2018-10-10 18:03:03 -04:00
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---
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id: 5900f52c1000cf542c51003f
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2020-12-16 00:37:30 -07:00
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title: 问题448:平均最小公倍数
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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-448-average-least-common-multiple
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2018-10-10 18:03:03 -04:00
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---
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2020-12-16 00:37:30 -07:00
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# --description--
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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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。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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令S(n)=ΣA(k)为1≤k≤n。 S(100)= 122726。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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找到S(99999999019)mod 999999017。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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# --hints--
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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`euler448()`应该返回106467648。
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2018-10-10 18:03:03 -04:00
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```js
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2020-12-16 00:37:30 -07:00
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assert.strictEqual(euler448(), 106467648);
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2018-10-10 18:03:03 -04:00
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```
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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 euler448() {
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return true;
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}
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euler448();
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```
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2020-12-16 00:37:30 -07:00
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# --solutions--
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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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```js
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// solution required
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```
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