<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 LIMIT = 100005;
    int primes[100005];
    void eratos() {
        for(int i = 1; i < LIMIT; i++) {
            primes[i] = 1;
        }

        for(int i = 2; i * i < LIMIT; i++) {
            if(primes[i] == 0) continue;
            for(int j = i * i; j < LIMIT; j+=i) primes[j] = 0;
        }
    }

    int solve(unordered_map<int, int>& um) {
        int sum = 0;
        for(int i = 2; i * i < LIMIT; i++) {
            if(primes[i] == 0) continue;
            for(int j = i * i; j < LIMIT; j+=i) {
                if(um.find(j) != um.end()) {
                    if((primes[j/i] && j/i != i) || (primes[j/i] == 0 && j/i == (i*i)) ) {
                        cout<<j<<", i:" <<i<<", j:"<<j/i<<endl;
                        sum += (1 + j + i + j/i) * um[j];
                    }
                }
            }
        }

        return sum;
    }

    int sumFourDivisors(vector<int>& nums) {
        eratos();
        unordered_map<int, int> um;
        for(int i = 0; i < nums.size(); i++) {
            um[nums[i]]++;
        }
        return solve(um);
    }
};
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