56 lines
1.6 KiB
Markdown
56 lines
1.6 KiB
Markdown
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---
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id: 5900f41c1000cf542c50ff2e
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challengeType: 5
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title: 'Problem 175: Fractions involving the number of different ways a number can be expressed as a sum of powers of 2'
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videoUrl: ''
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localeTitle: 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
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---
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## Description
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<section id="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。 <p>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 </p><p>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。 </p></section>
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## Instructions
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<section id="instructions">
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</section>
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## Tests
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<section id='tests'>
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```yml
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tests:
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- text: '<code>euler175()</code>应该返回1,13717420,8。'
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testString: 'assert.strictEqual(euler175(), 1, 13717420, 8, "<code>euler175()</code> should return 1, 13717420, 8.");'
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```
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</section>
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## Challenge Seed
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<section id='challengeSeed'>
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<div id='js-seed'>
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```js
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function euler175() {
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// Good luck!
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return true;
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}
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euler175();
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```
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</div>
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</section>
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## Solution
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<section id='solution'>
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```js
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// solution required
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```
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</section>
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