2018-10-10 18:03:03 -04:00
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id: 5900f3ac1000cf542c50febf
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2020-12-16 00:37:30 -07:00
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title: 问题64:奇数期平方根
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2018-10-10 18:03:03 -04:00
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challengeType: 5
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videoUrl: ''
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2021-01-13 03:31:00 +01:00
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dashedName: problem-64-odd-period-square-roots
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2018-10-10 18:03:03 -04:00
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---
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# --description--
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2018-10-10 18:03:03 -04:00
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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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2020-08-13 17:24:35 +02:00
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2021-01-13 03:31:00 +01:00
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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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2021-01-13 03:31:00 +01:00
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```js
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// solution required
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```
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