C”。*
-2. *“在我之前的C之前是C”。*
+Using the words provided, check if the two sub-clauses of the phrase are plausible individually:
-如果两个子短语都是合理的，则原始短语可以说是合理的。编写一个接受单词的函数，并检查单词是否遵循此规则。如果该函数遵循规则，则该函数应返回true，否则返回false。
+<ol>
+ <li>
+ <i>"I before E when not preceded by C".</i>
+ </li>
+ <li>
+ <i>"E before I when preceded by C".</i>
+ </li>
+</ol>
+
+If both sub-phrases are plausible then the original phrase can be said to be plausible.
+
+# --instructions--
+
+Write a function that accepts a word and check if the word follows this rule. The function should return true if the word follows the rule and false if it does not.
# --hints--
-`IBeforeExceptC`应该是一个函数。
+`IBeforeExceptC` should be a function.
```js
assert(typeof IBeforeExceptC == 'function');
```
-`IBeforeExceptC("receive")`应该返回一个布尔值。
+`IBeforeExceptC("receive")` should return a boolean.
```js
assert(typeof IBeforeExceptC('receive') == 'boolean');
```
-`IBeforeExceptC("receive")`应该返回`true` 。
+`IBeforeExceptC("receive")` should return `true`.
```js
assert.equal(IBeforeExceptC('receive'), true);
```
-`IBeforeExceptC("science")`应该返回`false` 。
+`IBeforeExceptC("science")` should return `false`.
```js
assert.equal(IBeforeExceptC('science'), false);
```
-`IBeforeExceptC("imperceivable")`应该返回`true` 。
+`IBeforeExceptC("imperceivable")` should return `true`.
```js
assert.equal(IBeforeExceptC('imperceivable'), true);
```
-`IBeforeExceptC("inconceivable")`应该返回`true` 。
+`IBeforeExceptC("inconceivable")` should return `true`.
```js
assert.equal(IBeforeExceptC('inconceivable'), true);
```
-`IBeforeExceptC("insufficient")`应返回`false` 。
+`IBeforeExceptC("insufficient")` should return `false`.
```js
assert.equal(IBeforeExceptC('insufficient'), false);
```
-`IBeforeExceptC("omniscient")`应该返回`false` 。
+`IBeforeExceptC("omniscient")` should return `false`.
```js
assert.equal(IBeforeExceptC('omniscient'), false);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/iban.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/iban.md
index 9fd6e2526d9..666c50a6be8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/iban.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/iban.md
@@ -1,60 +1,68 @@
---
id: 5a23c84252665b21eecc7eaf
-title: 他们
+title: IBAN
challengeType: 5
-videoUrl: ''
+forumTopicId: 302289
dashedName: iban
---
# --description--
-[国际银行账号（IBAN）](https://en.wikipedia.org/wiki/International_Bank_Account_Number)是一种国际商定的识别跨国界银行账户的方法，降低了传播[抄写错误的](https://en.wikipedia.org/wiki/Transcription_error)风险。 IBAN最多包含34个字母数字字符：
+The [International Bank Account Number (IBAN)](https://en.wikipedia.org/wiki/International_Bank_Account_Number) is an internationally agreed means of identifying bank accounts across national borders with a reduced risk of propagating [transcription errors](https://en.wikipedia.org/wiki/Transcription_error).
-- 首先是两个字母的ISO 3166-1 alpha-2国家代码
-- 然后是两个校验位，和
-- 最后是国家特定的基本银行帐号（BBAN）。
+The IBAN consists of up to 34 alphanumeric characters:
-通过检查数字，即使在提交交易之前，也可以对银行帐号进行健全性检查以确认其完整性。编写一个以IBAN字符串作为参数的函数。如果有效则返回true。否则，返回false。
+<ul>
+ <li>first the two-letter ISO 3166-1 alpha-2 country code</li>
+ <li>then two check digits, and</li>
+ <li>finally a country-specific Basic Bank Account Number (BBAN).</li>
+</ul>
+
+The check digits enable a sanity check of the bank account number to confirm its integrity even before submitting a transaction.
+
+# --instructions--
+
+Write a function that takes IBAN string as parameter. If it is valid return true. Otherwise, return false.
# --hints--
-`isValid`应该是一个函数。
+`isValid` should be a function.
```js
assert(typeof isValid == 'function');
```
-`isValid(""+tests[0]+"")`应该返回一个布尔值。
+`isValid("GB82 WEST 1234 5698 7654 32")` should return a boolean.
```js
assert(typeof isValid('GB82 WEST 1234 5698 7654 32') == 'boolean');
```
-`isValid(""+tests[0]+"")`应该返回`true` 。
+`isValid("GB82 WEST 1234 5698 7654 32")` should return `true`.
```js
assert.equal(isValid('GB82 WEST 1234 5698 7654 32'), true);
```
-`isValid(""+tests[1]+"")`应返回`false` 。
+`isValid("GB82 WEST 1.34 5698 7654 32")` should return `false`.
```js
assert.equal(isValid('GB82 WEST 1.34 5698 7654 32'), false);
```
-`isValid(""+tests[2]+"")`应返回`false` 。
+`isValid("GB82 WEST 1234 5698 7654 325")` should return `false`.
```js
assert.equal(isValid('GB82 WEST 1234 5698 7654 325'), false);
```
-`isValid(""+tests[3]+"")`应该返回`false` 。
+`isValid("GB82 TEST 1234 5698 7654 32")` should return `false`.
```js
assert.equal(isValid('GB82 TEST 1234 5698 7654 32'), false);
```
-`isValid(""+tests[4]+"")`应该返回`true` 。
+`isValid("SA03 8000 0000 6080 1016 7519")` should return `true`.
```js
assert.equal(isValid('SA03 8000 0000 6080 1016 7519'), true);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/identity-matrix.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/identity-matrix.md
index c9b640e8985..a8465da648c 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/identity-matrix.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/identity-matrix.md
@@ -1,48 +1,56 @@
---
id: 5a23c84252665b21eecc7eb1
-title: 身份矩阵
+title: Identity matrix
challengeType: 5
-videoUrl: ''
+forumTopicId: 302290
dashedName: identity-matrix
---
# --description--
-*单位矩阵*是大小为\\（n \\次n \\）的方阵，其中对角元素都是**1** s（1），所有其他元素都是**0** s（零）。 \\ begin {bmatrix} 1＆0＆0 \\ cr 0＆1＆0 \\ cr 0＆0＆1 \\ cr \\ end {bmatrix}编写一个以数字'n'作为参数并返回单位矩阵的函数订单nx n。
+An *identity matrix* is a square matrix of size \\( n \\times n \\), where the diagonal elements are all `1`s (ones), and all the other elements are all `0`s (zeroes).
+
+<ul>
+ <li style='list-style: none;'>\(\displaystyle I_{n}=\begin{bmatrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{bmatrix}\)</li>
+</ul>
+
+# --instructions--
+
+Write a function that takes a number `n` as a parameter and returns the identity matrix of order \\( n \\times n \\).
# --hints--
-`idMatrix`应该是一个功能。
+`idMatrix` should be a function.
```js
assert(typeof idMatrix == 'function');
```
-`idMatrix(1)`应该返回一个数组。
+`idMatrix(1)` should return an array.
```js
assert(Array.isArray(idMatrix(1)));
```
-`idMatrix(1)`应返回`"+JSON.stringify(results[0])+"` 。
+`idMatrix(1)` should return `[ [ 1 ] ]`.
```js
assert.deepEqual(idMatrix(1), results[0]);
```
-`idMatrix(2)`应返回`"+JSON.stringify(results[1])+"` 。
+`idMatrix(2)` should return `[ [ 1, 0 ], [ 0, 1 ] ]`.
```js
assert.deepEqual(idMatrix(2), results[1]);
```
-`idMatrix(3)`应返回`"+JSON.stringify(results[2])+"` 。
+`idMatrix(3)` should return `[ [ 1, 0, 0 ], [ 0, 1, 0 ], [ 0, 0, 1 ] ]`.
```js
assert.deepEqual(idMatrix(3), results[2]);
```
-`idMatrix(4)`应返回`"+JSON.stringify(results[3])+"` 。
+`idMatrix(4)` should return `[ [ 1, 0, 0, 0 ], [ 0, 1, 0, 0 ], [ 0, 0, 1, 0 ], [ 0, 0, 0, 1 ] ]`.
```js
assert.deepEqual(idMatrix(4), results[3]);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/iterated-digits-squaring.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/iterated-digits-squaring.md
index 304a24e2c07..bbed591151a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/iterated-digits-squaring.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/iterated-digits-squaring.md
@@ -1,67 +1,68 @@
---
id: 5a23c84252665b21eecc7ec1
-title: 迭代的数字平方
+title: Iterated digits squaring
challengeType: 5
-videoUrl: ''
+forumTopicId: 302291
dashedName: iterated-digits-squaring
---
# --description--
-如果添加自然数（大于零的整数）的数字的平方，则始终以1或89结尾：
+If you add the square of the digits of a Natural number (an integer bigger than zero), you always end with either 1 or 89:
-```
- 15 - > 26 - > 40 - > 16 - > 37 - > 58 - > 89
-7 - > 49 - > 97 - > 130 - > 10 - > 1
-```
+<pre>15 -> 26 -> 40 -> 16 -> 37 -> 58 -> 89
+7 -> 49 -> 97 -> 130 -> 10 -> 1
+</pre>
-编写一个函数，该函数将数字作为参数，并在执行上述过程后返回1或89。
+# --instructions--
+
+Write a function that takes a number as a parameter and returns 1 or 89 after performing the mentioned process.
# --hints--
-`iteratedSquare`应该是一个函数。
+`iteratedSquare` should be a function.
```js
assert(typeof iteratedSquare == 'function');
```
-`iteratedSquare(4)`应该返回一个数字。
+`iteratedSquare(4)` should return a number.
```js
assert(typeof iteratedSquare(4) == 'number');
```
-`iteratedSquare(4)`应该返回`89` 。
+`iteratedSquare(4)` should return `89`.
```js
assert.equal(iteratedSquare(4), 89);
```
-`iteratedSquare(7)`应该返回`1` 。
+`iteratedSquare(7)` should return `1`.
```js
assert.equal(iteratedSquare(7), 1);
```
-`iteratedSquare(15)`应该返回`89` 。
+`iteratedSquare(15)` should return `89`.
```js
assert.equal(iteratedSquare(15), 89);
```
-`iteratedSquare(20)`应该返回`89` 。
+`iteratedSquare(20)` should return `89`.
```js
assert.equal(iteratedSquare(20), 89);
```
-`iteratedSquare(70)`应该返回`1` 。
+`iteratedSquare(70)` should return `1`.
```js
assert.equal(iteratedSquare(70), 1);
```
-`iteratedSquare(100)`应该返回`1` 。
+`iteratedSquare(100)` should return `1`.
```js
assert.equal(iteratedSquare(100), 1);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/jaro-distance.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/jaro-distance.md
index b8f506b716e..d348acaee73 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/jaro-distance.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/jaro-distance.md
@@ -1,66 +1,88 @@
---
id: 5a23c84252665b21eecc7ec2
-title: Jaro距离
+title: Jaro distance
challengeType: 5
-videoUrl: ''
+forumTopicId: 302292
dashedName: jaro-distance
---
# --description--
-Jaro距离是两个弦之间相似性的度量。两个弦的Jaro距离越高，弦越相似。对得分进行归一化，使得**0**等于没有相似性， **1**等于完全匹配。定义两个给定字符串\\（s_1 \\）和\\（s_2 \\）的Jaro距离\\（d_j \\）是\\ begin {align} d_j = \\ begin {cases} 0 && \\ text {if} m = 0 \\\\\\ \\ {\\ frac {1} {3}} \\ left（{\\ frac {m} {| s\_ {1} |}} + {\\ frac {m} {| s\_ {2} |}} + {\\ frac { mt} {m}} \\ right）&& \\ text {otherwise} \\ end {cases} \\ end {align}其中：
+The Jaro distance is a measure of similarity between two strings. The higher the Jaro distance for two strings is, the more similar the strings are. The score is normalized such that `0` equates to no similarity and `1` is an exact match.
-- \\（m \\）是*匹配字符*的数量;
-- \\（t \\）是*换位*次数的一半。
+**Definition**
-分别来自\\（s_1 \\）和\\（s_2 \\）的两个字符只有在相同且不远于\\（\\ left \\ lfloor \\ frac {\\ max（| s_1 |，| s_2 |）}时才被认为是*匹配的* {2} \\右\\ rfloor-1 \\）。将\\（s_1 \\）的每个字符与\\（s_2 \\）中的所有匹配字符进行比较。匹配（但不同的序列顺序）字符除以2的数量定义了*转置*的数量。 **示例**给定字符串\\（s_1 \\） *DWAYNE*和\\（s_2 \\） *DUANE*我们发现：
+The Jaro distance \\( d_j \\) of two given strings \\(s_1\\) and \\(s_2\\) is
-- \\（m = 4 \\）
-- \\（| s_1 | = 6 \\）
-- \\（| s_2 | = 5 \\）
-- \\（t = 0 \\）
+\\begin{align}d_j = \\begin{cases}0& & \\text{if }m=0 \\\\\\\\{\\frac {1}{3}}\\left({\\frac {m}{|s\_{1}|}}+{\\frac {m}{|s\_{2}|}}+{\\frac {m-t}{m}}\\right)& & \\text{otherwise}\\end{cases}\\end{align}
-我们发现Jaro得分为：\\（d_j = \\ frac {1} {3} \\ left（\\ frac {4} {6} + \\ frac {4} {5} + \\ frac {4-0} {4} \\ right）= 0.822 \\）。编写一个函数a，它接受两个字符串作为参数并返回相关的Jaro距离。
+Where:
+
+<ul>
+ <li>\(m\) is the number of <i>matching characters</i>;</li>
+ <li> \(t\) is half the number of <i>transpositions</i>.</li>
+</ul>
+
+Two characters from \\(s_1\\) and \\(s_2\\) respectively, are considered *matching* only if they are the same and not farther than \\(\\left\\lfloor\\frac{\\max(|s_1|,|s_2|)}{2}\\right\\rfloor-1\\).
+
+Each character of \\(s_1\\) is compared with all its matching characters in \\(s_2\\) . The number of matching (but different sequence order) characters divided by 2 defines the number of *transpositions*.
+
+**Example**
+
+Given the strings \\(s_1\\) *DWAYNE* and \\(s_2\\) *DUANE* we find:
+
+<ul>
+ <li>\(m = 4\)</li>
+ <li>\(|s_1| = 6\)</li>
+ <li>\(|s_2| = 5\)</li>
+ <li>\(t = 0\)</li>
+</ul>
+
+We find a Jaro score of: \\(d_j = \\frac{1}{3}\\left(\\frac{4}{6} + \\frac{4}{5} + \\frac{4-0}{4}\\right) = 0.822\\).
+
+# --instructions--
+
+Write a function a that takes two strings as parameters and returns the associated Jaro distance.
# --hints--
-`jaro`应该是一个功能。
+`jaro` should be a function.
```js
assert(typeof jaro == 'function');
```
-`jaro(""+tests[0][0]+"",""+tests[0][1]+"")`应返回一个数字。
+`jaro("MARTHA", "MARHTA")` should return a number.
```js
assert(typeof jaro('MARTHA', 'MARHTA') == 'number');
```
-`jaro(""+tests[0][0]+"",""+tests[0][1]+"")`应该返回`"+results[0]+"` 。
+`jaro("MARTHA", "MARHTA")` should return `0.9444444444444445`.
```js
assert.equal(jaro('MARTHA', 'MARHTA'), 0.9444444444444445);
```
-`jaro(""+tests[1][0]+"",""+tests[1][1]+"")`应返回`"+results[1]+"` 。
+`jaro("DIXON", "DICKSONX")` should return `0.7666666666666666`.
```js
assert.equal(jaro('DIXON', 'DICKSONX'), 0.7666666666666666);
```
-`jaro(""+tests[2][0]+"",""+tests[2][1]+"")`应返回`"+results[2]+"` 。
+`jaro("JELLYFISH", "SMELLYFISH")` should return `0.8962962962962964`.
```js
assert.equal(jaro('JELLYFISH', 'SMELLYFISH'), 0.8962962962962964);
```
-`jaro(""+tests[3][0]+"",""+tests[3][1]+"")`应返回`"+results[3]+"` 。
+`jaro("HELLOS", "CHELLO")` should return `0.888888888888889`.
```js
assert.equal(jaro('HELLOS', 'CHELLO'), 0.888888888888889);
```
-`jaro(""+tests[4][0]+"",""+tests[4][1]+"")`应该返回`"+results[4]+"` 。
+`jaro("ABCD", "BCDA")` should return `0.8333333333333334`.
```js
assert.equal(jaro('ABCD', 'BCDA'), 0.8333333333333334);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/jortsort.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/jortsort.md
index 186cafa3d5f..0f121b99ee4 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/jortsort.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/jortsort.md
@@ -2,59 +2,61 @@
id: 5a23c84252665b21eecc7ec4
title: JortSort
challengeType: 5
-videoUrl: ''
+forumTopicId: 302293
dashedName: jortsort
---
# --description--
-jortSort是一个排序工具集，它使用户可以完成工作并保证效率，因为您不必再次排序。它最初是由着名的[JSConf的](https://www.youtube.com/watch?v=pj4U_W0OFoE) Jenn“Moneydollars”Schiffer [提出的](https://www.youtube.com/watch?v=pj4U_W0OFoE) 。 jortSort是一个函数，它将一个可比较对象数组作为其参数。然后它按升序对数组进行排序，并将排序后的数组与最初提供的数组进行比较。如果数组匹配（即原始数组已经排序），则该函数返回true。如果数组不匹配（即原始数组未排序），则该函数返回false。
+jortSort is a sorting toolset that makes the user do the work and guarantees efficiency because you don't have to sort ever again. It was originally presented by Jenn "Moneydollars" Schiffer at the prestigious [JSConf](https://www.youtube.com/watch?v=pj4U_W0OFoE).
+
+jortSort should be a function that takes a single array of comparable objects as its argument. It then sorts the array in ascending order and compares the sorted array to the originally provided array. If the arrays match (i.e. the original array was already sorted), the function returns true. If the arrays do not match (i.e. the original array was not sorted), the function returns false.
# --hints--
-`jortsort`应该是一个功能。
+`jortsort` should be a function.
```js
assert(typeof jortsort == 'function');
```
-`jortsort("+JSON.stringify(tests[0])+")`应该返回一个布尔值。
+`jortsort([1,2,3,4,5])` should return a boolean.
```js
assert(typeof jortsort([1, 2, 3, 4, 5]) == 'boolean');
```
-`jortsort("+JSON.stringify(tests[0])+")`应该返回`true` 。
+`jortsort([1,2,3,4,5])` should return `true`.
```js
assert.equal(jortsort([1, 2, 3, 4, 5]), true);
```
-`jortsort("+JSON.stringify(tests[1])+")`应该返回`false` 。
+`jortsort([1,2,13,4,5])` should return `false`.
```js
assert.equal(jortsort([1, 2, 13, 4, 5]), false);
```
-`jortsort("+JSON.stringify(tests[2])+")`应该返回`false` 。
+`jortsort([12,4,51,2,4])` should return `false`.
```js
assert.equal(jortsort([12, 4, 51, 2, 4]), false);
```
-`jortsort("+JSON.stringify(tests[3])+")`应该返回`true` 。
+`jortsort([1,2])` should return `true`.
```js
assert.equal(jortsort([1, 2]), true);
```
-`jortsort("+JSON.stringify(tests[4])+")`应该返回`false` 。
+`jortsort([5,4,3,2,1])` should return `false`.
```js
assert.equal(jortsort([5, 4, 3, 2, 1]), false);
```
-`jortsort("+JSON.stringify(tests[5])+")`应该返回`true` 。
+`jortsort([1,1,1,1,1])` should return `true`.
```js
assert.equal(jortsort([1, 1, 1, 1, 1]), true);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/josephus-problem.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/josephus-problem.md
index f467ebe8a71..93d7f7876b6 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/josephus-problem.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/josephus-problem.md
@@ -1,54 +1,72 @@
---
id: 5a23c84252665b21eecc7ec5
-title: 约瑟夫斯问题
+title: Josephus problem
challengeType: 5
-videoUrl: ''
+forumTopicId: 302294
dashedName: josephus-problem
---
# --description--
-[约瑟夫斯问题](<https://en.wikipedia.org/wiki/Josephus problem>)是一个严峻描述的数学难题：$ n $囚犯站在一个圆圈上，顺序编号从$ 0 $到$ n-1 $。一名刽子手沿着圈子走，从囚犯$ 0 $开始，移走每个$ k $囚犯并杀死他。随着过程的继续，圆圈变得越来越小，直到只剩下一个囚犯，然后被释放。例如，如果$ n = 5 $囚犯和$ k = 2 $，囚犯被杀的命令（我们称之为“杀戮序列”）将是1,3,0和4，幸存者将是＃2。鉴于任何$ n，k> 0 $ ，找出哪个囚犯将成为最后的幸存者。在一个这样的事件中，有41个囚犯和每3 <sup>次</sup>囚犯被杀死（$ K = 3 $）。其中有一个聪明的名字约瑟夫斯，他解决了这个问题，站在幸存的位置，并继续讲述这个故事。他是哪个号码？写一个函数，以囚犯的初始数量和'k'作为参数，并返回幸存的囚犯的数量。
+[Josephus problem](https://en.wikipedia.org/wiki/Josephus problem) is a math puzzle with a grim description: $n$ prisoners are standing on a circle, sequentially numbered from $0$ to $n-1$.
+
+An executioner walks along the circle, starting from prisoner $0$, removing every $k$-th prisoner and killing him.
+
+As the process goes on, the circle becomes smaller and smaller, until only one prisoner remains, who is then freed.
+
+For example, if there are $n=5$ prisoners and $k=2$, the order the prisoners are killed in (let's call it the "killing sequence") will be 1, 3, 0, and 4, and the survivor will be #2.
+
+Given any $n, k > 0$, find out which prisoner will be the final survivor.
+
+In one such incident, there were 41 prisoners and every 3<sup>rd</sup> prisoner was being killed ($k=3$).
+
+Among them was a clever chap name Josephus who worked out the problem, stood at the surviving position, and lived on to tell the tale.
+
+Which number was he?
+
+# --instructions--
+
+Write a function that takes the initial number of prisoners and 'k' as parameter and returns the number of the prisoner that survives.
# --hints--
-`josephus`应该是一个功能。
+`josephus` should be a function.
```js
assert(typeof josephus == 'function');
```
-`josephus(30,3)`应该返回一个数字。
+`josephus(30,3)` should return a number.
```js
assert(typeof josephus(30, 3) == 'number');
```
-`josephus(30,3)`应该回`29` 。
+`josephus(30,3)` should return `29`.
```js
assert.equal(josephus(30, 3), 29);
```
-`josephus(30,5)`应该返回`3` 。
+`josephus(30,5)` should return `3`.
```js
assert.equal(josephus(30, 5), 3);
```
-`josephus(20,2)`应该返回`9` 。
+`josephus(20,2)` should return `9`.
```js
assert.equal(josephus(20, 2), 9);
```
-`josephus(17,6)`应该回归`2` 。
+`josephus(17,6)` should return `2`.
```js
assert.equal(josephus(17, 6), 2);
```
-`josephus(29,4)`应该返回`2` 。
+`josephus(29,4)` should return `2`.
```js
assert.equal(josephus(29, 4), 2);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/kaprekar-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/kaprekar-numbers.md
index 59e368bc527..7b2530f4c62 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/kaprekar-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/kaprekar-numbers.md
@@ -8,7 +8,7 @@ dashedName: kaprekar-numbers
# --description--
-A positive integer is a [Kaprekar number](<https://en.wikipedia.org/wiki/Kaprekar number>) if:
+A positive integer is a [Kaprekar number](https://en.wikipedia.org/wiki/Kaprekar number) if:
<ul>
<li>It is 1, or,</li>
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/least-common-multiple.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/least-common-multiple.md
index 403e448d2f4..aa27446cbe9 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/least-common-multiple.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/least-common-multiple.md
@@ -8,7 +8,7 @@ dashedName: least-common-multiple
# --description--
-The least common multiple of 12 and 18 is 36, because 12 is a factor (12 × 3 = 36), and 18 is a factor (18 × 2 = 36), and there is no positive integer less than 36 that has both factors. As a special case, if either *m* or *n* is zero, then the least common multiple is zero. One way to calculate the least common multiple is to iterate all the multiples of *m*, until you find one that is also a multiple of *n*. If you already have *gcd* for [greatest common divisor](<https://rosettacode.org/wiki/greatest common divisor>), then this formula calculates *lcm*. ( \\operatorname{lcm}(m, n) = \\frac{|m \\times n|}{\\operatorname{gcd}(m, n)} )
+The least common multiple of 12 and 18 is 36, because 12 is a factor (12 × 3 = 36), and 18 is a factor (18 × 2 = 36), and there is no positive integer less than 36 that has both factors. As a special case, if either *m* or *n* is zero, then the least common multiple is zero. One way to calculate the least common multiple is to iterate all the multiples of *m*, until you find one that is also a multiple of *n*. If you already have *gcd* for [greatest common divisor](https://rosettacode.org/wiki/greatest common divisor), then this formula calculates *lcm*. ( \\operatorname{lcm}(m, n) = \\frac{|m \\times n|}{\\operatorname{gcd}(m, n)} )
# --instructions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/levenshtein-distance.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/levenshtein-distance.md
index 104543d9d00..2737ad5436a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/levenshtein-distance.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/levenshtein-distance.md
@@ -8,7 +8,7 @@ dashedName: levenshtein-distance
# --description--
-In information theory and computer science, the **Levenshtein distance** is a [metric](<https://en.wikipedia.org/wiki/string metric>) for measuring the amount of difference between two sequences (i.e. an [edit distance](<https://en.wikipedia.org/wiki/edit distance>)). The Levenshtein distance between two strings is defined as the minimum number of edits needed to transform one string into the other, with the allowable edit operations being insertion, deletion, or substitution of a single character.
+In information theory and computer science, the **Levenshtein distance** is a [metric](https://en.wikipedia.org/wiki/string metric) for measuring the amount of difference between two sequences (i.e. an [edit distance](https://en.wikipedia.org/wiki/edit distance)). The Levenshtein distance between two strings is defined as the minimum number of edits needed to transform one string into the other, with the allowable edit operations being insertion, deletion, or substitution of a single character.
Example:
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/linear-congruential-generator.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/linear-congruential-generator.md
index 24e89f42e62..0448399ccbe 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/linear-congruential-generator.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/linear-congruential-generator.md
@@ -8,9 +8,9 @@ dashedName: linear-congruential-generator
# --description--
-The [linear congruential generator](<https://en.wikipedia.org/wiki/linear congruential generator>) is a very simple example of a [random number generator](<http://rosettacode.org/wiki/random number generator>). All linear congruential generators use this formula:
+The [linear congruential generator](https://en.wikipedia.org/wiki/linear congruential generator) is a very simple example of a [random number generator](http://rosettacode.org/wiki/random number generator). All linear congruential generators use this formula:
-$$r\_{n + 1} = a \\times r_n + c \\pmod m$$
+$$r_{n + 1} = (a \times r_n + c) \bmod m$$
Where:
@@ -22,7 +22,7 @@ Where:
If one chooses the values of $a$, $c$ and $m$ with care, then the generator produces a uniform distribution of integers from $0$ to $m - 1$.
-LCG numbers have poor quality. $r_n$ and $r\_{n + 1}$ are not independent, as true random numbers would be. Anyone who knows $r_n$ can predict $r\_{n + 1}$, therefore LCG is not cryptographically secure. The LCG is still good enough for simple tasks like [Miller-Rabin primality test](<http://rosettacode.org/wiki/Miller-Rabin primality test>), or [FreeCell deals](<http://rosettacode.org/wiki/deal cards for FreeCell>). Among the benefits of the LCG, one can easily reproduce a sequence of numbers, from the same $r_0$. One can also reproduce such sequence with a different programming language, because the formula is so simple.
+LCG numbers have poor quality. $r_n$ and $r\_{n + 1}$ are not independent, as true random numbers would be. Anyone who knows $r_n$ can predict $r\_{n + 1}$, therefore LCG is not cryptographically secure. The LCG is still good enough for simple tasks like [Miller-Rabin primality test](http://rosettacode.org/wiki/Miller-Rabin primality test), or [FreeCell deals](http://rosettacode.org/wiki/deal cards for FreeCell). Among the benefits of the LCG, one can easily reproduce a sequence of numbers, from the same $r_0$. One can also reproduce such sequence with a different programming language, because the formula is so simple.
# --instructions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/long-multiplication.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/long-multiplication.md
index 431f88eeb8f..fb4e2456cf3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/long-multiplication.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/long-multiplication.md
@@ -8,7 +8,7 @@ dashedName: long-multiplication
# --description--
-Explicitly implement [long multiplication](<https://en.wikipedia.org/wiki/long multiplication>).
+Explicitly implement [long multiplication](https://en.wikipedia.org/wiki/long multiplication).
This is one possible approach to arbitrary-precision integer algebra.
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/longest-increasing-subsequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/longest-increasing-subsequence.md
index 4de5521d55f..0ba5661f6c8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/longest-increasing-subsequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/longest-increasing-subsequence.md
@@ -18,7 +18,7 @@ Longest increasing sequence is:
$\\{3, 10, 20\\}$
-For more information on this problem please see [Wikipedia](<https://en.wikipedia.org/wiki/Longest increasing subsequence>).
+For more information on this problem please see [Wikipedia](https://en.wikipedia.org/wiki/Longest increasing subsequence).
# --instructions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/look-and-say-sequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/look-and-say-sequence.md
index 073e41920d5..ed09269a091 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/look-and-say-sequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/look-and-say-sequence.md
@@ -8,7 +8,7 @@ dashedName: look-and-say-sequence
# --description--
-The [Look and say sequence](<https://en.wikipedia.org/wiki/Look and say sequence>) is a recursively defined sequence of numbers.
+The [Look and say sequence](https://en.wikipedia.org/wiki/Look and say sequence) is a recursively defined sequence of numbers.
Sequence Definition
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/lu-decomposition.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/lu-decomposition.md
index f7a900174ec..1b2442f3056 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/lu-decomposition.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/lu-decomposition.md
@@ -8,13 +8,13 @@ dashedName: lu-decomposition
# --description--
-Every square matrix $A$ can be decomposed into a product of a lower triangular matrix $L$ and a upper triangular matrix $U$, as described in [LU decomposition](<https://en.wikipedia.org/wiki/LU decomposition>).
+Every square matrix $A$ can be decomposed into a product of a lower triangular matrix $L$ and a upper triangular matrix $U$, as described in [LU decomposition](https://en.wikipedia.org/wiki/LU decomposition).
$A = LU$
It is a modified form of Gaussian elimination.
-While the [Cholesky decomposition](<http://rosettacode.org/wiki/Cholesky decomposition>) only works for symmetric, positive definite matrices, the more general LU decomposition works for any square matrix.
+While the [Cholesky decomposition](http://rosettacode.org/wiki/Cholesky decomposition) only works for symmetric, positive definite matrices, the more general LU decomposition works for any square matrix.
There are several algorithms for calculating $L$ and $U$.
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/luhn-test-of-credit-card-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/luhn-test-of-credit-card-numbers.md
index 0a44e1da794..4fcc90502ab 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/luhn-test-of-credit-card-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/luhn-test-of-credit-card-numbers.md
@@ -8,7 +8,7 @@ dashedName: luhn-test-of-credit-card-numbers
# --description--
-The [Luhn test](<https://en.wikipedia.org/wiki/Luhn algorithm>) is used by some credit card companies to distinguish valid credit card numbers from what could be a random selection of digits.
+The [Luhn test](https://en.wikipedia.org/wiki/Luhn algorithm) is used by some credit card companies to distinguish valid credit card numbers from what could be a random selection of digits.
Those companies using credit card numbers that can be validated by the Luhn test have numbers that pass the following test:
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/lzw-compression.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/lzw-compression.md
index d408984b6a6..ccde98aa946 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/lzw-compression.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/lzw-compression.md
@@ -14,7 +14,7 @@ You can read a complete description of it in the [Wikipedia article](https://en.
# --instructions--
-Write a function that takes two parameters. The first parameter is a boolean where \`true\` indicates compress and \`false\` indicates decompress. The second parameter is either a string or an array to be processed. If it is a string to be compressed, return an array of numbers. If it's an array of numbers to be decompressed, return a string.
+Write a function that takes two parameters. The first parameter is a boolean where `true` indicates compress and `false` indicates decompress. The second parameter is either a string or an array to be processed. If it is a string to be compressed, return an array of numbers. If it's an array of numbers to be decompressed, return a string.
# --hints--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/s-expressions.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/s-expressions.md
index 433e8929ee4..e7ed94741c6 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/s-expressions.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/s-expressions.md
@@ -1,32 +1,54 @@
---
id: 59667989bf71cf555dd5d2ff
-title: S-表达式
+title: S-Expressions
challengeType: 5
-videoUrl: ''
+forumTopicId: 302303
dashedName: s-expressions
---
# --description--
-<p> <a href='https://en.wikipedia.org/wiki/S-Expression' title='wp：S表达式'>S-Expressions</a>是一种解析和存储数据的便捷方式。 </p>任务： <p>为S-Expressions编写一个简单的读取器/解析器，处理引用的和不带引号的字符串，整数和浮点数。 </p><p>该函数应从字符串中读取单个但嵌套的S-Expression，并将其作为（嵌套）数组返回。 </p><p>除非包含在带引号的字符串中，否则可以忽略换行符和其他空格。 </p><p> “ <tt>（）</tt> ”内部引用的字符串不会被解释，但会被视为字符串的一部分。 </p><p>处理字符串中的转义引号是可选的;因此“ <tt>（foo”bar）</tt> “可能被视为字符串” <tt>foo“bar</tt> ”，或作为错误。 </p><p>为此，读者无需识别“ <tt>\</tt> ”以进行转义，但如果语言具有适当的数据类型，则还应识别数字。 </p><p>请注意，除了“ <tt>（）”</tt> “（” <tt>\</tt> “，如果支持转义）和空格，没有特殊字符。其他任何内容都是允许的，不带引号。 </p><p>读者应该能够阅读以下输入</p><p></p><pre> （（数据“引用数据”123 4.5）
- （数据（！@＃（4.5）“（更多”“数据”））））
-</pre><p></p><p>并将其转换为本机数据结构。 （有关本机数据结构的示例，请参阅<a href='http://rosettacode.org/wiki/#Pike' title='#Pike'>Pike</a> ， <a href='http://rosettacode.org/wiki/#Python' title='＃蟒蛇'>Python</a>和<a href='http://rosettacode.org/wiki/#Ruby' title='＃红宝石'>Ruby</a>实现。） </p>
+[S-Expressions](https://en.wikipedia.org/wiki/S-Expression "wp: S-Expression") are one convenient way to parse and store data.
+
+# --instructions--
+
+Write a simple reader/parser for S-Expressions that handles quoted and unquoted strings, integers and floats.
+
+The function should read a single but nested S-Expression from a string and return it as a (nested) array.
+
+Newlines and other whitespace may be ignored unless contained within a quoted string.
+
+"`()`" inside quoted strings are not interpreted, but treated as part of the string.
+
+Handling escaped quotes inside a string is optional; thus "`(foo"bar)`" may be treated as a string "`foo"bar`", or as an error.
+
+For this, the reader need not recognize "<code>\\</code>" for escaping, but should, in addition, recognize numbers if the language has appropriate data types.
+
+Note that with the exception of "`()"`" ("<code>\\</code>" if escaping is supported) and whitespace there are no special characters. Anything else is allowed without quotes.
+
+The reader should be able to read the following input
+
+<pre>((data "quoted data" 123 4.5)
+(data (!@# (4.5) "(more" "data)")))
+</pre>
+
+and turn it into a native data structure. (See the [Pike](https://rosettacode.org/wiki/S-Expressions#Pike "\#Pike"), [Python](https://rosettacode.org/wiki/S-Expressions#Python "\#Python") and [Ruby](https://rosettacode.org/wiki/S-Expressions#Ruby "\#Ruby") implementations for examples of native data structures.)
# --hints--
-`parseSexpr`是一个函数。
+`parseSexpr` should be a function.
```js
assert(typeof parseSexpr === 'function');
```
-`parseSexpr("(data1 data2 data3)")`应返回[“data1”，“data2”，“data3”]“）
+`parseSexpr('(data1 data2 data3)')` should return `['data1', 'data2', 'data3']`
```js
assert.deepEqual(parseSexpr(simpleSExpr), simpleSolution);
```
-`parseSexpr("(data1 data2 data3)")`应该返回一个包含3个元素的数组“）'
+`parseSexpr('(data1 data2 data3)')` should return an array with 3 elements.
```js
assert.deepEqual(parseSexpr(basicSExpr), basicSolution);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sailors-coconuts-and-a-monkey-problem.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sailors-coconuts-and-a-monkey-problem.md
index cf3a4fe252e..ec8a0c6e2db 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sailors-coconuts-and-a-monkey-problem.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sailors-coconuts-and-a-monkey-problem.md
@@ -1,36 +1,45 @@
---
id: 59da22823d04c95919d46269
-title: 水手，椰子和猴子问题
+title: 'Sailors, coconuts and a monkey problem'
challengeType: 5
-videoUrl: ''
+forumTopicId: 302304
dashedName: sailors-coconuts-and-a-monkey-problem
---
# --description--
-<p>五名水手在岛上遭遇海难，并在白天收集了一大堆椰子。 </p><p>那天晚上，第一个水手醒来并决定早点拿走他的第一份，所以试图将一堆椰子平分成五堆，但发现剩下一个椰子，所以他把它扔到一只猴子然后隐藏“他的”五个同样大小的椰子堆中的一个，并将其他四个桩推到一起，再次形成一堆可见的椰子并上床睡觉。 </p><p>长话短说，每个水手轮流在夜间起床，并执行将椰子堆分成五个的相同动作，发现剩下一个椰子并将剩下的椰子留给猴子。 </p><p>在早上（在夜间五个水手的暗中和分开行动之后），剩下的椰子被分成五个相等的堆，每个水手，然后发现一堆椰子在水手之间平分没有余数。 （早上猴子没什么。） </p><p>任务： </p><pre> <code> Create a function that returns the the minimum possible size of the initial pile of coconuts collected during the day for N sailors.</code> </pre><p>注意： </p><pre> <code> Of course the tale is told in a world where the collection of any amount of coconuts in a day and multiple divisions of the pile, etc can occur in time fitting the story line, so as not to affect the mathematics.</code> </pre><p> CF卡： </p><p> <a href='https://www.youtube.com/watch?v=U9qU20VmvaU' title='链接：https：//www.youtube.com/watch？v = U9qU20VmvaU'>猴子和椰子 - Numberphile</a> （视频）分析解决方案。 </p><pre> <code> <a href="http://oeis.org/A002021" title="link: http://oeis.org/A002021">A002021 Pile of coconuts problem</a> The On-Line Encyclopedia of Integer Sequences. (Although some of its references may use the alternate form of the tale).</code> </pre>
+Five sailors are shipwrecked on an island and collect a large pile of coconuts during the day. That night the first sailor wakes up and decides to take his first share early so tries to divide the pile of coconuts equally into five piles but finds that there is one coconut left over, so he tosses it to a monkey and then hides "his" one of the five equally sized piles of coconuts and pushes the other four piles together to form a single visible pile of coconuts again and goes to bed. To cut a long story short, each of the sailors in turn gets up once during the night and performs the same actions of dividing the coconut pile into five, finding that one coconut is left over and giving that single remainder coconut to the monkey. In the morning (after the surreptitious and separate action of each of the five sailors during the night), the remaining coconuts are divided into five equal piles for each of the sailors, whereupon it is found that the pile of coconuts divides equally amongst the sailors with no remainder. (Nothing for the monkey in the morning.)
+
+# --instructions--
+
+Create a function that returns the minimum possible size of the initial pile of coconuts collected during the day for `N` sailors. **Note:** Of course the tale is told in a world where the collection of any amount of coconuts in a day and multiple divisions of the pile, etc. can occur in time fitting the story line, so as not to affect the mathematics. **C.f:**
+
+<ul>
+ <li><a href="https://www.youtube.com/watch?v=U9qU20VmvaU" target="_blank"> Monkeys and Coconuts - Numberphile</a> (Video) Analytical solution.</li>
+ <li><a href="https://oeis.org/A002021" target="_blank">A002021 Pile of coconuts problem</a> The On-Line Encyclopedia of Integer Sequences. (Although some of its references may use the alternate form of the tale).</li>
+</ul>
# --hints--
-`splitCoconuts`是一个函数。
+`splitCoconuts` should be a function.
```js
assert(typeof splitCoconuts === 'function');
```
-`splitCoconuts(5)`应该返回3121。
+`splitCoconuts(5)` should return 3121.
```js
assert(splitCoconuts(5) === 3121);
```
-`splitCoconuts(6)`应返回233275。
+`splitCoconuts(6)` should return 233275.
```js
assert(splitCoconuts(6) === 233275);
```
-`splitCoconuts(7)`应该返回823537。
+`splitCoconuts(7)` should return 823537.
```js
assert(splitCoconuts(7) === 823537);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sedols.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sedols.md
index ab5b40bb42e..9d5ce33e9d3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sedols.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sedols.md
@@ -2,51 +2,57 @@
id: 59d9c6bc214c613ba73ff012
title: SEDOLs
challengeType: 5
-videoUrl: ''
+forumTopicId: 302305
dashedName: sedols
---
# --description--
-任务：
+For each number list of 6-digit [SEDOL](https://en.wikipedia.org/wiki/SEDOL "wp: SEDOL")s, calculate and append the checksum digit. That is, given the input string on the left, your function should return the corresponding string on the right:
-对于6位[SEDOL](https://en.wikipedia.org/wiki/SEDOL "wp：SEDOL")的每个数字列表，计算并附加校验和数字。
+<pre>
+710889 => 7108899
+B0YBKJ => B0YBKJ7
+406566 => 4065663
+B0YBLH => B0YBLH2
+228276 => 2282765
+B0YBKL => B0YBKL9
+557910 => 5579107
+B0YBKR => B0YBKR5
+585284 => 5852842
+B0YBKT => B0YBKT7
+B00030 => B000300
+</pre>
-也就是说，给定左侧的输入字符串，您的函数应返回右侧的相应字符串：
-
-```
- <pre> 710889 => 7108899 B0YBKJ => B0YBKJ7 406566 => 4065663 B0YBLH => B0YBLH2 228276 => 2282765 B0YBKL => B0YBKL9 557910 => 5579107 B0YBKR => B0YBKR5 585284 => 5852842 B0YBKT => B0YBKT7 B00030 => B000300 </pre>
-```
-
-还要检查每个输入是否正确形成，尤其是对于SEDOL字符串中允许的有效字符。您的函数应在无效输入时返回`null` 。
+Check that each input is correctly formed, especially with respect to valid characters allowed in a SEDOL string. Your function should return `null` on an invalid input.
# --hints--
-`sedol`是一个功能。
+`sedol` should be a function.
```js
assert(typeof sedol === 'function');
```
-`sedol('a')`应该返回null。
+`sedol('a')` should return null.
```js
assert(sedol('a') === null);
```
-`sedol('710889')`应返回'7108899'。
+`sedol('710889')` should return '7108899'.
```js
assert(sedol('710889') === '7108899');
```
-`sedol('BOATER')`应该返回null。
+`sedol('BOATER')` should return null.
```js
assert(sedol('BOATER') === null);
```
-`sedol('228276')`应该返回'228276'。
+`sedol('228276')` should return '2282765'.
```js
assert(sedol('228276') === '2282765');
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/self-describing-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/self-describing-numbers.md
index 5ff2f5c9923..df2a88508fb 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/self-describing-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/self-describing-numbers.md
@@ -89,7 +89,7 @@ function isSelfDescribing(n) {
if (digits.length != count.length) {
return false;
}
-
+
for (let i=0; i< digits.length; i++){
if (digits[i] !== count[i]) {
return false;
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/self-referential-sequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/self-referential-sequence.md
index 0cc45a42fff..11beda8ba0b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/self-referential-sequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/self-referential-sequence.md
@@ -8,7 +8,7 @@ dashedName: self-referential-sequence
# --description--
-There are several ways to generate a self-referential sequence. One very common one (the [Look-and-say sequence](<https://rosettacode.org/wiki/Look-and-say sequence>)) is to start with a positive integer, then generate the next term by concatenating enumerated groups of adjacent alike digits:
+There are several ways to generate a self-referential sequence. One very common one (the [Look-and-say sequence](https://rosettacode.org/wiki/Look-and-say sequence)) is to start with a positive integer, then generate the next term by concatenating enumerated groups of adjacent alike digits:
<pre>0, 10, 1110, 3110, 132110, 1113122110, 311311222110 ...</pre>
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-cocktail-sort.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-cocktail-sort.md
index 86aeeda4c83..f4c1b63f681 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-cocktail-sort.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-cocktail-sort.md
@@ -8,7 +8,7 @@ dashedName: sorting-algorithmscocktail-sort
# --description--
-The cocktail shaker sort is an improvement on the [Bubble Sort](<https://rosettacode.org/wiki/Bubble Sort>). The improvement is basically that values "bubble" both directions through the array, because on each iteration the cocktail shaker sort bubble sorts once forwards and once backwards. Pseudocode for the algorithm (from [wikipedia](<https://en.wikipedia.org/wiki/Cocktail sort>)):
+The cocktail shaker sort is an improvement on the [Bubble Sort](https://rosettacode.org/wiki/Bubble Sort). The improvement is basically that values "bubble" both directions through the array, because on each iteration the cocktail shaker sort bubble sorts once forwards and once backwards. Pseudocode for the algorithm (from [wikipedia](https://en.wikipedia.org/wiki/Cocktail sort)):
<pre><b>function</b> <i>cocktailSort</i>( A : list of sortable items )
<b>do</b>
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-comb-sort.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-comb-sort.md
index 2b31fb98db8..472ce20af5a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-comb-sort.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-comb-sort.md
@@ -10,9 +10,9 @@ dashedName: sorting-algorithmscomb-sort
Implement a *comb sort*.
-The **Comb Sort** is a variant of the [Bubble Sort](<https://rosettacode.org/wiki/Bubble Sort>).
+The **Comb Sort** is a variant of the [Bubble Sort](https://rosettacode.org/wiki/Bubble Sort).
-Like the [Shell sort](<https://rosettacode.org/wiki/Shell sort>), the Comb Sort increases the gap used in comparisons and exchanges.
+Like the [Shell sort](https://rosettacode.org/wiki/Shell sort), the Comb Sort increases the gap used in comparisons and exchanges.
Dividing the gap by $(1-e^{-\\varphi})^{-1} \\approx 1.247330950103979$ works best, but 1.3 may be more practical.
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-gnome-sort.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-gnome-sort.md
index 4aebdd0fe85..e3162404d9b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-gnome-sort.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-gnome-sort.md
@@ -8,7 +8,7 @@ dashedName: sorting-algorithmsgnome-sort
# --description--
-Gnome sort is a sorting algorithm which is similar to [Insertion sort](<https://rosettacode.org/wiki/Insertion sort>), except that moving an element to its proper place is accomplished by a series of swaps, as in [Bubble Sort](<https://rosettacode.org/wiki/Bubble Sort>).
+Gnome sort is a sorting algorithm which is similar to [Insertion sort](https://rosettacode.org/wiki/Insertion sort), except that moving an element to its proper place is accomplished by a series of swaps, as in [Bubble Sort](https://rosettacode.org/wiki/Bubble Sort).
The pseudocode for the algorithm is:
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-pancake-sort.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-pancake-sort.md
index 1a4a28928c8..f107df7b79a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-pancake-sort.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-pancake-sort.md
@@ -8,7 +8,7 @@ dashedName: sorting-algorithmspancake-sort
# --description--
-Write a function to sort an array of integers (of any convenient size) into ascending order using [Pancake sorting](<https://en.wikipedia.org/wiki/Pancake sorting>). The function should return the sorted array.
+Write a function to sort an array of integers (of any convenient size) into ascending order using [Pancake sorting](https://en.wikipedia.org/wiki/Pancake sorting). The function should return the sorted array.
In short, instead of individual elements being sorted, the only operation allowed is to "flip" one end of the list, like so:
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-shell-sort.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-shell-sort.md
index 3d69051f4d9..09403447320 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-shell-sort.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-shell-sort.md
@@ -8,7 +8,7 @@ dashedName: sorting-algorithmsshell-sort
# --description--
-Write a function to sort an array of elements using the [Shell sort](<https://en.wikipedia.org/wiki/Shell sort>) algorithm, a diminishing increment sort. The function should return the sorted array.
+Write a function to sort an array of elements using the [Shell sort](https://en.wikipedia.org/wiki/Shell sort) algorithm, a diminishing increment sort. The function should return the sorted array.
The Shell sort (also known as Shellsort or Shell's method) is named after its inventor, Donald Shell, who published the algorithm in 1959.
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-stooge-sort.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-stooge-sort.md
index 04dee734ead..ec108886c1f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-stooge-sort.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-stooge-sort.md
@@ -8,7 +8,7 @@ dashedName: sorting-algorithmsstooge-sort
# --description--
-Write a function to perform [Stooge Sort](<https://en.wikipedia.org/wiki/Stooge sort>) on an array of integers. The function should return a sorted array.
+Write a function to perform [Stooge Sort](https://en.wikipedia.org/wiki/Stooge sort) on an array of integers. The function should return a sorted array.
The Stooge Sort algorithm is as follows:
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-strand-sort.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-strand-sort.md
index a603bbab952..d8408782db5 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-strand-sort.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sorting-algorithms-strand-sort.md
@@ -8,7 +8,7 @@ dashedName: sorting-algorithmsstrand-sort
# --description--
-Write a function to sort an array using the [Strand sort](<https://en.wikipedia.org/wiki/Strand sort>). The function should return the sorted array.
+Write a function to sort an array using the [Strand sort](https://en.wikipedia.org/wiki/Strand sort). The function should return the sorted array.
This is a way of sorting numbers by extracting shorter sequences of already sorted numbers from an unsorted list.
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/stern-brocot-sequence.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/stern-brocot-sequence.md
index e8acb84c3c9..55580822989 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/stern-brocot-sequence.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/stern-brocot-sequence.md
@@ -8,7 +8,7 @@ dashedName: stern-brocot-sequence
# --description--
-For this task, the Stern-Brocot sequence is to be generated by an algorithm similar to that employed in generating the [Fibonacci sequence](<https://rosettacode.org/wiki/Fibonacci sequence>).
+For this task, the Stern-Brocot sequence is to be generated by an algorithm similar to that employed in generating the [Fibonacci sequence](https://rosettacode.org/wiki/Fibonacci sequence).
<ol>
<li>The first and second members of the sequence are both 1:</li>
@@ -36,7 +36,7 @@ For this task, the Stern-Brocot sequence is to be generated by an algorithm simi
# --instructions--
-Create a function that returns the $ n^{th} $ member of the sequence using the method outlined above.
+Create a function that returns the position in the Stern-Brocot sequence at which $ n $ is first encountered, where the sequence is generated with the method outlined above. Note that this sequence uses 1 based indexing.
# --hints--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sum-of-a-series.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sum-of-a-series.md
index 69fdff56995..35fee317742 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sum-of-a-series.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sum-of-a-series.md
@@ -8,7 +8,7 @@ dashedName: sum-of-a-series
# --description--
-Compute the **n**<sup>th</sup> term of a [series](<https://en.wikipedia.org/wiki/Series (mathematics)>), i.e. the sum of the **n** first terms of the corresponding [sequence](https://en.wikipedia.org/wiki/sequence). Informally this value, or its limit when **n** tends to infinity, is also called the *sum of the series*, thus the title of this task. For this task, use: $S*n = \\sum*{k=1}^n \\frac{1}{k^2}$ and compute $S\_{1000}$ This approximates the [zeta function](<https://en.wikipedia.org/wiki/Riemann zeta function>) for S=2, whose exact value $\\zeta(2) = {\\pi^2\\over 6}$ is the solution of the [Basel problem](<https://en.wikipedia.org/wiki/Basel problem>).
+Compute the **n**<sup>th</sup> term of a [series](https://en.wikipedia.org/wiki/Series (mathematics)), i.e. the sum of the **n** first terms of the corresponding [sequence](https://en.wikipedia.org/wiki/sequence). Informally this value, or its limit when **n** tends to infinity, is also called the *sum of the series*, thus the title of this task. For this task, use: $S*n = \\sum*{k=1}^n \\frac{1}{k^2}$ and compute $S\_{1000}$ This approximates the [zeta function](https://en.wikipedia.org/wiki/Riemann zeta function) for S=2, whose exact value $\\zeta(2) = {\\pi^2\\over 6}$ is the solution of the [Basel problem](https://en.wikipedia.org/wiki/Basel problem).
# --instructions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sutherland-hodgman-polygon-clipping.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sutherland-hodgman-polygon-clipping.md
index c3b671bf5a5..50fe64ed639 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sutherland-hodgman-polygon-clipping.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/sutherland-hodgman-polygon-clipping.md
@@ -8,7 +8,7 @@ dashedName: sutherland-hodgman-polygon-clipping
# --description--
-The [Sutherland-Hodgman clipping algorithm](<https://en.wikipedia.org/wiki/Sutherland-Hodgman clipping algorithm>) finds the polygon that is the intersection between an arbitrary polygon (the "subject polygon") and a convex polygon (the "clip polygon"). It is used in computer graphics (especially 2D graphics) to reduce the complexity of a scene being displayed by eliminating parts of a polygon that do not need to be displayed. Take the closed polygon defined by the points:
+The [Sutherland-Hodgman clipping algorithm](https://en.wikipedia.org/wiki/Sutherland-Hodgman clipping algorithm) finds the polygon that is the intersection between an arbitrary polygon (the "subject polygon") and a convex polygon (the "clip polygon"). It is used in computer graphics (especially 2D graphics) to reduce the complexity of a scene being displayed by eliminating parts of a polygon that do not need to be displayed. Take the closed polygon defined by the points:
<pre>[(50, 150), (200, 50), (350, 150), (350, 300), (250, 300), (200, 250), (150, 350), (100, 250), (100, 200)]</pre>
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/symmetric-difference.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/symmetric-difference.md
index 422cb77cc23..7009bce4617 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/symmetric-difference.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/symmetric-difference.md
@@ -8,7 +8,7 @@ dashedName: symmetric-difference
# --description--
-Given two [set](https://rosettacode.org/wiki/set)s *A* and *B*, compute $(A \\setminus B) \\cup (B \\setminus A).$ That is, enumerate the items that are in *A* or *B* but not both. This set is called the [symmetric difference](<https://en.wikipedia.org/wiki/Symmetric difference>) of *A* and *B*. In other words: $(A \\cup B) \\setminus (A \\cap B)$ (the set of items that are in at least one of *A* or *B* minus the set of items that are in both *A* and *B*).
+Given two [set](https://rosettacode.org/wiki/set)s *A* and *B*, compute $(A \\setminus B) \\cup (B \\setminus A).$ That is, enumerate the items that are in *A* or *B* but not both. This set is called the [symmetric difference](https://en.wikipedia.org/wiki/Symmetric difference) of *A* and *B*. In other words: $(A \\cup B) \\setminus (A \\cap B)$ (the set of items that are in at least one of *A* or *B* minus the set of items that are in both *A* and *B*).
# --instructions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/taxicab-numbers.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/taxicab-numbers.md
index 6aabd0e0aca..895b46ef050 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/taxicab-numbers.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/taxicab-numbers.md
@@ -1,48 +1,74 @@
---
id: 594ecc0d9a8cf816e3340187
-title: 出租车号码
+title: Taxicab numbers
challengeType: 5
-videoUrl: ''
+forumTopicId: 302337
dashedName: taxicab-numbers
---
# --description--
-[出租车编号](<https://en.wikipedia.org/wiki/Hardy–Ramanujan number> "wp：Hardy-Ramanujan号码") （此处使用的定义）是一个正整数，可以用多种方式表示为两个正立方体的总和。第一个出租车编号是1729，即：1 <sup>3</sup> + 12 <sup>3</sup>和9 <sup>3</sup> + 10 <sup>3</sup> 。出租车号码也被称为：\*出租车号码\*出租车号码\*出租车号码\* Hardy-Ramanujan号码任务：编写一个返回最低N个出租车号码的函数。对于每个出租车编号，显示数字以及它的构成立方体。另请参阅：在线整数序列百科全书上的\[`http://oeis.org/A001235` A001235出租车编号]。 MathWorld上的[Hardy-Ramanujan数字](http://mathworld.wolfram.com/Hardy-RamanujanNumber.html) 。 MathWorld上的[出租车编号](http://mathworld.wolfram.com/TaxicabNumber.html) 。维基百科上的[出租车号码](https://en.wikipedia.org/wiki/Taxicab_number) 。
+A [taxicab number](https://en.wikipedia.org/wiki/Hardy–Ramanujan number "wp: Hardy–Ramanujan number") (the definition that is being used here) is a positive integer that can be expressed as the sum of two positive cubes in more than one way.
+
+The first taxicab number is `1729`, which is:
+
+1<sup>3</sup> + 12<sup>3</sup> and
+
+9<sup>3</sup> + 10<sup>3</sup>.
+
+Taxicab numbers are also known as:
+
+<ul>
+ <li>taxi numbers</li>
+ <li>taxi-cab numbers</li>
+ <li>taxi cab numbers</li>
+ <li>Hardy-Ramanujan numbers</li>
+</ul>
+
+# --instructions--
+
+Write a function that returns the lowest `n` taxicab numbers. For each of the taxicab numbers, show the number as well as its constituent cubes.
+
+**See also:**
+
+<ul>
+ <li><a href='https://oeis.org/A001235' target='_blank'>A001235 taxicab numbers</a> on The On-Line Encyclopedia of Integer Sequences.</li>
+ <li><a href='https://en.wikipedia.org/wiki/Taxicab_number' target='_blank'>taxicab number</a> on Wikipedia.</li>
+</ul>
# --hints--
-`taxicabNumbers`是一个功能。
+`taxicabNumbers` should be a function.
```js
assert(typeof taxicabNumbers === 'function');
```
-`taxicabNumbers`应该返回一个数组。
+`taxicabNumbers` should return an array.
```js
assert(typeof taxicabNumbers(2) === 'object');
```
-`taxicabNumbers`应返回一组数字。
+`taxicabNumbers` should return an array of numbers.
```js
assert(typeof taxicabNumbers(100)[0] === 'number');
```
-`taxicabNumbers(4)`必须返回[1729,4104,13832,20683]。
+`taxicabNumbers(4)` should return [1729, 4104, 13832, 20683].
```js
assert.deepEqual(taxicabNumbers(4), res4);
```
-taxicabNumbers（25）应该返回[1729,4104,13832,20683,32832,39312,40033,46683,64232,65728,110656,110808,134379,149389,165464,171288,195841,216027,216125,262656,314696,320264 ，327763,373464,402597]
+`taxicabNumbers(25)` should return [1729, 4104, 13832, 20683, 32832, 39312, 40033, 46683, 64232, 65728, 110656, 110808, 134379, 149389, 165464, 171288, 195841, 216027, 216125, 262656, 314496, 320264, 327763, 373464, 402597]
```js
assert.deepEqual(taxicabNumbers(25), res25);
```
-taxicabNumbers（39）由20 - 29得到的数字应为[314496,320264,327763,373464,402597,439101,443889,513000,513856]。
+`taxicabNumbers(39)` resulting numbers from 20 - 29 should be [314496,320264,327763,373464,402597,439101,443889,513000,513856].
```js
assert.deepEqual(taxicabNumbers(39).slice(20, 29), res39From20To29);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/tokenize-a-string-with-escaping.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/tokenize-a-string-with-escaping.md
index d4c4f9f2724..24a65a163e8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/tokenize-a-string-with-escaping.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/tokenize-a-string-with-escaping.md
@@ -1,37 +1,72 @@
---
id: 594faaab4e2a8626833e9c3d
-title: 使用转义标记字符串
+title: Tokenize a string with escaping
challengeType: 5
-videoUrl: ''
+forumTopicId: 302338
dashedName: tokenize-a-string-with-escaping
---
# --description--
-<p>编写一个函数或程序，可以在分隔符的每个非转义事件中拆分字符串。 </p><p>它应该接受三个输入参数： </p> <b>字符串</b> <b>分隔符字符</b> <b>转义字符</b> <p>它应该输出一个字符串列表。 </p><p>拆分规则： </p>由分隔符分隔的字段将成为输出列表的元素。应保留空字段，即使在开始和结束时也是如此。 <p>转义规则： </p> “Escaped”意味着出现一个尚未自行转义的转义字符。当转义字符位于没有特殊含义的字符之前时，它仍然被视为转义符（但不会做任何特殊操作）。用于转义某些内容的每次出现的转义字符都不应成为输出的一部分。 <p>证明您的函数满足以下测试用例：给定字符串</p><pre>一个^ | UNO || 3 ^^^^ |四^^^ | ^夸| </pre>和使用<pre> | </pre>作为分隔符和<pre> ^ </pre>作为转义字符，您的函数应输出以下数组： <p></p><pre> ['one | uno'，“，'three ^^'，'four ^ | quatro'，”]
- </pre>
+Write a function or program that can split a string at each non-escaped occurrence of a separator character.
+
+It should accept three input parameters:
+
+<ul>
+ <li>The <strong>string</strong></li>
+ <li>The <strong>separator character</strong></li>
+ <li>The <strong>escape character</strong></li>
+</ul>
+
+It should output a list of strings.
+
+Rules for splitting:
+
+<ul>
+ <li>The fields that were separated by the separators, become the elements of the output list.</li>
+ <li>Empty fields should be preserved, even at the start and end.</li>
+</ul>
+
+Rules for escaping:
+
+<ul>
+ <li>"Escaped" means preceded by an occurrence of the escape character that is not already escaped itself.</li>
+ <li>When the escape character precedes a character that has no special meaning, it still counts as an escape (but does not do anything special).</li>
+ <li>Each occurrences of the escape character that was used to escape something, should not become part of the output.</li>
+</ul>
+
+Demonstrate that your function satisfies the following test-case:
+
+Given the string
+
+<pre>one^|uno||three^^^^|four^^^|^cuatro|</pre>
+
+and using `|` as a separator and `^` as escape character, your function should output the following array:
+
+<pre> ['one|uno', '', 'three^^', 'four^|cuatro', '']
+</pre>
# --hints--
-`tokenize`是一个函数。
+`tokenize` should be a function.
```js
assert(typeof tokenize === 'function');
```
-`tokenize`应该返回一个数组。
+`tokenize` should return an array.
```js
assert(typeof tokenize('a', 'b', 'c') === 'object');
```
-`tokenize("one^|uno||three^^^^|four^^^|^cuatro|", "|", "^")`应返回[“one | uno”，“”，“three ^^” ，“四个^ | cuatro”，“”]“）
+`tokenize('one^|uno||three^^^^|four^^^|^cuatro|', '|', '^')` should return `['one|uno', '', 'three^^', 'four^|cuatro', '']`
```js
assert.deepEqual(tokenize(testStr1, '|', '^'), res1);
```
-`tokenize("a@&bcd&ef&&@@hi", "&", "@")`应返回`["a&bcd", "ef", "", "@hi"]`
+`tokenize('a@&bcd&ef&&@@hi', '&', '@')` should return `['a&bcd', 'ef', '', '@hi']`
```js
assert.deepEqual(tokenize(testStr2, '&', '@'), res2);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/top-rank-per-group.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/top-rank-per-group.md
index 3f1b90ac5e8..c9405c8f3f3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/top-rank-per-group.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/top-rank-per-group.md
@@ -1,85 +1,94 @@
---
id: 595011cba5a81735713873bd
-title: 每组排名最高
+title: Top rank per group
challengeType: 5
-videoUrl: ''
+forumTopicId: 302339
dashedName: top-rank-per-group
---
# --description--
-任务：
+Find the top `n` ranked data in each group, where `n` is provided as a parameter. Name of the rank and the group are also provided as parameter.
-在每个组中查找排名前N位的数据，其中N作为参数提供。等级和组的名称也作为参数提供。
+Given the following data:
-鉴于以下数据：
+<pre>[
+ { name: 'Tyler Bennett', id: 'E10297', salary: 32000, dept: 'D101' },
+ { name: 'John Rappl', id: 'E21437', salary: 47000, dept: 'D050' },
+ { name: 'George Woltman', id: 'E00127', salary: 53500, dept: 'D101' },
+ { name: 'Adam Smith', id: 'E63535', salary: 18000, dept: 'D202' },
+ { name: 'Claire Buckman', id: 'E39876', salary: 27800, dept: 'D202' },
+ { name: 'David McClellan', id: 'E04242', salary: 41500, dept: 'D101' },
+ { name: 'Rich Holcomb', id: 'E01234', salary: 49500, dept: 'D202' },
+ { name: 'Nathan Adams', id: 'E41298', salary: 21900, dept: 'D050' },
+ { name: 'Richard Potter', id: 'E43128', salary: 15900, dept: 'D101' },
+ { name: 'David Motsinger', id: 'E27002', salary: 19250, dept: 'D202' },
+ { name: 'Tim Sampair', id: 'E03033', salary: 27000, dept: 'D101' },
+ { name: 'Kim Arlich', id: 'E10001', salary: 57000, dept: 'D190' },
+ { name: 'Timothy Grove', id: 'E16398', salary: 29900, dept: 'D190' }
+];
+</pre>
-```
- [
-{姓名：'Tyler Bennett'，id：'E10297'，薪水：32000，部门：'D101'}，
-{姓名：'John Rappl'，id：'E21437'，薪水：47000，部门：'D050'}，
-{姓名：'George Woltman'，id：'E00127'，薪水：53500，部门：'D101'}，
-{name：'Adam Smith'，id：'E63535'，薪水：18000，部门：'D202'}，
-{姓名：'Claire Buckman'，id：'E39876'，薪水：27800，部门：'D202'}，
-{姓名：'David McClellan'，id：'E04242'，薪水：41500，部门：'D101'}，
-{name：'Rich Holcomb'，id：'E01234'，薪水：49500，dept：'D202'}，
-{姓名：'Nathan Adams'，id：'E41298'，薪水：21900，部门：'D050'}，
-{姓名：'Richard Potter'，id：'E43128'，薪水：15900，部门：'D101'}，
-{姓名：'David Motsinger'，id：'E27002'，薪水：19250，dept：'D202'}，
-{姓名：'Tim Sampair'，id：'E03033'，薪水：27000，部门：'D101'}，
-{姓名：'Kim Arlich'，id：'E10001'，薪水：57000，部门：'D190'}，
-{name：'Timothy Grove'，id：'E16398'，薪水：29900，部门：'D190'}
-]。
-```
+one could rank top 10 employees in each department by calling
-</pre>通过调用<code>topRankPerGroup(10, data, 'dept', 'salary')</code>可以对每个部门的前10名员工进行排名。给出以下数据： <pre> [
- {name：'Friday 13th'，类型：'恐怖'，评分：9.9}，
- {姓名：“榆树街上的梦魇”，类型：'恐怖'，等级：5.7}，
- {name：'Titanic'，类型：'drama'，评分：7.3}，
- {name：'Maze Runner'，类型：'scifi'，评级：7.1}，
- {name：'Blade runner'，类型：'scifi'，评分：8.9}
-]。
-</pre>通过调用<code>topRankPerGroup(1, data, 'genre', 'rating')</code>可以在每个类型中对排名最高的电影进行排名
+`topRankPerGroup(10, data, 'dept', 'salary')`
+
+Given the following data:
+
+<pre>[
+ { name: 'Friday 13th', genre: 'horror', rating: 9.9 },
+ { name: "Nightmare on Elm's Street", genre: 'horror', rating: 5.7 },
+ { name: 'Titanic', genre: 'drama', rating: 7.3 },
+ { name: 'Maze Runner', genre: 'scifi', rating: 7.1 },
+ { name: 'Blade runner', genre: 'scifi', rating: 8.9 }
+];
+</pre>
+
+one could rank the top-rated movie in each genre by calling
+
+`topRankPerGroup(1, data, 'genre', 'rating')`
+
+The function should return an array with an array for each group containing the top `n` objects.
# --hints--
-`topRankPerGroup`是一个函数。
+`topRankPerGroup` should be a function.
```js
assert(typeof topRankPerGroup === 'function');
```
-`topRankPerGroup`在负n值上返回undefined。
+`topRankPerGroup` should return undefined on negative n values.
```js
assert(typeof topRankPerGroup(-1, []) === 'undefined');
```
-第一部门必须是D050
+First department should be D050
```js
assert.equal(res1[0][0].dept, 'D050');
```
-第一部门必须是D050
+First department should be D050
```js
assert.equal(res1[0][1].salary, 21900);
```
-最后一个部门必须是D202
+The last department should be D202
```js
assert.equal(res1[3][3].dept, 'D202');
```
-`topRankPerGroup(1, ...)`必须仅返回每组的排名最高的结果。
+`topRankPerGroup(1, ...)` should return only top ranking result per group.
```js
assert.equal(res2[2].length, 1);
```
-`topRankPerGroup(1, ...)`必须仅返回每组的排名最高的结果。
+`topRankPerGroup(1, ...)` should return only top ranking result per group.
```js
assert.equal(res3[2][1].name, 'Maze Runner');
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/topological-sort.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/topological-sort.md
index 608c67c7212..1f091c8bf9f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/topological-sort.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/topological-sort.md
@@ -1,57 +1,81 @@
---
id: 594fa2746886f41f7d8bf225
-title: 拓扑排序
+title: Topological sort
challengeType: 5
-videoUrl: ''
+forumTopicId: 302340
dashedName: topological-sort
---
# --description--
-<p>给定项目之间的映射以及它们所依赖的项目， <a href='https://en.wikipedia.org/wiki/Topological sorting' title='wp：拓扑排序'>拓扑排序</a>会对项目进行<a href='https://en.wikipedia.org/wiki/Topological sorting' title='wp：拓扑排序'>排序</a> ，以使项目不在其所依赖的项目之前。 </p><p>用<a href='https://en.wikipedia.org/wiki/VHDL' title='wp：VHDL'>VHDL</a>语言编译库有一个限制，即必须在它依赖的库之后编译库。 </p>任务： <p>编写一个函数，该函数将从其依赖项返回VHDL库的有效编译顺序。 </p>假设库名称是单个单词。仅作为家属提及的项目没有自己的家属，但必须给出他们的编制顺序。任何自我依赖都应该被忽略。应忽略任何不可订购的依赖项。 <p>使用以下数据作为示例： </p><pre>图书馆图书馆依赖
-======= ====================
-des_system_lib std synopsys std_cell_lib des_system_lib dw02 dw01 ramlib ieee
-dw01 ieee dw01 dware gtech
-dw02 ieee dw02 dware
-dw03 std synopsys dware dw03 dw02 dw01 ieee gtech
-dw04 dw04 ieee dw01 dware gtech
-dw05 dw05 ieee dware
-dw06 dw06 ieee dware
-dw07 ieee dware
-dware ieee dware
-gtech ieee gtech
-ramlib std ieee
-std_cell_lib ieee std_cell_lib
-新思
-</pre><p> <small>注意：如果将<code>dw04</code>添加到<code>dw01</code>的依赖项列表中，则上述数据将无法订购。</small> </p> CF卡： <pre> <code><a href="http://rosettacode.org/wiki/Topological sort/Extracted top item" title="Topological sort/Extracted top item">Topological sort/Extracted top item</a>.</code> </pre><p>拓扑排序有两种流行的算法： </p><p> Kahn的1962拓扑排序和深度优先搜索： <a href='https://en.wikipedia.org/wiki/Topological sorting' title='wp：拓扑排序'>拓扑排序</a> </p><p> Jason Sachs： <a href='http://www.embeddedrelated.com/showarticle/799.php' title='链接：http：//www.embeddedrelated.com/showarticle/799.php'>“十个小算法，第四部分：拓扑排序”</a> 。 </p>
+Given a mapping between items, and items they depend on, a [topological sort](https://en.wikipedia.org/wiki/Topological sorting "wp: Topological sorting") orders items so that no item precedes an item it depends upon. The compiling of a library in the [VHDL](https://en.wikipedia.org/wiki/VHDL "wp: VHDL") language has the constraint that a library must be compiled after any library it depends on.
+
+# --instructions--
+
+Write a function that will return a valid compile order of VHDL libraries from their dependencies.
+
+<ul>
+ <li>Assume library names are single words.</li>
+ <li>Items mentioned as only dependents have no dependents of their own, but their order of compiling must be given.</li>
+ <li>Any self dependencies should be ignored.</li>
+ <li>Any un-orderable dependencies should be ignored.</li>
+</ul>
+Use the following data as an example:
+<pre>
+LIBRARY LIBRARY DEPENDENCIES
+======= ====================
+des_system_lib std synopsys std_cell_lib des_system_lib dw02 dw01 ramlib ieee
+dw01 ieee dw01 dware gtech
+dw02 ieee dw02 dware
+dw03 std synopsys dware dw03 dw02 dw01 ieee gtech
+dw04 dw04 ieee dw01 dware gtech
+dw05 dw05 ieee dware
+dw06 dw06 ieee dware
+dw07 ieee dware
+dware ieee dware
+gtech ieee gtech
+ramlib std ieee
+std_cell_lib ieee std_cell_lib
+synopsys
+</pre>
+<small>Note: the above data would be un-orderable if, for example, <code>dw04</code> is added to the list of dependencies of <code>dw01</code>.</small>
+<strong>C.f.:</strong>
+<ul>
+ <li><a href="https://rosettacode.org/wiki/Topological sort/Extracted top item" title="Topological sort/Extracted top item" target="_blank">Topological sort/Extracted top item</a>.</li>
+</ul>
+There are two popular algorithms for topological sorting:
+<ul>
+ <li><a href="https://en.wikipedia.org/wiki/Topological sorting" title="wp: Topological sorting" target="_blank">Kahn's 1962 topological sort</a></li>
+ <li><a href="https://www.embeddedrelated.com/showarticle/799.php" target="_blank">depth-first search</a></li>
+</ul>
# --hints--
-`topologicalSort`是一个函数。
+`topologicalSort` should be a function.
```js
assert(typeof topologicalSort === 'function');
```
-`topologicalSort`必须返回正确的库顺序..
+`topologicalSort` should return correct library order.
```js
assert.deepEqual(topologicalSort(libsSimple), ['bbb', 'aaa']);
```
-`topologicalSort`必须返回正确的库顺序..
+`topologicalSort` should return correct library order.
```js
assert.deepEqual(topologicalSort(libsVHDL), solutionVHDL);
```
-`topologicalSort`必须返回正确的库顺序..
+`topologicalSort` should return correct library order.
```js
assert.deepEqual(topologicalSort(libsCustom), solutionCustom);
```
-`topologicalSort`必须忽略不可共享的依赖项。
+`topologicalSort` should ignore unorderable dependencies.
```js
assert.deepEqual(topologicalSort(libsUnorderable), solutionUnorderable);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/towers-of-hanoi.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/towers-of-hanoi.md
index 337c49dc469..4de9c851ba4 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/towers-of-hanoi.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/towers-of-hanoi.md
@@ -1,46 +1,48 @@
---
id: 5951ed8945deab770972ae56
-title: 河内的塔
+title: Towers of Hanoi
challengeType: 5
-videoUrl: ''
+forumTopicId: 302341
dashedName: towers-of-hanoi
---
# --description--
-任务：
+Solve the [Towers of Hanoi](https://en.wikipedia.org/wiki/Towers_of_Hanoi "wp: Towers_of_Hanoi") problem.
-解决[河内塔](https://en.wikipedia.org/wiki/Towers_of_Hanoi "wp：Towers_of_Hanoi")问题。
+Your solution should accept the number of discs as the first parameters, and three string used to identify each of the three stacks of discs, for example `towerOfHanoi(4, 'A', 'B', 'C')`. The function should return an array of arrays containing the list of moves, source -> destination.
-您的解决方案应该接受光盘数量作为第一个参数，并使用三个字符串来识别三个光盘堆栈中的每一个，例如`towerOfHanoi(4, 'A', 'B', 'C')` 。该函数应该返回一个包含移动列表的数组数组，source - > destination。例如，数组`[['A', 'C'], ['B', 'A']]`表示第一个移动是将光盘从堆栈A移动到C，第二个移动是移动一个从堆栈B到A的光盘
+For example, the array `[['A', 'C'], ['B', 'A']]` indicates that the 1st move was to move a disc from stack A to C, and the 2nd move was to move a disc from stack B to A.
+
+<p></p>
# --hints--
-`towerOfHanoi`是一个功能。
+`towerOfHanoi` should be a function.
```js
assert(typeof towerOfHanoi === 'function');
```
-`towerOfHanoi(3, ...)` 应该返回7招。
+`towerOfHanoi(3, ...)` should return 7 moves.
```js
assert(res3.length === 7);
```
-`towerOfHanoi(3, "A", "B", "C")`应返回\[[“A”，“B”]，[“A”，“C”]，[“B”，“C”]，[ “A”， “B”]，[ “C”， “A”]，[ “C”， “B”]，[ “A”， “B”]]“）。
+`towerOfHanoi(3, 'A', 'B', 'C')` should return `[['A','B'], ['A','C'], ['B','C'], ['A','B'], ['C','A'], ['C','B'], ['A','B']]`.
```js
assert.deepEqual(towerOfHanoi(3, 'A', 'B', 'C'), res3Moves);
```
-`towerOfHanoi(5, "X", "Y", "Z")`第10 `towerOfHanoi(5, "X", "Y", "Z")`应为Y - > X.
+`towerOfHanoi(5, "X", "Y", "Z")` 10th move should be Y -> X.
```js
assert.deepEqual(res5[9], ['Y', 'X']);
```
-`towerOfHanoi(7, "A", "B", "C")`前十个动作是\[[“A”，“B”]，[“A”，“C”]，[“B”，“C”][ “A”， “B”]，[ “C”， “A”]，[ “C”， “B”]，[ “A”， “B”]，[ “A”， “C”][ “B”， “C”]，[ “B”， “A”]]“）。
+`towerOfHanoi(7, 'A', 'B', 'C')` first ten moves should be `[['A','B'], ['A','C'], ['B','C'], ['A','B'], ['C','A'], ['C','B'], ['A','B'], ['A','C'], ['B','C'], ['B','A']]`
```js
assert.deepEqual(towerOfHanoi(7, 'A', 'B', 'C').slice(0, 10), res7First10Moves);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/vector-cross-product.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/vector-cross-product.md
index 7ebf88873be..b7a9d18be14 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/vector-cross-product.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/vector-cross-product.md
@@ -1,38 +1,34 @@
---
id: 594810f028c0303b75339ad2
-title: 矢量交叉产品
+title: Vector cross product
challengeType: 5
-videoUrl: ''
+forumTopicId: 302342
dashedName: vector-cross-product
---
# --description--
-矢量被定义为具有三个维度，由三个数字的有序集合表示：（X，Y，Z）。
+A vector is defined as having three dimensions as being represented by an ordered collection of three numbers: (X, Y, Z).
-任务：
+# --instructions--
-```
- Write a function that takes two vectors (arrays) as input and computes their cross product.
-```
-
-您的函数应在无效输入（即不同长度的向量）上返回`null` 。
+Write a function that takes two vectors (arrays) as input and computes their cross product. Your function should return `null` on invalid inputs such as vectors of different lengths.
# --hints--
-dotProduct必须是一个函数
+dotProduct should be a function.
```js
assert.equal(typeof crossProduct, 'function');
```
-dotProduct（）必须返回null
+dotProduct() should return null.
```js
assert.equal(crossProduct(), null);
```
-crossProduct（[1,2,3]，[4,5,6]）必须返回[-3,6，-3]。
+crossProduct([1, 2, 3], [4, 5, 6]) should return [-3, 6, -3].
```js
assert.deepEqual(res12, exp12);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/vector-dot-product.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/vector-dot-product.md
index 6c48d5258ce..fb8ae242aed 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/vector-dot-product.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/vector-dot-product.md
@@ -1,60 +1,61 @@
---
id: 594810f028c0303b75339ad3
-title: 矢量点积
+title: Vector dot product
challengeType: 5
-videoUrl: ''
+forumTopicId: 302343
dashedName: vector-dot-product
---
# --description--
-<p>矢量被定义为具有三个维度，由三个数字的有序集合表示：（X，Y，Z）。 </p><p>任务： </p><pre> <code>Write a function that takes any numbers of vectors (arrays) as input and computes their dot product.</code> </pre><p>您的函数应在无效输入（即不同长度的向量）上返回<code>null</code> 。 </p><p></p>
+A vector can have one or more values represented by an ordered collection. Examples could be (x), (x, y), or (x, y, z).
+
+# --instructions--
+
+Write a function that takes two vectors (represented as one-dimensional arrays) as input and computes their dot product. Your function should return `null` on invalid inputs such as vectors of different lengths or passing anything other than two vectors.
# --hints--
-dotProduct必须是一个函数
+`dotProduct` should be a function.
```js
assert.equal(typeof dotProduct, 'function');
```
-dotProduct（）必须返回null
+`dotProduct()` should return `null`.
```js
assert.equal(dotProduct(), null);
```
-dotProduct（\[[1]，[1]]）必须返回1。
+`dotProduct([1], [1])` should return `1`.
```js
assert.equal(dotProduct([1], [1]), 1);
```
-dotProduct（\[[1]，[1,2]]）必须返回null。
+`dotProduct([1], [1, 2])` should return `null`.
```js
assert.equal(dotProduct([1], [1, 2]), null);
```
-dotProduct（[1,3，-5]，[4，-2，-1]）必须返回3。
+`dotProduct([1, 3, -5], [4, -2, -1])` should return `3`.
```js
assert.equal(dotProduct([1, 3, -5], [4, -2, -1]), 3);
```
-`dotProduct(...nVectors)`应该返回`dotProduct(...nVectors)`
+`dotProduct([3, 2, 1], [2, 4, 2], [5, 3, 1])` should return `null`.
```js
-assert.equal(
- dotProduct(
- [0, 1, 2, 3, 4],
- [0, 2, 4, 6, 8],
- [0, 3, 6, 9, 12],
- [0, 4, 8, 12, 16],
- [0, 5, 10, 15, 20]
- ),
- 156000
-);
+assert.equal(dotProduct([3, 2, 1], [2, 4, 2], [5, 3, 1]), null);
+```
+
+`dotProduct([ 0, 3, 6, 9, 12 ], [ 0, 4, 8, 12, 16 ])` should return `360`.
