ee1e8abd87
* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
91 lines
1.8 KiB
Markdown
91 lines
1.8 KiB
Markdown
---
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id: 5900f3ac1000cf542c50febf
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title: 问题64：奇数期平方根
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challengeType: 5
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videoUrl: ''
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dashedName: problem-64-odd-period-square-roots
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---
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# --description--
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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：
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√N= a0 + 1
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a1 + 1
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a2 + 1
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a3 + ......
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例如，让我们考虑√23：
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√23= 4 +√23 - 4 = 4 + 1 = 4 + 1
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1√23-4
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1 +√23 - 37
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如果我们继续，我们将得到以下扩展：
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√23= 4 + 1
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1 + 1
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3 + 1
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1 + 1
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8 + ......
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该过程可归纳如下：
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a0 = 4，
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1√23-4=√23+ 47 = 1 +√23-37a1 = 1，
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7√23-3= 7（√23+ 3）14 = 3 +√23-32a2= 3，
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2√23-3= 2（√23+ 3）14 = 1 +√23-47a3 = 1，
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7√23-4= 7（√23+ 4）7 = 8 +√23-4a4= 8，
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1√23-4=√23+ 47 = 1 +√23-37a5 = 1，
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7√23-3= 7（√23+ 3）14 = 3 +√23-32a6= 3，
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2√23-3= 2（√23+ 3）14 = 1 +√23-47a7 = 1，
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7√23-4= 7（√23+ 4）7 = 8 +√23-4
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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。
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（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？
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# --hints--
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`euler64()`应返回1322。
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```js
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assert.strictEqual(euler64(), 1322);
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```
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# --seed--
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## --seed-contents--
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```js
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function oddPeriodSqrts() {
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return true;
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}
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oddPeriodSqrts();
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```
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# --solutions--
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```js
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// solution required
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```
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