ee1e8abd87
* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
39 lines
887 B
Markdown
39 lines
887 B
Markdown
---
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id: 5900f3ae1000cf542c50fec1
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title: 问题66：丢番图方程
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challengeType: 5
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videoUrl: ''
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dashedName: problem-66-diophantine-equation
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---
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# --description--
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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。
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# --hints--
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`euler66()`应返回661。
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```js
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assert.strictEqual(euler66(), 661);
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```
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# --seed--
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## --seed-contents--
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```js
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function diophantineEquation() {
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return true;
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}
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diophantineEquation();
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```
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# --solutions--
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```js
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// solution required
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```
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