<p>Given an integer array <code>nums</code>, return <em>the sum of divisors of the integers in that array that have exactly four divisors</em>. If there is no such integer in the array, return <code>0</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [21,4,7]
<strong>Output:</strong> 32
<strong>Explanation:</strong> 
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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [21,21]
<strong>Output:</strong> 64
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    int sumFourDivisors(vector<int>& nums) {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        int n = *max_element(nums.begin(), nums.end());
        vector<int> num_factors(n+1, 1);
        vector<long long> factors(n+1, 1);
        for(int i=2; i <= n; i++){
            factors[i] += i;
            num_factors[i]++;
            for(int j = 2*i; j <= n; j+=i){
                factors[j] += i;
                num_factors[j]++;
            }
        }
        int ans = 0;
        for(int x: nums){
            if(num_factors[x] == 4){
                ans += factors[x];
            }
        }
        return ans;
    }
};
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