freeCodeCamp/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-9-special-pythagorean-triplet.md

---
id: 5900f3761000cf542c50fe88
title: 问题9：特殊的毕达哥拉斯三重奏
challengeType: 5
videoUrl: ''
dashedName: problem-9-special-pythagorean-triplet
---

# --description--

毕达哥拉斯三元组是一组三个自然数， `a` < `b` < `c` ，其中，

`a` 

<sup>2</sup>

-   `b` 

<sup>2</sup>

 = `c` 

<sup>2</sup>

例如，3 

<sup>2</sup>

-   4 

<sup>2</sup>

 = 9 + 16 = 25 = 5 

<sup>2</sup>

 。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` 。

# --hints--

`specialPythagoreanTriplet(1000)`应返回31875000。

```js
assert.strictEqual(specialPythagoreanTriplet(1000), 31875000);
```

`specialPythagoreanTriplet(24)`应该返回480。

```js
assert.strictEqual(specialPythagoreanTriplet(24), 480);
```

`specialPythagoreanTriplet(120)`应该返回49920。

```js
assert([49920, 55080, 60000].includes(specialPythagoreanTriplet(120)));
```

# --seed--

## --seed-contents--

```js
function specialPythagoreanTriplet(n) {
 let sumOfabc = n;

 return true;
}

specialPythagoreanTriplet(1000);
```

# --solutions--

```js
const specialPythagoreanTriplet = (n)=>{
 let sumOfabc = n;
 let a,b,c;
 for(a = 1; a<=sumOfabc/3; a++){
 for(b = a+1; b<=sumOfabc/2; b++){
 c = Math.sqrt(a*a+b*b);
 if((a+b+c) == sumOfabc){
 return a*b*c;
 }
 }
 }
}
```