2018-10-10 18:03:03 -04:00
|
|
|
|
---
|
|
|
|
|
id: 5900f4cb1000cf542c50ffdd
|
2020-12-16 00:37:30 -07:00
|
|
|
|
title: 问题350:约束最小和最小
|
2018-10-10 18:03:03 -04:00
|
|
|
|
challengeType: 5
|
|
|
|
|
videoUrl: ''
|
2021-01-13 03:31:00 +01:00
|
|
|
|
dashedName: problem-350-constraining-the-least-greatest-and-the-greatest-least
|
2018-10-10 18:03:03 -04:00
|
|
|
|
---
|
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
# --description--
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
大小为n的列表是n个自然数的序列。实例是(2,4,6),(2,6,4),(10,6,15,6)和(11)。
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
列表的最大公约数(或gcd)是划分列表中所有条目的最大自然数。例子:gcd(2,6,4)= 2,gcd(10,6,15,6)= 1,gcd(11)= 11。
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
列表的最小公倍数或lcm是列表的每个条目可分割的最小自然数。实例:1cm(2,6,4)= 12,1cm(10,6,15,6)= 30和1cm(11)= 11。
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
设f(G,L,N)为大小为N的列表数,gcd≥G且lcm≤L。例如:
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
f(10,100,1)= 91.f(10,100,2)= 327.f(10,100,3)= 1135.f(10,100,1000)mod 1014 = 3286053。
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
找到f(106,1012,1018)mod 1014。
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
# --hints--
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
`euler350()`应该返回84664213。
|
2018-10-10 18:03:03 -04:00
|
|
|
|
|
|
|
|
|
```js
|
2020-12-16 00:37:30 -07:00
|
|
|
|
assert.strictEqual(euler350(), 84664213);
|
2018-10-10 18:03:03 -04:00
|
|
|
|
```
|
2020-08-13 17:24:35 +02:00
|
|
|
|
|
2021-01-13 03:31:00 +01:00
|
|
|
|
# --seed--
|
|
|
|
|
|
|
|
|
|
## --seed-contents--
|
|
|
|
|
|
|
|
|
|
```js
|
|
|
|
|
function euler350() {
|
|
|
|
|
|
|
|
|
|
return true;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
euler350();
|
|
|
|
|
```
|
|
|
|
|
|
2020-12-16 00:37:30 -07:00
|
|
|
|
# --solutions--
|
|
|
|
|
|
2021-01-13 03:31:00 +01:00
|
|
|
|
```js
|
|
|
|
|
// solution required
|
|
|
|
|
```
|