+
+```js
+assert.equal(dotProduct([ 0, 3, 6, 9, 12 ], [ 0, 4, 8, 12, 16 ]), 360);
```
# --seed--
@@ -71,29 +72,18 @@ function dotProduct(...vectors) {
```js
function dotProduct(...vectors) {
- if (!vectors || !vectors.length) {
- return null;
- }
- if (!vectors[0] || !vectors[0].length) {
+ if (!vectors || !vectors.length || vectors.length > 2 || vectors[0].length !== vectors[1].length) {
return null;
}
const vectorLen = vectors[0].length;
- const numVectors = vectors.length;
-
- // If all vectors not same length, return null
- for (let i = 0; i < numVectors; i++) {
- if (vectors[i].length !== vectorLen) {
- return null; // return undefined
- }
- }
let prod = 0;
let sum = 0;
let j = vectorLen;
- let i = numVectors;
+ let i = 2;
// Sum terms
while (j--) {
- i = numVectors;
+ i = 2;
prod = 1;
while (i--) {
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/word-wrap.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/word-wrap.md
index 3bc2a38c68c..344be87802e 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/word-wrap.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/word-wrap.md
@@ -1,56 +1,59 @@
---
id: 594810f028c0303b75339ad4
-title: 自动换行
+title: Word wrap
challengeType: 5
-videoUrl: ''
+forumTopicId: 302344
dashedName: word-wrap
---
# --description--
-<p>即使在今天，使用比例字体和复杂布局，仍然存在需要在指定列处包装文本的情况。基本任务是以简单的方式包装一段文本。示例文字： </p><pre>使用更复杂的算法（如Knuth和Plass TeX算法）包装文本。
-如果您的语言提供此功能，您可以获得额外的信用，
-但你“必须参考文档”表明该算法
-比简单的最小长度算法更好。
-</pre><p>任务： </p><pre> <code>Write a function that can wrap this text to any number of characters.</code> </pre><p>例如，包装为80个字符的文本应如下所示： </p><p></p><pre>使用更复杂的算法（如Knuth和Plass TeX）包装文本
-算法。如果您的语言提供此功能，您可以轻松获得额外的功劳
-必须参考文档，表明该算法更好
-而不是简单的最小长度算法。
+Even today, with proportional fonts and complex layouts, there are still cases where you need to wrap text at a specified column. The basic task is to wrap a paragraph of text in a simple way.
+
+# --instructions--
+
+Write a function that can wrap this text to any number of characters. As an example, the text wrapped to 80 characters should look like the following:
+
+<pre>
+Wrap text using a more sophisticated algorithm such as the Knuth and Plass TeX
+algorithm. If your language provides this, you get easy extra credit, but you
+must reference documentation indicating that the algorithm is something better
+than a simple minimum length algorithm.
</pre>
# --hints--
-`wrap`是一个功能。
+wrap should be a function.
```js
assert.equal(typeof wrap, 'function');
```
-`wrap("abc", 10)`必须返回一个字符串。
+wrap should return a string.
```js
assert.equal(typeof wrap('abc', 10), 'string');
```
-wrap（80）必须返回4行。
+wrap(80) should return 4 lines.
```js
assert(wrapped80.split('\n').length === 4);
```
-你的`wrap`函数应该返回我们期望的文本
+Your `wrap` function should return our expected text.
```js
assert.equal(wrapped80.split('\n')[0], firstRow80);
```
-wrap（42）必须返回7行。
+wrap(42) should return 7 lines.
```js
assert(wrapped42.split('\n').length === 7);
```
-你的`wrap`函数应该返回我们期望的文本
+Your `wrap` function should return our expected text.
```js
assert.equal(wrapped42.split('\n')[0], firstRow42);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/y-combinator.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/y-combinator.md
index 865fc78e050..f76ab91e3c8 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/y-combinator.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/y-combinator.md
@@ -1,48 +1,56 @@
---
id: 594810f028c0303b75339ad5
-title: 和组合
+title: Y combinator
challengeType: 5
-videoUrl: ''
+forumTopicId: 302345
dashedName: y-combinator
---
# --description--
-<p>在严格的<a href='https://en.wikipedia.org/wiki/Functional programming' title='wp：函数式编程'>函数编程</a>和<a href='https://en.wikipedia.org/wiki/lambda calculus' title='wp：lambda演算'>lambda演算中</a> ，函数（lambda表达式）没有状态，只允许引用封闭函数的参数。这排除了递归函数的通常定义，其中函数与变量的状态相关联，并且该变量的状态在函数体中使用。 </p><p> <a href='http://mvanier.livejournal.com/2897.html'>Y组合</a>器本身是一个无状态函数，当应用于另一个无状态函数时，它返回函数的递归版本。 Y组合器是这类函数中最简单的一种，称为<a href='https://en.wikipedia.org/wiki/Fixed-point combinator' title='wp：定点组合器'>定点组合器</a> 。 </p>任务： <pre> <code>Define the stateless Y combinator function and use it to compute <a href="https://en.wikipedia.org/wiki/Factorial" title="wp: factorial">factorial</a>.</code> </pre><p> <code>factorial(N)</code>功能已经给你了。另见<a href='http://vimeo.com/45140590'>Jim Weirich：功能编程中的冒险</a> 。 </p>
+In strict [functional programming](https://en.wikipedia.org/wiki/Functional programming "wp: functional programming") and the [lambda calculus](https://en.wikipedia.org/wiki/lambda calculus "wp: lambda calculus"), functions (lambda expressions) don't have state and are only allowed to refer to arguments of enclosing functions. This rules out the usual definition of a recursive function wherein a function is associated with the state of a variable and this variable's state is used in the body of the function. The [Y combinator](https://mvanier.livejournal.com/2897.html) is itself a stateless function that, when applied to another stateless function, returns a recursive version of the function. The Y combinator is the simplest of the class of such functions, called [fixed-point combinators](https://en.wikipedia.org/wiki/Fixed-point combinator "wp: fixed-point combinator").
+
+# --instructions--
+
+Define the stateless Y combinator function and use it to compute [factorial](https://en.wikipedia.org/wiki/Factorial "wp: factorial"). The `factorial(N)` function is already given to you. **See also:**
+
+<ul>
+ <li><a href="https://vimeo.com/45140590" target="_blank">Jim Weirich: Adventures in Functional Programming</a>.</li>
+</ul>
# --hints--
-Y必须返回一个函数
+Y should return a function.
```js
assert.equal(typeof Y((f) => (n) => n), 'function');
```
-factorial（1）必须返回1。
+factorial(1) should return 1.
```js
assert.equal(factorial(1), 1);
```
-factorial（2）必须返回2。
+factorial(2) should return 2.
```js
assert.equal(factorial(2), 2);
```
-factorial（3）必须返回6。
+factorial(3) should return 6.
```js
assert.equal(factorial(3), 6);
```
-factorial（4）必须返回24。
+factorial(4) should return 24.
```js
assert.equal(factorial(4), 24);
```
-factorial（10）必须返回3628800。
+factorial(10) should return 3628800.
```js
assert.equal(factorial(10), 3628800);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/zeckendorf-number-representation.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/zeckendorf-number-representation.md
index f69c43e5b67..35c7529bc40 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/zeckendorf-number-representation.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/zeckendorf-number-representation.md
@@ -1,24 +1,30 @@
---
id: 594810f028c0303b75339ad6
-title: 勾选村号码表示
+title: Zeckendorf number representation
challengeType: 5
-videoUrl: ''
+forumTopicId: 302346
dashedName: zeckendorf-number-representation
---
# --description--
-<p>正如数字可以用位置表示法表示为十（十进制）或二（二进制）的幂的倍数之和;所有正整数都可以表示为Fibonacci系列的不同成员的一次或零次的总和。 </p><p>回想一下，第一六个不同斐波那契数是： <code>1, 2, 3, 5, 8, 13</code> 。十进制数11可以用位置表示法写成<code>0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1</code>或<code>010100</code> ，其中列表示乘以序列的特定成员。前导零被丢弃，因此十进制11变为<code>10100</code> 。 </p><p> 10100不是从斐波那契数字中得到11的唯一方法，但是<code>0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1</code>或010011也代表十进制11。对于真正的Zeckendorf数字还有一个额外的限制，即“不能使用两个连续的Fibonacci数”，这导致了前一个独特的解决方案。 </p><p>任务：编写一个函数，按顺序生成并返回前N个Zeckendorf数的数组。 </p>
+Just as numbers can be represented in a positional notation as sums of multiples of the powers of ten (decimal) or two (binary); all the positive integers can be represented as the sum of one or zero times the distinct members of the Fibonacci series. Recall that the first six distinct Fibonacci numbers are: `1, 2, 3, 5, 8, 13`.
+
+The decimal number eleven can be written as `0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1` or `010100` in positional notation where the columns represent multiplication by a particular member of the sequence. Leading zeroes are dropped so that 11 decimal becomes `10100`. 10100 is not the only way to make 11 from the Fibonacci numbers however `0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1` or 010011 would also represent decimal 11. For a true Zeckendorf number there is the added restriction that *no two consecutive Fibonacci numbers can be used* which leads to the former unique solution.
+
+# --instructions--
+
+Write a function that generates and returns an array of the first `n` Zeckendorf numbers in order.
# --hints--
-zeckendorf必须是功能
+zeckendorf should be function.
```js
assert.equal(typeof zeckendorf, 'function');
```
-你的`zeckendorf`函数应该返回正确的答案
+Your `zeckendorf` function should return the correct answer.
```js
assert.deepEqual(answer, solution20);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/zhang-suen-thinning-algorithm.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/zhang-suen-thinning-algorithm.md
index adae08293ac..bad53c8710a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/zhang-suen-thinning-algorithm.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/zhang-suen-thinning-algorithm.md
@@ -1,6 +1,6 @@
---
id: 594810f028c0303b75339ad7
-title: Zhang-Suen 细化算法
+title: Zhang-Suen thinning algorithm
challengeType: 5
forumTopicId: 302347
dashedName: zhang-suen-thinning-algorithm
@@ -8,7 +8,7 @@ dashedName: zhang-suen-thinning-algorithm
# --description--
-这是一个黑白图像（二值图像）的细化算法。 例如，输入图像如下：
+This is an algorithm used to thin a black and white i.e. one bit per pixel images. For example, with an input image of:
<!-- TODO write fully in markdown>
<!-- markdownlint-disable -->
@@ -32,7 +32,7 @@ dashedName: zhang-suen-thinning-algorithm
######## ####### ###### ############# ######
</pre>
-细化后的输出图像为：
+It produces the thinned output:
<pre>
@@ -53,9 +53,9 @@ dashedName: zhang-suen-thinning-algorithm
</pre>
-<h2>算法</h2>
+<h2>Algorithm</h2>
-假设黑像素点为 1，白像素点为 0；则输入图像可以用一个 N \* M 的矩阵（或数组）来表示，其中，矩阵中的元素只能为 0 或 1。这个算法对所有黑像素点 P1 进行操作。每个点 P1 都可以有 8 个相邻的点，分别是：
+Assume black pixels are one and white pixels zero, and that the input image is a rectangular N by M array of ones and zeroes. The algorithm operates on all black pixels P1 that can have eight neighbours. The neighbours are, in order, arranged as:
<table border="3">
<tr><td style="text-align: center;">P9</td><td style="text-align: center;">P2</td><td style="text-align: center;">P3</td></tr>
@@ -63,70 +63,66 @@ dashedName: zhang-suen-thinning-algorithm
<tr><td style="text-align: center;">P7</td><td style="text-align: center;">P6</td><td style="text-align: center;">P5</td></tr>
</table>
-显然，对于图像边框上的像素点，与它们相邻的点的数量是小于 8 的。
+Obviously the boundary pixels of the image cannot have the full eight neighbours.
<ul>
- <li>令 $A(P1)$ 为需要变为黑点的白点数量，即在 P2, P3, P4, P5, P6, P7, P8, P9, P2 这一序列中，(0 -> 1) 的操作次数（注意：为了表示循环/闭环，我们在序列的结尾特地多加了一个 P2）</li>
- <li>令 $B(P1)$ 为与 P1 相邻的点中，黑点的数量（即 sum(P2 .. P9)）</li>
+ <li>Define $A(P1)$ = the number of transitions from white to black, (0 -> 1) in the sequence P2, P3, P4, P5, P6, P7, P8, P9, P2. (Note the extra P2 at the end - it is circular).</li>
+ <li>Define $B(P1)$ = the number of black pixel neighbours of P1. ( = sum(P2 .. P9) )</li>
</ul>
-<h3>步骤一：</h3>
+**Step 1:**
-选出同时满足以下列出条件的所有像素点：
-
- <ol>
- <li>像素点为黑色，且有 8 个点与之相邻</li>
+All pixels are tested and pixels satisfying all the following conditions (simultaneously) are just noted at this stage. <ol>
+ <li>The pixel is black and has eight neighbours</li>
<li>$2 <= B(P1) <= 6$</li>
<li>$A(P1) = 1$</li>
- <li><strong>P2, P4 and P6</strong> 中至少有一个是白点</li>
- <li><strong>P4, P6 and P8</strong> 中至少有一个是白点</li>
+ <li>At least one of <strong>P2, P4 and P6</strong> is white</li>
+ <li>At least one of <strong>P4, P6 and P8</strong> is white</li>
</ol>
-在对图像进行迭代并选出所有符合这一步所述条件的点后，把这些点都设置为白色。
+After iterating over the image and collecting all the pixels satisfying all step 1 conditions, all these condition satisfying pixels are set to white.
-<h3>步骤二：</h3>
+**Step 2:**
-选出同时满足以下列出条件的所有像素点：
-
- <ol>
- <li>像素点为黑色，且有 8 个点与之相邻</li>
+All pixels are again tested and pixels satisfying all the following conditions are just noted at this stage. <ol>
+ <li>The pixel is black and has eight neighbours</li>
<li>$2 <= B(P1) <= 6$</li>
<li>$A(P1) = 1$</li>
- <li><strong>P2, P4 and P6</strong> 中至少有一个是白点</li>
- <li><strong>P2, P6 and P8</strong> 中至少有一个是白点</li>
+ <li>At least one of <strong>P2, P4 and P8</strong> is white</li>
+ <li>At least one of <strong>P2, P6 and P8</strong> is white</li>
</ol>
-
-在对图像进行迭代并选出所有符合这一步所述条件的点后，把这些点都设置为白色。
-<h3>迭代：</h3>
+After iterating over the image and collecting all the pixels satisfying all step 2 conditions, all these condition satisfying pixels are again set to white.
-只要在步骤一或步骤二，有黑色点被选出并设置成了白色，则继续重复步骤一和步骤二，直到没有黑色点被选出并更改为止。
+**Iteration:**
+
+If any pixels were set in this round of either step 1 or step 2 then all steps are repeated until no image pixels are so changed.
# --instructions--
-基于输入图像（以矩阵的形式给出），实现 Zhang-suen 细化算法。
+Write a routine to perform Zhang-Suen thinning on the provided image matrix.
# --hints--
-`thinImage` 应为函数。
+`thinImage` should be a function.
```js
assert.equal(typeof thinImage, 'function');
```
-`thinImage` 应返回数组。
+`thinImage` should return an array.
```js
assert(Array.isArray(result));
```
-`thinImage` 应返回字符串数组。
+`thinImage` should return an array of strings.
```js
assert.equal(typeof result[0], 'string');
```
-`thinImage` 应返回预计的结果。
+`thinImage` should return an array of strings.
```js
assert.deepEqual(result, expected);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/zig-zag-matrix.md b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/zig-zag-matrix.md
index 6cdab4bf277..4f5277c28ad 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/zig-zag-matrix.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/zig-zag-matrix.md
@@ -1,58 +1,62 @@
---
id: 594810f028c0303b75339ad8
-title: 之字形矩阵
+title: Zig-zag matrix
challengeType: 5
-videoUrl: ''
+forumTopicId: 302348
dashedName: zig-zag-matrix
---
# --description--
-“zig-zag”数组是第一个$ N ^ 2 $整数的正方形排列，当数组沿着数组的[反对角线](https://en.wiktionary.org/wiki/antidiagonal)曲折时，数字会逐渐增加。例如，给定“'5”'，产生这个数组：
+A 'zig-zag' array is a square arrangement of the first $N^2$ integers, where the numbers increase sequentially as you zig-zag along the array's [anti-diagonals](https://en.wiktionary.org/wiki/antidiagonal).
-```
- 0 1 5 6 14
-2 4 7 13 15
-3 8 12 16 21
-9 11 17 20 22
+For example, for the input `5`, the following result should be produced:
+
+<pre>
+ 0 1 5 6 14
+ 2 4 7 13 15
+ 3 8 12 16 21
+ 9 11 17 20 22
10 18 19 23 24
-```
+</pre>
-编写一个采用Z字形矩阵大小的函数，并将相应的矩阵作为二维数组返回。
+# --instructions--
+
+Write a function that takes the size of the zig-zag matrix, and returns the corresponding matrix as two-dimensional array.
# --hints--
-ZigZagMatrix必须是一个功能
+ZigZagMatrix should be a function.
```js
assert.equal(typeof ZigZagMatrix, 'function');
```
-ZigZagMatrix应该返回数组
+ZigZagMatrix should return array.
```js
assert.equal(typeof ZigZagMatrix(1), 'object');
```
-ZigZagMatrix应该返回一个nestes数组的数组
+ZigZagMatrix should return an array of nested arrays.
```js
assert.equal(typeof ZigZagMatrix(1)[0], 'object');
```
-ZigZagMatrix（1）应返回\[[0]]
+ZigZagMatrix(1) should return \[[0]].
```js
assert.deepEqual(ZigZagMatrix(1), zm1);
```
-ZigZagMatrix（2）应返回\[[0,1]，[2,3]]
+ZigZagMatrix(2) should return \[[0, 1], [2, 3]].
```js
assert.deepEqual(ZigZagMatrix(2), zm2);
```
-ZigZagMatrix（5）必须返回指定的矩阵
+ZigZagMatrix(5) should return specified matrix.
```js
assert.deepEqual(ZigZagMatrix(5), zm5);
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-camper-leaderboard.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-camper-leaderboard.md
index ad9d097e450..35f9887785d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-camper-leaderboard.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-camper-leaderboard.md
@@ -1,14 +1,28 @@
---
id: bd7156d8c242eddfaeb5bd13
-title: 建立一个露营者排行榜
+title: Build a freeCodeCamp Forum Homepage
challengeType: 3
-videoUrl: ''
+forumTopicId: 302349
dashedName: build-a-freecodecamp-forum-homepage
---
# --description--
-**目标：** 构建一个功能类似于此的[CodePen.io](https://codepen.io)应用程序： [https](https://codepen.io/freeCodeCamp/full/eZGMjp/) **：** [//codepen.io/freeCodeCamp/full/eZGMjp/](https://codepen.io) 。完成以下[用户故事](https://en.wikipedia.org/wiki/User_story) 。使用您需要的任何库或API。给它你自己的个人风格。 **用户故事：** 我可以看到freeCodeCamp露营者的表格，他们在过去30天内获得了最多的布朗尼积分。 **用户故事：** 我可以看到他们在过去30天内获得了多少布朗尼积分，以及他们获得了多少积分。 **用户故事：** 我可以根据他们在过去30天内获得的布朗尼点数和他们总共获得的布朗尼点数来对列表进行排序。 **提示：** 要获得过去30天的前100名营员： [https](https://fcctop100.herokuapp.com/api/fccusers/top/recent) ： [//fcctop100.herokuapp.com/api/fccusers/top/recent](https://fcctop100.herokuapp.com/api/fccusers/top/recent) 。 **提示：** 要获得有史以来前100名的露营者： [https](https://fcctop100.herokuapp.com/api/fccusers/top/alltime) ： [//fcctop100.herokuapp.com/api/fccusers/top/alltime](https://fcctop100.herokuapp.com/api/fccusers/top/alltime) 。如果卡住，请记得使用[Read-Search-Ask](https://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck-coding/19514) 。完成后，单击“我已完成此挑战”按钮并包含指向CodePen的链接。您可以通过Facebook上的朋友分享您的项目反馈。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/JqdoMV>.
+
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story). Use whichever libraries or APIs you need. Give it your own personal style.
+
+**User Story:** I can see a list of the most recent posts on the freeCodeCamp forum.
+
+**User Story:** For each topic, I can see the title and a list of links to users who have recently posted in it.
+
+**User Story:** I can see the number of replies and views that each topic has had, and a timestamp of when the topic was last active.
+
+**Hint:** To get the 30 most recent forum posts: <https://forum-proxy.freecodecamp.rocks/latest>.
+
+When you are finished, include a link to your project on CodePen and click the "I've completed this challenge" button.
+
+You can get feedback on your project by sharing it on the [freeCodeCamp forum](https://forum.freecodecamp.org/c/project-feedback/409).
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-light-bright-app.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-light-bright-app.md
index afb2ba96d3c..5cf03cdd2a3 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-light-bright-app.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-light-bright-app.md
@@ -1,14 +1,34 @@
---
id: 5a5d02bd919fcf9ca8cf46cb
-title: 构建一个明亮的应用程序
+title: Build a Light-Bright App
challengeType: 3
-videoUrl: ''
+forumTopicId: 302350
dashedName: build-a-light-bright-app
---
# --description--
-**目标：** 构建一个功能类似于此的[CodePen.io](https://codepen.io)应用程序： [https](https://codepen.io/freeCodeCamp/full/eyLYXE) **：** [//codepen.io/freeCodeCamp/full/eyLYXE](https://codepen.io) 。 **规则＃1：** 不要查看示例项目的代码。自己搞清楚。 **规则＃2：** 完成以下[用户故事](https://en.wikipedia.org/wiki/User_story) 。使用您需要的任何库或API。给它你自己的个人风格。 **用户故事：** 我可以单击或拖动鼠标光标为圆圈着色。 **用户故事：** 我可以双击彩色圆圈去除颜色。 **用户故事：** 我可以点击彩色圆圈来改变它的颜色。 **用户故事：** 每次点击我都会得到一个不同颜色的圆圈。 **用户故事：** 我可以点击“重置”按钮删除最近的颜色。 **用户故事：** 我可以单击“全部重置”按钮删除圆圈中的所有颜色。如果卡住，请记得使用[Read-Search-Ask](https://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck-coding/19514) 。完成后，单击“我已完成此挑战”按钮并包含指向CodePen的链接。您可以通过Facebook上的朋友分享您的项目反馈。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/eyLYXE>.
+
+**Rule #1:** Don't look at the example project's code. Figure it out for yourself.
+
+**Rule #2:** Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story). Use whichever libraries or APIs you need. Give it your own personal style.
+
+**User Story:** I can click or drag the mouse cursor to color the circles.
+
+**User Story:** I can double-click on a colored circle to remove the color.
+
+**User Story:** I can click on a colored circle to change its color.
+
+**User Story:** I should get a circle of different color on each click.
+
+**User Story:** I can click on the 'Reset' button to remove the recent color.
+
+**User Story:** I can click on the 'Reset All' button to remove all the colors from the circles.
+
+When you are finished, include a link to your project on CodePen and click the "I've completed this challenge" button.
+
+You can get feedback on your project by sharing it on the [freeCodeCamp forum](https://forum.freecodecamp.org/c/project-feedback/409).
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-nightlife-coordination-app.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-nightlife-coordination-app.md
index 8cee9f8f35f..5d39e057a64 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-nightlife-coordination-app.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-nightlife-coordination-app.md
@@ -1,14 +1,28 @@
---
id: bd7158d8c443eddfaeb5bdff
-title: 建立夜生活协调应用程序
+title: Build a Nightlife Coordination App
challengeType: 4
-videoUrl: ''
+forumTopicId: 302351
dashedName: build-a-nightlife-coordination-app
---
# --description--
-**目标：** 构建一个功能类似于此的完整堆栈JavaScript应用程序： [http](http://whatsgoinontonight.herokuapp.com/) ： [//whatsgoinontonight.herokuapp.com/](http://whatsgoinontonight.herokuapp.com/)并将其部署到Heroku。请注意，对于每个项目，您应该创建一个新的GitHub存储库和一个新的Heroku项目。如果您不记得如何执行此操作，请重新访问[https://freecodecamp.com/challenges/get-set-for-our-dynamic-web-application-projects](/challenges/get-set-for-our-dynamic-web-application-projects) 。以下是您应为此项目实施的特定用户故事： **用户故事：** 作为未经身份验证的用户，我可以查看我所在区域的所有条形图。 **用户故事：** 作为一名经过身份验证的用户，我可以将自己添加到一个栏中，以表明我今晚要去那里。 **用户故事：** 作为一名经过身份验证的用户，如果我不想再去那里，我可以将自己从酒吧中删除。 **用户故事：** 作为未经身份验证的用户，当我登录时，我不应该再次搜索。 **提示：** 尝试使用[Yelp API](https://www.yelp.com/developers/documentation/v2/overview)查找用户搜索的城市中的场地。如果您使用Yelp的API，请务必在您的应用中提及。完成这些用户故事的实现后，单击“我已完成此挑战”按钮，然后输入GitHub存储库和在Heroku上运行的实时应用程序的URL。您可以通过Facebook上的朋友分享您的项目反馈。
+Build a full stack JavaScript app that is functionally similar to this: <https://yoyo44.herokuapp.com/>. Use a site builder of your choice to complete the project.
+
+Here are the specific user stories you should implement for this project:
+
+**User Story:** As an unauthenticated user, you can view all bars in my area.
+
+**User Story:** As an authenticated user, you can add myself to a bar to indicate you am going there tonight.
+
+**User Story:** As an authenticated user, you can remove myself from a bar if you no longer want to go there.
+
+**User Story:** As an unauthenticated user, when you login you should not have to search again.
+
+**Hint:** Try using the [Yelp API](https://www.yelp.com/developers/documentation/v3) to find venues in the cities your users search for. If you use Yelp's API, be sure to mention so in your app.
+
+When you are done, make sure a working demo of your project is hosted somewhere public. Then submit the URL to it in the `Solution Link` field. Optionally, also submit a link to your project's source code in the `GitHub Link` field.
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-pinterest-clone.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-pinterest-clone.md
index c05d0d83854..0cf1a3d2f7d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-pinterest-clone.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-pinterest-clone.md
@@ -1,14 +1,34 @@
---
id: bd7158d8c443eddfaeb5bdee
-title: 建立一个Pinterest克隆
+title: Build a Pinterest Clone
challengeType: 4
-videoUrl: ''
+forumTopicId: 302352
dashedName: build-a-pinterest-clone
---
# --description--
-**目标：** 构建一个功能类似于此的[Glitch](https://glitch.com)应用程序： [https](https://wild-song.glitch.me/) ： [//wild-song.glitch.me](https://wild-song.glitch.me/) 。完成以下[用户故事](https://en.wikipedia.org/wiki/User_story) 。使用您需要的任何库或API。给它你自己的个人风格。 **用户故事：** 作为未经身份验证的用户，我可以使用GitHub登录。 **用户故事：** 作为经过身份验证的用户，我可以链接到图像。 **用户故事：** 作为经过身份验证的用户，我可以删除已链接到的图像。 **用户故事：** 作为一个经过身份验证的用户，我可以看到我所链接的所有图像的Pinterest风格的墙。 **用户故事：** 作为未经身份验证的用户，我可以浏览其他用户的图像墙。 **用户故事：** 作为经过身份验证的用户，如果我上传的图片已损坏，则会被占位符图片替换。 （可以使用jQuery破碎图像检测） **提示：** [Masonry.js](http://masonry.desandro.com/)是一个允许Pinterest风格图像网格的库。完成这些用户故事的实现后，单击“我已完成此挑战”按钮，然后输入GitHub存储库和在Heroku上运行的实时应用程序的URL。您可以通过Facebook上的朋友分享您的项目反馈。
+**Objective:** Build a [Repl.it](https://repl.it/) app that is functionally similar to this: <https://build-a-pinterest-clone.freecodecamp.rocks/>.
+
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story). Use whichever libraries or APIs you need. Give it your own personal style.
+
+**User Story:** As an unauthenticated user, I can login with GitHub.
+
+**User Story:** As an authenticated user, I can link to images.
+
+**User Story:** As an authenticated user, I can delete images that I've linked to.
+
+**User Story:** As an authenticated user, I can see a Pinterest-style wall of all the images I've linked to.
+
+**User Story:** As an unauthenticated user, I can browse other users' walls of images.
+
+**User Story:** As an authenticated user, if I upload an image that is broken, it will be replaced by a placeholder image. (can use jQuery broken image detection)
+
+**Hint:** [Masonry.js](https://masonry.desandro.com/) is a library that allows for Pinterest-style image grids.
+
+Once you've finished implementing these user stories, enter the URL to your live app and, optionally, your GitHub repository. Then click the "I've completed this challenge" button.
+
+You can get feedback on your project by sharing it on the [freeCodeCamp forum](https://forum.freecodecamp.org/c/project-feedback/409).
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-pong-game.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-pong-game.md
index 722dec9683f..cfa0cc598b6 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-pong-game.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-pong-game.md
@@ -1,14 +1,30 @@
---
id: 5a4b7fcdb66f799f199e11db
-title: 建立乒乓球比赛
+title: Build a Pong Game
challengeType: 3
-videoUrl: ''
+forumTopicId: 302353
dashedName: build-a-pong-game
---
# --description--
-**目标：** 构建一个功能类似于此的[CodePen.io](https://codepen.io)应用程序： [https](https://codepen.io/satyamdev/full/pdMmBp) **：** [//codepen.io/satyamdev/full/pdMmBp](https://codepen.io) 。 **规则＃1：** 不要查看示例项目的代码。自己搞清楚。 **规则＃2：** 完成以下[用户故事](https://en.wikipedia.org/wiki/User_story) 。使用您需要的任何库或API。给它你自己的个人风格。 **用户故事：** 我可以控制一个球拍。 **用户故事：** 计算机可以控制另一个拨片。 **用户故事：** 计算机的桨是无与伦比的。它绝不应该错过球。 **用户故事：** 游戏记录玩家和计算机的分数。如果卡住，请记得使用[Read-Search-Ask](https://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck-coding/19514) 。完成后，单击“我已完成此挑战”按钮并包含指向CodePen的链接。您可以通过Facebook上的朋友分享您的项目反馈。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/satyamdev/full/pdMmBp>.
+
+**Rule #1:** Don't look at the example project's code. Figure it out for yourself.
+
+**Rule #2:** Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story). Use whichever libraries or APIs you need. Give it your own personal style.
+
+**User Story:** I can control a paddle.
+
+**User Story:** The computer can control the other paddle.
+
+**User Story:** The computer's paddle is unbeatable. It should never miss the ball.
+
+**User Story:** The game keeps track of the player and computer's score.
+
+When you are finished, include a link to your project on CodePen and click the "I've completed this challenge" button.
+
+You can get feedback on your project by sharing it on the [freeCodeCamp forum](https://forum.freecodecamp.org/c/project-feedback/409).
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-recipe-box.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-recipe-box.md
index 0a5ff5846db..7817e26951d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-recipe-box.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-recipe-box.md
@@ -1,14 +1,34 @@
---
id: bd7155d8c242eddfaeb5bd13
-title: 建立配方盒
+title: Build a Recipe Box
challengeType: 3
-videoUrl: ''
+forumTopicId: 302354
dashedName: build-a-recipe-box
---
# --description--
-**目标：** 构建一个功能类似于此的[CodePen.io](https://codepen.io)应用程序： [https](https://codepen.io/freeCodeCamp/full/dNVazZ/) **：** [//codepen.io/freeCodeCamp/full/dNVazZ/](https://codepen.io) 。完成以下[用户故事](https://en.wikipedia.org/wiki/User_story) 。使用您需要的任何库或API。给它你自己的个人风格。 **用户故事：** 我可以创建具有名称和成分的食谱。 **用户故事：** 我可以看到一个索引视图，其中所有配方的名称都是可见的。 **用户故事：** 我可以点击任何这些食谱来查看它。 **用户故事：** 我可以编辑这些食谱。 **用户故事：** 我可以删除这些食谱。 **用户故事：** 我添加的所有新食谱都保存在浏览器的本地存储中。如果我刷新页面，这些配方仍然会在那里。 **提示：** 您应该在CodePen上为本地存储密钥添加前缀，即`_username_recipes`如果卡住，请记住使用[Read-Search-Ask](//github.com/FreeCodeCamp/freecodecamp/wiki/FreeCodeCamp-Get-Help) 。完成后，单击“我已完成此挑战”按钮并包含指向CodePen的链接。您可以通过Facebook上的朋友分享您的项目反馈。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/dNVazZ/>.
+
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story). Use whichever libraries or APIs you need. Give it your own personal style.
+
+**User Story:** I can create recipes that have names and ingredients.
+
+**User Story:** I can see an index view where the names of all the recipes are visible.
+
+**User Story:** I can click into any of those recipes to view it.
+
+**User Story:** I can edit these recipes.
+
+**User Story:** I can delete these recipes.
+
+**User Story:** All new recipes I add are saved in my browser's local storage. If I refresh the page, these recipes will still be there.
+
+**Hint:** You should prefix your local storage keys on CodePen, i.e. `_username_recipes`
+
+When you are finished, include a link to your project on CodePen and click the "I've completed this challenge" button.
+
+You can get feedback on your project by sharing it on the [freeCodeCamp forum](https://forum.freecodecamp.org/c/project-feedback/409).
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-roguelike-dungeon-crawler-game.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-roguelike-dungeon-crawler-game.md
index 8bddb998a86..de57859782b 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-roguelike-dungeon-crawler-game.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-roguelike-dungeon-crawler-game.md
@@ -1,14 +1,38 @@
---
id: bd7153d8c242eddfaeb5bd13
-title: 建立一个Roguelike Dungeon Crawler游戏
+title: Build a Roguelike Dungeon Crawler Game
challengeType: 3
-videoUrl: ''
+forumTopicId: 302355
dashedName: build-a-roguelike-dungeon-crawler-game
---
# --description--
-**目标：** 构建一个功能类似于此的[CodePen.io](https://codepen.io)应用程序： [https](https://codepen.io/freeCodeCamp/full/apLXEJ/) **：** [//codepen.io/freeCodeCamp/full/apLXEJ/](https://codepen.io) 。完成以下[用户故事](https://en.wikipedia.org/wiki/User_story) 。使用您需要的任何库或API。给它你自己的个人风格。 **用户故事：** 我有健康，水平和武器。我可以拿起更好的武器。我可以拿起健康食品。 **用户故事：** 地图上的所有物品和敌人都是随机排列的。 **用户故事：** 我可以在整个地图中移动，发现项目。 **用户故事：** 我可以在地图边界内的任何地方移动，但是在我击败它之前我无法穿越敌人。 **用户故事：** 大部分地图都是隐藏的。当我迈出一步时，所有与我在一定数量空间内的空间都会被揭示出来。 **用户故事：** 当我击败敌人时，敌人消失，我得到了XP，最终提高了我的等级。 **用户故事：** 当我与敌人作战时，我们会轮流造成伤害，直到我们其中一人失败为止。我根据自己的等级和武器造成伤害。敌人根据其等级造成伤害。损伤在一定范围内有些随机。 **用户故事：** 当我找到并击败老板时，我赢了。 **用户故事：** 游戏应该具有挑战性，但理论上可以获胜。如果卡住，请记得使用[Read-Search-Ask](https://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck-coding/19514) 。完成后，单击“我已完成此挑战”按钮并包含指向CodePen的链接。您可以通过Facebook上的朋友分享您的项目反馈。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/apLXEJ/>.
+
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story). Use whichever libraries or APIs you need. Give it your own personal style.
+
+**User Story:** I have health, a level, and a weapon. I can pick up a better weapon. I can pick up health items.
+
+**User Story:** All the items and enemies on the map are arranged at random.
+
+**User Story:** I can move throughout a map, discovering items.
+
+**User Story:** I can move anywhere within the map's boundaries, but I can't move through an enemy until I've beaten it.
+
+**User Story:** Much of the map is hidden. When I take a step, all spaces that are within a certain number of spaces from me are revealed.
+
+**User Story:** When I beat an enemy, the enemy goes away and I get XP, which eventually increases my level.
+
+**User Story:** When I fight an enemy, we take turns damaging each other until one of us loses. I do damage based off of my level and my weapon. The enemy does damage based off of its level. Damage is somewhat random within a range.
+
+**User Story:** When I find and beat the boss, I win.
+
+**User Story:** The game should be challenging, but theoretically winnable.
+
+When you are finished, include a link to your project on CodePen and click the "I've completed this challenge" button.
+
+You can get feedback on your project by sharing it on the [freeCodeCamp forum](https://forum.freecodecamp.org/c/project-feedback/409).
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-simon-game.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-simon-game.md
index 156e995bc5d..9de29f15668 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-simon-game.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-simon-game.md
@@ -1,14 +1,38 @@
---
id: bd7158d8c442eddfaeb5bd1c
-title: 建立一个西蒙游戏
+title: Build a Simon Game
challengeType: 3
-videoUrl: ''
+forumTopicId: 302357
dashedName: build-a-simon-game
---
# --description--
-**目标：** 构建一个功能类似于此的[CodePen.io](https://codepen.io)应用程序： [https](https://codepen.io/Em-Ant/full/QbRyqq/) **：** [//codepen.io/freeCodeCamp/full/obYBjE](https://codepen.io) 。完成以下[用户故事](https://en.wikipedia.org/wiki/User_story) 。使用您需要的任何库或API。给它你自己的个人风格。 **用户故事：** 我看到一系列随机按钮。 **用户故事：** 每当我正确输入一系列按钮时，我会看到同一系列的按钮按下，但还有一个额外的步骤。 **用户故事：** 当按下一系列按钮时，以及当我亲自按下按钮时，我都会听到与每个按钮相对应的声音。 **用户故事：** 如果我按下了错误的按钮，我会收到通知，告知我已经这样做了，然后再按一系列按钮开始提醒我模式，这样我就可以再试一次。 **用户故事：** 我可以看到当前按钮按下的步数。 **用户故事：** 如果我想重新启动，我可以点击一个按钮来执行此操作，游戏将返回一个步骤。 **用户故事：** 我可以在严格的模式下玩，如果我按下按钮错误，它会通知我我已经这样做了，并且游戏重新开始按下一系列按钮。 **用户故事：** 我可以通过一系列正确的20个步骤来赢得游戏。我被告知我的胜利，然后比赛重新开始。 **提示：** 在这里**，**您可以使用每个按钮的MP3： `https://s3.amazonaws.com/freecodecamp/simonSound1.mp3` ， `https://s3.amazonaws.com/freecodecamp/simonSound2.mp3` ， `https://s3.amazonaws.com/freecodecamp/simonSound3.mp3` ， `https://s3.amazonaws.com/freecodecamp/simonSound4.mp3` 。如果卡住，请记得使用[Read-Search-Ask](https://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck-coding/19514) 。完成后，单击“我已完成此挑战”按钮并包含指向CodePen的链接。您可以通过Facebook上的朋友分享您的项目反馈。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: [https://codepen.io/freeCodeCamp/full/obYBjE](https://codepen.io/Em-Ant/full/QbRyqq/).
+
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story). Use whichever libraries or APIs you need. Give it your own personal style.
+
+**User Story:** I am presented with a random series of button presses.
+
+**User Story:** Each time I input a series of button presses correctly, I see the same series of button presses but with an additional step.
+
+**User Story:** I hear a sound that corresponds to each button both when the series of button presses plays, and when I personally press a button.
+
+**User Story:** If I press the wrong button, I am notified that I have done so, and that series of button presses starts again to remind me of the pattern so I can try again.
+
+**User Story:** I can see how many steps are in the current series of button presses.
+
+**User Story:** If I want to restart, I can hit a button to do so, and the game will return to a single step.
+
+**User Story:** I can play in strict mode where if I get a button press wrong, it notifies me that I have done so, and the game restarts at a new random series of button presses.
+
+**User Story:** I can win the game by getting a series of 20 steps correct. I am notified of my victory, then the game starts over.
+
+**Hint:** Here are mp3s you can use for each button: `https://s3.amazonaws.com/freecodecamp/simonSound1.mp3`, `https://s3.amazonaws.com/freecodecamp/simonSound2.mp3`, `https://s3.amazonaws.com/freecodecamp/simonSound3.mp3`, `https://s3.amazonaws.com/freecodecamp/simonSound4.mp3`.
+
+When you are finished, include a link to your project on CodePen and click the "I've completed this challenge" button.
+
+You can get feedback on your project by sharing it on the [freeCodeCamp forum](https://forum.freecodecamp.org/c/project-feedback/409).
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-tic-tac-toe-game.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-tic-tac-toe-game.md
index bf5857e52e6..bf79eae711d 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-tic-tac-toe-game.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-tic-tac-toe-game.md
@@ -1,14 +1,26 @@
---
id: bd7158d8c442eedfaeb5bd1c
-title: 建立一个Tic Tac Toe游戏
+title: Build a Tic Tac Toe Game
challengeType: 3
-videoUrl: ''
+forumTopicId: 302358
dashedName: build-a-tic-tac-toe-game
---
# --description--
-**目标：** 构建一个功能类似于此的[CodePen.io](https://codepen.io)应用程序： [https](https://codepen.io/freeCodeCamp/full/KzXQgy/) **：** [//codepen.io/freeCodeCamp/full/KzXQgy/](https://codepen.io) 。完成以下[用户故事](https://en.wikipedia.org/wiki/User_story) 。使用您需要的任何库或API。给它你自己的个人风格。 **用户故事：** 我可以用电脑玩Tic Tac Toe游戏。 **用户故事：** 我的游戏一旦结束就会重置，所以我可以再玩一次。 **用户故事：** 我可以选择是否要播放为X或O.如果卡住，请记得使用[Read-Search-Ask](https://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck-coding/19514) 。完成后，单击“我已完成此挑战”按钮并包含指向CodePen的链接。您可以通过Facebook上的朋友分享您的项目反馈。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/KzXQgy/>.
+
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story). Use whichever libraries or APIs you need. Give it your own personal style.
+
+**User Story:** I can play a game of Tic Tac Toe with the computer.
+
+**User Story:** My game will reset as soon as it's over so I can play again.
+
+**User Story:** I can choose whether I want to play as X or O.
+
+When you are finished, include a link to your project on CodePen and click the "I've completed this challenge" button.
+
+You can get feedback on your project by sharing it on the [freeCodeCamp forum](https://forum.freecodecamp.org/c/project-feedback/409).
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-voting-app.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-voting-app.md
index 63b62467e76..3e6e257ce88 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-voting-app.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-voting-app.md
@@ -1,14 +1,34 @@
---
id: bd7158d8c443eddfaeb5bdef
-title: 建立一个投票应用程序
+title: Build a Voting App
challengeType: 4
-videoUrl: ''
+forumTopicId: 302359
dashedName: build-a-voting-app
---
# --description--
-**目标：** 构建一个功能类似于此的完整堆栈JavaScript应用程序： [https](https://fcc-voting-arthow4n.herokuapp.com/) ： [//fcc-voting-arthow4n.herokuapp.com/](https://fcc-voting-arthow4n.herokuapp.com/)并将其部署到Heroku。请注意，对于每个项目，您应该创建一个新的GitHub存储库和一个新的Heroku项目。如果您不记得如何执行此操作，请重新访问[https://freecodecamp.com/challenges/get-set-for-our-dynamic-web-application-projects](/challenges/get-set-for-our-dynamic-web-application-projects) 。以下是您应该为此项目实现的特定用户故事： **用户故事：** 作为经过身份验证的用户，我可以保留我的民意调查，然后再回来访问它们。 **用户故事：** 作为经过身份验证的用户，我可以与朋友分享我的民意调查。 **用户故事：** 作为经过身份验证的用户，我可以看到我的民意调查的汇总结果。 **用户故事：** 作为经过身份验证的用户，我可以删除我认为不再需要的民意调查。 **用户故事：** 作为经过身份验证的用户，我可以创建包含任意数量项目的民意调查。 **用户故事：** 作为未经身份验证或经过身份验证的用户，我可以查看并对每个人的民意调查进行投票。 **用户故事：** 作为未经身份验证或经过身份验证的用户，我可以以图表形式查看民意调查的结果。 （这可以使用Chart.js或Google Charts实现。） **用户故事：** 作为经过身份验证的用户，如果我不喜欢民意调查中的选项，我可以创建一个新选项。完成这些用户故事的实现后，单击“我已完成此挑战”按钮，然后输入GitHub存储库和在Heroku上运行的实时应用程序的URL。您可以通过Facebook上的朋友分享您的项目反馈。
+Build a full stack JavaScript app that is functionally similar to this: <https://voting-app.freecodecamp.rocks/>. Use a site builder of your choice to complete the project.
+
+Here are the specific user stories you should implement for this project:
+
+**User Story:** As an authenticated user, you can keep my polls and come back later to access them.
+
+**User Story:** As an authenticated user, you can share my polls with my friends.
+
+**User Story:** As an authenticated user, you can see the aggregate results of my polls.
+
+**User Story:** As an authenticated user, you can delete polls that I decide I don't want anymore.
+
+**User Story:** As an authenticated user, you can create a poll with any number of possible items.
+
+**User Story:** As an unauthenticated or authenticated user, you can see and vote on everyone's polls.
+
+**User Story:** As an unauthenticated or authenticated user, you can see the results of polls in chart form. (This could be implemented using Chart.js or Google Charts.)
+
+**User Story:** As an authenticated user, if you don't like the options on a poll, you can create a new option.
+
+When you are done, make sure a working demo of your project is hosted somewhere public. Then submit the URL to it in the `Solution Link` field. Optionally, also submit a link to your project's source code in the `GitHub Link` field.
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-wikipedia-viewer.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-wikipedia-viewer.md
index b0ac10bbcf9..d4cf3b350f7 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-wikipedia-viewer.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-a-wikipedia-viewer.md
@@ -1,14 +1,30 @@
---
id: bd7158d8c442eddfaeb5bd19
-title: 构建Wikipedia查看器
+title: Build a Wikipedia Viewer
challengeType: 3
-videoUrl: ''
+forumTopicId: 302360
dashedName: build-a-wikipedia-viewer
---
# --description--
-**目标：** 构建一个功能类似于此的[CodePen.io](https://codepen.io)应用程序： [https](https://codepen.io/freeCodeCamp/full/wGqEga/) **：** [//codepen.io/freeCodeCamp/full/wGqEga/](https://codepen.io) 。完成以下[用户故事](https://en.wikipedia.org/wiki/User_story) 。使用您需要的任何库或API。给它你自己的个人风格。 **用户故事：** 我可以在搜索框中搜索维基百科条目，并查看生成的维基百科条目。 **用户故事：** 我可以单击按钮查看随机的Wikipedia条目。 提示＃1：这是一个可用于获取随机维基百科文章的URL： `https://en.wikipedia.org/wiki/Special:Random` ： `https://en.wikipedia.org/wiki/Special:Random` :Random。 提示＃2：这是使用维基百科API的条目： `https://www.mediawiki.org/wiki/API:Main_page` ： `https://www.mediawiki.org/wiki/API:Main_page` ： `https://www.mediawiki.org/wiki/API:Main_page` 。 提示＃3：使用此[链接](https://en.wikipedia.org/wiki/Special:ApiSandbox#action=query&titles=Main%20Page&prop=revisions&rvprop=content&format=jsonfm)试验维基百科的API。如果卡住，请记得使用[Read-Search-Ask](https://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck-coding/19514) 。完成后，单击“我已完成此挑战”按钮并包含指向CodePen的链接。您可以通过Facebook上的朋友分享您的项目反馈。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/wGqEga/>.
+
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story). Use whichever libraries or APIs you need. Give it your own personal style.
+
+**User Story:** I can search Wikipedia entries in a search box and see the resulting Wikipedia entries.
+
+**User Story:** I can click a button to see a random Wikipedia entry.
+
+Hint #1: Here's a URL you can use to get a random Wikipedia article: `https://en.wikipedia.org/wiki/Special:Random`.
+
+Hint #2: Here's an entry on using Wikipedia's API: `https://www.mediawiki.org/wiki/API:Main_page`.
+
+Hint #3: Use this [link](https://en.wikipedia.org/wiki/Special:ApiSandbox#action=query&titles=Main%20Page&prop=revisions&rvprop=content&format=jsonfm) to experiment with Wikipedia's API.
+
+When you are finished, include a link to your project on CodePen and click the "I've completed this challenge" button.
+
+You can get feedback on your project by sharing it on the [freeCodeCamp forum](https://forum.freecodecamp.org/c/project-feedback/409).
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-an-image-search-abstraction-layer.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-an-image-search-abstraction-layer.md
index ba2f5892d4f..a17732de47f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-an-image-search-abstraction-layer.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-an-image-search-abstraction-layer.md
@@ -1,14 +1,24 @@
---
id: bd7158d8c443edefaeb5bdee
-title: 构建图像搜索抽象层
+title: Build an Image Search Abstraction Layer
challengeType: 4
-videoUrl: ''
+forumTopicId: 302361
dashedName: build-an-image-search-abstraction-layer
---
# --description--
-**目标：** 构建一个完整的堆栈JavaScript应用程序，允许您搜索如下图像： [https](https://cryptic-ridge-9197.herokuapp.com/api/imagesearch/lolcats%20funny?offset=10) ： [//cryptic-ridge-9197.herokuapp.com/api/imagesearch/lolcats%20funny？offset = 10](https://cryptic-ridge-9197.herokuapp.com/api/imagesearch/lolcats%20funny?offset=10)并浏览此类最近的搜索查询： [https](https://cryptic-ridge-9197.herokuapp.com/api/latest/imagesearch/) ： [//cryptic-ridge-9197.herokuapp.com/api/latest/imagesearch/](https://cryptic-ridge-9197.herokuapp.com/api/latest/imagesearch/) 。然后将其部署到Glitch。请注意，对于每个项目，您应该创建一个新的GitHub存储库和一个新的Glitch项目。如果您不记得如何执行此操作，请重新访问[https://freecodecamp.org/challenges/get-set-for-our-api-development-projects](/challenges/get-set-for-our-api-development-projects) 。以下是您应该为此项目实现的特定用户故事： **用户故事：** 我可以获取与给定搜索字符串相关的一组图像的图像URL，替代文本和页面URL。 **用户故事：** 我可以通过在URL中添加？offset = 2参数来对响应进行分页。 **用户故事：** 我可以获得最近提交的搜索字符串列表。完成这些用户故事的实现后，单击“我已完成此挑战”按钮，然后输入GitHub存储库和在Glitch上运行的实时应用程序的URL。您可以通过Facebook上的朋友分享您的项目反馈。
+Build a full stack JavaScript app that allows you to search for images like this: <https://image-search-abstraction-layer.freecodecamp.rocks/query/lolcats%20funny?page=10> and browse recent search queries like this: <https://image-search-abstraction-layer.freecodecamp.rocks/recent/>. Use a site builder of your choice to complete the project.
+
+Here are the specific user stories you should implement for this project:
+
+**User Story:** You can get the image URLs, description and page URLs for a set of images relating to a given search string.
+
+**User Story:** You can paginate through the responses by adding a `?page=2` parameter to the URL.
+
+**User Story:** You can get a list of the most recently submitted search strings.
+
+When you are done, make sure a working demo of your project is hosted somewhere public. Then submit the URL to it in the `Solution Link` field. Optionally, also submit a link to your project's source code in the `GitHub Link` field.
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-the-game-of-life.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-the-game-of-life.md
index b5b1360a4cb..498b2a027a2 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-the-game-of-life.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/build-the-game-of-life.md
@@ -1,14 +1,36 @@
---
id: bd7154d8c242eddfaeb5bd13
-title: 打造生命游戏
+title: Build the Game of Life
challengeType: 3
-videoUrl: ''
+forumTopicId: 302362
dashedName: build-the-game-of-life
---
# --description--
-**目标：** 构建一个功能类似于此的[CodePen.io](https://codepen.io)应用程序： [https](https://codepen.io/freeCodeCamp/full/BpwMZv/) **：** [//codepen.io/freeCodeCamp/full/BpwMZv/](https://codepen.io) 。完成以下[用户故事](https://en.wikipedia.org/wiki/User_story) 。使用您需要的任何库或API。给它你自己的个人风格。 **用户故事：** 当我第一次到达游戏时，它将随机生成一个棋盘并开始播放。 **用户故事：** 我可以开始和停止董事会。 **用户故事：** 我可以设置电路板。 **用户故事：** 我可以清除董事会。 **用户故事：** 当我按下开始时，游戏将播出。 **用户故事：** 每次董事会改变，我都能看到已经过了多少代。 **提示：** 以下是对Conway的生命游戏的解释，来自John Conway本人： [https](https://www.youtube.com/watch?v=E8kUJL04ELA) ： [//www.youtube.com/watch？v =](https://www.youtube.com/watch?v=E8kUJL04ELA) **E8kUJL04ELA提示：** 这里是Conway生命游戏的概述，规则供您参考： [https：// en.wikipedia.org/wiki/Conway%27s_Game_of_Life](https://en.wikipedia.org/wiki/Conway%27s_Game_of_Life)如果卡住，请记得使用[Read-Search-Ask](//github.com/FreeCodeCamp/freecodecamp/wiki/FreeCodeCamp-Get-Help) 。完成后，单击“我已完成此挑战”按钮并包含指向CodePen的链接。您可以通过Facebook上的朋友分享您的项目反馈。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/BpwMZv/>.
+
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story). Use whichever libraries or APIs you need. Give it your own personal style.
+
+**User Story:** When I first arrive at the game, it will randomly generate a board and start playing.
+
+**User Story:** I can start and stop the board.
+
+**User Story:** I can set up the board.
+
+**User Story:** I can clear the board.
+
+**User Story:** When I press start, the game will play out.
+
+**User Story:** Each time the board changes, I can see how many generations have gone by.
+
+**Hint:** Here's an explanation of Conway's Game of Life from John Conway himself: <https://www.youtube.com/watch?v=E8kUJL04ELA>
+
+**Hint:** Here's an overview of Conway's Game of Life with rules for your reference: [https://en.wikipedia.org/wiki/Conway%27s_Game_of_Life](https://en.wikipedia.org/wiki/Conway%27s_Game_of_Life)
+
+When you are finished, include a link to your project on CodePen and click the "I've completed this challenge" button.
+
+You can get feedback on your project by sharing it on the [freeCodeCamp forum](https://forum.freecodecamp.org/c/project-feedback/409).
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/chart-the-stock-market.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/chart-the-stock-market.md
index a4f38f22d76..fbae3e461ad 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/chart-the-stock-market.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/chart-the-stock-market.md
@@ -1,14 +1,26 @@
---
id: bd7158d8c443eddfaeb5bd0e
-title: 图表股市
+title: Chart the Stock Market
challengeType: 4
-videoUrl: ''
+forumTopicId: 302363
dashedName: chart-the-stock-market
---
# --description--
-**目标：** 构建一个功能类似于此的完整堆栈JavaScript应用程序： [http](http://watchstocks.herokuapp.com/) ： [//watchstocks.herokuapp.com/](http://watchstocks.herokuapp.com/)并将其部署到Heroku。请注意，对于每个项目，您应该创建一个新的GitHub存储库和一个新的Heroku项目。如果您不记得如何执行此操作，请重新访问[https://freecodecamp.com/challenges/get-set-for-our-dynamic-web-application-projects](/challenges/get-set-for-our-dynamic-web-application-projects) 。以下是您应为此项目实施的特定用户故事： **用户故事：** 我可以查看显示每个添加的股票的最近趋势线的图表。 **用户故事：** 我可以按符号名称添加新股票。 **用户故事：** 我可以删除股票。 **用户故事：** 当任何其他用户添加或删除股票时，我可以实时看到变化。为此，您需要使用Web套接字。完成这些用户故事的实现后，单击“我已完成此挑战”按钮，然后输入GitHub存储库和在Heroku上运行的实时应用程序的URL。您可以通过Facebook上的朋友分享您的项目反馈。
+Build a full stack JavaScript app that is functionally similar to this: <https://chart-the-stock-market.freecodecamp.rocks/>. Use a site builder of your choice to complete the project.
+
+Here are the specific user stories you should implement for this project:
+
+**User Story:** You can view a graph displaying the recent trend lines for each added stock.
+
+**User Story:** You can add new stocks by their symbol name.
+
+**User Story:** You can remove stocks.
+
+**User Story:** You can see changes in real-time when any other user adds or removes a stock. For this you will need to use Web Sockets.
+
+When you are done, make sure a working demo of your project is hosted somewhere public. Then submit the URL to it in the `Solution Link` field. Optionally, also submit a link to your project's source code in the `GitHub Link` field.
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/manage-a-book-trading-club.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/manage-a-book-trading-club.md
index 12b3c2fd52d..3a0f2e0cd95 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/manage-a-book-trading-club.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/manage-a-book-trading-club.md
@@ -1,14 +1,28 @@
---
id: bd7158d8c443eddfaeb5bd0f
-title: 管理图书交易俱乐部
+title: Manage a Book Trading Club
challengeType: 4
-videoUrl: ''
+forumTopicId: 302364
dashedName: manage-a-book-trading-club
---
# --description--
-**目标：** 构建一个功能类似于此的[Glitch](https://glitch.com)应用程序： [https](<https://chrome-delivery.glitch.me/ />) ： [//chrome-delivery.glitch.me](<https://chrome-delivery.glitch.me/ />) 。完成以下[用户故事](https://en.wikipedia.org/wiki/User_story) 。使用您需要的任何库或API。给它你自己的个人风格。 **用户故事：** 我可以查看每个用户发布的所有图书。 **用户故事：** 我可以添加一本新书。 **用户故事：** 我可以更新我的设置以存储我的全名，城市和州。 **用户故事：** 我可以提议交易并等待其他用户接受交易。完成这些用户故事的实现后，单击“我已完成此挑战”按钮，然后输入GitHub存储库和在Heroku上运行的实时应用程序的URL。您可以通过Facebook上的朋友分享您的项目反馈。
+**Objective:** Build a [Repl.it](https://repl.it/) app that is functionally similar to this: <https://manage-a-book-trading-club.freecodecamp.rocks/>.
+
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story). Use whichever libraries or APIs you need. Give it your own personal style.
+
+**User Story:** I can view all books posted by every user.
+
+**User Story:** I can add a new book.
+
+**User Story:** I can update my settings to store my full name, city, and state.
+
+**User Story:** I can propose a trade and wait for the other user to accept the trade.
+
+Once you've finished implementing these user stories, enter the URL to your live app and, optionally, your GitHub repository. Then click the "I've completed this challenge" button.
+
+You can get feedback on your project by sharing it on the [freeCodeCamp forum](https://forum.freecodecamp.org/c/project-feedback/409).
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/map-data-across-the-globe.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/map-data-across-the-globe.md
index 8e813a4b03e..aa1dcc3a84a 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/map-data-across-the-globe.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/map-data-across-the-globe.md
@@ -1,14 +1,28 @@
---
id: bd7108d8c242eddfaeb5bd13
-title: 全球地图数据
+title: Map Data Across the Globe
challengeType: 3
-videoUrl: ''
+forumTopicId: 302365
dashedName: map-data-across-the-globe
---
# --description--
-**目标：** 构建一个功能类似于此的[CodePen.io](https://codepen.io)应用程序： [https](https://codepen.io/freeCodeCamp/full/mVEJag) **：** [//codepen.io/freeCodeCamp/full/mVEJag](https://codepen.io) 。完成以下[用户故事](https://en.wikipedia.org/wiki/User_story) 。使用您需要的任何库或API。给它你自己的个人风格。 **用户故事：** 我可以看到所有陨石落在世界地图上的位置。 **用户故事：** 我可以通过查看陨石在地图上的表示方式来判断陨石的相对大小。 **用户故事：** 我可以将鼠标悬停在陨石的数据点上以获取更多数据。 **提示：** 这是一个可用于构建此数据集的数据集： [https](https://raw.githubusercontent.com/freeCodeCamp/ProjectReferenceData/master/meteorite-strike-data.json) ： [//raw.githubusercontent.com/freeCodeCamp/ProjectReferenceData/master/meteorite-strike-data.json](https://raw.githubusercontent.com/freeCodeCamp/ProjectReferenceData/master/meteorite-strike-data.json)如果卡住，请记住使用[Read-Search-Ask](https://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck-coding/19514) 。完成后，单击“我已完成此挑战”按钮并包含指向CodePen的链接。您可以通过Facebook上的朋友分享您的项目反馈。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/mVEJag>.
+
+Fulfill the following [user stories](https://en.wikipedia.org/wiki/User_story). Use whichever libraries or APIs you need. Give it your own personal style.
+
+**User Story:** I can see where all Meteorites landed on a world map.
+
+**User Story:** I can tell the relative size of the meteorite, just by looking at the way it's represented on the map.
+
+**User Story:** I can mouse over the meteorite's data point for additional data.
+
+**Hint:** Here's a dataset you can use to build this: <https://raw.githubusercontent.com/freeCodeCamp/ProjectReferenceData/master/meteorite-strike-data.json>
+
+When you are finished, include a link to your project on CodePen and click the "I've completed this challenge" button.
+
+You can get feedback on your project by sharing it on the [freeCodeCamp forum](https://forum.freecodecamp.org/c/project-feedback/409).
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/p2p-video-chat-application.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/p2p-video-chat-application.md
index 623f560d566..0f7eaa434e1 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/p2p-video-chat-application.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/p2p-video-chat-application.md
@@ -1,14 +1,38 @@
---
id: bd7150d8c442eddfafb5bd1c
-title: P2P视频聊天应用
-challengeType: 3
-videoUrl: ''
+title: P2P Video Chat Application
+challengeType: 4
+forumTopicId: 302366
dashedName: p2p-video-chat-application
---
# --description--
-**目标：** 构建一个功能类似于此的[Glitch](https://glitch.com)应用程序： [https](https://grove-voice.glitch.me/) ： [//grove-voice.glitch.me](https://grove-voice.glitch.me/) 。完成以下[用户故事](https://en.wikipedia.org/wiki/User_story) 。使用您需要的任何库或API。给它你自己的个人风格。 **用户故事：** 到达后，浏览器将提示我访问我的相机和麦克风。 **用户故事：** 在我给予许可后，系统会提示我输入房间名称。 **用户故事：** 输入房间名称后，如果之前没有该房间的房间，则会创建房间。 **用户故事：我的** 一个朋友可以随后访问同一个网站，输入我输入的同一个房间，然后加入同一个房间，然后与我进行视频聊天。 **用户故事：** 如果我输入房间名称，并且该房间已经有两个人，我会收到房间已满的通知。 **用户故事：** 任何人都可以创建或加入任何房间。并且可以有任意数量的房间，但所有房间都必须具有唯一的名称。 **用户故事：** 我可以选择不允许该网站访问我的麦克风和网络摄像头。如果我选择不这样做，如果出现其他一些驱动程序问题，我会看到一条错误消息，说明这些是必需的。 **用户故事：** 当我选择取消房间名称输入步骤，或者如果我没有输入姓名或仅输入空格时，它应该再次要求我输入有效的房间名称。 **用户故事：** 如果房间中的两个人中的一个断开连接，他们可以重新连接到同一个房间并继续聊天。如果卡住，请记得使用[Read-Search-Ask](https://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck-coding/19514) 。完成后，单击“我已完成此挑战”按钮，并包含指向您的Glitch应用程序的链接。您可以通过Facebook上的朋友分享您的项目反馈。
+**Objective:** Build a [Repl.it](https://repl.it/) app that is functionally similar to this: <https://p2p-video-chat-application.freecodecamp.rocks/>.
+
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story). Use whichever libraries or APIs you need. Give it your own personal style.
+
+**User Story:** Upon arriving, the browser will prompt me to access my camera and microphone.
+
+**User Story:** After I give it permission, I am prompted to type in a room name.
+
+**User Story:** Once I type in the room name, a room will be created if no room of that name existed before.
+
+**User Story:** A friend of mine can subsequently go to the same website, type in the same room I entered, and join the same room, then enter into a video chat with me.
+
+**User Story:** If I type in a room name, and there are already two people in that room, I get a notification that the room is full.
+
+**User Story:** Anyone can create or join any room. And there can be any number of rooms, but all of them must have unique names.
+
+**User Story:** I can choose to not permit the site to access my microphone and webcam. If I choose not to do this, or if some other driver problem occurs, I see an error message saying these are required.
+
+**User Story:** When I choose to cancel the room name input step, or if I type in no name, or just spaces, it should again ask me again to type in a valid room name.
+
+**User Story:** If one of the two people in the room get disconnected, they can reconnect to the same room and continue chatting.
+
+Once you've finished implementing these user stories, enter the URL to your live app and, optionally, your GitHub repository. Then click the "I've completed this challenge" button.
+
+You can get feedback on your project by sharing it on the [freeCodeCamp forum](https://forum.freecodecamp.org/c/project-feedback/409).
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/show-national-contiguity-with-a-force-directed-graph.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/show-national-contiguity-with-a-force-directed-graph.md
index 12da26123d5..9e83b937f40 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/show-national-contiguity-with-a-force-directed-graph.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/show-national-contiguity-with-a-force-directed-graph.md
@@ -1,14 +1,28 @@
---
id: bd7198d8c242eddfaeb5bd13
-title: 用力导向图显示全国连续性
+title: Show National Contiguity with a Force Directed Graph
challengeType: 3
-videoUrl: ''
+forumTopicId: 302367
dashedName: show-national-contiguity-with-a-force-directed-graph
---
# --description--
-**目标：** 构建一个功能类似于此的[CodePen.io](https://codepen.io)应用程序： [https](https://codepen.io/freeCodeCamp/full/xVopBo) **：** [//codepen.io/freeCodeCamp/full/xVopBo](https://codepen.io) 。完成以下[用户故事](https://en.wikipedia.org/wiki/User_story) 。使用您需要的任何库或API。给它你自己的个人风格。 **用户故事：** 我可以看到一个强制导向图，显示哪些国家/地区共享边界。 **用户故事：** 我可以在其节点上看到每个国家/地区的旗帜。 **提示：** 这是一个可用于构建此数据集的数据集： [https](https://raw.githubusercontent.com/DealPete/forceDirected/master/countries.json) **：** //raw.githubusercontent.com/DealPete/forceDirected/master/countries.json **提示：** 您可以在[https：//www.flag-](https://www.flag-sprites.com)创建国旗的spritesheet [sprites.com](https://www.flag-sprites.com) 。如果卡住，请记得使用[Read-Search-Ask](https://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck-coding/19514) 。完成后，单击“我已完成此挑战”按钮并包含指向CodePen的链接。您可以通过Facebook上的朋友分享您的项目反馈。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/xVopBo>.
+
+Fulfill the following [user stories](https://en.wikipedia.org/wiki/User_story). Use whichever libraries or APIs you need. Give it your own personal style.
+
+**User Story:** I can see a Force-directed Graph that shows which countries share borders.
+
+**User Story:** I can see each country's flag on its node.
+
+**Hint:** Here's a dataset you can use to build this: <https://raw.githubusercontent.com/DealPete/forceDirected/master/countries.json>
+
+**Hint:** You can create a spritesheet of national flags at <https://www.flag-sprites.com>.
+
+When you are finished, include a link to your project on CodePen and click the "I've completed this challenge" button.
+
+You can get feedback on your project by sharing it on the [freeCodeCamp forum](https://forum.freecodecamp.org/c/project-feedback/409).
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/show-the-local-weather.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/show-the-local-weather.md
index a4668f22c92..9d470d6b1cf 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/show-the-local-weather.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/show-the-local-weather.md
@@ -1,14 +1,30 @@
---
id: bd7158d8c442eddfaeb5bd10
-title: 显示当地天气
+title: Show the Local Weather
challengeType: 3
-videoUrl: ''
+forumTopicId: 302368
dashedName: show-the-local-weather
---
# --description--
-**目标：** 构建一个功能类似于此的[CodePen.io](https://codepen.io)应用程序： [https](https://codepen.io/freeCodeCamp/full/bELRjV) **：** [//codepen.io/freeCodeCamp/full/bELRjV](https://codepen.io) 。 **规则＃1：** 不要查看示例项目的代码。自己搞清楚。 **规则＃2：** 完成以下[用户故事](https://en.wikipedia.org/wiki/User_story) 。使用您需要的任何库或API。给它你自己的个人风格。 **用户故事：** 我可以在当前位置看到天气。 **用户故事：** 根据天气情况，我可以看到不同的图标或背景图像（例如雪山，炎热的沙漠）。 **用户故事：** 我可以按一个按钮在华氏温度和摄氏温度之间切换。 **注意：** 许多Internet浏览器现在需要HTTP Secure（ `https://` ）连接才能通过HTML5 Geolocation获取用户的语言环境。因此，我们建议使用HTML5 Geolocation获取用户位置，然后使用freeCodeCamp Weather API <https://fcc-weather-api.glitch.me> ，该API使用HTTP安全连接进行天气预报。此外，请务必通过`https://`连接到[CodePen.io](https://codepen.io) 。如果卡住，请记得使用[Read-Search-Ask](https://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck-coding/19514) 。完成后，单击“我已完成此挑战”按钮并包含指向CodePen的链接。您可以通过Facebook上的朋友分享您的项目反馈。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/bELRjV>.
+
+**Rule #1:** Don't look at the example project's code. Figure it out for yourself.
+
+**Rule #2:** Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story). Use whichever libraries or APIs you need. Give it your own personal style.
+
+**User Story:** I can see the weather in my current location.
+
+**User Story:** I can see a different icon or background image (e.g. snowy mountain, hot desert) depending on the weather.
+
+**User Story:** I can push a button to toggle between Fahrenheit and Celsius.
+
+**Note:** Many internet browsers now require an HTTP Secure (`https://`) connection to obtain a user's locale via HTML5 Geolocation. For this reason, we recommend using HTML5 Geolocation to get user location and then use the freeCodeCamp Weather API <https://weather-proxy.freecodecamp.rocks/> which uses an HTTP Secure connection for the weather. Also, be sure to connect to [CodePen.io](https://codepen.io) via `https://`.
+
+When you are finished, include a link to your project on CodePen and click the "I've completed this challenge" button.
+
+You can get feedback on your project by sharing it on the [freeCodeCamp forum](https://forum.freecodecamp.org/c/project-feedback/409).
# --solutions--
diff --git a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/use-the-twitch-json-api.md b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/use-the-twitch-json-api.md
index bd21769958d..55d17b1745f 100644
--- a/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/use-the-twitch-json-api.md
+++ b/curriculum/challenges/chinese/10-coding-interview-prep/take-home-projects/use-the-twitch-json-api.md
@@ -1,14 +1,32 @@
---
id: bd7158d8c442eddfaeb5bd1f
-title: 使用Twitch JSON API
+title: Use the Twitch JSON API
challengeType: 3
-videoUrl: ''
+forumTopicId: 19541
dashedName: use-the-twitch-json-api
---
# --description--
-**目标：** 构建一个功能类似于此的[CodePen.io](https://codepen.io)应用程序： [https](https://codepen.io/freeCodeCamp/full/Myvqmo/) **：** [//codepen.io/freeCodeCamp/full/Myvqmo/](https://codepen.io) 。完成以下[用户故事](https://en.wikipedia.org/wiki/User_story) 。使用您需要的任何库或API。给它你自己的个人风格。 **用户故事：** 我可以看到freeCodeCamp当前是否在Twitch.tv上流式传输。 **用户故事：** 我可以点击状态输出并直接发送到freeCodeCamp的Twitch.tv频道。 **用户故事：** 如果Twitch用户当前正在播放，我可以看到有关他们正在播放的内容的其他详细信息。 **提示：** 请参阅[http://forum.freecodecamp.org/t/use-the-twitchtv-json-api/19541上](http://forum.freecodecamp.org/t/use-the-twitchtv-json-api/19541)对Twitch.tv的JSONP API的示例调用。 **提示：** 有关此API调用的相关文档位于： [https](https://dev.twitch.tv/docs/v5/reference/streams/#get-stream-by-user) ： [//dev.twitch.tv/docs/v5/reference/streams/#get-stream-by-user](https://dev.twitch.tv/docs/v5/reference/streams/#get-stream-by-user) 。 **提示：** 这是定期传输的人的Twitch.tv用户名数组： `["ESL_SC2", "OgamingSC2", "cretetion", "freecodecamp", "storbeck", "habathcx", "RobotCaleb", "noobs2ninjas"]` **更新：** 由于[此处](https://blog.twitch.tv/client-id-required-for-kraken-api-calls-afbb8e95f843#.f8hipkht1)解释的API使用条件发生变化，Twitch.tv现在需要API密钥，但我们已经构建了一种解决方法。使用[https://wind-bow.glitch.me/twitch-api](https://wind-bow.glitch.me)而不是twitch的API基本URL（即`https://api.twitch.tv/kraken），您仍然可以获取帐户信息，而无需注册API密钥。如果卡住，请记得使用`[Read-Search-Ask](https://forum.freecodecamp.org/t/how-to-get-help-when-you-are-stuck-coding/19514) 。完成后，单击“我已完成此挑战”按钮并包含指向CodePen的链接。您可以通过Facebook上的朋友分享您的项目反馈。
+**Objective:** Build a [CodePen.io](https://codepen.io) app that is functionally similar to this: <https://codepen.io/freeCodeCamp/full/Myvqmo/>.
+
+Fulfill the below [user stories](https://en.wikipedia.org/wiki/User_story). Use whichever libraries or APIs you need. Give it your own personal style.
+
+**User Story:** I can see whether freeCodeCamp is currently streaming on Twitch.tv.
+
+**User Story:** I can click the status output and be sent directly to the freeCodeCamp's Twitch.tv channel.
+
+**User Story:** if a Twitch user is currently streaming, I can see additional details about what they are streaming.
+
+**Hint:** The relevant documentation about Twitch.tv's JSON API is here: <https://dev.twitch.tv/docs/api/reference/#get-streams>.
+
+**Hint:** Here's an array of the Twitch.tv usernames of people who regularly stream: `["ESL_SC2", "OgamingSC2", "cretetion", "freecodecamp", "storbeck", "habathcx", "RobotCaleb", "noobs2ninjas"]`
+
+**UPDATE:** Due to a change in conditions on API usage, Twitch.tv requires an API key, but we've built a workaround. Use <https://twitch-proxy.freecodecamp.rocks/> instead of Twitch's API base URL (i.e. `https://api.twitch.tv/helix` ) and you'll still be able to get account information, without needing to sign up for an API key.
+
+When you are finished, include a link to your project on CodePen and click the "I've completed this challenge" button.
+
+You can get feedback on your project by sharing it on the [freeCodeCamp forum](https://forum.freecodecamp.org/c/project-feedback/409).
# --solutions--
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/how-neural-networks-work/deep-learning-demystified.md b/curriculum/challenges/chinese/11-machine-learning-with-python/how-neural-networks-work/deep-learning-demystified.md
index 81e2c2e0cc5..7d7338ee653 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/how-neural-networks-work/deep-learning-demystified.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/how-neural-networks-work/deep-learning-demystified.md
@@ -1,5 +1,6 @@
---
id: 5e9a0e9ef99a403d019610cc
+title: Deep Learning Demystified
challengeType: 11
videoId: bejQ-W9BGJg
dashedName: deep-learning-demystified
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/how-neural-networks-work/how-convolutional-neural-networks-work.md b/curriculum/challenges/chinese/11-machine-learning-with-python/how-neural-networks-work/how-convolutional-neural-networks-work.md
index 1620a377c6a..b06b2fbed8a 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/how-neural-networks-work/how-convolutional-neural-networks-work.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/how-neural-networks-work/how-convolutional-neural-networks-work.md
@@ -1,5 +1,6 @@
---
id: 5e9a0e9ef99a403d019610cd
+title: How Convolutional Neural Networks work
challengeType: 11
videoId: Y5M7KH4A4n4
dashedName: how-convolutional-neural-networks-work
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/how-neural-networks-work/how-deep-neural-networks-work.md b/curriculum/challenges/chinese/11-machine-learning-with-python/how-neural-networks-work/how-deep-neural-networks-work.md
index d43c63b5c93..ae41c9be52a 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/how-neural-networks-work/how-deep-neural-networks-work.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/how-neural-networks-work/how-deep-neural-networks-work.md
@@ -1,5 +1,6 @@
---
id: 5e9a0e9ef99a403d019610ca
+title: How Deep Neural Networks Work
challengeType: 11
videoId: zvalnHWGtx4
dashedName: how-deep-neural-networks-work
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/how-neural-networks-work/recurrent-neural-networks-rnn-and-long-short-term-memory-lstm.md b/curriculum/challenges/chinese/11-machine-learning-with-python/how-neural-networks-work/recurrent-neural-networks-rnn-and-long-short-term-memory-lstm.md
index b544873fa78..ab702e9b666 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/how-neural-networks-work/recurrent-neural-networks-rnn-and-long-short-term-memory-lstm.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/how-neural-networks-work/recurrent-neural-networks-rnn-and-long-short-term-memory-lstm.md
@@ -1,5 +1,6 @@
---
id: 5e9a0e9ef99a403d019610cb
+title: Recurrent Neural Networks RNN and Long Short Term Memory LSTM
challengeType: 11
videoId: UVimlsy9eW0
dashedName: recurrent-neural-networks-rnn-and-long-short-term-memory-lstm
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/book-recommendation-engine-using-knn.md b/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/book-recommendation-engine-using-knn.md
index 7000574b925..b7c54579550 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/book-recommendation-engine-using-knn.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/book-recommendation-engine-using-knn.md
@@ -1,5 +1,6 @@
---
id: 5e46f8e3ac417301a38fb92f
+title: Book Recommendation Engine using KNN
challengeType: 10
dashedName: book-recommendation-engine-using-knn
---
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/cat-and-dog-image-classifier.md b/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/cat-and-dog-image-classifier.md
index 552c604cc1c..76e92dcfb2b 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/cat-and-dog-image-classifier.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/cat-and-dog-image-classifier.md
@@ -1,5 +1,6 @@
---
id: 5e46f8dcac417301a38fb92e
+title: Cat and Dog Image Classifier
challengeType: 10
dashedName: cat-and-dog-image-classifier
---
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/linear-regression-health-costs-calculator.md b/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/linear-regression-health-costs-calculator.md
index 162b4df3f11..d0b92385ed1 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/linear-regression-health-costs-calculator.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/linear-regression-health-costs-calculator.md
@@ -1,5 +1,6 @@
---
id: 5e46f8edac417301a38fb930
+title: Linear Regression Health Costs Calculator
challengeType: 10
dashedName: linear-regression-health-costs-calculator
---
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/neural-network-sms-text-classifier.md b/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/neural-network-sms-text-classifier.md
index 83790f74258..f938843a59e 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/neural-network-sms-text-classifier.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/neural-network-sms-text-classifier.md
@@ -1,5 +1,6 @@
---
id: 5e46f8edac417301a38fb931
+title: Neural Network SMS Text Classifier
challengeType: 10
dashedName: neural-network-sms-text-classifier
---
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/rock-paper-scissors.md b/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/rock-paper-scissors.md
index 5cc0f1cf719..5685211ac41 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/rock-paper-scissors.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/machine-learning-with-python-projects/rock-paper-scissors.md
@@ -1,5 +1,6 @@
---
id: 5e46f8d6ac417301a38fb92d
+title: Rock Paper Scissors
challengeType: 10
dashedName: rock-paper-scissors
---
@@ -8,7 +9,7 @@ dashedName: rock-paper-scissors
For this challenge, you will create a program to play Rock, Paper, Scissors. A program that picks at random will usually win 50% of the time. To pass this challenge your program must play matches against four different bots, winning at least 60% of the games in each match.
-You can access [the full project description and starter code on repl.it](https://repl.it/@freeCodeCamp/fcc-rock-paper-scissors).
+You can access [the full project description and starter code on repl.it](https://repl.it/github/freeCodeCamp/boilerplate-rock-paper-scissors).
After going to that link, fork the project. Once you complete the project based on the instructions in 'README.md', submit your project link below.
