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curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/hofstadter-q-sequence.md

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+---
+title: Hofstadter Q sequence
+id: 59637c4d89f6786115efd814
+challengeType: 5
+videoUrl: ''
+localeTitle: Hofstadter Q序列
+---
+
+## Description
+<section id="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&gt; 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 </section>
+
+## Instructions
+<section id="instructions">
+</section>
+
+## Tests
+<section id='tests'>
+
+```yml
+tests:
+  - text: <code>hofstadterQ</code>是一个函数。
+    testString: assert(typeof hofstadterQ === 'function');
+  - text: <code>hofstadterQ()</code>应该返回<code>integer</code>
+    testString: assert(Number.isInteger(hofstadterQ(1000)));
+  - text: <code>hofstadterQ(1000)</code>应该返回<code>502</code>
+    testString: assert.equal(hofstadterQ(testCase[0]), res[0]);
+  - text: <code>hofstadterQ(1500)</code>应该返回<code>755</code>
+    testString: assert.equal(hofstadterQ(testCase[1]), res[1]);
+  - text: <code>hofstadterQ(2000)</code>应该返回<code>1005</code>
+    testString: assert.equal(hofstadterQ(testCase[2]), res[2]);
+  - text: <code>hofstadterQ(2500)</code>应该返回<code>1261</code>
+    testString: assert.equal(hofstadterQ(testCase[3]), res[3]);
+
+```
+
+</section>
+
+## Challenge Seed
+<section id='challengeSeed'>
+
+<div id='js-seed'>
+
+```js
+function hofstadterQ (n) {
+  // Good luck!
+  return n;
+}
+
+```
+
+</div>
+
+
+### After Test
+<div id='js-teardown'>
+
+```js
+console.info('after the test');
+```
+
+</div>
+
+</section>
+
+## Solution
+<section id='solution'>
+
+```js
+// solution required
+```
+
+/section>