2018-10-10 22:03:03 +00:00
|
|
|
|
---
|
|
|
|
|
id: 5900f3ae1000cf542c50fec1
|
2020-12-16 07:37:30 +00:00
|
|
|
|
title: 问题66:丢番图方程
|
2018-10-10 22:03:03 +00:00
|
|
|
|
challengeType: 5
|
|
|
|
|
videoUrl: ''
|
2021-01-13 02:31:00 +00:00
|
|
|
|
dashedName: problem-66-diophantine-equation
|
2018-10-10 22:03:03 +00:00
|
|
|
|
---
|
|
|
|
|
|
2020-12-16 07:37:30 +00:00
|
|
|
|
# --description--
|
2018-10-10 22:03:03 +00:00
|
|
|
|
|
2020-12-16 07:37:30 +00:00
|
|
|
|
考虑形式的二次丢番图方程:x2 - Dy2 = 1例如,当D = 13时,x中的最小解是6492 - 13×1802 = 1.可以假设当D是正整数时没有解广场。通过在D中找到D = {2,3,5,6,7}的最小解,我们得到以下结果:32 - 2×22 = 1 22 - 3×12 = 192 - 5×42 = 1 52 - 6× 22 = 1 82 - 7×32 = 1因此,通过考虑D中对于D≤7的最小解,当D = 5时获得最大的x。在x的最小解中找到D≤1000的值,其中获得x的最大值。
|
2018-10-10 22:03:03 +00:00
|
|
|
|
|
2020-12-16 07:37:30 +00:00
|
|
|
|
# --hints--
|
2018-10-10 22:03:03 +00:00
|
|
|
|
|
2020-12-16 07:37:30 +00:00
|
|
|
|
`euler66()`应返回661。
|
2018-10-10 22:03:03 +00:00
|
|
|
|
|
|
|
|
|
```js
|
2020-12-16 07:37:30 +00:00
|
|
|
|
assert.strictEqual(euler66(), 661);
|
2018-10-10 22:03:03 +00:00
|
|
|
|
```
|
|
|
|
|
|
2021-01-13 02:31:00 +00:00
|
|
|
|
# --seed--
|
|
|
|
|
|
|
|
|
|
## --seed-contents--
|
|
|
|
|
|
|
|
|
|
```js
|
|
|
|
|
function diophantineEquation() {
|
|
|
|
|
|
|
|
|
|
return true;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
diophantineEquation();
|
|
|
|
|
```
|
|
|
|
|
|
2020-12-16 07:37:30 +00:00
|
|
|
|
# --solutions--
|
2020-08-13 15:24:35 +00:00
|
|
|
|
|
2021-01-13 02:31:00 +00:00
|
|
|
|
```js
|
|
|
|
|
// solution required
|
|
|
|
|
```
|