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/conclusion.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/conclusion.md
index 25a0a52d3d4..b87ca8022b9 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/conclusion.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/conclusion.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72da6
+title: Conclusion
challengeType: 11
videoId: LMNub5frQi4
dashedName: conclusion
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/convolutional-neural-networks-evaluating-the-model.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/convolutional-neural-networks-evaluating-the-model.md
index c389c08e748..71a466e2b59 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/convolutional-neural-networks-evaluating-the-model.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/convolutional-neural-networks-evaluating-the-model.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d99
+title: 'Convolutional Neural Networks: Evaluating the Model'
challengeType: 11
videoId: eCATNvwraXg
dashedName: convolutional-neural-networks-evaluating-the-model
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/convolutional-neural-networks-picking-a-pretrained-model.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/convolutional-neural-networks-picking-a-pretrained-model.md
index a96357dc7d3..9d34c0b25aa 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/convolutional-neural-networks-picking-a-pretrained-model.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/convolutional-neural-networks-picking-a-pretrained-model.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d9a
+title: 'Convolutional Neural Networks: Picking a Pretrained Model'
challengeType: 11
videoId: h1XUt1AgIOI
dashedName: convolutional-neural-networks-picking-a-pretrained-model
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/convolutional-neural-networks-the-convolutional-layer.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/convolutional-neural-networks-the-convolutional-layer.md
index 0791370a747..87c0b9ef9be 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/convolutional-neural-networks-the-convolutional-layer.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/convolutional-neural-networks-the-convolutional-layer.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d97
+title: 'Convolutional Neural Networks: The Convolutional Layer'
challengeType: 11
videoId: LrdmcQpTyLw
dashedName: convolutional-neural-networks-the-convolutional-layer
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/convolutional-neural-networks.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/convolutional-neural-networks.md
index 187f224de4e..616ef134768 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/convolutional-neural-networks.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/convolutional-neural-networks.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d96
+title: Convolutional Neural Networks
challengeType: 11
videoId: _1kTP7uoU9E
dashedName: convolutional-neural-networks
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-building-the-model.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-building-the-model.md
index c260cb2519d..4b792df4cd1 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-building-the-model.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-building-the-model.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d8e
+title: 'Core Learning Algorithms: Building the Model'
challengeType: 11
videoId: 5wHw8BTd2ZQ
dashedName: core-learning-algorithms-building-the-model
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-classification.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-classification.md
index 4bcec13bb87..245cf7febe3 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-classification.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-classification.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d8d
+title: 'Core Learning Algorithms: Classification'
challengeType: 11
videoId: qFF7ZQNvK9E
dashedName: core-learning-algorithms-classification
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-clustering.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-clustering.md
index b5fac932611..0f8065dedcf 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-clustering.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-clustering.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d8f
+title: 'Core Learning Algorithms: Clustering'
challengeType: 11
videoId: 8sqIaHc9Cz4
dashedName: core-learning-algorithms-clustering
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-hidden-markov-models.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-hidden-markov-models.md
index 6de64a817f2..b829085f33a 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-hidden-markov-models.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-hidden-markov-models.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d90
+title: 'Core Learning Algorithms: Hidden Markov Models'
challengeType: 11
videoId: IZg24y4wEPY
dashedName: core-learning-algorithms-hidden-markov-models
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-the-training-process.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-the-training-process.md
index 2aa83b408c9..29ca4b48963 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-the-training-process.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-the-training-process.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d8c
+title: 'Core Learning Algorithms: The Training Process'
challengeType: 11
videoId: _cEwvqVoBhI
dashedName: core-learning-algorithms-the-training-process
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-training-and-testing-data.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-training-and-testing-data.md
index 4e4078a91da..d31d7425eb7 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-training-and-testing-data.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-training-and-testing-data.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d8b
+title: 'Core Learning Algorithms: Training and Testing Data'
challengeType: 11
videoId: wz9J1slsi7I
dashedName: core-learning-algorithms-training-and-testing-data
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-using-probabilities-to-make-predictions.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-using-probabilities-to-make-predictions.md
index 1127c7104d3..2f77b3c6ea2 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-using-probabilities-to-make-predictions.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-using-probabilities-to-make-predictions.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d91
+title: 'Core Learning Algorithms: Using Probabilities to make Predictions'
challengeType: 11
videoId: fYAYvLUawnc
dashedName: core-learning-algorithms-using-probabilities-to-make-predictions
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-working-with-data.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-working-with-data.md
index 74476574928..bd2c0c44f11 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-working-with-data.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms-working-with-data.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d8a
+title: 'Core Learning Algorithms: Working with Data'
challengeType: 11
videoId: u85IOSsJsPI
dashedName: core-learning-algorithms-working-with-data
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms.md
index 5e2f51184f9..470a5db663b 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/core-learning-algorithms.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d89
+title: Core Learning Algorithms
challengeType: 11
videoId: u5lZURgcWnU
dashedName: core-learning-algorithms
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/creating-a-convolutional-neural-network.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/creating-a-convolutional-neural-network.md
index 620618f323f..68999197b25 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/creating-a-convolutional-neural-network.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/creating-a-convolutional-neural-network.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d98
+title: Creating a Convolutional Neural Network
challengeType: 11
videoId: kfv0K8MtkIc
dashedName: creating-a-convolutional-neural-network
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/introduction-machine-learning-fundamentals.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/introduction-machine-learning-fundamentals.md
index 6bedc4c51d7..185278bb1e7 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/introduction-machine-learning-fundamentals.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/introduction-machine-learning-fundamentals.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d87
+title: 'Introduction: Machine Learning Fundamentals'
challengeType: 11
videoId: KwL1qTR5MT8
dashedName: introduction-machine-learning-fundamentals
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/introduction-to-tensorflow.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/introduction-to-tensorflow.md
index d7811e36738..cefca2bfe26 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/introduction-to-tensorflow.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/introduction-to-tensorflow.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d88
+title: Introduction to TensorFlow
challengeType: 11
videoId: r9hRyGGjOgQ
dashedName: introduction-to-tensorflow
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-building-the-model.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-building-the-model.md
index 4abd3765561..7b7546ba33f 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-building-the-model.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-building-the-model.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72da1
+title: 'Natural Language Processing With RNNs: Building the Model'
challengeType: 11
videoId: 32WBFS7lfsw
dashedName: natural-language-processing-with-rnns-building-the-model
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-create-a-play-generator.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-create-a-play-generator.md
index 1f790351523..9ea1c8ae2a7 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-create-a-play-generator.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-create-a-play-generator.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72da0
+title: 'Natural Language Processing With RNNs: Create a Play Generator'
challengeType: 11
videoId: j5xsxjq_Xk8
dashedName: natural-language-processing-with-rnns-create-a-play-generator
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-making-predictions.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-making-predictions.md
index d3a5af66ff0..29dbc05f8d2 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-making-predictions.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-making-predictions.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d9f
+title: 'Natural Language Processing With RNNs: Making Predictions'
challengeType: 11
videoId: WO1hINnBj20
dashedName: natural-language-processing-with-rnns-making-predictions
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-part-2.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-part-2.md
index 3e975bfd2d4..c572d154e16 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-part-2.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-part-2.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d9c
+title: 'Natural Language Processing With RNNs: Part 2'
challengeType: 11
videoId: mUU9YXOFbZg
dashedName: natural-language-processing-with-rnns-part-2
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-recurring-neural-networks.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-recurring-neural-networks.md
index b0da055afbc..e3d8d7e2638 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-recurring-neural-networks.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-recurring-neural-networks.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d9d
+title: 'Natural Language Processing With RNNs: Recurring Neural Networks'
challengeType: 11
videoId: bX5681NPOcA
dashedName: natural-language-processing-with-rnns-recurring-neural-networks
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-sentimental-analysis.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-sentimental-analysis.md
index 5d2e8abc043..60118b25c81 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-sentimental-analysis.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-sentimental-analysis.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d9e
+title: 'Natural Language Processing With RNNs: Sentiment Analysis'
challengeType: 11
videoId: lYeLtu8Nq7c
dashedName: natural-language-processing-with-rnns-sentiment-analysis
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-training-the-model.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-training-the-model.md
index 5394a7e5d16..2ae0cab5b74 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-training-the-model.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns-training-the-model.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72da2
+title: 'Natural Language Processing With RNNs: Training the Model'
challengeType: 11
videoId: hEUiK7j9UI8
dashedName: natural-language-processing-with-rnns-training-the-model
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns.md
index 8ed003d0528..f4bde454837 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/natural-language-processing-with-rnns.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d9b
+title: Natural Language Processing With RNNs
challengeType: 11
videoId: ZyCaF5S-lKg
dashedName: natural-language-processing-with-rnns
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/neural-networks-activation-functions.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/neural-networks-activation-functions.md
index 0bc16616ca9..635e29a35db 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/neural-networks-activation-functions.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/neural-networks-activation-functions.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d93
+title: 'Neural Networks: Activation Functions'
challengeType: 11
videoId: S45tqW6BqRs
dashedName: neural-networks-activation-functions
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/neural-networks-creating-a-model.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/neural-networks-creating-a-model.md
index 087677cf28b..05b3df57e0a 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/neural-networks-creating-a-model.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/neural-networks-creating-a-model.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d95
+title: 'Neural Networks: Creating a Model'
challengeType: 11
videoId: K8bz1bmOCTw
dashedName: neural-networks-creating-a-model
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/neural-networks-optimizers.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/neural-networks-optimizers.md
index d283ecb4cdb..216b851ab51 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/neural-networks-optimizers.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/neural-networks-optimizers.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d94
+title: 'Neural Networks: Optimizers'
challengeType: 11
videoId: hdOtRPQe1o4
dashedName: neural-networks-optimizers
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/neural-networks-with-tensorflow.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/neural-networks-with-tensorflow.md
index a610ee38acc..d9d07b00fcc 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/neural-networks-with-tensorflow.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/neural-networks-with-tensorflow.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72d92
+title: Neural Networks with TensorFlow
challengeType: 11
videoId: uisdfrNrZW4
dashedName: neural-networks-with-tensorflow
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/reinforcement-learning-with-q-learning-example.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/reinforcement-learning-with-q-learning-example.md
index 59ade950b65..1b9f6923a90 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/reinforcement-learning-with-q-learning-example.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/reinforcement-learning-with-q-learning-example.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72da5
+title: 'Reinforcement Learning With Q-Learning: Example'
challengeType: 11
videoId: RBBSNta234s
dashedName: reinforcement-learning-with-q-learning-example
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/reinforcement-learning-with-q-learning-part-2.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/reinforcement-learning-with-q-learning-part-2.md
index a63484c6fc6..089f1440386 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/reinforcement-learning-with-q-learning-part-2.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/reinforcement-learning-with-q-learning-part-2.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72da4
+title: 'Reinforcement Learning With Q-Learning: Part 2'
challengeType: 11
videoId: DX7hJuaUZ7o
dashedName: reinforcement-learning-with-q-learning-part-2
diff --git a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/reinforcement-learning-with-q-learning.md b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/reinforcement-learning-with-q-learning.md
index 406e5bd2132..3cbeedcfb3f 100644
--- a/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/reinforcement-learning-with-q-learning.md
+++ b/curriculum/challenges/chinese/11-machine-learning-with-python/tensorflow/reinforcement-learning-with-q-learning.md
@@ -1,5 +1,6 @@
---
id: 5e8f2f13c4cdbe86b5c72da3
+title: Reinforcement Learning With Q-Learning
challengeType: 11
videoId: Cf7DSU0gVb4
dashedName: reinforcement-learning-with-q-learning
diff --git a/curriculum/challenges/chinese/12-certificates/apis-and-microservices-certificate/apis-and-microservices-certificate.yml b/curriculum/challenges/chinese/12-certificates/apis-and-microservices-certificate/apis-and-microservices-certificate.yml
index 801577063d8..b691e98afac 100644
--- a/curriculum/challenges/chinese/12-certificates/apis-and-microservices-certificate/apis-and-microservices-certificate.yml
+++ b/curriculum/challenges/chinese/12-certificates/apis-and-microservices-certificate/apis-and-microservices-certificate.yml
@@ -1,50 +1,21 @@
---
id: 561add10cb82ac38a17523bc
+title: APIs and Microservices Certificate
challengeType: 7
isPrivate: true
-title: API和微服务证书
----
-
-## Description
-<section id='description'>
-
-</section>
-
-## Instructions
-<section id='instructions'>
-
-</section>
-
-## Tests
-<section id='tests'>
-
-```yml
tests:
- - id: bd7158d8c443edefaeb5bdef
+ -
+ id: bd7158d8c443edefaeb5bdef
title: Timestamp Microservice
- - id: bd7158d8c443edefaeb5bdff
+ -
+ id: bd7158d8c443edefaeb5bdff
title: Request Header Parser Microservice
- - id: bd7158d8c443edefaeb5bd0e
+ -
+ id: bd7158d8c443edefaeb5bd0e
title: URL Shortener Microservice
- - id: 5a8b073d06fa14fcfde687aa
+ -
+ id: 5a8b073d06fa14fcfde687aa
title: Exercise Tracker
- - id: bd7158d8c443edefaeb5bd0f
+ -
+ id: bd7158d8c443edefaeb5bd0f
title: File Metadata Microservice
-
-```
-
-</section>
-
-## Challenge Seed
-<section id='challengeSeed'>
-
-</section>
-
-## Solution
-<section id='solution'>
-
-```js
-// solution required
-```
-
-</section>
diff --git a/curriculum/challenges/chinese/12-certificates/data-analysis-with-python-certificate/data-analysis-with-python-certificate.yml b/curriculum/challenges/chinese/12-certificates/data-analysis-with-python-certificate/data-analysis-with-python-certificate.yml
index 996e60d9aad..238899200c5 100644
--- a/curriculum/challenges/chinese/12-certificates/data-analysis-with-python-certificate/data-analysis-with-python-certificate.yml
+++ b/curriculum/challenges/chinese/12-certificates/data-analysis-with-python-certificate/data-analysis-with-python-certificate.yml
@@ -1,49 +1,22 @@
---
id: 5e46fc95ac417301a38fb934
+title: Data Analysis with Python Certificate
challengeType: 7
isPrivate: true
-title: ''
----
-
-## Description
-<section id='description'>
-
-</section>
-
-## Instructions
-<section id='instructions'>
-
-</section>
-
-## Tests
-<section id='tests'>
-
-```yml
tests:
- - id: 5e46f7e5ac417301a38fb928
+ -
+ id: 5e46f7e5ac417301a38fb928
title: Mean-Variance-Standard Deviation Calculator
- - id: 5e46f7e5ac417301a38fb929
+ -
+ id: 5e46f7e5ac417301a38fb929
title: Demographic Data Analyzer
- - id: 5e46f7f8ac417301a38fb92a
+ -
+ id: 5e46f7f8ac417301a38fb92a
title: Medical Data Visualizer
- - id: 5e46f802ac417301a38fb92b
+ -
+ id: 5e46f802ac417301a38fb92b
title: Page View Time Series Visualizer
- - id: 5e4f5c4b570f7e3a4949899f
+ -
+ id: 5e4f5c4b570f7e3a4949899f
title: Sea Level Predictor
-```
-</section>
-
-## Challenge Seed
-<section id='challengeSeed'>
-
-</section>
-
-## Solution
-<section id='solution'>
-
-```js
-// solution required
-```
-
-</section>
diff --git a/curriculum/challenges/chinese/12-certificates/data-visualization-certificate/data-visualization-certificate.yml b/curriculum/challenges/chinese/12-certificates/data-visualization-certificate/data-visualization-certificate.yml
index fd728a4df28..589eebe67f9 100644
--- a/curriculum/challenges/chinese/12-certificates/data-visualization-certificate/data-visualization-certificate.yml
+++ b/curriculum/challenges/chinese/12-certificates/data-visualization-certificate/data-visualization-certificate.yml
@@ -1,50 +1,21 @@
---
id: 5a553ca864b52e1d8bceea14
+title: Data Visualization Certificate
challengeType: 7
isPrivate: true
-title: 数据可视化证书
----
-
-## Description
-<section id='description'>
-
-</section>
-
-## Instructions
-<section id='instructions'>
-
-</section>
-
-## Tests
-<section id='tests'>
-
-```yml
tests:
- - id: bd7168d8c242eddfaeb5bd13
+ -
+ id: bd7168d8c242eddfaeb5bd13
title: Visualize Data with a Bar Chart
- - id: bd7178d8c242eddfaeb5bd13
+ -
+ id: bd7178d8c242eddfaeb5bd13
title: Visualize Data with a Scatterplot Graph
- - id: bd7188d8c242eddfaeb5bd13
+ -
+ id: bd7188d8c242eddfaeb5bd13
title: Visualize Data with a Heat Map
- - id: 587d7fa6367417b2b2512bbf
+ -
+ id: 587d7fa6367417b2b2512bbf
title: Visualize Data with a Choropleth Map
- - id: 587d7fa6367417b2b2512bc0
+ -
+ id: 587d7fa6367417b2b2512bc0
title: Visualize Data with a Treemap Diagram
-
-```
-
-</section>
-
-## Challenge Seed
-<section id='challengeSeed'>
-
-</section>
-
-## Solution
-<section id='solution'>
-
-```js
-// solution required
-```
-
-</section>
diff --git a/curriculum/challenges/chinese/12-certificates/front-end-libraries-certificate/front-end-libraries-certificate.yml b/curriculum/challenges/chinese/12-certificates/front-end-libraries-certificate/front-end-libraries-certificate.yml
index 6c6a0c30368..1f0a8f26894 100644
--- a/curriculum/challenges/chinese/12-certificates/front-end-libraries-certificate/front-end-libraries-certificate.yml
+++ b/curriculum/challenges/chinese/12-certificates/front-end-libraries-certificate/front-end-libraries-certificate.yml
@@ -1,50 +1,21 @@
---
id: 561acd10cb82ac38a17513bc
+title: Front End Libraries Certificate
challengeType: 7
isPrivate: true
-title: 前端库证书
----
-
-## Description
-<section id='description'>
-
-</section>
-
-## Instructions
-<section id='instructions'>
-
-</section>
-
-## Tests
-<section id='tests'>
-
-```yml
tests:
- - id: bd7158d8c442eddfaeb5bd13
+ -
+ id: bd7158d8c442eddfaeb5bd13
title: Build a Random Quote Machine
- - id: bd7157d8c242eddfaeb5bd13
+ -
+ id: bd7157d8c242eddfaeb5bd13
title: Build a Markdown Previewer
- - id: 587d7dbc367417b2b2512bae
+ -
+ id: 587d7dbc367417b2b2512bae
title: Build a Drum Machine
- - id: bd7158d8c442eddfaeb5bd17
+ -
+ id: bd7158d8c442eddfaeb5bd17
title: Build a JavaScript Calculator
- - id: bd7158d8c442eddfaeb5bd0f
+ -
+ id: bd7158d8c442eddfaeb5bd0f
title: Build a 25 + 5 Clock
-
-```
-
-</section>
-
-## Challenge Seed
-<section id='challengeSeed'>
-
-</section>
-
-## Solution
-<section id='solution'>
-
-```js
-// solution required
-```
-
-</section>
diff --git a/curriculum/challenges/chinese/12-certificates/information-security-certificate/information-security-certificate.yml b/curriculum/challenges/chinese/12-certificates/information-security-certificate/information-security-certificate.yml
index 6a2b82a735d..a6339cfb861 100644
--- a/curriculum/challenges/chinese/12-certificates/information-security-certificate/information-security-certificate.yml
+++ b/curriculum/challenges/chinese/12-certificates/information-security-certificate/information-security-certificate.yml
@@ -1,49 +1,21 @@
---
id: 5e6021435ac9d0ecd8b94b00
+title: Information Security Certificate
challengeType: 7
isPrivate: true
-title: ''
----
-
-## Description
-<section id='description'>
-
-</section>
-
-## Instructions
-<section id='instructions'>
-
-</section>
-
-## Tests
-<section id='tests'>
-
-```yml
tests:
- - id: 587d824a367417b2b2512c44
+ -
+ id: 587d824a367417b2b2512c44
title: Stock Price Checker
- - id: 587d824a367417b2b2512c45
+ -
+ id: 587d824a367417b2b2512c45
title: Anonymous Message Board
- - id: 5e46f979ac417301a38fb932
+ -
+ id: 5e46f979ac417301a38fb932
title: Port Scanner
- - id: 5e46f983ac417301a38fb933
+ -
+ id: 5e46f983ac417301a38fb933
title: SHA-1 Password Cracker
- - id: 5e601c775ac9d0ecd8b94aff
+ -
+ id: 5e601c775ac9d0ecd8b94aff
title: Secure Real Time Multiplayer Game
-```
-
-</section>
-
-## Challenge Seed
-<section id='challengeSeed'>
-
-</section>
-
-## Solution
-<section id='solution'>
-
-```js
-// solution required
-```
-
-</section>
diff --git a/curriculum/challenges/chinese/12-certificates/javascript-algorithms-and-data-structures-certificate/javascript-algorithms-and-data-structures-certificate.yml b/curriculum/challenges/chinese/12-certificates/javascript-algorithms-and-data-structures-certificate/javascript-algorithms-and-data-structures-certificate.yml
index 7d5046ff378..aebebf678cf 100644
--- a/curriculum/challenges/chinese/12-certificates/javascript-algorithms-and-data-structures-certificate/javascript-algorithms-and-data-structures-certificate.yml
+++ b/curriculum/challenges/chinese/12-certificates/javascript-algorithms-and-data-structures-certificate/javascript-algorithms-and-data-structures-certificate.yml
@@ -1,50 +1,21 @@
---
id: 561abd10cb81ac38a17513bc
+title: JavaScript Algorithms and Data Structures Certificate
challengeType: 7
isPrivate: true
-title: JavaScript算法和数据结构证书
----
-
-## Description
-<section id='description'>
-
-</section>
-
-## Instructions
-<section id='instructions'>
-
-</section>
-
-## Tests
-<section id='tests'>
-
-```yml
tests:
- - id: aaa48de84e1ecc7c742e1124
+ -
+ id: aaa48de84e1ecc7c742e1124
title: Palindrome Checker
- - id: a7f4d8f2483413a6ce226cac
+ -
+ id: a7f4d8f2483413a6ce226cac
title: Roman Numeral Converter
- - id: 56533eb9ac21ba0edf2244e2
+ -
+ id: 56533eb9ac21ba0edf2244e2
title: Caesars Cipher
- - id: aff0395860f5d3034dc0bfc9
+ -
+ id: aff0395860f5d3034dc0bfc9
title: Telephone Number Validator
- - id: aa2e6f85cab2ab736c9a9b24
+ -
+ id: aa2e6f85cab2ab736c9a9b24
title: Cash Register
-
-```
-
-</section>
-
-## Challenge Seed
-<section id='challengeSeed'>
-
-</section>
-
-## Solution
-<section id='solution'>
-
-```js
-// solution required
-```
-
-</section>
diff --git a/curriculum/challenges/chinese/12-certificates/legacy-back-end-certificate/legacy-back-end-certificate.yml b/curriculum/challenges/chinese/12-certificates/legacy-back-end-certificate/legacy-back-end-certificate.yml
index 20235b31cac..d48ecd7921b 100644
--- a/curriculum/challenges/chinese/12-certificates/legacy-back-end-certificate/legacy-back-end-certificate.yml
+++ b/curriculum/challenges/chinese/12-certificates/legacy-back-end-certificate/legacy-back-end-certificate.yml
@@ -1,60 +1,36 @@
---
id: 660add10cb82ac38a17513be
+title: Legacy Back End Certificate
challengeType: 7
isPrivate: true
-title: 旧版后端证书
----
-
-## Description
-<section id='description'>
-
-</section>
-
-## Instructions
-<section id='instructions'>
-
-</section>
-
-## Tests
-<section id='tests'>
-
-```yml
tests:
- - id: bd7158d8c443edefaeb5bdef
+ -
+ id: bd7158d8c443edefaeb5bdef
title: Timestamp Microservice
- - id: bd7158d8c443edefaeb5bdff
+ -
+ id: bd7158d8c443edefaeb5bdff
title: Request Header Parser Microservice
- - id: bd7158d8c443edefaeb5bd0e
+ -
+ id: bd7158d8c443edefaeb5bd0e
title: URL Shortener Microservice
- - id: bd7158d8c443edefaeb5bdee
+ -
+ id: bd7158d8c443edefaeb5bdee
title: Image Search Abstraction Layer
- - id: bd7158d8c443edefaeb5bd0f
+ -
+ id: bd7158d8c443edefaeb5bd0f
title: File Metadata Microservice
- - id: bd7158d8c443eddfaeb5bdef
+ -
+ id: bd7158d8c443eddfaeb5bdef
title: Build a Voting App
- - id: bd7158d8c443eddfaeb5bdff
+ -
+ id: bd7158d8c443eddfaeb5bdff
title: Build a Nightlife Coordination App
- - id: bd7158d8c443eddfaeb5bd0e
+ -
+ id: bd7158d8c443eddfaeb5bd0e
title: Chart the Stock Market
- - id: bd7158d8c443eddfaeb5bd0f
+ -
+ id: bd7158d8c443eddfaeb5bd0f
title: Manage a Book Trading Club
- - id: bd7158d8c443eddfaeb5bdee
+ -
+ id: bd7158d8c443eddfaeb5bdee
title: Build a Pinterest Clone
-
-```
-
-</section>
-
-## Challenge Seed
-<section id='challengeSeed'>
-
-</section>
-
-## Solution
-<section id='solution'>
-
-```js
-// solution required
-```
-
-</section>
diff --git a/curriculum/challenges/chinese/12-certificates/legacy-data-visualization-certificate/legacy-data-visualization-certificate.yml b/curriculum/challenges/chinese/12-certificates/legacy-data-visualization-certificate/legacy-data-visualization-certificate.yml
index dd040b0160e..1508cf30d86 100644
--- a/curriculum/challenges/chinese/12-certificates/legacy-data-visualization-certificate/legacy-data-visualization-certificate.yml
+++ b/curriculum/challenges/chinese/12-certificates/legacy-data-visualization-certificate/legacy-data-visualization-certificate.yml
@@ -1,60 +1,36 @@
---
id: 561add10cb82ac39a17513bc
+title: Legacy Data Visualization Certificate
challengeType: 7
isPrivate: true
-title: 旧数据可视化证书
----
-
-## Description
-<section id='description'>
-
-</section>
-
-## Instructions
-<section id='instructions'>
-
-</section>
-
-## Tests
-<section id='tests'>
-
-```yml
tests:
- - id: bd7157d8c242eddfaeb5bd13
+ -
+ id: bd7157d8c242eddfaeb5bd13
title: Build a Markdown Previewer
- - id: bd7156d8c242eddfaeb5bd13
+ -
+ id: bd7156d8c242eddfaeb5bd13
title: Build a Camper Leaderboard
- - id: bd7155d8c242eddfaeb5bd13
+ -
+ id: bd7155d8c242eddfaeb5bd13
title: Build a Recipe Box
- - id: bd7154d8c242eddfaeb5bd13
+ -
+ id: bd7154d8c242eddfaeb5bd13
title: Build the Game of Life
- - id: bd7153d8c242eddfaeb5bd13
+ -
+ id: bd7153d8c242eddfaeb5bd13
title: Build a Roguelike Dungeon Crawler Game
- - id: bd7168d8c242eddfaeb5bd13
+ -
+ id: bd7168d8c242eddfaeb5bd13
title: Visualize Data with a Bar Chart
- - id: bd7178d8c242eddfaeb5bd13
+ -
+ id: bd7178d8c242eddfaeb5bd13
title: Visualize Data with a Scatterplot Graph
- - id: bd7188d8c242eddfaeb5bd13
+ -
+ id: bd7188d8c242eddfaeb5bd13
title: Visualize Data with a Heat Map
- - id: bd7198d8c242eddfaeb5bd13
+ -
+ id: bd7198d8c242eddfaeb5bd13
title: Show National Contiguity with a Force Directed Graph
- - id: bd7108d8c242eddfaeb5bd13
+ -
+ id: bd7108d8c242eddfaeb5bd13
title: Map Data Across the Globe
-
-```
-
-</section>
-
-## Challenge Seed
-<section id='challengeSeed'>
-
-</section>
-
-## Solution
-<section id='solution'>
-
-```js
-// solution required
-```
-
-</section>
diff --git a/curriculum/challenges/chinese/12-certificates/legacy-front-end-certificate/legacy-front-end-certificate.yml b/curriculum/challenges/chinese/12-certificates/legacy-front-end-certificate/legacy-front-end-certificate.yml
index 83598a605fb..93d72c9b7f8 100644
--- a/curriculum/challenges/chinese/12-certificates/legacy-front-end-certificate/legacy-front-end-certificate.yml
+++ b/curriculum/challenges/chinese/12-certificates/legacy-front-end-certificate/legacy-front-end-certificate.yml
@@ -1,60 +1,36 @@
---
id: 561add10cb82ac38a17513be
+title: Legacy Front End Certificate
challengeType: 7
isPrivate: true
-title: 传统前端证书
----
-
-## Description
-<section id='description'>
-
-</section>
-
-## Instructions
-<section id='instructions'>
-
-</section>
-
-## Tests
-<section id='tests'>
-
-```yml
tests:
- - id: bd7158d8c242eddfaeb5bd13
+ -
+ id: bd7158d8c242eddfaeb5bd13
title: Build a Personal Portfolio Webpage
- - id: bd7158d8c442eddfaeb5bd13
+ -
+ id: bd7158d8c442eddfaeb5bd13
title: Build a Random Quote Machine
- - id: bd7158d8c442eddfaeb5bd0f
+ -
+ id: bd7158d8c442eddfaeb5bd0f
title: Build a 25 + 5 Clock
- - id: bd7158d8c442eddfaeb5bd17
+ -
+ id: bd7158d8c442eddfaeb5bd17
title: Build a JavaScript Calculator
- - id: bd7158d8c442eddfaeb5bd10
+ -
+ id: bd7158d8c442eddfaeb5bd10
title: Show the Local Weather
- - id: bd7158d8c442eddfaeb5bd1f
+ -
+ id: bd7158d8c442eddfaeb5bd1f
title: Use the Twitch JSON API
- - id: bd7158d8c442eddfaeb5bd18
+ -
+ id: bd7158d8c442eddfaeb5bd18
title: Stylize Stories on Camper News
- - id: bd7158d8c442eddfaeb5bd19
+ -
+ id: bd7158d8c442eddfaeb5bd19
title: Build a Wikipedia Viewer
- - id: bd7158d8c442eedfaeb5bd1c
+ -
+ id: bd7158d8c442eedfaeb5bd1c
title: Build a Tic Tac Toe Game
- - id: bd7158d8c442eddfaeb5bd1c
+ -
+ id: bd7158d8c442eddfaeb5bd1c
title: Build a Simon Game
-
-```
-
-</section>
-
-## Challenge Seed
-<section id='challengeSeed'>
-
-</section>
-
-## Solution
-<section id='solution'>
-
-```js
-// solution required
-```
-
-</section>
diff --git a/curriculum/challenges/chinese/12-certificates/legacy-full-stack-certificate/legacy-full-stack-certificate.yml b/curriculum/challenges/chinese/12-certificates/legacy-full-stack-certificate/legacy-full-stack-certificate.yml
index a3be1a12a4d..97e2fcfc9d3 100644
--- a/curriculum/challenges/chinese/12-certificates/legacy-full-stack-certificate/legacy-full-stack-certificate.yml
+++ b/curriculum/challenges/chinese/12-certificates/legacy-full-stack-certificate/legacy-full-stack-certificate.yml
@@ -1,52 +1,24 @@
---
id: 561add10cb82ac38a17213bd
+title: Legacy Full Stack Certificate
challengeType: 7
isPrivate: true
-title: ''
----
-
-## Description
-<section id='description'>
-
-</section>
-
-## Instructions
-<section id='instructions'>
-
-</section>
-
-## Tests
-<section id='tests'>
-
-```yml
tests:
- - id: 561add10cb82ac38a17513bc
+ -
+ id: 561add10cb82ac38a17513bc
title: Responsive Web Design Certificate
- - id: 561abd10cb81ac38a17513bc
+ -
+ id: 561abd10cb81ac38a17513bc
title: JavaScript Algorithms and Data Structures Certificate
- - id: 561acd10cb82ac38a17513bc
+ -
+ id: 561acd10cb82ac38a17513bc
title: Front End Libraries Certificate
- - id: 5a553ca864b52e1d8bceea14
+ -
+ id: 5a553ca864b52e1d8bceea14
title: Data Visualization Certificate
- - id: 561add10cb82ac38a17523bc
+ -
+ id: 561add10cb82ac38a17523bc
title: API's and Microservices Certificate
- - id: 561add10cb82ac38a17213bc
+ -
+ id: 561add10cb82ac38a17213bc
title: Legacy Information Security and Quality Assurance Certificate
-
-```
-
-</section>
-
-## Challenge Seed
-<section id='challengeSeed'>
-
-</section>
-
-## Solution
-<section id='solution'>
-
-```js
-// solution required
-```
-
-</section>
diff --git a/curriculum/challenges/chinese/12-certificates/legacy-information-security-and-quality-assurance-certificate/legacy-information-security-and-quality-assurance-certificate.yml b/curriculum/challenges/chinese/12-certificates/legacy-information-security-and-quality-assurance-certificate/legacy-information-security-and-quality-assurance-certificate.yml
index 8bb7b2c97f2..1990a343edc 100644
--- a/curriculum/challenges/chinese/12-certificates/legacy-information-security-and-quality-assurance-certificate/legacy-information-security-and-quality-assurance-certificate.yml
+++ b/curriculum/challenges/chinese/12-certificates/legacy-information-security-and-quality-assurance-certificate/legacy-information-security-and-quality-assurance-certificate.yml
@@ -1,50 +1,21 @@
---
id: 561add10cb82ac38a17213bc
+title: Legacy Information Security and Quality Assurance Certificate
challengeType: 7
isPrivate: true
-title: ''
----
-
-## Description
-<section id='description'>
-
-</section>
-
-## Instructions
-<section id='instructions'>
-
-</section>
-
-## Tests
-<section id='tests'>
-
-```yml
tests:
- - id: 587d8249367417b2b2512c41
+ -
+ id: 587d8249367417b2b2512c41
title: Metric-Imperial Converter
- - id: 587d8249367417b2b2512c42
+ -
+ id: 587d8249367417b2b2512c42
title: Issue Tracker
- - id: 587d824a367417b2b2512c43
+ -
+ id: 587d824a367417b2b2512c43
title: Personal Library
- - id: 587d824a367417b2b2512c44
+ -
+ id: 587d824a367417b2b2512c44
title: Stock Price Checker
- - id: 587d824a367417b2b2512c45
+ -
+ id: 587d824a367417b2b2512c45
title: Anonymous Message Board
-
-```
-
-</section>
-
-## Challenge Seed
-<section id='challengeSeed'>
-
-</section>
-
-## Solution
-<section id='solution'>
-
-```js
-// solution required
-```
-
-</section>
diff --git a/curriculum/challenges/chinese/12-certificates/machine-learning-with-python-certificate/machine-learning-with-python-certificate.yml b/curriculum/challenges/chinese/12-certificates/machine-learning-with-python-certificate/machine-learning-with-python-certificate.yml
index d3535151cdb..266eb639481 100644
--- a/curriculum/challenges/chinese/12-certificates/machine-learning-with-python-certificate/machine-learning-with-python-certificate.yml
+++ b/curriculum/challenges/chinese/12-certificates/machine-learning-with-python-certificate/machine-learning-with-python-certificate.yml
@@ -1,49 +1,22 @@
---
id: 5e46fc95ac417301a38fb935
+title: Machine Learning with Python Certificate
challengeType: 7
isPrivate: true
-title: ''
----
-
-## Description
-<section id='description'>
-
-</section>
-
-## Instructions
-<section id='instructions'>
-
-</section>
-
-## Tests
-<section id='tests'>
-
-```yml
tests:
- - id: 5e46f8d6ac417301a38fb92d
+ -
+ id: 5e46f8d6ac417301a38fb92d
title: Rock Paper Scissors
- - id: 5e46f8dcac417301a38fb92e
+ -
+ id: 5e46f8dcac417301a38fb92e
title: Cat and Dog Image Classifier
- - id: 5e46f8e3ac417301a38fb92f
+ -
+ id: 5e46f8e3ac417301a38fb92f
title: Book Recommendation Engine using KNN
- - id: 5e46f8edac417301a38fb930
+ -
+ id: 5e46f8edac417301a38fb930
title: Linear Regression Health Costs Calculator
- - id: 5e46f8edac417301a38fb931
+ -
+ id: 5e46f8edac417301a38fb931
title: Neural Network SMS Text Classifier
-```
-</section>
-
-## Challenge Seed
-<section id='challengeSeed'>
-
-</section>
-
-## Solution
-<section id='solution'>
-
-```js
-// solution required
-```
-
-</section>
diff --git a/curriculum/challenges/chinese/12-certificates/quality-assurance-certificate/quality-assurance-certificate.yml b/curriculum/challenges/chinese/12-certificates/quality-assurance-certificate/quality-assurance-certificate.yml
index 302a11e1034..5e229ba55c0 100644
--- a/curriculum/challenges/chinese/12-certificates/quality-assurance-certificate/quality-assurance-certificate.yml
+++ b/curriculum/challenges/chinese/12-certificates/quality-assurance-certificate/quality-assurance-certificate.yml
@@ -1,49 +1,22 @@
---
id: 5e611829481575a52dc59c0e
+title: Quality Assurance Certificate
challengeType: 7
isPrivate: true
-title: ''
----
-
-## Description
-<section id='description'>
-
-</section>
-
-## Instructions
-<section id='instructions'>
-
-</section>
-
-## Tests
-<section id='tests'>
-
-```yml
tests:
- - id: 587d8249367417b2b2512c41
+ -
+ id: 587d8249367417b2b2512c41
title: Metric-Imperial Converter
- - id: 587d8249367417b2b2512c42
+ -
+ id: 587d8249367417b2b2512c42
title: Issue Tracker
- - id: 587d824a367417b2b2512c43
+ -
+ id: 587d824a367417b2b2512c43
title: Personal Library
- - id: 5e601bf95ac9d0ecd8b94afd
+ -
+ id: 5e601bf95ac9d0ecd8b94afd
title: Sudoku Solver
- - id: 5e601c0d5ac9d0ecd8b94afe
+ -
+ id: 5e601c0d5ac9d0ecd8b94afe
title: American British Translator
-```
-</section>
-
-## Challenge Seed
-<section id='challengeSeed'>
-
-</section>
-
-## Solution
-<section id='solution'>
-
-```js
-// solution required
-```
-
-</section>
diff --git a/curriculum/challenges/chinese/12-certificates/responsive-web-design-certificate/responsive-web-design-certificate.yml b/curriculum/challenges/chinese/12-certificates/responsive-web-design-certificate/responsive-web-design-certificate.yml
index fb9a60f6753..8fc00be7be7 100644
--- a/curriculum/challenges/chinese/12-certificates/responsive-web-design-certificate/responsive-web-design-certificate.yml
+++ b/curriculum/challenges/chinese/12-certificates/responsive-web-design-certificate/responsive-web-design-certificate.yml
@@ -1,50 +1,21 @@
---
id: 561add10cb82ac38a17513bc
+title: 响应式网页设计认证项目
challengeType: 7
isPrivate: true
-title: 响应式网页设计证书
----
-
-## Description
-<section id='description'>
-
-</section>
-
-## Instructions
-<section id='instructions'>
-
-</section>
-
-## Tests
-<section id='tests'>
-
-```yml
tests:
- - id: bd7158d8c442eddfaeb5bd18
- title: Build a Tribute Page
- - id: 587d78af367417b2b2512b03
- title: Build a Survey Form
- - id: 587d78af367417b2b2512b04
- title: Build a Product Landing Page
- - id: 587d78b0367417b2b2512b05
- title: Build a Technical Documentation Page
- - id: bd7158d8c242eddfaeb5bd13
- title: Build a Personal Portfolio Webpage
-
-```
-
-</section>
-
-## Challenge Seed
-<section id='challengeSeed'>
-
-</section>
-
-## Solution
-<section id='solution'>
-
-```js
-// solution required
-```
-
-</section>
+ -
+ id: bd7158d8c442eddfaeb5bd18
+ title: 制作一个致敬页
+ -
+ id: 587d78af367417b2b2512b03
+ title: 制作一个调查表格
+ -
+ id: 587d78af367417b2b2512b04
+ title: 制作一个产品登录页
+ -
+ id: 587d78b0367417b2b2512b05
+ title: 制作一个技术文档页面
+ -
+ id: bd7158d8c242eddfaeb5bd13
+ title: 制作一个个人作品集展示页
diff --git a/curriculum/challenges/chinese/12-certificates/scientific-computing-with-python-certificate/scientific-computing-with-python-certificate.yml b/curriculum/challenges/chinese/12-certificates/scientific-computing-with-python-certificate/scientific-computing-with-python-certificate.yml
index 91e964dcdf1..1f2f103e94b 100644
--- a/curriculum/challenges/chinese/12-certificates/scientific-computing-with-python-certificate/scientific-computing-with-python-certificate.yml
+++ b/curriculum/challenges/chinese/12-certificates/scientific-computing-with-python-certificate/scientific-computing-with-python-certificate.yml
@@ -1,50 +1,21 @@
---
id: 5e44431b903586ffb414c951
+title: Scientific Computing with Python Certificate
challengeType: 7
isPrivate: true
-title: ''
----
-
-## Description
-<section id='description'>
-
-</section>
-
-## Instructions
-<section id='instructions'>
-
-</section>
-
-## Tests
-<section id='tests'>
-
-```yml
tests:
- - id: 5e44412c903586ffb414c94c
+ -
+ id: 5e44412c903586ffb414c94c
title: Arithmetic Formatter
- - id: 5e444136903586ffb414c94d
+ -
+ id: 5e444136903586ffb414c94d
title: Time Calculator
- - id: 5e44413e903586ffb414c94e
+ -
+ id: 5e44413e903586ffb414c94e
title: Budget App
- - id: 5e444147903586ffb414c94f
+ -
+ id: 5e444147903586ffb414c94f
title: Polygon Area Calculator
- - id: 5e44414f903586ffb414c950
+ -
+ id: 5e44414f903586ffb414c950
title: Probability Calculator
-
-```
-
-</section>
-
-## Challenge Seed
-<section id='challengeSeed'>
-
-</section>
-
-## Solution
-<section id='solution'>
-
-```js
-// solution required
-```
-
-</section>
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/add-a-text-alternative-to-images-for-visually-impaired-accessibility.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/add-a-text-alternative-to-images-for-visually-impaired-accessibility.md
new file mode 100644
index 00000000000..719b77d1db7
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/add-a-text-alternative-to-images-for-visually-impaired-accessibility.md
@@ -0,0 +1,44 @@
+---
+id: 587d774c367417b2b2512a9c
+title: Agregar un texto alternativo a las imágenes para accesibilidad de usuarios con dificultades de la vista
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cPp7VfD'
+forumTopicId: 16628
+dashedName: add-a-text-alternative-to-images-for-visually-impaired-accessibility
+---
+
+# --description--
+
+Es probable que en otros desafíos hayas visto un atributo `alt` en una etiqueta `img`. El texto `alt` describe el contenido de la imagen y proporciona una descripción alternativa en formato de texto. Esto ayuda en casos en que la imagen no se carga o que el usuario no pueda verla. También es aprovechado por los motores de búsqueda para interpretar el contenido de la imagen e incluirlo en los resultados de búsqueda. Aquí hay un ejemplo:
+
+`<img src="importantLogo.jpeg" alt="Company logo">`
+
+Las personas con dificultades visuales dependen de lectores de pantalla para convertir el contenido web a una interfaz de audio. Por esta razón, no podrán recibir la información si solo se les presenta de manera visual. En el caso de las imágenes, los lectores de pantalla pueden acceder el atributo `alt` y leer su contenido para proporcionar información clave.
+
+Un buen texto `alt` le brinda al lector una breve descripción de la imagen. Siempre deberías incluir el atributo `alt` en tus imágenes. Además, según la especificación de HTML5, esto ahora se considera obligatorio.
+
+# --instructions--
+
+Camper Cat es tanto un ninja de la programación como un ninja de la vida real, y está construyendo un sitio web para compartir sus conocimientos. La foto de perfil que quiere usar es una muestra de sus habilidades y debería poder ser apreciada por todos los visitantes del sitio. Agrega un atributo `alt` en la etiqueta `img` para explicar que Camper Cat está practicando karate. (El atributo `src` de esta imagen no está enlazado a un archivo real, por lo que podrás ver el texto `alt` en la página.)
+
+# --hints--
+
+Tu etiqueta `img` debe tener un atributo `alt`, y el mismo no debería estar vacío.
+
+```js
+assert($('img').attr('alt'));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<img src="doingKarateWow.jpeg">
+```
+
+# --solutions--
+
+```html
+<img src="doingKarateWow.jpeg" alt="Someone doing karate">
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/add-an-accessible-date-picker.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/add-an-accessible-date-picker.md
new file mode 100644
index 00000000000..c63eaaa9d61
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/add-an-accessible-date-picker.md
@@ -0,0 +1,105 @@
+---
+id: 587d778b367417b2b2512aa8
+title: Agregar un selector de fechas accesible
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cR3bRbCV'
+forumTopicId: 301008
+dashedName: add-an-accessible-date-picker
+---
+
+# --description--
+
+Los formularios suelen incluir el campo `input`, que puede usarse para crear diferentes tipos de controles en los formularios. El atributo `type` en este elemento indica el tipo de elemento `input` a crear.
+
+Puede que hayas visto los tipos de campo `text` y `submit` en desafíos anteriores. HTML5 además introdujo una opción para especificar un campo `date` para fechas. Dependiendo del soporte de los navegadores, un selector de fechas debería aparecer cuando el campo `input` esté en foco, y esto hace mucho más sencillo para los usuarios cargar información en el formulario.
+
+Para los navegadores más antiguos, el tipo será por defecto `text`, por lo que ayuda a mostrar a los usuarios el formato de fecha(date) esperado en el texto `label` o `placeholder` por si acaso.
+
+Aquí hay un ejemplo:
+
+```html
+<label for="input1">Enter a date:</label>
+<input type="date" id="input1" name="input1">
+```
+
+# --instructions--
+
+Camper Cat está organizando un torneo de Mortal Kombat y quiere pedir a los participantes que consideren cuál fecha les resultaría mejor. Agrega una etiqueta `input` con un atributo `type` de `date`, un atributo `id` de `pickdate` y un atributo `name` con valor `date`.
+
+# --hints--
+
+Tu código debería añadir una etiqueta `input` para el campo de selección de fecha.
+
+```js
+assert($('input').length == 2);
+```
+
+Tu etiqueta `input` debe tener un atributo `type` con un valor de `date`.
+
+```js
+assert($('input').attr('type') == 'date');
+```
+
+Tu etiqueta `input` debe tener un atributo `id` con un valor de `pickdate`.
+
+```js
+assert($('input').attr('id') == 'pickdate');
+```
+
+Tu etiqueta `input` debe tener un atributo `name` con un valor de `date`.
+
+```js
+assert($('input').attr('name') == 'date');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<body>
+ <header>
+ <h1>Tournaments</h1>
+ </header>
+ <main>
+ <section>
+ <h2>Mortal Kombat Tournament Survey</h2>
+ <form>
+ <p>Tell us the best date for the competition</p>
+ <label for="pickdate">Preferred Date:</label>
+
+ <!-- Only change code below this line -->
+
+
+
+ <!-- Only change code above this line -->
+
+ <input type="submit" name="submit" value="Submit">
+ </form>
+ </section>
+ </main>
+ <footer>© 2018 Camper Cat</footer>
+</body>
+```
+
+# --solutions--
+
+```html
+<body>
+ <header>
+ <h1>Tournaments</h1>
+ </header>
+ <main>
+ <section>
+ <h2>Mortal Kombat Tournament Survey</h2>
+ <form>
+ <p>Tell us the best date for the competition</p>
+ <label for="pickdate">Preferred Date:</label>
+ <input type="date" id="pickdate" name="date">
+ <input type="submit" name="submit" value="Submit">
+ </form>
+ </section>
+ </main>
+ <footer>© 2018 Camper Cat</footer>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/avoid-colorblindness-issues-by-carefully-choosing-colors-that-convey-information.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/avoid-colorblindness-issues-by-carefully-choosing-colors-that-convey-information.md
new file mode 100644
index 00000000000..3b620dc5925
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/avoid-colorblindness-issues-by-carefully-choosing-colors-that-convey-information.md
@@ -0,0 +1,74 @@
+---
+id: 587d778f367417b2b2512aad
+title: >-
+ Evita problemas de color para usuarios daltónicos eligiendo cuidadosamente los colores que transmiten información
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c437as3'
+forumTopicId: 301011
+dashedName: >-
+ avoid-colorblindness-issues-by-carefully-choosing-colors-that-convey-information
+---
+
+# --description--
+
+Existen varias formas de daltonismo o ceguera al color. Estos pueden ir desde tener una sensibilidad reducida a una longitud de onda de luz específica, hasta la incapacidad total de ver algún color. La forma más común en que se presenta es una sensibilidad reducida para detectar tonos de verde.
+
+Por ejemplo, si dos colores verdes similares son el color usado en el primer plano y el fondo de tu contenido, un usuario daltónico podría no ser capaz de distinguirlos entre sí. Podemos pensar en los colores cercanos como aquellos colores que son vecinos o adyacentes en la rueda de color. Por este motivo, las combinaciones de estos colores deben evitarse cuando se usan para transmitir información importante.
+
+**Nota:** Algunas herramientas de selección de colores disponibles en Internet incluyen simulaciones visuales de cómo serían vistos por usuarios con diversos tipos de daltonismo. Estos son excelentes recursos que puedes aprovechar, sumados a las calculadoras para verificar el contraste de color que puedes encontrar en Internet.
+
+# --instructions--
+
+Camper Cat está probando diferentes estilos para un botón importante, pero el `background-color` amarillo (`#FFF33`) y el `color` de texto verde (`#33FF33`) son tonos vecinos en la rueda de color, por lo que resultan prácticamente indistinguibles para algunos usuarios daltónicos. (Además, como tienen un nivel de luminosidad similar, no pasan la verificación de índice de contraste o contrast ratio). Cambia el `color` del texto a azul oscuro (`#003366`) para resolver ambos problemas.
+
+# --hints--
+
+Tu código debe cambiar el `color` del texto en el `button` al color azul oscuro.
+
+```js
+assert($('button').css('color') == 'rgb(0, 51, 102)');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<head>
+ <style>
+ button {
+ color: #33FF33;
+ background-color: #FFFF33;
+ font-size: 14px;
+ padding: 10px;
+ }
+ </style>
+</head>
+<body>
+ <header>
+ <h1>Danger!</h1>
+ </header>
+ <button>Delete Internet</button>
+</body>
+```
+
+# --solutions--
+
+```html
+<head>
+ <style>
+ button {
+ color: #003366;
+ background-color: #FFFF33;
+ font-size: 14px;
+ padding: 10px;
+ }
+ </style>
+</head>
+<body>
+ <header>
+ <h1>Danger!</h1>
+ </header>
+ <button>Delete Internet</button>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/avoid-colorblindness-issues-by-using-sufficient-contrast.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/avoid-colorblindness-issues-by-using-sufficient-contrast.md
new file mode 100644
index 00000000000..5a1a4632c7b
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/avoid-colorblindness-issues-by-using-sufficient-contrast.md
@@ -0,0 +1,84 @@
+---
+id: 587d778f367417b2b2512aac
+title: Evita problemas de percepción del color usando el suficiente contraste
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cmzMEUw'
+forumTopicId: 301012
+dashedName: avoid-colorblindness-issues-by-using-sufficient-contrast
+---
+
+# --description--
+
+El color es una parte importante del diseño visual, pero su aplicación presenta dos problemas de accesibilidad. El primer problema es que no se debe utilizar el color como la única forma de transmitir información importante, porque los lectores de pantalla no lo distinguen. En segundo lugar, los colores de primer plano y de fondo necesitan tener suficiente contraste para que los usuarios sean capaces de distinguirlos.
+
+Los desafíos anteriores explicaron la existencia de texto alternativo para remediar el primer problema. El último desafío introdujo herramientas de comprobación de contraste para ayudar con el segundo problema. El índice de contraste de 4.5:1 recomendado por WCAG se aplica tanto para los colores como para combinaciones de escalas de grises.
+
+Los usuarios daltónicos tienen problemas para distinguir algunos colores de otros. Esto generalmente depende del tono pero también depende a veces de su luminosidad. Posiblemente recuerdes que el índice de contraste se calcula utilizando los valores de luminancia (o luminosidad) relativa de los colores de primer plano y del fondo.
+
+En la práctica, la relación de contraste de 4.5:1 puede alcanzarse oscureciendo el color más oscuro (o sea, añadiendo negro) y aclarando el color más claro (añadiéndole blanco). Se considera que los tonos más oscuros en la rueda de color son los azules, violetas, magentas y los rojos, mientras que los colores más claros son los naranjas, amarillos, verdes y azul-verdosos.
+
+# --instructions--
+
+Camper Cat está experimentando con usar color en el texto y fondo de su blog, pero su combinación actual de `background-color` verdoso con `color` de texto granate tiene un índice de contraste de 2.5:1. Puedes ajustar fácilmente la luminosidad de los colores, ya que los declaró usando la propiedad CSS `hsl()` (que significa tono, saturación, luminosidad o "hue, saturation, lightness") cambiando solo el tercer argumento. Aumenta el valor de luminosidad del `background-color` de 35% a 55%, y disminuye el valor de luminosidad del `color` del texto del 20% al 15%. Esto mejora el índice de contraste llevándolo a 5.9:1.
+
+# --hints--
+
+Tu código solo debe cambiar el valor de luminosidad para la propiedad de `color` del texto a un valor del 15%.
+
+```js
+assert(code.match(/color:\s*?hsl\(0,\s*?55%,\s*?15%\)/gi));
+```
+
+Tu código solo debe cambiar el valor de luminosidad para la propiedad de `background-color` a un valor del 55%.
+
+```js
+assert(code.match(/background-color:\s*?hsl\(120,\s*?25%,\s*?55%\)/gi));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<head>
+ <style>
+ body {
+ color: hsl(0, 55%, 20%);
+ background-color: hsl(120, 25%, 35%);
+ }
+ </style>
+</head>
+<body>
+ <header>
+ <h1>Deep Thoughts with Master Camper Cat</h1>
+ </header>
+ <article>
+ <h2>A Word on the Recent Catnip Doping Scandal</h2>
+ <p>The influence that catnip has on feline behavior is well-documented, and its use as an herbal supplement in competitive ninja circles remains controversial. Once again, the debate to ban the substance is brought to the public's attention after the high-profile win of Kittytron, a long-time proponent and user of the green stuff, at the Claw of Fury tournament.</p>
+ <p>As I've stated in the past, I firmly believe a true ninja's skills must come from within, with no external influences. My own catnip use shall continue as purely recreational.</p>
+ </article>
+</body>
+```
+
+# --solutions--
+
+```html
+<head>
+ <style>
+ body {
+ color: hsl(0, 55%, 15%);
+ background-color: hsl(120, 25%, 55%);
+ }
+ </style>
+</head>
+<body>
+ <header>
+ <h1>Deep Thoughts with Master Camper Cat</h1>
+ </header>
+ <article>
+ <h2>A Word on the Recent Catnip Doping Scandal</h2>
+ <p>The influence that catnip has on feline behavior is well-documented, and its use as an herbal supplement in competitive ninja circles remains controversial. Once again, the debate to ban the substance is brought to the public's attention after the high-profile win of Kittytron, a long-time proponent and user of the green stuff, at the Claw of Fury tournament.</p>
+ <p>As I've stated in the past, I firmly believe a true ninja's skills must come from within, with no external influences. My own catnip use shall continue as purely recreational.</p>
+ </article>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/give-links-meaning-by-using-descriptive-link-text.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/give-links-meaning-by-using-descriptive-link-text.md
new file mode 100644
index 00000000000..cf95bb0779c
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/give-links-meaning-by-using-descriptive-link-text.md
@@ -0,0 +1,75 @@
+---
+id: 587d778f367417b2b2512aae
+title: Dar significado a los enlaces agregando un texto descriptivo
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c437DcV'
+forumTopicId: 301013
+dashedName: give-links-meaning-by-using-descriptive-link-text
+---
+
+# --description--
+
+Los lectores de pantalla tienen diferentes opciones según el tipo de contenido que lea el dispositivo del usuario. Esto incluye saltar a (o pasar por alto) elementos importantes, saltar al contenido principal u obtener un resumen de la página a partir de los títulos. Otra opción es escuchar la narración de los enlaces disponibles en una página.
+
+Los lectores de pantalla hacen esto leyendo el texto del enlace, o lo que haya entre las etiquetas anchor (`a`). Tener una lista de enlaces que solo digan "clic aquí" o "leer más" no ayuda. En lugar de eso, debes utilizar un texto breve pero descriptivo dentro de las etiquetas `a` para proporcionar más significado a estos usuarios.
+
+# --instructions--
+
+El texto de los enlaces que utiliza el Camper Cat no es muy descriptivo si se lo toma fuera de su contexto. Mueve las etiquetas anchor (`a`) para que envuelvan el texto `information about batteries` en lugar de `Click here`.
+
+# --hints--
+
+Tu código debe mover las etiquetas anchor `a` de las palabras `Click here` para envolver las palabras `information about batteries`.
+
+```js
+assert(
+ $('a')
+ .text()
+ .match(/^(information about batteries)$/g)
+);
+```
+
+El elemento `a` debe tener un atributo `href` con un valor de cadena vacía `""`.
+
+```js
+assert($('a').attr('href') === '');
+```
+
+Tu elemento `a` debe tener una etiqueta de cierre.
+
+```js
+assert(
+ code.match(/<\/a>/g) &&
+ code.match(/<\/a>/g).length === code.match(/<a href=(''|"")>/g).length
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<body>
+ <header>
+ <h1>Deep Thoughts with Master Camper Cat</h1>
+ </header>
+ <article>
+ <h2>Defeating your Foe: the Red Dot is Ours!</h2>
+ <p>Felines the world over have been waging war on the most persistent of foes. This red nemesis combines both cunning stealth and lightning speed. But chin up, fellow fighters, our time for victory may soon be near. <a href="">Click here</a> for information about batteries</p>
+ </article>
+</body>
+```
+
+# --solutions--
+
+```html
+<body>
+ <header>
+ <h1>Deep Thoughts with Master Camper Cat</h1>
+ </header>
+ <article>
+ <h2>Defeating your Foe: the Red Dot is Ours!</h2>
+ <p>Felines the world over have been waging war on the most persistent of foes. This red nemesis combines both cunning stealth and lightning speed. But chin up, fellow fighters, our time for victory may soon be near. Click here for <a href="">information about batteries</a></p>
+ </article>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/improve-accessibility-of-audio-content-with-the-audio-element.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/improve-accessibility-of-audio-content-with-the-audio-element.md
new file mode 100644
index 00000000000..8655ecd0203
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/improve-accessibility-of-audio-content-with-the-audio-element.md
@@ -0,0 +1,115 @@
+---
+id: 587d7789367417b2b2512aa4
+title: Mejorar la accesibilidad del contenido de audio con el elemento de audio
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cVJVkcZ'
+forumTopicId: 301014
+dashedName: improve-accessibility-of-audio-content-with-the-audio-element
+---
+
+# --description--
+
+El elemento `audio` de HTLM le da un significado semántico cuando contiene sonido o contenido de flujo de audio en el código. El contenido del audio también necesita un texto alternativo para que las personas sordas o con dificultad para escuchar puedan acceder al mismo. Para esto debe tener un texto cercano en la página o un enlace a una transcripción.
+
+La etiqueta de `audio` suporta los atributos `controls`. Esto muestra los controles por defecto de reproducir, pausar, entre otros controles, y soporta la funcionalidad del teclado. Este es un atributo booleano por lo que no necesita un valor, su presencia en la etiqueta activa la configuración.
+
+Acá tenemos un ejemplo:
+
+```html
+<audio id="meowClip" controls>
+ <source src="audio/meow.mp3" type="audio/mpeg" />
+ <source src="audio/meow.ogg" type="audio/ogg" />
+</audio>
+```
+
+**Note:** El contenido multimedia generalmente tiene componentes visuales y de audio. Se necesita sincronizar los subtítulos y una transcripción para que los usuarios con dificultades de vista o con problemas para escuchar puedan tener acceso a las mismas. Por lo general, un desarrollador de web no se responsabiliza de la creación de subtítulos o transcripciones pero necesita saber para incluirlos.
+
+# --instructions--
+
+Es hora de salir del Camper Cat y conocer a nuestro compañero Zersiax (@zersiax), un campión en accesibilidad y un usuario de lector de pantalla. Para escuchar un clip de su lector de pantalla en acción y añadir el elemento `audio` después de `p`. Incluye el atributo `controls`. A continuación, coloca una etiqueta `source` dentro de las etiquetas `audio` con el atributo `src` establecido en `https://s3.amazonaws.com/freecodecamp/screen-reader.mp3` y el atributo `type` establecido en `"audio/mpeg"`.
+
+**Note:** El clip del audio puede sonar rápido y quizás sea difícil de entender pero esta velocidad es normal para los lectores de pantalla.
+
+# --hints--
+
+Tu código debe tener una etiqueta `audio`.
+
+```js
+assert($('audio').length === 1);
+```
+
+El elemento `audio` debe tener una etiqueta de cierre.
+
+```js
+assert(
+ code.match(/<\/audio>/g).length === 1 &&
+ code.match(/<audio.*>[\s\S]*<\/audio>/g)
+);
+```
+
+La etiqueta `audio` debe tener el atributo `controls`.
+
+```js
+assert($('audio').attr('controls'));
+```
+
+Tu código debe tener una etiqueta `source`.
+
+```js
+assert($('source').length === 1);
+```
+
+La etiqueta `source` debe estar adentro de las etiquetas `audio`.
+
+```js
+assert($('audio').children('source').length === 1);
+```
+
+El valor para el atributo `src` en `source` debe coincidir exactamente con el link en las instrucciones.
+
+```js
+assert(
+ $('source').attr('src') ===
+ 'https://s3.amazonaws.com/freecodecamp/screen-reader.mp3'
+);
+```
+
+El código debe incluir un atributo `type` en la etiqueta `source` con un valor de audio/mpeg.
+
+```js
+assert($('source').attr('type') === 'audio/mpeg');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<body>
+ <header>
+ <h1>Real Coding Ninjas</h1>
+ </header>
+ <main>
+ <p>A sound clip of Zersiax's screen reader in action.</p>
+
+
+
+ </main>
+</body>
+```
+
+# --solutions--
+
+```html
+<body>
+ <header>
+ <h1>Real Coding Ninjas</h1>
+ </header>
+ <main>
+ <p>A sound clip of Zersiax's screen reader in action.</p>
+ <audio controls>
+ <source src="https://s3.amazonaws.com/freecodecamp/screen-reader.mp3" type="audio/mpeg"/>
+ </audio>
+ </main>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/improve-chart-accessibility-with-the-figure-element.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/improve-chart-accessibility-with-the-figure-element.md
new file mode 100644
index 00000000000..58e9c95ec5f
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/improve-chart-accessibility-with-the-figure-element.md
@@ -0,0 +1,161 @@
+---
+id: 587d778a367417b2b2512aa5
+title: Mejora la accesibilidad de gráficos con el elemento figure
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cGJMqtE'
+forumTopicId: 301015
+dashedName: improve-chart-accessibility-with-the-figure-element
+---
+
+# --description--
+
+HTML5 introdujo el elemento `figure` junto con el elemento relacionado `figcaption`. Cuando se usan juntos, estos elementos envuelven una representación visual (como puede ser una imagen, diagrama o gráfico) junto con su leyenda. Esto representa una mejora doble de accesibilidad, porque agrupa contenido semánticamente relacionado y además proporciona un texto alternativo que explica el contenido visual del elemento `figure`.
+
+En visualizaciones de datos como gráficos, la leyenda o "figcaption" puede ser utilizada para resumir en formato de texto las tendencias o conclusiones, para beneficio de usuarios con discapacidades visuales. Otro de los desafíos se ocupa cómo mover fuera de la pantalla una versión en formato de tabla con los datos del gráfico (usando CSS) para ayudar a usuarios de lectores de pantalla.
+
+Aquí hay un ejemplo: Ten en cuenta que el elemento `figcaption` va dentro de las etiquetas `figure` y se puede combinar con otros elementos:
+
+```html
+<figure>
+ <img src="roundhouseDestruction.jpeg" alt="Photo of Camper Cat executing a roundhouse kick">
+ <br>
+ <figcaption>
+ Master Camper Cat demonstrates proper form of a roundhouse kick.
+ </figcaption>
+</figure>
+```
+
+# --instructions--
+
+Camper Cat está trabajando para crear un gráfico de barras apiladas que muestre la cantidad de tiempo semanal a dedicar en entrenamiento en sigilo, combate y armas. Ayúdalo a estructurar mejor su página cambiando la etiqueta `div` que usó por una etiqueta `figure`, y cambiando la etiqueta `p` que rodea la leyenda por una etiqueta `figcaption`.
+
+# --hints--
+
+Tu código debe tener una etiqueta `figure`.
+
+```js
+assert($('figure').length == 1);
+```
+
+Tu código debe tener una etiqueta `figcaption`.
+
+```js
+assert($('figcaption').length == 1);
+```
+
+Tu código no debe tener ninguna etiqueta `div`.
+
+```js
+assert($('div').length == 0);
+```
+
+Tu código no debe tener ninguna etiqueta `p`.
+
+```js
+assert($('p').length == 0);
+```
+
+La etiqueta `figcaption` debe ser un hijo de la etiqueta `figure`.
+
+```js
+assert($('figure').children('figcaption').length == 1);
+```
+
+Tu elemento `figure` debe tener una etiqueta de cierre.
+
+```js
+assert(
+ code.match(/<\/figure>/g) &&
+ code.match(/<\/figure>/g).length === code.match(/<figure>/g).length
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<body>
+ <header>
+ <h1>Training</h1>
+ <nav>
+ <ul>
+ <li><a href="#stealth">Stealth & Agility</a></li>
+ <li><a href="#combat">Combat</a></li>
+ <li><a href="#weapons">Weapons</a></li>
+ </ul>
+ </nav>
+ </header>
+ <main>
+ <section>
+
+ <!-- Only change code below this line -->
+ <div>
+ <!-- Stacked bar chart will go here -->
+ <br>
+ <p>Breakdown per week of time to spend training in stealth, combat, and weapons.</p>
+ </div>
+ <!-- Only change code above this line -->
+
+ </section>
+ <section id="stealth">
+ <h2>Stealth & Agility Training</h2>
+ <article><h3>Climb foliage quickly using a minimum spanning tree approach</h3></article>
+ <article><h3>No training is NP-complete without parkour</h3></article>
+ </section>
+ <section id="combat">
+ <h2>Combat Training</h2>
+ <article><h3>Dispatch multiple enemies with multithreaded tactics</h3></article>
+ <article><h3>Goodbye world: 5 proven ways to knock out an opponent</h3></article>
+ </section>
+ <section id="weapons">
+ <h2>Weapons Training</h2>
+ <article><h3>Swords: the best tool to literally divide and conquer</h3></article>
+ <article><h3>Breadth-first or depth-first in multi-weapon training?</h3></article>
+ </section>
+ </main>
+ <footer>© 2018 Camper Cat</footer>
+</body>
+```
+
+# --solutions--
+
+```html
+<body>
+ <header>
+ <h1>Training</h1>
+ <nav>
+ <ul>
+ <li><a href="#stealth">Stealth & Agility</a></li>
+ <li><a href="#combat">Combat</a></li>
+ <li><a href="#weapons">Weapons</a></li>
+ </ul>
+ </nav>
+ </header>
+ <main>
+ <section>
+ <figure>
+ <!-- Stacked bar chart will go here -->
+ <br>
+ <figcaption>Breakdown per week of time to spend training in stealth, combat, and weapons.</figcaption>
+ </figure>
+ </section>
+ <section id="stealth">
+ <h2>Stealth & Agility Training</h2>
+ <article><h3>Climb foliage quickly using a minimum spanning tree approach</h3></article>
+ <article><h3>No training is NP-complete without parkour</h3></article>
+ </section>
+ <section id="combat">
+ <h2>Combat Training</h2>
+ <article><h3>Dispatch multiple enemies with multithreaded tactics</h3></article>
+ <article><h3>Goodbye world: 5 proven ways to knock out an opponent</h3></article>
+ </section>
+ <section id="weapons">
+ <h2>Weapons Training</h2>
+ <article><h3>Swords: the best tool to literally divide and conquer</h3></article>
+ <article><h3>Breadth-first or depth-first in multi-weapon training?</h3></article>
+ </section>
+ </main>
+ <footer>© 2018 Camper Cat</footer>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/improve-form-field-accessibility-with-the-label-element.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/improve-form-field-accessibility-with-the-label-element.md
new file mode 100644
index 00000000000..044ca09f7ac
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/improve-form-field-accessibility-with-the-label-element.md
@@ -0,0 +1,119 @@
+---
+id: 587d778a367417b2b2512aa6
+title: Mejorar la accesibilidad del campo de formulario con la etiqueta element
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cGJMMAN'
+forumTopicId: 301016
+dashedName: improve-form-field-accessibility-with-the-label-element
+---
+
+# --description--
+
+La mejora de la accesibilidad con el marcado HTML semántico se aplica tanto al uso de nombres de etiquetas apropiados como de atributos. Los próximos desafíos cubren algunos escenarios importantes utilizando atributos en formularios.
+
+La etiqueta `label` contiene el texto para un elemento específico de control de formulario, por lo general el nombre o etiqueta para una elección. Esto vincula el significado al elemento y hace que el formulario se lea mejor. El atributo `for` en una etiqueta `label` asocia de manera explícita dicho `label` con el control de formulario utilizado por los lectores de pantalla.
+
+Ya aprendiste sobre botones de radio y sus etiquetas en una lección de la sección de HTML básico. En esa lección, colocamos el elemento de entrada del botón de radio dentro de la etiqueta `label` junto con la etiqueta del texto para hacer que el texto se pueda cliquear. También, se puede utilizar el atributo `for` como se explicó en esta lección.
+
+El valor del atributo `for` debe ser igual al valor del atributo `id` del formulario de control. Por ejemplo:
+
+```html
+<form>
+ <label for="name">Name:</label>
+ <input type="text" id="name" name="name">
+</form>
+```
+
+# --instructions--
+
+Camper Cat espera que haya mucho interés en sus publicaciones bien armadas en el blog y quiere incluir un formulario de registro por correo electrónico. Añade un atributo `for` en el correo electrónico `label` que coincida con el `id` en su campo `input`.
+
+# --hints--
+
+Tu código debe tener un atributo `for` en una etiqueta `label` que no esté vacía.
+
+```js
+assert($('label').attr('for'));
+```
+
+El valor del atributo `for` debe coincidir con el valor `id` en el correo electrónico `input`.
+
+```js
+assert($('label').attr('for') == 'email');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<body>
+ <header>
+ <h1>Deep Thoughts with Master Camper Cat</h1>
+ </header>
+ <section>
+ <form>
+ <p>Sign up to receive Camper Cat's blog posts by email here!</p>
+
+
+ <label>Email:</label>
+ <input type="text" id="email" name="email">
+
+
+ <input type="submit" name="submit" value="Submit">
+ </form>
+ </section>
+ <article>
+ <h2>The Garfield Files: Lasagna as Training Fuel?</h2>
+ <p>The internet is littered with varying opinions on nutritional paradigms, from catnip paleo to hairball cleanses. But let's turn our attention to an often overlooked fitness fuel, and examine the protein-carb-NOM trifecta that is lasagna...</p>
+ </article>
+ <img src="samuraiSwords.jpeg" alt="">
+ <article>
+ <h2>Defeating your Foe: the Red Dot is Ours!</h2>
+ <p>Felines the world over have been waging war on the most persistent of foes. This red nemesis combines both cunning stealth and lightning speed. But chin up, fellow fighters, our time for victory may soon be near...</p>
+ </article>
+ <img src="samuraiSwords.jpeg" alt="">
+ <article>
+ <h2>Is Chuck Norris a Cat Person?</h2>
+ <p>Chuck Norris is widely regarded as the premier martial artist on the planet, and it's a complete coincidence anyone who disagrees with this fact mysteriously disappears soon after. But the real question is, is he a cat person?...</p>
+ </article>
+ <footer>© 2018 Camper Cat</footer>
+</body>
+```
+
+# --solutions--
+
+```html
+<body>
+ <header>
+ <h1>Deep Thoughts with Master Camper Cat</h1>
+ </header>
+ <section>
+ <form>
+ <p>Sign up to receive Camper Cat's blog posts by email here!</p>
+
+
+ <label for="email">Email:</label>
+ <input type="text" id="email" name="email">
+
+
+ <input type="submit" name="submit" value="Submit">
+ </form>
+ </section>
+ <article>
+ <h2>The Garfield Files: Lasagna as Training Fuel?</h2>
+ <p>The internet is littered with varying opinions on nutritional paradigms, from catnip paleo to hairball cleanses. But let's turn our attention to an often overlooked fitness fuel, and examine the protein-carb-NOM trifecta that is lasagna...</p>
+ </article>
+ <img src="samuraiSwords.jpeg" alt="">
+ <article>
+ <h2>Defeating your Foe: the Red Dot is Ours!</h2>
+ <p>Felines the world over have been waging war on the most persistent of foes. This red nemesis combines both cunning stealth and lightning speed. But chin up, fellow fighters, our time for victory may soon be near...</p>
+ </article>
+ <img src="samuraiSwords.jpeg" alt="">
+ <article>
+ <h2>Is Chuck Norris a Cat Person?</h2>
+ <p>Chuck Norris is widely regarded as the premier martial artist on the planet, and it's a complete coincidence anyone who disagrees with this fact mysteriously disappears soon after. But the real question is, is he a cat person?...</p>
+ </article>
+ <footer>© 2018 Camper Cat</footer>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/improve-readability-with-high-contrast-text.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/improve-readability-with-high-contrast-text.md
new file mode 100644
index 00000000000..e7ba230d298
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/improve-readability-with-high-contrast-text.md
@@ -0,0 +1,80 @@
+---
+id: 587d778e367417b2b2512aab
+title: Mejora la legibilidad con texto de alto contraste
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cKb3nCq'
+forumTopicId: 301017
+dashedName: improve-readability-with-high-contrast-text
+---
+
+# --description--
+
+El bajo contraste entre los colores de primer plano y de fondo puede dificultar la lectura del texto. El contraste suficiente mejora la legibilidad de tu contenido, pero ¿qué significa exactamente "suficiente"?
+
+Las Directrices de Accesibilidad al Contenido Web (WCAG) recomiendan al menos una relación de contraste de 4.5 a 1 para el texto normal. La relación se calcula comparando los valores de luminancia relativa de dos colores. Esto equilibra entre 1:1 para el mismo color, o sin contraste, y 21:1 para el blanco contra el negro, el contraste más fuerte. Hay muchas herramientas de comprobación de contraste disponibles en línea que calculan esta relación por ti.
+
+# --instructions--
+
+La elección del Gato Campero de texto gris claro sobre un fondo blanco para su reciente publicación de blog tiene una relación de contraste de 1.5:1, por lo que es difícil de leer. Cambie el `color` del texto del gris, actual (`#D3D3D3`) a un gris más oscuro (`#636363`) para mejorar la relación de contraste a 6:1.
+
+# --hints--
+
+Tu código debe cambiar el texto `color` para el `body` al gris más oscuro.
+
+```js
+assert($('body').css('color') == 'rgb(99, 99, 99)');
+```
+
+Tu código no debe cambiar el `background-color` para el `body`.
+
+```js
+assert($('body').css('background-color') == 'rgb(255, 255, 255)');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<head>
+ <style>
+ body {
+ color: #D3D3D3;
+ background-color: #FFF;
+ }
+ </style>
+</head>
+<body>
+ <header>
+ <h1>Deep Thoughts with Master Camper Cat</h1>
+ </header>
+ <article>
+ <h2>A Word on the Recent Catnip Doping Scandal</h2>
+ <p>The influence that catnip has on feline behavior is well-documented, and its use as an herbal supplement in competitive ninja circles remains controversial. Once again, the debate to ban the substance is brought to the public's attention after the high-profile win of Kittytron, a long-time proponent and user of the green stuff, at the Claw of Fury tournament.</p>
+ <p>As I've stated in the past, I firmly believe a true ninja's skills must come from within, with no external influences. My own catnip use shall continue as purely recreational.</p>
+ </article>
+</body>
+```
+
+# --solutions--
+
+```html
+<head>
+ <style>
+ body {
+ color: #636363;
+ background-color: #FFF;
+ }
+ </style>
+</head>
+<body>
+ <header>
+ <h1>Deep Thoughts with Master Camper Cat</h1>
+ </header>
+ <article>
+ <h2>A Word on the Recent Catnip Doping Scandal</h2>
+ <p>The influence that catnip has on feline behavior is well-documented, and its use as an herbal supplement in competitive ninja circles remains controversial. Once again, the debate to ban the substance is brought to the public's attention after the high-profile win of Kittytron, a long-time proponent and user of the green stuff, at the Claw of Fury tournament.</p>
+ <p>As I've stated in the past, I firmly believe a true ninja's skills must come from within, with no external influences. My own catnip use shall continue as purely recreational.</p>
+ </article>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/jump-straight-to-the-content-using-the-main-element.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/jump-straight-to-the-content-using-the-main-element.md
new file mode 100644
index 00000000000..01cbfb7efae
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/jump-straight-to-the-content-using-the-main-element.md
@@ -0,0 +1,62 @@
+---
+id: 587d774e367417b2b2512a9f
+title: Salta directamente al contenido usando el elemento principal (main)
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cPp7zuE'
+forumTopicId: 301018
+dashedName: jump-straight-to-the-content-using-the-main-element
+---
+
+# --description--
+
+HTML5 introdujo una serie de nuevos elementos que ofrecen a los desarrolladores más opciones al tiempo que incorporan características de accesibilidad. Estas etiquetas incluyen `main`, `header`, `footer`, `nav`, `article`, y `section`, entre otros.
+
+Por defecto, un navegador representa estos elementos de manera similar a la humilde `div`. Sin embargo, usarlos cuando sea apropiado da un significado adicional en tu lenguaje de marcado. El nombre de la etiqueta solo puede indicar el tipo de información que contiene, lo que agrega significado semántico a ese contenido. Las tecnologías de asistencia pueden acceder a esta información para proporcionar mejores opciones de resumen de páginas o de navegación a sus usuarios.
+
+El elemento `main` se usa para envolver (lo adivinaste) el contenido principal, y solo debe haber uno por página. Está destinado a rodear la información que está relacionada con el tema central de su página. No está destinado a incluir elementos que se repiten en todas las páginas, como enlaces de navegación o banners.
+
+La etiqueta `main` también tiene una característica de referencia insertada que la tecnología de asistencia puede usar para navegar rápidamente al contenido principal. Si alguna vez has visto un enlace "Salta al contenido principal" en la parte superior de una página, usando una etiqueta principal proporciona automáticamente a los dispositivos de asistencia esa funcionalidad.
+
+# --instructions--
+
+Camper Cat tiene algunas grandes ideas para su página de armas ninja. Ayúdelo a configurar su marcado agregado etiquetas de apertura y cierre `main` entre el `header` y el `footer` (cubierto en otros desafíos). Mantenga las etiquetas `main` vacías por ahora.
+
+# --hints--
+
+Tu código debe tener una etiqueta `main`.
+
+```js
+assert($('main').length == 1);
+```
+
+Las etiquetas `main` deben estar entre la etiqueta `header` de cierre y la etiqueta `footer` de apertura.
+
+```js
+assert(code.match(/<\/header>\s*?<main>\s*?<\/main>/gi));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<header>
+ <h1>Weapons of the Ninja</h1>
+</header>
+
+
+
+<footer></footer>
+```
+
+# --solutions--
+
+```html
+<header>
+ <h1>Weapons of the Ninja</h1>
+</header>
+<main>
+
+</main>
+<footer></footer>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/know-when-alt-text-should-be-left-blank.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/know-when-alt-text-should-be-left-blank.md
new file mode 100644
index 00000000000..a1ec60b73c9
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/know-when-alt-text-should-be-left-blank.md
@@ -0,0 +1,74 @@
+---
+id: 587d774c367417b2b2512a9d
+title: Saber Cuando el Texto Alt Debe Dejarse en Blanco
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cM9P4t2'
+forumTopicId: 301019
+dashedName: know-when-alt-text-should-be-left-blank
+---
+
+# --description--
+
+En el último desafío, aprendiste que es obligatorio incluir un atributo `alt` al usar etiquetas `img`. Sin embargo, a veces las imágenes se agrupan con un título que ya las describe, o se usan solo para decoración. En estos casos `alt` el texto puede parecer redundante o innecesario.
+
+En situaciones en las que una ya está explicada con contenido de texto, o no agrega significado a una página, el `img` todavía necesita un atributo `alt`, pero se puede establecer en una cadena vacía. Aquí hay un ejemplo:
+
+`<img src="visualDecoration.jpeg" alt="">`
+
+Las imágenes de fondo generalmente también caen bajo la etiqueta "decorativa". Sin embargo, normalmente se aplican con reglas CSS y, por lo tanto, no forman parte del proceso de lectores de pantalla del lenguaje de marcado.
+
+**Nota:** Para imágenes con subtítulos, es posible que desees incluir texto `alt`, ya que ayuda a los motores de búsqueda a catalogar el contenido de la imagen.
+
+# --instructions--
+
+Camper Cat ha programado una página esqueleto para la parte del blog de su sitio web. Está planeando agregar una ruptura visual entre sus dos artículos con una imagen decorativa de una espada samurái. Agrega un atributo `alt` a la etiqueta `img` y establezca una cadena vacía. (Ten en cuenta que la imagen `src` no enlace a un archivo real - no te preocupes de que no hay espadas que aparece en la pantalla.)
+
+# --hints--
+
+Tu etiqueta `img` debe tener un atributo `alt`.
+
+```js
+assert(!($('img').attr('alt') == undefined));
+```
+
+El atributo `alt` debe establecerse en una cadena vacía.
+
+```js
+assert($('img').attr('alt') == '');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<h1>Deep Thoughts with Master Camper Cat</h1>
+<article>
+ <h2>Defeating your Foe: the Red Dot is Ours!</h2>
+ <p>To Come...</p>
+</article>
+
+<img src="samuraiSwords.jpeg">
+
+<article>
+ <h2>Is Chuck Norris a Cat Person?</h2>
+ <p>To Come...</p>
+</article>
+```
+
+# --solutions--
+
+```html
+<h1>Deep Thoughts with Master Camper Cat</h1>
+<article>
+ <h2>Defeating your Foe: the Red Dot is Ours!</h2>
+ <p>To Come...</p>
+</article>
+
+<img src="samuraiSwords.jpeg" alt="">
+
+<article>
+ <h2>Is Chuck Norris a Cat Person?</h2>
+ <p>To Come...</p>
+</article>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/make-elements-only-visible-to-a-screen-reader-by-using-custom-css.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/make-elements-only-visible-to-a-screen-reader-by-using-custom-css.md
new file mode 100644
index 00000000000..848f5f6f1fd
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/make-elements-only-visible-to-a-screen-reader-by-using-custom-css.md
@@ -0,0 +1,244 @@
+---
+id: 587d778d367417b2b2512aaa
+title: Haz que los elementos solo sean visibles para un lector de pantalla mediante CSS personalizado
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cJ8QGkhJ'
+forumTopicId: 301020
+dashedName: make-elements-only-visible-to-a-screen-reader-by-using-custom-css
+---
+
+# --description--
+
+¿Has notado que todos los desafíos de accesibilidad aplicados hasta ahora no han usado ningún CSS? Esto es para mostrar la importancia de un esquema de documento lógico, y el uso de etiquetas semánticamente significativas alrededor de tu contenido antes de introducir el aspecto de diseño visual.
+
+Sin embargo, la magia de CSS también puede mejorar la accesibilidad en tu página cuando deseas ocultar visualmente contenido destinado solo para lectores de pantalla. Esto sucede cuando la información está en un formato visual (como un gráfico), pero los usuarios del lector de pantalla necesitan una presentación alternativa (como una tabla) para acceder a los datos. CSS se utiliza para colocar los elementos exclusivos de lector de pantalla fuera del área visual de la ventana del navegador.
+
+Aquí hay un ejemplo de las reglas de CSS que logran esto:
+
+```css
+.sr-only {
+ position: absolute;
+ left: -10000px;
+ width: 1px;
+ height: 1px;
+ top: auto;
+ overflow: hidden;
+}
+```
+
+**Nota:** Los siguientes enfoques CSS No harán lo mismo:
+
+<ul>
+<li><code>display: none;</code> o <code>visibility: hidden;</code> oculta el contenido para todos, incluidos los usuarios del lector de pantalla</li>
+<li>Los valores cero para los tamaños del píxel, como <code>width: 0px; height: 0px;</code> eliminan ese elemento del flujo de tu documento, lo que significa que los lectores de pantalla lo ignorarán</li>
+</ul>
+
+# --instructions--
+
+Camper Cat creó un gráfico de barras apiladas realmente genial para su página de entrenamiento y coloco los datos en una tabla para sus usuarios con dificultad visual. La tabla ya tiene una clase `sr-only`, pero las reglas de CSS aún no se han completado. Asigna a `position` un valor `absolute`, a `left` un valor de `-10000px`, y tanto a `width` como a `height` un valor de `1px`.
+
+# --hints--
+
+Tu código debe establecer la propiedad `position` de la clase `sr-only` en un valor de `absolute`.
+
+```js
+assert($('.sr-only').css('position') == 'absolute');
+```
+
+Tu código debe establecer la propiedad `left` de la clase `sr-only` en un valor de `-10000px`.
+
+```js
+assert($('.sr-only').css('left') == '-10000px');
+```
+
+Tu código debe establecer la propiedad `width` de la clase `sr-only` en un valor de `1` píxel.
+
+```js
+assert(code.match(/width:\s*?1px/gi));
+```
+
+Tu código debe establecer la propiedad `height` de la clase `sr-only` en un valor de `1` píxel.
