60 lines
916 B
Markdown
60 lines
916 B
Markdown
---
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id: 5900f3761000cf542c50fe88
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title: 问题9：特殊的毕达哥拉斯三重奏
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challengeType: 5
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videoUrl: ''
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---
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# --description--
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毕达哥拉斯三元组是一组三个自然数， `a` < `b` < `c` ，其中，
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`a`
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<sup>2</sup>
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- `b`
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<sup>2</sup>
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= `c`
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<sup>2</sup>
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例如，3
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<sup>2</sup>
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- 4
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<sup>2</sup>
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= 9 + 16 = 25 = 5
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<sup>2</sup>
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。恰好存在一个毕达哥拉斯三元组，其中`a` + `b` + `c` = 1000.求产品`abc`使得`a` + `b` + `c` = `n` 。
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# --hints--
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`specialPythagoreanTriplet(1000)`应返回31875000。
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```js
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assert.strictEqual(specialPythagoreanTriplet(1000), 31875000);
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```
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`specialPythagoreanTriplet(24)`应该返回480。
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```js
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assert.strictEqual(specialPythagoreanTriplet(24), 480);
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```
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`specialPythagoreanTriplet(120)`应该返回49920。
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```js
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assert([49920, 55080, 60000].includes(specialPythagoreanTriplet(120)));
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```
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# --solutions--
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