【leetcode】1390. Four Divisors

题目如下：

Given an integer array nums, return the sum of divisors of the integers in that array that have exactly four divisors.

If there is no such integer in the array, return 0.

Example 1:

Input: nums = [21,4,7]
Output: 32
Explanation:
21 has 4 divisors: 1, 3, 7, 21
4 has 3 divisors: 1, 2, 4
7 has 2 divisors: 1, 7
The answer is the sum of divisors of 21 only.

Constraints:

• 1 <= nums.length <= 10^4
• 1 <= nums[i] <= 10^5

解题思路：把nums中每个元素有几个除数都算出来就好了。

代码如下：

class Solution(object):
    def sumFourDivisors(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        import math
        res = 0
        for num in nums:
            div = []
            for i in range(1,int(math.sqrt(num))+1):
                if num % i == 0 and i not in div:
                    div.append(i)
                    if num / i != i:div.append(num/i)
                if len(div) > 4:break
            if len(div) == 4:res += sum(div)
        return res