+
+```js
+assert(code.match(/height:\s*?1px/gi));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<head>
+ <style>
+ .sr-only {
+ position: ;
+ left: ;
+ width: ;
+ height: ;
+ top: auto;
+ overflow: hidden;
+ }
+ </style>
+</head>
+<body>
+ <header>
+ <h1>Training</h1>
+ <nav>
+ <ul>
+ <li><a href="#stealth">Stealth & Agility</a></li>
+ <li><a href="#combat">Combat</a></li>
+ <li><a href="#weapons">Weapons</a></li>
+ </ul>
+ </nav>
+ </header>
+ <section>
+ <h2>Master Camper Cat's Beginner Three Week Training Program</h2>
+ <figure>
+ <!-- Stacked bar chart of weekly training -->
+ <p>[Stacked bar chart]</p>
+ <br />
+ <figcaption>Breakdown per week of time to spend training in stealth, combat, and weapons.</figcaption>
+ </figure>
+ <table class="sr-only">
+ <caption>Hours of Weekly Training in Stealth, Combat, and Weapons</caption>
+ <thead>
+ <tr>
+ <th></th>
+ <th scope="col">Stealth & Agility</th>
+ <th scope="col">Combat</th>
+ <th scope="col">Weapons</th>
+ <th scope="col">Total</th>
+ </tr>
+ </thead>
+ <tbody>
+ <tr>
+ <th scope="row">Week One</th>
+ <td>3</td>
+ <td>5</td>
+ <td>2</td>
+ <td>10</td>
+ </tr>
+ <tr>
+ <th scope="row">Week Two</th>
+ <td>4</td>
+ <td>5</td>
+ <td>3</td>
+ <td>12</td>
+ </tr>
+ <tr>
+ <th scope="row">Week Three</th>
+ <td>4</td>
+ <td>6</td>
+ <td>3</td>
+ <td>13</td>
+ </tr>
+ </tbody>
+ </table>
+ </section>
+ <section id="stealth">
+ <h2>Stealth & Agility Training</h2>
+ <article><h3>Climb foliage quickly using a minimum spanning tree approach</h3></article>
+ <article><h3>No training is NP-complete without parkour</h3></article>
+ </section>
+ <section id="combat">
+ <h2>Combat Training</h2>
+ <article><h3>Dispatch multiple enemies with multithreaded tactics</h3></article>
+ <article><h3>Goodbye, world: 5 proven ways to knock out an opponent</h3></article>
+ </section>
+ <section id="weapons">
+ <h2>Weapons Training</h2>
+ <article><h3>Swords: the best tool to literally divide and conquer</h3></article>
+ <article><h3>Breadth-first or depth-first in multi-weapon training?</h3></article>
+ </section>
+ <footer>© 2018 Camper Cat</footer>
+</body>
+```
+
+# --solutions--
+
+```html
+<head>
+ <style>
+ .sr-only {
+ position: absolute;
+ left: -10000px;
+ width: 1px;
+ height: 1px;
+ top: auto;
+ overflow: hidden;
+ }
+ </style>
+</head>
+<body>
+ <header>
+ <h1>Training</h1>
+ <nav>
+ <ul>
+ <li><a href="#stealth">Stealth & Agility</a></li>
+ <li><a href="#combat">Combat</a></li>
+ <li><a href="#weapons">Weapons</a></li>
+ </ul>
+ </nav>
+ </header>
+ <section>
+ <h2>Master Camper Cat's Beginner Three Week Training Program</h2>
+ <figure>
+ <!-- Stacked bar chart of weekly training -->
+ <p>[Stacked bar chart]</p>
+ <br />
+ <figcaption>Breakdown per week of time to spend training in stealth, combat, and weapons.</figcaption>
+ </figure>
+ <table class="sr-only">
+ <caption>Hours of Weekly Training in Stealth, Combat, and Weapons</caption>
+ <thead>
+ <tr>
+ <th></th>
+ <th scope="col">Stealth & Agility</th>
+ <th scope="col">Combat</th>
+ <th scope="col">Weapons</th>
+ <th scope="col">Total</th>
+ </tr>
+ </thead>
+ <tbody>
+ <tr>
+ <th scope="row">Week One</th>
+ <td>3</td>
+ <td>5</td>
+ <td>2</td>
+ <td>10</td>
+ </tr>
+ <tr>
+ <th scope="row">Week Two</th>
+ <td>4</td>
+ <td>5</td>
+ <td>3</td>
+ <td>12</td>
+ </tr>
+ <tr>
+ <th scope="row">Week Three</th>
+ <td>4</td>
+ <td>6</td>
+ <td>3</td>
+ <td>13</td>
+ </tr>
+ </tbody>
+ </table>
+ </section>
+ <section id="stealth">
+ <h2>Stealth & Agility Training</h2>
+ <article><h3>Climb foliage quickly using a minimum spanning tree approach</h3></article>
+ <article><h3>No training is NP-complete without parkour</h3></article>
+ </section>
+ <section id="combat">
+ <h2>Combat Training</h2>
+ <article><h3>Dispatch multiple enemies with multithreaded tactics</h3></article>
+ <article><h3>Goodbye, world: 5 proven ways to knock out an opponent</h3></article>
+ </section>
+ <section id="weapons">
+ <h2>Weapons Training</h2>
+ <article><h3>Swords: the best tool to literally divide and conquer</h3></article>
+ <article><h3>Breadth-first or depth-first in multi-weapon training?</h3></article>
+ </section>
+ <footer>© 2018 Camper Cat</footer>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/make-links-navigable-with-html-access-keys.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/make-links-navigable-with-html-access-keys.md
new file mode 100644
index 00000000000..341085ea856
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/make-links-navigable-with-html-access-keys.md
@@ -0,0 +1,104 @@
+---
+id: 587d7790367417b2b2512aaf
+title: Haz que los enlaces sean navegables con claves de acceso HTML
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cQvmaTp'
+forumTopicId: 301021
+dashedName: make-links-navigable-with-html-access-keys
+---
+
+# --description--
+
+HTML ofrece el atributo `accesskey` para especificar una tecla de acceso directo para activar o enfocar un elemento. Esto puede hacer que la navegación sea más eficiente para los usuarios de solo usan el teclado.
+
+HTML5 permite que este atributo se use en cualquier elemento, pero es particularmente útil cuando se usa con elementos interactivos. Esto incluye enlaces, botones y controles de formulario.
+
+Aquí hay un ejemplo:
+
+`<button accesskey="b">Important Button</button>`
+
+# --instructions--
+
+El Gato Campero quiere que los enlaces alrededor de los dos títulos de artículos de blog tengan atajos de teclado para que los usuarios de su sitio puedan navegar rápidamente a la historia completa. Agrega un atributo `accesskey` a ambos enlaces y establece el primero en `g` (para Garfield) y el segundo en `c` (para Chuck Norris).
+
+# --hints--
+
+Tu código debe agregar un atributo `accesskey` a la etiqueta `a` con el `id` de `first`.
+
+```js
+assert($('#first').attr('accesskey'));
+```
+
+Tu código debe agregar un atributo `accesskey` a la etiqueta `a` con el `id` de `second`.
+
+```js
+assert($('#second').attr('accesskey'));
+```
+
+Tu código debe establecer el atributo `accesskey` en la etiqueta `a` con el `id` de `first` en `g`. Ten en cuenta que las mayúsculas y minúsculas importan.
+
+```js
+assert($('#first').attr('accesskey') == 'g');
+```
+
+Tu código debe establecer `accesskey` en la etiqueta `a` con el `id` de `second` en `c`. Ten en cuenta que las mayúsculas y minúsculas importan.
+
+```js
+assert($('#second').attr('accesskey') == 'c');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<body>
+ <header>
+ <h1>Deep Thoughts with Master Camper Cat</h1>
+ </header>
+ <article>
+
+
+ <h2><a id="first" href="#">The Garfield Files: Lasagna as Training Fuel?</a></h2>
+
+
+ <p>The internet is littered with varying opinions on nutritional paradigms, from catnip paleo to hairball cleanses. But let's turn our attention to an often overlooked fitness fuel, and examine the protein-carb-NOM trifecta that is lasagna...</p>
+ </article>
+ <article>
+
+
+ <h2><a id="second" href="#">Is Chuck Norris a Cat Person?</a></h2>
+
+
+ <p>Chuck Norris is widely regarded as the premier martial artist on the planet, and it's a complete coincidence anyone who disagrees with this fact mysteriously disappears soon after. But the real question is, is he a cat person?...</p>
+ </article>
+ <footer>© 2018 Camper Cat</footer>
+</body>
+```
+
+# --solutions--
+
+```html
+<body>
+ <header>
+ <h1>Deep Thoughts with Master Camper Cat</h1>
+ </header>
+ <article>
+
+
+ <h2><a id="first" accesskey="g" href="#">The Garfield Files: Lasagna as Training Fuel?</a></h2>
+
+
+ <p>The internet is littered with varying opinions on nutritional paradigms, from catnip paleo to hairball cleanses. But let's turn our attention to an often overlooked fitness fuel, and examine the protein-carb-NOM trifecta that is lasagna...</p>
+ </article>
+ <article>
+
+
+ <h2><a id="second" accesskey="c" href="#">Is Chuck Norris a Cat Person?</a></h2>
+
+
+ <p>Chuck Norris is widely regarded as the premier martial artist on the planet, and it's a complete coincidence anyone who disagrees with this fact mysteriously disappears soon after. But the real question is, is he a cat person?...</p>
+ </article>
+ <footer>© 2018 Camper Cat</footer>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/make-screen-reader-navigation-easier-with-the-footer-landmark.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/make-screen-reader-navigation-easier-with-the-footer-landmark.md
new file mode 100644
index 00000000000..169c2ae9d8c
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/make-screen-reader-navigation-easier-with-the-footer-landmark.md
@@ -0,0 +1,116 @@
+---
+id: 587d7788367417b2b2512aa3
+title: Haz que la navegación del lector de pantalla sea más fácil con el footer Landmark
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/crVrDh8'
+forumTopicId: 301022
+dashedName: make-screen-reader-navigation-easier-with-the-footer-landmark
+---
+
+# --description--
+
+Similar a `header` y `nav`, el elemento `footer` tiene una característica de referencia incorporada que permite a los dispositivos de asistencia navegar rápidamente hacia él. Se utiliza principalmente para contener información sobre derechos de autor o enlaces a documentos relacionados que normalmente se encuentran en la parte inferior de una página.
+
+# --instructions--
+
+La página de entrenamiento del Camper Cat está progresando bien. Cambia el `div` que utilizó para envolver su información de copyright en la parte inferior de la página a un elemento `footer`.
+
+# --hints--
+
+Tu código debe tener una etiqueta `footer`.
+
+```js
+assert($('footer').length == 1);
+```
+
+Tu código no debe tener ninguna etiqueta `div`.
+
+```js
+assert($('div').length == 0);
+```
+
+Tu código debe tener una etiqueta `footer` de apertura y cierre.
+
+```js
+assert(code.match(/<footer>\s*© 2018 Camper Cat\s*<\/footer>/g));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<body>
+ <header>
+ <h1>Training</h1>
+ <nav>
+ <ul>
+ <li><a href="#stealth">Stealth & Agility</a></li>
+ <li><a href="#combat">Combat</a></li>
+ <li><a href="#weapons">Weapons</a></li>
+ </ul>
+ </nav>
+ </header>
+ <main>
+ <section id="stealth">
+ <h2>Stealth & Agility Training</h2>
+ <article><h3>Climb foliage quickly using a minimum spanning tree approach</h3></article>
+ <article><h3>No training is NP-complete without parkour</h3></article>
+ </section>
+ <section id="combat">
+ <h2>Combat Training</h2>
+ <article><h3>Dispatch multiple enemies with multithreaded tactics</h3></article>
+ <article><h3>Goodbye world: 5 proven ways to knock out an opponent</h3></article>
+ </section>
+ <section id="weapons">
+ <h2>Weapons Training</h2>
+ <article><h3>Swords: the best tool to literally divide and conquer</h3></article>
+ <article><h3>Breadth-first or depth-first in multi-weapon training?</h3></article>
+ </section>
+ </main>
+
+
+ <div>© 2018 Camper Cat</div>
+
+
+</body>
+```
+
+# --solutions--
+
+```html
+<body>
+ <header>
+ <h1>Training</h1>
+ <nav>
+ <ul>
+ <li><a href="#stealth">Stealth & Agility</a></li>
+ <li><a href="#combat">Combat</a></li>
+ <li><a href="#weapons">Weapons</a></li>
+ </ul>
+ </nav>
+ </header>
+ <main>
+ <section id="stealth">
+ <h2>Stealth & Agility Training</h2>
+ <article><h3>Climb foliage quickly using a minimum spanning tree approach</h3></article>
+ <article><h3>No training is NP-complete without parkour</h3></article>
+ </section>
+ <section id="combat">
+ <h2>Combat Training</h2>
+ <article><h3>Dispatch multiple enemies with multithreaded tactics</h3></article>
+ <article><h3>Goodbye world: 5 proven ways to knock out an opponent</h3></article>
+ </section>
+ <section id="weapons">
+ <h2>Weapons Training</h2>
+ <article><h3>Swords: the best tool to literally divide and conquer</h3></article>
+ <article><h3>Breadth-first or depth-first in multi-weapon training?</h3></article>
+ </section>
+ </main>
+
+
+ <footer>© 2018 Camper Cat</footer>
+
+
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/make-screen-reader-navigation-easier-with-the-header-landmark.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/make-screen-reader-navigation-easier-with-the-header-landmark.md
new file mode 100644
index 00000000000..99b75762a8e
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/make-screen-reader-navigation-easier-with-the-header-landmark.md
@@ -0,0 +1,111 @@
+---
+id: 587d7787367417b2b2512aa1
+title: Haz que la navegación del lector de pantalla sea más fácil con el encabezado Landmark
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cB76vtv'
+forumTopicId: 301023
+dashedName: make-screen-reader-navigation-easier-with-the-header-landmark
+---
+
+# --description--
+
+El siguiente elemento HTML5 que agrega significado semántico y mejora la accesibilidad es la etiqueta `header`. Se usa para envolver información introductoria o enlaces de navegación para su etiqueta principal y funciona bien con el contenido que se repite en la parte superior en varias páginas.
+
+`header` comparte la función landmark integrada que viste con `main`, lo que permite a las tecnologías de asistencia navegar rápidamente a ese contenido.
+
+**Nota:** El `header` está diseñado para usarse dentro de la etiqueta `body` de tu documento HTML. Esto es diferente al elemento `head`, que contiene el título de la página, metainformación, etc.
+
+# --instructions--
+
+Camper Cat está escribiendo algunos grandes artículos sobre el entrenamiento ninja, y quiere añadir una página para ellos a su sitio. Cambia la parte superior `div` que actualmente contiene el `h1` a una etiqueta `header`.
+
+# --hints--
+
+Tu código debe tener una etiqueta `header`.
+
+```js
+assert($('header').length == 1);
+```
+
+Tus etiquetas `header` deben envolver `h1`.
+
+```js
+assert($('header').children('h1').length == 1);
+```
+
+Tu código no debe tener ninguna etiqueta `div`.
+
+```js
+assert($('div').length == 0);
+```
+
+Tu elemento `header` debe tener una etiqueta de cierre.
+
+```js
+assert(
+ code.match(/<\/header>/g) &&
+ code.match(/<\/header>/g).length === code.match(/<header>/g).length
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<body>
+
+ <div>
+ <h1>Training with Camper Cat</h1>
+ </div>
+
+
+ <main>
+ <section id="stealth">
+ <h2>Stealth & Agility Training</h2>
+ <article><h3>Climb foliage quickly using a minimum spanning tree approach</h3></article>
+ <article><h3>No training is NP-complete without parkour</h3></article>
+ </section>
+ <section id="combat">
+ <h2>Combat Training</h2>
+ <article><h3>Dispatch multiple enemies with multithreaded tactics</h3></article>
+ <article><h3>Goodbye world: 5 proven ways to knock out an opponent</h3></article>
+ </section>
+ <section id="weapons">
+ <h2>Weapons Training</h2>
+ <article><h3>Swords: the best tool to literally divide and conquer</h3></article>
+ <article><h3>Breadth-first or depth-first in multi-weapon training?</h3></article>
+ </section>
+ </main>
+</body>
+```
+
+# --solutions--
+
+```html
+<body>
+
+ <header>
+ <h1>Training with Camper Cat</h1>
+ </header>
+
+
+ <main>
+ <section id="stealth">
+ <h2>Stealth & Agility Training</h2>
+ <article><h3>Climb foliage quickly using a minimum spanning tree approach</h3></article>
+ <article><h3>No training is NP-complete without parkour</h3></article>
+ </section>
+ <section id="combat">
+ <h2>Combat Training</h2>
+ <article><h3>Dispatch multiple enemies with multithreaded tactics</h3></article>
+ <article><h3>Goodbye world: 5 proven ways to knock out an opponent</h3></article>
+ </section>
+ <section id="weapons">
+ <h2>Weapons Training</h2>
+ <article><h3>Swords: the best tool to literally divide and conquer</h3></article>
+ <article><h3>Breadth-first or depth-first in multi-weapon training?</h3></article>
+ </section>
+ </main>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/make-screen-reader-navigation-easier-with-the-nav-landmark.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/make-screen-reader-navigation-easier-with-the-nav-landmark.md
new file mode 100644
index 00000000000..33fcd678178
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/make-screen-reader-navigation-easier-with-the-nav-landmark.md
@@ -0,0 +1,121 @@
+---
+id: 587d7788367417b2b2512aa2
+title: Haz que la navegación del lector de pantalla sea más fácil con el navegador Landmark
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/czVwWSv'
+forumTopicId: 301024
+dashedName: make-screen-reader-navigation-easier-with-the-nav-landmark
+---
+
+# --description--
+
+El elemento `nav` es otro elemento de HTML5 con la función de punto de referencia insertado para facilitar la navegación del lector de pantalla. Esta etiqueta esta destinada a envolver los principales enlaces de navegación en tu página.
+
+Si hay enlaces internos repetidos en la parte de la página, no es necesario usar el lenguaje de marcado con aquellos con una etiqueta `nav` también. Usando una etiqueta `footer` (cubierto en el próximo desafío) es suficiente.
+
+# --instructions--
+
+Camper Cat incluyó enlaces de navegación en la parte superior de su página de entrenamiento, pero los envolvió en un `div`. Cambia la etiqueta `div` a una etiqueta `nav` para mejorar la accesibilidad en su página.
+
+# --hints--
+
+Tu código debe tener una etiqueta `nav`.
+
+```js
+assert($('nav').length == 1);
+```
+
+Tus etiquetas `nav` deben envolver la etiqueta `ul` y sus elementos de la lista.
+
+```js
+assert($('nav').children('ul').length == 1);
+```
+
+Tu código no debe tener ninguna etiqueta `div`.
+
+```js
+assert($('div').length == 0);
+```
+
+Tu elemento `nav` debería tener un cierre.
+
+```js
+assert(
+ code.match(/<\/nav>/g) &&
+ code.match(/<\/nav>/g).length === code.match(/<nav>/g).length
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<body>
+ <header>
+ <h1>Training with Camper Cat</h1>
+
+ <div>
+ <ul>
+ <li><a href="#stealth">Stealth & Agility</a></li>
+ <li><a href="#combat">Combat</a></li>
+ <li><a href="#weapons">Weapons</a></li>
+ </ul>
+ </div>
+
+ </header>
+ <main>
+ <section id="stealth">
+ <h2>Stealth & Agility Training</h2>
+ <article><h3>Climb foliage quickly using a minimum spanning tree approach</h3></article>
+ <article><h3>No training is NP-complete without parkour</h3></article>
+ </section>
+ <section id="combat">
+ <h2>Combat Training</h2>
+ <article><h3>Dispatch multiple enemies with multithreaded tactics</h3></article>
+ <article><h3>Goodbye world: 5 proven ways to knock out an opponent</h3></article>
+ </section>
+ <section id="weapons">
+ <h2>Weapons Training</h2>
+ <article><h3>Swords: the best tool to literally divide and conquer</h3></article>
+ <article><h3>Breadth-first or depth-first in multi-weapon training?</h3></article>
+ </section>
+ </main>
+</body>
+```
+
+# --solutions--
+
+```html
+<body>
+ <header>
+ <h1>Training with Camper Cat</h1>
+
+ <nav>
+ <ul>
+ <li><a href="#stealth">Stealth & Agility</a></li>
+ <li><a href="#combat">Combat</a></li>
+ <li><a href="#weapons">Weapons</a></li>
+ </ul>
+ </nav>
+
+ </header>
+ <main>
+ <section id="stealth">
+ <h2>Stealth & Agility Training</h2>
+ <article><h3>Climb foliage quickly using a minimum spanning tree approach</h3></article>
+ <article><h3>No training is NP-complete without parkour</h3></article>
+ </section>
+ <section id="combat">
+ <h2>Combat Training</h2>
+ <article><h3>Dispatch multiple enemies with multithreaded tactics</h3></article>
+ <article><h3>Goodbye world: 5 proven ways to knock out an opponent</h3></article>
+ </section>
+ <section id="weapons">
+ <h2>Weapons Training</h2>
+ <article><h3>Swords: the best tool to literally divide and conquer</h3></article>
+ <article><h3>Breadth-first or depth-first in multi-weapon training?</h3></article>
+ </section>
+ </main>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/standardize-times-with-the-html5-datetime-attribute.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/standardize-times-with-the-html5-datetime-attribute.md
new file mode 100644
index 00000000000..d84cffbedde
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/standardize-times-with-the-html5-datetime-attribute.md
@@ -0,0 +1,130 @@
+---
+id: 587d778c367417b2b2512aa9
+title: Estandarizar horas con el atributo datetime HTML5
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cmzMgtz'
+forumTopicId: 301025
+dashedName: standardize-times-with-the-html5-datetime-attribute
+---
+
+# --description--
+
+Continuando con el tema de fecha, HTML5 también introdujo el elemento `time` junto con un atributo `datetime` para estandarizar los tiempos. Este es un elemento inline que puede ajustar una fecha u hora en una página. Un formato válido de esa fecha se mantiene mediante el atributo `datetime`. Este es el valor al que acceden los dispositivos de asistencia. Ayuda a evitar confusiones al indicar una versión estandarizada de un tiempo, incluso si está escrito de manera informal o coloquial en el texto.
+
+Aquí hay un ejemplo:
+
+`<p>Master Camper Cat officiated the cage match between Goro and Scorpion <time datetime="2013-02-13">last Wednesday</time>, which ended in a draw.</p>`
+
+# --instructions--
+
+¡Ya tenemos los resultados de la encuesta de Mortal Kombat de Camper Cat! Envuelva una etiqueta `time` alrededor del texto `Thursday, September 15<sup>th<sup>` y agregue un atributo `datetime` establecido en `2016-09-15`.
+
+# --hints--
+
+Tu código debe tener un elemento `p` que incluya el texto `Thank you to everyone for responding to Master Camper Cat's survey.` e incluye un elemento `time`.
+
+```js
+assert(timeElement.length);
+```
+
+Las etiquetas `time` añadidas deben envolver el texto `Thursday, September 15<sup>th</sup>`.
+
+```js
+assert(
+ timeElement.length &&
+ $(timeElement).html().trim() === 'Thursday, September 15<sup>th</sup>'
+);
+```
+
+Tu etiqueta `time` agregada debe tener un atributo `datetime` que no esté vacío.
+
+```js
+assert(datetimeAttr && datetimeAttr.length);
+```
+
+Tu atributo `datetime` agregado debe establecerse en un valor de `2016-09-15`.
+
+```js
+assert(datetimeAttr === '2016-09-15');
+```
+
+# --seed--
+
+## --after-user-code--
+
+```html
+<script>
+const pElement = $("article > p")
+ .filter((_, elem) => $(elem).text().includes("Thank you to everyone for responding to Master Camper Cat's survey."));
+const timeElement = pElement[0] ? $(pElement[0]).find("time") : null;
+const datetimeAttr = $(timeElement).attr("datetime");
+</script>
+```
+
+## --seed-contents--
+
+```html
+<body>
+ <header>
+ <h1>Tournaments</h1>
+ </header>
+ <article>
+ <h2>Mortal Kombat Tournament Survey Results</h2>
+
+ <!-- Only change code below this line -->
+
+ <p>Thank you to everyone for responding to Master Camper Cat's survey. The best day to host the vaunted Mortal Kombat tournament is Thursday, September 15<sup>th</sup>. May the best ninja win!</p>
+
+ <!-- Only change code above this line -->
+
+ <section>
+ <h3>Comments:</h3>
+ <article>
+ <p>Posted by: Sub-Zero on <time datetime="2016-08-13T20:01Z">August 13<sup>th</sup></time></p>
+ <p>Johnny Cage better be there, I'll finish him!</p>
+ </article>
+ <article>
+ <p>Posted by: Doge on <time datetime="2016-08-15T08:12Z">August 15<sup>th</sup></time></p>
+ <p>Wow, much combat, so mortal.</p>
+ </article>
+ <article>
+ <p>Posted by: The Grim Reaper on <time datetime="2016-08-16T00:00Z">August 16<sup>th</sup></time></p>
+ <p>Looks like I'll be busy that day.</p>
+ </article>
+ </section>
+ </article>
+ <footer>© 2018 Camper Cat</footer>
+</body>
+```
+
+# --solutions--
+
+```html
+<body>
+ <header>
+ <h1>Tournaments</h1>
+ </header>
+ <article>
+ <h2>Mortal Kombat Tournament Survey Results</h2>
+
+ <p>Thank you to everyone for responding to Master Camper Cat's survey. The best day to host the vaunted Mortal Kombat tournament is <time datetime="2016-09-15">Thursday, September 15<sup>th</sup></time>. May the best ninja win!</p>
+
+ <section>
+ <h3>Comments:</h3>
+ <article>
+ <p>Posted by: Sub-Zero on <time datetime="2016-08-13T20:01Z">August 13<sup>th</sup></time></p>
+ <p>Johnny Cage better be there, I'll finish him!</p>
+ </article>
+ <article>
+ <p>Posted by: Doge on <time datetime="2016-08-15T08:12Z">August 15<sup>th</sup></time></p>
+ <p>Wow, much combat, so mortal.</p>
+ </article>
+ <article>
+ <p>Posted by: The Grim Reaper on <time datetime="2016-08-16T00:00Z">August 16<sup>th</sup></time></p>
+ <p>Looks like I'll be busy that day.</p>
+ </article>
+ </section>
+ </article>
+ <footer>© 2018 Camper Cat</footer>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/use-headings-to-show-hierarchical-relationships-of-content.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/use-headings-to-show-hierarchical-relationships-of-content.md
new file mode 100644
index 00000000000..0fde73a9ebe
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/use-headings-to-show-hierarchical-relationships-of-content.md
@@ -0,0 +1,92 @@
+---
+id: 587d774d367417b2b2512a9e
+title: Usa títulos para mostrar relaciones jerárquicas de contenido
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cqVEktm'
+forumTopicId: 301026
+dashedName: use-headings-to-show-hierarchical-relationships-of-content
+---
+
+# --description--
+
+Los títulos (`h1` a `h6` elementos) son etiquetas de caballo de batalla que ayudan a proporcionar estructura y etiquetado a su contenido. Los lectores de pantalla se pueden configurar para leer solo los títulos de una página para que el usuario obtenga un resumen. Esto significa que es importante que las etiquetas de los títulos en tu lenguaje de marcado tengan un significado semántico y se relacionen entre sí, no se elijan simplemente por sus valores de tamaño.
+
+* Significado semántico* significa que la etiqueta que usas alrededor del contenido indica el tipo de información que contiene.
+
+Si estuvieras escribiendo un documento con una introducción, un cuerpo y una conclusión, no tendría mucho sentido poner la conclusión como una subsección del cuerpo en tu esquema. Debería ser su propia sección. Del mismo modo, las etiquetas de los títulos en una página web deben ir en orden e indicar las relaciones jerárquicas de su contenido.
+
+Los títulos con rango igual (o superior) comienzan nuevas secciones implícitas, los títulos con rango inferior comienzan subsecciones de la anterior.
+
+Como ejemplo, una página con un elemento `h2` seguido de varias subsecciones marcadas con etiquetas `h4` confundirá a un lector de pantalla. Con seis opciones, es tentador usar una etiqueta porque se ve mejor en un navegador, pero puede usar CSS para editar el tamaño relativo.
+
+Un punto final, cada página siempre debe tener un (y solo un) elemento `h1`, que es el tema principal de tu contenido. Este y los otros títulos son utilizados en parte por los motores de búsqueda para comprender el tema de la página.
+
+# --instructions--
+
+Camper Cat quiere una página en su sitio dedicada a convertirse en un ninja. Ayúdelo a arreglar los títulos para que su lenguaje de marcado de un significado semántico al contenido y muestre las relaciones padre-hijo adecuadas de sus secciones. Cambia todas las etiquetas `h5` al nivel de título adecuado para indicar que son subsecciones de las `h2`. Utiliza etiquetas `h3` para este propósito.
+
+# --hints--
+
+Tu código debe tener 6 etiquetas `h3`.
+
+```js
+assert($('h3').length === 6);
+```
+
+Tu código debe tener 6 etiquetas de cierre `h3`.
+
+```js
+assert((code.match(/\/h3/g) || []).length === 6);
+```
+
+Tu código no debe tener ninguna etiqueta `h5`.
+
+```js
+assert($('h5').length === 0);
+```
+
+Tu código no debe tener ninguna etiqueta de cierre `h5`.
+
+```js
+assert(/\/h5/.test(code) === false);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<h1>How to Become a Ninja</h1>
+<main>
+ <h2>Learn the Art of Moving Stealthily</h2>
+ <h5>How to Hide in Plain Sight</h5>
+ <h5>How to Climb a Wall</h5>
+
+ <h2>Learn the Art of Battle</h2>
+ <h5>How to Strengthen your Body</h5>
+ <h5>How to Fight like a Ninja</h5>
+
+ <h2>Learn the Art of Living with Honor</h2>
+ <h5>How to Breathe Properly</h5>
+ <h5>How to Simplify your Life</h5>
+</main>
+```
+
+# --solutions--
+
+```html
+<h1>How to Become a Ninja</h1>
+<main>
+ <h2>Learn the Art of Moving Stealthily</h2>
+ <h3>How to Hide in Plain Sight</h3>
+ <h3>How to Climb a Wall</h3>
+
+ <h2>Learn the Art of Battle</h2>
+ <h3>How to Strengthen your Body</h3>
+ <h3>How to Fight like a Ninja</h3>
+
+ <h2>Learn the Art of Living with Honor</h2>
+ <h3>How to Breathe Properly</h3>
+ <h3>How to Simplify your Life</h3>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/use-tabindex-to-add-keyboard-focus-to-an-element.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/use-tabindex-to-add-keyboard-focus-to-an-element.md
new file mode 100644
index 00000000000..07b63968c51
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/use-tabindex-to-add-keyboard-focus-to-an-element.md
@@ -0,0 +1,142 @@
+---
+id: 587d7790367417b2b2512ab0
+title: Usa tabindex para agregar enfoque de teclado a un elemento
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cmzMDHW'
+forumTopicId: 301027
+dashedName: use-tabindex-to-add-keyboard-focus-to-an-element
+---
+
+# --description--
+
+El atributo HTML `tabindex` tiene tres funciones distintas relacionadas con el foco del teclado de un elemento. Cuando está en una etiqueta, indica que el elemento se puede enfocar en él. El valor (un número entero que es positivo, negativo o cero) determina el comportamiento.
+
+Ciertos elementos, como los vínculos y los controles de formulario, reciben automáticamente el foco del teclado cuando un usuario pestañas a través de una página. Está en el mismo orden en que los elementos vienen en la fuente del lenguaje de marcado de HTML. Esta misma funcionalidad se puede dar a otros elementos, como `div`, `span` y `p`, colocando un atributo `tabindex="0"`. Aquí hay un ejemplo:
+
+`<div tabindex="0">I need keyboard focus!</div>`
+
+**Nota:** Un valor negativo de `tabindex` (normalmente -1) indica que un elemento es enfocable, pero no es accesible por el teclado. Este método generalmente se usa para enfocar el contenido mediante programación (como cuando se activa un `div` utilizando para una ventana emergente), y esta más allá del alcance de estos desafíos.
+
+# --instructions--
+
+Camper Cat creó una nueva encuesta para recopilar información sobre sus usuarios. Él sabe que los campos de entrada obtienen automáticamente el enfoque del teclado, pero quiere asegurarse de que los usuarios de su teclado hagan una pausa en las instrucciones mientras tabulan los elementos. Agrega un atributo `tabindex` a la etiqueta `p` y establezca su valor en `0`. Extra - el uso de `tabindex` también permite que la pseudo-clase CSS `:focus` funcione en la etiqueta `p`.
+
+# --hints--
+
+Tu código debe agregar un atributo `tabindex` a la etiqueta `p` que contiene las instrucciones del formulario.
+
+```js
+assert($('p').attr('tabindex'));
+```
+
+Tu código debe establecer el atributo `tabindex` en la etiqueta `p` en un valor de 0.
+
+```js
+assert($('p').attr('tabindex') == '0');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<head>
+ <style>
+ p:focus {
+ background-color: yellow;
+ }
+ </style>
+</head>
+<body>
+ <header>
+ <h1>Ninja Survey</h1>
+ </header>
+ <section>
+ <form>
+
+
+ <p>Instructions: Fill in ALL your information then click <b>Submit</b></p>
+
+
+ <label for="username">Username:</label>
+ <input type="text" id="username" name="username"><br>
+ <fieldset>
+ <legend>What level ninja are you?</legend>
+ <input id="newbie" type="radio" name="levels" value="newbie">
+ <label for="newbie">Newbie Kitten</label><br>
+ <input id="intermediate" type="radio" name="levels" value="intermediate">
+ <label for="intermediate">Developing Student</label><br>
+ <input id="master" type="radio" name="levels" value="master">
+ <label for="master">9th Life Master</label>
+ </fieldset>
+ <br>
+ <fieldset>
+ <legend>Select your favorite weapons:</legend>
+ <input id="stars" type="checkbox" name="weapons" value="stars">
+ <label for="stars">Throwing Stars</label><br>
+ <input id="nunchucks" type="checkbox" name="weapons" value="nunchucks">
+ <label for="nunchucks">Nunchucks</label><br>
+ <input id="sai" type="checkbox" name="weapons" value="sai">
+ <label for="sai">Sai Set</label><br>
+ <input id="sword" type="checkbox" name="weapons" value="sword">
+ <label for="sword">Sword</label>
+ </fieldset>
+ <br>
+ <input type="submit" name="submit" value="Submit">
+ </form><br>
+ </section>
+ <footer>© 2018 Camper Cat</footer>
+</body>
+```
+
+# --solutions--
+
+```html
+<head>
+ <style>
+ p:focus {
+ background-color: yellow;
+ }
+ </style>
+</head>
+<body>
+ <header>
+ <h1>Ninja Survey</h1>
+ </header>
+ <section>
+ <form>
+
+
+ <p tabindex="0">Instructions: Fill in ALL your information then click <b>Submit</b></p>
+
+
+ <label for="username">Username:</label>
+ <input type="text" id="username" name="username"><br>
+ <fieldset>
+ <legend>What level ninja are you?</legend>
+ <input id="newbie" type="radio" name="levels" value="newbie">
+ <label for="newbie">Newbie Kitten</label><br>
+ <input id="intermediate" type="radio" name="levels" value="intermediate">
+ <label for="intermediate">Developing Student</label><br>
+ <input id="master" type="radio" name="levels" value="master">
+ <label for="master">9th Life Master</label>
+ </fieldset>
+ <br>
+ <fieldset>
+ <legend>Select your favorite weapons:</legend>
+ <input id="stars" type="checkbox" name="weapons" value="stars">
+ <label for="stars">Throwing Stars</label><br>
+ <input id="nunchucks" type="checkbox" name="weapons" value="nunchucks">
+ <label for="nunchucks">Nunchucks</label><br>
+ <input id="sai" type="checkbox" name="weapons" value="sai">
+ <label for="sai">Sai Set</label><br>
+ <input id="sword" type="checkbox" name="weapons" value="sword">
+ <label for="sword">Sword</label>
+ </fieldset>
+ <br>
+ <input type="submit" name="submit" value="Submit">
+ </form><br>
+ </section>
+ <footer>© 2018 Camper Cat</footer>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/use-tabindex-to-specify-the-order-of-keyboard-focus-for-several-elements.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/use-tabindex-to-specify-the-order-of-keyboard-focus-for-several-elements.md
new file mode 100644
index 00000000000..c08f5a05703
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/use-tabindex-to-specify-the-order-of-keyboard-focus-for-several-elements.md
@@ -0,0 +1,126 @@
+---
+id: 587d7790367417b2b2512ab1
+title: Utiliza tabindex para especificar el orden de enfoque del teclado para múltiples elementos
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cmzRRcb'
+forumTopicId: 301028
+dashedName: use-tabindex-to-specify-the-order-of-keyboard-focus-for-several-elements
+---
+
+# --description--
+
+El atributo `tabindex` también especifica el orden de tabulación exacto de los elementos. Esto se logra cuando el valor del atributo se establece en un número positivo de 1 o superior.
+
+Establecer un `tabindex="1"` hará que el teclado se enfoque primero en ese elemento. Luego, recorre la secuencia de valores `tabindex` especificados (2, 3, etc.), antes de pasar a los elementos predeterminados y `tabindex="0"`.
+
+Es importante tener en cuenta que cuando el orden de tabulación se establece de esta manera, anula el orden predeterminado (que usa el código fuente HTML). Esto puede confundir a los usuarios que esperan comenzar la navegación desde la parte superior de la página. Esta técnica puede ser necesaria en algunas circunstancias pero en términos de accesibilidad, ten cuidado antes de aplicarla.
+
+Aquí hay un ejemplo:
+
+`<div tabindex="1">I get keyboard focus, and I get it first!</div>`
+
+`<div tabindex="2">I get keyboard focus, and I get it second!</div>`
+
+# --instructions--
+
+Camper Cat tiene un campo de búsqueda en su página de Citas Inspiradoras que planea colocar en la esquina superior derecha con CSS. Él quiere que los controles de formulario de búsqueda `input` y envío `input` sean los dos primeros elementos en el orden de tabulación. Agrega un atributo `tabindex` establecido en `1` al `search` `input`, y un atributo `tabindex` establecido en `2` al `submit` `input`.
+
+# --hints--
+
+Tu código debe agregar un atributo `tabindex` a la etiqueta `search` `input`.
+
+```js
+assert($('#search').attr('tabindex'));
+```
+
+Tu código debe agregar un atributo `tabindex` a la etiqueta `submit` `input`.
+
+```js
+assert($('#submit').attr('tabindex'));
+```
+
+Tu código debe establecer el atributo `tabindex` en la etiqueta `search` `input` a un valor de 1.
+
+```js
+assert($('#search').attr('tabindex') == '1');
+```
+
+Tu código debe establecer el atributo `tabindex` en la etiqueta `submit` `input` a un valor de 2.
+
+```js
+assert($('#submit').attr('tabindex') == '2');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<body>
+ <header>
+ <h1>Even Deeper Thoughts with Master Camper Cat</h1>
+ <nav>
+ <ul>
+ <li><a href="">Home</a></li>
+ <li><a href="">Blog</a></li>
+ <li><a href="">Training</a></li>
+ </ul>
+ </nav>
+ </header>
+ <form>
+ <label for="search">Search:</label>
+
+
+ <input type="search" name="search" id="search">
+ <input type="submit" name="submit" value="Submit" id="submit">
+
+
+ </form>
+ <h2>Inspirational Quotes</h2>
+ <blockquote>
+ <p>“There's no Theory of Evolution, just a list of creatures I've allowed to live.”<br>
+ - Chuck Norris</p>
+ </blockquote>
+ <blockquote>
+ <p>“Wise men say forgiveness is divine, but never pay full price for late pizza.”<br>
+ - TMNT</p>
+ </blockquote>
+ <footer>© 2018 Camper Cat</footer>
+</body>
+```
+
+# --solutions--
+
+```html
+<body>
+ <header>
+ <h1>Even Deeper Thoughts with Master Camper Cat</h1>
+ <nav>
+ <ul>
+ <li><a href="">Home</a></li>
+ <li><a href="">Blog</a></li>
+ <li><a href="">Training</a></li>
+ </ul>
+ </nav>
+ </header>
+ <form>
+ <label for="search">Search:</label>
+
+
+ <input tabindex="1" type="search" name="search" id="search">
+ <input tabindex="2" type="submit" name="submit" value="Submit" id="submit">
+
+
+ </form>
+ <h2>Inspirational Quotes</h2>
+ <blockquote>
+ <p>“There's no Theory of Evolution, just a list of creatures I've allowed to live.”<br>
+ - Chuck Norris</p>
+ </blockquote>
+ <blockquote>
+ <p>“Wise men say forgiveness is divine, but never pay full price for late pizza.”<br>
+ - TMNT</p>
+ </blockquote>
+ <footer>© 2018 Camper Cat</footer>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/wrap-content-in-the-article-element.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/wrap-content-in-the-article-element.md
new file mode 100644
index 00000000000..234750e6097
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/wrap-content-in-the-article-element.md
@@ -0,0 +1,92 @@
+---
+id: 587d774e367417b2b2512aa0
+title: Envuelve el contenido en el elemento article
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cPp79S3'
+forumTopicId: 301029
+dashedName: wrap-content-in-the-article-element
+---
+
+# --description--
+
+`article` es otro de los nuevos elementos HTML5 que agrega significado semántico a tu lenguaje de marcado. `article` es un elemento de seccionamiento, y se utiliza para envolver contenido independiente y autónomo. La etiqueta funciona bien con entradas de blog, publicaciones en el foro o artículos de noticias.
+
+Determinar si el contenido puede estar solo suele ser una decisión complicada, pero hay un par de pruebas simples que puedes usar. Pregúntate, si elimino todo el contexto circundante, ¿ese contenido aún tendría sentido? Del mismo modo para el texto, ¿el contenido se mantendría si estuviera en una fuente RSS?
+
+Recuerda que las personas que usan tecnologías de asistencia dependen de un lenguaje de marcado organizado y semánticamente significativo para comprender mejor su trabajo.
+
+**Nota:** El elemento `section` también es nuevo HTML5, y tiene significado semántico ligeramente diferente al de `article`. Un `article` es para contenido independiente, y una `section` es para agrupar contenido relacionado temáticamente. Se pueden usar uno dentro del otro, según sea necesario. Por ejemplo, si un libro es el `article`, entonces cada capítulo es una `section`. Cuando no haya relación entre grupos de contenido, usa un `div`.
+
+`<div>` - grupos de contenido `<section>` - grupos de contenido relacionado `<article>` - grupos independientes, contenido autónomo
+
+# --instructions--
+
+Camper Cat usó etiquetas `article` para envolver las publicaciones en la página de su blog, pero olvido usarlas en la parte superior. Cambia la etiqueta `div` para usar una etiqueta `article`.
+
+# --hints--
+
+Tu código debe tener tres etiquetas `article`.
+
+```js
+assert($('article').length == 3);
+```
+
+Tu código no debe tener ninguna etiqueta `div`.
+
+```js
+assert($('div').length == 0);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<h1>Deep Thoughts with Master Camper Cat</h1>
+<main>
+ <div>
+ <h2>The Garfield Files: Lasagna as Training Fuel?</h2>
+ <p>The internet is littered with varying opinions on nutritional paradigms, from catnip paleo to hairball cleanses. But let's turn our attention to an often overlooked fitness fuel, and examine the protein-carb-NOM trifecta that is lasagna...</p>
+ </div>
+
+ <img src="samuraiSwords.jpeg" alt="">
+
+ <article>
+ <h2>Defeating your Foe: the Red Dot is Ours!</h2>
+ <p>Felines the world over have been waging war on the most persistent of foes. This red nemesis combines both cunning stealth and lightning speed. But chin up, fellow fighters, our time for victory may soon be near...</p>
+ </article>
+
+ <img src="samuraiSwords.jpeg" alt="">
+
+ <article>
+ <h2>Is Chuck Norris a Cat Person?</h2>
+ <p>Chuck Norris is widely regarded as the premier martial artist on the planet, and it's a complete coincidence anyone who disagrees with this fact mysteriously disappears soon after. But the real question is, is he a cat person?...</p>
+ </article>
+</main>
+```
+
+# --solutions--
+
+```html
+<h1>Deep Thoughts with Master Camper Cat</h1>
+<main>
+ <article>
+ <h2>The Garfield Files: Lasagna as Training Fuel?</h2>
+ <p>The internet is littered with varying opinions on nutritional paradigms, from catnip paleo to hairball cleanses. But let's turn our attention to an often overlooked fitness fuel, and examine the protein-carb-NOM trifecta that is lasagna...</p>
+ </article>
+
+ <img src="samuraiSwords.jpeg" alt="">
+
+ <article>
+ <h2>Defeating your Foe: the Red Dot is Ours!</h2>
+ <p>Felines the world over have been waging war on the most persistent of foes. This red nemesis combines both cunning stealth and lightning speed. But chin up, fellow fighters, our time for victory may soon be near...</p>
+ </article>
+
+ <img src="samuraiSwords.jpeg" alt="">
+
+ <article>
+ <h2>Is Chuck Norris a Cat Person?</h2>
+ <p>Chuck Norris is widely regarded as the premier martial artist on the planet, and it's a complete coincidence anyone who disagrees with this fact mysteriously disappears soon after. But the real question is, is he a cat person?...</p>
+ </article>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/wrap-radio-buttons-in-a-fieldset-element-for-better-accessibility.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/wrap-radio-buttons-in-a-fieldset-element-for-better-accessibility.md
new file mode 100644
index 00000000000..7e256d157d1
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-accessibility/wrap-radio-buttons-in-a-fieldset-element-for-better-accessibility.md
@@ -0,0 +1,165 @@
+---
+id: 587d778b367417b2b2512aa7
+title: Envuelve los botones de radio en un elemento fieldset para una mejor accesibilidad
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cVJVefw'
+forumTopicId: 301030
+dashedName: wrap-radio-buttons-in-a-fieldset-element-for-better-accessibility
+---
+
+# --description--
+
+El siguiente tema de formulario cubre la accesibilidad de los botones de radio. A cada opción se le da una `label` (etiqueta) con un atributo `for` vinculado al `id` del elemento correspondiente como se cubrió en el último desafío. Dado que los botones de radio a menudo vienen en un grupo donde el usuario debe elegir uno, hay una manera de mostrar semánticamente que las opciones son parte de un conjunto.
+
+La etiqueta `fieldset` rodea toda la agrupación de botones de radio para lograr esto. A menudo utiliza una etiqueta `legend` para proporcionar una descripción para la agrupación, la cual es leída por los lectores de pantalla para cada elección en el elemento `fieldset`.
+
+El contenedor `fieldset` y la etiqueta `legend` no son necesarias cuando las opciones se explican por sí mismas, como selección de género. Usar `label` con el atributo `for` para cada botón de radio es suficiente.
+
+Aquí hay un ejemplo:
+
+```html
+<form>
+ <fieldset>
+ <legend>Choose one of these three items:</legend>
+ <input id="one" type="radio" name="items" value="one">
+ <label for="one">Choice One</label><br>
+ <input id="two" type="radio" name="items" value="two">
+ <label for="two">Choice Two</label><br>
+ <input id="three" type="radio" name="items" value="three">
+ <label for="three">Choice Three</label>
+ </fieldset>
+</form>
+```
+
+# --instructions--
+
+Camper Cat quiere información sobre el nivel ninja de sus usuarios cuando se registran en su lista de correo electrónico. Agregó un conjunto de botones de radio y aprendió de nuestra última lección a usar etiquetas `label` con atributos `for` para cada opción. ¡Vamos Camper Cat! Sin embargo, su código todavía necesita ayuda. Cambia la etiqueta `div` que rodea los botones de radio a una etiqueta `fieldset` y cambia la etiqueta `p` dentro de ella a una `legend`.
+
+# --hints--
+
+Tu código debe tener una etiqueta `fieldset` alrededor del conjunto de botones de radio.
+
+```js
+assert($('fieldset').length == 1);
+```
+
+El elemento `fieldset` debe tener una etiqueta de cierre.
+
+```js
+assert(
+ code.match(/<\/fieldset>/g) &&
+ code.match(/<\/fieldset>/g).length === code.match(/<fieldset>/g).length
+);
+```
+
+Tu código debe tener una etiqueta `legend` alrededor del texto preguntando que nivel ninja es un usuario.
+
+```js
+assert($('legend').length == 1);
+```
+
+Tu código no debe tener ninguna etiqueta `div`.
+
+```js
+assert($('div').length == 0);
+```
+
+Tu código ya no debería tener una etiqueta `p` alrededor del texto preguntando que nivel ninja es un usuario.
+
+```js
+assert($('p').length == 4);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<body>
+ <header>
+ <h1>Deep Thoughts with Master Camper Cat</h1>
+ </header>
+ <section>
+ <form>
+ <p>Sign up to receive Camper Cat's blog posts by email here!</p>
+ <label for="email">Email:</label>
+ <input type="text" id="email" name="email">
+
+
+ <!-- Only change code below this line -->
+ <div>
+ <p>What level ninja are you?</p>
+ <input id="newbie" type="radio" name="levels" value="newbie">
+ <label for="newbie">Newbie Kitten</label><br>
+ <input id="intermediate" type="radio" name="levels" value="intermediate">
+ <label for="intermediate">Developing Student</label><br>
+ <input id="master" type="radio" name="levels" value="master">
+ <label for="master">Master</label>
+ </div>
+ <!-- Only change code above this line -->
+
+
+ <input type="submit" name="submit" value="Submit">
+ </form>
+ </section>
+ <article>
+ <h2>The Garfield Files: Lasagna as Training Fuel?</h2>
+ <p>The internet is littered with varying opinions on nutritional paradigms, from catnip paleo to hairball cleanses. But let's turn our attention to an often overlooked fitness fuel, and examine the protein-carb-NOM trifecta that is lasagna...</p>
+ </article>
+ <img src="samuraiSwords.jpeg" alt="">
+ <article>
+ <h2>Defeating your Foe: the Red Dot is Ours!</h2>
+ <p>Felines the world over have been waging war on the most persistent of foes. This red nemesis combines both cunning stealth and lightning speed. But chin up, fellow fighters, our time for victory may soon be near...</p>
+ </article>
+ <img src="samuraiSwords.jpeg" alt="">
+ <article>
+ <h2>Is Chuck Norris a Cat Person?</h2>
+ <p>Chuck Norris is widely regarded as the premier martial artist on the planet, and it's a complete coincidence anyone who disagrees with this fact mysteriously disappears soon after. But the real question is, is he a cat person?...</p>
+ </article>
+ <footer>© 2018 Camper Cat</footer>
+</body>
+```
+
+# --solutions--
+
+```html
+<body>
+ <header>
+ <h1>Deep Thoughts with Master Camper Cat</h1>
+ </header>
+ <section>
+ <form>
+ <p>Sign up to receive Camper Cat's blog posts by email here!</p>
+ <label for="email">Email:</label>
+ <input type="text" id="email" name="email">
+
+ <fieldset>
+ <legend>What level ninja are you?</legend>
+ <input id="newbie" type="radio" name="levels" value="newbie">
+ <label for="newbie">Newbie Kitten</label><br>
+ <input id="intermediate" type="radio" name="levels" value="intermediate">
+ <label for="intermediate">Developing Student</label><br>
+ <input id="master" type="radio" name="levels" value="master">
+ <label for="master">Master</label>
+ </fieldset>
+
+ <input type="submit" name="submit" value="Submit">
+ </form>
+ </section>
+ <article>
+ <h2>The Garfield Files: Lasagna as Training Fuel?</h2>
+ <p>The internet is littered with varying opinions on nutritional paradigms, from catnip paleo to hairball cleanses. But let's turn our attention to an often overlooked fitness fuel, and examine the protein-carb-NOM trifecta that is lasagna...</p>
+ </article>
+ <img src="samuraiSwords.jpeg" alt="">
+ <article>
+ <h2>Defeating your Foe: the Red Dot is Ours!</h2>
+ <p>Felines the world over have been waging war on the most persistent of foes. This red nemesis combines both cunning stealth and lightning speed. But chin up, fellow fighters, our time for victory may soon be near...</p>
+ </article>
+ <img src="samuraiSwords.jpeg" alt="">
+ <article>
+ <h2>Is Chuck Norris a Cat Person?</h2>
+ <p>Chuck Norris is widely regarded as the premier martial artist on the planet, and it's a complete coincidence anyone who disagrees with this fact mysteriously disappears soon after. But the real question is, is he a cat person?...</p>
+ </article>
+ <footer>© 2018 Camper Cat</footer>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/add-a-box-shadow-to-a-card-like-element.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/add-a-box-shadow-to-a-card-like-element.md
new file mode 100644
index 00000000000..15f6ceae790
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/add-a-box-shadow-to-a-card-like-element.md
@@ -0,0 +1,154 @@
+---
+id: 587d781b367417b2b2512abe
+title: Agrega box-shadow a un elemento tipo tarjeta
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cvVZdUd'
+forumTopicId: 301031
+dashedName: add-a-box-shadow-to-a-card-like-element
+---
+
+# --description--
+
+La propiedad `box-shadow` aplica una o más sombras a un elemento.
+
+La propiedad `box-shadow` toma valores para
+
+<ul>
+ <li><code>offset-x</code> (qué tan lejos extender la sombra horizontalmente desde el elemento),</li>
+ <li><code>offset-y</code> (qué tan lejos extender la sombra verticalmente desde el elemento),</li>
+ <li><code>blur-radius</code>,</li>
+ <li><code>spread-radius</code> y</li>
+ <li><code>color</code>, en ese orden.</li>
+</ul>
+
+Los valores `blur-radius` y `spread-radius` son opcionales.
+
+Se pueden crear múltiples box-shadows usando comas para separar las propiedades de cada elemento `box-shadow`.
+
+A continuación un ejemplo de CSS para crear múltiples sombras con un poco de desenfoque, con colores negros casi transparentes:
+
+```css
+box-shadow: 0 10px 20px rgba(0,0,0,0.19), 0 6px 6px rgba(0,0,0,0.23);
+```
+
+# --instructions--
+
+El elemento ahora tiene un id de `thumbnail`. Con este selector, usa el los valores del ejemplo CSS anterior para aplicar `box-shadow` sobre la tarjeta.
+
+# --hints--
+
+Tu código debe agregar la propiedad `box-shadow` para el id `thumbnail`.
+
+```js
+assert(code.match(/#thumbnail\s*?{\s*?box-shadow/g));
+```
+
+Debes usar el CSS proporcionado para el valor `box-shadow`.
+
+```js
+assert(
+ code.match(
+ /box-shadow:\s*?0\s+?10px\s+?20px\s+?rgba\(\s*?0\s*?,\s*?0\s*?,\s*?0\s*?,\s*?0?\.19\)\s*?,\s*?0\s+?6px\s+?6px\s+?rgba\(\s*?0\s*?,\s*?0\s*?,\s*?0\s*?,\s*?0?\.23\)/gi
+ )
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ background-color: rgba(45, 45, 45, 0.1);
+ padding: 10px;
+ font-size: 27px;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ }
+
+
+
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard" id="thumbnail">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Alphabet</h4>
+ <hr>
+ <p><em>Google was founded by Larry Page and Sergey Brin while they were <u>Ph.D. students</u> at <strong>Stanford University</strong>.</em></p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ background-color: rgba(45, 45, 45, 0.1);
+ padding: 10px;
+ font-size: 27px;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ }
+ #thumbnail {
+ box-shadow: 0 10px 20px rgba(0,0,0,0.19), 0 6px 6px rgba(0,0,0,0.23);
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard" id="thumbnail">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Alphabet</h4>
+ <hr>
+ <p><em>Google was founded by Larry Page and Sergey Brin while they were <u>Ph.D. students</u> at <strong>Stanford University</strong>.</em></p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-background-color-property-of-text.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-background-color-property-of-text.md
new file mode 100644
index 00000000000..22e9eb57dec
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-background-color-property-of-text.md
@@ -0,0 +1,146 @@
+---
+id: 587d781b367417b2b2512abc
+title: Ajusta la propiedad de background-color del texto
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cEDqwA6'
+forumTopicId: 301032
+dashedName: adjust-the-background-color-property-of-text
+---
+
+# --description--
+
+En lugar de ajustar el fondo general o el color del texto para que el primer plano sea fácilmente legible, puedes agregar un `background-color` al elemento que contiene el texto que deseas destacar. Este reto utiliza `rgba()` en lugar de códigos `hex` o `rgb()` normal.
+
+<blockquote>rgba significa:<br> r = red<br> g = green<br> b = blue<br> a = alfa/nivel de opacidad</blockquote>
+
+Los valores RGB pueden variar de 0 a 255. El valor alfa puede variar de 1, que es completamente opaco o un color sólido, a 0, que es completamente transparente o claro. `rgba()` es ideal para usar en este caso, ya que te permite ajustar la opacidad. Esto significa que no tienes que bloquear completamente el fondo.
+
+Utilizarás `background-color: rgba(45, 45, 45, 0.1)` para este desafío. Produce un color gris oscuro que es casi transparente dado el bajo valor de opacidad de 0.1.
+
+# --instructions--
+
+Para que el texto destaque más, ajusta el `background-color` del elemento `h4` al valor `rgba()` dado.
+
+También para el `h4`, elimina la propiedad `height` y agrega `padding` de 10px.
+
+# --hints--
+
+Tu código debe agregar una propiedad `background-color` al elemento `h4` establecido en `rgba(45, 45, 45, 0.1)`.
+
+```js
+assert(
+ /(background-color|background):rgba\(45,45,45,0?\.1\)(;?}|;)/gi.test(
+ code.replace(/\s/g, '')
+ )
+);
+```
+
+Tu código debe agregar una propiedad `padding` al elemento `h4` y establecerlo en 10 pixeles.
+
+```js
+assert(
+ $('h4').css('padding-top') == '10px' &&
+ $('h4').css('padding-right') == '10px' &&
+ $('h4').css('padding-bottom') == '10px' &&
+ $('h4').css('padding-left') == '10px'
+);
+```
+
+La propiedad `height` del elemento `h4` debe eliminarse.
+
+```js
+assert(!($('h4').css('height') == '25px'));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ height: 25px;
+
+
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Alphabet</h4>
+ <hr>
+ <p><em>Google was founded by Larry Page and Sergey Brin while they were <u>Ph.D. students</u> at <strong>Stanford University</strong>.</em></p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ padding: 10px;
+ background-color: rgba(45, 45, 45, 0.1);
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Alphabet</h4>
+ <hr>
+ <p><em>Google was founded by Larry Page and Sergey Brin while they were <u>Ph.D. students</u> at <strong>Stanford University</strong>.</em></p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-color-of-various-elements-to-complementary-colors.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-color-of-various-elements-to-complementary-colors.md
new file mode 100644
index 00000000000..71757fb4676
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-color-of-various-elements-to-complementary-colors.md
@@ -0,0 +1,130 @@
+---
+id: 587d78a4367417b2b2512ad3
+title: Adjustar los colores de varios elementos para colores complementarios
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cWmPpud'
+forumTopicId: 301033
+dashedName: adjust-the-color-of-various-elements-to-complementary-colors
+---
+
+# --description--
+
+En el desafío de colores complementarios vimos que al colocar dos colores opuestos del círculo cromático, parecen más vivos. Sin embargo, el contraste visual fuerte puede ser molesto si se utiliza en un sitio web y algunos veces pueden hacer que el texto sea difícil de leer si está dentro de un complementary-color background. En la práctica, usualmente se usa uno de los colores como dominante y los complementarios se usan para atraer atención visual a cierto contenido dentro de la página.
+
+# --instructions--
+
+Está página utilizará una sombra de cerceta (`#09A7A1`) como color dominante y el complementario naranja (`#FF790E`) para resaltar los botones de inicio de sesión. Cambia el `background-color` del `header` y `footer` de negro a cerceta. Después, cambia `h2` texto `color` también a cerceta. Por último, pon naranja el `background-color` del `button`.
+
+# --hints--
+
+El elemento `header` debe tener un `background-color` de #09A7A1.
+
+```js
+assert($('header').css('background-color') == 'rgb(9, 167, 161)');
+```
+
+El elemento `footer` debe tener un `background-color` de #09A7A1.
+
+```js
+assert($('footer').css('background-color') == 'rgb(9, 167, 161)');
+```
+
+El elemento `h2` debe tener un `color` de #09A7A1.
+
+```js
+assert($('h2').css('color') == 'rgb(9, 167, 161)');
+```
+
+El elemento `button` debe tener un `background-color` de #FF790E.
+
+```js
+assert($('button').css('background-color') == 'rgb(255, 121, 14)');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ body {
+ background-color: white;
+ }
+ header {
+ background-color: black;
+ color: white;
+ padding: 0.25em;
+ }
+ h2 {
+ color: black;
+ }
+ button {
+ background-color: white;
+ }
+ footer {
+ background-color: black;
+ color: white;
+ padding: 0.5em;
+ }
+</style>
+<header>
+ <h1>Cooking with FCC!</h1>
+</header>
+<main>
+ <article>
+ <h2>Machine Learning in the Kitchen</h2>
+ <p>Join this two day workshop that walks through how to implement cutting-edge snack-getting algorithms with a command line interface. Coding usually involves writing exact instructions, but sometimes you need your computer to execute flexible commands, like <code>fetch Pringles</code>.</p>
+ <button>Sign Up</button>
+ </article>
+ <article>
+ <h2>Bisection Vegetable Chopping</h2>
+ <p>This week-long retreat will level-up your coding ninja skills to actual ninja skills. No longer is the humble bisection search limited to sorted arrays or coding interview questions, applying its concepts in the kitchen will have you chopping carrots in O(log n) time before you know it.</p>
+ <button>Sign Up</button>
+ </article>
+</main>
+<br>
+<footer>© 2018 FCC Kitchen</footer>
+```
+
+# --solutions--
+
+```html
+<style>
+ body {
+ background-color: white;
+ }
+ header {
+ background-color: #09A7A1;
+ color: white;
+ padding: 0.25em;
+ }
+ h2 {
+ color: #09A7A1;
+ }
+ button {
+ background-color: #FF790E;
+ }
+ footer {
+ background-color: #09A7A1;
+ color: white;
+ padding: 0.5em;
+ }
+</style>
+<header>
+ <h1>Cooking with FCC!</h1>
+</header>
+<main>
+ <article>
+ <h2>Machine Learning in the Kitchen</h2>
+ <p>Join this two day workshop that walks through how to implement cutting-edge snack-getting algorithms with a command line interface. Coding usually involves writing exact instructions, but sometimes you need your computer to execute flexible commands, like <code>fetch Pringles</code>.</p>
+ <button>Sign Up</button>
+ </article>
+ <article>
+ <h2>Bisection Vegetable Chopping</h2>
+ <p>This week-long retreat will level-up your coding ninja skills to actual ninja skills. No longer is the humble bisection search limited to sorted arrays or coding interview questions, applying its concepts in the kitchen will have you chopping carrots in O(log n) time before you know it.</p>
+ <button>Sign Up</button>
+ </article>
+</main>
+<br>
+<footer>© 2018 FCC Kitchen</footer>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-height-of-an-element-using-the-height-property.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-height-of-an-element-using-the-height-property.md
new file mode 100644
index 00000000000..696529a0ca0
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-height-of-an-element-using-the-height-property.md
@@ -0,0 +1,118 @@
+---
+id: 587d7791367417b2b2512ab5
+title: Ajustar la altura de un elemento con la propiedad height
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cEDaDTN'
+forumTopicId: 301034
+dashedName: adjust-the-height-of-an-element-using-the-height-property
+---
+
+# --description--
+
+Puedes especificar la altura de un elemento con la propiedad `height` en CSS como lo haces con la propiedad `width` (ancho). En el siguiente ejemplo, se cambia la altura de una imagen a 20px:
+
+```css
+img {
+ height: 20px;
+}
+```
+
+# --instructions--
+
+Añadir la propiedad `height` a la etiqueta `h4` y establecerlo a 25px.
+
+**Nota:** Puede que necesites tener un zoom de 100% para pasar la prueba de este desafío.
+
+# --hints--
+
+El código debe cambiar la propiedad `height` del `h4` a un valor de 25 píxeles.
+
+```js
+assert(
+ Math.round(document.querySelector('h4').getBoundingClientRect().height) ===
+ 25 &&
+ /h4{\S*height:25px(;\S*}|})/.test($('style').text().replace(/\s/g, ''))
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ margin-right: 20px;
+ text-align: left;
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Google</h4>
+ <p>Google was founded by Larry Page and Sergey Brin while they were Ph.D. students at Stanford University.</p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ height: 25px;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ margin-right: 20px;
+ text-align: left;
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Google</h4>
+ <p>Google was founded by Larry Page and Sergey Brin while they were Ph.D. students at Stanford University.</p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-hover-state-of-an-anchor-tag.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-hover-state-of-an-anchor-tag.md
new file mode 100644
index 00000000000..7c5691b4ed9
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-hover-state-of-an-anchor-tag.md
@@ -0,0 +1,72 @@
+---
+id: 587d781d367417b2b2512ac8
+title: Ajustar el hover de una etiqueta ancla
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cakRGcm'
+forumTopicId: 301035
+dashedName: adjust-the-hover-state-of-an-anchor-tag
+---
+
+# --description--
+
+En este desafío, se incluye el uso de pseudo-clases. Una pseudo-clase es una palabra clave que se puede agregar a los selectores para seleccionar un estado específico de un elemento.
+
+Por ejemplo, el estilo de una etiqueta "anchor" puede ser cambiado para su estado hover usando el selector de pseudo-clase `:hover`. Aquí está el CSS para cambiar el color `color` de la etiqueta de "anchor" a rojo durante el estado hover:
+
+```css
+a:hover {
+ color: red;
+}
+```
+
+# --instructions--
+
+El editor de código tiene una regla CSS para dar estilo a todas las etiquetas `a` negras. Añade una regla para que cuando el usuario pase sobre la etiqueta `a`, el color `` sea azul.
+
+# --hints--
+
+La etiqueta anchor `color` debe permanecer en negro, solo añade reglas CSS para el estado `:hover`.
+
+```js
+assert($('a').css('color') == 'rgb(0, 0, 0)');
+```
+
+La etiqueta "anchor" debe tener un `color` al pasar el ratón.
+
+```js
+assert(
+ code.match(
+ /a:hover\s*?{\s*?color:\s*?(blue|rgba\(\s*?0\s*?,\s*?0\s*?,\s*?255\s*?,\s*?1\s*?\)|#00F|rgb\(\s*?0\s*?,\s*?0\s*?,\s*?255\s*?\))\s*?;\s*?}/gi
+ )
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ a {
+ color: #000;
+ }
+
+
+
+</style>
+<a href="https://freecatphotoapp.com/" target="_blank">CatPhotoApp</a>
+```
+
+# --solutions--
+
+```html
+<style>
+ a {
+ color: #000;
+ }
+ a:hover {
+ color: rgba(0,0,255,1);
+ }
+</style>
+<a href="https://freecatphotoapp.com/" target="_blank">CatPhotoApp</a>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-hue-of-a-color.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-hue-of-a-color.md
new file mode 100644
index 00000000000..66817ad0b9c
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-hue-of-a-color.md
@@ -0,0 +1,129 @@
+---
+id: 587d78a4367417b2b2512ad4
+title: Ajustar el tono de un color
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cPp38TZ'
+forumTopicId: 301036
+dashedName: adjust-the-hue-of-a-color
+---
+
+# --description--
+
+Los colores tienen varias características tales como el matiz, la saturación y la ligereza. CSS3 introdujo la propiedad `hsl()` como una forma alternativa de elegir un color indicando directamente estas características.
+
+Se suele pensar que **Hue** es el "color". Si imaginas un espectro de colores con un rojo en la izquierda que se torna verde en el medio y azul en la derecha, el tono es donde cabe un color a lo largo de esta línea. En `hsl()`, el tono usa un concepto de círculo cromático en lugar del espectro, donde el ángulo del color en el círculo se da como un valor entre 0 y 360.
+
+**Saturation** es la cantidad de gris en un color. Un color totalmente saturado no tiene gris y un color mínimamente saturado es casi completamente gris. Esto se da como un porcentaje con 100% de saturación.
+
+**Lightness** es la cantidad de blanco en un color. Un porcentaje se da desde 0% (negro) hasta 100% (blanco), donde 50% es el color normal.
+
+Aquí hay algunos ejemplos de `hsl()` con colores de iluminación normales y completamente saturados:
+
+<table class='table table-striped'><thead><tr><th>Color</th><th>HSL</th></tr></thead><tbody><tr><td>rojo</td><td>hsl(0, 100 %, 50 %)</td></tr><tr><td>amarillo</td><td>hsl(60, 100 %, 50 %)</td></tr><tr><td>verde</td><td>hsl(120, 100 %, 50 %)</td></tr><tr><td>cian</td><td>hsl(180, 100 %, 50 %)</td></tr><tr><td>azul</td><td>hsl(240, 100 %, 50 %)</td></tr><tr><td>magenta</td><td>hsl(300, 100 %, 50 %)</td></tr></tbody></table>
+
+# --instructions--
+
+Cambia el `background-color` de cada elemento `div` sobre la base de los nombres de las clases (`green`, `cyan` o `blue`) utilizando `hsl()`. Los tres deben tener una saturación completa y una iluminación normal.
+
+# --hints--
+
+Tu código debe usar la propiedad `hsl()` para declarar el color `green`.
+
+```js
+assert(code.match(/\.green\s*?{\s*?background-color:\s*?hsl/gi));
+```
+
+Tu código debe usar la propiedad `hsl()` para declarar el color `cyan`.
+
+```js
+assert(code.match(/\.cyan\s*?{\s*?background-color:\s*?hsl/gi));
+```
+
+Tu código debe usar la propiedad `hsl()` para declarar el color `blue`.
+
+```js
+assert(code.match(/\.blue\s*?{\s*?background-color:\s*?hsl/gi));
+```
+
+El elemento `div` con clase `green` debe tener un `background-color` verde.
+
+```js
+assert($('.green').css('background-color') == 'rgb(0, 255, 0)');
+```
+
+El elemento `div` con clase `cyan` debe tener un `background-color` cian.
+
+```js
+assert($('.cyan').css('background-color') == 'rgb(0, 255, 255)');
+```
+
+El elemento `div` con clase `blue` debe tener un `background-color` azul.
+
+```js
+assert($('.blue').css('background-color') == 'rgb(0, 0, 255)');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ body {
+ background-color: #FFFFFF;
+ }
+
+ .green {
+ background-color: #000000;
+ }
+
+ .cyan {
+ background-color: #000000;
+ }
+
+ .blue {
+ background-color: #000000;
+ }
+
+ div {
+ display: inline-block;
+ height: 100px;
+ width: 100px;
+ }
+</style>
+
+<div class="green"></div>
+<div class="cyan"></div>
+<div class="blue"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ body {
+ background-color: #FFFFFF;
+ }
+
+ .green {
+ background-color: hsl(120, 100%, 50%);
+ }
+
+ .cyan {
+ background-color: hsl(180, 100%, 50%);
+ }
+
+ .blue {
+ background-color: hsl(240, 100%, 50%);
+ }
+
+ div {
+ display: inline-block;
+ height: 100px;
+ width: 100px;
+ }
+</style>
+<div class="green"></div>
+<div class="cyan"></div>
+<div class="blue"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-size-of-a-header-versus-a-paragraph-tag.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-size-of-a-header-versus-a-paragraph-tag.md
new file mode 100644
index 00000000000..ec4f71e9527
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-size-of-a-header-versus-a-paragraph-tag.md
@@ -0,0 +1,118 @@
+---
+id: 587d781b367417b2b2512abd
+title: Ajustar el tamaño de un encabezado contra una etiqueta de párrafo
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c3bRPTz'
+forumTopicId: 301037
+dashedName: adjust-the-size-of-a-header-versus-a-paragraph-tag
+---
+
+# --description--
+
+El tamaño de fuente de las etiquetas de encabezado (`h1` a `h6`) generalmente debería ser mayor que el tamaño de fuente de las etiquetas de párrafos. Esto hace que sea más sencillo para que el usuario entienda visualmente el diseño y el nivel de importancia de cada elemento en la página. Utiliza la propiedad `font-size` para ajustar el tamaño del texto en un elemento.
+
+# --instructions--
+
+Para que el encabezado sea significativamente más grande que el párrafo, cambia la etiqueta `font-size` de la etiqueta `h4` a 27 píxeles.
+
+# --hints--
+
+Tu código debe agregar una propiedad `font-size` al elemento `h4` y establecerlo en 27 píxeles.
+
+```js
+assert($('h4').css('font-size') == '27px');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ background-color: rgba(45, 45, 45, 0.1);
+ padding: 10px;
+
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Alphabet</h4>
+ <hr>
+ <p><em>Google was founded by Larry Page and Sergey Brin while they were <u>Ph.D. students</u> at <strong>Stanford University</strong>.</em></p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ background-color: rgba(45, 45, 45, 0.1);
+ padding: 10px;
+ font-size: 27px;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Alphabet</h4>
+ <hr>
+ <p><em>Google was founded by Larry Page and Sergey Brin while they were <u>Ph.D. students</u> at <strong>Stanford University</strong>.</em></p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-tone-of-a-color.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-tone-of-a-color.md
new file mode 100644
index 00000000000..a90ae565c2f
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-tone-of-a-color.md
@@ -0,0 +1,119 @@
+---
+id: 587d78a4367417b2b2512ad5
+title: Ajustar el tono de un color
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cEDJvT7'
+forumTopicId: 301038
+dashedName: adjust-the-tone-of-a-color
+---
+
+# --description--
+
+La opción `hsl()` en CSS también hace que sencillo ajustar el tono de un color. Mezclar blanco con un tono puro crea un tinte de ese color y añadir negro hará una sombra. De forma alternativa, un tono se produce al añadir gris o tintes y sombras. Recuerda que la 's' y 'l' del `hsl()` representan saturación y ligereza, respectivamente. El porcentaje de saturación cambia la cantidad de gris y el porcentaje de luz determina el porcentaje de blanco o de negro que hay en el color. Esto es útil cuando se tiene un tono base que se quiere, pero necesita variaciones diferentes del mismo.
+
+# --instructions--
+
+Todos los elementos tienen un `background-color` predeterminado de `transparent`. Nuestro elemento `nav` parece tener un fondo `cyan` ya que el elemento detrás del mismo tiene un `background-color` establecido a `cyan`. Añade un `background-color` al elemento `nav` para que use el mismo tono `cyan`, pero tiene `80% saturation` y `25% lightness` para cambiar su tono y sombreado.
+
+# --hints--
+
+El elemento `nav` debe tener un `background-color` del tono del cyan ajustado con la propiedad `hsl()`.
+
+```js
+assert(
+ code.match(/nav\s*?{\s*?background-color:\s*?hsl\(180,\s*?80%,\s*?25%\)/gi)
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ header {
+ background-color: hsl(180, 90%, 35%);
+ color: #FFFFFF;
+ }
+
+ nav {
+
+ }
+
+ h1 {
+ text-indent: 10px;
+ padding-top: 10px;
+ }
+
+ nav ul {
+ margin: 0px;
+ padding: 5px 0px 5px 30px;
+ }
+
+ nav li {
+ display: inline;
+ margin-right: 20px;
+ }
+
+ a {
+ text-decoration: none;
+ color: inherit;
+ }
+</style>
+
+<header>
+ <h1>Cooking with FCC!</h1>
+ <nav>
+ <ul>
+ <li><a href="#">Home</a></li>
+ <li><a href="#">Classes</a></li>
+ <li><a href="#">Contact</a></li>
+ </ul>
+ </nav>
+</header>
+```
+
+# --solutions--
+
+```html
+<style>
+ header {
+ background-color: hsl(180, 90%, 35%);
+ color: #FFFFFF;
+ }
+
+ nav {
+ background-color: hsl(180, 80%, 25%);
+ }
+
+ h1 {
+ text-indent: 10px;
+ padding-top: 10px;
+ }
+
+ nav ul {
+ margin: 0px;
+ padding: 5px 0px 5px 30px;
+ }
+
+ nav li {
+ display: inline;
+ margin-right: 20px;
+ }
+
+ a {
+ text-decoration: none;
+ color: inherit;
+ }
+</style>
+<header>
+ <h1>Cooking with FCC!</h1>
+ <nav>
+ <ul>
+ <li><a href="#">Home</a></li>
+ <li><a href="#">Classes</a></li>
+ <li><a href="#">Contact</a></li>
+ </ul>
+ </nav>
+</header>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-width-of-an-element-using-the-width-property.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-width-of-an-element-using-the-width-property.md
new file mode 100644
index 00000000000..12930be17d9
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/adjust-the-width-of-an-element-using-the-width-property.md
@@ -0,0 +1,115 @@
+---
+id: 587d7791367417b2b2512ab4
+title: Ajustar el ancho de un elemento utilizando la propiedad width
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cvVLPtN'
+forumTopicId: 301039
+dashedName: adjust-the-width-of-an-element-using-the-width-property
+---
+
+# --description--
+
+Puedes especificar el ancho de un elemento con la propiedad `width` en CSS. Los valores se pueden dar en unidades de longitud relativa (como `em`, unidades de longitud absoluta (como `px`, o como un porcentaje de su elemento padre contenedor. El siguiente ejemplo cambia el ancho de una imagen a 220px:
+
+```css
+img {
+ width: 220px;
+}
+```
+
+# --instructions--
+
+Añade la propiedad `width` a toda la tarjeta y establécelo en un valor absoluto de 245px. Utiliza la clase `fullCard` para seleccionar el elemento.
+
+# --hints--
+
+El código debe cambiar la propiedad `width` de la tarjeta a 245 píxeles con el selector de clase `fullCard`.
+
+```js
+const fullCard = code.match(/\.fullCard\s*{[\s\S]+?[^}]}/g);
+assert(
+ fullCard &&
+ /width\s*:\s*245px\s*(;|})/gi.test(fullCard[0]) &&
+ $('.fullCard').css('maxWidth') === 'none'
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ margin-right: 20px;
+ text-align: left;
+ }
+ .fullCard {
+
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Google</h4>
+ <p>Google was founded by Larry Page and Sergey Brin while they were Ph.D. students at Stanford University.</p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ margin-right: 20px;
+ text-align: left;
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Google</h4>
+ <p>Google was founded by Larry Page and Sergey Brin while they were Ph.D. students at Stanford University.</p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/animate-elements-at-variable-rates.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/animate-elements-at-variable-rates.md
new file mode 100644
index 00000000000..54bf6dc485b
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/animate-elements-at-variable-rates.md
@@ -0,0 +1,141 @@
+---
+id: 587d78a8367417b2b2512ae5
+title: Elementos animados con fluctuaciones
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cZ89WA4'
+forumTopicId: 301040
+dashedName: animate-elements-at-variable-rates
+---
+
+# --description--
+
+Hay muchas formas de alterar la cantidad de animaciones de elementos similares con animaciones. Hasta ahora, esto se ha logrado al aplicar una propiedad `animation-iteration-count` y estableciendo reglas `@keyframes`.
+
+A modo de ilustración, la animación de la derecha consta de dos estrellas, cada una de las cuales disminuye en tamaño y opacidad en la marca del 20% en la regla `@keyframes`, que crea la animación centelleante. Puede cambiar la regla `@keyframes` para uno de los elementos, así las estrellas titilan con diferentes ritmos.
+
+# --instructions--
+
+Cambia la velocidad de animación del elemento con el nombre de la clase `star-1` cambiando su regla `@keyframes` al 50%.
+
+# --hints--
+
+La regla `@keyframes` para la clase `star-1` debe ser del 50%.
+
+```js
+assert(code.match(/twinkle-1\s*?{\s*?50%/g));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .stars {
+ background-color: white;
+ height: 30px;
+ width: 30px;
+ border-radius: 50%;
+ animation-iteration-count: infinite;
+ }
+
+ .star-1 {
+ margin-top: 15%;
+ margin-left: 60%;
+ animation-name: twinkle-1;
+ animation-duration: 1s;
+ }
+
+ .star-2 {
+ margin-top: 25%;
+ margin-left: 25%;
+ animation-name: twinkle-2;
+ animation-duration: 1s;
+ }
+
+ @keyframes twinkle-1 {
+ 20% {
+ transform: scale(0.5);
+ opacity: 0.5;
+ }
+ }
+
+ @keyframes twinkle-2 {
+ 20% {
+ transform: scale(0.5);
+ opacity: 0.5;
+ }
+ }
+
+ #back {
+ position: fixed;
+ padding: 0;
+ margin: 0;
+ top: 0;
+ left: 0;
+ width: 100%;
+ height: 100%;
+ background: linear-gradient(black, #000099, #66c2ff, #ffcccc, #ffeee6);
+ }
+</style>
+
+<div id="back"></div>
+<div class="star-1 stars"></div>
+<div class="star-2 stars"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ .stars {
+ background-color: white;
+ height: 30px;
+ width: 30px;
+ border-radius: 50%;
+ animation-iteration-count: infinite;
+ }
+
+ .star-1 {
+ margin-top: 15%;
+ margin-left: 60%;
+ animation-name: twinkle-1;
+ animation-duration: 1s;
+ }
+
+ .star-2 {
+ margin-top: 25%;
+ margin-left: 25%;
+ animation-name: twinkle-2;
+ animation-duration: 1s;
+ }
+
+ @keyframes twinkle-1 {
+ 50% {
+ transform: scale(0.5);
+ opacity: 0.5;
+ }
+ }
+
+ @keyframes twinkle-2 {
+ 20% {
+ transform: scale(0.5);
+ opacity: 0.5;
+ }
+ }
+
+ #back {
+ position: fixed;
+ padding: 0;
+ margin: 0;
+ top: 0;
+ left: 0;
+ width: 100%;
+ height: 100%;
+ background: linear-gradient(black, #000099, #66c2ff, #ffcccc, #ffeee6);
+ }
+</style>
+<div id="back"></div>
+<div class="star-1 stars"></div>
+<div class="star-2 stars"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/animate-elements-continually-using-an-infinite-animation-count.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/animate-elements-continually-using-an-infinite-animation-count.md
new file mode 100644
index 00000000000..68de50cb770
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/animate-elements-continually-using-an-infinite-animation-count.md
@@ -0,0 +1,105 @@
+---
+id: 587d78a8367417b2b2512ae3
+title: Animar los elementos continuamente utilizando un contador de animaciones infinitas
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cVJDVfq'
+forumTopicId: 301041
+dashedName: animate-elements-continually-using-an-infinite-animation-count
+---
+
+# --description--
+
+En los desafíos anteriores, vimos cómo utilizar algunas de las propiedades de la animación y la regla `@keyframes`. Otra propiedad de animación es la `animation-iteration-count` la cual te permite controlar cuántas veces te gustaría hacer un bucle a través de la animación. Por ejemplo:
+
+`animation-iteration-count: 3;`
+
+En este caso, la animación se detendrá después de ejecutarse 3 veces, pero es posible hacer que la animación se ejecute continuamente estableciendo ese valor en `infinite`.
+
+# --instructions--
+
+Para mantener la bola rebotando a la derecha en un bucle continuo, cambia la propiedad `animation-iteration-count` a `infinite`.
+
+# --hints--
+
+La propiedad `animation-iteration-count` debe tener un valor de `infinite`.
+
+```js
+assert($('#ball').css('animation-iteration-count') == 'infinite');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+
+ #ball {
+ width: 100px;
+ height: 100px;
+ margin: 50px auto;
+ position: relative;
+ border-radius: 50%;
+ background: linear-gradient(
+ 35deg,
+ #ccffff,
+ #ffcccc
+ );
+ animation-name: bounce;
+ animation-duration: 1s;
+ animation-iteration-count: 3;
+ }
+
+ @keyframes bounce{
+ 0% {
+ top: 0px;
+ }
+ 50% {
+ top: 249px;
+ width: 130px;
+ height: 70px;
+ }
+ 100% {
+ top: 0px;
+ }
+ }
+</style>
+<div id="ball"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ #ball {
+ width: 100px;
+ height: 100px;
+ margin: 50px auto;
+ position: relative;
+ border-radius: 50%;
+ background: linear-gradient(
+ 35deg,
+ #ccffff,
+ #ffcccc
+ );
+ animation-name: bounce;
+ animation-duration: 1s;
+ animation-iteration-count: infinite;
+ }
+
+ @keyframes bounce{
+ 0% {
+ top: 0px;
+ }
+ 50% {
+ top: 249px;
+ width: 130px;
+ height: 70px;
+ }
+ 100% {
+ top: 0px;
+ }
+ }
+</style>
+<div id="ball"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/animate-multiple-elements-at-variable-rates.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/animate-multiple-elements-at-variable-rates.md
new file mode 100644
index 00000000000..99c01a80fc1
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/animate-multiple-elements-at-variable-rates.md
@@ -0,0 +1,155 @@
+---
+id: 587d78a8367417b2b2512ae6
+title: Animar múltiples elementos con ritmos diferentes
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cnpWZc9'
+forumTopicId: 301042
+dashedName: animate-multiple-elements-at-variable-rates
+---
+
+# --description--
+
+En el desafío anterior, cambiaste el ritmo de la animación para dos elementos similares animados alterando sus reglas `@keyframes`. Puedes lograr el mismo objetivo manipulando la `animation-duration` de múltiples elementos.
