freeCodeCamp/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-12-highly-divisible-triangular-number.md

---
id: 5900f3781000cf542c50fe8b
title: 问题12：高度可分的三角数
challengeType: 5
videoUrl: ''
dashedName: problem-12-highly-divisible-triangular-number
---

# --description--

通过添加自然数生成三角数的序列。所以第7个三角形数字是1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.前十个术语是：

1,3,6,10,15,21,28,36,45,55 ......

让我们列出前七个三角形数字的因子：

**1：** 1

**3：** 1,3

**6：** 1,2,3,6

**10：** 1,2,5,10

**15：** 1,3,5,15

**21：** 1,3,7,21

**28：** 1,2,4,7,14,28

我们可以看到28是第一个超过五个除数的三角形数。超过`n`除数的第一个三角形数的值是多少？

# --hints--

`divisibleTriangleNumber(5)`应该返回28。

```js
assert.strictEqual(divisibleTriangleNumber(5), 28);
```

`divisibleTriangleNumber(23)`应该返回630。

```js
assert.strictEqual(divisibleTriangleNumber(23), 630);
```

divisibleTriangleNumber `divisibleTriangleNumber(167)`应该返回1385280。

```js
assert.strictEqual(divisibleTriangleNumber(167), 1385280);
```

divisibleTriangleNumber `divisibleTriangleNumber(374)`应该返回17907120。

```js
assert.strictEqual(divisibleTriangleNumber(374), 17907120);
```

divisibleTriangleNumber `divisibleTriangleNumber(500)`应该返回76576500。

```js
assert.strictEqual(divisibleTriangleNumber(500), 76576500);
```

# --seed--

## --seed-contents--

```js
function divisibleTriangleNumber(n) {

  return true;
}

divisibleTriangleNumber(500);
```

# --solutions--

```js
function divisibleTriangleNumber(n) {
  if (n === 1) return 3;
  let counter = 1;
  let triangleNumber = counter++;


 while (noOfFactors(triangleNumber) < n) {
   triangleNumber += counter++;
 }
return triangleNumber;
}

function noOfFactors(num) {
  const primeFactors = getPrimeFactors(num);
  let prod = 1;
  for(let p in primeFactors) {
    prod *= (primeFactors[p] + 1)
  }
  return prod;
}

function getPrimeFactors(num) {
  let n = num;
  let primes = {};

  let p = 2;
  let sqrt = Math.sqrt(num);

  function checkAndUpdate(inc) {
    if (n % p === 0) {
      const curr = primes[p];
      if (curr) {
        primes[p]++
      } else {
        primes[p] = 1;
      }
      n /= p;
    } else {
      p += inc;
    }
  }

  while(p === 2 && p <= n) {
    checkAndUpdate(1);
  }

  while (p <= n && p <= sqrt) {
    checkAndUpdate(2);
  }
  if(Object.keys(primes).length === 0) {
    primes[num] = 1;
  } else if(n !== 1) {
    primes[n] = 1;
  }
  return primes;
}
```