1390. Four Divisors

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

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给定一个数组，问其中哪些数只有4个因子，把只有4个因子的数对应的所有因子加起来。

也是很无聊的一个题，按题意照做即可，注意求解因子的时候只需要循环到sqrt(v)即可，因为如果一个因子大于sqrt(v)的话，其对应的另一个因子肯定小于sqrt(v)，也就是之前已经遍历过了。

完整代码如下：

class Solution {
public:
	void FindDivisors(int v, vector<int>& ans) {
		int mid = sqrt(v);
		for (int i = 1; i <= mid; ++i) {
			if (v%i == 0) {
				ans.push_back(i);
				if (v / i != i)ans.push_back(v / i);
			}
		}
	}
	int sumFourDivisors(vector<int>& nums) {
		int ans = 0;
		for (int i = 0; i < nums.size(); ++i) {
			vector<int> divs;
			FindDivisors(nums[i], divs);
			if (divs.size() == 4) {
				for (int j = 0; j < divs.size(); ++j)ans += divs[j];
			}
		}
		return ans;
	}
};

本代码提交AC，用时440MS。