+
+En la animación que se ejecuta en el editor de código, hay tres estrellas en el cielo que brillan a la misma velocidad en un ciclo continuo. Para hacerlos titilar con diferentes ritmos, puedes establecer la propiedad `animation-duration` con diferentes valores para cada elemento.
+
+# --instructions--
+
+Establece la `animation-duration` de los elementos con las clases `star-1`, `star-2`y `star-3` a 1s, 0.9s y 1.1s, respectivamente.
+
+# --hints--
+
+La propiedad `animation-duration` para la estrella con clase `star-1` debe permanecer en 1s.
+
+```js
+assert($('.star-1').css('animation-duration') == '1s');
+```
+
+La propiedad `animation-duration` para la estrella con clase `star-2` debe ser de 0.9s.
+
+```js
+assert($('.star-2').css('animation-duration') == '0.9s');
+```
+
+La propiedad `animation-duration` para la estrella con clase `star-3` debe ser de 1.1s.
+
+```js
+assert($('.star-3').css('animation-duration') == '1.1s');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .stars {
+ background-color: white;
+ height: 30px;
+ width: 30px;
+ border-radius: 50%;
+ animation-iteration-count: infinite;
+ }
+
+ .star-1 {
+ margin-top: 15%;
+ margin-left: 60%;
+ animation-duration: 1s;
+ animation-name: twinkle;
+ }
+
+ .star-2 {
+ margin-top: 25%;
+ margin-left: 25%;
+ animation-duration: 1s;
+ animation-name: twinkle;
+ }
+
+ .star-3 {
+ margin-top: 10%;
+ margin-left: 50%;
+ animation-duration: 1s;
+ animation-name: twinkle;
+ }
+
+ @keyframes twinkle {
+ 20% {
+ transform: scale(0.5);
+ opacity: 0.5;
+ }
+ }
+
+ #back {
+ position: fixed;
+ padding: 0;
+ margin: 0;
+ top: 0;
+ left: 0;
+ width: 100%;
+ height: 100%;
+ background: linear-gradient(black, #000099, #66c2ff, #ffcccc, #ffeee6);
+ }
+</style>
+
+<div id="back"></div>
+<div class="star-1 stars"></div>
+<div class="star-2 stars"></div>
+<div class="star-3 stars"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ .stars {
+ background-color: white;
+ height: 30px;
+ width: 30px;
+ border-radius: 50%;
+ animation-iteration-count: infinite;
+ }
+
+ .star-1 {
+ margin-top: 15%;
+ margin-left: 60%;
+ animation-duration: 1s;
+ animation-name: twinkle;
+ }
+
+ .star-2 {
+ margin-top: 25%;
+ margin-left: 25%;
+ animation-duration: 0.9s;
+ animation-name: twinkle;
+ }
+
+ .star-3 {
+ margin-top: 10%;
+ margin-left: 50%;
+ animation-duration: 1.1s;
+ animation-name: twinkle;
+ }
+
+ @keyframes twinkle {
+ 20% {
+ transform: scale(0.5);
+ opacity: 0.5;
+ }
+ }
+
+ #back {
+ position: fixed;
+ padding: 0;
+ margin: 0;
+ top: 0;
+ left: 0;
+ width: 100%;
+ height: 100%;
+ background: linear-gradient(black, #000099, #66c2ff, #ffcccc, #ffeee6);
+ }
+</style>
+<div id="back"></div>
+<div class="star-1 stars"></div>
+<div class="star-2 stars"></div>
+<div class="star-3 stars"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/center-an-element-horizontally-using-the-margin-property.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/center-an-element-horizontally-using-the-margin-property.md
new file mode 100644
index 00000000000..24ffb95716c
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/center-an-element-horizontally-using-the-margin-property.md
@@ -0,0 +1,56 @@
+---
+id: 587d78a3367417b2b2512ad0
+title: Centra un elemento horizontalmente usando la propiedad margin
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cyLJqU4'
+forumTopicId: 301043
+dashedName: center-an-element-horizontally-using-the-margin-property
+---
+
+# --description--
+
+Otra técnica de posicionamiento consiste en centrar un elemento de bloque horizontalmente. Una manera de hacer esto es que `margin` tenga valor auto.
+
+Este método también funciona para imágenes. Las imágenes son elementos en línea de forma predeterminada, pero se pueden cambiar a elementos de bloque cuando se establece la propiedad `display` en `block`.
+
+# --instructions--
+
+Centra el `div` en la página agregando una propiedad `margin` con un valor de `auto`.
+
+# --hints--
+
+El `div` debe tener un `margin` establecido en `auto`.
+
+```js
+assert(code.match(/margin:\s*?auto;/g));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ div {
+ background-color: blue;
+ height: 100px;
+ width: 100px;
+
+ }
+</style>
+<div></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ div {
+ background-color: blue;
+ height: 100px;
+ width: 100px;
+ margin: auto;
+ }
+</style>
+<div></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/change-an-elements-relative-position.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/change-an-elements-relative-position.md
new file mode 100644
index 00000000000..5e9c8e04229
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/change-an-elements-relative-position.md
@@ -0,0 +1,77 @@
+---
+id: 587d781e367417b2b2512ac9
+title: Cambia la posición relativa de un elemento
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/czVmMtZ'
+forumTopicId: 301044
+dashedName: change-an-elements-relative-position
+---
+
+# --description--
+
+CSS trata cada elemento HTML como su propia caja, esto es a lo que usualmente se refiere como el <dfn>Modelo de Caja de CSS</dfn>. Los elementos bloque automáticamente empiezan en una nueva línea (piensa en las etiquetas título, párrafos y divs) mientras que los elementos en línea se ubican entre el contenido al rededor (como imágenes o spans). El diseño por defecto de los elementos en esta manera se llama el <dfn>flujo normal</dfn> de un documento, pero CSS ofrece la propiedad position para sobreescribirlo.
+
+Cuando la posición de un elemento se establece a `relative`, te permite especificar como CSS lo moverá *relativo* a su posición actual dentro del flujo normal de la página. Se empareja con las propiedades de desplazamiento CSS de `left` o `right`, y `top` o `bottom`. Estas dicen cuántos pixeles, porcentajes, o ems se debe mover el elemento *lejos* de donde esté normalmente posicionado. El siguiente ejemplo mueve el párrafo 10 pixeles lejos de la parte inferior:
+
+```css
+p {
+ position: relative;
+ bottom: 10px;
+}
+```
+
+Cambiando la posición de un elemento a relative no lo quita del flujo normal; otros elementos a su alrededor aún se comportarán como si dicho elemento estuviera en su posición predeterminada.
+
+**Nota:** el posicionamiento te da mucha flexibilidad y poder sobre el diseño visual de una página. Es bueno recordar que no importa la posición de los elementos, el lenguaje de marcado HTML subyacente debe organizarse y tener sentido cuando se lee de arriba a abajo. Así es como los usuarios con discapacidades visuales (que dependen de dispositivos de asistencia como lectores de pantalla) acceden a tu contenido.
+
+# --instructions--
+
+Cambia la propiedad `position` de `h2` a `relative`, y usa CSS desplazamiento para moverlo 15 pixeles lejos de la parte superior `top` donde se ubica dentro del flujo normal. Observa que no hay impacto en las posiciones de los elementos h1 y p que están al rededor.
+
+# --hints--
+
+El elemento `h2` debe tener una propiedad `position` con valor `relative`.
+
+```js
+assert($('h2').css('position') == 'relative');
+```
+
+Tu código debe usar un desplazamiento CSS para posicionar relativamente el elemento `h2` a 15px lejos de la parte superior `top` donde se ubica normalmente.
+
+```js
+assert($('h2').css('top') == '15px');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ h2 {
+
+
+ }
+</style>
+<body>
+ <h1>On Being Well-Positioned</h1>
+ <h2>Move me!</h2>
+ <p>I still think the h2 is where it normally sits.</p>
+</body>
+```
+
+# --solutions--
+
+```html
+<style>
+ h2 {
+ position: relative;
+ top: 15px;
+ }
+</style>
+<body>
+ <h1>On Being Well-Positioned</h1>
+ <h2>Move me!</h2>
+ <p>I still think the h2 is where it normally sits.</p>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/change-animation-timing-with-keywords.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/change-animation-timing-with-keywords.md
new file mode 100644
index 00000000000..0577dafe36e
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/change-animation-timing-with-keywords.md
@@ -0,0 +1,127 @@
+---
+id: 587d78a8367417b2b2512ae7
+title: Cambia la duración de las animaciones con palabras clave
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cJKvwCM'
+forumTopicId: 301045
+dashedName: change-animation-timing-with-keywords
+---
+
+# --description--
+
+En las animaciones CSS, la propiedad `animation-timing-function` controla qué tan rápido un elemento animado cambia sobre la duración total de la animación. Si la animación es un carro moviéndose de un punto A a un punto B en un tiempo establecido (tu `animation-duration`), la `animation-timing-function` dicta cómo el carro acelera y desacelera durante el transcurso en marcha.
+
+Existe un número de palabras clave predefinidas disponibles para las opciones populares. Por ejemplo, el valor por defecto es `ease`, que empieza despacio, acelera en el medio y luego reduce la velocidad de nuevo al final. Otras opciones incluye `ease-out`, que es rápida al inicio y luego reduce la velocidad, `ease-in`, que es lenta al inicio y luego acelera al final, o `linear`, que aplica una velocidad constante a lo largo de la animación.
+
+# --instructions--
+
+Para los elementos con id `ball1` y `ball2`, agrega una propiedad `animation-timing-function` a cada uno y asigna `linear` a `#ball1` y `ease-out` a `#ball2`. Nota la diferencia entre cómo los elementos se mueven durante la animación pero terminan juntos, dado que comparten la misma `animation-duration` de 2 segundos.
+
+# --hints--
+
+El valor de la propiedad `animation-timing-function` para el elemento con el id `ball1` debe ser `linear`.
+
+```js
+const ball1Animation = __helpers.removeWhiteSpace(
+ $('#ball1').css('animation-timing-function')
+);
+assert(ball1Animation == 'linear' || ball1Animation == 'cubic-bezier(0,0,1,1)');
+```
+
+El valor de la propiedad `animation-timing-function` para el elemento con el id `ball2` debe ser `ease-out`.
+
+```js
+const ball2Animation = __helpers.removeWhiteSpace(
+ $('#ball2').css('animation-timing-function')
+);
+assert(
+ ball2Animation == 'ease-out' || ball2Animation == 'cubic-bezier(0,0,0.58,1)'
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+
+ .balls {
+ border-radius: 50%;
+ background: linear-gradient(
+ 35deg,
+ #ccffff,
+ #ffcccc
+ );
+ position: fixed;
+ width: 50px;
+ height: 50px;
+ margin-top: 50px;
+ animation-name: bounce;
+ animation-duration: 2s;
+ animation-iteration-count: infinite;
+ }
+ #ball1 {
+ left:27%;
+
+ }
+ #ball2 {
+ left:56%;
+
+ }
+
+ @keyframes bounce {
+ 0% {
+ top: 0px;
+ }
+ 100% {
+ top: 249px;
+ }
+ }
+
+</style>
+
+<div class="balls" id="ball1"></div>
+<div class="balls" id="ball2"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ .balls {
+ border-radius: 50%;
+ background: linear-gradient(
+ 35deg,
+ #ccffff,
+ #ffcccc
+ );
+ position: fixed;
+ width: 50px;
+ height: 50px;
+ margin-top: 50px;
+ animation-name: bounce;
+ animation-duration: 2s;
+ animation-iteration-count: infinite;
+ }
+ #ball1 {
+ left:27%;
+ animation-timing-function: linear;
+ }
+ #ball2 {
+ left:56%;
+ animation-timing-function: ease-out;
+ }
+
+ @keyframes bounce {
+ 0% {
+ top: 0px;
+ }
+ 100% {
+ top: 249px;
+ }
+ }
+</style>
+<div class="balls" id="ball1"></div>
+<div class="balls" id="ball2"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/change-the-position-of-overlapping-elements-with-the-z-index-property.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/change-the-position-of-overlapping-elements-with-the-z-index-property.md
new file mode 100644
index 00000000000..825dc90be4e
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/change-the-position-of-overlapping-elements-with-the-z-index-property.md
@@ -0,0 +1,81 @@
+---
+id: 587d78a3367417b2b2512acf
+title: Cambia la posición de los elementos superpuestos con la propiedad z-index
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cM94aHk'
+forumTopicId: 301046
+dashedName: change-the-position-of-overlapping-elements-with-the-z-index-property
+---
+
+# --description--
+
+Cuando los elementos son posicionados para superponerse (i.e. usando `position: absolute | relative | fixed | sticky`), el elemento que viene después dentro del marcado HTML aparecerá, por defecto, encima de los otros elementos. Sin embargo, la propiedad `z-index` puede especificar el orden de cómo los elementos están apilados unos sobre otros. Debe ser un entero (i.e. un número entero y no un decimal), y los elementos que mayor valor tengan en `z-index` serán movidos más arriba en la pila de elementos que aquellos con valores menores.
+
+# --instructions--
+
+Agrega una propiedad `z-index` al elemento con la clase `first` (el rectángulo rojo) y asígnale un valor de 2 para que cubra al otro elemento (rectángulo azul).
+
+# --hints--
+
+El elemento con clase `first` debe tener un valor `z-index` de 2.
+
+```js
+assert($('.first').css('z-index') == '2');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ div {
+ width: 60%;
+ height: 200px;
+ margin-top: 20px;
+ }
+
+ .first {
+ background-color: red;
+ position: absolute;
+
+ }
+ .second {
+ background-color: blue;
+ position: absolute;
+ left: 40px;
+ top: 50px;
+ z-index: 1;
+ }
+</style>
+
+<div class="first"></div>
+<div class="second"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ div {
+ width: 60%;
+ height: 200px;
+ margin-top: 20px;
+ }
+
+ .first {
+ background-color: red;
+ position: absolute;
+ z-index: 2;
+ }
+ .second {
+ background-color: blue;
+ position: absolute;
+ left: 40px;
+ top: 50px;
+ z-index: 1;
+ }
+</style>
+<div class="first"></div>
+<div class="second"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-a-gradual-css-linear-gradient.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-a-gradual-css-linear-gradient.md
new file mode 100644
index 00000000000..420c6796d50
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-a-gradual-css-linear-gradient.md
@@ -0,0 +1,72 @@
+---
+id: 587d78a5367417b2b2512ad6
+title: Crea un gradiente lineal de CSS gradual
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cg4dpt9'
+forumTopicId: 301047
+dashedName: create-a-gradual-css-linear-gradient
+---
+
+# --description--
+
+La aplicación de un color en elementos HTML no se limita a un tono plano. CSS proporciona la capacidad de usar transiciones de color, también conocidas como degradados, en los elementos. Esto se accede a través de la función `linear-gradient()` de la propiedad `background`. Aquí está la sintaxis general:
+
+`background: linear-gradient(gradient_direction, color 1, color 2, color 3, ...);`
+
+El primer argumento especifica la dirección desde la que comienza la transición de color - se puede establecer como un grado, donde `90deg` hace un gradiente horizontal (de izquierda a derecha) y `45deg` hace un gradiente diagonal (de abajo a izquierda hacia arriba a la derecha). Los siguientes argumentos especifican el orden de los colores utilizados en el degradado.
+
+Ejemplo:
+
+`background: linear-gradient(90deg, red, yellow, rgb(204, 204, 255));`
+
+# --instructions--
+
+Utiliza un `linear-gradient()` para el `div` del elemento `background` y configúralo desde una dirección de 35 grados para cambiar el color de `#CCFFFF` a `#FFCCCC`.
+
+# --hints--
+
+El elemento `div` debe tener un `linear-gradient` `background` con la dirección y los colores especificados.
+
+```js
+assert(
+ $('div')
+ .css('background-image')
+ .match(
+ /linear-gradient\(35deg, rgb\(204, 255, 255\), rgb\(255, 204, 204\)\)/gi
+ )
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ div {
+ border-radius: 20px;
+ width: 70%;
+ height: 400px;
+ margin: 50px auto;
+
+ }
+
+</style>
+
+<div></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ div {
+ border-radius: 20px;
+ width: 70%;
+ height: 400px;
+ margin: 50px auto;
+ background: linear-gradient(35deg, #CCFFFF, #FFCCCC);
+ }
+</style>
+<div></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-a-graphic-using-css.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-a-graphic-using-css.md
new file mode 100644
index 00000000000..61019e61b22
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-a-graphic-using-css.md
@@ -0,0 +1,89 @@
+---
+id: 587d78a6367417b2b2512add
+title: Crea un gráfico usando CSS
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cEDWPs6'
+forumTopicId: 301048
+dashedName: create-a-graphic-using-css
+---
+
+# --description--
+
+Al manipular diferentes selectores y propiedades, puedes hacer figuras interesantes. Una de las figuras más fáciles de intentar es la luna creciente. Para este desafío necesitas trabajar con la propiedad `box-shadow` que aplica la sombra de un elemento, junto con la propiedad `border-radius` que controla la redondez de las esquinas del elemento.
+
+Crearás un objeto redondo y transparente con una sombra nítida que está ligeramente desplazada hacia un lado - la sombra en realidad va a ser la figura de luna que verás.
+
+Para crear un objeto redondo, la propiedad `border-radius` se le debe asignar un valor de 50%.
+
+Puede que recuerdes de un desafío anterior que la propiedad `box-shadow` toma valores para `offset-x`, `offset-y`, `blur-radius`, `spread-radius` y un valor para el color, en ese orden. Los valores `blur-radius` y `spread-radius` son opcionales.
+
+# --instructions--
+
+Manipula el elemento cuadrado en el editor para crear la figura de luna. Primero, cambia el `background-color` a `transparent`, luego establece la propiedad `border-radius` en 50% para hacer la forma circular. Finalmente, cambia la propiedad `box-shadow` para asignar `offset-x` a 25px, `offset-y` a 10px, `blur-radius` a 0, `spread-radius` a 0 y el color a `blue`.
+
+# --hints--
+
+El valor de la propiedad `background-color` debe establecerse como `transparent`.
+
+```js
+assert(code.match(/background-color:\s*?transparent;/gi));
+```
+
+El valor de la propiedad `border-radius` debe establecerse como `50%`.
+
+```js
+assert(code.match(/border-radius:\s*?50%;/gi));
+```
+
+El valor de la propiedad `box-shadow` debe establecerse en 25px para `offset-x`, 10px para `offset-y`, 0 para `blur-radius`, 0 para `spread-radius` y finalmente `blue` para el color.
+
+```js
+assert(
+ code.match(/box-shadow:\s*?25px\s+?10px\s+?0(px)?\s+?0(px)?\s+?blue\s*?;/gi)
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .center {
+ position: absolute;
+ margin: auto;
+ top: 0;
+ right: 0;
+ bottom: 0;
+ left: 0;
+ width: 100px;
+ height: 100px;
+ background-color: blue;
+ border-radius: 0px;
+ box-shadow: 25px 10px 10px 10px green;
+ }
+
+</style>
+<div class="center"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ .center {
+ position: absolute;
+ margin: auto;
+ top: 0;
+ right: 0;
+ bottom: 0;
+ left: 0;
+ width: 100px;
+ height: 100px;
+ background-color: transparent;
+ border-radius: 50%;
+ box-shadow: 25px 10px 0 0 blue;
+ }
+</style>
+<div class="center"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-a-horizontal-line-using-the-hr-element.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-a-horizontal-line-using-the-hr-element.md
new file mode 100644
index 00000000000..fb372b3c30f
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-a-horizontal-line-using-the-hr-element.md
@@ -0,0 +1,122 @@
+---
+id: 587d781b367417b2b2512abb
+title: Crea una línea horizontal usando el elemento hr
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c3bR8t7'
+forumTopicId: 301049
+dashedName: create-a-horizontal-line-using-the-hr-element
+---
+
+# --description--
+
+Puedes usar la etiqueta `hr` para agregar una línea horizontal a través del ancho de su elemento contenedor. Esto se puede usar para definir un cambio de tema o para separa grupos de contenido visualmente.
+
+# --instructions--
+
+Agrega una etiqueta `hr` debajo de `h4` que contiene el título de la tarjeta.
+
+**Nota:** en HTML, la etiqueta `hr` se cierra sola, por lo tanto no necesita una etiqueta de cierre por separado.
+
+# --hints--
+
+Tu código debe agregar una etiqueta `hr` al lenguaje de marcado.
+
+```js
+assert($('hr').length == 1);
+```
+
+La etiqueta `hr` debe estar entre el título y el párrafo.
+
+```js
+assert(code.match(/<\/h4>\s*?<hr(>|\s*?\/>)\s*?<p>/gi));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ height: 25px;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4><s>Google</s>Alphabet</h4>
+
+ <p><em>Google was founded by Larry Page and Sergey Brin while they were <u>Ph.D. students</u> at <strong>Stanford University</strong>.</em></p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ height: 25px;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4><s>Google</s>Alphabet</h4>
+ <hr>
+ <p><em>Google was founded by Larry Page and Sergey Brin while they were <u>Ph.D. students</u> at <strong>Stanford University</strong>.</em></p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-a-more-complex-shape-using-css-and-html.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-a-more-complex-shape-using-css-and-html.md
new file mode 100644
index 00000000000..27af1f914b7
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-a-more-complex-shape-using-css-and-html.md
@@ -0,0 +1,146 @@
+---
+id: 587d78a6367417b2b2512ade
+title: Crea una forma más compleja usando CSS y HTML
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cPpz4fr'
+forumTopicId: 301050
+dashedName: create-a-more-complex-shape-using-css-and-html
+---
+
+# --description--
+
+Una de las formas más populares en el mundo es la forma del corazón, y en este desafío crearás una usando CSS puro. Pero primero, debes comprender los pseudo-elements `::before` y `::after`. Estos pseudo-elements se utilizan para agregar algo antes o después de un elemento seleccionado. En el siguiente ejemplo, se utiliza un pseudo-elemento `::before` para agregar un rectángulo a un elemento con la clase `heart`:
+
+```css
+.heart::before {
+ content: "";
+ background-color: yellow;
+ border-radius: 25%;
+ position: absolute;
+ height: 50px;
+ width: 70px;
+ top: -50px;
+ left: 5px;
+}
+```
+
+Para que los pseudo-elements `::before` y `::after` funcionen correctamente, deben tener una propiedad `content` definida. Esta propiedad generalmente se usa para agregar cosas como una foto o texto al elemento seleccionado. Cuando se utilizan los pseudo-elements `::before` y `::after` para crear formas, la propiedad `content` sigue siendo necesaria, pero se establece en una cadena vacía. En el ejemplo anterior, el elemento con la clase de `heart` tiene un pseudo-element `::before` que produce un rectángulo amarillo con una altura y un ancho de `50px` y `70px`, respectivamente. Este rectángulo tiene esquinas redondeadas debido a su 25% `border-radius` y está posicionado absolutamente a `5px` desde la izquierda y `50px` por encima de la parte superior del elemento.
+
+# --instructions--
+
+Transforma el elemento en la pantalla en un corazón. En el selector `heart::after`, cambia `background-color` a `pink` y `border-radius` a 50%.
+
+A continuación, apunta el elemento con la clase `heart` (solo `heart`) y fill en la propiedad `transform`. Utiliza la función `rotate()` con -45 grados.
+
+Finalmente, en el selector `heart::before`, establece su propiedad `content` en una cadena vacía.
+
+# --hints--
+
+La propiedad `background-color` del selector `heart::after` debe ser `pink`.
+
+```js
+const heartAfter = code.match(/\.heart::after\s*{[\s\S]+?[^\}]}/g)[0];
+assert(
+ /({|;)\s*background-color\s*:\s*pink\s*(;|})/g.test(heartAfter)
+);
+```
+
+El `border-radius` del selector `heart::after` debe ser 50%.
+
+```js
+assert(code.match(/border-radius\s*?:\s*?50%/gi).length == 2);
+```
+
+La propiedad `transform` para la clase `heart` debe usar una función `rotate()` configurada en -45 grados.
+
+```js
+assert(code.match(/transform\s*?:\s*?rotate\(\s*?-45deg\s*?\)/gi));
+```
+
+El `content` del selector `heart::before` debe ser una cadena vacía.
+
+```js
+assert(code.match(/\.heart::before\s*?{\s*?content\s*?:\s*?("|')\1\s*?;/gi));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .heart {
+ position: absolute;
+ margin: auto;
+ top: 0;
+ right: 0;
+ bottom: 0;
+ left: 0;
+ background-color: pink;
+ height: 50px;
+ width: 50px;
+ transform: ;
+ }
+ .heart::after {
+ background-color: blue;
+ content: "";
+ border-radius: 25%;
+ position: absolute;
+ width: 50px;
+ height: 50px;
+ top: 0px;
+ left: 25px;
+ }
+ .heart::before {
+ content: ;
+ background-color: pink;
+ border-radius: 50%;
+ position: absolute;
+ width: 50px;
+ height: 50px;
+ top: -25px;
+ left: 0px;
+ }
+</style>
+<div class="heart"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ .heart {
+ position: absolute;
+ margin: auto;
+ top: 0;
+ right: 0;
+ bottom: 0;
+ left: 0;
+ background-color: pink;
+ height: 50px;
+ width: 50px;
+ transform: rotate(-45deg);
+ }
+ .heart::after {
+ background-color: pink;
+ content: "";
+ border-radius: 50%;
+ position: absolute;
+ width: 50px;
+ height: 50px;
+ top: 0px;
+ left: 25px;
+ }
+ .heart::before {
+ content: "";
+ background-color: pink;
+ border-radius: 50%;
+ position: absolute;
+ width: 50px;
+ height: 50px;
+ top: -25px;
+ left: 0px;
+ }
+</style>
+<div class="heart"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-movement-using-css-animation.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-movement-using-css-animation.md
new file mode 100644
index 00000000000..a5a6de66fb0
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-movement-using-css-animation.md
@@ -0,0 +1,136 @@
+---
+id: 587d78a7367417b2b2512ae1
+title: Crear movimiento usando animación CSS
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c7amZfW'
+forumTopicId: 301051
+dashedName: create-movement-using-css-animation
+---
+
+# --description--
+
+Cuando los elementos tienen una `position`, como `fixed` o `relative`, las propiedades de desplazamiento CSS `right`, `left`, `top` y `bottom` se pueden usar en las reglas de animación para crear movimiento.
+
+Como se muestra en el siguiente ejemplo, puedes empujar el elemento hacia abajo y luego hacia arriba estableciendo la propiedad `top` fotograma clave (keyframe) a `50%` en 50px, pero estableciéndolo en 0px para el primer fotograma clave (`0%`) y el último a (`100%`).
+
+```css
+@keyframes rainbow {
+ 0% {
+ background-color: blue;
+ top: 0px;
+ }
+ 50% {
+ background-color: green;
+ top: 50px;
+ }
+ 100% {
+ background-color: yellow;
+ top: 0px;
+ }
+}
+```
+
+# --instructions--
+
+Agrega un movimiento horizontal a la animación `div`. Usando la propiedad desplazamiento `left`, agrega a la regla `@keyframes` para que el arcoíris (rainbow) comience en 0 píxeles en `0%`, se mueva a 25 píxeles en `50%`, y termine en -25 píxeles en `100%`. No reemplaces la propiedad `top` en el editor; la animación debe tener movimiento vertical y horizontal.
+
+# --hints--
+
+La regla `@keyframes` para `0%` debe usar el desplazamiento `left` de 0px.
+
+```js
+assert(code.match(/[^50]0%\s*?{[\s\S]*?left:\s*?0px(;[\s\S]*?|\s*?)}/gi));
+```
+
+La regla `@keyframes` para `50%` debe usar el desplazamiento `left` de 25px.
+
+```js
+assert(code.match(/50%\s*?{[\s\S]*?left:\s*?25px(;[\s\S]*?|\s*?)}/gi));
+```
+
+La regla `@keyframes` para `100%` debe usar el desplazamiento `left` de -25px.
+
+```js
+assert(code.match(/100%\s*?{[\s\S]*?left:\s*?-25px(;[\s\S]*?|\s*?)}/gi));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ div {
+ height: 40px;
+ width: 70%;
+ background: black;
+ margin: 50px auto;
+ border-radius: 5px;
+ position: relative;
+ }
+
+ #rect {
+ animation-name: rainbow;
+ animation-duration: 4s;
+ }
+
+ @keyframes rainbow {
+ 0% {
+ background-color: blue;
+ top: 0px;
+
+ }
+ 50% {
+ background-color: green;
+ top: 50px;
+
+ }
+ 100% {
+ background-color: yellow;
+ top: 0px;
+
+ }
+ }
+</style>
+
+<div id="rect"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ div {
+ height: 40px;
+ width: 70%;
+ background: black;
+ margin: 50px auto;
+ border-radius: 5px;
+ position: relative;
+ }
+
+ #rect {
+ animation-name: rainbow;
+ animation-duration: 4s;
+ }
+
+ @keyframes rainbow {
+ 0% {
+ background-color: blue;
+ top: 0px;
+ left: 0px;
+ }
+ 50% {
+ background-color: green;
+ top: 50px;
+ left: 25px;
+ }
+ 100% {
+ background-color: yellow;
+ top: 0px;
+ left: -25px;
+ }
+ }
+</style>
+<div id="rect"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-texture-by-adding-a-subtle-pattern-as-a-background-image.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-texture-by-adding-a-subtle-pattern-as-a-background-image.md
new file mode 100644
index 00000000000..b7450f9d52b
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-texture-by-adding-a-subtle-pattern-as-a-background-image.md
@@ -0,0 +1,50 @@
+---
+id: 587d78a5367417b2b2512ad8
+title: Crea textura agregando un patrón sutil como imagen de fondo
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cQdwJC8'
+forumTopicId: 301052
+dashedName: create-texture-by-adding-a-subtle-pattern-as-a-background-image
+---
+
+# --description--
+
+Una forma de agregar textura e interés a un fondo y hacer que se destaque más es agregar un patrón sutil. La clave está en el balance, dado que no querrás que el fondo destaque demasiado y quite atención al primer plano. La propiedad `background` acepta la función `url()` para enlazar una imagen de la textura o patrón elegido. El enlace es cubierto entre comillas dentro del paréntesis.
+
+# --instructions--
+
+Usando el enlace `https://cdn-media-1.freecodecamp.org/imgr/MJAkxbh.png`, asigna el fondo `background` de toda la página con el selector `body`.
+
+# --hints--
+
+Tu elemento `body` debe tener una propiedad `background` asignado con `url()` que tenga el enlace dado.
+
+```js
+assert(
+ code.match(
+ /background:\s*?url\(\s*("|'|)https:\/\/cdn-media-1\.freecodecamp\.org\/imgr\/MJAkxbh\.png\1\s*\)/gi
+ )
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ body {
+
+ }
+</style>
+```
+
+# --solutions--
+
+```html
+<style>
+ body {
+ background: url("https://cdn-media-1.freecodecamp.org/imgr/MJAkxbh.png");
+ }
+</style>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-visual-balance-using-the-text-align-property.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-visual-balance-using-the-text-align-property.md
new file mode 100644
index 00000000000..9a3ab89e2f9
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-visual-balance-using-the-text-align-property.md
@@ -0,0 +1,118 @@
+---
+id: 587d7791367417b2b2512ab3
+title: Crea balance visual usando la propiedad text-align
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c3b4EAp'
+forumTopicId: 301053
+dashedName: create-visual-balance-using-the-text-align-property
+---
+
+# --description--
+
+Esta sección del currículo se enfoca en Diseño Visual Aplicado. El primer grupo de desafíos están construidos sobre el diseño de la tarjeta dada para mostrar un número de principios fundamentales.
+
+El texto es frecuentemente una gran parte del contenido web. CSS tiene múltiples opciones para alinearlo con la propiedad `text-align`.
+
+`text-align: justify;` hace que todas las líneas de texto, excepto la última línea, se encuentren con los lados izquierdo y derecho de la caja.
+
+`text-align: center;` centra el texto
+
+`text-align: right;` alinea el texto hacia la derecha
+
+Y `text-align: left;` (opción por defecto) alinea el texto hacia la izquierda.
+
+# --instructions--
+
+Alinea el texto de la etiqueta `h4`, que dice "Google", al centro. Luego justifica la etiqueta párrafo que contiene información sobre cómo Google fue fundado.
+
+# --hints--
+
+Tu código debe usar la propiedad text-align en la etiqueta `h4` para establecerlo en `center`.
+
+```js
+assert($('h4').css('text-align') == 'center');
+```
+
+Tu código debe usar la propiedad text-align en la etiqueta `p` para establecerlo en `justify`.
+
+```js
+assert($('p').css('text-align') == 'justify');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ h4 {
+
+ }
+ p {
+
+ }
+ .links {
+ margin-right: 20px;
+
+ }
+ .fullCard {
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Google</h4>
+ <p>Google was founded by Larry Page and Sergey Brin while they were Ph.D. students at Stanford University.</p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ margin-right: 20px;
+
+ }
+ .fullCard {
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Google</h4>
+ <p>Google was founded by Larry Page and Sergey Brin while they were Ph.D. students at Stanford University.</p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-visual-direction-by-fading-an-element-from-left-to-right.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-visual-direction-by-fading-an-element-from-left-to-right.md
new file mode 100644
index 00000000000..4c8fae54674
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/create-visual-direction-by-fading-an-element-from-left-to-right.md
@@ -0,0 +1,95 @@
+---
+id: 587d78a7367417b2b2512ae2
+title: Crear dirección visual desvaneciendo un elemento de izquierda a derecha
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cGJqqAE'
+forumTopicId: 301054
+dashedName: create-visual-direction-by-fading-an-element-from-left-to-right
+---
+
+# --description--
+
+Para este desafío, cambiarás la opacidad (`opacity`) de un elemento animado para que se desvanezca gradualmente a medida que llega al lado derecho de la pantalla.
+
+En la animación mostrada, el elemento redondo con el fondo degradado se mueve hacia la derecha con la marca del 50% de la animación según la regla `@keyframes`.
+
+# --instructions--
+
+Apunta al elemento con el id de `ball` y agrega la propiedad `opacity` establecida en 0.1 en `50%`, de modo que el elemento se desvanezca a medida que se mueve hacia la derecha.
+
+# --hints--
+
+La regla `keyframes` para desvanecimiento debe establecer la propiedad `opacity` en 0.1 al 50%.
+
+```js
+assert(
+ code.match(
+ /@keyframes fade\s*?{\s*?50%\s*?{\s*?(?:left:\s*?60%;\s*?opacity:\s*?0?\.1;|opacity:\s*?0?\.1;\s*?left:\s*?60%;)/gi
+ )
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+
+ #ball {
+ width: 70px;
+ height: 70px;
+ margin: 50px auto;
+ position: fixed;
+ left: 20%;
+ border-radius: 50%;
+ background: linear-gradient(
+ 35deg,
+ #ccffff,
+ #ffcccc
+ );
+ animation-name: fade;
+ animation-duration: 3s;
+ }
+
+ @keyframes fade {
+ 50% {
+ left: 60%;
+
+ }
+ }
+
+</style>
+
+<div id="ball"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ #ball {
+ width: 70px;
+ height: 70px;
+ margin: 50px auto;
+ position: fixed;
+ left: 20%;
+ border-radius: 50%;
+ background: linear-gradient(
+ 35deg,
+ #ccffff,
+ #ffcccc
+ );
+ animation-name: fade;
+ animation-duration: 3s;
+ }
+
+ @keyframes fade {
+ 50% {
+ left: 60%;
+ opacity: 0.1;
+ }
+ }
+</style>
+<div id="ball"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/decrease-the-opacity-of-an-element.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/decrease-the-opacity-of-an-element.md
new file mode 100644
index 00000000000..3b972ba8490
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/decrease-the-opacity-of-an-element.md
@@ -0,0 +1,134 @@
+---
+id: 587d781c367417b2b2512abf
+title: Disminuir la opacidad de un elemento
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c7aKqu4'
+forumTopicId: 301055
+dashedName: decrease-the-opacity-of-an-element
+---
+
+# --description--
+
+La propiedad `opacity` en CSS se usa para ajustar la opacidad o, por el contrario, la transparencia de un elemento.
+
+<blockquote>Un valor de 1 es opaco, que no es transparente en absoluto.<br>Un valor de 0.5 es la mitad transparente.<br>Un valor de 0 es completamente transparente.</blockquote>
+
+El valor dado se aplicará a todo el elemento, ya sea una imagen con cierta transparencia, o los colores de primer plano y fondo para un bloque de texto.
+
+# --instructions--
+
+Establece la `opacity` de las etiquetas de anclajes en 0.7 usando la clase `links` para seleccionarlas.
+
+# --hints--
+
+Tu código debe establecer la propiedad `opacity` en 0.7 en las etiquetas de anclajes seleccionando la clase de `links`.
+
+```js
+assert(
+ /\.links\s*{([\s\S]*?;)*\s*opacity\s*:\s*0*\.70*\s*(;[\s\S]*?|\s*)}/.test(
+ $('style').text()
+ )
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ background-color: rgba(45, 45, 45, 0.1);
+ padding: 10px;
+ font-size: 27px;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+
+ }
+ #thumbnail {
+ box-shadow: 0 10px 20px rgba(0,0,0,0.19), 0 6px 6px rgba(0,0,0,0.23);
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard" id="thumbnail">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Alphabet</h4>
+ <hr>
+ <p><em>Google was founded by Larry Page and Sergey Brin while they were <u>Ph.D. students</u> at <strong>Stanford University</strong>.</em></p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ background-color: rgba(45, 45, 45, 0.1);
+ padding: 10px;
+ font-size: 27px;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ opacity: 0.7;
+ }
+ #thumbnail {
+ box-shadow: 0 10px 20px rgba(0,0,0,0.19), 0 6px 6px rgba(0,0,0,0.23);
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard" id="thumbnail">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Alphabet</h4>
+ <hr>
+ <p><em>Google was founded by Larry Page and Sergey Brin while they were <u>Ph.D. students</u> at <strong>Stanford University</strong>.</em></p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/learn-about-complementary-colors.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/learn-about-complementary-colors.md
new file mode 100644
index 00000000000..97442840617
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/learn-about-complementary-colors.md
@@ -0,0 +1,90 @@
+---
+id: 587d78a3367417b2b2512ad1
+title: Aprende sobre colores complementarios
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c2MD3Tr'
+forumTopicId: 301056
+dashedName: learn-about-complementary-colors
+---
+
+# --description--
+
+La teoría del color y su impacto en el diseño es un tema pesado y solo cubriremos los aspectos básicos en los próximos desafíos. En un sitio web, los colores llaman la atención, provocan emociones y crean una armonía visual. Con diferentes combinaciones de colores se puede cambiar el aspecto de un sitio web y requiere una planificación extensa decidirse por una paleta de color que se integre con nuestro contenido.
+
+El círculo cromático es una herramienta útil para observar cómo los colores están relacionados entre sí - es un círculo donde los tonos similares están juntos y los tonos diferentes alejados. Cuando dos colores opuestos están juntos en el círculo, se los llama colores complementarios. Se caracterizan porque si se combinan, se cancelan así mismos y crean un color gris. Sin embargo, al ubicarse juntos, estos colores parecen más brillantes y producen un contraste visual fuerte.
+
+A continuación hay algunos ejemplos de colores con sus códigos hexadecimales:
+
+<blockquote>rojo (#FF0000) y cian (#00FFFF)<br>verde(#00FF00) y magenta (#FF00FF)<br>azul (#0000FF) y amarillo (#FFF00)</blockquote>
+
+Esto es diferente del anticuado modelo de color RYB que muchos de nosotros aprendimos en la escuela, que tiene diferentes colores primarios y complementarios. La teoría moderna del color utiliza el modelo aditivo RGB (como en una pantalla de computadora) y el modelo restante CMY(K) (como en la impresión). Lee [aquí](https://en.wikipedia.org/wiki/Color_model) para obtener más información sobre este complejo tema.
+
+Hay muchas herramientas de selección de color disponibles en línea que tienen la opción de encontrar el complemento de un color.
+
+**Nota:** El uso del color puede ser una forma poderosa de agregar interés visual a una página. Sin embargo, el color por sí solo no debe utilizarse como la única manera de transmitir información importante porque los usuarios con deficiencias visuales pueden no entender ese contenido. Esta cuestión se tratará con más detalle en los desafíos de accesibilidad aplicada.
+
+# --instructions--
+
+Cambia la propiedad `background-color` de las clases `blue` y `yellow` a sus colores respectivos. Observa cómo se ven los colores unos a otros distintos de los que se comparan con el fondo blanco.
+
+# --hints--
+
+El elemento `div` con clase `blue` debe tener un `background-color` azul.
+
+```js
+assert($('.blue').css('background-color') == 'rgb(0, 0, 255)');
+```
+
+El elemento `div` con clase `yellow` debe tener un `background-color` amarillo.
+
+```js
+assert($('.yellow').css('background-color') == 'rgb(255, 255, 0)');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ body {
+ background-color: #FFFFFF;
+ }
+ .blue {
+ background-color: #000000;
+ }
+ .yellow {
+ background-color: #000000;
+ }
+ div {
+ display: inline-block;
+ height: 100px;
+ width: 100px;
+ }
+</style>
+<div class="blue"></div>
+<div class="yellow"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ body {
+ background-color: #FFFFFF;
+ }
+ .blue {
+ background-color: blue;
+ }
+ .yellow {
+ background-color: yellow;
+ }
+ div {
+ display: inline-block;
+ height: 100px;
+ width: 100px;
+ }
+</style>
+<div class="blue"></div>
+<div class="yellow"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/learn-about-tertiary-colors.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/learn-about-tertiary-colors.md
new file mode 100644
index 00000000000..c28f412dad7
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/learn-about-tertiary-colors.md
@@ -0,0 +1,114 @@
+---
+id: 587d78a4367417b2b2512ad2
+title: Aprende sobre colores terciarios
+challengeType: 0
+forumTopicId: 301057
+dashedName: learn-about-tertiary-colors
+---
+
+# --description--
+
+Los monitores y las pantallas crean diferentes colores al combinar cantidades de luz roja, verde y azul. Esto se conoce como modelo aditivo de color RGB en la teoría de moderna de color. Rojo (R), verde (G) y azul (B) «por sus siglas en inglés» son colores primarios. Al combinar dos colores primarios se los colores secundarios cian (G + B), magenta (R + B) y amarillo (R + G). Ya viste estos colores en los desafíos de colores complementarios. Estos colores secundarios son el complemento del color primario no utilizado en su creación y están frente a ese color primario en el círculo cromático. Por ejemplo, el magenta está compuesto de rojo y azul y es el complemento del verde.
+
+Los colores terciarios se forman al combinar dos colores primarios con uno de sus vecinos de color secundario. Por ejemplo, entre el modelo de color RGB, rojo (primario) y amarillo (secundario) forman naranja (terciario). Esto añade seis colores a un círculo cromático simple para un total de doce.
+
+Hay varios métodos para seleccionar colores diferentes que resultan de una combinación armoniosa en diseño. Un ejemplo que puede usar colores terciarios es el esquema de color complementario dividido. Este esquema comienza con un color base, luego lo empareja con los dos colores que están adyacentes a su complemento. Los tres colores proporcionan un fuerte contraste visual en un diseño, pero son más sutiles que utilizar dos colores complementarios.
+
+Aquí hay tres colores creados usando el esquema de dividir-complemento:
+
+<table class='table table-striped'><thead><tr><th>Color</th><th>Código hexadecimal</th></tr></thead><thead></thead><tbody><tr><td>anaranjado</td><td>#F7F00</td></tr><tr><td>cian</td><td>#00FFFF</td></tr><tr><td>frambuesa</td><td>#FF007F</td></tr></tbody></table>
+
+# --instructions--
+
+Cambia la propiedad `background-color` de las clases `orange`, `cyan` y `raspberry` a sus colores respectivos. Asegúrate de usar los códigos hexadecimales y no los nombres de colores.
+
+# --hints--
+
+El elemento `div` con clase `orange` debe tener un `background-color` naranja.
+
+```js
+assert($('.orange').css('background-color') == 'rgb(255, 127, 0)');
+```
+
+El elemento `div` con clase `cyan` debe tener un `background-color` cian.
+
+```js
+assert($('.cyan').css('background-color') == 'rgb(0, 255, 255)');
+```
+
+El elemento `div` con clase `raspberry` debe tener un `background-color` naranja.
+
+```js
+assert($('.raspberry').css('background-color') == 'rgb(255, 0, 127)');
+```
+
+Todos los valores de `background-color` para las clases de color deben ser códigos hexadecimales y no nombres de color.
+
+```js
+assert(!/background-color:\s(orange|cyan|raspberry)/.test(code));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ body {
+ background-color: #FFFFFF;
+ }
+
+ .orange {
+ background-color: #000000;
+ }
+
+ .cyan {
+ background-color: #000000;
+ }
+
+ .raspberry {
+ background-color: #000000;
+ }
+
+ div {
+ height: 100px;
+ width: 100px;
+ margin-bottom: 5px;
+ }
+</style>
+
+<div class="orange"></div>
+<div class="cyan"></div>
+<div class="raspberry"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ body {
+ background-color: #FFFFFF;
+ }
+
+ .orange {
+ background-color: #FF7F00;
+ }
+
+ .cyan {
+ background-color: #00FFFF;
+ }
+
+ .raspberry {
+ background-color: #FF007F;
+ }
+
+ div {
+ height: 100px;
+ width: 100px;
+ margin-bottom: 5px;
+ }
+</style>
+<div class="orange"></div>
+<div class="cyan"></div>
+<div class="raspberry"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/learn-how-bezier-curves-work.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/learn-how-bezier-curves-work.md
new file mode 100644
index 00000000000..cd33d45a38a
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/learn-how-bezier-curves-work.md
@@ -0,0 +1,134 @@
+---
+id: 587d78a9367417b2b2512ae8
+title: Descubre como funcionan las curvas de Bezier
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c9bDrs8'
+forumTopicId: 301058
+dashedName: learn-how-bezier-curves-work
+---
+
+# --description--
+
+El último desafío introdujo la propiedad `animation-timing-function` y algunas palabras clave que cambian la velocidad de una animación a lo largo de su duración. CSS ofrece una opción distinta a las palabras clave que proporciona un control aún más fino sobre cómo se desarrolla la animación, a través del uso de curvas Bezier.
+
+En las animaciones CSS, las curvas Bezier se utilizan con la función `cubic-bezier`. La forma de la curva representa cómo se desarrolla la animación. La curva vive en un sistema de coordenadas de 1 a 1. El eje X de este sistema de coordenadas es la duración de la animación (piénsa en una escala de tiempo), y el eje Y es el cambio en la animación.
+
+La función `cubic-bezier` consta de cuatro puntos principales que se encuentran en esta cuadrícula de 1 por 1: `p0`, `p1`, `p2` y `p3`. `p0` y `p3` están configurados para ti: son los puntos inicial y final que siempre se encuentran respectivamente en el origen (0, 0) y (1, 1). Establece los valores x e y para los otros dos puntos, y donde los coloca en la cuadrícula determina la forma de la curva para la animación que debe seguir. Esto se hace en CSS declarando los valores x e y de los puntos "anchor" `p1` y `p2` de la siguiente forma:`(x1, y1, x2, y2)`. Juntando todo, aquí hay un ejemplo de una curva de Bezier en codigo CSS:
+
+`animation-timing-function: cubic-bezier(0.25, 0.25, 0.75, 0.75);`
+
+En el ejemplo anterior, los valores x e y son equivalentes para cada punto (x1 = 0.25 = y1 y x2 = 0.75 = y2), que si recuerdas la clase de geometría, da como resultado una línea que se extiende desde el origen hasta el punto (1, 1). Esta animación es un cambio lineal de un elemento durante la duración de una animación, y es lo mismo que usar la palabra clave `linear`. En otras palabras, cambia a una velocidad constante.
+
+# --instructions--
+
+Para el elemento con el id de `ball1`, cambia el valor de la propiedad `animation-timing-function` de `linear` a su valor equivalente de la función `cubic-bezier`. Utiliza los valores de puntos dados en el ejemplo anterior.
+
+# --hints--
+
+El valor de la propiedad `animation-timing-function` para el elemento con el id `ball1` debe ser el equivalente lineal `cubic-bezier` función.
+
+```js
+assert(
+ $('#ball1').css('animation-timing-function') ==
+ 'cubic-bezier(0.25, 0.25, 0.75, 0.75)'
+);
+```
+
+El valor de la propiedad `animation-timing-function` para el elemento con el id `ball2` no debería cambiar.
+
+```js
+const ball2Animation = __helpers.removeWhiteSpace(
+ $('#ball2').css('animation-timing-function')
+);
+assert(
+ ball2Animation == 'ease-out' || ball2Animation == 'cubic-bezier(0,0,0.58,1)'
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+
+ .balls{
+ border-radius: 50%;
+ background: linear-gradient(
+ 35deg,
+ #ccffff,
+ #ffcccc
+ );
+ position: fixed;
+ width: 50px;
+ height: 50px;
+ margin-top: 50px;
+ animation-name: bounce;
+ animation-duration: 2s;
+ animation-iteration-count: infinite;
+ }
+ #ball1 {
+ left: 27%;
+ animation-timing-function: linear;
+ }
+ #ball2 {
+ left: 56%;
+ animation-timing-function: ease-out;
+ }
+
+ @keyframes bounce {
+ 0% {
+ top: 0px;
+ }
+ 100% {
+ top: 249px;
+ }
+ }
+
+</style>
+
+<div class="balls" id="ball1"></div>
+<div class="balls" id="ball2"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+
+ .balls{
+ border-radius: 50%;
+ background: linear-gradient(
+ 35deg,
+ #ccffff,
+ #ffcccc
+ );
+ position: fixed;
+ width: 50px;
+ height: 50px;
+ margin-top: 50px;
+ animation-name: bounce;
+ animation-duration: 2s;
+ animation-iteration-count: infinite;
+ }
+ #ball1 {
+ left: 27%;
+ animation-timing-function: cubic-bezier(0.25, 0.25, 0.75, 0.75);
+ }
+ #ball2 {
+ left: 56%;
+ animation-timing-function: ease-out;
+ }
+
+ @keyframes bounce {
+ 0% {
+ top: 0px;
+ }
+ 100% {
+ top: 249px;
+ }
+ }
+</style>
+<div class="balls" id="ball1"></div>
+<div class="balls" id="ball2"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/learn-how-the-css-keyframes-and-animation-properties-work.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/learn-how-the-css-keyframes-and-animation-properties-work.md
new file mode 100644
index 00000000000..6ac7d078be7
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/learn-how-the-css-keyframes-and-animation-properties-work.md
@@ -0,0 +1,136 @@
+---
+id: 587d78a7367417b2b2512adf
+title: Aprende como funcionan las propiedades de CSS @keyframes y animación
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cakprhv'
+forumTopicId: 301059
+dashedName: learn-how-the-css-keyframes-and-animation-properties-work
+---
+
+# --description--
+
+Para animar un elemento, necesitas conocer las propiedades de animación y la regla `@keyframes`. Las propiedades de animación controlan como debe comportarse la animación y la regla `@keyframes` controla lo que sucede durante esa animación. Hay ocho propiedades de animación en total. Este desafío lo mantendrá simple y cubrirá primero los dos más importantes:
+
+`animation-name` establece el nombre de la animación, que luego es utilizada por `@keyframes` para decirle a CSS que reglas van con que animaciones.
+
+`animation-duration` establece el tiempo de la animación.
+
+`@keyframes` es como especificar exactamente lo que sucede dentro de la animación durante la duración. Esto se hace dando propiedades CSS para "marcos" específicos durante la animación, con porcentajes que van del 0% al 100%. Si comparas esto con una película, las propiedades de CSS de 0% es como se muestra el elemento en la escena inicial. Las propiedades de CSS con 100% es como aparece el elemento al final, justo antes de que rueden los créditos. Luego, CSS aplica la magia para hacer la transición del elemento durante la duración dada para representar la escena. Aquí hay un ejemplo para ilustrar el uso de `@keyframes` y las propiedades de animación:
+
+```css
+#anim {
+ animation-name: colorful;
+ animation-duration: 3s;
+}
+
+@keyframes colorful {
+ 0% {
+ background-color: blue;
+ }
+ 100% {
+ background-color: yellow;
+ }
+}
+```
+
+Para el elemento `anim` id, el fragmento de codigo anterior establece el `animation-name` para `colorful` y establece el `animation-duration` a 3 segundos. A continuación, la regla `@keyframes` vincula a las propiedades de animación con el nombre `colorful`. Establece el color en azul al principio de la animación (0%) que pasara a amarillo al final de la animación (100%). No estás limitado solo a las transiciones de principio a fin, puedes establecer propiedades para el elemento para cualquier porcentaje entre 0% y 100%.
+
+# --instructions--
+
+Crea una animación para el elemento con el id `rect`, estableciendo `animation-name` en `rainbow` y `animation-duration` a 4 segundos. A continuación, declara una regla `@keyframes` y estableca el `background-color` al principio de la animación (`0%`) en azul, el centro de la animación (`50%`) en verde y el final de la animación (`100%`) en amarillo.
+
+# --hints--
+
+El elemento con id de `rect` debe tener una propiedad `animation-name` con un valor de `rainbow`.
+
+```js
+assert($('#rect').css('animation-name') == 'rainbow');
+```
+
+El elemento con id de `rect` debe tener una propiedad `animation-duration` con un valor de 4s.
+
+```js
+assert($('#rect').css('animation-duration') == '4s');
+```
+
+La regla `@keyframes` debe usar el `animation-name` de `rainbow`.
+
+```js
+assert(code.match(/@keyframes\s+?rainbow\s*?{/g));
+```
+
+La regla `@keyframes` para `rainbow` debe usar un `background-color` de `blue` al 0%.
+
+```js
+assert(code.match(/0%\s*?{\s*?background-color:\s*?blue;\s*?}/gi));
+```
+
+La regla `@keyframes` para `rainbow` debe usar un `background-color` de `green` al 50%.
+
+```js
+assert(code.match(/50%\s*?{\s*?background-color:\s*?green;\s*?}/gi));
+```
+
+La regla `@keyframes` para el rainbow debe usar un `background-color` de `yellow` al 100%.
+
+```js
+assert(code.match(/100%\s*?{\s*?background-color:\s*?yellow;\s*?}/gi));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ div {
+ height: 40px;
+ width: 70%;
+ background: black;
+ margin: 50px auto;
+ border-radius: 5px;
+ }
+
+ #rect {
+
+
+ }
+
+
+
+
+</style>
+<div id="rect"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ div {
+ height: 40px;
+ width: 70%;
+ background: black;
+ margin: 50px auto;
+ border-radius: 5px;
+ }
+
+ #rect {
+ animation-name: rainbow;
+ animation-duration: 4s;
+ }
+
+ @keyframes rainbow {
+ 0% {
+ background-color: blue;
+ }
+ 50% {
+ background-color: green;
+ }
+ 100% {
+ background-color: yellow;
+ }
+ }
+</style>
+<div id="rect"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/lock-an-element-to-its-parent-with-absolute-positioning.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/lock-an-element-to-its-parent-with-absolute-positioning.md
new file mode 100644
index 00000000000..d07adac549b
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/lock-an-element-to-its-parent-with-absolute-positioning.md
@@ -0,0 +1,90 @@
+---
+id: 587d781e367417b2b2512acb
+title: Bloquea un elemento a su padre con posicionamiento absoluto
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cyLJ7c3'
+forumTopicId: 301060
+dashedName: lock-an-element-to-its-parent-with-absolute-positioning
+---
+
+# --description--
+
+La siguiente opción para la propiedad CSS `position` es `absolute`, que bloquea el elemento en su lugar en relación con su contenedor principal. A diferencia de la posición `relative`, esto elimina el elemento del flujo normal del documento, por lo que los elementos circundantes lo ignoran. Las propiedades de desplazamiento de CSS (superior o inferior e izquierda o derecha) se utilizan para ajustar la posición.
+
+Un matiz del posicionamiento absoluto es que estará bloqueado en relación con su antepasado *posicionado* más cercano. Si olvidas agregar una regla de posición al elemento principal, (esto generalmente se hace usando `position: relative;`), el navegador seguirá buscando en la jerarquía de elementos y, en última instancia tomará por defecto la etiqueta `body`.
+
+# --instructions--
+
+Bloquea el elemento `#searchbar` en al parte superior derecha de su `section` padre declarando su `position`como `absolute`. Dale desplazamientos `top` y `right` de 50 píxeles cada uno.
+
+# --hints--
+
+El elemento `#searchbar` debe tener una `position` establecida en `absolute`.
+
+```js
+assert($('#searchbar').css('position') == 'absolute');
+```
+
+Tu código debe usar el desplazamiento CSS `top` de 50 píxeles en el elemento `#searchbar`.
+
+```js
+assert($('#searchbar').css('top') == '50px');
+```
+
+Tu código debe usar el desplazamiento CSS `right` de 50 píxeles en el elemento `#searchbar`.
+
+```js
+assert($('#searchbar').css('right') == '50px');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ #searchbar {
+
+
+
+ }
+ section {
+ position: relative;
+ }
+</style>
+<body>
+ <h1>Welcome!</h1>
+ <section>
+ <form id="searchbar">
+ <label for="search">Search:</label>
+ <input type="search" id="search" name="search">
+ <input type="submit" name="submit" value="Go!">
+ </form>
+ </section>
+</body>
+```
+
+# --solutions--
+
+```html
+<style>
+ #searchbar {
+ position: absolute;
+ top: 50px;
+ right: 50px;
+ }
+ section {
+ position: relative;
+ }
+</style>
+<body>
+ <h1>Welcome!</h1>
+ <section>
+ <form id="searchbar">
+ <label for="search">Search:</label>
+ <input type="search" id="search" name="search">
+ <input type="submit" name="submit" value="Go!">
+ </form>
+ </section>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/lock-an-element-to-the-browser-window-with-fixed-positioning.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/lock-an-element-to-the-browser-window-with-fixed-positioning.md
new file mode 100644
index 00000000000..2b3999fc862
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/lock-an-element-to-the-browser-window-with-fixed-positioning.md
@@ -0,0 +1,120 @@
+---
+id: 587d781e367417b2b2512acc
+title: Bloquear un elemento en la ventana del explorador con posicionamiento fijo (fixed)
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c2MDNUR'
+forumTopicId: 301061
+dashedName: lock-an-element-to-the-browser-window-with-fixed-positioning
+---
+
+# --description--
+
+El siguiente esquema de diseño que ofrece CSS es la posición `fixed`, que es un tipo de posicionamiento absoluto que bloquea un elemento relativo a la ventana del navegador. Similar al posicionamiento absoluto, se usa con las propiedades de desplazamiento CSS y también elimina el elemento del flujo normal del documento. Otros elementos ya no "se dan cuenta" de donde se coloca, lo que puede requerir algunos ajustes de diseño en otros lugares.
+
+Una diferencia clave entre las posiciones `fixed` y `absolute` es que un elemento con una posición fija (fixed) no se moverá cuando el usuario se desplace.
+
+# --instructions--
+
+La barra de navegación en el código está etiquetada con un id de `navbar`. Cambia su `position` para `fixed`, y el desplazamiento 0 píxeles del `top` y 0 píxeles de la `left`. Después de agregar el código, desplázate por la ventana de vista previa para ver como la navegación permanece en su lugar.
+
+# --hints--
+
+El elemento `#navbar` debe tener una `position` establecida en `fixed`.
+
+```js
+assert($('#navbar').css('position') == 'fixed');
+```
+
+Tu código debe usar el desplazamiento CSS `top` de 0 pixeles en el elemento `#navbar`.
+
+```js
+assert($('#navbar').css('top') == '0px');
+```
+
+Tu código debe usar el desplazamiento CSS `left` de 0 pixeles en el elemento `#navbar`.
+
+```js
+assert($('#navbar').css('left') == '0px');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ body {
+ min-height: 150vh;
+ }
+ #navbar {
+
+
+
+ width: 100%;
+ background-color: #767676;
+ }
+ nav ul {
+ margin: 0px;
+ padding: 5px 0px 5px 30px;
+ }
+ nav li {
+ display: inline;
+ margin-right: 20px;
+ }
+ a {
+ text-decoration: none;
+ }
+</style>
+<body>
+ <header>
+ <h1>Welcome!</h1>
+ <nav id="navbar">
+ <ul>
+ <li><a href="">Home</a></li>
+ <li><a href="">Contact</a></li>
+ </ul>
+ </nav>
+ </header>
+ <p>I shift up when the #navbar is fixed to the browser window.</p>
+</body>
+```
+
+# --solutions--
+
+```html
+<style>
+ body {
+ min-height: 150vh;
+ }
+ #navbar {
+ position: fixed;
+ top: 0;
+ left: 0;
+ width: 100%;
+ background-color: #767676;
+ }
+ nav ul {
+ margin: 0px;
+ padding: 5px 0px 5px 30px;
+ }
+ nav li {
+ display: inline;
+ margin-right: 20px;
+ }
+ a {
+ text-decoration: none;
+ }
+</style>
+<body>
+ <header>
+ <h1>Welcome!</h1>
+ <nav id="navbar">
+ <ul>
+ <li><a href="">Home</a></li>
+ <li><a href="">Contact</a></li>
+ </ul>
+ </nav>
+ </header>
+ <p>I shift up when the #navbar is fixed to the browser window.</p>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/make-a-css-heartbeat-using-an-infinite-animation-count.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/make-a-css-heartbeat-using-an-infinite-animation-count.md
new file mode 100644
index 00000000000..fa41766b18e
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/make-a-css-heartbeat-using-an-infinite-animation-count.md
@@ -0,0 +1,181 @@
+---
+id: 587d78a8367417b2b2512ae4
+title: Haz latir un corazón con CSS usando un recuento de animación infinita
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cDZpDUr'
+forumTopicId: 301062
+dashedName: make-a-css-heartbeat-using-an-infinite-animation-count
+---
+
+# --description--
+
+Aquí hay un ejemplo más de animación con la propiedad `animation-iteration-count` que usa el corazón que diseñaste en un desafío anterior.
+
+La animación del latido de un segundo consta de dos piezas animadas. Los elementos `heart` (incluyendo las piezas `:before` y `:after`) se animan para cambiar el tamaño usando la propiedad `transform`, y el fondo `div` se anima para cambiar su color usando la propiedad `background`.
+
+# --instructions--
+
+Mantén el corazón latiendo agregando la propiedad `animation-iteration-count` tanto para la clase `back` como para la clase `heart` y estableciendo el valor en `infinite`. Los selectores `heart:before` y `heart:after` no necesitan ninguna propiedad de animación.
+
+# --hints--
+
+La propiedad `animation-iteration-count` para la clase `heart` debe tener un valor de `infinite`.
+
+```js
+assert($('.heart').css('animation-iteration-count') == 'infinite');
+```
+
+La propiedad `animation-iteration-count` para la clase `back` debe tener un valor de `infinite`.
+
+```js
+assert($('.back').css('animation-iteration-count') == 'infinite');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .back {
+ position: fixed;
+ padding: 0;
+ margin: 0;
+ top: 0;
+ left: 0;
+ width: 100%;
+ height: 100%;
+ background: white;
+ animation-name: backdiv;
+ animation-duration: 1s;
+
+ }
+
+ .heart {
+ position: absolute;
+ margin: auto;
+ top: 0;
+ right: 0;
+ bottom: 0;
+ left: 0;
+ background-color: pink;
+ height: 50px;
+ width: 50px;
+ transform: rotate(-45deg);
+ animation-name: beat;
+ animation-duration: 1s;
+
+ }
+ .heart:after {
+ background-color: pink;
+ content: "";
+ border-radius: 50%;
+ position: absolute;
+ width: 50px;
+ height: 50px;
+ top: 0px;
+ left: 25px;
+ }
+ .heart:before {
+ background-color: pink;
+ content: "";
+ border-radius: 50%;
+ position: absolute;
+ width: 50px;
+ height: 50px;
+ top: -25px;
+ left: 0px;
+ }
+
+ @keyframes backdiv {
+ 50% {
+ background: #ffe6f2;
+ }
+ }
+
+ @keyframes beat {
+ 0% {
+ transform: scale(1) rotate(-45deg);
+ }
+ 50% {
+ transform: scale(0.6) rotate(-45deg);
+ }
+ }
+
+</style>
+<div class="back"></div>
+<div class="heart"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ .back {
+ position: fixed;
+ padding: 0;
+ margin: 0;
+ top: 0;
+ left: 0;
+ width: 100%;
+ height: 100%;
+ background: white;
+ animation-name: backdiv;
+ animation-duration: 1s;
+ animation-iteration-count: infinite;
+ }
+
+ .heart {
+ position: absolute;
+ margin: auto;
+ top: 0;
+ right: 0;
+ bottom: 0;
+ left: 0;
+ background-color: pink;
+ height: 50px;
+ width: 50px;
+ transform: rotate(-45deg);
+ animation-name: beat;
+ animation-duration: 1s;
+ animation-iteration-count: infinite;
+ }
+ .heart:after {
+ background-color: pink;
+ content: "";
+ border-radius: 50%;
+ position: absolute;
+ width: 50px;
+ height: 50px;
+ top: 0px;
+ left: 25px;
+ }
+ .heart:before {
+ background-color: pink;
+ content: "";
+ border-radius: 50%;
+ position: absolute;
+ width: 50px;
+ height: 50px;
+ top: -25px;
+ left: 0px;
+ }
+
+ @keyframes backdiv {
+ 50% {
+ background: #ffe6f2;
+ }
+ }
+
+ @keyframes beat {
+ 0% {
+ transform: scale(1) rotate(-45deg);
+ }
+ 50% {
+ transform: scale(0.6) rotate(-45deg);
+ }
+ }
+</style>
+<div class="back"></div>
+<div class="heart"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/make-motion-more-natural-using-a-bezier-curve.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/make-motion-more-natural-using-a-bezier-curve.md
new file mode 100644
index 00000000000..501270e64c6
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/make-motion-more-natural-using-a-bezier-curve.md
@@ -0,0 +1,119 @@
+---
+id: 587d78a9367417b2b2512aea
+title: Haz que el movimiento sea más natural usando una curva de Bezier
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c7akWUv'
+forumTopicId: 301063
+dashedName: make-motion-more-natural-using-a-bezier-curve
+---
+
+# --description--
+
+Este desafío anima un elemento para replicar el movimiento de una pelota que se hace rebotes. Los desafíos anteriores cubrirán las curvas cúbicas de Bezier `linear` y `ease-out`, sin embargo, ninguna representa el movimiento de rebotes con precisión. Necesitas personalizar una curva de Bezier para esto.
+
+La función `animation-timing-function` se realiza automáticamente en cada fotograma clave (keyframe) cuando el `animation-iteration-count` se establece en infinito. Dado que hay una regla de fotogramas clave establecida en el medio de la duración de la animación (en `50%`, da como resultado dos progresiones de animación idénticas en el movimiento hacia arriba y hacia abajo de la pelota.
+
+La siguiente curva cúbica de Bezier simula el movimiento de rebotes:
+
+`cubic-bezier(0.3, 0.4, 0.5, 1.6);`
+
+Observa que el valor de y2 es mayor que 1. Aunque la curva cúbica de Bezier se mapea en un sistema de coordenadas 1 por 1, y solo puede aceptar valores x de 0 a 1, el valor y se puede establecer en números mayores que uno. Esto da como resultado un movimiento de rebote que es ideal para simular la pelota rebotando.
+
+# --instructions--
+
+Cambia el valor de la función `animation-timing-function` del elemento con el id de `green` a una función `cubic-bezier` con valores x1, y1, x2, y2 establecidos respectivamente en 0.311, 0.441, 0.444, 1.649.
+
+# --hints--
+
+El valor de la propiedad `animation-timing-function` para el elemento con el id `green` debe ser una función `cubic-bezier` con valores x1, y1, x2, y2 como se especifica.
+
+```js
+assert(
+ $('#green').css('animation-timing-function') ==
+ 'cubic-bezier(0.311, 0.441, 0.444, 1.649)'
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .balls {
+ border-radius: 50%;
+ position: fixed;
+ width: 50px;
+ height: 50px;
+ top: 60%;
+ animation-name: jump;
+ animation-duration: 2s;
+ animation-iteration-count: infinite;
+ }
+ #red {
+ background: red;
+ left: 25%;
+ animation-timing-function: linear;
+ }
+ #blue {
+ background: blue;
+ left: 50%;
+ animation-timing-function: ease-out;
+ }
+ #green {
+ background: green;
+ left: 75%;
+ animation-timing-function: cubic-bezier(0.69, 0.1, 1, 0.1);
+ }
+
+ @keyframes jump {
+ 50% {
+ top: 10%;
+ }
+ }
+</style>
+<div class="balls" id="red"></div>
+<div class="balls" id="blue"></div>
+<div class="balls" id="green"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ .balls {
+ border-radius: 50%;
+ position: fixed;
+ width: 50px;
+ height: 50px;
+ top: 60%;
+ animation-name: jump;
+ animation-duration: 2s;
+ animation-iteration-count: infinite;
+ }
+ #red {
+ background: red;
+ left: 25%;
+ animation-timing-function: linear;
+ }
+ #blue {
+ background: blue;
+ left: 50%;
+ animation-timing-function: ease-out;
+ }
+ #green {
+ background: green;
+ left: 75%;
+ animation-timing-function: cubic-bezier(0.311, 0.441, 0.444, 1.649);
+ }
+
+ @keyframes jump {
+ 50% {
+ top: 10%;
+ }
+ }
+</style>
+<div class="balls" id="red"></div>
+<div class="balls" id="blue"></div>
+<div class="balls" id="green"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/modify-fill-mode-of-an-animation.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/modify-fill-mode-of-an-animation.md
new file mode 100644
index 00000000000..134445ea6e5
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/modify-fill-mode-of-an-animation.md
@@ -0,0 +1,90 @@
+---
+id: 58a7a6ebf9a6318348e2d5aa
+title: Modifica el modo de relleno de una animación (animation-fill-mode)
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cVJDmcE'
+forumTopicId: 301064
+dashedName: modify-fill-mode-of-an-animation
+---
+
+# --description--
+
+Eso es genial, pero aún no funciona bien. Observa como la animación se restablece después de que haya pasado `500ms`, haciendo que el botón vuelva al color original. Lo que quieres es que el botón permanezca resaltado.
+
+Esto se puede hacer estableciendo la propiedad `animation-fill-mode` en `forwards`. El `animation-fill-mode` especifica el estilo aplicado a un elemento cuando la animación ha finalizado. Puedes configurarlo así:
+
+`animation-fill-mode: forwards;`
+
+# --instructions--
+
+Establece la propiedad `animation-fill-mode` de `button:hover` en `forwards` para que botón permanezca resaltado cuando un usuario pase sobre el.
+
+# --hints--
+
+`button:hover` debe tener una propiedad `animation-fill-mode` con un valor de `forwards`.
+
+```js
+assert(
+ code.match(
+ /button\s*?:\s*?hover\s*?{[\s\S]*animation-fill-mode\s*?:\s*?forwards\s*?;[\s\S]*}/gi
+ ) &&
+ code.match(
+ /button\s*?:\s*?hover\s*?{[\s\S]*animation-name\s*?:\s*?background-color\s*?;[\s\S]*}/gi
+ ) &&
+ code.match(
+ /button\s*?:\s*?hover\s*?{[\s\S]*animation-duration\s*?:\s*?500ms\s*?;[\s\S]*}/gi
+ )
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ button {
+ border-radius: 5px;
+ color: white;
+ background-color: #0F5897;
+ padding: 5px 10px 8px 10px;
+ }
+ button:hover {
+ animation-name: background-color;
+ animation-duration: 500ms;
+ /* Only change code below this line */
+
+ /* Only change code above this line */
+ }
+ @keyframes background-color {
+ 100% {
+ background-color: #4791d0;
+ }
+ }
+</style>
+<button>Register</button>
+```
+
+# --solutions--
+
+```html
+<style>
+ button {
+ border-radius: 5px;
+ color: white;
+ background-color: #0F5897;
+ padding: 5px 10px 8px 10px;
+ }
+ button:hover {
+ animation-name: background-color;
+ animation-duration: 500ms;
+ animation-fill-mode: forwards;
+ }
+ @keyframes background-color {
+ 100% {
+ background-color: #4791d0;
+ }
+ }
+</style>
+<button>Register</button>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/move-a-relatively-positioned-element-with-css-offsets.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/move-a-relatively-positioned-element-with-css-offsets.md
new file mode 100644
index 00000000000..24287a5cc3f
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/move-a-relatively-positioned-element-with-css-offsets.md
@@ -0,0 +1,72 @@
+---
+id: 587d781e367417b2b2512aca
+title: Mueve un elemento posicionado relativamente con desplazamientos de CSS
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c9bQEA4'
+forumTopicId: 301065
+dashedName: move-a-relatively-positioned-element-with-css-offsets
+---
+
+# --description--
+
+Los desplazamientos CSS de `top` o `bottom` y `left` o `right` indican al navegador hasta que punto debe compensar un elemento en relación con el lugar donde se ubicara en el flujo normal del documento. Está compensando un elemento lejos de un punto dado, lo que aleja el elemento del lado al que se hace referencia (efectivamente, en la dirección opuesta). Como viste en el último desafío, usando el desplazamiento `top` movió el `h2` hacia abajo. Del mismo modo, usando un desplazamiento `left` mueve un elemento hacia la derecha.
+
+<img src='https://cdn-media-1.freecodecamp.org/imgr/eWWi3gZ.gif' alt='' />
+
+# --instructions--
+
+Utiliza los desplazamientos CSS para mover los `h2` 15 píxeles a la derecha y 10 píxeles hacia arriba.
+
+# --hints--
+
+Tu código debe usar un desplazamiento CSS para posicionar relativamente el `h2` 10px hacia arriba. En otras palabras, alejarlo 10px de la `bottom` de donde normalmente se encuentra.
+
+```js
+assert($('h2').css('bottom') == '10px');
+```
+
+Tu código debe usar un desplazamiento CSS para posicionar relativamente el `h2` 15px hacia la derecha. En otras palabras, alejarlo 15px de la `left` de donde normalmente se encuentra.
+
+```js
+assert($('h2').css('left') == '15px');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<head>
+<style>
+ h2 {
+ position: relative;
+
+
+ }
+</style>
+</head>
+<body>
+ <h1>On Being Well-Positioned</h1>
+ <h2>Move me!</h2>
+ <p>I still think the h2 is where it normally sits.</p>
+</body>
+```
+
+# --solutions--
+
+```html
+<head>
+<style>
+ h2 {
+ position: relative;
+ left: 15px;
+ bottom: 10px;
+ }
+</style>
+</head>
+<body>
+ <h1>On Being Well-Positioned</h1>
+ <h2>Move me!</h2>
+ <p>I still think the h2 is where it normally sits.</p>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/push-elements-left-or-right-with-the-float-property.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/push-elements-left-or-right-with-the-float-property.md
new file mode 100644
index 00000000000..1cadb283926
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/push-elements-left-or-right-with-the-float-property.md
@@ -0,0 +1,100 @@
+---
+id: 587d78a3367417b2b2512ace
+title: Empuja elementos hacia la izquierda o hacia la derecha con la propiedad float
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c2MDqu2'
+forumTopicId: 301066
+dashedName: push-elements-left-or-right-with-the-float-property
+---
+
+# --description--
+
+La siguiente herramienta de posicionamiento en realidad no usa `position`, sino que establece la propiedad `float` de un elemento. Los elementos flotantes se eliminan del flujo normal de un documento y se empujan a `left` o `right` de su elemento padre contenedor. Se usa comúnmente con la propiedad `width` para especificar cuanto espacio horizontal requiere el elemento flotante.
+
+# --instructions--
+
+El lenguaje de marcado dado funcionaria bien como un diseño de dos columnas, con los elementos `section` y `aside` uno al lado del otro. Da el `#left` elemento `float` de `left` y `#right` elemento `float` de `right`.
+
+# --hints--
+
+El elemento con id `left` debe tener un valor `float` de `left`.
+
+```js
+assert($('#left').css('float') == 'left');
+```
+
+El elemento con id `right` debe tener un valor `float` de `right`.
+
+```js
+assert($('#right').css('float') == 'right');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<head>
+ <style>
+ #left {
+
+ width: 50%;
+ }
+ #right {
+
+ width: 40%;
+ }
+ aside, section {
+ padding: 2px;
+ background-color: #ccc;
+ }
+ </style>
+</head>
+<body>
+ <header>
+ <h1>Welcome!</h1>
+ </header>
+ <section id="left">
+ <h2>Content</h2>
+ <p>Good stuff</p>
+ </section>
+ <aside id="right">
+ <h2>Sidebar</h2>
+ <p>Links</p>
+ </aside>
+</body>
+```
+
+# --solutions--
+
+```html
+<head>
+ <style>
+ #left {
+ float: left;
+ width: 50%;
+ }
+ #right {
+ float: right;
+ width: 40%;
+ }
+ aside, section {
+ padding: 2px;
+ background-color: #ccc;
+ }
+ </style>
+</head>
+<body>
+ <header>
+ <h1>Welcome!</h1>
+ </header>
+ <section id="left">
+ <h2>Content</h2>
+ <p>Good stuff</p>
+ </section>
+ <aside id="right">
+ <h2>Sidebar</h2>
+ <p>Links</p>
+ </aside>
+</body>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/set-the-font-size-for-multiple-heading-elements.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/set-the-font-size-for-multiple-heading-elements.md
new file mode 100644
index 00000000000..33307a14854
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/set-the-font-size-for-multiple-heading-elements.md
@@ -0,0 +1,114 @@
+---
+id: 587d781c367417b2b2512ac2
+title: Establece el tamaño de fuente para varios elementos de títulos
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cPpQNT3'
+forumTopicId: 301067
+dashedName: set-the-font-size-for-multiple-heading-elements
+---
+
+# --description--
+
+La propiedad `font-size` se usa para especificar que tan grande es el texto en un elemento dado. Esta regla se puede utilizar para varios elementos para crear coherencia visual del texto en una página. En este desafío, establecerá los valores para todas las etiquetas `h1` a `h6` para equilibrar los tamaños de los títulos.
+
+# --instructions-- <p>En las etiquetas <code>style</code>, establece el <code>font-size</code> de:</p>
+
+ <ul>
+ <li>Etiqueta <code>h1</code> a 68px.</li>
+ <li>Etiqueta <code>h2</code> a 52px.</li>
+ <li>Etiqueta <code>h3</code> a 40px.</li>
+ <li>Etiqueta <code>h4</code> a 32px.</li>
+ <li>Etiqueta <code>h5</code> a 21px.</li>
+ <li>Etiqueta <code>h6</code> a 14px.</li>
+ </ul>
+
+# --hints--
+
+Tu código debe establecer la propiedad `font-size` para la etiqueta `h1` en 68 píxeles.
+
+```js
+assert($('h1').css('font-size') == '68px');
+```
+
+Tu código debe establecer la propiedad `font-size` para la etiqueta `h2` en 52 píxeles.
+
+```js
+assert($('h2').css('font-size') == '52px');
+```
+
+Tu código debe establecer la propiedad `font-size` para la etiqueta `h3` en 40 píxeles.
+
+```js
+assert($('h3').css('font-size') == '40px');
+```
+
+Tu código debe establecer la propiedad `font-size` para la etiqueta `h4` en 32 píxeles.
+
+```js
+assert($('h4').css('font-size') == '32px');
+```
+
+Tu código debe establecer la propiedad `font-size` para la etiqueta `h5` en 21 píxeles.
+
+```js
+assert($('h5').css('font-size') == '21px');
+```
+
+Tu código debe establecer la propiedad `font-size` para la etiqueta `h6` en 14 píxeles.
+
+```js
+const regex = /h6\s*\{\s*font-size\s*:\s*14px\s*(;\s*\}|\})/i;
+assert.strictEqual(true, regex.test(code));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+
+
+
+
+
+
+</style>
+<h1>This is h1 text</h1>
+<h2>This is h2 text</h2>
+<h3>This is h3 text</h3>
+<h4>This is h4 text</h4>
+<h5>This is h5 text</h5>
+<h6>This is h6 text</h6>
+```
+
+# --solutions--
+
+```html
+<style>
+ h1 {
+ font-size: 68px;
+ }
+ h2 {
+ font-size: 52px;
+ }
+ h3 {
+ font-size: 40px;
+ }
+ h4 {
+ font-size: 32px;
+ }
+ h5 {
+ font-size: 21px;
+ }
+ h6 {
+ font-size: 14px;
+ }
+</style>
+<h1>This is h1 text</h1>
+<h2>This is h2 text</h2>
+<h3>This is h3 text</h3>
+<h4>This is h4 text</h4>
+<h5>This is h5 text</h5>
+<h6>This is h6 text</h6>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/set-the-font-size-of-paragraph-text.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/set-the-font-size-of-paragraph-text.md
new file mode 100644
index 00000000000..2fe8fec3ba3
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/set-the-font-size-of-paragraph-text.md
@@ -0,0 +1,52 @@
+---
+id: 587d781c367417b2b2512ac4
+title: Establece el tamaño de fuente del texto del párrafo
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cVJ36Cr'
+forumTopicId: 301068
+dashedName: set-the-font-size-of-paragraph-text
+---
+
+# --description--
+
+La propiedad `font-size` en CSS no se limita a los títulos, se puede aplicar a cualquier elemento que contenga texto.
+
+# --instructions--
+
+Cambia el valor de la propiedad `font-size` para el párrafo a 16px para hacerlo más visible.
+
+# --hints--
+
+Tu etiqueta `p` debe tener un `font-size` de 16 píxeles.
+
+```js
+assert($('p').css('font-size') == '16px');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ p {
+ font-size: 10px;
+ }
+</style>
+<p>
+ Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur.
+</p>
+```
+
+# --solutions--
+
+```html
+<style>
+ p {
+ font-size: 16px;
+ }
+</style>
+<p>
+ Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur.
+</p>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/set-the-font-weight-for-multiple-heading-elements.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/set-the-font-weight-for-multiple-heading-elements.md
new file mode 100644
index 00000000000..c57d48a1f88
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/set-the-font-weight-for-multiple-heading-elements.md
@@ -0,0 +1,132 @@
+---
+id: 587d781c367417b2b2512ac3
+title: Establece el font-weight para varios elementos de títulos
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/crVWRHq'
+forumTopicId: 301069
+dashedName: set-the-font-weight-for-multiple-heading-elements
+---
+
+# --description--
+
+Se establece el `font-size` de cada etiqueta de título en el último desafío, aquí podrás ajustar el `font-weight`.
+
+La propiedad `font-weight` establece que tan gruesos o delgados son los caracteres en una sección de texto.
+
+# --instructions--
+
+<ul><li>Establece el <code>font-weight</code> de la etiqueta <code>h1</code> en 800.</li><li>Establece el <code>font-weight</code> de la etiqueta <code>h2</code> en 600.</li><li>Establece el <code>font-weight</code> de la etiqueta <code>h3</code> en 500.</li><li>Establece el <code>font-weight</code> de la etiqueta <code>h4</code> en 400.</li><li>Establece el <code>font-weight</code> de la etiqueta <code>h5</code> en 300.</li><li>Establece <code>font-weight</code> de la etiqueta <code>h6</code> en 200.</li></ul>
+
+# --hints--
+
+Tu código debe establecer la propiedad `font-weight` para la etiqueta `h1` en 800.
