<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>

auto _ = [](){ios_base::sync_with_stdio(false); cin.tie(NULL);};
class Solution {
    
public:
    int sumFourDivisors(vector<int>& nums) {
        int sum = 0, n = 100000;
        vector<int> lp(n+1), pr;
  for (int i = 2; i <= n; ++i)
  {
    if (!lp[i])
    {
      lp[i] = i;
      pr.push_back(i);
    }
    for (int j = 0; i*pr[j] <= n; ++j)
    {
      int p = pr[j];
      lp[i*p] = p;
      if (p == lp[i]) break;
    }
  }
        int p, q;
        for (auto x : nums) {
            if (x == 1) continue;
            p = lp[x];
            q = lp[x/p];
            if (p*q == x && p != q) sum += 1+p+q+x;
            else if (p <= 43 && p*p*p == x) sum += ((1+p)*p+1)*p+1;
        }
        return sum;
    }
};
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