2018-10-10 18:03:03 -04:00
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---
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id: 5900f5201000cf542c510032
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2020-12-16 00:37:30 -07:00
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title: 问题435:斐波那契数的多项式
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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-435-polynomials-of-fibonacci-numbers
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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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斐波纳契数{fn,n≥0}被递归定义为fn = fn-1 + fn-2,基本情况为f0 = 0和f1 = 1。
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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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对于0≤i≤n,将多项式{Fn,n≥0}定义为Fn(x)= ∑fixi。
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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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例如,F7(x)= x + x2 + 2x3 + 3x4 + 5x5 + 8x6 + 13x7,而F7(11)= 268357683。
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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n =1015。求和\[∑0≤x≤100 Fn(x)] mod 1307674368000(= 15!)。
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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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`euler435()`252541322550。
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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(euler435(), 252541322550);
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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 euler435() {
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return true;
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}
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euler435();
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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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