+
+```js
+assert($('h1').css('font-weight') == '800');
+```
+
+Tu código debe establecer la propiedad `font-weight` para la etiqueta `h2` en 600.
+
+```js
+assert($('h2').css('font-weight') == '600');
+```
+
+Tu código debe establecer la propiedad `font-weight` para la etiqueta `h3` en 500.
+
+```js
+assert($('h3').css('font-weight') == '500');
+```
+
+Tu código debe establecer la propiedad `font-weight` para la etiqueta `h4` en 400.
+
+```js
+assert($('h4').css('font-weight') == '400');
+```
+
+Tu código debe establecer la propiedad `font-weight` para la etiqueta `h5` en 300.
+
+```js
+assert($('h5').css('font-weight') == '300');
+```
+
+Tu código debe establecer la propiedad `font-weight` para la etiqueta `h6` en 200.
+
+```js
+assert($('h6').css('font-weight') == '200');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ h1 {
+ font-size: 68px;
+
+ }
+ h2 {
+ font-size: 52px;
+
+ }
+ h3 {
+ font-size: 40px;
+
+ }
+ h4 {
+ font-size: 32px;
+
+ }
+ h5 {
+ font-size: 21px;
+
+ }
+ h6 {
+ font-size: 14px;
+
+ }
+</style>
+<h1>This is h1 text</h1>
+<h2>This is h2 text</h2>
+<h3>This is h3 text</h3>
+<h4>This is h4 text</h4>
+<h5>This is h5 text</h5>
+<h6>This is h6 text</h6>
+```
+
+# --solutions--
+
+```html
+<style>
+ h1 {
+ font-size: 68px;
+ font-weight: 800;
+ }
+ h2 {
+ font-size: 52px;
+ font-weight: 600;
+ }
+ h3 {
+ font-size: 40px;
+ font-weight: 500;
+ }
+ h4 {
+ font-size: 32px;
+ font-weight: 400;
+ }
+ h5 {
+ font-size: 21px;
+ font-weight: 300;
+ }
+ h6 {
+ font-size: 14px;
+ font-weight: 200;
+ }
+</style>
+<h1>This is h1 text</h1>
+<h2>This is h2 text</h2>
+<h3>This is h3 text</h3>
+<h4>This is h4 text</h4>
+<h5>This is h5 text</h5>
+<h6>This is h6 text</h6>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/set-the-line-height-of-paragraphs.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/set-the-line-height-of-paragraphs.md
new file mode 100644
index 00000000000..ef109836386
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/set-the-line-height-of-paragraphs.md
@@ -0,0 +1,54 @@
+---
+id: 587d781d367417b2b2512ac5
+title: Establece la line-height de los párrafos
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/crVWdcv'
+forumTopicId: 301070
+dashedName: set-the-line-height-of-paragraphs
+---
+
+# --description--
+
+CSS ofrece la propiedad `line-height` para cambiar la altura de cada línea en un bloque de texto. Como sugiere el nombre, cambia la cantidad de espacio vertical que recibe cada línea de texto.
+
+# --instructions--
+
+Agrega una propiedad `line-height` a la etiqueta `p` y establezca en 25px.
+
+# --hints--
+
+Tu código debe establecer el `line-height` de la etiqueta `p` en 25 píxeles.
+
+```js
+assert($('p').css('line-height') == '25px');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ p {
+ font-size: 16px;
+
+ }
+</style>
+<p>
+ Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur.
+</p>
+```
+
+# --solutions--
+
+```html
+<style>
+ p {
+ font-size: 16px;
+ line-height: 25px;
+ }
+</style>
+<p>
+ Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur.
+</p>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-a-bezier-curve-to-move-a-graphic.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-a-bezier-curve-to-move-a-graphic.md
new file mode 100644
index 00000000000..6da0c1d9dee
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-a-bezier-curve-to-move-a-graphic.md
@@ -0,0 +1,126 @@
+---
+id: 587d78a9367417b2b2512ae9
+title: Usa una curva de Bezier para mover un gráfico
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c6bnRCK'
+forumTopicId: 301071
+dashedName: use-a-bezier-curve-to-move-a-graphic
+---
+
+# --description--
+
+Un desafío anterior discutió la palabra clave `ease-out` que describe un cambio de animación que se acelera primero y luego se ralentiza al final de la animación. A la derecha, se muestra la diferencia entre la palabra clave `ease-out` (para el elemento azul) y la palabra clave `linear` (para el elemento rojo). Se pueden lograr progresiones de animación similares a la palabra clave `ease-out` utilizando una función de curva Bezier cúbica personalizada.
+
+En general, el cambio de los puntos de anclaje `p1` y `p2` impulsa la creación de diferentes curvas Bezier, que controlan como la animación progresa a través del tiempo. Aquí hay un ejemplo de una curva de Bezier que usa valores para imitar el estilo de facilidad:
+
+`animation-timing-function: cubic-bezier(0, 0, 0.58, 1);`
+
+Recuerda que todas las funciones `cubic-bezier` comienzan con `p0` en (0, 0) y terminan con `p3` en (1, 1). En este ejemplo, la curva se mueve más rápido a través del eje Y (comienza en 0, va a `p1` y valor de 0, luego va a `p2` y valor de 1) que se mueve a través del eje X (0 para iniciar, luego 0 para `p1`, hasta 0.58 para `p2`. Como resultado, el cambio en el elemento animado progresa más rápido que el tiempo de la animación para ese segmento. Hacia el final de la curva, la relación entre el cambio en los valores X e Y se invierte: el valor y se mueve de 1 a 1 (sin cambios), y los valores X se mueven de 0.58 a 1, lo que hace que los cambios de animación progresen más lentamente en comparación con la duración de la animación.
+
+# --instructions--
+
+Para ver el efecto de esta curva Bezier en acción, cambia la `animation-timing-function` del elemento con id de `red` a una función `cubic-bezier` con valores x1, y1, x2, y2 establecidos respectivamente en 0, 0, 0.58, 1. Esto hará que ambos elementos progresen a través de la animación de manera similar.
+
+# --hints--
+
+El valor de la propiedad `animation-timing-function` del elemento con el id `red` debe ser una función `cubic-bezier` con valores x1, y1, x2, y2 establecidos respectivamente en 0, 0, 0.58, 1.
+
+```js
+assert(
+ $('#red').css('animation-timing-function') == 'cubic-bezier(0, 0, 0.58, 1)'
+);
+```
+
+El elemento con el id `red` ya no debería tener la propiedad `animation-timing-function` de `linear`.
+
+```js
+assert($('#red').css('animation-timing-function') !== 'linear');
+```
+
+El valor de la propiedad `animation-timing-function` para el elemento con el id `blue` no debe cambiar.
+
+```js
+const blueBallAnimation = __helpers.removeWhiteSpace(
+ $('#blue').css('animation-timing-function')
+);
+assert(
+ blueBallAnimation == 'ease-out' ||
+ blueBallAnimation == 'cubic-bezier(0,0,0.58,1)'
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .balls{
+ border-radius: 50%;
+ position: fixed;
+ width: 50px;
+ height: 50px;
+ margin-top: 50px;
+ animation-name: bounce;
+ animation-duration: 2s;
+ animation-iteration-count: infinite;
+ }
+ #red {
+ background: red;
+ left: 27%;
+ animation-timing-function: linear;
+ }
+ #blue {
+ background: blue;
+ left: 56%;
+ animation-timing-function: ease-out;
+ }
+ @keyframes bounce {
+ 0% {
+ top: 0px;
+ }
+ 100% {
+ top: 249px;
+ }
+ }
+</style>
+<div class="balls" id= "red"></div>
+<div class="balls" id= "blue"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ .balls{
+ border-radius: 50%;
+ position: fixed;
+ width: 50px;
+ height: 50px;
+ margin-top: 50px;
+ animation-name: bounce;
+ animation-duration: 2s;
+ animation-iteration-count: infinite;
+ }
+ #red {
+ background: red;
+ left: 27%;
+ animation-timing-function: cubic-bezier(0, 0, 0.58, 1);
+ }
+ #blue {
+ background: blue;
+ left: 56%;
+ animation-timing-function: ease-out;
+ }
+ @keyframes bounce {
+ 0% {
+ top: 0px;
+ }
+ 100% {
+ top: 249px;
+ }
+ }
+</style>
+<div class="balls" id= "red"></div>
+<div class="balls" id= "blue"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-a-css-linear-gradient-to-create-a-striped-element.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-a-css-linear-gradient-to-create-a-striped-element.md
new file mode 100644
index 00000000000..21cc545cd8f
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-a-css-linear-gradient-to-create-a-striped-element.md
@@ -0,0 +1,111 @@
+---
+id: 587d78a5367417b2b2512ad7
+title: Utiliza un degradado lineal CSS para crear un elemento rayado
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c6bmQh2'
+forumTopicId: 301072
+dashedName: use-a-css-linear-gradient-to-create-a-striped-element
+---
+
+# --description--
+
+La función `repeating-linear-gradient()` es muy similar a `linear-gradient()` con la principal diferencia de que repite el patrón de degradado especificado. `repeating-linear-gradient()` acepta una variedad de valores, pero para simplificar, trabajarás con un valor de ángulo y valores de parada de color en este desafío.
+
+El valor del ángulo es la dirección del gradiente. Las paradas de color son como valores de ancho que marcan donde tiene lugar una transición, y se dan con un porcentaje o un número de píxeles.
+
+En el ejemplo demostrado en el editor de código, el degradado comienza con el color `yellow` a 0 píxeles que se funde en el segundo color `blue` a 40 píxeles de distancia desde el principio. Puesto que la siguiente parada de color también es de 40 píxeles, el degradado cambia inmediatamente al tercer color `green`, el cual se funde en el cuarto valor de color `red` ya que está a 80 píxeles del principio del degradado.
+
+Para este ejemplo, ayuda a pensar en las paradas de color como pares donde cada dos colores se mezclan juntos.
+
+`0px [yellow -- blend -- blue] 40px [green -- blend -- red] 80px`
+
+Si cada dos valores de parada de color son del mismo color, la mezcla no es visible porque está entre el mismo color, seguido de una dura transición hacia el siguiente color, así que terminas con rayas.
+
+# --instructions--
+
+Haz rayas cambiando el `repeating-linear-gradient()` para usar un ángulo gradiente de `45deg`, luego ajusta las dos primeras paradas de color en `yellow`, y finalmente las dos segundas paradas de color en `black`.
+
+# --hints--
+
+El ángulo del `repeating-linear-gradient()` debe ser 45deg.
+
+```js
+assert(code.match(/background:\s*?repeating-linear-gradient\(\s*?45deg/gi));
+```
+
+El ángulo del `repeating-linear-gradient()` ya no debe ser 90deg
+
+```js
+assert(!code.match(/90deg/gi));
+```
+
+La parada de color en 0 píxeles debe ser `yellow`.
+
+```js
+assert(code.match(/yellow\s+?0(px)?/gi));
+```
+
+Una parada de color a 40 píxeles debe ser `yellow`.
+
+```js
+assert(code.match(/yellow\s+?40px/gi));
+```
+
+La segunda parada de color en 40 píxeles debe ser `black`.
+
+```js
+assert(code.match(/yellow\s+?40px,\s*?black\s+?40px/gi));
+```
+
+La última parada de color en 80 píxeles debe ser `black`.
+
+```js
+assert(code.match(/black\s+?80px/gi));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+
+ div{
+ border-radius: 20px;
+ width: 70%;
+ height: 400px;
+ margin: 50 auto;
+ background: repeating-linear-gradient(
+ 90deg,
+ yellow 0px,
+ blue 40px,
+ green 40px,
+ red 80px
+ );
+ }
+
+</style>
+
+<div></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ div{
+ border-radius: 20px;
+ width: 70%;
+ height: 400px;
+ margin: 50 auto;
+ background: repeating-linear-gradient(
+ 45deg,
+ yellow 0px,
+ yellow 40px,
+ black 40px,
+ black 80px
+ );
+ }
+</style>
+<div></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-css-animation-to-change-the-hover-state-of-a-button.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-css-animation-to-change-the-hover-state-of-a-button.md
new file mode 100644
index 00000000000..40b0b75ad1b
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-css-animation-to-change-the-hover-state-of-a-button.md
@@ -0,0 +1,100 @@
+---
+id: 587d78a7367417b2b2512ae0
+title: Usa animación CSS para cambiar el estado del desplazamiento de un botón
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cg4vZAa'
+forumTopicId: 301073
+dashedName: use-css-animation-to-change-the-hover-state-of-a-button
+---
+
+# --description--
+
+Puedes usar CSS `@keyframes` para cambiar el color de un botón en su estado de desplazamiento.
+
+Aquí hay un ejemplo de como cambiar el ancho de una imagen al pasar sobre ella:
+
+```html
+<style>
+ img:hover {
+ animation-name: width;
+ animation-duration: 500ms;
+ }
+
+ @keyframes width {
+ 100% {
+ width: 40px;
+ }
+ }
+</style>
+
+<img src="https://bit.ly/smallgooglelogo" alt="Google's Logo" />
+```
+
+# --instructions--
+
+Ten en cuenta que `ms` significa milisegundos, donde 1000ms es igual a 1s.
+
+Utiliza CSS `@keyframes` para cambiar el `background-color` del elemento `button` para que se convierta en `#4791d0` cuando un usuario pase sobre él. La regla `@keyframes` solo debe tener una entrada para `100%`.
+
+# --hints--
+
+La regla @keyframes debe usar el `animation-name` background-color.
+
+```js
+assert(code.match(/@keyframes\s+?background-color\s*?{/g));
+```
+
+Debe haber una regla en `@keyframes` que cambie el `background-color` a `#4791d0` al 100%.
+
+```js
+assert(code.match(/100%\s*?{\s*?background-color:\s*?#4791d0;\s*?}/gi));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ button {
+ border-radius: 5px;
+ color: white;
+ background-color: #0F5897;
+ padding: 5px 10px 8px 10px;
+ }
+
+ button:hover {
+ animation-name: background-color;
+ animation-duration: 500ms;
+ }
+
+
+</style>
+
+<button>Register</button>
+```
+
+# --solutions--
+
+```html
+<style>
+ button {
+ border-radius: 5px;
+ color: white;
+ background-color: #0F5897;
+ padding: 5px 10px 8px 10px;
+ }
+
+ button:hover {
+ animation-name: background-color;
+ animation-duration: 500ms;
+ }
+
+ @keyframes background-color {
+ 100% {
+ background-color: #4791d0;
+ }
+ }
+</style>
+<button>Register</button>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-css-transform-property-skewx-to-skew-an-element-along-the-x-axis.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-css-transform-property-skewx-to-skew-an-element-along-the-x-axis.md
new file mode 100644
index 00000000000..a9fcae4fbdc
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-css-transform-property-skewx-to-skew-an-element-along-the-x-axis.md
@@ -0,0 +1,77 @@
+---
+id: 587d78a6367417b2b2512adb
+title: Utiliza la propiedad de transformación CSS skewX para inclinar un elemento a lo largo del eje X
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cyLP8Sr'
+forumTopicId: 301074
+dashedName: use-the-css-transform-property-skewx-to-skew-an-element-along-the-x-axis
+---
+
+# --description--
+
+La siguiente función de la propiedad `transform` es `skewX()`, que inclinar el elemento seleccionado a lo largo de su eje X (horizontal).
+
+El siguiente código inclina el elemento de párrafo en -32 grados a lo largo del eje X.
+
+```css
+p {
+ transform: skewX(-32deg);
+}
+```
+
+# --instructions--
+
+Inclina el elemento con el id de `bottom` en 24 grados a lo largo del eje X utilizando la propiedad `transform`.
+
+# --hints--
+
+El elemento con id `bottom` debe estar inclinado por 24 grados a lo largo de su eje X.
+
+```js
+assert(code.match(/#bottom\s*?{\s*?.*?\s*?transform:\s*?skewX\(24deg\);/g));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ div {
+ width: 70%;
+ height: 100px;
+ margin: 50px auto;
+ }
+ #top {
+ background-color: red;
+ }
+ #bottom {
+ background-color: blue;
+
+ }
+</style>
+
+<div id="top"></div>
+<div id="bottom"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ div {
+ width: 70%;
+ height: 100px;
+ margin: 50px auto;
+ }
+ #top {
+ background-color: red;
+ }
+ #bottom {
+ background-color: blue;
+ transform: skewX(24deg);
+ }
+</style>
+<div id="top"></div>
+<div id="bottom"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-css-transform-property-skewy-to-skew-an-element-along-the-y-axis.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-css-transform-property-skewy-to-skew-an-element-along-the-y-axis.md
new file mode 100644
index 00000000000..083424608a5
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-css-transform-property-skewy-to-skew-an-element-along-the-y-axis.md
@@ -0,0 +1,71 @@
+---
+id: 587d78a6367417b2b2512adc
+title: Utiliza la propiedad de transformación CSS skewY para inclinar un elemento a lo largo del eje Y
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c2MZ2uB'
+forumTopicId: 301075
+dashedName: use-the-css-transform-property-skewy-to-skew-an-element-along-the-y-axis
+---
+
+# --description--
+
+Dado que la función `skewX()` inclina el elemento seleccionado a lo largo del eje X en un grado dado, no sorprende que la propiedad `skewY()` incline un elemento a lo largo del eje Y (vertical).
+
+# --instructions--
+
+Inclina el elemento con el id de `top` -10 grados a lo largo del eje Y utilizando la propiedad `transform`.
+
+# --hints--
+
+El elemento con id `top` debe estar inclinado en -10 grados a lo largo de su eje Y.
+
+```js
+assert(code.match(/#top\s*?{\s*?.*?\s*?transform:\s*?skewY\(-10deg\);/g));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ div {
+ width: 70%;
+ height: 100px;
+ margin: 50px auto;
+ }
+ #top {
+ background-color: red;
+
+ }
+ #bottom {
+ background-color: blue;
+ transform: skewX(24deg);
+ }
+</style>
+
+<div id="top"></div>
+<div id="bottom"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ div {
+ width: 70%;
+ height: 100px;
+ margin: 50px auto;
+ }
+ #top {
+ background-color: red;
+ transform: skewY(-10deg);
+ }
+ #bottom {
+ background-color: blue;
+ transform: skewX(24deg);
+ }
+</style>
+<div id="top"></div>
+<div id="bottom"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-css-transform-scale-property-to-change-the-size-of-an-element.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-css-transform-scale-property-to-change-the-size-of-an-element.md
new file mode 100644
index 00000000000..032c6ab9382
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-css-transform-scale-property-to-change-the-size-of-an-element.md
@@ -0,0 +1,95 @@
+---
+id: 587d78a5367417b2b2512ad9
+title: Utiliza la propiedad de escala de transformación CSS para cambiar el tamaño de un elemento
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c2MZVSg'
+forumTopicId: 301076
+dashedName: use-the-css-transform-scale-property-to-change-the-size-of-an-element
+---
+
+# --description--
+
+Para cambiar la escala de un elemento, CSS tiene la propiedad `transform`, junto con su función `scale()`. En el ejemplo de código siguiente se duplica el tamaño de todos los elementos de párrafo de la página:
+
+```css
+p {
+ transform: scale(2);
+}
+```
+
+# --instructions--
+
+Aumenta el tamaño del elemento con el id de `ball2` a 1.5 veces su tamaño original.
+
+# --hints--
+
+La propiedad `transform` para `#ball2` debe establecerse para escalarla a 1.5 veces su tamaño.
+
+```js
+assert(
+ code.match(
+ /#ball2\s*?{\s*?left:\s*?65%;\s*?transform:\s*?scale\(1\.5\);\s*?}|#ball2\s*?{\s*?transform:\s*?scale\(1\.5\);\s*?left:\s*?65%;\s*?}/gi
+ )
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .ball {
+ width: 40px;
+ height: 40px;
+ margin: 50 auto;
+ position: fixed;
+ background: linear-gradient(
+ 35deg,
+ #ccffff,
+ #ffcccc
+ );
+ border-radius: 50%;
+ }
+ #ball1 {
+ left: 20%;
+ }
+ #ball2 {
+ left: 65%;
+
+ }
+
+
+</style>
+
+<div class="ball" id= "ball1"></div>
+<div class="ball" id= "ball2"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ .ball {
+ width: 40px;
+ height: 40px;
+ margin: 50 auto;
+ position: fixed;
+ background: linear-gradient(
+ 35deg,
+ #ccffff,
+ #ffcccc
+ );
+ border-radius: 50%;
+ }
+ #ball1 {
+ left: 20%;
+ }
+ #ball2 {
+ left: 65%;
+ transform: scale(1.5);
+ }
+</style>
+<div class="ball" id= "ball1"></div>
+<div class="ball" id= "ball2"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-css-transform-scale-property-to-scale-an-element-on-hover.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-css-transform-scale-property-to-scale-an-element-on-hover.md
new file mode 100644
index 00000000000..72f8457d409
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-css-transform-scale-property-to-scale-an-element-on-hover.md
@@ -0,0 +1,79 @@
+---
+id: 587d78a5367417b2b2512ada
+title: Utiliza la propiedad de escala de transformación CSS para escalar un elemento al desplazarse
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cyLPJuM'
+forumTopicId: 301077
+dashedName: use-the-css-transform-scale-property-to-scale-an-element-on-hover
+---
+
+# --description--
+
+La propiedad `transform` tiene una variedad de funciones que el permiten escalar, mover, rotar, sesgar, etc., sus elementos. Cuando se usa con pseudo-classes como `:hover` que especifican un cierto estado de un elemento, la propiedad `transform` puede agregar fácilmente interactividad a sus elementos.
+
+Aquí hay un ejemplo para escalar los elementos de párrafo a 2.1 veces su tamaño original:
+
+```css
+p:hover {
+ transform: scale(2.1);
+}
+```
+
+<strong x=id="1">Nota:</strong> La aplicación de una transformación a un elemento `div` también afectará a cualquier elemento secundario contenido de el div.
+
+# --instructions--
+
+Agrega una regla CSS para el estado `hover` del `div` y usa la propiedad `transform` para escalar el elemento `div` a 1.1 veces su tamaño original cuando un usuario pasa sobre él.
+
+# --hints--
+
+El tamaño del elemento `div` debe escalar 1.1 veces cuando el usuario pase el cursor sobre él.
+
+```js
+assert(code.match(/div:hover\s*?{\s*?transform:\s*?scale\(1\.1\);/gi));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ div {
+ width: 70%;
+ height: 100px;
+ margin: 50px auto;
+ background: linear-gradient(
+ 53deg,
+ #ccfffc,
+ #ffcccf
+ );
+ }
+
+
+
+</style>
+
+<div></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ div {
+ width: 70%;
+ height: 100px;
+ margin: 50px auto;
+ background: linear-gradient(
+ 53deg,
+ #ccfffc,
+ #ffcccf
+ );
+ }
+ div:hover {
+ transform: scale(1.1);
+ }
+</style>
+<div></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-em-tag-to-italicize-text.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-em-tag-to-italicize-text.md
new file mode 100644
index 00000000000..7d695ed67a1
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-em-tag-to-italicize-text.md
@@ -0,0 +1,118 @@
+---
+id: 587d781a367417b2b2512ab9
+title: Usa la etiqueta em para poner texto en cursiva
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cVJRBtp'
+forumTopicId: 301078
+dashedName: use-the-em-tag-to-italicize-text
+---
+
+# --description--
+
+Para enfatizar el texto, puedes usar la etiqueta `em`. Esto muestra el texto en cursiva, ya que el navegador aplica el CSS de `font-style: italic;` al elemento.
+
+# --instructions--
+
+Envuelve una etiqueta `em` alrededor del contenido de la etiqueta de párrafo para darle énfasis.
+
+# --hints--
+
+Tu código debe agregar una etiqueta `em` al lenguaje de marcado.
+
+```js
+assert($('em').length == 1);
+```
+
+La etiqueta `em` debe envolver el contenido de la etiqueta `p` pero no la etiqueta `p` en sí.
+
+```js
+assert($('p').children().length == 1 && $('em').children().length == 2);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ height: 25px;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Google</h4>
+ <p>Google was founded by Larry Page and Sergey Brin while they were <u>Ph.D. students</u> at <strong>Stanford University</strong>.</p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ height: 25px;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Google</h4>
+ <p><em>Google was founded by Larry Page and Sergey Brin while they were <u>Ph.D. students</u> at <strong>Stanford University</strong>.</em></p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-s-tag-to-strikethrough-text.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-s-tag-to-strikethrough-text.md
new file mode 100644
index 00000000000..13b0494f562
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-s-tag-to-strikethrough-text.md
@@ -0,0 +1,135 @@
+---
+id: 587d781b367417b2b2512aba
+title: Usa la etiqueta s para tachar texto
+challengeType: 0
+videoUrl: ''
+forumTopicId: 301079
+dashedName: use-the-s-tag-to-strikethrough-text
+---
+
+# --description--
+
+Para tachar el texto, que es cuando una línea horizontal atraviesa los caracteres, puede usar la etiqueta `s`. Muestra que una sección de texto ya no es válida. Con la etiqueta `s`, el navegador aplica el CSS de `text-decoration: line-through;` al elemento.
+
+# --instructions--
+
+Envuelve la etiqueta `s` alrededor de `Google` dentro de la etiqueta `h4` y luego agregue la palabra `Alphabet` al lado, que no debería tienen el formato tachado.
+
+# --hints--
+
+Tu código debe agregar una etiqueta `s` al lenguaje de marcado.
+
+```js
+assert($('s').length == 1);
+```
+
+Una etiqueta `s` debería envolver alrededor del texto `Google` en la etiqueta `h4`. No debe contener la palabra `Alphabet`.
+
+```js
+assert(
+ $('h4 > s')
+ .text()
+ .match(/Google/gi) &&
+ !$('h4 > s')
+ .text()
+ .match(/Alphabet/gi)
+);
+```
+
+Debes incluir la palabra `Alphabet` en la etiqueta `h4`, sin formato de tachado.
+
+```js
+assert(
+ $('h4')
+ .html()
+ .match(/Alphabet/gi)
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ height: 25px;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Google</h4>
+ <p><em>Google was founded by Larry Page and Sergey Brin while they were <u>Ph.D. students</u> at <strong>Stanford University</strong>.</em></p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ height: 25px;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4><s>Google</s> Alphabet</h4>
+ <p><em>Google was founded by Larry Page and Sergey Brin while they were <u>Ph.D. students</u> at <strong>Stanford University</strong>.</em></p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-strong-tag-to-make-text-bold.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-strong-tag-to-make-text-bold.md
new file mode 100644
index 00000000000..99844c4d1b5
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-strong-tag-to-make-text-bold.md
@@ -0,0 +1,128 @@
+---
+id: 587d781a367417b2b2512ab7
+title: Usa la etiqueta fuerte para hacer que el texto sea negrita
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/ceJNBSb'
+forumTopicId: 301080
+dashedName: use-the-strong-tag-to-make-text-bold
+---
+
+# --description--
+
+Para poner el texto en negrita, puedes usar la etiqueta `strong`. Esto se usa a menudo para llamar la atención sobre el texto y simbolizar que es importante. Con la etiqueta `strong`, el navegador aplica el CSS de `font-weight: bold;` al elemento.
+
+# --instructions--
+
+Envuelve una etiqueta `strong` alrededor del texto de `Stanford University` dentro de la etiqueta `p` (no incluyas el punto).
+
+# --hints--
+
+Tu código debe agregar una etiqueta `strong` al lenguaje de marcado.
+
+```js
+assert($('strong').length == 1);
+```
+
+La etiqueta `strong` debe estar dentro de la etiqueta `p`.
+
+```js
+assert($('p').children('strong').length == 1);
+```
+
+The `strong` tag should wrap around the words `Stanford University`.
+
+```js
+assert(
+ $('strong')
+ .text()
+ .match(/^Stanford University\.?$/gi)
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ height: 25px;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Google</h4>
+ <p>Google was founded by Larry Page and Sergey Brin while they were Ph.D. students at Stanford University.</p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ height: 25px;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Google</h4>
+ <p>Google was founded by Larry Page and Sergey Brin while they were Ph.D. students at <strong>Stanford University</strong>.</p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-text-transform-property-to-make-text-uppercase.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-text-transform-property-to-make-text-uppercase.md
new file mode 100644
index 00000000000..f0895fac032
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-text-transform-property-to-make-text-uppercase.md
@@ -0,0 +1,138 @@
+---
+id: 587d781c367417b2b2512ac0
+title: Usar la propiedad de text-transform para hacer texto mayúscula
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cvVZQSP'
+forumTopicId: 301081
+dashedName: use-the-text-transform-property-to-make-text-uppercase
+---
+
+# --description--
+
+La propiedad `text-transform` en CSS se utiliza para cambiar la apariencia del texto. Es una forma conveniente de asegurarse de que el texto en una página web aparezca de manera consistente, sin tener que cambiar el contenido del texto de los elementos HTML reales.
+
+La siguiente tabla muestra como los diferentes valores de `text-transform` cambian el texto de ejemplo "Transformame".
+
+<table class='table table-striped'><thead><tr><th>Valor</th><th>Resultado</th></tr></thead><tbody><tr><td><code>lowercase</code></td><td>"Transformame"</td></tr><tr><td><code>uppercase</code></td><td>"TRANSFORMAME"</td></tr><tr><td><code>capitalize</code></td><td>"Transformame"</td></tr><tr><td><code>initial</code></td><td>Usa el valor predeterminado</td></tr><tr><td><code>inherit</code></td><td>Utiliza el valor <code>text-transform</code> del elemento principal</td></tr><tr><td><code>none</code></td><td><strong>Predeterminado:</strong> Usa el texto original</td></tr></tbody></table>
+
+# --instructions--
+
+Transforma el texto de `h4` en mayúsculas usando la propiedad `text-transform`.
+
+# --hints--
+
+El texto `h4` debe ser `uppercase`.
+
+```js
+assert($('h4').css('text-transform') === 'uppercase');
+```
+
+El texto original del h4 no debe cambiarse.
+
+```js
+assert($('h4').text() !== $('h4').text().toUpperCase());
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ background-color: rgba(45, 45, 45, 0.1);
+ padding: 10px;
+ font-size: 27px;
+
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ opacity: 0.7;
+ }
+ #thumbnail {
+ box-shadow: 0 10px 20px rgba(0,0,0,0.19), 0 6px 6px rgba(0,0,0,0.23);
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard" id="thumbnail">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Alphabet</h4>
+ <hr>
+ <p><em>Google was founded by Larry Page and Sergey Brin while they were <u>Ph.D. students</u> at <strong>Stanford University</strong>.</em></p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ background-color: rgba(45, 45, 45, 0.1);
+ padding: 10px;
+ font-size: 27px;
+ text-transform: uppercase;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ opacity: 0.7;
+ }
+ #thumbnail {
+ box-shadow: 0 10px 20px rgba(0,0,0,0.19), 0 6px 6px rgba(0,0,0,0.23);
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard" id="thumbnail">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Alphabet</h4>
+ <hr>
+ <p><em>Google was founded by Larry Page and Sergey Brin while they were <u>Ph.D. students</u> at <strong>Stanford University</strong>.</em></p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-u-tag-to-underline-text.md b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-u-tag-to-underline-text.md
new file mode 100644
index 00000000000..d983bf5d182
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/applied-visual-design/use-the-u-tag-to-underline-text.md
@@ -0,0 +1,120 @@
+---
+id: 587d781a367417b2b2512ab8
+title: Utiliza la etiqueta u para subrayar texto
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cN6aQCL'
+forumTopicId: 301082
+dashedName: use-the-u-tag-to-underline-text
+---
+
+# --description--
+
+Para subrayar el texto, puedes utilizar la etiqueta `u`. Esto se usa a menudo para indicar que una sección del texto es importante, o algo que hay que recordar. Con la etiqueta `u`, el navegador aplica el CSS de `text-decoration: underline;` al elemento.
+
+# --instructions--
+
+Envuelve la etiqueta `u` solo alrededor del texto `Ph.D. students`.
+
+**Nota:** Trata de evitar usar la etiqueta `u` cuando se pueda confundir con un enlace. Las etiquetas Anchor también tienen un formato subrayado por defecto.
+
+# --hints--
+
+Tu código debe agregar una etiqueta `u` al lenguaje de marcado.
+
+```js
+assert($('u').length === 1);
+```
+
+Envuelve la etiqueta `u` solo alrededor del texto `Ph.D. students`.
+
+```js
+assert($('u').text() === 'Ph.D. students');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ height: 25px;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Google</h4>
+ <p>Google was founded by Larry Page and Sergey Brin while they were Ph.D. students at <strong>Stanford University</strong>.</p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ h4 {
+ text-align: center;
+ height: 25px;
+ }
+ p {
+ text-align: justify;
+ }
+ .links {
+ text-align: left;
+ color: black;
+ }
+ .fullCard {
+ width: 245px;
+ border: 1px solid #ccc;
+ border-radius: 5px;
+ margin: 10px 5px;
+ padding: 4px;
+ }
+ .cardContent {
+ padding: 10px;
+ }
+ .cardText {
+ margin-bottom: 30px;
+ }
+</style>
+<div class="fullCard">
+ <div class="cardContent">
+ <div class="cardText">
+ <h4>Google</h4>
+ <p>Google was founded by Larry Page and Sergey Brin while they were <u>Ph.D. students</u> at <strong>Stanford University</strong>.</p>
+ </div>
+ <div class="cardLinks">
+ <a href="https://en.wikipedia.org/wiki/Larry_Page" target="_blank" class="links">Larry Page</a><br><br>
+ <a href="https://en.wikipedia.org/wiki/Sergey_Brin" target="_blank" class="links">Sergey Brin</a>
+ </div>
+ </div>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/add-a-negative-margin-to-an-element.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/add-a-negative-margin-to-an-element.md
new file mode 100644
index 00000000000..0bcbb297d54
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/add-a-negative-margin-to-an-element.md
@@ -0,0 +1,115 @@
+---
+id: bad87fee1348bd9aedf08823
+title: Añade un margen negativo a un elemento
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cnpyGs3'
+forumTopicId: 16166
+dashedName: add-a-negative-margin-to-an-element
+---
+
+# --description--
+
+El `margin` (margen) de un elemento controla la cantidad de espacio entre su `border` y los elementos que lo rodean.
+
+Si estableces el `margin` de un elemento a un valor negativo, el elemento crecerá de tamaño.
+
+# --instructions--
+
+Intenta establecer el `margin` a un valor negativo como el de la caja roja.
+
+Cambia el `margin` de la caja azul a `-15px` para que llene todo el ancho horizontal de la caja amarilla que lo rodea.
+
+# --hints--
+
+Tu clase `blue-box` debería dar a los elementos un `margin` de `-15px`.
+
+```js
+assert($('.blue-box').css('margin-top') === '-15px');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .injected-text {
+ margin-bottom: -25px;
+ text-align: center;
+ }
+
+ .box {
+ border-style: solid;
+ border-color: black;
+ border-width: 5px;
+ text-align: center;
+ }
+
+ .yellow-box {
+ background-color: yellow;
+ padding: 10px;
+ }
+
+ .red-box {
+ background-color: crimson;
+ color: #fff;
+ padding: 20px;
+ margin: -15px;
+ }
+
+ .blue-box {
+ background-color: blue;
+ color: #fff;
+ padding: 20px;
+ margin: 20px;
+ }
+</style>
+
+<div class="box yellow-box">
+ <h5 class="box red-box">padding</h5>
+ <h5 class="box blue-box">padding</h5>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ .injected-text {
+ margin-bottom: -25px;
+ text-align: center;
+ }
+
+ .box {
+ border-style: solid;
+ border-color: black;
+ border-width: 5px;
+ text-align: center;
+ }
+
+ .yellow-box {
+ background-color: yellow;
+ padding: 10px;
+ }
+
+ .red-box {
+ background-color: crimson;
+ color: #fff;
+ padding: 20px;
+ margin: -15px;
+ }
+
+ .blue-box {
+ background-color: blue;
+ color: #fff;
+ padding: 20px;
+ margin: 20px;
+ margin-top: -15px;
+ }
+</style>
+
+<div class="box yellow-box">
+ <h5 class="box red-box">padding</h5>
+ <h5 class="box blue-box">padding</h5>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/add-borders-around-your-elements.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/add-borders-around-your-elements.md
new file mode 100644
index 00000000000..e0a34a6d558
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/add-borders-around-your-elements.md
@@ -0,0 +1,188 @@
+---
+id: bad87fee1348bd9bedf08813
+title: Añade bordes alrededor de tus elementos
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c2MvnHZ'
+forumTopicId: 16630
+dashedName: add-borders-around-your-elements
+---
+
+# --description--
+
+Los bordes CSS tienen propiedades como `style` (estilo), `color` y `width` (ancho).
+
+Por ejemplo, si quisiéramos crear un borde rojo de 5 píxeles alrededor de un elemento HTML, podríamos usar la siguiente clase:
+
+```html
+<style>
+ .thin-red-border {
+ border-color: red;
+ border-width: 5px;
+ border-style: solid;
+ }
+</style>
+```
+
+# --instructions--
+
+Crea una clase llamada `thick-green-border` (borde grueso verde). Esta clase debería añadir un borde verde de 10px, "solid", alrededor de un elemento HTML. Aplica esa clase a tu foto de gato.
+
+Recuerda que puedes aplicar múltiples clases a un elemento usando su atributo `class`, separando cada nombre de clase con un espacio. Por ejemplo:
+
+`<img class="class1 class2">`
+
+# --hints--
+
+Tu elemento `img` debe incluir la "class" `smaller-image`.
+
+```js
+assert($('img').hasClass('smaller-image'));
+```
+
+Tu elemento `img` debe incluir la "class" `thick-green-border`.
+
+```js
+assert($('img').hasClass('thick-green-border'));
+```
+
+Tu imagen debe tener un ancho de borde o "border-width" de `10px`.
+
+```js
+assert(
+ $('img').hasClass('thick-green-border') &&
+ parseInt($('img').css('border-top-width'), 10) >= 8 &&
+ parseInt($('img').css('border-top-width'), 10) <= 12
+);
+```
+
+Tu imagen debe tener un estilo de borde o "border-style" `solid`.
+
+```js
+assert($('img').css('border-right-style') === 'solid');
+```
+
+El borde alrededor de tu elemento `img` debe ser verde.
+
+```js
+assert($('img').css('border-left-color') === 'rgb(0, 128, 0)');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">
+<style>
+ .red-text {
+ color: red;
+ }
+
+ h2 {
+ font-family: Lobster, monospace;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+
+ .smaller-image {
+ width: 100px;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img class="smaller-image" src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">
+<style>
+ .red-text {
+ color: red;
+ }
+
+ h2 {
+ font-family: Lobster, monospace;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+
+ .smaller-image {
+ width: 100px;
+ }
+
+ .thick-green-border {
+ border-width: 10px;
+ border-color: green;
+ border-style: solid;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img class="smaller-image thick-green-border" src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/add-different-margins-to-each-side-of-an-element.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/add-different-margins-to-each-side-of-an-element.md
new file mode 100644
index 00000000000..5b6920b9895
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/add-different-margins-to-each-side-of-an-element.md
@@ -0,0 +1,136 @@
+---
+id: bad87fee1248bd9aedf08824
+title: Añade márgenes diferentes a cada lado de un elemento
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cg4RWh4'
+forumTopicId: 16633
+dashedName: add-different-margins-to-each-side-of-an-element
+---
+
+# --description--
+
+En ocasiones, querrás personalizar un elemento para que tenga un `margin` diferente en cada uno de sus lados.
+
+CSS te permite controlar por separado el `margin` de los cuatro lados individuales de un elemento por medio de las propiedades `margin-top`, `margin-right`, `margin-bottom` y `margin-left`.
+
+# --instructions--
+
+Asigna a la caja azul un `margin` de `40px` en sus lados superior e izquierdo, pero sólo de `20px` en sus lados inferior y derecho.
+
+# --hints--
+
+Tu clase `blue-box` debe asignar al lado superior (top) de los elementos `40px` de `margin`.
+
+```js
+assert($('.blue-box').css('margin-top') === '40px');
+```
+
+Tu clase `blue-box` debe asignar al lado derecho (right) de los elementos `20px` de `margin`.
+
+```js
+assert($('.blue-box').css('margin-right') === '20px');
+```
+
+Tu clase `blue-box` debe asignar al lado inferior (bottom) de los elementos `20px` de `margin`.
+
+```js
+assert($('.blue-box').css('margin-bottom') === '20px');
+```
+
+Tu clase `blue-box` debe asignar al lado izquierdo (left) de los elementos `40px` de `margin`.
+
+```js
+assert($('.blue-box').css('margin-left') === '40px');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .injected-text {
+ margin-bottom: -25px;
+ text-align: center;
+ }
+
+ .box {
+ border-style: solid;
+ border-color: black;
+ border-width: 5px;
+ text-align: center;
+ }
+
+ .yellow-box {
+ background-color: yellow;
+ padding: 10px;
+ }
+
+ .red-box {
+ background-color: crimson;
+ color: #fff;
+ margin-top: 40px;
+ margin-right: 20px;
+ margin-bottom: 20px;
+ margin-left: 40px;
+ }
+
+ .blue-box {
+ background-color: blue;
+ color: #fff;
+ }
+</style>
+<h5 class="injected-text">margin</h5>
+
+<div class="box yellow-box">
+ <h5 class="box red-box">padding</h5>
+ <h5 class="box blue-box">padding</h5>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ .injected-text {
+ margin-bottom: -25px;
+ text-align: center;
+ }
+
+ .box {
+ border-style: solid;
+ border-color: black;
+ border-width: 5px;
+ text-align: center;
+ }
+
+ .yellow-box {
+ background-color: yellow;
+ padding: 10px;
+ }
+
+ .red-box {
+ background-color: crimson;
+ color: #fff;
+ margin-top: 40px;
+ margin-right: 20px;
+ margin-bottom: 20px;
+ margin-left: 40px;
+ }
+
+ .blue-box {
+ background-color: blue;
+ color: #fff;
+ margin-top: 40px;
+ margin-right: 20px;
+ margin-bottom: 20px;
+ margin-left: 40px;
+ }
+</style>
+<h5 class="injected-text">margin</h5>
+
+<div class="box yellow-box">
+ <h5 class="box red-box">padding</h5>
+ <h5 class="box blue-box">padding</h5>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/add-different-padding-to-each-side-of-an-element.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/add-different-padding-to-each-side-of-an-element.md
new file mode 100644
index 00000000000..958faea4fd9
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/add-different-padding-to-each-side-of-an-element.md
@@ -0,0 +1,136 @@
+---
+id: bad87fee1348bd9aedf08824
+title: Añade un "padding" o relleno diferente a cada lado de un elemento
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cB7mwUw'
+forumTopicId: 16634
+dashedName: add-different-padding-to-each-side-of-an-element
+---
+
+# --description--
+
+En ocasiones, querrás personalizar un elemento para que tenga un `padding` o relleno diferente en cada uno de sus lados.
+
+CSS te permite controlar por separado el `padding` de los cuatro lados individuales de un elemento por medio de las propiedades `padding-top`, `padding-right`, `padding-bottom` y `padding-left`.
+
+# --instructions--
+
+Asigna a la caja azul un `padding` de `40px` en sus lados superior e izquierdo, pero sólo de `20px` en sus lados inferior y derecho.
+
+# --hints--
+
+Tu clase `blue-box` debe asignar al lado superior (top) de los elementos `40px` de `padding`.
+
+```js
+assert($('.blue-box').css('padding-top') === '40px');
+```
+
+Tu clase `blue-box` debe asignar al lado derecho (right) de los elementos `20px` de `padding`.
+
+```js
+assert($('.blue-box').css('padding-right') === '20px');
+```
+
+Tu clase `blue-box` debe asignar al lado inferior (bottom) de los elementos `20px` de `padding`.
+
+```js
+assert($('.blue-box').css('padding-bottom') === '20px');
+```
+
+Tu clase `blue-box` debe asignar al lado izquierlo (left) de los elementos `40px` de `padding`.
+
+```js
+assert($('.blue-box').css('padding-left') === '40px');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .injected-text {
+ margin-bottom: -25px;
+ text-align: center;
+ }
+
+ .box {
+ border-style: solid;
+ border-color: black;
+ border-width: 5px;
+ text-align: center;
+ }
+
+ .yellow-box {
+ background-color: yellow;
+ padding: 10px;
+ }
+
+ .red-box {
+ background-color: crimson;
+ color: #fff;
+ padding-top: 40px;
+ padding-right: 20px;
+ padding-bottom: 20px;
+ padding-left: 40px;
+ }
+
+ .blue-box {
+ background-color: blue;
+ color: #fff;
+ }
+</style>
+<h5 class="injected-text">margin</h5>
+
+<div class="box yellow-box">
+ <h5 class="box red-box">padding</h5>
+ <h5 class="box blue-box">padding</h5>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ .injected-text {
+ margin-bottom: -25px;
+ text-align: center;
+ }
+
+ .box {
+ border-style: solid;
+ border-color: black;
+ border-width: 5px;
+ text-align: center;
+ }
+
+ .yellow-box {
+ background-color: yellow;
+ padding: 10px;
+ }
+
+ .red-box {
+ background-color: crimson;
+ color: #fff;
+ padding-top: 40px;
+ padding-right: 20px;
+ padding-bottom: 20px;
+ padding-left: 40px;
+ }
+
+ .blue-box {
+ background-color: blue;
+ color: #fff;
+ padding-top: 40px;
+ padding-right: 20px;
+ padding-bottom: 20px;
+ padding-left: 40px;
+ }
+</style>
+<h5 class="injected-text">margin</h5>
+
+<div class="box yellow-box">
+ <h5 class="box red-box">padding</h5>
+ <h5 class="box blue-box">padding</h5>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/add-rounded-corners-with-border-radius.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/add-rounded-corners-with-border-radius.md
new file mode 100644
index 00000000000..358863bec6f
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/add-rounded-corners-with-border-radius.md
@@ -0,0 +1,164 @@
+---
+id: bad87fee1348bd9aedf08814
+title: Añade esquinas redondeadas usando border-radius
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cbZm2hg'
+forumTopicId: 16649
+dashedName: add-rounded-corners-with-border-radius
+---
+
+# --description--
+
+Actualmente, las esquinas de tu foto de gato son puntiagudas. Podemos redondear esas esquinas con una propiedad CSS llamada `border-radius` (radio del borde).
+
+# --instructions--
+
+Puedes especificar un `border-radius` usando píxeles como unidad de medida. Dale a tu foto de gato un `border-radius` de `10px`.
+
+**Nota:** Este desafío acepta múltiples soluciones posibles. Por ejemplo, puedes añadir `border-radius` a la clase `.thick-green-border` o a la clase `.smaller-image`.
+
+# --hints--
+
+Tu elemento img debe incluir la clase `thick-green-border`.
+
+```js
+assert($('img').hasClass('thick-green-border'));
+```
+
+Tu imagen debe tener un "border-radius" (radio del borde) de `10px`.
+
+```js
+assert(
+ $('img').css('border-top-left-radius') === '10px' &&
+ $('img').css('border-top-right-radius') === '10px' &&
+ $('img').css('border-bottom-left-radius') === '10px' &&
+ $('img').css('border-bottom-right-radius') === '10px'
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">
+<style>
+ .red-text {
+ color: red;
+ }
+
+ h2 {
+ font-family: Lobster, monospace;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+
+ .thick-green-border {
+ border-color: green;
+ border-width: 10px;
+ border-style: solid;
+ }
+
+ .smaller-image {
+ width: 100px;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img class="smaller-image thick-green-border" src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">
+<style>
+ .red-text {
+ color: red;
+ }
+
+ h2 {
+ font-family: Lobster, monospace;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+
+ .thick-green-border {
+ border-color: green;
+ border-width: 10px;
+ border-style: solid;
+ }
+
+ .smaller-image {
+ width: 100px;
+ border-radius: 10px;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img class="smaller-image thick-green-border" src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/adjust-the-margin-of-an-element.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/adjust-the-margin-of-an-element.md
new file mode 100644
index 00000000000..6a3dea82e46
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/adjust-the-margin-of-an-element.md
@@ -0,0 +1,116 @@
+---
+id: bad87fee1348bd9aedf08822
+title: Ajusta el margen de un elemento
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cVJarHW'
+forumTopicId: 16654
+dashedName: adjust-the-margin-of-an-element
+---
+
+# --description--
+
+El `margin` (margen) de un elemento controla la cantidad de espacio entre su `border` y los elementos que lo rodean.
+
+Aquí podemos ver que la caja azul y la caja roja están anidadas dentro de la caja amarilla. Ten en cuenta que la caja roja tiene un `margin` más grande que el de la caja azul, lo que hace que aparezca más pequeña.
+
+Cuando aumentas el `margin` de la caja azul, esto aumenta la distancia entre su borde y los elementos que la rodean.
+
+# --instructions--
+
+Cambia el `margin` de la caja azul para que coincida con el de la caja roja.
+
+# --hints--
+
+Tu clase `blue-box` debe asignar a los elementos `20px` de `margin`.
+
+```js
+assert($('.blue-box').css('margin-top') === '20px');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .injected-text {
+ margin-bottom: -25px;
+ text-align: center;
+ }
+
+ .box {
+ border-style: solid;
+ border-color: black;
+ border-width: 5px;
+ text-align: center;
+ }
+
+ .yellow-box {
+ background-color: yellow;
+ padding: 10px;
+ }
+
+ .red-box {
+ background-color: crimson;
+ color: #fff;
+ padding: 20px;
+ margin: 20px;
+ }
+
+ .blue-box {
+ background-color: blue;
+ color: #fff;
+ padding: 20px;
+ margin: 10px;
+ }
+</style>
+<h5 class="injected-text">margin</h5>
+
+<div class="box yellow-box">
+ <h5 class="box red-box">padding</h5>
+ <h5 class="box blue-box">padding</h5>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ .injected-text {
+ margin-bottom: -25px;
+ text-align: center;
+ }
+
+ .box {
+ border-style: solid;
+ border-color: black;
+ border-width: 5px;
+ text-align: center;
+ }
+
+ .yellow-box {
+ background-color: yellow;
+ padding: 10px;
+ }
+
+ .red-box {
+ background-color: crimson;
+ color: #fff;
+ padding: 20px;
+ margin: 20px;
+ }
+
+ .blue-box {
+ background-color: blue;
+ color: #fff;
+ padding: 20px;
+ margin: 20px;
+ }
+</style>
+<h5 class="injected-text">margin</h5>
+
+<div class="box yellow-box">
+ <h5 class="box red-box">padding</h5>
+ <h5 class="box blue-box">padding</h5>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/adjust-the-padding-of-an-element.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/adjust-the-padding-of-an-element.md
new file mode 100644
index 00000000000..3780c026e08
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/adjust-the-padding-of-an-element.md
@@ -0,0 +1,118 @@
+---
+id: bad88fee1348bd9aedf08825
+title: Ajusta el padding (relleno) de un elemento
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cED8ZC2'
+forumTopicId: 301083
+dashedName: adjust-the-padding-of-an-element
+---
+
+# --description--
+
+Dejaremos de lado por un momento nuestra Cat Photo App y aprenderemos un poco más sobre como aplicar estilos al HTML.
+
+Quizás ya te hayas dado cuenta, pero todos los elementos HTML son, esencialmente, pequeños rectángulos.
+
+Hay tres propiedades importantes que controlan el espacio que rodea cada elemento HTML: `padding` (relleno), `border` (borde) y `margin` (margen).
+
+El `padding` o relleno de un elemento controla la cantidad de espacio entre su contenido y su `border`.
+
+Aquí podemos ver que la caja azul y la caja roja están anidadas dentro de la caja amarilla. Fíjate que la caja roja tiene más `padding` (relleno) que la caja azul.
+
+Cuando aumentas el `padding` de la caja azul, esto aumenta la distancia (`padding`) entre el texto y el borde que lo rodea.
+
+# --instructions--
+
+Cambia el `padding` de la caja azul para que coincida con el de la caja roja.
+
+# --hints--
+
+Tu clase `blue-box` debe asignar a los elementos `20px` de `padding`.
+
+```js
+assert($('.blue-box').css('padding-top') === '20px');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .injected-text {
+ margin-bottom: -25px;
+ text-align: center;
+ }
+
+ .box {
+ border-style: solid;
+ border-color: black;
+ border-width: 5px;
+ text-align: center;
+ }
+
+ .yellow-box {
+ background-color: yellow;
+ padding: 10px;
+ }
+
+ .red-box {
+ background-color: crimson;
+ color: #fff;
+ padding: 20px;
+ }
+
+ .blue-box {
+ background-color: blue;
+ color: #fff;
+ padding: 10px;
+ }
+</style>
+<h5 class="injected-text">margin</h5>
+
+<div class="box yellow-box">
+ <h5 class="box red-box">padding</h5>
+ <h5 class="box blue-box">padding</h5>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ .injected-text {
+ margin-bottom: -25px;
+ text-align: center;
+ }
+
+ .box {
+ border-style: solid;
+ border-color: black;
+ border-width: 5px;
+ text-align: center;
+ }
+
+ .yellow-box {
+ background-color: yellow;
+ padding: 10px;
+ }
+
+ .red-box {
+ background-color: crimson;
+ color: #fff;
+ padding: 20px;
+ }
+
+ .blue-box {
+ background-color: blue;
+ color: #fff;
+ padding: 20px;
+ }
+</style>
+<h5 class="injected-text">margin</h5>
+
+<div class="box yellow-box">
+ <h5 class="box red-box">padding</h5>
+ <h5 class="box blue-box">padding</h5>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/attach-a-fallback-value-to-a-css-variable.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/attach-a-fallback-value-to-a-css-variable.md
new file mode 100644
index 00000000000..edd8d97bd64
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/attach-a-fallback-value-to-a-css-variable.md
@@ -0,0 +1,272 @@
+---
+id: 5a9d7286424fe3d0e10cad13
+title: Agrega un valor de respaldo (fallback) a una variable CSS
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c6bDNfp'
+forumTopicId: 301084
+dashedName: attach-a-fallback-value-to-a-css-variable
+---
+
+# --description--
+
+Cuando utilices tu variable como valor de una propiedad CSS, puedes adjuntar un valor de respaldo o "fallback", que será utilizado por el navegador si la variable dada no es válida.
+
+**Nota:** Este valor de respaldo no se utiliza para aumentar la compatibilidad con otras versiones de navegadores y no funcionará en navegadores IE (Internet Explorer). Más bien, se utiliza para que el navegador tenga un color para mostrar si no encuentra tu variable.
+
+A continuación te mostramos como se hace:
+
+```css
+background: var(--penguin-skin, black);
+```
+
+Esto establecerá el "background" (color de fondo) a `black` (negro) si tu variable no está establecida. Fíjate que esto puede ser útil para la depuración de errores.
+
+# --instructions--
+
+Aparentemente hay un problema con las variables proporcionadas a las clases `.penguin-top` y `.penguin-bottom`. En lugar de corregir el error de tipeo, añade `black` como un valor de reserva o "fallback" a la propiedad `background` de las clases `.penguin-top` y `.penguin-bottom`.
+
+# --hints--
+
+El valor de reserva `black` debe utilizarse en la propiedad `background` de la clase `penguin-top`.
+
+```js
+assert(
+ code.match(
+ /.penguin-top\s*?{[\s\S]*background\s*?:\s*?var\(\s*?--pengiun-skin\s*?,\s*?black\s*?\)\s*?;[\s\S]*}[\s\S]*.penguin-bottom\s{/gi
+ )
+);
+```
+
+El valor de reserva `black` debe utilizarse en la propiedad `background` de la clase `penguin-bottom`.
+
+```js
+assert(
+ code.match(
+ /.penguin-bottom\s*?{[\s\S]*background\s*?:\s*?var\(\s*?--pengiun-skin\s*?,\s*?black\s*?\)\s*?;[\s\S]*}/gi
+ )
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .penguin {
+ --penguin-skin: black;
+ --penguin-belly: gray;
+ --penguin-beak: yellow;
+ position: relative;
+ margin: auto;
+ display: block;
+ margin-top: 5%;
+ width: 300px;
+ height: 300px;
+ }
+
+ .penguin-top {
+ top: 10%;
+ left: 25%;
+
+ /* Change code below this line */
+ background: var(--pengiun-skin);
+ /* Change code above this line */
+
+ width: 50%;
+ height: 45%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .penguin-bottom {
+ top: 40%;
+ left: 23.5%;
+
+ /* Change code below this line */
+ background: var(--pengiun-skin);
+ /* Change code above this line */
+
+ width: 53%;
+ height: 45%;
+ border-radius: 70% 70% 100% 100%;
+ }
+
+ .right-hand {
+ top: 0%;
+ left: -5%;
+ background: var(--penguin-skin, black);
+ width: 30%;
+ height: 60%;
+ border-radius: 30% 30% 120% 30%;
+ transform: rotate(45deg);
+ z-index: -1;
+ }
+
+ .left-hand {
+ top: 0%;
+ left: 75%;
+ background: var(--penguin-skin, black);
+ width: 30%;
+ height: 60%;
+ border-radius: 30% 30% 30% 120%;
+ transform: rotate(-45deg);
+ z-index: -1;
+ }
+
+ .right-cheek {
+ top: 15%;
+ left: 35%;
+ background: var(--penguin-belly, white);
+ width: 60%;
+ height: 70%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .left-cheek {
+ top: 15%;
+ left: 5%;
+ background: var(--penguin-belly, white);
+ width: 60%;
+ height: 70%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .belly {
+ top: 60%;
+ left: 2.5%;
+ background: var(--penguin-belly, white);
+ width: 95%;
+ height: 100%;
+ border-radius: 120% 120% 100% 100%;
+ }
+
+ .right-feet {
+ top: 85%;
+ left: 60%;
+ background: var(--penguin-beak, orange);
+ width: 15%;
+ height: 30%;
+ border-radius: 50% 50% 50% 50%;
+ transform: rotate(-80deg);
+ z-index: -2222;
+ }
+
+ .left-feet {
+ top: 85%;
+ left: 25%;
+ background: var(--penguin-beak, orange);
+ width: 15%;
+ height: 30%;
+ border-radius: 50% 50% 50% 50%;
+ transform: rotate(80deg);
+ z-index: -2222;
+ }
+
+ .right-eye {
+ top: 45%;
+ left: 60%;
+ background: black;
+ width: 15%;
+ height: 17%;
+ border-radius: 50%;
+ }
+
+ .left-eye {
+ top: 45%;
+ left: 25%;
+ background: black;
+ width: 15%;
+ height: 17%;
+ border-radius: 50%;
+ }
+
+ .sparkle {
+ top: 25%;
+ left: 15%;
+ background: white;
+ width: 35%;
+ height: 35%;
+ border-radius: 50%;
+ }
+
+ .blush-right {
+ top: 65%;
+ left: 15%;
+ background: pink;
+ width: 15%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .blush-left {
+ top: 65%;
+ left: 70%;
+ background: pink;
+ width: 15%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .beak-top {
+ top: 60%;
+ left: 40%;
+ background: var(--penguin-beak, orange);
+ width: 20%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .beak-bottom {
+ top: 65%;
+ left: 42%;
+ background: var(--penguin-beak, orange);
+ width: 16%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ body {
+ background: #c6faf1;
+ }
+
+ .penguin * {
+ position: absolute;
+ }
+</style>
+<div class="penguin">
+ <div class="penguin-bottom">
+ <div class="right-hand"></div>
+ <div class="left-hand"></div>
+ <div class="right-feet"></div>
+ <div class="left-feet"></div>
+ </div>
+ <div class="penguin-top">
+ <div class="right-cheek"></div>
+ <div class="left-cheek"></div>
+ <div class="belly"></div>
+ <div class="right-eye">
+ <div class="sparkle"></div>
+ </div>
+ <div class="left-eye">
+ <div class="sparkle"></div>
+ </div>
+ <div class="blush-right"></div>
+ <div class="blush-left"></div>
+ <div class="beak-top"></div>
+ <div class="beak-bottom"></div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ .penguin-top {
+ background: var(--pengiun-skin, black);
+ }
+ .penguin-bottom {
+ background: var(--pengiun-skin, black);
+ }
+</style>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/change-a-variable-for-a-specific-area.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/change-a-variable-for-a-specific-area.md
new file mode 100644
index 00000000000..2810ec2c616
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/change-a-variable-for-a-specific-area.md
@@ -0,0 +1,245 @@
+---
+id: 5a9d72a1424fe3d0e10cad15
+title: Cambia una variable para un área específica
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cdRwbuW'
+forumTopicId: 301085
+dashedName: change-a-variable-for-a-specific-area
+---
+
+# --description--
+
+Cuando creas tus variables en `:root`, el valor de esa variable quedará establecido para toda la página.
+
+Luego podrás sobreescribir estas variables, configurándolas de nuevo dentro de un elemento específico.
+
+# --instructions--
+
+Cambia el valor de `--penguin-belly` a `white` en la clase `penguin`.
+
+# --hints--
+
+La clase `penguin` debe reasignar la variable `--penguin-belly` a `white`.
+
+```js
+assert(
+ code.match(/.penguin\s*?{[\s\S]*--penguin-belly\s*?:\s*?white\s*?;[\s\S]*}/gi)
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ :root {
+ --penguin-skin: gray;
+ --penguin-belly: pink;
+ --penguin-beak: orange;
+ }
+
+ body {
+ background: var(--penguin-belly, #c6faf1);
+ }
+
+ .penguin {
+ /* Only change code below this line */
+
+ /* Only change code above this line */
+ position: relative;
+ margin: auto;
+ display: block;
+ margin-top: 5%;
+ width: 300px;
+ height: 300px;
+ }
+
+ .right-cheek {
+ top: 15%;
+ left: 35%;
+ background: var(--penguin-belly, pink);
+ width: 60%;
+ height: 70%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .left-cheek {
+ top: 15%;
+ left: 5%;
+ background: var(--penguin-belly, pink);
+ width: 60%;
+ height: 70%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .belly {
+ top: 60%;
+ left: 2.5%;
+ background: var(--penguin-belly, pink);
+ width: 95%;
+ height: 100%;
+ border-radius: 120% 120% 100% 100%;
+ }
+
+ .penguin-top {
+ top: 10%;
+ left: 25%;
+ background: var(--penguin-skin, gray);
+ width: 50%;
+ height: 45%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .penguin-bottom {
+ top: 40%;
+ left: 23.5%;
+ background: var(--penguin-skin, gray);
+ width: 53%;
+ height: 45%;
+ border-radius: 70% 70% 100% 100%;
+ }
+
+ .right-hand {
+ top: 0%;
+ left: -5%;
+ background: var(--penguin-skin, gray);
+ width: 30%;
+ height: 60%;
+ border-radius: 30% 30% 120% 30%;
+ transform: rotate(45deg);
+ z-index: -1;
+ }
+
+ .left-hand {
+ top: 0%;
+ left: 75%;
+ background: var(--penguin-skin, gray);
+ width: 30%;
+ height: 60%;
+ border-radius: 30% 30% 30% 120%;
+ transform: rotate(-45deg);
+ z-index: -1;
+ }
+
+ .right-feet {
+ top: 85%;
+ left: 60%;
+ background: var(--penguin-beak, orange);
+ width: 15%;
+ height: 30%;
+ border-radius: 50% 50% 50% 50%;
+ transform: rotate(-80deg);
+ z-index: -2222;
+ }
+
+ .left-feet {
+ top: 85%;
+ left: 25%;
+ background: var(--penguin-beak, orange);
+ width: 15%;
+ height: 30%;
+ border-radius: 50% 50% 50% 50%;
+ transform: rotate(80deg);
+ z-index: -2222;
+ }
+
+ .right-eye {
+ top: 45%;
+ left: 60%;
+ background: black;
+ width: 15%;
+ height: 17%;
+ border-radius: 50%;
+ }
+
+ .left-eye {
+ top: 45%;
+ left: 25%;
+ background: black;
+ width: 15%;
+ height: 17%;
+ border-radius: 50%;
+ }
+
+ .sparkle {
+ top: 25%;
+ left: 15%;
+ background: white;
+ width: 35%;
+ height: 35%;
+ border-radius: 50%;
+ }
+
+ .blush-right {
+ top: 65%;
+ left: 15%;
+ background: pink;
+ width: 15%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .blush-left {
+ top: 65%;
+ left: 70%;
+ background: pink;
+ width: 15%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .beak-top {
+ top: 60%;
+ left: 40%;
+ background: var(--penguin-beak, orange);
+ width: 20%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .beak-bottom {
+ top: 65%;
+ left: 42%;
+ background: var(--penguin-beak, orange);
+ width: 16%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .penguin * {
+ position: absolute;
+ }
+</style>
+<div class="penguin">
+ <div class="penguin-bottom">
+ <div class="right-hand"></div>
+ <div class="left-hand"></div>
+ <div class="right-feet"></div>
+ <div class="left-feet"></div>
+ </div>
+ <div class="penguin-top">
+ <div class="right-cheek"></div>
+ <div class="left-cheek"></div>
+ <div class="belly"></div>
+ <div class="right-eye">
+ <div class="sparkle"></div>
+ </div>
+ <div class="left-eye">
+ <div class="sparkle"></div>
+ </div>
+ <div class="blush-right"></div>
+ <div class="blush-left"></div>
+ <div class="beak-top"></div>
+ <div class="beak-bottom"></div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+.penguin {--penguin-belly: white;}
+</style>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/change-the-color-of-text.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/change-the-color-of-text.md
new file mode 100644
index 00000000000..41793829ae1
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/change-the-color-of-text.md
@@ -0,0 +1,120 @@
+---
+id: bad87fee1348bd9aedf08803
+title: Cambia el color del texto
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cRkVmSm'
+forumTopicId: 16775
+dashedName: change-the-color-of-text
+---
+
+# --description--
+
+Ahora cambiemos el color de parte de nuestro texto.
+
+Podemos hacer esto cambiando el `style` de tu elemento `h2`.
+
+La propiedad que es responsable del color del texto de un elemento es la propiedad de estilo `color`.
+
+Aquí te mostramos como establecerías el color de tu elemento `h2` para que sea azul:
+
+`<h2 style="color: blue;">CatPhotoApp</h2>`
+
+Ten en cuenta que es una buena práctica terminar las declaraciones de inline `style` usando punto y coma (`;`).
+
+# --instructions--
+
+Cambia el estilo de tu elemento `h2` para que el texto de ese elemento sea de color rojo.
+
+# --hints--
+
+Tu elemento `h2` debe incluir una declaración `style`.
+
+```js
+assert($('h2').attr('style'));
+```
+
+Tu elemento `h2` debe tener "color" configurado como `red`.
+
+```js
+assert($('h2')[0].style.color === 'red');
+```
+
+Tu declaración `style` debe terminar con punto y coma (`;`).
+
+```js
+assert($('h2').attr('style') && $('h2').attr('style').endsWith(';'));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<h2 style="color: red;">CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/change-the-font-size-of-an-element.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/change-the-font-size-of-an-element.md
new file mode 100644
index 00000000000..e81d4c1178d
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/change-the-font-size-of-an-element.md
@@ -0,0 +1,119 @@
+---
+id: bad87fee1348bd9aedf08806
+title: Cambia el tamaño de fuente de un elemento
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c3bvDc8'
+forumTopicId: 16777
+dashedName: change-the-font-size-of-an-element
+---
+
+# --description--
+
+El tamaño de la fuente es controlado por la propiedad CSS `font-size`, como se muestra a continuación:
+
+```css
+h1 {
+ font-size: 30px;
+}
+```
+
+# --instructions--
+
+Dentro de la misma etiqueta `<style>` que contiene tu clase `red-text`, crea una entrada para elementos `p` y establece su tamaño de fuente o `font-size` a 16 píxeles (`16px`).
+
+# --hints--
+
+Entre las etiquetas `style`, asigna a los elementos `p` un `font-size` de `16px`. El zoom del navegador y del texto deben estar al 100%.
+
+```js
+assert(code.match(/p\s*{\s*font-size\s*:\s*16\s*px\s*;\s*}/i));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .red-text {
+ color: red;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<style>
+ .red-text {
+ color: red;
+ }
+ p {
+ font-size: 16px;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/create-a-custom-css-variable.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/create-a-custom-css-variable.md
new file mode 100644
index 00000000000..6923015ed61
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/create-a-custom-css-variable.md
@@ -0,0 +1,241 @@
+---
+id: 5a9d726c424fe3d0e10cad11
+title: Crea una variable de CSS personalizada
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cQd27Hr'
+forumTopicId: 301086
+dashedName: create-a-custom-css-variable
+---
+
+# --description--
+
+Para crear una variable CSS, solo tienes que darle un nombre que comience con dos guiones (--) y asignarle un valor, como se muestra a continuación:
+
+```css
+--penguin-skin: gray;
+```
+
+Esto creará una variable llamada `--penguin-skin` y le asignará el valor `gray`. Ahora puedes usar esa variable en cualquier otro lugar de tu código CSS para cambiar el valor de otros elementos a "gray" (gris).
+
+# --instructions--
+
+En la clase `penguin`, crea un nombre de variable `--penguin-skin` y asígnale el valor `gray`.
+
+# --hints--
+
+La clase `penguin` debe declarar la variable `--penguin-skin` y asignarle el valor `gray`.
+
+```js
+assert(
+ code.match(/\.penguin\s*\{[^{}]*?--penguin-skin\s*:\s*gr[ae]y\s*;[^{}]*?\}/gi)
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .penguin {
+ /* Only change code below this line */
+
+ /* Only change code above this line */
+ position: relative;
+ margin: auto;
+ display: block;
+ margin-top: 5%;
+ width: 300px;
+ height: 300px;
+ }
+
+ .penguin-top {
+ top: 10%;
+ left: 25%;
+ background: black;
+ width: 50%;
+ height: 45%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .penguin-bottom {
+ top: 40%;
+ left: 23.5%;
+ background: black;
+ width: 53%;
+ height: 45%;
+ border-radius: 70% 70% 100% 100%;
+ }
+
+ .right-hand {
+ top: 0%;
+ left: -5%;
+ background: black;
+ width: 30%;
+ height: 60%;
+ border-radius: 30% 30% 120% 30%;
+ transform: rotate(45deg);
+ z-index: -1;
+ }
+
+ .left-hand {
+ top: 0%;
+ left: 75%;
+ background: black;
+ width: 30%;
+ height: 60%;
+ border-radius: 30% 30% 30% 120%;
+ transform: rotate(-45deg);
+ z-index: -1;
+ }
+
+ .right-cheek {
+ top: 15%;
+ left: 35%;
+ background: white;
+ width: 60%;
+ height: 70%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .left-cheek {
+ top: 15%;
+ left: 5%;
+ background: white;
+ width: 60%;
+ height: 70%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .belly {
+ top: 60%;
+ left: 2.5%;
+ background: white;
+ width: 95%;
+ height: 100%;
+ border-radius: 120% 120% 100% 100%;
+ }
+
+ .right-feet {
+ top: 85%;
+ left: 60%;
+ background: orange;
+ width: 15%;
+ height: 30%;
+ border-radius: 50% 50% 50% 50%;
+ transform: rotate(-80deg);
+ z-index: -2222;
+ }
+
+ .left-feet {
+ top: 85%;
+ left: 25%;
+ background: orange;
+ width: 15%;
+ height: 30%;
+ border-radius: 50% 50% 50% 50%;
+ transform: rotate(80deg);
+ z-index: -2222;
+ }
+
+ .right-eye {
+ top: 45%;
+ left: 60%;
+ background: black;
+ width: 15%;
+ height: 17%;
+ border-radius: 50%;
+ }
+
+ .left-eye {
+ top: 45%;
+ left: 25%;
+ background: black;
+ width: 15%;
+ height: 17%;
+ border-radius: 50%;
+ }
+
+ .sparkle {
+ top: 25%;
+ left: 15%;
+ background: white;
+ width: 35%;
+ height: 35%;
+ border-radius: 50%;
+ }
+
+ .blush-right {
+ top: 65%;
+ left: 15%;
+ background: pink;
+ width: 15%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .blush-left {
+ top: 65%;
+ left: 70%;
+ background: pink;
+ width: 15%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .beak-top {
+ top: 60%;
+ left: 40%;
+ background: orange;
+ width: 20%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .beak-bottom {
+ top: 65%;
+ left: 42%;
+ background: orange;
+ width: 16%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ body {
+ background:#c6faf1;
+ }
+
+ .penguin * {
+ position: absolute;
+ }
+</style>
+<div class="penguin">
+ <div class="penguin-bottom">
+ <div class="right-hand"></div>
+ <div class="left-hand"></div>
+ <div class="right-feet"></div>
+ <div class="left-feet"></div>
+ </div>
+ <div class="penguin-top">
+ <div class="right-cheek"></div>
+ <div class="left-cheek"></div>
+ <div class="belly"></div>
+ <div class="right-eye">
+ <div class="sparkle"></div>
+ </div>
+ <div class="left-eye">
+ <div class="sparkle"></div>
+ </div>
+ <div class="blush-right"></div>
+ <div class="blush-left"></div>
+ <div class="beak-top"></div>
+ <div class="beak-bottom"></div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>.penguin {--penguin-skin: gray;}</style>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/give-a-background-color-to-a-div-element.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/give-a-background-color-to-a-div-element.md
new file mode 100644
index 00000000000..510c614da80
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/give-a-background-color-to-a-div-element.md
@@ -0,0 +1,176 @@
+---
+id: bad87fed1348bd9aede07836
+title: Establece el color de fondo de un elemento div
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cdRKMCk'
+forumTopicId: 18190
+dashedName: give-a-background-color-to-a-div-element
+---
+
+# --description--
+
+Puedes establecer el color de fondo de un elemento con la propiedad `background-color`.
+
+Por ejemplo, si quieres que el color de fondo de un elemento sea `green`, lo harías dentro de tu elemento `style`:
+
+```css
+.green-background {
+ background-color: green;
+}
+```
+
+# --instructions--
+
+Crea una clase llamada `silver-background` con un `background-color` `silver` (plateado). Asigna esta clase a tu elemento `div`.
+
+# --hints--
+
+Tu elemento `div` debe incluir la "class" `silver-background`.
+
+```js
+assert($('div').hasClass('silver-background'));
+```
+
+Tu elemento `div` debe tener un color de fondo "silver" (plateado).
+
+```js
+assert($('div').css('background-color') === 'rgb(192, 192, 192)');
+```
+
+Una clase llamada `silver-background` debe estar definida dentro del elemento `style` y el valor `silver` debe ser asignado a la propiedad `background-color`.
+
+```js
+assert(code.match(/\.silver-background\s*{\s*background-color:\s*silver;\s*}/));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">
+<style>
+ .red-text {
+ color: red;
+ }
+
+ h2 {
+ font-family: Lobster, monospace;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+
+ .thick-green-border {
+ border-color: green;
+ border-width: 10px;
+ border-style: solid;
+ border-radius: 50%;
+ }
+
+ .smaller-image {
+ width: 100px;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img class="smaller-image thick-green-border" src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">
+<style>
+ .red-text {
+ color: red;
+ }
+
+ h2 {
+ font-family: Lobster, monospace;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+
+ .thick-green-border {
+ border-color: green;
+ border-width: 10px;
+ border-style: solid;
+ border-radius: 50%;
+ }
+
+ .smaller-image {
+ width: 100px;
+ }
+
+ .silver-background {
+ background-color: silver;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img class="smaller-image thick-green-border" src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div class="silver-background">
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/import-a-google-font.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/import-a-google-font.md
new file mode 100644
index 00000000000..b6c52eb3716
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/import-a-google-font.md
@@ -0,0 +1,170 @@
+---
+id: bad87fee1348bd9aedf08807
+title: Importa una fuente de Google Fonts
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cM9MRsJ'
+forumTopicId: 18200
+dashedName: import-a-google-font
+---
+
+# --description--
+
+Además de especificar fuentes comunes disponibles en la mayoría de los sistemas operativos, también podemos especificar fuentes web no estándar y personalizadas para usarlas en nuestro sitio web. En Internet hay muchos lugares donde podemos conseguir web fonts. Para este ejemplo nos centraremos en la biblioteca de Google Fonts.
+
+[Google Fonts](https://fonts.google.com/) es una biblioteca gratuita de web fonts que puedes utilizar en tu código CSS, haciendo referencia a la URL de la fuente.
+
+Entonces, importemos y apliquemos una fuente de Google (ten en cuenta que si Google está bloqueado en tu país, tendrás que pasar por alto este desafío).
+
+Para importar una fuente de Google, puedes copiar la URL de la fuente desde la librería de Google Fonts y luego pegarla en tu código HTML. Para este desafío, importaremos la fuente `Lobster`. Para ello, copia el siguiente fragmento de código y pégalo en la parte superior de tu editor de código (antes de abrir el elemento `style`):
+
+`<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">`
+
+Ahora puedes usar la fuente `Lobster` en tu código CSS usando `Lobster` como FAMILY_NAME, siguiendo la sintaxis del siguiente ejemplo (aquí, FAMILY_NAME significa "nombre de la familia de fuentes", y GENERIC_NAME significa "nombre genérico"):
+`font-family: FAMILY_NAME, GENERIC_NAME;`.
+
+GENERIC_NAME es opcional, y es el modo de especificar una fuente de reserva o "fallback font" en caso de que la otra fuente especificada no esté disponible. Veremos esto en el siguiente desafío.
+
+Los nombres de familia de fuentes son sensibles a mayúsculas y minúsculas, y si incluyen espacios deben estar envueltos entre comillas. Por ejemplo, necesitas comillas para usar la fuente `"Open Sans"`, pero no las necesitas para usar la fuente `Lobster`.
+
+# --instructions--
+
+Importa la fuente `Lobster` a tu página web. Luego, usa un selector de elementos para establecer `Lobster` como `font-family` en tu elemento `h2`.
+
+# --hints--
+
+Debes importar la fuente `Lobster`.
+
+```js
+assert(new RegExp('googleapis', 'gi').test(code));
+```
+
+Tu elemento `h2` debe usar la fuente `Lobster`.
+
+```js
+assert(
+ $('h2')
+ .css('font-family')
+ .match(/lobster/i)
+);
+```
+
+Solo debes utilizar un selector de elementos `h2` para cambiar la fuente.
+
+```js
+assert(
+ /\s*[^\.]h2\s*\{\s*font-family\:\s*(['"]?)Lobster\1\s*(;\s*\}|\})/gi.test(
+ code
+ )
+);
+```
+
+Tu elemento `p` debe seguir usando la fuente `monospace`.
+
+```js
+assert(
+ $('p')
+ .css('font-family')
+ .match(/monospace/i)
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .red-text {
+ color: red;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">
+<style>
+ .red-text {
+ color: red;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+
+ h2 {
+ font-family: Lobster;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/improve-compatibility-with-browser-fallbacks.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/improve-compatibility-with-browser-fallbacks.md
new file mode 100644
index 00000000000..1f021c6835d
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/improve-compatibility-with-browser-fallbacks.md
@@ -0,0 +1,70 @@
+---
+id: 5b7d72c338cd7e35b63f3e14
+title: Mejora la compatibilidad con navegadores por medio de configuraciones de respaldo o "browser fallbacks"
+challengeType: 0
+videoUrl: ''
+forumTopicId: 301087
+dashedName: improve-compatibility-with-browser-fallbacks
+---
+
+# --description--
+
+Cuando trabajes con código CSS, posiblemente te enfrentarás en algún momento con problemas de compatibilidad con otros navegadores web. Por esta razón es importante proporcionar configuraciones de respaldo para otros navegadores o "browser fallbacks" para lidiar con posibles problemas de compatiblidad.
+
+Cuando tu navegador analiza el código CSS de una página web, ignora cualquier propiedad que no reconozca o soporte. Por ejemplo, si utilizas una variable CSS para asignar un color de fondo en un sitio, Internet Explorer ignorará el color de fondo establecido porque no soporta el uso de variables CSS. En ese caso, el navegador utilizará cualquier valor que tenga establecido como valor de esa propiedad. Si no puede encontrar ningún otro valor establecido para esa propiedad en el código, se revertirá al valor por defecto de ese navegador, lo que habitualmente no será lo ideal.
+
+Esto significa que si quieres proporcionar un valor de respaldo para el navegador, esto es tan sencillo como incluir otro valor más comúnmente soportado inmediatamente antes de tu declaración. De este modo, un navegador antiguo tendrá ún valor que sí pueda soportar, mientras que un navegador más nuevo interpretará cualquier declaración que incluyas más adelante en la cascada.
+
+# --instructions--
+
+Parece que en este código se utiliza una variable para establecer el color de fondo de la clase `.red-box`. Mejoremos la compatibilidad de nuestro código con otros navegadores, añadiendo otra declaración de `background` justo antes de la declaración existente, y estableciendo este valor de respaldo como `red` (rojo).
+
+# --hints--
+
+Tu regla `.red-box` debe incluir un valor de respaldo o "fallback" con el `background` con un valor de `red`, inmediatamente antes de la declaración `background` existente.
+
+```js
+assert(
+ code
+ .replace(/\s/g, '')
+ .match(
+ /\.red-box{background:(red|#ff0000|#f00|rgb\(255,0,0\)|rgb\(100%,0%,0%\)|hsl\(0,100%,50%\));background:var\(--red-color\);height:200px;width:200px;}/gi
+ )
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ :root {
+ --red-color: red;
+ }
+ .red-box {
+
+ background: var(--red-color);
+ height: 200px;
+ width:200px;
+ }
+</style>
+<div class="red-box"></div>
+```
+
+# --solutions--
+
+```html
+<style>
+ :root {
+ --red-color: red;
+ }
+ .red-box {
+ background: red;
+ background: var(--red-color);
+ height: 200px;
+ width:200px;
+ }
+</style>
+<div class="red-box"></div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/inherit-css-variables.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/inherit-css-variables.md
new file mode 100644
index 00000000000..e5f0c7128b9
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/inherit-css-variables.md
@@ -0,0 +1,244 @@
+---
+id: 5a9d7295424fe3d0e10cad14
+title: Hereda variables CSS
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cyLZZhZ'
+forumTopicId: 301088
+dashedName: inherit-css-variables
+---
+
+# --description--
+
+Cuando creas una variable, queda disponible para que la utilices dentro del selector en el que la hayas creado. Esa variable también estará disponible en cualquiera de los descendientes de ese selector. Esto ocurre porque las variables CSS son heredadas, al igual que las propiedades comunes.
+
+Para hacer uso de la herencia, las variables CSS suelen ser definidas en el elemento <dfn>:root</dfn>.
+
+`:root` es un "<dfn>pseudo-class</dfn> selector" (selector de pseudo-clase) que corresponde al elemento raíz o "root" del documento, que generalmente es el elemento `html`. Al crear tus variables en `:root`, estarán disponibles globalmente y se podrán acceder desde cualquier otro selector en la hoja de estilo.
+
+# --instructions--
+
+Define una variable llamada `--penguin-belly` en el selector `:root` y asígnale el valor `pink` (rosado). Luego podrás ver que la variable es heredada y que todos los elementos secundarios que la utilizan tendrán el color de fondo rosado.
+
+# --hints--
+
+La variable `--penguin-belly` debe ser declarada en `:root` y se le debe asignar el valor `pink`.
+
+```js
+assert(
+ code.match(/:root\s*?{[\s\S]*--penguin-belly\s*?:\s*?pink\s*?;[\s\S]*}/gi)
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ :root {
+ /* Only change code below this line */
+
+ /* Only change code above this line */
+ }
+
+ body {
+ background: var(--penguin-belly, #c6faf1);
+ }
+
+ .penguin {
+ --penguin-skin: gray;
+ --penguin-beak: orange;
+ position: relative;
+ margin: auto;
+ display: block;
+ margin-top: 5%;
+ width: 300px;
+ height: 300px;
+ }
+
+ .right-cheek {
+ top: 15%;
+ left: 35%;
+ background: var(--penguin-belly, white);
+ width: 60%;
+ height: 70%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .left-cheek {
+ top: 15%;
+ left: 5%;
+ background: var(--penguin-belly, white);
+ width: 60%;
+ height: 70%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .belly {
+ top: 60%;
+ left: 2.5%;
+ background: var(--penguin-belly, white);
+ width: 95%;
+ height: 100%;
+ border-radius: 120% 120% 100% 100%;
+ }
+
+ .penguin-top {
+ top: 10%;
+ left: 25%;
+ background: var(--penguin-skin, gray);
+ width: 50%;
+ height: 45%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .penguin-bottom {
+ top: 40%;
+ left: 23.5%;
+ background: var(--penguin-skin, gray);
+ width: 53%;
+ height: 45%;
+ border-radius: 70% 70% 100% 100%;
+ }
+
+ .right-hand {
+ top: 0%;
+ left: -5%;
+ background: var(--penguin-skin, gray);
+ width: 30%;
+ height: 60%;
+ border-radius: 30% 30% 120% 30%;
+ transform: rotate(45deg);
+ z-index: -1;
+ }
+
+ .left-hand {
+ top: 0%;
+ left: 75%;
+ background: var(--penguin-skin, gray);
+ width: 30%;
+ height: 60%;
+ border-radius: 30% 30% 30% 120%;
+ transform: rotate(-45deg);
+ z-index: -1;
+ }
+
+ .right-feet {
+ top: 85%;
+ left: 60%;
+ background: var(--penguin-beak, orange);
+ width: 15%;
+ height: 30%;
+ border-radius: 50% 50% 50% 50%;
+ transform: rotate(-80deg);
+ z-index: -2222;
+ }
+
+ .left-feet {
+ top: 85%;
+ left: 25%;
+ background: var(--penguin-beak, orange);
+ width: 15%;
+ height: 30%;
+ border-radius: 50% 50% 50% 50%;
+ transform: rotate(80deg);
+ z-index: -2222;
+ }
+
+ .right-eye {
+ top: 45%;
+ left: 60%;
+ background: black;
+ width: 15%;
+ height: 17%;
+ border-radius: 50%;
+ }
+
+ .left-eye {
+ top: 45%;
+ left: 25%;
+ background: black;
+ width: 15%;
+ height: 17%;
+ border-radius: 50%;
+ }
+
+ .sparkle {
+ top: 25%;
+ left: 15%;
+ background: white;
+ width: 35%;
+ height: 35%;
+ border-radius: 50%;
+ }
+
+ .blush-right {
+ top: 65%;
+ left: 15%;
+ background: pink;
+ width: 15%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .blush-left {
+ top: 65%;
+ left: 70%;
+ background: pink;
+ width: 15%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .beak-top {
+ top: 60%;
+ left: 40%;
+ background: var(--penguin-beak, orange);
+ width: 20%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .beak-bottom {
+ top: 65%;
+ left: 42%;
+ background: var(--penguin-beak, orange);
+ width: 16%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .penguin * {
+ position: absolute;
+ }
+</style>
+<div class="penguin">
+ <div class="penguin-bottom">
+ <div class="right-hand"></div>
+ <div class="left-hand"></div>
+ <div class="right-feet"></div>
+ <div class="left-feet"></div>
+ </div>
+ <div class="penguin-top">
+ <div class="right-cheek"></div>
+ <div class="left-cheek"></div>
+ <div class="belly"></div>
+ <div class="right-eye">
+ <div class="sparkle"></div>
+ </div>
+ <div class="left-eye">
+ <div class="sparkle"></div>
+ </div>
+ <div class="blush-right"></div>
+ <div class="blush-left"></div>
+ <div class="beak-top"></div>
+ <div class="beak-bottom"></div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>:root {--penguin-belly: pink;}</style>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/inherit-styles-from-the-body-element.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/inherit-styles-from-the-body-element.md
new file mode 100644
index 00000000000..3ac886731b2
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/inherit-styles-from-the-body-element.md
@@ -0,0 +1,111 @@
+---
+id: bad87fee1348bd9aedf08746
+title: Hereda estilos del elemento body
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c9bmdtR'
+forumTopicId: 18204
+dashedName: inherit-styles-from-the-body-element
+---
+
+# --description--
+
+Ahora hemos demostrado que cada página HTML tiene un elemento `body`, y que a este elemento `body` también se le puede dar estilo con CSS.
+
+Recuerda, puedes dar estilo a tu elemento `body` como a cualquier otro elemento HTML, y todos los demás elementos heredarán los estilos del elemento `body`.
+
+# --instructions--
+
+Primero, crea un elemento `h1` con el texto `Hello World`
+
+Luego, demos el color `green` (verde) a todos los elementos de tu página, añadiendo `color: green;` a tu declaración de estilo del elemento `body`.
+
+Finalmente, da a tu elemento `body` un valor para font-family de `monospace` añadiendo `font-family: monospace;` a la declaración de estilo del elemento `body`.
+
+# --hints--
+
+Debes crear un elemento `h1`.
+
+```js
+assert($('h1').length > 0);
+```
+
+Tu elemento `h1` debe contener el texto `Hello World`.
+
+```js
+assert(
+ $('h1').length > 0 &&
+ $('h1')
+ .text()
+ .match(/hello world/i)
+);
+```
+
+Tu elemento `h1` debe tener una etiqueta de cierre.
+
+```js
+assert(
+ code.match(/<\/h1>/g) &&
+ code.match(/<h1/g) &&
+ code.match(/<\/h1>/g).length === code.match(/<h1/g).length
+);
+```
+
+Tu elemento `body` debe tener la propiedad `color` con el valor `green`.
+
+```js
+assert($('body').css('color') === 'rgb(0, 128, 0)');
+```
+
+Tu elemento `body` debe tener la propiedad `font-family` con el valor `monospace`.
+
+```js
+assert(
+ $('body')
+ .css('font-family')
+ .match(/monospace/i)
+);
+```
+
+Tu elemento `h1` debe heredar la fuente `monospace` de tu elemento `body`.
+
+```js
+assert(
+ $('h1').length > 0 &&
+ $('h1')
+ .css('font-family')
+ .match(/monospace/i)
+);
+```
+
+Tu elemento `h1` debe heredar el color "green" de tu elemento `body`.
+
+```js
+assert($('h1').length > 0 && $('h1').css('color') === 'rgb(0, 128, 0)');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ body {
+ background-color: black;
+ }
+
+</style>
+```
+
+# --solutions--
+
+```html
+<style>
+ body {
+ background-color: black;
+ font-family: monospace;
+ color: green;
+ }
+
+</style>
+<h1>Hello World!</h1>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/make-circular-images-with-a-border-radius.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/make-circular-images-with-a-border-radius.md
new file mode 100644
index 00000000000..29a27109cd9
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/make-circular-images-with-a-border-radius.md
@@ -0,0 +1,159 @@
+---
+id: bad87fee1348bd9aedf08815
+title: Haz imágenes circulares usando border-radius (radio de borde)
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c2MvrcB'
+forumTopicId: 18229
+dashedName: make-circular-images-with-a-border-radius
+---
+
+# --description--
+
+Además de los píxeles, también puedes especificar el `border-radius` usando un porcentaje.
+
+# --instructions--
+
+Dale a tu foto de gato un `border-radius` de `50%`.
+
+# --hints--
+
+Tu imagen debe tener un "border-radius" de `50%`, lo que hará que el borde sea perfectamente circular.
+
+```js
+assert(parseInt($('img').css('border-top-left-radius')) > 48);
+```
+
+El valor `border-radius` debe usar un valor porcentual de `50%`.
+
+```js
+assert(code.match(/50%/g));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">
+<style>
+ .red-text {
+ color: red;
+ }
+
+ h2 {
+ font-family: Lobster, monospace;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+
+ .thick-green-border {
+ border-color: green;
+ border-width: 10px;
+ border-style: solid;
+ border-radius: 10px;
+ }
+
+ .smaller-image {
+ width: 100px;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img class="smaller-image thick-green-border" src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">
+<style>
+ .red-text {
+ color: red;
+ }
+
+ h2 {
+ font-family: Lobster, monospace;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+
+ .thick-green-border {
+ border-color: green;
+ border-width: 10px;
+ border-style: solid;
+ border-radius: 10px;
+ }
+
+ .smaller-image {
+ width: 100px;
+ border-radius: 50%;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img class="smaller-image thick-green-border" src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/override-all-other-styles-by-using-important.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/override-all-other-styles-by-using-important.md
new file mode 100644
index 00000000000..7ccd376e902
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/override-all-other-styles-by-using-important.md
@@ -0,0 +1,112 @@
+---
+id: bad87fee1348bd9aedf07756
+title: Sobreescribe todos los demás estilos usando !important
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cm24rcp'
+forumTopicId: 18249
+dashedName: override-all-other-styles-by-using-important
+---
+
+# --description--
+
+¡Excelente! Acabamos de demostrar que los estilos en línea sobrescribirán todas las declaraciones CSS en tu elemento `style`.
+
+Pero, ¡aguarda! Existe una última forma de sobreescribir CSS. Este es el método más poderoso de todos. Pero antes de utilizarlo, consideremos por qué querrías sobreescribir una regla CSS.
+
+En muchas situaciones usarás librerías de CSS. Estas librerías pueden sobreescribir accidentalmente tu propio código CSS. Entonces, cuando necesites asegurarte de que a un elemento se le aplique un código CSS específico, puedes usar `!important`.
+
+Volvamos a nuestra declaración de la clase `pink-text`. Recuerda que nuestra clase `pink-text` fue sobreescrita por declaraciones posteriores de clases, declaraciones de id e "inline styles".
+
+# --instructions--
+
+Añade la palabra clave `!important` a la declaración de color de pink-text para asegurarte completamente de que tu elemento `h1` será de color rosado.
+
+Aquí te mostramos un ejemplo de cómo hacerlo:
+
+`color: red !important;`
+
+# --hints--
+
+Tu elemento `h1` debe incluir la clase `pink-text`.
+
+```js
+assert($('h1').hasClass('pink-text'));
+```
+
+Tu elemento `h1` debe incluir la clase `blue-text`.
+
+```js
+assert($('h1').hasClass('blue-text'));
+```
+
+Tu elemento `h1` debe incluir el `id` `orange-text`.
+
+```js
+assert($('h1').attr('id') === 'orange-text');
+```
+
+Tu elemento `h1` debe incluir el "inline style" `color: white`.
+
+```js
+assert(code.match(/<h1.*style/gi) && code.match(/<h1.*style.*color\s*?:/gi));
+```
+
+Tu elemento `pink-text` debe tener la palabra clave `!important` para sobrescribir todas las demás declaraciones.
+
+```js
+assert(
+ code.match(/\.pink-text\s*?\{[\s\S]*?color:.*pink.*!important\s*;?[^\.]*\}/g)
+);
+```
+
+Tu elemento `h1` debe ser rosado.
+
+```js
+assert($('h1').css('color') === 'rgb(255, 192, 203)');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ body {
+ background-color: black;
+ font-family: monospace;
+ color: green;
+ }
+ #orange-text {
+ color: orange;
+ }
+ .pink-text {
+ color: pink;
+ }
+ .blue-text {
+ color: blue;
+ }
+</style>
+<h1 id="orange-text" class="pink-text blue-text" style="color: white">Hello World!</h1>
+```
+
+# --solutions--
+
+```html
+<style>
+ body {
+ background-color: black;
+ font-family: monospace;
+ color: green;
+ }
+ #orange-text {
+ color: orange;
+ }
+ .pink-text {
+ color: pink !important;
+ }
+ .blue-text {
+ color: blue;
+ }
+</style>
+<h1 id="orange-text" class="pink-text blue-text" style="color: white">Hello World!</h1>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/override-class-declarations-by-styling-id-attributes.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/override-class-declarations-by-styling-id-attributes.md
new file mode 100644
index 00000000000..2e460d472f7
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/override-class-declarations-by-styling-id-attributes.md
@@ -0,0 +1,121 @@
+---
+id: bad87fee1348bd8aedf06756
+title: Sobreescribe las declaraciones de clase dando estilo a atributos ID
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cRkpDhB'
+forumTopicId: 18251
+dashedName: override-class-declarations-by-styling-id-attributes
+---
+
+# --description--
+
+Acabamos de comprobar que los navegadores leen CSS desde arriba hacia abajo siguiendo el orden de las declaraciones. Por lo tanto, si llega a ocasionarse un conflicto, el navegador utilizará la última declaración CSS. Ten en cuenta que incluso si hubiésemos puesto `blue-text` antes que `pink-text` en nuestro elemento de clase `h1`, seguiría buscando al orden de la declaración y no al orden de su uso.
+
+Pero todavía no terminamos. Hay otras formas de sobreescribir CSS. ¿Recuerdas los atributos id?
+
+Sobrescribamos tus clases `pink-text` y `blue-text`, y haz que el `h1` sea naranja al darle al elemento `h1` un id y luego estilizando ese id.
+
+# --instructions--
+
+Dale al elemento `h1` el atributo `id` de `orange-text`. Recuerda, los estilos id se ven así:
+
+`<h1 id="orange-text">`
+
+Deja las clases `blue-text` y `pink-text` en tu elemento `h1`.
+
+Crea una declaración CSS para el id `orange-text` en el elemento `style`. Aquí hay un ejemplo de cómo se debería ver:
+
+```css
+#brown-text {
+ color: brown;
+}
+```
+
+**Nota:** No importa si declaras el CSS arriba o debajo de la clase pink-text, ya que el atributo id siempre tendrá precedencia.
+
+# --hints--
+
+Tu elemento `h1` debe incluir la clase `pink-text`.
+
+```js
+assert($('h1').hasClass('pink-text'));
+```
+
+Tu elemento `h1` debe incluir la clase `blue-text`.
+
+```js
+assert($('h1').hasClass('blue-text'));
+```
+
+Tu elemento `h1` debe incluir el id `orange-text`.
+
+```js
+assert($('h1').attr('id') === 'orange-text');
+```
+
+Debe haber un solo elemento `h1`.
+
+```js
+assert($('h1').length === 1);
+```
+
+El id `orange-text` debe tener una declaración CSS.
+
+```js
+assert(code.match(/#orange-text\s*{/gi));
+```
+
+Tu `h1` no debe tener ningún atributo `style`.
+
+```js
+assert(!code.match(/<h1.*style.*>/gi));
+```
+
+Tu elemento `h1` debe ser naranja.
+
+```js
+assert($('h1').css('color') === 'rgb(255, 165, 0)');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ body {
+ background-color: black;
+ font-family: monospace;
+ color: green;
+ }
+ .pink-text {
+ color: pink;
+ }
+ .blue-text {
+ color: blue;
+ }
+</style>
+<h1 class="pink-text blue-text">Hello World!</h1>
+```
+
+# --solutions--
+
+```html
+<style>
+ body {
+ background-color: black;
+ font-family: monospace;
+ color: green;
+ }
+ .pink-text {
+ color: pink;
+ }
+ .blue-text {
+ color: blue;
+ }
+ #orange-text {
+ color: orange;
+ }
+</style>
+<h1 id="orange-text" class="pink-text blue-text">Hello World!</h1>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/override-class-declarations-with-inline-styles.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/override-class-declarations-with-inline-styles.md
new file mode 100644
index 00000000000..f6cd25b59f9
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/override-class-declarations-with-inline-styles.md
@@ -0,0 +1,100 @@
+---
+id: bad87fee1348bd9aedf06756
+title: Sobreescribe declaraciones de clase con inline styles (estilos en línea)
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cGJDRha'
+forumTopicId: 18252
+dashedName: override-class-declarations-with-inline-styles
+---
+
+# --description--
+
+Hemos demostrado que las declaraciones de id tienen prioridad por sobre las declaraciones de clase, independientemente de dónde hayan sido declaradas en el código CSS del elemento `style`.
+
+Pero existen otras formas de sobreescribir código CSS. ¿Recuerdas los "inline styles" (estilos en línea)?
+
+# --instructions--
+
+Usa un inline style para hacer que nuestro elemento `h1` sea de color blanco. Recuerda que los inline styles se ven así:
+
+`<h1 style="color: green;">`
+
+Deja las clases `blue-text` y `pink-text` en tu elemento `h1`.
+
+# --hints--
+
+Tu elemento `h1` debe incluir la clase `pink-text`.
+
+```js
+assert($('h1').hasClass('pink-text'));
+```
+
+Tu elemento `h1` debe incluir la clase `blue-text`.
+
+```js
+assert($('h1').hasClass('blue-text'));
+```
+
+Tu elemento `h1` debe incluir el id `orange-text`.
+
+```js
+assert($('h1').attr('id') === 'orange-text');
+```
+
+Tu elemento `h1` debe tener un inline style.
+
+```js
+assert(document.querySelector('h1[style]'));
+```
+
+Tu elemento `h1` debe ser de color blanco ("white").
+
+```js
+assert($('h1').css('color') === 'rgb(255, 255, 255)');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ body {
+ background-color: black;
+ font-family: monospace;
+ color: green;
+ }
+ #orange-text {
+ color: orange;
+ }
+ .pink-text {
+ color: pink;
+ }
+ .blue-text {
+ color: blue;
+ }
+</style>
+<h1 id="orange-text" class="pink-text blue-text">Hello World!</h1>
+```
+
+# --solutions--
+
+```html
+<style>
+ body {
+ background-color: black;
+ font-family: monospace;
+ color: green;
+ }
+ #orange-text {
+ color: orange;
+ }
+ .pink-text {
+ color: pink;
+ }
+ .blue-text {
+ color: blue;
+ }
+</style>
+<h1 id="orange-text" class="pink-text blue-text" style="color: white">Hello World!</h1>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/override-styles-in-subsequent-css.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/override-styles-in-subsequent-css.md
new file mode 100644
index 00000000000..144ee6154c3
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/override-styles-in-subsequent-css.md
@@ -0,0 +1,92 @@
+---
+id: bad87fee1348bd9aedf04756
+title: Sobreescribe estilos en código CSS posterior
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cGJDQug'
+forumTopicId: 18253
+dashedName: override-styles-in-subsequent-css
+---
+
+# --description--
+
+¡Nuestra clase `pink-text` sobrescribió la declaración CSS de nuestro elemento `body`!
+
+Acabamos de demostrar que nuestras clases sobrescribirán el CSS del elemento `body`. Entonces, la siguiente pregunta lógica es: ¿qué podemos hacer para sobrescribir nuestra clase `pink-text`?
+
+# --instructions--
+
+Crea una clase CSS adicional llamada `blue-text` que asigne a un elemento el color azul ("blue"). Asegúrate de que esté debajo de tu declaración de `pink-text`.
+
+Aplica la clase `blue-text` a tu elemento `h1` además de tu clase `pink-text`, y veamos cuál de las dos gana.
+
+Para aplicar múltiples atributos de clase a un elemento HTML debes dejar un espacio entre ellos, como se muestra a continuación:
+
+`class="class1 class2"`
+
+**Nota:** No importa el orden en que las clases estén enlistadas dentro del elemento HTML.
+
+Sin embargo, lo importante es el orden de las declaraciones de `class` clases en la sección `<style>`. La segunda declaración siempre tendrá prioridad sobre la primera. Debido a que `.blue-text` ha sido declarada en segundo lugar, sobrescribirá los atributos de `.pink-text`
+
+# --hints--
+
+Tu elemento `h1` debe incluir la clase `pink-text`.
+
+```js
+assert($('h1').hasClass('pink-text'));
+```
+
+Tu elemento `h1` debe incluir la clase `blue-text`.
+
+```js
+assert($('h1').hasClass('blue-text'));
+```
+
+Tanto `blue-text` como `pink-text` deben pertenecer al mismo elemento `h1`.
+
+```js
+assert($('.pink-text').hasClass('blue-text'));
+```
+
+Tu elemento `h1` debe ser de color azul ("blue").
+
+```js
+assert($('h1').css('color') === 'rgb(0, 0, 255)');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ body {
+ background-color: black;
+ font-family: monospace;
+ color: green;
+ }
+ .pink-text {
+ color: pink;
+ }
+</style>
+<h1 class="pink-text">Hello World!</h1>
+```
+
+# --solutions--
+
+```html
+<style>
+ body {
+ background-color: black;
+ font-family: monospace;
+ color: green;
+ }
+ .pink-text {
+ color: pink;
+ }
+
+ .blue-text {
+ color: blue;
+ }
+</style>
+<h1 class="pink-text blue-text">Hello World!</h1>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/prioritize-one-style-over-another.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/prioritize-one-style-over-another.md
new file mode 100644
index 00000000000..24561797a84
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/prioritize-one-style-over-another.md
@@ -0,0 +1,73 @@
+---
+id: bad87fee1348bd9aedf08756
+title: Prioriza un estilo por sobre otro
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cZ8wnHv'
+forumTopicId: 18258
+dashedName: prioritize-one-style-over-another
+---
+
+# --description--
+
+A veces los elementos HTML reciben múltiples estilos que entran en conflicto entre sí.
+
+Por ejemplo, tu elemento `h1` no puede ser verde y rosado al mismo tiempo.
+
+Veamos qué ocurre cuando creamos una clase que hace que el texto sea rosado ("pink"), y luego se la aplicamos a un elemento. ¿*Sobreescribirá* nuestra clase la prioridad CSS `color: green;` del elemento `body`?
+
+# --instructions--
+
+Crea una clase CSS adicional llamada `pink-text` que asigne a un elemento el color rosado ("pink").
+
+Asigna a tu elemento `h1` la clase `pink-text`.
+
+# --hints--
+
+Tu elemento `h1` debe incluir la clase `pink-text`.
+
+```js
+assert($('h1').hasClass('pink-text'));
+```
+
+Tu `<style>` debería tener una clase CSS `pink-text` que cambie su `color`.
+
+```js
+assert(code.match(/\.pink-text\s*\{\s*color\s*:\s*.+\s*;\s*\}/g));
+```
+
+Tu elemento `h1` debe ser de color rosado ("pink").
+
+```js
+assert($('h1').css('color') === 'rgb(255, 192, 203)');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ body {
+ background-color: black;
+ font-family: monospace;
+ color: green;
+ }
+</style>
+<h1>Hello World!</h1>
+```
+
+# --solutions--
+
+```html
+<style>
+ body {
+ background-color: black;
+ font-family: monospace;
+ color: green;
+ }
+ .pink-text {
+ color: pink;
+ }
+</style>
+<h1 class="pink-text">Hello World!</h1>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/set-the-font-family-of-an-element.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/set-the-font-family-of-an-element.md
new file mode 100644
index 00000000000..88feb30ea94
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/set-the-font-family-of-an-element.md
@@ -0,0 +1,132 @@
+---
+id: bad87fee1348bd9aede08807
+title: Establece la familia de fuentes de un elemento
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c3bvpCg'
+forumTopicId: 18278
+dashedName: set-the-font-family-of-an-element
+---
+
+# --description--
+
+Puedes utilizar la propiedad `font-family` (familia de fuente) para establecer qué fuente debe usar un elemento.
+
+Por ejemplo, si quieres establecer la fuente de tu elemento `h2` a `sans-serif`, usarías el siguiente CSS:
+
+```css
+h2 {
+ font-family: sans-serif;
+}
+```
+
+# --instructions--
+
+Haz que todos tus elementos `p` usen la fuente `monospace`.
+
+# --hints--
+
+Tus elementos `p` deben usar la fuente `monospace`.
+
+```js
+assert(
+ $('p')
+ .not('.red-text')
+ .css('font-family')
+ .match(/monospace/i)
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .red-text {
+ color: red;
+ }
+
+ p {
+ font-size: 16px;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<style>
+ .red-text {
+ color: red;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/set-the-id-of-an-element.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/set-the-id-of-an-element.md
new file mode 100644
index 00000000000..0f65d273ffe
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/set-the-id-of-an-element.md
@@ -0,0 +1,168 @@
+---
+id: bad87eee1348bd9aede07836
+title: Establece el id de un elemento
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cN6MEc2'
+forumTopicId: 18279
+dashedName: set-the-id-of-an-element
+---
+
+# --description--
+
+Además de las clases, cada elemento HTML también puede tener un atributo `id`.
+
+Existen varios beneficios de usar atributos de `id`: Puedes usar un `id` para dar estilo a un elemento específico; más adelante aprenderás que puedes usar atributos "id" para seleccionar y modificar elementos específicos por medio de JavaScript.
+
+Los atributos `id` deben ser únicos. Los navegadores no verifican esta regla, pero es una buena práctica ampliamente aceptada. Entonces, por favor ten cuidado de no asignar a más de un elemento el mismo atributo `id`.
+
+A continuación te mostramos un ejemplo de cómo asignar a tu elemento `h2` el id `cat-photo-app`:
+
+`<h2 id="cat-photo-app">`
+
+# --instructions--
+
+Asigna a tu elemento `form` el id `cat-photo-form`.
+
+# --hints--
+
+Tu elemento `form` debe contener el id `cat-photo-form`.
+
+```js
+assert($('form').attr('id') === 'cat-photo-form');
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">
+<style>
+ .red-text {
+ color: red;
+ }
+
+ h2 {
+ font-family: Lobster, monospace;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+
+ .thick-green-border {
+ border-color: green;
+ border-width: 10px;
+ border-style: solid;
+ border-radius: 50%;
+ }
+
+ .smaller-image {
+ width: 100px;
+ }
+
+ .silver-background {
+ background-color: silver;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img class="smaller-image thick-green-border" src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div class="silver-background">
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">
+<style>
+ .red-text {
+ color: red;
+ }
+
+ h2 {
+ font-family: Lobster, monospace;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+
+ .thick-green-border {
+ border-color: green;
+ border-width: 10px;
+ border-style: solid;
+ border-radius: 50%;
+ }
+
+ .smaller-image {
+ width: 100px;
+ }
+
+ .silver-background {
+ background-color: silver;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img class="smaller-image thick-green-border" src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div class="silver-background">
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo" id="cat-photo-form">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/size-your-images.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/size-your-images.md
new file mode 100644
index 00000000000..a5eaccfe8ab
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/size-your-images.md
@@ -0,0 +1,156 @@
+---
+id: bad87fee1348bd9acdf08812
+title: Cambia el tamaño de tus imágenes
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cM9MmCP'
+forumTopicId: 18282
+dashedName: size-your-images
+---
+
+# --description--
+
+CSS tiene una propiedad llamada `width` que controla el ancho de un elemento. Al igual que con las fuentes, usaremos la unidad de medida `px` (píxeles) para especificar el ancho de la imagen.
+
+Por ejemplo, si queremos crear una clase CSS llamada `larger-image` que le asigne a los elementos HTML un ancho de 500 píxeles, usamos el siguiente código:
+
+```html
+<style>
+ .larger-image {
+ width: 500px;
+ }
+</style>
+```
+
+# --instructions--
+
+Crea una clase llamada `smaller-image` y úsala para redimensionar la imagen de modo que solo tenga 100 píxeles de ancho.
+
+# --hints--
+
+Tu elemento `img` debe incluir la "class" `smaller-image`.
+
+```js
+assert(
+ $("img[src='https://bit.ly/fcc-relaxing-cat']").attr('class') ===
+ 'smaller-image'
+);
+```
+
+Tu imagen debe tener 100 píxeles de ancho.
+
+```js
+assert(
+ $('img').width() < 200 &&
+ code.match(/\.smaller-image\s*{\s*width\s*:\s*100px\s*(;\s*}|})/i)
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">
+<style>
+ .red-text {
+ color: red;
+ }
+
+ h2 {
+ font-family: Lobster, monospace;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">
+<style>
+ .red-text {
+ color: red;
+ }
+
+ h2 {
+ font-family: Lobster, monospace;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+
+ .smaller-image {
+ width: 100px;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img class="smaller-image" src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/specify-how-fonts-should-degrade.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/specify-how-fonts-should-degrade.md
new file mode 100644
index 00000000000..3a7094e1154
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/specify-how-fonts-should-degrade.md
@@ -0,0 +1,172 @@
+---
+id: bad87fee1348bd9aedf08808
+title: Especifica cómo deben degradarse las fuentes
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cpVKBfQ'
+forumTopicId: 18304
+dashedName: specify-how-fonts-should-degrade
+---
+
+# --description--
+
+Hay varias fuentes por defecto disponibles en todos los navegadores. Estas familias de fuentes genéricas incluyen a `monospace`, `serif` y `sans-serif`.
+
+Cuando una fuente no está disponible en el sistema, puedes indicarle al navegador que "degrade" a otra fuente alternativa.
+
+Por ejemplo, si quieres que un elemento utilice la fuente `Helvetica` pero que degrade a la fuente `sans-serif` en caso que `Helvetica` no esté disponible, se especifica de la siguiente manera:
+
+```css
+p {
+ font-family: Helvetica, sans-serif;
+}
+```
+
+Los nombres de las fuentes genéricas no distinguen entre mayúsculas y minúsculas. Además, no necesitan comillas porque son palabras clave CSS.
+
+# --instructions--
+
+Para comenzar, aplica la fuente `monospace` al elemento `h2`, para que ahora tenga dos fuentes: `Lobster` y `monospace`.
+
+En el último desafío, habías importado la fuente `Lobster` usando la etiqueta `link`. Ahora comenta esa importación de la fuente `Lobster` de Google Fonts (usando los comentarios HTML que aprendiste antes) para que ya no esté disponible. Verás que todos tus elementos `h2` degradan a la fuente `monospace` font.
+
+**Nota:** Si tienes la fuente `Lobster` instalada en tu computadora, no verás la degradación porque tu navegador puede encontrar la fuente en tu sistema.
+
+# --hints--
+
+Tu elemento h2 debe usar la fuente `Lobster`.
+
+```js
+assert(
+ $('h2')
+ .css('font-family')
+ .match(/^"?lobster/i)
+);
+```
+
+El elemento h2 debe degradarse a la fuente `monospace` cuando `Lobster` no está disponible.
+
+```js
+assert(
+ /\s*h2\s*\{\s*font-family\:\s*(\'|")?Lobster(\'|")?,\s*monospace\s*;\s*\}/gi.test(
+ code
+ )
+);
+```
+
+Debes comentar tu llamada a Google para importar la fuente `Lobster` agregando `<!--` delante de ella.
+
+```js
+assert(new RegExp('<!--[^fc]', 'gi').test(code));
+```
+
+Debes cerrar tu comentario agregando `-->` al final.
+
+```js
+assert(new RegExp('[^fc]-->', 'gi').test(code));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">
+<style>
+ .red-text {
+ color: red;
+ }
+
+ h2 {
+ font-family: Lobster;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<!--<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">-->
+<style>
+ .red-text {
+ color: red;
+ }
+
+ h2 {
+ font-family: Lobster, monospace;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/style-multiple-elements-with-a-css-class.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/style-multiple-elements-with-a-css-class.md
new file mode 100644
index 00000000000..16b005c5025
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/style-multiple-elements-with-a-css-class.md
@@ -0,0 +1,133 @@
+---
+id: bad87fee1348bd9aefe08806
+title: Aplica un estilo a múltiples elementos usando una clase CSS
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cRkVbsQ'
+forumTopicId: 18311
+dashedName: style-multiple-elements-with-a-css-class
+---
+
+# --description--
+
+Las clases te permiten usar los mismos estilos CSS en múltiples elementos HTML. Puedes ver esto aplicando tu clase de `red-text` al primer elemento `p`.
+
+# --hints--
+
+Tu elemento `h2` debe ser de color rojo ("red").
+
+```js
+assert($('h2').css('color') === 'rgb(255, 0, 0)');
+```
+
+Tu elemento `h2` debe incluir la clase `red-text`.
+
+```js
+assert($('h2').hasClass('red-text'));
+```
+
+Tu primer elemento `p` debe ser de color rojo ("red").
+
+```js
+assert($('p:eq(0)').css('color') === 'rgb(255, 0, 0)');
+```
+
+Tu segundo y tercer elemento `p` no deben ser de color rojo ("red").
+
+```js
+assert(
+ !($('p:eq(1)').css('color') === 'rgb(255, 0, 0)') &&
+ !($('p:eq(2)').css('color') === 'rgb(255, 0, 0)')
+);
+```
+
+Tu primer elemento `p` debe incluir la clase `red-text`.
+
+```js
+assert($('p:eq(0)').hasClass('red-text'));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .red-text {
+ color: red;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<style>
+ .red-text {
+ color: red;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/style-the-html-body-element.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/style-the-html-body-element.md
new file mode 100644
index 00000000000..1d4d4c7ba5a
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/style-the-html-body-element.md
@@ -0,0 +1,70 @@
+---
+id: bad87fee1348bd9aedf08736
+title: Aplica un estilo al elemento HTML body
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cB77PHW'
+forumTopicId: 18313
+dashedName: style-the-html-body-element
+---
+
+# --description--
+
+Ahora vamos comenzar de cero y hablaremos sobre la herencia CSS (en inglés: "CSS inheritance").
+
+Toda página HTML tiene un elemento `body`.
+
+# --instructions--
+
+Para demostrar que el elemento `body` existe aquí, podemos asignarle un `background-color` black (negro).
+
+Para ello, agregamos la siguiente declaración a nuestro elemento `style`:
+
+```css
+body {
+ background-color: black;
+}
+```
+
+# --hints--
+
+Tu elemento `body` debe tener un `background-color` black (negro).
+
+```js
+assert($('body').css('background-color') === 'rgb(0, 0, 0)');
+```
+
+Tu regla CSS debe tener el formato correcto, usando llaves de apertura y de cierre.
+
+```js
+assert(
+ code.match(/<style>\s*body\s*\{\s*background.*\s*:\s*.*;\s*\}\s*<\/style>/i)
+);
+```
+
+Tu regla CSS debe terminar con un punto y coma.
+
+```js
+assert(
+ code.match(/<style>\s*body\s*\{\s*background.*\s*:\s*.*;\s*\}\s*<\/style>/i)
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+
+</style>
+```
+
+# --solutions--
+
+```html
+<style>
+body {
+ background-color: black;
+}
+</style>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/understand-absolute-versus-relative-units.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/understand-absolute-versus-relative-units.md
new file mode 100644
index 00000000000..3e8d014cfe3
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/understand-absolute-versus-relative-units.md
@@ -0,0 +1,123 @@
+---
+id: bad82fee1322bd9aedf08721
+title: Unidades absolutas y relativas de medida
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cN66JSL'
+forumTopicId: 301089
+dashedName: understand-absolute-versus-relative-units
+---
+
+# --description--
+
+Varios de los últimos desafíos establecen el "margin" o "padding" de un elemento usando píxeles (`px`). Los píxeles son un tipo de unidad de longitud que le indica al navegador qué tamaño o cuánto espaciado asignarle a un elemento. Además de `px`, CSS cuenta con variedad de opciones de unidades de longitud que puedes utilizar.
+
+Los dos tipos principales de unidades de longitud son las unidades absolutas y relativas. Las unidades absolutas están relacionadas con unidades físicas de longitud. Por ejemplo, `in` y `mm` se refieren a pulgadas y milímetros, respectivamente. Las unidades de longitud absoluta aproximan la medición real sobre una pantalla, pero existen cierta variación que depende de la resolución de la pantalla utilizada.
+
+Las unidades relativas, como `em` o `rem` son relativas a otro valor de longitud. Por ejemplo, `em` se basa en el tamaño de fuente de un elemento. Si la utilizas para establecer la propiedad `font-size`, es relativa al `font-size` del elemento padre.
+
+**Nota:** Hay varias opciones de unidades relativas que están vinculadas al tamaño del viewport. Veremos estas unidades relativas de medida en la sección de principios de diseño web responsivo.
+
+# --instructions--
+
+Añade una propiedad `padding` al elemento con clase `red-box` y establécelo en `1.5em`.
+
+# --hints--
+
+La clase `red-box` debe tener una propiedad `padding`.
+
+```js
+assert(
+ $('.red-box').css('padding-top') != '0px' &&
+ $('.red-box').css('padding-right') != '0px' &&
+ $('.red-box').css('padding-bottom') != '0px' &&
+ $('.red-box').css('padding-left') != '0px'
+);
+```
+
+Tu clase `red-box` debe asignar a los elementos 1.5em de `padding`.
+
+```js
+assert(code.match(/\.red-box\s*?{[\s\S]*padding:\s*?1\.5em/gi));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .injected-text {
+ margin-bottom: -25px;
+ text-align: center;
+ }
+
+ .box {
+ border-style: solid;
+ border-color: black;
+ border-width: 5px;
+ text-align: center;
+ }
+
+ .yellow-box {
+ background-color: yellow;
+ padding: 20px 40px 20px 40px;
+ }
+
+ .red-box {
+ background-color: red;
+ margin: 20px 40px 20px 40px;
+
+ }
+
+ .green-box {
+ background-color: green;
+ margin: 20px 40px 20px 40px;
+ }
+</style>
+<h5 class="injected-text">margin</h5>
+
+<div class="box yellow-box">
+ <h5 class="box red-box">padding</h5>
+ <h5 class="box green-box">padding</h5>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ .injected-text {
+ margin-bottom: -25px;
+ text-align: center;
+ }
+
+ .box {
+ border-style: solid;
+ border-color: black;
+ border-width: 5px;
+ text-align: center;
+ }
+
+ .yellow-box {
+ background-color: yellow;
+ padding: 20px 40px 20px 40px;
+ }
+
+ .red-box {
+ background-color: red;
+ margin: 20px 40px 20px 40px;
+ padding: 1.5em;
+ }
+
+ .green-box {
+ background-color: green;
+ margin: 20px 40px 20px 40px;
+ }
+</style>
+<h5 class="injected-text">margin</h5>
+
+<div class="box yellow-box">
+ <h5 class="box red-box">padding</h5>
+ <h5 class="box green-box">padding</h5>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-a-css-class-to-style-an-element.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-a-css-class-to-style-an-element.md
new file mode 100644
index 00000000000..9321f17aaa4
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-a-css-class-to-style-an-element.md
@@ -0,0 +1,142 @@
+---
+id: bad87fee1348bd9aecf08806
+title: Usa una clase CSS para aplicar estilos a un elemento
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c2MvDtV'
+forumTopicId: 18337
+dashedName: use-a-css-class-to-style-an-element
+---
+
+# --description--
+
+Las clases son estilos reutilizables que se pueden agregar a los elementos HTML.
+
+A continuación te presentamos un ejemplo de cómo declarar una clase CSS:
+
+```html
+<style>
+ .blue-text {
+ color: blue;
+ }
+</style>
+```
+
+Puedes ver que hemos creado una clase CSS llamada `blue-text` dentro de la etiqueta `<style>`. Puedes aplicar una clase a un elemento HTML de este modo: `<h2 class="blue-text">CatPhotoApp</h2>` Ten en cuenta que en tu elemento `style` CSS, los nombres de clase comienzan con un punto. En el atributo de clase de tus elementos HTML, el nombre de la clase no lleva punto delante.
+
+# --instructions--
+
+Dentro de tu elemento `style`, cambia el selector `h2` a `.red-text` y actualiza el valor del color de `blue` a `red`.
+
+Asigna al elemento `h2` el atributo `class` con el valor `'red-text'`.
+
+# --hints--
+
+Tu elemento `h2` debe ser de color rojo ("red").
+
+```js
+assert($('h2').css('color') === 'rgb(255, 0, 0)');
+```
+
+Tu elemento `h2` debe incluir la clase `red-text`.
+
+```js
+assert($('h2').hasClass('red-text'));
+```
+
+Tu hoja de estilos debe declarar una clase `red-text` y su color debe ser `red` (rojo).
+
+```js
+assert(code.match(/\.red-text\s*\{\s*color\s*:\s*red;\s*\}/g));
+```
+
+No debes usar declaraciones de tipo inline style como `style="color: red"` en tu elemento `h2`.
+
+```js
+assert($('h2').attr('style') === undefined);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ h2 {
+ color: blue;
+ }
+</style>
+
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<style>
+ .red-text {
+ color: red;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-a-custom-css-variable.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-a-custom-css-variable.md
new file mode 100644
index 00000000000..afc02008f06
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-a-custom-css-variable.md
@@ -0,0 +1,287 @@
+---
+id: 5a9d727a424fe3d0e10cad12
+title: Utiliza una variable de CSS personalizada
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cM989ck'
+forumTopicId: 301090
+dashedName: use-a-custom-css-variable
+---
+
+# --description--
+
+Luego de crear tu variable, puedes asignar su valor a otras propiedades CSS haciendo referencia a su nombre.
+
+```css
+background: var(--penguin-skin);
+```
+
+Esto cambiará el fondo de cualquier elemento que utilice esta variable a "gray" (gris) porque ese es el valor de la variable `--penguin-skin`. Ten en cuenta que los estilos no se aplicarán a menos que los nombres de las variables utilizados estén escritos exactamente igual.
+
+# --instructions--
+
+Aplica la variable `--penguin-skin` a la propiedad `background` de las clases `penguin-top`, `penguin-bottom`, `right-hand` y `left-hand`.
+
+# --hints--
+
+La variable `--penguin-skin` debe aplicarse a la propiedad `background` de la clase `penguin-top`.
+
+```js
+assert(
+ code.match(
+ /.penguin-top\s*?{[\s\S]*background\s*?:\s*?var\s*?\(\s*?--penguin-skin\s*?\)\s*?;[\s\S]*}[\s\S]*.penguin-bottom\s{/gi
+ )
+);
+```
+
+La variable `--penguin-skin` debe aplicarse a la propiedad `background` de la clase `penguin-bottom`.
+
+```js
+assert(
+ code.match(
+ /.penguin-bottom\s*?{[\s\S]*background\s*?:\s*?var\s*?\(\s*?--penguin-skin\s*?\)\s*?;[\s\S]*}[\s\S]*.right-hand\s{/gi
+ )
+);
+```
+
+La variable `--penguin-skin` debe aplicarse a la propiedad `background` de la clase `right-hand`.
+
+```js
+assert(
+ code.match(
+ /.right-hand\s*?{[\s\S]*background\s*?:\s*?var\s*?\(\s*?--penguin-skin\s*?\)\s*?;[\s\S]*}[\s\S]*.left-hand\s{/gi
+ )
+);
+```
+
+La variable `--penguin-skin` debe aplicarse a la propiedad `background` de la clase `left-hand`.
+
+```js
+assert(
+ code.match(
+ /.left-hand\s*?{[\s\S]*background\s*?:\s*?var\s*?\(\s*?--penguin-skin\s*?\)\s*?;[\s\S]*}/gi
+ )
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .penguin {
+ --penguin-skin: gray;
+ position: relative;
+ margin: auto;
+ display: block;
+ margin-top: 5%;
+ width: 300px;
+ height: 300px;
+ }
+
+ .penguin-top {
+ top: 10%;
+ left: 25%;
+
+ /* Change code below this line */
+ background: black;
+ /* Change code above this line */
+
+ width: 50%;
+ height: 45%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .penguin-bottom {
+ top: 40%;
+ left: 23.5%;
+
+ /* Change code below this line */
+ background: black;
+ /* Change code above this line */
+
+ width: 53%;
+ height: 45%;
+ border-radius: 70% 70% 100% 100%;
+ }
+
+ .right-hand {
+ top: 0%;
+ left: -5%;
+
+ /* Change code below this line */
+ background: black;
+ /* Change code above this line */
+
+ width: 30%;
+ height: 60%;
+ border-radius: 30% 30% 120% 30%;
+ transform: rotate(45deg);
+ z-index: -1;
+ }
+
+ .left-hand {
+ top: 0%;
+ left: 75%;
+
+ /* Change code below this line */
+ background: black;
+ /* Change code above this line */
+
+ width: 30%;
+ height: 60%;
+ border-radius: 30% 30% 30% 120%;
+ transform: rotate(-45deg);
+ z-index: -1;
+ }
+
+ .right-cheek {
+ top: 15%;
+ left: 35%;
+ background: white;
+ width: 60%;
+ height: 70%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .left-cheek {
+ top: 15%;
+ left: 5%;
+ background: white;
+ width: 60%;
+ height: 70%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .belly {
+ top: 60%;
+ left: 2.5%;
+ background: white;
+ width: 95%;
+ height: 100%;
+ border-radius: 120% 120% 100% 100%;
+ }
+
+ .right-feet {
+ top: 85%;
+ left: 60%;
+ background: orange;
+ width: 15%;
+ height: 30%;
+ border-radius: 50% 50% 50% 50%;
+ transform: rotate(-80deg);
+ z-index: -2222;
+ }
+
+ .left-feet {
+ top: 85%;
+ left: 25%;
+ background: orange;
+ width: 15%;
+ height: 30%;
+ border-radius: 50% 50% 50% 50%;
+ transform: rotate(80deg);
+ z-index: -2222;
+ }
+
+ .right-eye {
+ top: 45%;
+ left: 60%;
+ background: black;
+ width: 15%;
+ height: 17%;
+ border-radius: 50%;
+ }
+
+ .left-eye {
+ top: 45%;
+ left: 25%;
+ background: black;
+ width: 15%;
+ height: 17%;
+ border-radius: 50%;
+ }
+
+ .sparkle {
+ top: 25%;
+ left: 15%;
+ background: white;
+ width: 35%;
+ height: 35%;
+ border-radius: 50%;
+ }
+
+ .blush-right {
+ top: 65%;
+ left: 15%;
+ background: pink;
+ width: 15%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .blush-left {
+ top: 65%;
+ left: 70%;
+ background: pink;
+ width: 15%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .beak-top {
+ top: 60%;
+ left: 40%;
+ background: orange;
+ width: 20%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .beak-bottom {
+ top: 65%;
+ left: 42%;
+ background: orange;
+ width: 16%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ body {
+ background:#c6faf1;
+ }
+
+ .penguin * {
+ position: absolute;
+ }
+</style>
+<div class="penguin">
+ <div class="penguin-bottom">
+ <div class="right-hand"></div>
+ <div class="left-hand"></div>
+ <div class="right-feet"></div>
+ <div class="left-feet"></div>
+ </div>
+ <div class="penguin-top">
+ <div class="right-cheek"></div>
+ <div class="left-cheek"></div>
+ <div class="belly"></div>
+ <div class="right-eye">
+ <div class="sparkle"></div>
+ </div>
+ <div class="left-eye">
+ <div class="sparkle"></div>
+ </div>
+ <div class="blush-right"></div>
+ <div class="blush-left"></div>
+ <div class="beak-top"></div>
+ <div class="beak-bottom"></div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>.penguin-top {background: var(--penguin-skin);} .penguin-bottom {background: var(--penguin-skin);} .right-hand {background: var(--penguin-skin);} .left-hand {background: var(--penguin-skin);}</style>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-a-media-query-to-change-a-variable.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-a-media-query-to-change-a-variable.md
new file mode 100644
index 00000000000..1444f5f9033
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-a-media-query-to-change-a-variable.md
@@ -0,0 +1,281 @@
+---
+id: 5a9d72ad424fe3d0e10cad16
+title: Usa un media query para cambiar una variable
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cWmL8UP'
+forumTopicId: 301091
+dashedName: use-a-media-query-to-change-a-variable
+---
+
+# --description--
+
+Las variables CSS pueden simplificar la forma en que utilizas "media queries" (consultas sobre el tipo de dispositivo donde se muestra el documento HTML).
+
+Por ejemplo, cuando la pantalla es más pequeña o más grande que el breakpoint de tu media query, puedes cambiar el valor de una variable, y su estilo se aplicará dondequiera que la utilices.
+
+# --instructions--
+
+Modifica el selector `:root` de la `media query` para que `--penguin-size` sea redefinido y reciba un valor de `200px`. Además, redefine `--penguin-skin` y asígnale el valor `black` (negro). Luego, cambia el tamaño de la vista previa para ver este cambio en acción.
+
+# --hints--
+
+`:root` debería reasignar a la variable `--penguin-size` el valor `200px`.
+
+```js
+assert(
+ code.match(
+ /media\s*?\(\s*?max-width\s*?:\s*?350px\s*?\)\s*?{[\s\S]*:root\s*?{[\s\S]*--penguin-size\s*?:\s*?200px\s*?;[\s\S]*}[\s\S]*}/gi
+ )
+);
+```
+
+`:root` debería reasignar a la variable `--penguin-skin` el valor `black`.
+
+```js
+assert(
+ code.match(
+ /media\s*?\(\s*?max-width\s*?:\s*?350px\s*?\)\s*?{[\s\S]*:root\s*?{[\s\S]*--penguin-skin\s*?:\s*?black\s*?;[\s\S]*}[\s\S]*}/gi
+ )
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ :root {
+ --penguin-size: 300px;
+ --penguin-skin: gray;
+ --penguin-belly: white;
+ --penguin-beak: orange;
+ }
+
+ @media (max-width: 350px) {
+ :root {
+ /* Only change code below this line */
+
+ /* Only change code above this line */
+ }
+ }
+
+ .penguin {
+ position: relative;
+ margin: auto;
+ display: block;
+ margin-top: 5%;
+ width: var(--penguin-size, 300px);
+ height: var(--penguin-size, 300px);
+ }
+
+ .right-cheek {
+ top: 15%;
+ left: 35%;
+ background: var(--penguin-belly, white);
+ width: 60%;
+ height: 70%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .left-cheek {
+ top: 15%;
+ left: 5%;
+ background: var(--penguin-belly, white);
+ width: 60%;
+ height: 70%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .belly {
+ top: 60%;
+ left: 2.5%;
+ background: var(--penguin-belly, white);
+ width: 95%;
+ height: 100%;
+ border-radius: 120% 120% 100% 100%;
+ }
+
+ .penguin-top {
+ top: 10%;
+ left: 25%;
+ background: var(--penguin-skin, gray);
+ width: 50%;
+ height: 45%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .penguin-bottom {
+ top: 40%;
+ left: 23.5%;
+ background: var(--penguin-skin, gray);
+ width: 53%;
+ height: 45%;
+ border-radius: 70% 70% 100% 100%;
+ }
+
+ .right-hand {
+ top: 5%;
+ left: 25%;
+ background: var(--penguin-skin, black);
+ width: 30%;
+ height: 60%;
+ border-radius: 30% 30% 120% 30%;
+ transform: rotate(130deg);
+ z-index: -1;
+ animation-duration: 3s;
+ animation-name: wave;
+ animation-iteration-count: infinite;
+ transform-origin:0% 0%;
+ animation-timing-function: linear;
+ }
+
+ @keyframes wave {
+ 10% {
+ transform: rotate(110deg);
+ }
+ 20% {
+ transform: rotate(130deg);
+ }
+ 30% {
+ transform: rotate(110deg);
+ }
+ 40% {
+ transform: rotate(130deg);
+ }
+ }
+
+ .left-hand {
+ top: 0%;
+ left: 75%;
+ background: var(--penguin-skin, gray);
+ width: 30%;
+ height: 60%;
+ border-radius: 30% 30% 30% 120%;
+ transform: rotate(-45deg);
+ z-index: -1;
+ }
+
+ .right-feet {
+ top: 85%;
+ left: 60%;
+ background: var(--penguin-beak, orange);
+ width: 15%;
+ height: 30%;
+ border-radius: 50% 50% 50% 50%;
+ transform: rotate(-80deg);
+ z-index: -2222;
+ }
+
+ .left-feet {
+ top: 85%;
+ left: 25%;
+ background: var(--penguin-beak, orange);
+ width: 15%;
+ height: 30%;
+ border-radius: 50% 50% 50% 50%;
+ transform: rotate(80deg);
+ z-index: -2222;
+ }
+
+ .right-eye {
+ top: 45%;
+ left: 60%;
+ background: black;
+ width: 15%;
+ height: 17%;
+ border-radius: 50%;
+ }
+
+ .left-eye {
+ top: 45%;
+ left: 25%;
+ background: black;
+ width: 15%;
+ height: 17%;
+ border-radius: 50%;
+ }
+
+ .sparkle {
+ top: 25%;
+ left:-23%;
+ background: white;
+ width: 150%;
+ height: 100%;
+ border-radius: 50%;
+ }
+
+ .blush-right {
+ top: 65%;
+ left: 15%;
+ background: pink;
+ width: 15%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .blush-left {
+ top: 65%;
+ left: 70%;
+ background: pink;
+ width: 15%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .beak-top {
+ top: 60%;
+ left: 40%;
+ background: var(--penguin-beak, orange);
+ width: 20%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .beak-bottom {
+ top: 65%;
+ left: 42%;
+ background: var(--penguin-beak, orange);
+ width: 16%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ body {
+ background:#c6faf1;
+ }
+
+ .penguin * {
+ position: absolute;
+ }
+</style>
+<div class="penguin">
+ <div class="penguin-bottom">
+ <div class="right-hand"></div>
+ <div class="left-hand"></div>
+ <div class="right-feet"></div>
+ <div class="left-feet"></div>
+ </div>
+ <div class="penguin-top">
+ <div class="right-cheek"></div>
+ <div class="left-cheek"></div>
+ <div class="belly"></div>
+ <div class="right-eye">
+ <div class="sparkle"></div>
+ </div>
+ <div class="left-eye">
+ <div class="sparkle"></div>
+ </div>
+ <div class="blush-right"></div>
+ <div class="blush-left"></div>
+ <div class="beak-top"></div>
+ <div class="beak-bottom"></div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>@media (max-width: 350px) {:root {--penguin-size: 200px; --penguin-skin: black;}}</style>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-abbreviated-hex-code.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-abbreviated-hex-code.md
new file mode 100644
index 00000000000..95be2cb3f24
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-abbreviated-hex-code.md
@@ -0,0 +1,128 @@
+---
+id: bad87fee1348bd9aedf08719
+title: Usa código hexadecimal (hex code) abreviado
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cRkpKAm'
+forumTopicId: 18338
+dashedName: use-abbreviated-hex-code
+---
+
+# --description--
+
+Muchas personas se sienten abrumadas por tener más de 16 millones de colores posibles. Además, los códigos hexadecimales resultan difíciles de recordar. Afortunadamente, puedes abreviar gran parte de ellos.
+
+Por ejemplo, el código hexadecimal `#FF0000` del color rojo puede acortarse a `#F00`. Esta forma abreviada utiliza un dígito para el rojo, un dígito para el verde, y un dígito para el azul.
+
+Esto reduce el número total de colores posibles a alrededor de 4.000. Sin embargo, los navegadores interpretarán que `#FF0000` y `#F00` son exactamente el mismo color.
+
+# --instructions--
+
+¡Pruébalo! Intenta usar los códigos hexadecimales abreviados para asignar colores a los elementos como se indica a continuación.
+
+<table class='table table-striped'><tbody><tr><th>Color</th><th>Hex code abreviado</th></tr><tr><td>Cyan (cian)</td><td><code>#0FF</code></td></tr><tr><td>Green (verde)</td><td><code>#0F0</code></td></tr><tr><td>Red (rojo)</td><td><code>#F00</code></td></tr><tr><td>Fuchsia (fucsia)</td><td><code>#F0F</code></td></tr></tbody></table>
+
+# --hints--
+
+Debes asignar al elemento `h1` que tiene el texto `I am red!` ("¡Soy de color rojo!) el `color` rojo.
+
+```js
+assert($('.red-text').css('color') === 'rgb(255, 0, 0)');
+```
+
+Debes usar el `hex code` abreviado para el color rojo en lugar del código hexadecimal `#FF0000`.
+
+```js
+assert(code.match(/\.red-text\s*?{\s*?color:\s*?#F00\s*?;\s*?}/gi));
+```
+
+Debes asignar al elemento `h1` que tiene el texto `I am green!` ("¡Soy de color verde!) el `color` verde.
+
+```js
+assert($('.green-text').css('color') === 'rgb(0, 255, 0)');
+```
+
+Debes usar el `hex code` abreviado para el color verde en lugar del código hexadecimal `#00FF00`.
+
+```js
+assert(code.match(/\.green-text\s*?{\s*?color:\s*?#0F0\s*?;\s*?}/gi));
+```
+
+Debes asignar al elemento `h1` que tiene el texto `I am cyan!` ("¡Soy de color cian!) el `color` cian.
+
+```js
+assert($('.cyan-text').css('color') === 'rgb(0, 255, 255)');
+```
+
+Debes usar el `hex code` abreviado para el color cian en lugar del código hexadecimal `#00FFFF`.
+
+```js
+assert(code.match(/\.cyan-text\s*?{\s*?color:\s*?#0FF\s*?;\s*?}/gi));
+```
+
+Debes asignar al elemento `h1` que tiene el texto `I am fuchsia!` ("¡Soy de color fucsia!) el `color` fucsia.
+
+```js
+assert($('.fuchsia-text').css('color') === 'rgb(255, 0, 255)');
+```
+
+Debes usar el `hex code` abreviado para el color fucsia en lugar del código hexadecimal `#FF00FF`.
+
+```js
+assert(code.match(/\.fuchsia-text\s*?{\s*?color:\s*?#F0F\s*?;\s*?}/gi));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .red-text {
+ color: #000000;
+ }
+ .fuchsia-text {
+ color: #000000;
+ }
+ .cyan-text {
+ color: #000000;
+ }
+ .green-text {
+ color: #000000;
+ }
+</style>
+
+<h1 class="red-text">I am red!</h1>
+
+<h1 class="fuchsia-text">I am fuchsia!</h1>
+
+<h1 class="cyan-text">I am cyan!</h1>
+
+<h1 class="green-text">I am green!</h1>
+```
+
+# --solutions--
+
+```html
+<style>
+ .red-text {
+ color: #F00;
+ }
+ .fuchsia-text {
+ color: #F0F;
+ }
+ .cyan-text {
+ color: #0FF;
+ }
+ .green-text {
+ color: #0F0;
+ }
+</style>
+
+<h1 class="red-text">I am red!</h1>
+
+<h1 class="fuchsia-text">I am fuchsia!</h1>
+
+<h1 class="cyan-text">I am cyan!</h1>
+
+<h1 class="green-text">I am green!</h1>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-an-id-attribute-to-style-an-element.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-an-id-attribute-to-style-an-element.md
new file mode 100644
index 00000000000..bcdde64e5f3
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-an-id-attribute-to-style-an-element.md
@@ -0,0 +1,197 @@
+---
+id: bad87dee1348bd9aede07836
+title: Usa un atributo de id para aplicar estilo a un elemento
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cakyZfL'
+forumTopicId: 18339
+dashedName: use-an-id-attribute-to-style-an-element
+---
+
+# --description--
+
+Una cosa genial de los atributos `id` es que, al igual que las clases, puedes aplicarles estilos con CSS.
+
+Sin embargo, un `id` no es reutilizable y solo debe aplicarse a un único elemento. Un `id` también tiene mayor especificidad (importancia) que una clase, así que si aplicas un id y una clase al mismo elemento y sus estilos se contradicen, se aplicarán los estilos del `id`.
+
+A continuación te mostramos un ejemplo de cómo puedes tomar tu elemento con el atributo `id` llamado `cat-photo-element` y asignarle el color de fondo verde. En tu elemento `style` agrega la siguiente declaración:
+
+```css
+#cat-photo-element {
+ background-color: green;
+}
+```
+
+Ten en cuenta que dentro de tu elemento `style` siempre debes hacer referencia a las clases agregándoles un punto `.` adelante del nombre. Para hacer referencia a un id, debes agregar `#` delante de su nombre.
+
+# --instructions--
+
+Intenta asignar a tu formulario, que ahora tiene el atributo `id` llamado `cat-photo-form`, un fondo verde.
+
+# --hints--
+
+Tu elemento `form` debe tener el id `cat-photo-form`.
+
+```js
+assert($('form').attr('id') === 'cat-photo-form');
+```
+
+Tu elemento `form` debe tener `background-color` de color verde (green).
+
+```js
+assert($('#cat-photo-form').css('background-color') === 'rgb(0, 128, 0)');
+```
+
+Tu elemento `form` debe tener un atributo `id`.
+
+```js
+assert(
+ code.match(/<form.*cat-photo-form.*>/gi) &&
+ code.match(/<form.*cat-photo-form.*>/gi).length > 0
+);
+```
+
+No debes asignar a tu `form` ningún atributo `class` o `style`.
+
+```js
+assert(!code.match(/<form.*style.*>/gi) && !code.match(/<form.*class.*>/gi));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">
+<style>
+ .red-text {
+ color: red;
+ }
+
+ h2 {
+ font-family: Lobster, monospace;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+
+ .thick-green-border {
+ border-color: green;
+ border-width: 10px;
+ border-style: solid;
+ border-radius: 50%;
+ }
+
+ .smaller-image {
+ width: 100px;
+ }
+
+ .silver-background {
+ background-color: silver;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img class="smaller-image thick-green-border" src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div class="silver-background">
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo" id="cat-photo-form">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">
+<style>
+ .red-text {
+ color: red;
+ }
+
+ h2 {
+ font-family: Lobster, monospace;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+
+ .thick-green-border {
+ border-color: green;
+ border-width: 10px;
+ border-style: solid;
+ border-radius: 50%;
+ }
+
+ .smaller-image {
+ width: 100px;
+ }
+
+ .silver-background {
+ background-color: silver;
+ }
+
+ #cat-photo-form {
+ background-color: green;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img class="smaller-image thick-green-border" src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div class="silver-background">
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo" id="cat-photo-form">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-attribute-selectors-to-style-elements.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-attribute-selectors-to-style-elements.md
new file mode 100644
index 00000000000..b719e1f9e53
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-attribute-selectors-to-style-elements.md
@@ -0,0 +1,210 @@
+---
+id: 58c383d33e2e3259241f3076
+title: Usa selectores de atributos para dar estilo a elementos
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cnpymfJ'
+forumTopicId: 301092
+dashedName: use-attribute-selectors-to-style-elements
+---
+
+# --description--
+
+Hasta ahora, has añadido atributos `id` o `class` a elementos para aplicarles estilos específicos. Estos se conocen también como selectores de ID y de clase. Sin embargo, existen otros selectores CSS que puedes utilizar para seleccionar grupos personalizados de elementos a los que quieras aplicar el mismo estilo.
+
+Volvamos a nuestra CatPhotoApp para practicar el uso de selectores CSS.
+
+Para este desafío, usarás el selector de atributos `[attr=value]` para aplicar estilo a las casillas de verificación (checkboxes) en CatPhotoApp. Este selector busca estilos que tengan un valor de atributo específico. Por ejemplo, el código a continuación cambia los márgenes de todos los elementos que tengan el atributo `type` con el valor `radio`:
+
+```css
+[type='radio'] {
+ margin: 20px 0px 20px 0px;
+}
+```
+
+# --instructions--
+
+Usando el selector de atributos `type`, intenta asignar a las casillas de verificación de CatPhoto App un margen superior (margin-top) de 10px y un margen inferior (margin-bottom) de 15px.
+
+# --hints--
+
+El selector de atributos `type` debe utilizarse para seleccionar los checkboxes.
+
+```js
+assert(
+ code.match(
+ /<style>[\s\S]*?\[\s*?type\s*?=\s*?("|')checkbox\1\s*?\]\s*?{[\s\S]*?}[\s\S]*?<\/style>/gi
+ )
+);
+```
+
+El margen superior de las casillas de verificación debe ser 10px.
+
+```js
+assert(
+ (function () {
+ var count = 0;
+ $("[type='checkbox']").each(function () {
+ if ($(this).css('marginTop') === '10px') {
+ count++;
+ }
+ });
+ return count === 3;
+ })()
+);
+```
+
+El margen inferior de las casillas de verificación debe ser 15px.
+
+```js
+assert(
+ (function () {
+ var count = 0;
+ $("[type='checkbox']").each(function () {
+ if ($(this).css('marginBottom') === '15px') {
+ count++;
+ }
+ });
+ return count === 3;
+ })()
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">
+<style>
+ .red-text {
+ color: red;
+ }
+
+ h2 {
+ font-family: Lobster, monospace;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+
+ .thick-green-border {
+ border-color: green;
+ border-width: 10px;
+ border-style: solid;
+ border-radius: 50%;
+ }
+
+ .smaller-image {
+ width: 100px;
+ }
+
+ .silver-background {
+ background-color: silver;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img class="smaller-image thick-green-border" src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div class="silver-background">
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo" id="cat-photo-form">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<link href="https://fonts.googleapis.com/css?family=Lobster" rel="stylesheet" type="text/css">
+<style>
+ .red-text {
+ color: red;
+ }
+
+ h2 {
+ font-family: Lobster, monospace;
+ }
+
+ p {
+ font-size: 16px;
+ font-family: monospace;
+ }
+
+ .thick-green-border {
+ border-color: green;
+ border-width: 10px;
+ border-style: solid;
+ border-radius: 50%;
+ }
+
+ .smaller-image {
+ width: 100px;
+ }
+
+ .silver-background {
+ background-color: silver;
+ }
+ [type='checkbox'] {
+ margin-top: 10px;
+ margin-bottom: 15px;
+ }
+</style>
+
+<h2 class="red-text">CatPhotoApp</h2>
+<main>
+ <p class="red-text">Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img class="smaller-image thick-green-border" src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div class="silver-background">
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo" id="cat-photo-form">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-clockwise-notation-to-specify-the-margin-of-an-element.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-clockwise-notation-to-specify-the-margin-of-an-element.md
new file mode 100644
index 00000000000..0b38f008448
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-clockwise-notation-to-specify-the-margin-of-an-element.md
@@ -0,0 +1,141 @@
+---
+id: bad87fee1348bd9afdf08726
+title: Utiliza clockwise notation para especificar el margen de un elemento
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cnpybAd'
+forumTopicId: 18345
+dashedName: use-clockwise-notation-to-specify-the-margin-of-an-element
+---
+
+# --description--
+
+Intentemos de nuevo, pero esta vez con `margin`.
+
+En lugar de especificar las propiedades `margin-top`, `margin-right`, `margin-bottom`, y `margin-left` individualmente, puedes especificarlas todas en una sola línea, como se muestra a continuación:
+
+`margin: 10px 20px 10px 20px;`
+
+Estos cuatro valores se leen en el sentido de las agujas del reloj: arriba, derecha, abajo, izquierda, (top, right, bottom, left) y producirán exactamente el mismo resultado que usar las instrucciones específicas de margen.
+
+# --instructions--
+
+Usa clockwise notation (notación en el sentido de las agujas del reloj) para asignarle al elemento con la clase `blue-box` un margen de `40px` en su lado superior e izquierdo pero un margen de solo `20px` en su lado inferior y derecho.
+
+# --hints--
+
+Tu clase `blue-box` debe asignar al lado superior (top) de los elementos `40px` de `margin`.
+
+```js
+assert($('.blue-box').css('margin-top') === '40px');
+```
+
+Tu clase `blue-box` debe asignar al lado derecho (right) de los elementos `20px` de `margin`.
+
+```js
+assert($('.blue-box').css('margin-right') === '20px');
+```
+
+Tu clase `blue-box` debe asignar al lado inferior (bottom) de los elementos `20px` de `margin`.
+
+```js
+assert($('.blue-box').css('margin-bottom') === '20px');
+```
+
+Tu clase `blue-box` debe asignar al lado izquierdo (left) de los elementos `40px` de `margin`.
+
+```js
+assert($('.blue-box').css('margin-left') === '40px');
+```
+
+Debes usar clockwise notation para establecer el margen de la clase `blue-box`.
+
+```js
+assert(
+ /\.blue-box\s*{[\s\S]*margin[\s]*:\s*\d+px\s+\d+px\s+\d+px\s+\d+px(;\s*[^}]+\s*}|;?\s*})/.test(
+ __helpers.removeCssComments($('style').text())
+ )
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .injected-text {
+ margin-bottom: -25px;
+ text-align: center;
+ }
+
+ .box {
+ border-style: solid;
+ border-color: black;
+ border-width: 5px;
+ text-align: center;
+ }
+
+ .yellow-box {
+ background-color: yellow;
+ padding: 20px 40px 20px 40px;
+ }
+
+ .red-box {
+ background-color: crimson;
+ color: #fff;
+ margin: 20px 40px 20px 40px;
+ }
+
+ .blue-box {
+ background-color: blue;
+ color: #fff;
+ }
+</style>
+<h5 class="injected-text">margin</h5>
+
+<div class="box yellow-box">
+ <h5 class="box red-box">padding</h5>
+ <h5 class="box blue-box">padding</h5>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ .injected-text {
+ margin-bottom: -25px;
+ text-align: center;
+ }
+
+ .box {
+ border-style: solid;
+ border-color: black;
+ border-width: 5px;
+ text-align: center;
+ }
+
+ .yellow-box {
+ background-color: yellow;
+ padding: 20px 40px 20px 40px;
+ }
+
+ .red-box {
+ background-color: crimson;
+ color: #fff;
+ margin: 20px 40px 20px 40px;
+ }
+
+ .blue-box {
+ background-color: blue;
+ color: #fff;
+ margin: 40px 20px 20px 40px;
+ }
+</style>
+<h5 class="injected-text">margin</h5>
+
+<div class="box yellow-box">
+ <h5 class="box red-box">padding</h5>
+ <h5 class="box blue-box">padding</h5>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-clockwise-notation-to-specify-the-padding-of-an-element.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-clockwise-notation-to-specify-the-padding-of-an-element.md
new file mode 100644
index 00000000000..f397984686a
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-clockwise-notation-to-specify-the-padding-of-an-element.md
@@ -0,0 +1,139 @@
+---
+id: bad87fee1348bd9aedf08826
+title: Utiliza clockwise notation para especificar el relleno (padding) de un elemento
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cB7mBS9'
+forumTopicId: 18346
+dashedName: use-clockwise-notation-to-specify-the-padding-of-an-element
+---
+
+# --description--
+
+En lugar de especificar las propiedades `padding-top`, `padding-right`, `padding-bottom`, y `padding-left` individualmente, puedes especificarlas todas en una sola línea, como se muestra a continuación:
+
+`padding: 10px 20px 10px 20px;`
+
+Estos cuatro valores se leen en el sentido de las agujas del reloj: arriba, derecha, abajo, izquierda, (top, right, bottom, left) y producirán exactamente el mismo resultado que usar las instrucciones específicas de padding.
+
+# --instructions--
+
+Utiliza clockwise notation (notación en el sentido de las agujas del reloj) para asignarle al elemento con la clase `.blue-box` un `padding` de `40px` en su lado superior e izquierdo pero solo `20px` en su lado inferior y derecho.
+
+# --hints--
+
+Tu clase `blue-box` debe asignar al lado superior (top) de los elementos `40px` de `padding`.
+
+```js
+assert($('.blue-box').css('padding-top') === '40px');
+```
+
+Tu clase `blue-box` debe asignar al lado derecho (right) de los elementos `20px` de `padding`.
+
+```js
+assert($('.blue-box').css('padding-right') === '20px');
+```
+
+Tu clase `blue-box` debe asignar al lado inferior (bottom) de los elementos `20px` de `padding`.
+
+```js
+assert($('.blue-box').css('padding-bottom') === '20px');
+```
+
+Tu clase `blue-box` debe asignar al lado izquierlo (left) de los elementos `40px` de `padding`.
+
+```js
+assert($('.blue-box').css('padding-left') === '40px');
+```
+
+Debes usar clockwise notation para establecer el padding de la clase `blue-box`.
+
+```js
+assert(
+ /\.blue-box\s*{[\s\S]*padding[\s]*:\s*\d+px\s+\d+px\s+\d+px\s+\d+px(;\s*[^}]+\s*}|;?\s*})/.test(
+ __helpers.removeCssComments($('style').text())
+ )
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .injected-text {
+ margin-bottom: -25px;
+ text-align: center;
+ }
+
+ .box {
+ border-style: solid;
+ border-color: black;
+ border-width: 5px;
+ text-align: center;
+ }
+
+ .yellow-box {
+ background-color: yellow;
+ padding: 20px 40px 20px 40px;
+ }
+
+ .red-box {
+ background-color: crimson;
+ color: #fff;
+ padding: 20px 40px 20px 40px;
+ }
+
+ .blue-box {
+ background-color: blue;
+ color: #fff;
+ }
+</style>
+<h5 class="injected-text">margin</h5>
+
+<div class="box yellow-box">
+ <h5 class="box red-box">padding</h5>
+ <h5 class="box blue-box">padding</h5>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>
+ .injected-text {
+ margin-bottom: -25px;
+ text-align: center;
+ }
+
+ .box {
+ border-style: solid;
+ border-color: black;
+ border-width: 5px;
+ text-align: center;
+ }
+
+ .yellow-box {
+ background-color: yellow;
+ padding: 20px 40px 20px 40px;
+ }
+
+ .red-box {
+ background-color: crimson;
+ color: #fff;
+ padding: 20px 40px 20px 40px;
+ }
+
+ .blue-box {
+ background-color: blue;
+ color: #fff;
+ padding: 40px 20px 20px 40px;
+ }
+</style>
+<h5 class="injected-text">margin</h5>
+
+<div class="box yellow-box">
+ <h5 class="box red-box">padding</h5>
+ <h5 class="box blue-box">padding</h5>
+</div>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-css-selectors-to-style-elements.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-css-selectors-to-style-elements.md
new file mode 100644
index 00000000000..0c34976d0e5
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-css-selectors-to-style-elements.md
@@ -0,0 +1,158 @@
+---
+id: bad87fee1348bd9aedf08805
+title: Utiliza selectores CSS para dar estilo a elementos
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cJKMBT2'
+forumTopicId: 18349
+dashedName: use-css-selectors-to-style-elements
+---
+
+# --description--
+
+En CSS tienes a tu disposición cientos de propiedades CSS que puedes usar para cambiar el aspecto de un elemento de tu página.
+
+Cuando escribiste `<h2 style="color: red;">CatPhotoApp</h2>`, estabas aplicando un estilo a ese elemento `h2` específico usando CSS en línea; CSS significa hojas de estilo en cascada (Cascading Style Sheets).
+
+Inline CSS es una de las formas de especificar el estilo de un elemento, pero existe una mejor forma de aplicar estilos CSS.
+
+En la parte superior de tu código, crea un bloque `style` como el que se ve a continuación:
+
+```html
+<style>
+</style>
+```
+
+Dentro de ese bloque de estilo (style), puedes crear un <dfn>selector CSS</dfn> para todos los elementos `h2`. Por ejemplo, para que todos los elementos `h2` sean de color rojo puedes añadir una regla de estilo como la siguiente:
+
+```html
+<style>
+ h2 {
+ color: red;
+ }
+</style>
+```
+
+Ten en cuenta que es importante tener tanto llaves de apertura como de cierre (`{` y `}`) alrededor de la(s) regla(s) de estilo de cada elemento. También debes asegurarte de que tu definición de estilo para ese elemento esté dentro de las etiquetas de apertura y cierre de estilo. Por último, asegúrate de añadir un punto y coma (;) al final de cada una de las reglas de estilo de tu elemento.
+
+# --instructions--
+
+Elimina el atributo de estilo de tu elemento `h2` y crea un bloque `style` para agregar tu código CSS. Añade el código CSS necesario para hacer que todos los elementos `h2` sean de color azul (blue).
+
+# --hints--
+
+Debes eliminar el atributo `style` que está aplicado directamente a tu elemento `h2`.
+
+```js
+assert(!$('h2').attr('style'));
+```
+
+Debes crear un elemento `style`.
+
+```js
+assert($('style') && $('style').length >= 1);
+```
+
+Tu elemento `h2` debe ser de color azul.
+
+```js
+assert($('h2').css('color') === 'rgb(0, 0, 255)');
+```
+
+Tu la declaración de estilo para `h2` debe ser válida, con punto y coma y llave de cierre.
+
+```js
+assert(code.match(/h2\s*\{\s*color\s*:.*;\s*\}/g));
+```
+
+Todos tus elementos `style` deben ser válidos y tener etiquetas de cierre.
+
+```js
+assert(
+ code.match(/<\/style>/g) &&
+ code.match(/<\/style>/g).length ===
+ (
+ code.match(
+ /<style((\s)*((type|media|scoped|title|disabled)="[^"]*")?(\s)*)*>/g
+ ) || []
+ ).length
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<h2 style="color: red;">CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<style>
+ h2 {
+ color: blue;
+ }
+</style>
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <div>
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ </div>
+
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label><input type="radio" name="indoor-outdoor" checked> Indoor</label>
+ <label><input type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label><input type="checkbox" name="personality" checked> Loving</label>
+ <label><input type="checkbox" name="personality"> Lazy</label>
+ <label><input type="checkbox" name="personality"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-css-variables-to-change-several-elements-at-once.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-css-variables-to-change-several-elements-at-once.md
new file mode 100644
index 00000000000..4953aa64da1
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-css-variables-to-change-several-elements-at-once.md
@@ -0,0 +1,257 @@
+---
+id: 5a9d725e424fe3d0e10cad10
+title: Usa variables CSS para modificar varios elementos a la vez
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c6bDECm'
+forumTopicId: 301093
+dashedName: use-css-variables-to-change-several-elements-at-once
+---
+
+# --description--
+
+Las <dfn>Variables CSS</dfn> son una manera poderosa de modificar varias propiedades de estilos CSS a la vez, cambiando su valor en un único sitio.
+
+Sigue las instrucciones a continuación para ver cómo puedes cambiar el estilo de varios elementos modificando únicamente tres valores.
+
+# --instructions--
+
+En la clase `penguin`, cambia el valor `black` a `gray`, el valor `gray` a `white` y el valor `yellow` a `orange`.
+
+# --hints--
+
+La clase `penguin` debe declarar la variable `--penguin-skin` y asignarle el valor `gray`.
+
+```js
+assert(
+ code.match(/.penguin\s*?{[\s\S]*--penguin-skin\s*?:\s*?gray\s*?;[\s\S]*}/gi)
+);
+```
+
+La clase `penguin` debe declarar la variable `--penguin-belly` y asignarle el valor `white`.
+
+```js
+assert(
+ code.match(/.penguin\s*?{[\s\S]*--penguin-belly\s*?:\s*?white\s*?;[\s\S]*}/gi)
+);
+```
+
+La clase `penguin` debe declarar la variable `--penguin-beak` y asignarle el valor `orange`.
+
+```js
+assert(
+ code.match(/.penguin\s*?{[\s\S]*--penguin-beak\s*?:\s*?orange\s*?;[\s\S]*}/gi)
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .penguin {
+
+ /* Only change code below this line */
+ --penguin-skin: black;
+ --penguin-belly: gray;
+ --penguin-beak: yellow;
+ /* Only change code above this line */
+
+ position: relative;
+ margin: auto;
+ display: block;
+ margin-top: 5%;
+ width: 300px;
+ height: 300px;
+ }
+
+ .penguin-top {
+ top: 10%;
+ left: 25%;
+ background: var(--penguin-skin, gray);
+ width: 50%;
+ height: 45%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .penguin-bottom {
+ top: 40%;
+ left: 23.5%;
+ background: var(--penguin-skin, gray);
+ width: 53%;
+ height: 45%;
+ border-radius: 70% 70% 100% 100%;
+ }
+
+ .right-hand {
+ top: 0%;
+ left: -5%;
+ background: var(--penguin-skin, gray);
+ width: 30%;
+ height: 60%;
+ border-radius: 30% 30% 120% 30%;
+ transform: rotate(45deg);
+ z-index: -1;
+ }
+
+ .left-hand {
+ top: 0%;
+ left: 75%;
+ background: var(--penguin-skin, gray);
+ width: 30%;
+ height: 60%;
+ border-radius: 30% 30% 30% 120%;
+ transform: rotate(-45deg);
+ z-index: -1;
+ }
+
+ .right-cheek {
+ top: 15%;
+ left: 35%;
+ background: var(--penguin-belly, white);
+ width: 60%;
+ height: 70%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .left-cheek {
+ top: 15%;
+ left: 5%;
+ background: var(--penguin-belly, white);
+ width: 60%;
+ height: 70%;
+ border-radius: 70% 70% 60% 60%;
+ }
+
+ .belly {
+ top: 60%;
+ left: 2.5%;
+ background: var(--penguin-belly, white);
+ width: 95%;
+ height: 100%;
+ border-radius: 120% 120% 100% 100%;
+ }
+
+ .right-feet {
+ top: 85%;
+ left: 60%;
+ background: var(--penguin-beak, orange);
+ width: 15%;
+ height: 30%;
+ border-radius: 50% 50% 50% 50%;
+ transform: rotate(-80deg);
+ z-index: -2222;
+ }
+
+ .left-feet {
+ top: 85%;
+ left: 25%;
+ background: var(--penguin-beak, orange);
+ width: 15%;
+ height: 30%;
+ border-radius: 50% 50% 50% 50%;
+ transform: rotate(80deg);
+ z-index: -2222;
+ }
+
+ .right-eye {
+ top: 45%;
+ left: 60%;
+ background: black;
+ width: 15%;
+ height: 17%;
+ border-radius: 50%;
+ }
+
+ .left-eye {
+ top: 45%;
+ left: 25%;
+ background: black;
+ width: 15%;
+ height: 17%;
+ border-radius: 50%;
+ }
+
+ .sparkle {
+ top: 25%;
+ left: 15%;
+ background: white;
+ width: 35%;
+ height: 35%;
+ border-radius: 50%;
+ }
+
+ .blush-right {
+ top: 65%;
+ left: 15%;
+ background: pink;
+ width: 15%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .blush-left {
+ top: 65%;
+ left: 70%;
+ background: pink;
+ width: 15%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .beak-top {
+ top: 60%;
+ left: 40%;
+ background: var(--penguin-beak, orange);
+ width: 20%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ .beak-bottom {
+ top: 65%;
+ left: 42%;
+ background: var(--penguin-beak, orange);
+ width: 16%;
+ height: 10%;
+ border-radius: 50%;
+ }
+
+ body {
+ background:#c6faf1;
+ }
+
+ .penguin * {
+ position: absolute;
+ }
+</style>
+<div class="penguin">
+ <div class="penguin-bottom">
+ <div class="right-hand"></div>
+ <div class="left-hand"></div>
+ <div class="right-feet"></div>
+ <div class="left-feet"></div>
+ </div>
+ <div class="penguin-top">
+ <div class="right-cheek"></div>
+ <div class="left-cheek"></div>
+ <div class="belly"></div>
+ <div class="right-eye">
+ <div class="sparkle"></div>
+ </div>
+ <div class="left-eye">
+ <div class="sparkle"></div>
+ </div>
+ <div class="blush-right"></div>
+ <div class="blush-left"></div>
+ <div class="beak-top"></div>
+ <div class="beak-bottom"></div>
+ </div>
+</div>
+```
+
+# --solutions--
+
+```html
+<style>.penguin {--penguin-skin: gray; --penguin-belly: white; --penguin-beak: orange;}</style>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-hex-code-for-specific-colors.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-hex-code-for-specific-colors.md
new file mode 100644
index 00000000000..9d72b770bf6
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-hex-code-for-specific-colors.md
@@ -0,0 +1,66 @@
+---
+id: bad87fee1348bd9aedf08726
+title: Utiliza código hexadecima (hex code) para indicar colores específicos
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/c8W9mHM'
+forumTopicId: 18350
+dashedName: use-hex-code-for-specific-colors
+---
+
+# --description--
+
+¿Sabías que hay otras maneras de representar colores en CSS? Una de estas formas se llama código hexadecimal, o código hex (hex code) para abreviar.
+
+Generalmente usamos <dfn>decimales</dfn>, o números de base 10, los cuales utilizan los símbolos del 0 al 9 para representar cada dígito. Los números <dfn>hexadecimales</dfn> (o <dfn>hex</dfn>) son números de base 16. Esto significa que utilizan dieciseis símbolos distintos. Al igual que en los números decimales, en hexadecimal los símbolos 0-9 representan los valores de cero a nueve. Luego A,B,C,D,E,F representan los valores de diez a quince. Todos estos símbolos juntos, de 0 a F, pueden representar un dígito en hexadecimal, lo que da un total de 16 valores posibles. Puedes encontrar más información sobre [los números hexadecimales aquí](https://es.wikipedia.org/wiki/Sistema_hexadecimal).
+
+En CSS, podemos representar colores utilizando 6 dígitos hexadecimales, dos para los componentes de rojo (R), verde (G) y azul (B). Por ejemplo, `#000000` corresponde al color negro, y también es el valor más bajo posible. Puedes encontrar más información sobre [el sistema de color RGB aquí](https://es.wikipedia.org/wiki/RGB).
+
+```css
+body {
+ color: #000000;
+}
+```
+
+# --instructions--
+
+Reemplaza la palabra `black` en el "background-color" (color de fondo) de nuestro elemento `body` con su código hexadecimal equivalente, `#000000`.
+
+# --hints--
+
+Tu elemento `body` debe tener "background-color" black (negro).
+
+```js
+assert($('body').css('background-color') === 'rgb(0, 0, 0)');
+```
+
+Debes usar el `hex code` correspondiente al color negro en lugar de la palabra reservada `black`.
+
+```js
+assert(
+ code.match(
+ /body\s*{(([\s\S]*;\s*?)|\s*?)background.*\s*:\s*?#000(000)?((\s*})|(;[\s\S]*?}))/gi
+ )
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ body {
+ background-color: black;
+ }
+</style>
+```
+
+# --solutions--
+
+```html
+<style>
+ body {
+ background-color: #000000;
+ }
+</style>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-hex-code-to-mix-colors.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-hex-code-to-mix-colors.md
new file mode 100644
index 00000000000..b63a39de9bf
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-hex-code-to-mix-colors.md
@@ -0,0 +1,132 @@
+---
+id: bad87fee1348bd9aedf08721
+title: Utiliza código hexadecimal (hex code) para mezclar colores
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cK89PhP'
+forumTopicId: 18359
+dashedName: use-hex-code-to-mix-colors
+---
+
+# --description--
+
+Repasemos: En código hexadecimal se utilizan 6 dígitos hexadecimales para representar los colores, dos para el componente rojo (R), verde (G) y azul (B).
+
+¡A partir de estos tres colores puros (rojo, verde y azul) podemos variar las cantidades de cada componente de color para crear más de 16 millones de colores distintos!
+
+Por ejemplo, el naranja es rojo puro mezclado con algo de verde y nada de azul. En hex code, esto se traduce como `#FFA500`.
+
+El dígito `0` es el número más bajo en hex code, y representa la ausencia total de color.
+
+El dígito `F` es el número más alto en hex code, y representa el brillo máximo.
+
+# --instructions--
+
+Reemplaza las palabras clave de color en nuestro elemento `style` con los códigos hexadecimales correctos.
+
+<table class='table table-striped'><tbody><tr><th>Color</th><th>Hex Code</th></tr><tr><td>Dodger Blue (azul dodger)</td><td><code>#1E90FF</code></td></tr><tr><td>Green (verde)</td><td><code>#00FF00</code></td></tr><tr><td>Orange (naranja)</td><td><code>#FFA500</code></td></tr><tr><td>Red (rojo)</td><td><code>#FF0000</code></td></tr></tbody></table>
+
+# --hints--
+
+Debes asignar al elemento `h1` que tiene el texto `I am red!` ("¡Soy de color rojo!) el `color` rojo.
+
+```js
+assert($('.red-text').css('color') === 'rgb(255, 0, 0)');
+```
+
+Debes usar el `hex code` correspondiente al color rojo en lugar de la palabra reservada `red`.
+
+```js
+assert(code.match(/\.red-text\s*?{\s*?color:\s*?(#FF0000|#F00)\s*?;\s*?}/gi));
+```
+
+Debes asignar al elemento `h1` que tiene el texto `I am green!` ("¡Soy de color verde!) el `color` verde.
+
+```js
+assert($('.green-text').css('color') === 'rgb(0, 255, 0)');
+```
+
+Debes usar el `hex code` correspondiente al color verde en lugar de la palabra reservada `green`.
+
+```js
+assert(code.match(/\.green-text\s*?{\s*?color:\s*?(#00FF00|#0F0)\s*?;\s*?}/gi));
+```
+
+Debes asignar al elemento `h1` que tiene el texto `I am dodger blue!` ("¡Soy de color azul dodger!) el `color` azul dodger.
+
+```js
+assert($('.dodger-blue-text').css('color') === 'rgb(30, 144, 255)');
+```
+
+Debes usar el `hex code` correspondiente al color azul dodger en lugar de la palabra reservada `dodgerblue`.
+
+```js
+assert(code.match(/\.dodger-blue-text\s*?{\s*?color:\s*?#1E90FF\s*?;\s*?}/gi));
+```
+
+Debes asignar al elemento `h1` que tiene el texto `I am orange!` ("¡Soy de color naranja!) el `color` naranja.
+
+```js
+assert($('.orange-text').css('color') === 'rgb(255, 165, 0)');
+```
+
+Debes usar el `hex code` correspondiente al color naranja en lugar de la palabra reservada `orange`.
+
+```js
+assert(code.match(/\.orange-text\s*?{\s*?color:\s*?#FFA500\s*?;\s*?}/gi));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .red-text {
+ color: black;
+ }
+ .green-text {
+ color: black;
+ }
+ .dodger-blue-text {
+ color: black;
+ }
+ .orange-text {
+ color: black;
+ }
+</style>
+
+<h1 class="red-text">I am red!</h1>
+
+<h1 class="green-text">I am green!</h1>
+
+<h1 class="dodger-blue-text">I am dodger blue!</h1>
+
+<h1 class="orange-text">I am orange!</h1>
+```
+
+# --solutions--
+
+```html
+<style>
+ .red-text {
+ color: #FF0000;
+ }
+ .green-text {
+ color: #00FF00;
+ }
+ .dodger-blue-text {
+ color: #1E90FF;
+ }
+ .orange-text {
+ color: #FFA500;
+ }
+</style>
+
+<h1 class="red-text">I am red!</h1>
+
+<h1 class="green-text">I am green!</h1>
+
+<h1 class="dodger-blue-text">I am dodger blue!</h1>
+
+<h1 class="orange-text">I am orange!</h1>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-rgb-to-mix-colors.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-rgb-to-mix-colors.md
new file mode 100644
index 00000000000..13990991537
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-rgb-to-mix-colors.md
@@ -0,0 +1,140 @@
+---
+id: bad82fee1348bd9aedf08721
+title: Utiliza RGB para mezclar colores
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cm24JU6'
+forumTopicId: 18368
+dashedName: use-rgb-to-mix-colors
+---
+
+# --description--
+
+Al igual que con el código hexadecimal, puedes mezclar colores combinando valores RGB.
+
+# --instructions--
+
+Reemplaza los hex codes en nuestro elemento `style` con los valores RGB correctos.
+
+<table class='table table-striped'><tbody><tr><th>Color</th><th>RGB</th></tr><tr><td>Blue (azul)</td><td><code>rgb(0, 0, 255)</code></td></tr><tr><td>Red (rojo)</td><td><code>rgb(255, 0, 0)</code></td></tr><tr><td>Orchid (color orquídea)</td><td><code>rgb(218, 112, 214)</code></td></tr><tr><td>Sienna (siena)</td><td><code>rgb(160, 82, 45)</code></td></tr></tbody></table>
+
+# --hints--
+
+Debes asignar al elemento `h1` que tiene el texto `I am red!` ("¡Soy de color rojo!) el `color` rojo.
+
+```js
+assert($('.red-text').css('color') === 'rgb(255, 0, 0)');
+```
+
+Debes usar el valor `rgb` que corresponde al color rojo.
+
+```js
+assert(
+ code.match(
+ /\.red-text\s*?{\s*?color:\s*?rgb\(\s*?255\s*?,\s*?0\s*?,\s*?0\s*?\)\s*?;\s*?}/gi
+ )
+);
+```
+
+Debes asignar al elemento `h1` que tiene el texto `I am orchid!` ("¡Soy de color orquídea!) el `color` orquídea.
+
+```js
+assert($('.orchid-text').css('color') === 'rgb(218, 112, 214)');
+```
+
+Debes usar el valor `rgb` que corresponde al color orchid (orquídea).
+
+```js
+assert(
+ code.match(
+ /\.orchid-text\s*?{\s*?color:\s*?rgb\(\s*?218\s*?,\s*?112\s*?,\s*?214\s*?\)\s*?;\s*?}/gi
+ )
+);
+```
+
+Debes asignar al elemento `h1` que tiene el texto `I am blue!` ("¡Soy de color azul!) el `color` azul.
+
+```js
+assert($('.blue-text').css('color') === 'rgb(0, 0, 255)');
+```
+
+Debes usar el valor `rgb` que corresponde al color azul.
+
+```js
+assert(
+ code.match(
+ /\.blue-text\s*?{\s*?color:\s*?rgb\(\s*?0\s*?,\s*?0\s*?,\s*?255\s*?\)\s*?;\s*?}/gi
+ )
+);
+```
+
+Debes asignar al elemento `h1` que tiene el texto `I am sienna!` ("¡Soy de color siena!) el `color` sienna (siena).
+
+```js
+assert($('.sienna-text').css('color') === 'rgb(160, 82, 45)');
+```
+
+Debes usar el valor `rgb` que corresponde al color sienna (siena).
+
+```js
+assert(
+ code.match(
+ /\.sienna-text\s*?{\s*?color:\s*?rgb\(\s*?160\s*?,\s*?82\s*?,\s*?45\s*?\)\s*?;\s*?}/gi
+ )
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ .red-text {
+ color: #000000;
+ }
+ .orchid-text {
+ color: #000000;
+ }
+ .sienna-text {
+ color: #000000;
+ }
+ .blue-text {
+ color: #000000;
+ }
+</style>
+
+<h1 class="red-text">I am red!</h1>
+
+<h1 class="orchid-text">I am orchid!</h1>
+
+<h1 class="sienna-text">I am sienna!</h1>
+
+<h1 class="blue-text">I am blue!</h1>
+```
+
+# --solutions--
+
+```html
+<style>
+ .red-text {
+ color: rgb(255, 0, 0);
+ }
+ .orchid-text {
+ color: rgb(218, 112, 214);
+ }
+ .sienna-text {
+ color: rgb(160, 82, 45);
+ }
+ .blue-text {
+ color:rgb(0, 0, 255);
+ }
+</style>
+
+<h1 class="red-text">I am red!</h1>
+
+<h1 class="orchid-text">I am orchid!</h1>
+
+<h1 class="sienna-text">I am sienna!</h1>
+
+<h1 class="blue-text">I am blue!</h1>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-rgb-values-to-color-elements.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-rgb-values-to-color-elements.md
new file mode 100644
index 00000000000..6b674f9cd05
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-css/use-rgb-values-to-color-elements.md
@@ -0,0 +1,72 @@
+---
+id: bad87fee1348bd9aede08718
+title: Utiliza valores RGB para asignar color a los elementos
+challengeType: 0
+videoUrl: 'https://scrimba.com/c/cRkp2fr'
+forumTopicId: 18369
+dashedName: use-rgb-values-to-color-elements
+---
+
+# --description--
+
+Otra forma de representar colores en CSS es utilizar valores `RGB`.
+
+El valor `RGB` del color negro se ve así:
+
+`rgb(0, 0, 0)`
+
+El valor `RGB` del color blanco se ve así:
+
+`rgb(255, 255, 255)`
+
+En lugar de usar seis dígitos hexadecimales, como hacemos con el código hexadecimal, en `RGB` se especifica el brillo de cada color con un número que va de 0 a 255.
+
+Si haces el cálculo, cada uno de los dos dígitos para un color representa 16 combinaciones, lo que nos da 256 valores posibles. Entonces, `RGB`, que comienza a contar desde cero, tiene el mismo número exacto de valores posibles que el código hexadecimal.
+
+A continuación puedes ver un ejemplo de cómo cambiar el color de fondo de "body" a naranja usando su código RGB.
+
+```css
+body {
+ background-color: rgb(255, 165, 0);
+}
+```
+
+# --instructions--
+
+Reemplacemos el código hexadecimal en el color de fondo de nuestro elemento `body` por el valor RGB correspondiente al color negro: `rgb(0, 0, 0)`
+
+# --hints--
+
+Tu elemento `body` debe tener un color de fondo "black" (negro).
+
+```js
+assert($('body').css('background-color') === 'rgb(0, 0, 0)');
+```
+
+Debes usar `rgb` para asignar a tu elemento `body` el color negro.
+
+```js
+assert(code.match(/rgb\s*\(\s*0\s*,\s*0\s*,\s*0\s*\)/gi));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<style>
+ body {
+ background-color: #F00;
+ }
+</style>
+```
+
+# --solutions--
+
+```html
+<style>
+ body {
+ background-color: rgb(0, 0, 0);
+ }
+</style>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/add-a-submit-button-to-a-form.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/add-a-submit-button-to-a-form.md
new file mode 100644
index 00000000000..c82b5a8acc2
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/add-a-submit-button-to-a-form.md
@@ -0,0 +1,111 @@
+---
+id: bad87fee1348bd9aedd08830
+title: Agrega un botón de envío a un formulario
+challengeType: 0
+videoUrl: 'https://scrimba.com/p/pVMPUv/cp2Nkhz'
+forumTopicId: 16627
+dashedName: add-a-submit-button-to-a-form
+---
+
+# --description--
+
+Agreguemos un botón `submit` a tu formulario. Al hacer clic en este botón se enviarán los datos de tu formulario a la URL especificada con el atributo `action` de tu formulario.
+
+Aquí hay un ejemplo de botón de envío:
+
+`<button type="submit">this button submits the form</button>`
+
+# --instructions--
+
+Agrega un botón del tipo `submit` como último elemento de tu formulario `form`, y usa `Submit` como su texto.
+
+# --hints--
+
+Tu formulario `form` debe tener un botón `button` dentro de él.
+
+```js
+assert($('form').children('button').length > 0);
+```
+
+Tu botón de envío debe tener el atributo `type` establecido como `submit`.
+
+```js
+assert($('button').attr('type') === 'submit');
+```
+
+Tu botón de envío solo debe contener el texto `Submit`.
+
+```js
+assert(
+ $('button')
+ .text()
+ .match(/^\s*submit\s*$/gi)
+);
+```
+
+Tu elemento `button` debe tener una etiqueta de cierre.
+
+```js
+assert(
+ code.match(/<\/button>/g) &&
+ code.match(/<button/g) &&
+ code.match(/<\/button>/g).length === code.match(/<button/g).length
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <input type="text" placeholder="cat photo URL">
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <input type="text" placeholder="cat photo URL">
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/add-images-to-your-website.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/add-images-to-your-website.md
new file mode 100644
index 00000000000..603a006758f
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/add-images-to-your-website.md
@@ -0,0 +1,91 @@
+---
+id: bad87fee1348bd9aedf08812
+title: Agrega imágenes a tu sitio web
+challengeType: 0
+videoUrl: 'https://scrimba.com/p/pVMPUv/c8EbJf2'
+forumTopicId: 16640
+dashedName: add-images-to-your-website
+---
+
+# --description--
+
+Puedes agregar imágenes a tu sitio web utilizando el elemento `img`, y apuntar a la URL de una imagen específica usando el atributo `src`.
+
+Un ejemplo de esto sería:
+
+`<img src="https://www.freecatphotoapp.com/your-image.jpg">`
+
+Ten en cuenta que los elementos `img` se cierran automáticamente.
+
+Todos los elementos `img` **deben** tener un atributo `alt`. El texto dentro de un atributo `alt` es utilizado por los lectores de pantalla para mejorar la accesibilidad y se muestra si la imagen falla en cargar.
+
+**Note:** Si la imagen es puramente decorativa, usar un atributo `alt` vacío es una buena práctica.
+
+Idealmente, el atributo `alt` no debe contener caracteres especiales a menos que sea necesario.
+
+Agreguemos un atributo `alt` a nuestro ejemplo `img` anterior:
+
+`<img src="https://www.freecatphotoapp.com/your-image.jpg" alt="A business cat wearing a necktie.">`
+
+# --instructions--
+
+Intentemos agregar una imagen a nuestro sitio web:
+
+Dentro del elemento `main`, inserta un elemento `img` antes de los elementos `p` existentes.
+
+Ahora establece el atributo `src` para que apunte a esta url:
+
+`https://bit.ly/fcc-relaxing-cat`
+
+Finalmente, no olvides darle a tu elemento `img` un atributo `alt` con texto descriptivo.
+
+# --hints--
+
+Tu página debe tener un elemento imagen.
+
+```js
+assert($('img').length);
+```
+
+Tu imagen debe tener un atributo `src` que apunte a la imagen del gatito.
+
+```js
+assert(/^https:\/\/bit\.ly\/fcc-relaxing-cat$/i.test($('img').attr('src')));
+```
+
+El atributo `alt` de tu elemento imagen no debe estar vacío.
+
+```js
+assert(
+ $('img').attr('alt') &&
+ $('img').attr('alt').length &&
+ /<img\S*alt=(['"])(?!\1|>)\S+\1\S*\/?>/.test(
+ __helpers.removeWhiteSpace(code)
+ )
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+
+
+ <p>Kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff.</p>
+ <p>Purr jump eat the grass rip the couch scratched sunbathe, shed everywhere rip the couch sleep in the sink fluffy fur catnip scratched.</p>
+</main>
+```
+
+# --solutions--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+ <p>Kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff.</p>
+ <p>Purr jump eat the grass rip the couch scratched sunbathe, shed everywhere rip the couch sleep in the sink fluffy fur catnip scratched.</p>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/add-placeholder-text-to-a-text-field.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/add-placeholder-text-to-a-text-field.md
new file mode 100644
index 00000000000..a5b4949b1af
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/add-placeholder-text-to-a-text-field.md
@@ -0,0 +1,106 @@
+---
+id: bad87fee1348bd9aedf08830
+title: Agrega texto provisional a un campo de texto
+challengeType: 0
+videoUrl: 'https://scrimba.com/p/pVMPUv/cKdJDhg'
+forumTopicId: 16647
+dashedName: add-placeholder-text-to-a-text-field
+---
+
+# --description--
+
+El texto provisional es lo que se muestra en tu elemento de entrada `input` antes de que el usuario haya ingresado nada.
+
+Puedes crear texto provisional así:
+
+`<input type="text" placeholder="this is placeholder text">`
+
+**Note:** Recuerda que los elementos `input` se cierran automáticamente.
+
+# --instructions--
+
+Establece el valor provisional `placeholder` de tu entrada de texto `input` como "cat photo URL".
+
+# --hints--
+
+Debes agregar un atributo `placeholder` al elemento de entrada de texto `input` existente.
+
+```js
+assert($('input[placeholder]').length > 0);
+```
+
+Debes establecer el valor de tu atributo provisional `placeholder` como `cat photo URL`.
+
+```js
+assert(
+ $('input') &&
+ $('input').attr('placeholder') &&
+ $('input')
+ .attr('placeholder')
+ .match(/cat\s+photo\s+URL/gi)
+);
+```
+
+El elemento `input` final no debe tener una etiqueta de cierre.
+
+```js
+assert(!code.match(/<input.*\/?>.*<\/input>/gi));
+```
+
+El elemento `input` final debe tener una sintaxis válida.
+
+```js
+assert($('input[type=text]').length > 0);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ <input type="text">
+</main>
+```
+
+# --solutions--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ <input type="text" placeholder="cat photo URL">
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/check-radio-buttons-and-checkboxes-by-default.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/check-radio-buttons-and-checkboxes-by-default.md
new file mode 100644
index 00000000000..caedaae0cc1
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/check-radio-buttons-and-checkboxes-by-default.md
@@ -0,0 +1,102 @@
+---
+id: bad87fee1348bd9aedd08835
+title: Marca botones de radio y casillas de verificación por defecto
+challengeType: 0
+videoUrl: 'https://scrimba.com/p/pVMPUv/cWk3Qh6'
+forumTopicId: 301094
+dashedName: check-radio-buttons-and-checkboxes-by-default
+---
+
+# --description--
+
+Puedes hacer que una casilla de verificación o botón de radio se marque por defecto usando el atributo `checked`.
+
+Para hacer esto, simplemente agrega la palabra `checked` al interior de un elemento de entrada. Por ejemplo:
+
+`<input type="radio" name="test-name" checked>`
+
+# --instructions--
+
+Establece el primero de tus botones de radio y la primera de tus casillas de verificación para que ambos sean marcados por defecto.
+
+# --hints--
+
+Tu primer botón de radio en tu formulario debe ser marcado por defecto.
+
+```js
+assert($('input[type="radio"]').prop('checked'));
+```
+
+Tu primera casilla de verificación en tu formulario debe ser marcado por defecto.
+
+```js
+assert($('input[type="checkbox"]').prop('checked'));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label for="indoor"><input id="indoor" type="radio" name="indoor-outdoor" value="indoor"> Indoor</label>
+ <label for="outdoor"><input id="outdoor" type="radio" name="indoor-outdoor" value="outdoor"> Outdoor</label><br>
+ <label for="loving"><input id="loving" type="checkbox" name="personality" value="loving"> Loving</label>
+ <label for="lazy"><input id="lazy" type="checkbox" name="personality" value="lazy"> Lazy</label>
+ <label for="energetic"><input id="energetic" type="checkbox" name="personality" value="energetic"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label for="indoor"><input id="indoor" type="radio" name="indoor-outdoor" value="indoor" checked> Indoor</label>
+ <label for="outdoor"><input id="outdoor" type="radio" name="indoor-outdoor" value="outdoor"> Outdoor</label><br>
+ <label for="loving"><input id="loving" type="checkbox" name="personality" value="loving" checked> Loving</label>
+ <label for="lazy"><input id="lazy" type="checkbox" name="personality" value="lazy"> Lazy</label>
+ <label for="energetic"><input id="energetic" type="checkbox" name="personality" value="energetic"> Energetic</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/comment-out-html.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/comment-out-html.md
new file mode 100644
index 00000000000..bef4c894593
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/comment-out-html.md
@@ -0,0 +1,76 @@
+---
+id: bad87fee1348bd9aedf08804
+title: Comenta HTML
+challengeType: 0
+videoUrl: 'https://scrimba.com/p/pVMPUv/cGyGbca'
+forumTopicId: 16782
+dashedName: comment-out-html
+---
+
+# --description--
+
+Recuerda que para iniciar un comentario, necesitas usar `<!--` y para terminar un comentario, necesitas usar `-->`
+
+Aquí tendrás que terminar el comentario antes de que tu elemento `h2` comience.
+
+# --instructions--
+
+Comenta tu elemento `h1` y tu elemento `p`, pero no tu elemento `h2`.
+
+# --hints--
+
+Tu elemento `h1` debe ser comentado para que no sea visible en la página.
+
+```js
+assert($('h1').length === 0);
+```
+
+Tu elemento `h2` no debe ser comentado para que sea visible en la página.
+
+```js
+assert($('h2').length > 0);
+```
+
+Tu elemento `p` debe ser comentado para que no sea visible en la página.
+
+```js
+assert($('p').length === 0);
+```
+
+Cada uno de tus comentarios debe cerrarse con `-->`.
+
+```js
+assert(code.match(/[^fc]-->/g).length > 1);
+```
+
+No debes cambiar el orden de los elementos `h1` `h2` o `p` en el código.
+
+```js
+assert(
+ code.match(/<([a-z0-9]){1,2}>/g)[0] === '<h1>' &&
+ code.match(/<([a-z0-9]){1,2}>/g)[1] === '<h2>' &&
+ code.match(/<([a-z0-9]){1,2}>/g)[2] === '<p>'
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<!--
+<h1>Hello World</h1>
+
+<h2>CatPhotoApp</h2>
+
+<p>Kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff.</p>
+-->
+```
+
+# --solutions--
+
+```html
+<!--<h1>Hello World</h1>-->
+<h2>CatPhotoApp</h2>
+<!--<p>Kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff.</p> -->
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/create-a-bulleted-unordered-list.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/create-a-bulleted-unordered-list.md
new file mode 100644
index 00000000000..5e9ba729b82
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/create-a-bulleted-unordered-list.md
@@ -0,0 +1,102 @@
+---
+id: bad87fee1348bd9aedf08827
+title: Crea una lista no ordenada
+challengeType: 0
+videoUrl: 'https://scrimba.com/p/pVMPUv/cDKVPuv'
+forumTopicId: 16814
+dashedName: create-a-bulleted-unordered-list
+---
+
+# --description--
+
+HTML tiene un elemento especial para crear <dfn>listas no ordenadas</dfn>, o listas con estilo de viñetas.
+
+Las listas no ordenadas comienzan con un elemento `<ul>` de apertura, seguido de cualquier número de elementos `<li>`. Por último, las listas no ordenadas cierran con un `</ul>`
+
+Por ejemplo:
+
+```html
+<ul>
+ <li>milk</li>
+ <li>cheese</li>
+</ul>
+```
+
+crearía una lista con estilo de viñetas de `milk` y `cheese`.
+
+# --instructions--
+
+Elimina los últimos dos elementos `p` y crea una lista no ordenada con tres cosas que amen los gatos en la parte inferior de la página.
+
+# --hints--
+
+Crea un elemento `ul`.
+
+```js
+assert($('ul').length > 0);
+```
+
+Debes tener tres elementos `li` dentro de tu elemento `ul`.
+
+```js
+assert($('ul li').length > 2);
+```
+
+Tu elemento `ul` debe tener una etiqueta de cierre.
+
+```js
+assert(
+ code.match(/<\/ul>/gi) &&
+ code.match(/<ul/gi) &&
+ code.match(/<\/ul>/gi).length === code.match(/<ul/gi).length
+);
+```
+
+Tus elementos `li` deben tener etiquetas de cierre.
+
+```js
+assert(
+ code.match(/<\/li>/gi) &&
+ code.match(/<li[\s>]/gi) &&
+ code.match(/<\/li>/gi).length === code.match(/<li[\s>]/gi).length
+);
+```
+
+Tus elementos `li` no deben contener una cadena vacía o solo espacios en blanco.
+
+```js
+assert($('ul li').filter((_, item) => !$(item).text().trim()).length === 0);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <p>Kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff.</p>
+ <p>Purr jump eat the grass rip the couch scratched sunbathe, shed everywhere rip the couch sleep in the sink fluffy fur catnip scratched.</p>
+</main>
+```
+
+# --solutions--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <ul>
+ <li>milk</li>
+ <li>mice</li>
+ <li>catnip</li>
+ </ul>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/create-a-form-element.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/create-a-form-element.md
new file mode 100644
index 00000000000..bc4e9e6a45d
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/create-a-form-element.md
@@ -0,0 +1,107 @@
+---
+id: bad87fee1348bd9aede08830
+title: Crea un elemento de formulario
+challengeType: 0
+forumTopicId: 16817
+dashedName: create-a-form-element
+---
+
+# --description--
+
+Puedes construir formularios web que realmente envíen datos a un servidor usando sólo HTML puro. Puedes hacer esto especificando un atributo `action` en tu elemento `form`.
+
+Por ejemplo:
+
+```html
+<form action="/url-where-you-want-to-submit-form-data">
+ <input>
+</form>
+```
+
+# --instructions--
+
+Anida el elemento `input` existente dentro de un elemento `form` y asigna `"https://freecatphotoapp.com/submit-cat-photo"` al atributo `action` del elemento `form`.
+
+# --hints--
+
+El elemento `input` existente debe anidarse dentro de un elemento `form`.
+
+```js
+const inputElem = document.querySelector('form input');
+assert(
+ inputElem.getAttribute('type') === 'text' &&
+ inputElem.getAttribute('placeholder') === 'cat photo URL'
+);
+```
+
+Tu formulario `form` debe tener un atributo `action` que esté establecido como `https://freecatphotoapp.com/submit-cat-photo`
+
+```js
+assert(
+ $('form').attr('action') === 'https://freecatphotoapp.com/submit-cat-photo'
+);
+```
+
+Tu elemento `form` debe tener etiquetas correctamente abiertas y cerradas.
+
+```js
+assert(
+ code.match(/<\/form>/g) &&
+ code.match(/<form [^<]*>/g) &&
+ code.match(/<\/form>/g).length === code.match(/<form [^<]*>/g).length
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ <input type="text" placeholder="cat photo URL">
+</main>
+```
+
+# --solutions--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <input type="text" placeholder="cat photo URL">
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/create-a-set-of-checkboxes.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/create-a-set-of-checkboxes.md
new file mode 100644
index 00000000000..f20bad512c9
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/create-a-set-of-checkboxes.md
@@ -0,0 +1,133 @@
+---
+id: bad87fee1348bd9aedf08835
+title: Crea un conjunto de casillas de verificación
+challengeType: 0
+videoUrl: 'https://scrimba.com/p/pVMPUv/cqrkJsp'
+forumTopicId: 16821
+dashedName: create-a-set-of-checkboxes
+---
+
+# --description--
+
+Los formularios suelen usar casillas de verificación (<dfn>checkboxes</dfn>) para preguntas que puedan tener más de una respuesta.
+
+Las casillas de verificación son un tipo de `input` (entrada).
+
+Cada una de tus casillas de verificación puede anidarse dentro de su propio elemento `label`. Envolver un elemento `input` dentro de un elemento `label` asociará automáticamente ese input de tipo checkbox con el elemento label que lo rodea.
+
+Todos los inputs de tipo casilla de verificación que están relacionados entre sí deben tener el mismo atributo `name`.
+
+Se considera buena práctica definir explícitamente la relación entre un `input` de tipo checkbox (casilla de verificación) y su correspondiente `label` (etiqueta), estableciendo el atributo `for` en el elemento `label` para que coincida con el atributo `id` del `input` asociado.
+
+A continuación te presentamos un ejemplo de una casilla de verificación:
+
+`<label for="loving"><input id="loving" type="checkbox" name="personality"> Loving</label>`
+
+# --instructions--
+
+Agrega a tu formulario un conjunto de tres casillas de verificación. Cada casilla de verificación debe anidarse dentro de su propio elemento `label`. Las tres deben compartir el atributo `name` con el valor `personality` (personalidad).
+
+# --hints--
+
+Tu página debe tener tres elementos de casilla de verificación.
+
+```js
+assert($('input[type="checkbox"]').length > 2);
+```
+
+Cada uno de tus tres elementos de casilla de verificación debe anidarse en su propio elemento `label`.
+
+```js
+assert($('label > input[type="checkbox"]:only-child').length > 2);
+```
+
+Asegúrate de que cada uno de tus elementos `label` tenga una etiqueta de cierre.
+
+```js
+assert(
+ code.match(/<\/label>/g) &&
+ code.match(/<label/g) &&
+ code.match(/<\/label>/g).length === code.match(/<label/g).length
+);
+```
+
+Tus casillas de verificación deben tener el atributo `name` establecido con el valor `personality`.
+
+```js
+assert(
+ $('label > input[type="checkbox"]').filter('[name="personality"]').length > 2
+);
+```
+
+Debes agregar cada una de tus casillas de verificación dentro de la etiqueta `form`.
+
+```js
+assert($('label').parent().get(0).tagName.match('FORM'));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label for="indoor"><input id="indoor" type="radio" name="indoor-outdoor"> Indoor</label>
+ <label for="outdoor"><input id="outdoor" type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label for="indoor"><input id="indoor" type="radio" name="indoor-outdoor"> Indoor</label>
+ <label for="outdoor"><input id="outdoor" type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <label for="playful"><input id="playful" type="checkbox" name="personality">Playful</label>
+ <label for="lazy"><input id="lazy" type="checkbox"
+name="personality">Lazy</label>
+ <label for="evil"><input id="evil" type="checkbox"
+name="personality">Evil</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/create-a-set-of-radio-buttons.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/create-a-set-of-radio-buttons.md
new file mode 100644
index 00000000000..bb180061ba6
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/create-a-set-of-radio-buttons.md
@@ -0,0 +1,153 @@
+---
+id: bad87fee1348bd9aedf08834
+title: Crea un conjunto de botones de radio
+challengeType: 0
+forumTopicId: 16822
+dashedName: create-a-set-of-radio-buttons
+---
+
+# --description--
+
+Puedes usar <dfn>botones de radio</dfn> para preguntas en las que quieres que el usuario solo te dé una respuesta a partir de múltiples opciones.
+
+Los botones de radio son un tipo de entrada `input`.
+
+Cada uno de tus botones de radio puede anidarse dentro de su propio elemento `label`. Envolver un elemento `input` dentro de un elemento `label` asociará automáticamente la entrada del botón de radio con el elemento label que lo rodea.
+
+Todos los botones de radio relacionados deben tener el mismo atributo `name` para crear un grupo de botones de radio. Al crear un grupo de radio, si se selecciona cualquier botón de radio se deselecciona automáticamente los otros botones dentro del mismo grupo, asegurándose que el usuario proporcione solo una respuesta.
+
+Aquí hay un ejemplo de un botón de radio:
+
+```html
+<label>
+ <input type="radio" name="indoor-outdoor">Indoor
+</label>
+```
+
+Se considera buena práctica establecer un atributo `for` en el elemento `label`, con un valor que coincida con el valor del atributo `id` del elemento `input`. Esto permite a las tecnologías asistivas crear una relación vinculada entre la etiqueta y el elemento hijo `input`. Por ejemplo:
+
+```html
+<label for="indoor">
+ <input id="indoor" type="radio" name="indoor-outdoor">Indoor
+</label>
+```
+
+# --instructions--
+
+Agrega un par de botones de radio a tu formulario, cada uno anidado en su propio elemento `label`. Uno debe tener la opción `indoor` y el otro debe tener la opción `outdoor`. Ambos deben compartir el atributo `name` establecido como `indoor-outdoor` para crear un grupo de radio.
+
+# --hints--
+
+Tu página debe tener dos elementos de botón de `radio`.
+
+```js
+assert($('input[type="radio"]').length > 1);
+```
+
+Tus botones de radio deben recibir el atributo `name` establecido como `indoor-outdoor`.
+
+```js
+assert($('input[type="radio"]').filter("[name='indoor-outdoor']").length > 1);
+```
+
+Cada uno de tus elementos de botón de radio debe anidarse en su propio elemento `label`.
+
+```js
+assert($('label > input[type="radio"]:only-child').length > 1);
+```
+
+Cada uno de tus elementos `label` debe tener una etiqueta de cierre.
+
+```js
+assert(
+ code.match(/<\/label>/g) &&
+ code.match(/<label/g) &&
+ code.match(/<\/label>/g).length === code.match(/<label/g).length
+);
+```
+
+Uno de tus botones de radio debe tener la etiqueta `indoor`.
+
+```js
+assert(
+ $('label')
+ .text()
+ .match(/indoor/gi)
+);
+```
+
+Uno de tus botones de radio debe tener la etiqueta `outdoor`.
+
+```js
+assert(
+ $('label')
+ .text()
+ .match(/outdoor/gi)
+);
+```
+
+Cada uno de los elementos de botón de radio debe ser agregado dentro de la etiqueta `form`.
+
+```js
+assert($('label').parent().get(0).tagName.match('FORM'));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
+
+# --solutions--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ <form action="https://freecatphotoapp.com/submit-cat-photo">
+ <label for="indoor"><input id="indoor" type="radio" name="indoor-outdoor"> Indoor</label>
+ <label for="outdoor"><input id="outdoor" type="radio" name="indoor-outdoor"> Outdoor</label><br>
+ <input type="text" placeholder="cat photo URL" required>
+ <button type="submit">Submit</button>
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/create-a-text-field.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/create-a-text-field.md
new file mode 100644
index 00000000000..575a53c06d6
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/create-a-text-field.md
@@ -0,0 +1,87 @@
+---
+id: bad87fee1348bd9aedf08829
+title: Crea un campo de texto
+challengeType: 0
+videoUrl: 'https://scrimba.com/p/pVMPUv/c2EVnf6'
+forumTopicId: 16823
+dashedName: create-a-text-field
+---
+
+# --description--
+
+Ahora crearemos un formulario web.
+
+Los elementos de entrada `input` son una forma conveniente de obtener información de tu usuario.
+
+Puedes crear una entrada de texto de esta manera:
+
+`<input type="text">`
+
+Ten en cuenta que los elementos `input` se cierran automáticamente.
+
+# --instructions--
+
+Crea un elemento `input` de tipo `text` debajo de tus listas.
+
+# --hints--
+
+Tu aplicación debe tener un elemento `input` de tipo `text`.
+
+```js
+assert($('input[type=text]').length > 0);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+
+
+</main>
+```
+
+# --solutions--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>flea treatment</li>
+ <li>thunder</li>
+ <li>other cats</li>
+ </ol>
+ <form>
+ <input type="text">
+ </form>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/create-an-ordered-list.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/create-an-ordered-list.md
new file mode 100644
index 00000000000..fb87c8e2252
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/create-an-ordered-list.md
@@ -0,0 +1,157 @@
+---
+id: bad87fee1348bd9aedf08828
+title: Crea una lista ordenada
+challengeType: 0
+videoUrl: 'https://scrimba.com/p/pVMPUv/cQ3B8TM'
+forumTopicId: 16824
+dashedName: create-an-ordered-list
+---
+
+# --description--
+
+HTML tiene otro elemento especial para crear <dfn>listas ordenadas</dfn>, o listas numeradas.
+
+Las listas ordenadas comienzan con un elemento de apertura `<ol>`, seguido de cualquier número de elementos `<li>`. Por último, las listas ordenadas se cierran con la etiqueta `</ol>`.
+
+Por ejemplo:
+
+```html
+<ol>
+ <li>Garfield</li>
+ <li>Sylvester</li>
+</ol>
+```
+
+crearía una lista numerada de `Garfield` y `Sylvester`.
+
+# --instructions--
+
+Crea una lista ordenada de las 3 cosas que los gatos odian (Top 3 things cats hate) más.
+
+# --hints--
+
+Debes tener una lista ordenada para `Top 3 things cats hate:`
+
+```js
+assert(/Top 3 things cats hate:/i.test($('ol').prev().text()));
+```
+
+Debes tener una lista no ordenada para `Things cats love:`
+
+```js
+assert(/Things cats love:/i.test($('ul').prev().text()));
+```
+
+Solo debes tener un elemento `ul`.
+
+```js
+assert.equal($('ul').length, 1);
+```
+
+Solo debes tener un elemento `ol`.
+
+```js
+assert.equal($('ol').length, 1);
+```
+
+Debes tener tres elementos `li` dentro de tu elemento `ul`.
+
+```js
+assert.equal($('ul li').length, 3);
+```
+
+Debes tener tres elementos `li` dentro de tu elemento `ol`.
+
+```js
+assert.equal($('ol li').length, 3);
+```
+
+Tu elemento `ul` debe tener una etiqueta de cierre.
+
+```js
+assert(
+ code.match(/<\/ul>/g) &&
+ code.match(/<\/ul>/g).length === code.match(/<ul>/g).length
+);
+```
+
+Tu elemento `ol` debe tener una etiqueta de cierre.
+
+```js
+assert(
+ code.match(/<\/ol>/g) &&
+ code.match(/<\/ol>/g).length === code.match(/<ol>/g).length
+);
+```
+
+Tu elemento `li` debe tener una etiqueta de cierre.
+
+```js
+assert(
+ code.match(/<\/li>/g) &&
+ code.match(/<li>/g) &&
+ code.match(/<\/li>/g).length === code.match(/<li>/g).length
+);
+```
+
+Los elementos `li` de tu lista no ordenada no deben estar vacíos.
+
+```js
+$('ul li').each((i, val) =>
+ assert(__helpers.removeWhiteSpace(val.textContent))
+);
+```
+
+Los elementos `li` de tu lista ordenada no deben estar vacíos.
+
+```js
+$('ol li').each((i, val) =>
+ assert(!!__helpers.removeWhiteSpace(val.textContent))
+);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+
+</main>
+```
+
+# --solutions--
+
+```html
+<h2>CatPhotoApp</h2>
+<main>
+ <p>Click here to view more <a href="#">cat photos</a>.</p>
+
+ <a href="#"><img src="https://bit.ly/fcc-relaxing-cat" alt="A cute orange cat lying on its back."></a>
+
+ <p>Things cats love:</p>
+ <ul>
+ <li>cat nip</li>
+ <li>laser pointers</li>
+ <li>lasagna</li>
+ </ul>
+ <p>Top 3 things cats hate:</p>
+ <ol>
+ <li>hate 1</li>
+ <li>hate 2</li>
+ <li>hate 3</li>
+ </ol>
+</main>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/declare-the-doctype-of-an-html-document.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/declare-the-doctype-of-an-html-document.md
new file mode 100644
index 00000000000..8d007f8920a
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/declare-the-doctype-of-an-html-document.md
@@ -0,0 +1,70 @@
+---
+id: 587d78aa367417b2b2512aed
+title: Declara el Doctype de un documento HTML
+challengeType: 0
+videoUrl: 'https://scrimba.com/p/pVMPUv/cra98AJ'
+forumTopicId: 301095
+dashedName: declare-the-doctype-of-an-html-document
+---
+
+# --description--
+
+Hasta el momento, los desafíos han cubierto elementos HTML específicos y sus usos. Sin embargo, hay algunos elementos que dan una estructura general a tu página, y deben incluirse en cada documento HTML.
+
+En la parte superior de tu documento, necesitas decirle al navegador qué versión de HTML está utilizando tu página. HTML es un lenguaje en evolución, y se actualiza regularmente. La mayoría de los navegadores principales soportan la última especificación, que es HTML5. Sin embargo, páginas web más antiguas puede que hagan uso de versiones anteriores del lenguaje.
+
+Proporcionas al navegador esta información agregando la etiqueta `<!DOCTYPE ...>` en la primera línea, donde la parte `...` es la versión de HTML. Para HTML5, utilizas `<!DOCTYPE html>`.
+
+El `!` y `DOCTYPE` en mayúsculas es importante, especialmente para los navegadores más antiguos. El `html` no es sensible a mayúsculas y minúsculas.
+
+A continuación, el resto de tu código HTML necesita ser envuelto en etiquetas `html`. La apertura `<html>` va directamente debajo de la línea `<!DOCTYPE html>`, y el cierre `</html>` va en el final de la página.
+
+Aquí hay un ejemplo de estructura de página:
+
+```html
+<!DOCTYPE html>
+<html>
+ <!-- Your HTML code goes here -->
+</html>
+```
+
+# --instructions--
+
+Agrega una etiqueta `DOCTYPE` para HTML5 a la parte superior del documento en el editor de código. Debajo de el, agrega etiquetas de apertura y cierre de `html`, que envuelvan alrededor un elemento `h1`. El título puede incluir cualquier texto.
+
+# --hints--
+
+Tu código debe incluir una etiqueta `<!DOCTYPE html>`.
+
+```js
+assert(code.match(/<!DOCTYPE\s+?html\s*?>/gi));
+```
+
+Debe haber un elemento `html`.
+
+```js
+assert($('html').length == 1);
+```
+
+Las etiquetas `html` deben envolver alrededor un elemento `h1`.
+
+```js
+assert(code.match(/<html>\s*?<h1>\s*?.*?\s*?<\/h1>\s*?<\/html>/gi));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+
+```
+
+# --solutions--
+
+```html
+<!DOCTYPE html>
+<html>
+ <h1> Hello world </h1>
+</html>
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/define-the-head-and-body-of-an-html-document.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/define-the-head-and-body-of-an-html-document.md
new file mode 100644
index 00000000000..3a6dc222842
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/define-the-head-and-body-of-an-html-document.md
@@ -0,0 +1,139 @@
+---
+id: 587d78aa367417b2b2512aec
+title: Define el encabezado y el cuerpo de un Documento HTML
+challengeType: 0
+videoUrl: 'https://scrimba.com/p/pVMPUv/cra9bfP'
+forumTopicId: 301096
+dashedName: define-the-head-and-body-of-an-html-document
+---
+
+# --description--
+
+Puedes agregar otro nivel de organización en tu documento HTML dentro de las etiquetas `html` con los elementos `head` y `body`. Cualquier código con información sobre tu página iría dentro de la etiqueta `head`. Entonces, cualquier código con el contenido de la página (lo que se muestra para un usuario) iría dentro de la etiqueta `body`.
+
+Elementos de metadatos, tales como `link`, `meta`, `title`, y `style`, típicamente van dentro del elemento `head`.
+
+Aquí hay un ejemplo de la disposición de una página:
+
+```html
+<!DOCTYPE html>
+<html>
+ <head>
+ <!-- metadata elements -->
+ </head>
+ <body>
+ <!-- page contents -->
+ </body>
+</html>
+```
+
+# --instructions--
+
+Edita el código para que haya un `head` y un `body`. El elemento `head` solo debe incluir `title`, y el elemento `body` solo debe incluir `h1` y `p`.
+
+# --hints--
+
+Solo debe haber un elemento `head` en la página.
+
+```js
+const headElems = code.replace(/\n/g, '').match(/\<head\s*>.*?\<\/head\s*>/g);
+assert(headElems && headElems.length === 1);
+```
+
+Solo debe haber un elemento `body` en la página.
+
+```js
+const bodyElems = code.replace(/\n/g, '').match(/<body\s*>.*?<\/body\s*>/g);
+assert(bodyElems && bodyElems.length === 1);
+```
+
+El elemento `head` debe ser hijo del elemento `html`.
+
+```js
+const htmlChildren = code
+ .replace(/\n/g, '')
+ .match(/<html\s*>(?<children>.*)<\/html\s*>/);
+let foundHead;
+if (htmlChildren) {
+ const { children } = htmlChildren.groups;
+
+ foundHead = children.match(/<head\s*>.*<\/head\s*>/);
+}
+assert(foundHead);
+```
+
+El elemento `body` debe ser hijo del elemento `html`.
+
+```js
+const htmlChildren = code
+ .replace(/\n/g, '')
+ .match(/<html\s*>(?<children>.*?)<\/html\s*>/);
+let foundBody;
+if (htmlChildren) {
+ const { children } = htmlChildren.groups;
+ foundBody = children.match(/<body\s*>.*<\/body\s*>/);
+}
+assert(foundBody);
+```
+
+El elemento `head` debe envolver alrededor un elemento `title`.
+
+```js
+const headChildren = code
+ .replace(/\n/g, '')
+ .match(/<head\s*>(?<children>.*?)<\/head\s*>/);
+let foundTitle;
+if (headChildren) {
+ const { children } = headChildren.groups;
+ foundTitle = children.match(/<title\s*>.*?<\/title\s*>/);
+}
+assert(foundTitle);
+```
+
+El elemento `body` debe envolver ambos elementos `h1` y `p`.
+
+```js
+const bodyChildren = code
+ .replace(/\n/g, '')
+ .match(/<body\s*>(?<children>.*?)<\/body\s*>/);
+let foundElems;
+if (bodyChildren) {
+ const { children } = bodyChildren.groups;
+ const h1s = children.match(/<h1\s*>.*<\/h1\s*>/g);
+ const ps = children.match(/<p\s*>.*<\/p\s*>/g);
+ const numH1s = h1s ? h1s.length : 0;
+ const numPs = ps ? ps.length : 0;
+ foundElems = numH1s === 1 && numPs === 1;
+}
+assert(foundElems);
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+<!DOCTYPE html>
+<html>
+ <title>The best page ever
+
+ The best page ever
+ Cat ipsum dolor sit amet, jump launch to pounce upon little yarn mouse, bare fangs at toy run hide in litter box until treats are fed. Go into a room to decide you didn't want to be in there anyway. I like big cats and i can not lie kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff. Meow i could pee on this if i had the energy for slap owner's face at 5am until human fills food dish yet scamper. Knock dish off table head butt cant eat out of my own dish scratch the furniture. Make meme, make cute face. Sleep in the bathroom sink chase laser but pee in the shoe. Paw at your fat belly licks your face and eat grass, throw it back up kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff.
+
+
+```
+
+# --solutions--
+
+```html
+
+
+
+ The best page ever
+
+
+ The best page ever
+ Cat ipsum dolor sit amet, jump launch to pounce upon little yarn mouse, bare fangs at toy run hide in litter box until treats are fed. Go into a room to decide you didn't want to be in there anyway. I like big cats and i can not lie kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff. Meow i could pee on this if i had the energy for slap owner's face at 5am until human fills food dish yet scamper. Knock dish off table head butt cant eat out of my own dish scratch the furniture. Make meme, make cute face. Sleep in the bathroom sink chase laser but pee in the shoe. Paw at your fat belly licks your face and eat grass, throw it back up kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff.
+
+
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/delete-html-elements.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/delete-html-elements.md
new file mode 100644
index 00000000000..7891fd5db4c
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/delete-html-elements.md
@@ -0,0 +1,56 @@
+---
+id: bad87fed1348bd9aedf08833
+title: Elimina elementos HTML
+challengeType: 0
+videoUrl: 'https://scrimba.com/p/pVMPUv/ckK73C9'
+forumTopicId: 17559
+dashedName: delete-html-elements
+---
+
+# --description--
+
+Nuestro teléfono no tiene mucho espacio vertical.
+
+Eliminemos los elementos innecesarios para que podamos empezar a construir nuestra CatPhotoApp.
+
+# --instructions--
+
+Elimina tu elemento `h1` para que podamos simplificar nuestra vista.
+
+# --hints--
+
+Tu elemento `h1` debe ser eliminado.
+
+```js
+assert(!code.match(//gi) && !code.match(/<\/h1>/gi));
+```
+
+Tu elemento `h2` debe estar en la página.
+
+```js
+assert(code.match(/[\w\W]*<\/h2>/gi));
+```
+
+Tu elemento `p` debe estar en la página.
+
+```js
+assert(code.match(/
[\w\W]*<\/p>/gi));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+
Hello World
+
+CatPhotoApp
+
+Kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff.
+```
+
+# --solutions--
+
+```html
+CatPhotoApp
Kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff.
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/fill-in-the-blank-with-placeholder-text.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/fill-in-the-blank-with-placeholder-text.md
new file mode 100644
index 00000000000..d82d070fd47
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/fill-in-the-blank-with-placeholder-text.md
@@ -0,0 +1,50 @@
+---
+id: bad87fee1348bd9aedf08833
+title: Rellena el espacio en blanco con texto provisional
+challengeType: 0
+videoUrl: 'https://scrimba.com/p/pVMPUv/cgR7Dc7'
+forumTopicId: 18178
+dashedName: fill-in-the-blank-with-placeholder-text
+---
+
+# --description--
+
+Los desarrolladores web tradicionalmente usan texto lorem ipsum como texto provisional. El texto lorem ipsum se extrajo aleatoriamente de un famoso pasaje de Cicerón de la Antigua Roma.
+
+El texto lorem ipsum ha sido utilizado como texto provisional por los tipógrafos desde el siglo XVI, y esta tradición continúa en la web.
+
+Pues bien, 5 siglos es suficiente. Puesto que estamos construyendo una aplicación de fotos de gatos, utilicemos algo llamado "kitty ipsum text".
+
+# --instructions--
+
+Reemplaza el texto dentro de tu elemento `p` con las primeras palabras de este kitty ipsum text: `Kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff.`
+
+# --hints--
+
+Tu elemento `p` debe contener las primeras palabras del "kitty ipsum text" que te proporcionamos aquí.
+
+```js
+assert.isTrue(/Kitty(\s)+ipsum/gi.test($('p').text()));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+Hello World
+
+CatPhotoApp
+
+Hello Paragraph
+```
+
+# --solutions--
+
+```html
+Hello World
+
+CatPhotoApp
+
+Kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/headline-with-the-h2-element.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/headline-with-the-h2-element.md
new file mode 100644
index 00000000000..ef05d8395ac
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/headline-with-the-h2-element.md
@@ -0,0 +1,70 @@
+---
+id: bad87fee1348bd9aedf0887a
+title: Título con el elemento h2
+challengeType: 0
+videoUrl: 'https://scrimba.com/p/pVMPUv/cE8Gqf3'
+forumTopicId: 18196
+dashedName: headline-with-the-h2-element
+---
+
+# --description--
+
+Durante las próximas lecciones, construiremos pieza a pieza una aplicación web en HTML5 de fotos de gatos.
+
+El elemento `h2` que añadirás en este paso agregará un título nivel dos a la página web.
+
+Este elemento le informa al navegador sobre la estructura de tu sitio web. Los elementos `h1` se utilizan a menudo para títulos principales, mientras que los elementos `h2` se utilizan generalmente para subtítulos. También hay elementos `h3`, `h4`, `h5` y `h6` para indicar diferentes niveles de subtítulos.
+
+# --instructions--
+
+Añade una etiqueta `h2` que diga "CatPhotoApp" para crear un segundo elemento HTML debajo de tu elemento `h1` "Hello World".
+
+# --hints--
+
+Debes crear un elemento `h2`.
+
+```js
+assert($('h2').length > 0);
+```
+
+Tu elemento `h2` debe tener una etiqueta de cierre.
+
+```js
+assert(
+ code.match(/<\/h2>/g) &&
+ code.match(/<\/h2>/g).length === code.match(//g).length
+);
+```
+
+Tu elemento `h2` debe contener el texto `CatPhotoApp`.
+
+```js
+assert.isTrue(/cat(\s)?photo(\s)?app/gi.test($('h2').text()));
+```
+
+Tu elemento `h1` debe contener el texto `Hello World`.
+
+```js
+assert.isTrue(/hello(\s)+world/gi.test($('h1').text()));
+```
+
+Tu elemento `h1` debe estar antes que tu elemento `h2`.
+
+```js
+assert(code.match(/\s*?.*?\s*?<\/h1>\s*\s*?.*?\s*?<\/h2>/gi));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+Hello World
+```
+
+# --solutions--
+
+```html
+Hello World
+CatPhotoApp
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/inform-with-the-paragraph-element.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/inform-with-the-paragraph-element.md
new file mode 100644
index 00000000000..c712d639f31
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/inform-with-the-paragraph-element.md
@@ -0,0 +1,62 @@
+---
+id: bad87fee1348bd9aedf08801
+title: Informa con el elemento párrafo
+challengeType: 0
+videoUrl: 'https://scrimba.com/p/pVMPUv/ceZ7DtN'
+forumTopicId: 18202
+dashedName: inform-with-the-paragraph-element
+---
+
+# --description--
+
+Los elementos `p` son el elemento preferido para el texto de los párrafos en los sitios web. `p` es la abreviatura de "párrafo" (paragraph).
+
+Puedes crear un elemento párrafo de esta manera:
+
+`
I'm a p tag!
`
+
+# --instructions--
+
+Crea un elemento `p` debajo de tu elemento `h2` y dale como texto `Hello Paragraph`.
+
+**Note:** Por convención, todas las etiquetas HTML son escritas en minúsculas, por ejemplo `` y no ``.
+
+# --hints--
+
+Tu código debe tener un elemento `p` válido.
+
+```js
+assert($('p').length > 0);
+```
+
+Tu elemento `p` debe contener el texto `Hello Paragraph`.
+
+```js
+assert.isTrue(/hello(\s)+paragraph/gi.test($('p').text()));
+```
+
+Tu elemento `p` debe tener una etiqueta de cierre.
+
+```js
+assert(
+ code.match(/<\/p>/g) &&
+ code.match(/<\/p>/g).length === code.match(/Hello World
+CatPhotoApp
+```
+
+# --solutions--
+
+```html
+Hello World
+CatPhotoApp
+Hello Paragraph
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/introduction-to-html5-elements.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/introduction-to-html5-elements.md
new file mode 100644
index 00000000000..f8f954e21de
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/introduction-to-html5-elements.md
@@ -0,0 +1,98 @@
+---
+id: bad87fee1348bd9aecf08801
+title: Introducción a los elementos HTML5
+challengeType: 0
+videoUrl: 'https://scrimba.com/p/pVMPUv/cBkZGpt7'
+forumTopicId: 301097
+dashedName: introduction-to-html5-elements
+---
+
+# --description--
+
+HTML5 introduce etiquetas HTML más descriptivas. Estas incluyen `main`, `header`, `footer`, `nav`, `video`, `article`, `section`, entre otras.
+
+Estas etiquetas dan una estructura descriptiva a tu HTML, hacen que tu HTML sea más fácil de leer, ayudan con la Optimización de Motores de Búsqueda (SEO) y mejoran la accesibilidad. La etiqueta HTML5 `main` ayuda a los motores de búsqueda y otros desarrolladores a encontrar el contenido principal de tu página.
+
+Por ejemplo, un elemento `main` con dos elementos hijos anidados en su interior:
+
+```html
+
+ Hello World
+ Hello Paragraph
+
+```
+
+**Note:** Muchas de las nuevas etiquetas HTML5 y sus beneficios están cubiertos en la sección Accesibilidad Aplicada.
+
+# --instructions--
+
+Crea un segundo elemento `p` después del elemento `p` existente con el siguiente texto ipsum gatuno: `Purr jump eat the grass rip the couch scratched sunbathe, shed everywhere rip the couch sleep in the sink fluffy fur catnip scratched.`
+
+Luego, crea un elemento `main` y anida los dos elementos `p` dentro del elemento `main`.
+
+# --hints--
+
+Debes tener 2 elementos `p` con texto Ipsum Gatuno.
+
+```js
+assert($('p').length > 1);
+```
+
+Cada uno de tus elementos `p` debe tener una etiqueta de cierre.
+
+```js
+assert(
+ code.match(/<\/p>/g) &&
+ code.match(/<\/p>/g).length === code.match(/\s*?
/g));
+```
+
+La etiquete de cierre `main` debe aparecer después de la segunda etiqueta de cierre párrafo.
+
+```js
+assert(code.match(/<\/p>\s*?<\/main>/g));
+```
+
+# --seed--
+
+## --seed-contents--
+
+```html
+
CatPhotoApp
+
+Kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff.
+```
+
+# --solutions--
+
+```html
+CatPhotoApp
+
+ Kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff.
+ Purr jump eat the grass rip the couch scratched sunbathe, shed everywhere rip the couch sleep in the sink fluffy fur catnip scratched.
+
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/link-to-external-pages-with-anchor-elements.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/link-to-external-pages-with-anchor-elements.md
new file mode 100644
index 00000000000..af63d5b3f38
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/link-to-external-pages-with-anchor-elements.md
@@ -0,0 +1,76 @@
+---
+id: bad87fee1348bd9aedf08816
+title: Enlaza hacia páginas externas con los elementos "anchor"
+challengeType: 0
+videoUrl: 'https://scrimba.com/p/pVMPUv/c8EkncB'
+forumTopicId: 18226
+dashedName: link-to-external-pages-with-anchor-elements
+---
+
+# --description--
+
+Puedes usar los elementos `a` (*anchor*) para enlazar a contenido fuera de tu página web.
+
+Los elementos `a` requieren un atributo `href` con la dirección web de destino. También necesitan un texto anchor. Por ejemplo:
+
+`this links to freecodecamp.org`
+
+Entonces tu navegador mostrará el texto `this links to freecodecamp.org` como un enlace que puedes hacer clic. Y ese enlace te llevará a la dirección web `https://www.freecodecamp.org`.
+
+# --instructions--
+
+Crea un elemento `a` que enlace a `https://freecatphotoapp.com` y tenga "cat photos" como su texto anchor.
+
+# --hints--
+
+Tu elemento `a` debe contener el texto anchor: `cat photos`.
+
+```js
+assert(/cat photos/gi.test($('a').text()));
+```
+
+Necesitas un elemento `a` que enlace `https://freecatphotoapp.com`
+
+```js
+assert(/https:\/\/(www\.)?freecatphotoapp\.com/gi.test($('a').attr('href')));
+```
+
+Tu elemento `a` debe tener una etiqueta de cierre.
+
+```js
+assert(
+ code.match(/<\/a>/g) &&
+ code.match(/<\/a>/g).length === code.match(/CatPhotoApp
+
+
+
+
+ 
+
+ Kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff.
+ Purr jump eat the grass rip the couch scratched sunbathe, shed everywhere rip the couch sleep in the sink fluffy fur catnip scratched.
+
+```
+
+# --solutions--
+
+```html
+CatPhotoApp
+
+
+
+
+ cat photos
+ Kitty ipsum dolor sit amet, shed everywhere shed everywhere stretching attack your ankles chase the red dot, hairball run catnip eat the grass sniff.
+ Purr jump eat the grass rip the couch scratched sunbathe, shed everywhere rip the couch sleep in the sink fluffy fur catnip scratched.
+
+```
diff --git a/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/link-to-internal-sections-of-a-page-with-anchor-elements.md b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/link-to-internal-sections-of-a-page-with-anchor-elements.md
new file mode 100644
index 00000000000..abd4643ab15
--- /dev/null
+++ b/curriculum/challenges/espanol/01-responsive-web-design/basic-html-and-html5/link-to-internal-sections-of-a-page-with-anchor-elements.md
@@ -0,0 +1,120 @@
+---
+id: bad88fee1348bd9aedf08816
+title: Enlaza hacia secciones internas de una página con los elementos anchor
+challengeType: 0
+videoUrl: 'https://scrimba.com/p/pVMPUv/cyrDRUL'
+forumTopicId: 301098
+dashedName: link-to-internal-sections-of-a-page-with-anchor-elements
+---
+
+# --description--
+
+Los elementos `a` (*anchor*) también pueden utilizarse para crear enlaces internos para saltar a diferentes secciones dentro de una página web.
+
+Para crear un enlace interno, asignas el atributo `href` de un enlace con un símbolo hash `#` más el valor del atributo `id` para el elemento al que deseas enlazar internamente, normalmente más abajo de la página. Luego necesitas agregar el mismo atributo `id` al elemento que estás enlazando. Un `id` es un atributo que describe un elemento de forma única.
+
+A continuación hay un ejemplo de un enlace interno y su elemento destino:
+
+```html
+Contacts
+...
+
+```
+
+Cuando los usuarios hagan clic en el enlace `Contacts`, serán llevados a la sección de la página con el elemento título **Contacts**.
+
+# --instructions--
+
+Cambia tu enlace externo a un enlace interno cambiando el atributo `href` a `"#footer"` y el texto de `cat photos` a `Jump to Bottom`.
+
+Elimina el atributo `target="_blank"` de la etiqueta anchor ya que esto provoca que el documento enlazado se abra en una nueva pestaña.
+
+Luego agrega un atributo `id` con un valor de `footer` al elemento `
`
-视觉障碍用户无法通过视觉获取信息，而是通过屏幕阅读器将网页内容转换为音频以获取信息。屏幕阅读器可以识别 `alt` 属性，朗读其中的内容，来告知用户图片包含的关键信息。
+视觉障碍用户无法通过视觉获取信息，而是通过屏幕阅读器将网页内容转换为音频以获取信息。 他们无法通过直观的呈现理解信息。 屏幕阅读器可以识别 `alt` 属性，朗读其中的内容，来告知用户图片包含的关键信息。
-`alt` 文本可以为屏幕阅读器提供图片的描述信息，所以你应始终为图片添加 `alt` 属性。另外，根据最新的 HTML5 标准，为图片添加 `alt` 属性是必需的。
+`alt` 文本可以为屏幕阅读器提供图片的描述信息， 所以你应始终为图片添加 `alt` 属性。 另外，根据最新的 HTML5 标准，为图片添加这个属性是必须的。
# --instructions--
-碰巧，Camper Cat 是忍者中写代码最厉害的，他正在建立一个可以分享忍者知识的网站。在这个网站中，他添加了一张展示技能的资料图片，并希望这张图片可以让访问网站的所有用户看到。请给 `img` 标签添加一个 `alt` 属性，说明 Camper Cat 在学习空手道（图片的 `src` 属性指向的是一个不存在的文件，因此你可以看到 `alt` 属性中的文本出现在页面上）。
+碰巧，Camper Cat 是忍者中写代码最厉害的，他正在建立一个可以分享忍者知识的网站。 他想要使用的个人资料图片显示了他的技能，应该得到所有网站访问者的赞赏。 请给 `img` 标签添加一个 `alt` 属性，说明 Camper Cat 在学习空手道 （图片的 `src` 属性指向的是一个不存在的文件，因此你可以看到 `alt` 属性中的文本出现在页面上）。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/add-an-accessible-date-picker.md b/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/add-an-accessible-date-picker.md
index 4318298274c..b5f187bd034 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/add-an-accessible-date-picker.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/add-an-accessible-date-picker.md
@@ -9,11 +9,11 @@ dashedName: add-an-accessible-date-picker
# --description--
-表单中经常出现 `input` 标签，它可以用来创建多种表单控件。它的 `type` 属性指定了所要创建的 `input` 标签类型。
+表单中经常出现 `input` 标签，它可以用来创建多种表单控件。 它的 `type` 属性指定了所要创建的 `input` 标签类型。
-在以前的挑战中，我们已经见过 `text` 与 `submit` 类型的 `input` 标签。HTML5 规范添加了 `date` 类型来创建日期选择器。如果浏览器支持，在点击 `input` 标签时，日期选择器会显示出来，这让用户填写表单变得更加容易。
+在以前的挑战中，我们已经见过 `text` 与 `submit` 类型的 input 标签。 HTML5 规范添加了 `date` 类型来创建日期选择器。 如果浏览器支持，在点击 `input` 标签时，日期选择器会显示出来，这让用户填写表单变得更加容易。
-对于旧版本的浏览器，由于不支持 `date`，此时 `type` 属性会被浏览器设置为 `text`。这种情况下，我们可以利用 `label` 标签或者占位符文本来提示用户需要输入的日期格式。
+对于较老的浏览器，类型将默认为 `text`， 这样它可以通过 `label` 或 `placeholder` 文本向用户显示预期的日期格式。
举个例子：
@@ -24,29 +24,29 @@ dashedName: add-an-accessible-date-picker
# --instructions--
-Camper Cat 想举办一场比武大会，他想收集参赛者的最佳参赛时间。请为 Camper Cat 的页面添加一个 `input` 标签，其 `type` 属性值为 "date"，`id` 为 "pickdate"，`name` 属性值为 "date"。
+Camper Cat 想举办一场比武大会，他想收集参赛者的最佳参赛时间。 请为 Camper Cat 的页面添加一个`input` 标签，起 `type` 属性值为 `date`，`id` 属性为 `pickdate`，`name` 属性为 `date`。
# --hints--
-应有 1 个 `input` 标签。
+日期选择器应有一个 `input` 标签。
```js
assert($('input').length == 2);
```
-`input` 标签的 `type` 属性值应该为 date。
+`input` 标签应有一个值为 `date` 的 `type` 属性。
```js
assert($('input').attr('type') == 'date');
```
-`input` 标签的 `id` 应为 pickdate。
+`input` 标签应有一个值为 `pickdate` 的 `id` 属性。
```js
assert($('input').attr('id') == 'pickdate');
```
-`input` 标签的 `name` 属性值应该为 date。
+`input` 标签应有一个值为 `date` 的 `name` 属性。
```js
assert($('input').attr('name') == 'date');
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/avoid-colorblindness-issues-by-carefully-choosing-colors-that-convey-information.md b/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/avoid-colorblindness-issues-by-carefully-choosing-colors-that-convey-information.md
index 5e2410ee994..41348ab426f 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/avoid-colorblindness-issues-by-carefully-choosing-colors-that-convey-information.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/avoid-colorblindness-issues-by-carefully-choosing-colors-that-convey-information.md
@@ -1,6 +1,7 @@
---
id: 587d778f367417b2b2512aad
-title: 考虑色盲用户的需求仔细选择传达信息的颜色
+title: >-
+ 考虑色盲用户的需求仔细选择传达信息的颜色
challengeType: 0
videoUrl: 'https://scrimba.com/c/c437as3'
forumTopicId: 301011
@@ -10,15 +11,15 @@ dashedName: >-
# --description--
-色盲的形式有很多种，它的表现可以从对特定波长光波的感知度较低，到完全无法看到颜色。最常见的形式是对绿色的低感知度。
+色盲的形式有很多种， 它的表现可以从对特定波长光波的感知度较低，到完全无法看到颜色。 最常见的形式是对绿色的低感知度。
-例如，如果内容的前景色与背景色是两种相近的绿色，那么色盲用户可能会无法区分它们。色轮上距离较近的颜色，特别是相邻的颜色，看起来都会很难区分。在表示重要信息的时候应避免使用这类相近颜色的组合。
+例如，如果内容的前景色与背景色是两种相近的绿色，那么色盲用户可能会无法区分它们。 色轮上距离较近的颜色，特别是相邻的颜色，看起来都会很难区分。 在表示重要信息的时候应避免使用这类相近颜色的组合。
-**注意：**一些在线颜色选择器有色盲模拟功能，可以模拟颜色在不同形式的色盲中所呈现的效果。它们和在线对比度检查器一样，都是很好的工具。
+**注意：**一些在线颜色选择器有色盲模拟功能，可以模拟颜色在不同形式的色盲中所呈现的效果。 它们和在线对比度检查器一样，都是很好的工具。
# --instructions--
-Camper Cat 正在测试一个重要按钮的不同样式。在色轮上，用于背景色的黄色（`#FFFF33`）和文字颜色绿色（`#33FF33`）是相邻的色调，一些色盲用户几乎无法区分它们，而且这两个颜色的亮度相近，对比度太小。为了解决这两个问题，请将文本的 `color` 修改为深蓝色（`#003366`）。
+Camper Cat 正在测试一个重要按钮的不同样式。 在色轮上，用于 `background-color` 的黄色（`#FFFF33`）和用于 `color` 的绿色（`#33FF33`）是相邻的色调，一些色盲用户几乎无法区分它们 （而且这两个颜色的亮度相近，对比度太小。） 为了解决这两个问题，请将文本的 `color` 修改为深蓝色（`#003366`）。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/avoid-colorblindness-issues-by-using-sufficient-contrast.md b/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/avoid-colorblindness-issues-by-using-sufficient-contrast.md
index b8f0d1a6c24..1d0981a37eb 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/avoid-colorblindness-issues-by-using-sufficient-contrast.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/avoid-colorblindness-issues-by-using-sufficient-contrast.md
@@ -9,17 +9,17 @@ dashedName: avoid-colorblindness-issues-by-using-sufficient-contrast
# --description--
-颜色是可视化设计的重要组成部分，但是使用颜色也引入了两个可访问性问题。首先，不能仅仅使用颜色作为传达重要信息的唯一方式，因为屏幕阅读器无法获取这些信息。其次，前景色与背景色需要有足够的对比度，这样色盲用户才可以区分它们。
+颜色是可视化设计的重要组成部分，但是使用颜色也引入了两个可访问性问题。 首先，不能仅仅使用颜色作为传达重要信息的唯一方式，因为屏幕阅读器无法获取这些信息。 其次，前景色与背景色需要有足够的对比度，这样色盲用户才可以区分它们。
-在之前的挑战中，我们用文本备用方案解决了第一个问题。在上一个挑战中，我们使用对比度检测工具解决了第二个问题。WCAG 建议为颜色及灰度组合使用 4.5 : 1 的对比度。
+在之前的挑战中，我们用文本备用方案解决了第一个问题。 在上一个挑战中，我们使用对比度检测工具解决了第二个问题。 WCAG 建议为颜色及灰度组合使用 4.5:1 的对比度。
-色盲用户无法将一些颜色与另一些颜色区分出来，这通常是因为色调，有时候是因为亮度。你可能还记得，对比度是用前景色与背景色的相对亮度计算的。
+色盲用户无法将一些颜色与另一些颜色区分出来，这通常是因为色调，有时候是因为亮度。 你可能还记得，对比度是用前景色与背景色的相对亮度计算的。
-实践中，在对比度检测工具的帮助下，我们可以通过将较暗的颜色变暗、将较淡的颜色变淡的方法来使对比度达到 4.5 : 1。在色轮中，较暗的颜色通常是蓝色、紫色、洋红和红色，而较淡的颜色通常是橙色、黄色、绿色和蓝绿色。
+实践中，在对比度检测工具的帮助下，我们可以通过将较暗的颜色变暗、将较淡的颜色变淡的方法来使对比度达到 4.5:1。 在色轮中，较暗的颜色通常是蓝色、紫色、洋红和红色，而较淡的颜色通常是橙色、黄色、绿色和蓝绿色。
# --instructions--
-Camper Cat 正在尝试为他的博客文本与背景配置颜色。他目前使用的组合是绿色的 `background-color` 与栗色的 `color`，它们的对比度为 2.5 : 1。Camper Cat 使用了 CSS 的 `hsl()`（hsl 为 hue、saturation、lightness 的缩写）属性来设置颜色。这样，通过修改 `hsl()` 属性的第三个参数，我们可以很轻松地调整颜色的亮度。请将 `background-color` 的亮度从 35% 增加到 55%，将 `color` 的亮度从 20% 减少到 15%，这样可以使对比度达到 5.9 : 1。
+Camper Cat 正在尝试为他的博客文本与背景配置颜色。 他目前使用的组合是绿色的 `background-color` 与栗色的 `color`，它们的对比度为 2.5:1。 这样，通过修改 `hsl()` 属性的第三个参数，我们可以很轻松地调整颜色的亮度。 请将 `background-color` 的亮度从 35% 增加到 55%，将 `color` 的亮度从 20% 减少到 15%， 这样可以使对比度达到 5.9:1。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/give-links-meaning-by-using-descriptive-link-text.md b/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/give-links-meaning-by-using-descriptive-link-text.md
index c06520f3fbe..2546524cd50 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/give-links-meaning-by-using-descriptive-link-text.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/give-links-meaning-by-using-descriptive-link-text.md
@@ -9,17 +9,17 @@ dashedName: give-links-meaning-by-using-descriptive-link-text
# --description--
-屏幕阅读器用户可以选择其设备读取的内容的类型，包括表示“跳转到”（或“跳过”）的元素，跳转到主要内容，或者从标题中获取页面摘要。用户还可以选择只听取页面中的超链接内容。
+屏幕阅读器用户可以选择其设备读取的内容的类型， 包括表示“跳转到”（或“跳过”）的元素，跳转到主要内容，或者从标题中获取页面摘要。 用户还可以选择只听取页面中的超链接内容。
-屏幕阅读器通过阅读链接文本（即 `a` 标签的内容文本）来完成这个操作。如果我们只在链接中写上 "click here"（点击这里）或者 "read more"（阅读更多），显然帮助有限。相反地，我们应该在 `a` 标签中使用简洁的描述性语言来为用户提供更多信息。
+屏幕阅读器通过阅读链接文本（即 `a` 标签的内容文本）来完成这个操作。 如果我们只在链接中写上 "click here"（点击这里）或者 "read more"（阅读更多），显然帮助有限。 相反地，我们应该在 `a` 标签中使用简洁的描述性语言来为用户提供更多信息。
# --instructions--
-Camper Cat 在链接中使用的文本在脱离上下文的情况下，描述性不是很好。请修改 `a` 标签，将其包含的文本从 "click here" 改为 "information about batteries"。
+Camper Cat 在链接中使用的文本在脱离上下文的情况下，描述性不是很好。 请修改锚点标签（`a`），将其包含的文本从 `Click here` 改为 `information about batteries`。
# --hints--
-你应修改 `a` 标签，将其包含的文本从 "click here" 改为 "information about batteries"。
+应修改 `a` 标签，将其包含的文本从 `Click here` 改为 `information about batteries`。
```js
assert(
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/improve-accessibility-of-audio-content-with-the-audio-element.md b/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/improve-accessibility-of-audio-content-with-the-audio-element.md
index 0509c1c8e98..48a28e55b6c 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/improve-accessibility-of-audio-content-with-the-audio-element.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/improve-accessibility-of-audio-content-with-the-audio-element.md
@@ -9,9 +9,9 @@ dashedName: improve-accessibility-of-audio-content-with-the-audio-element
# --description--
-HTML5 的 `audio` 标签用于呈现音频内容或音频流，它也具有语义化特性。可以在 `audio` 上下文中为音频内容添加描述文字或字幕链接，使听觉障碍用户也能获取音频中的信息。
+HTML5 的 `audio` 标签用于呈现音频内容或音频流，它也具有语义化特性。 音频内容也需要备用文本，供聋哑人或听力困难的人使用。 这可以通过页面上的文本或与字幕链接来实现。
-`audio` 支持 `controls` 属性，可以使浏览器为音频提供具有开始、暂停等功能的播放控件，而且还支持键盘操作。`controls` 属性是一个布尔属性，即不需要提供属性值。只要这个属性出现在 `audio` 标签中，浏览器就会开启播放控件。
+`audio` 标签支持 `controls` 属性， 用于显示浏览器默认播放、停止和其他控制，以及支持键盘功能。 这是一个布尔值属性，意味着它不需要一个值，它在标签上存在即开启设置。
举个例子：
@@ -22,13 +22,13 @@ HTML5 的 `audio` 标签用于呈现音频内容或音频流，它也具有语
```
-**注意：**多媒体内容通常同时包含音频与视频部分，它需要同步的字幕与逐字稿，以使视觉或听觉障碍用户可以获取它的内容。一般情况下，网页开发者不负责创建字幕或逐字稿，但是需要将它们添加到多媒体中。
+**注意：**多媒体内容通常同时包含音频与视频部分， 它需要同步的字幕，使视觉或听觉障碍用户可以获取它的内容。 一般情况下，网页开发者不负责创建字幕或逐字稿，但是需要将它们添加到多媒体中。
# --instructions--
-是时候让 Camper Cat 休息一下，并与朋友 Zersiax (@zersiax) 会面。Zersiax 是一位屏幕阅读器用户，同时也是无障碍设计的高手。为了体验屏幕阅读器的朗读效果，请在 `p` 标签之后添加一个具有 `controls` 属性的 `audio` 标签。然后在 `audio` 标签内添加一个 `source` 标签并设置 `src` 属性值为 "`https://s3.amazonaws.com/freecodecamp/screen-reader.mp3`"。同时，请将 `source` 的 `type` 属性值设置为 "audio/mpeg"。
+是时候让 Camper Cat 休息一下，并与朋友 Zersiax (@zersiax) 会面了。 Zersiax 是一位屏幕阅读器用户，同时也是无障碍设计的高手。 为了体验屏幕阅读器的朗读效果，请在 `p` 标签之后添加一个 `audio` 标签， 具有 `controls` 属性。 然后在 `audio` 标签里面放一个带有 `src` 属性的 `source` 标签，属性值为 `https://s3.amazonaws.com/freecodecamp/screen-reader.mp3`。将 `type` 属性设置为 `"audio/mpeg"`.
-**注意：**音频片段的播放速度可能会快到另我们难以理解，但是对于屏幕阅读器用户来说这是正常速度。
+**注意：**音频片段的播放速度可能会快到令我们难以理解，但是对于屏幕阅读器用户来说这是正常速度。
# --hints--
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/improve-chart-accessibility-with-the-figure-element.md b/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/improve-chart-accessibility-with-the-figure-element.md
index b319a534013..fb8bc80aaf8 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/improve-chart-accessibility-with-the-figure-element.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/improve-chart-accessibility-with-the-figure-element.md
@@ -9,9 +9,9 @@ dashedName: improve-chart-accessibility-with-the-figure-element
# --description--
-HTML5 引入了 `figure` 标签以及与之相关的 `figcaption` 标签。它们一起用于展示可视化信息（如：图片、图表）及其标题。通过语义化对内容进行分组并配以用于解释 `figure` 的文字，可以极大地提升内容的可访问性。
+HTML5 引入了 `figure` 标签以及与之相关的 `figcaption` 标签。 它们一起用于展示可视化信息（如：图片、图表）及其标题。 通过语义化对内容进行分组并配以用于解释 `figure` 的文字，可以极大地提升内容的可访问性。
-对于图表之类的可视化数据，标题可以为屏幕阅读器用户提供简要的说明。但是这里有一个难点，如何为屏幕阅读器用户展示那些超出屏幕可视范围（使用 CSS）的表格所表现的图表数据。
+对于图表之类的可视化数据，标题可以为屏幕阅读器用户提供简要的说明。 但是这里有一个难点，如何为屏幕阅读器用户展示那些超出屏幕可视范围（使用 CSS）的表格所表现的图表数据。
举个例子，注意 `figcaption` 包含在 `figure` 标签中，并且可以与其他标签组合使用：
@@ -27,17 +27,17 @@ HTML5 引入了 `figure` 标签以及与之相关的 `figcaption` 标签。它
# --instructions--
-Camper Cat 正在努力创建一张条形图，用来显示每周用于隐形、战斗、武器训练的时间。请帮助完善他的页面，将他用于呈现图表的 `div` 标签修改为 `figure` 标签；将用于呈现图表标题的 `p` 标签改为 `figcaption` 标签。
+Camper Cat 正在努力创建一张条形图，用来显示每周用于隐形、战斗、武器训练的时间。 请帮助完善他的页面，将他用于呈现图表的 `div` 标签修改为 `figure` 标签；将用于呈现图表标题的 `p` 标签改为 `figcaption` 标签。
# --hints--
-应存在 1 个 `figure` 标签。
+应存在一个 `figure` 标签。
```js
assert($('figure').length == 1);
```
-应存在 1 个 `figcaption` 标签。
+应存在一个 `figcaption` 标签。
```js
assert($('figcaption').length == 1);
diff --git a/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/improve-form-field-accessibility-with-the-label-element.md b/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/improve-form-field-accessibility-with-the-label-element.md
index 38d7899fa48..1dfe460a65a 100644
--- a/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/improve-form-field-accessibility-with-the-label-element.md
+++ b/curriculum/challenges/chinese/01-responsive-web-design/applied-accessibility/improve-form-field-accessibility-with-the-label-element.md
@@ -9,13 +9,13 @@ dashedName: improve-form-field-accessibility-with-the-label-element
# --description--
-合理地使用语义化 HTML 标签和属性可以提升页面的可访问性。在接下来的挑战中，你将会看到在表单中使用属性的场景。
+合理地使用语义化 HTML 标签和属性可以提升页面的可访问性。 在接下来的挑战中，你将会看到在表单中使用属性的场景。
-`label` 标签的文本内容通常会是表单组件的名称或标签。这些文本表明了组件的意义，也提升了表单的可访问性。`label` 标签的 `for` 属性将标签与表单组件绑定；同时，屏幕阅读器也会读取 `for` 属性的属性值。
+`label` 标签的文本内容通常会是表单组件的名称或标签。 这些文本表明了组件的意义，也提升了表单的可访问性。 `label` 标签的 `for` 属性将标签与表单组件绑定；同时，屏幕阅读器也会读取 `for` 属性的属性值。
-在 HTML 基础章节中，我们已经学习使用了单选按钮标签。在那次挑战中，为了让标签可以在点击的时候也选中输入框，我们将 `input` 标签嵌套在了 `label` 标签里面。在本节课程中，我们会了解到另外一种实现这个功能的方法，那就是使用 `for` 属性。
+在 HTML 基础章节中，我们已经学习使用了单选按钮标签。 在那个挑战中，为了让标签可以在点击的时候也选中输入框，我们将单选按钮 input 标签嵌套在了 `label` 标签里面。 在本节课程中，我们会了解到另外一种实现这个功能的方法，那就是使用 `for` 属性。
-`for` 的属性值必须与表单组件的 `id` 属性值相同。举个例子：
+`for` 的属性值必须与表单组件的 `id` 属性值相同。 举个例子：
```html