<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>


int m=1e5;
vector<int> prime(m+1,1);
vector<int> res(m+1,0);

int main2 = [] {
    ios::sync_with_stdio(0); cin.tie(0);
	prime[0]=prime[1]=0;
	for(int i=2;i*i<=m;i++){
		if(prime[i]==1){
			prime[i*i]=0;
			for(int j=i*i+i;j<=m;j+=i) {
				if(prime[j]==1) prime[j]=i;
			}
		}
	}
	for(int i=2;i<=m;i++){
		if(prime[i]!=1 and prime[i]!=0){
			int num=(i/prime[i]);
			int sum=0;
			if(prime[num]==1 or prime[i]*prime[i]==num){
				sum=1+num+i+prime[i];
				res[i]=sum;
			}
		}
	}
    return 0;
}();

class Solution {
public:
    int sumFourDivisors(vector<int>& nums) {

    int n=nums.size();
	int sum=0;
    for(int i=0;i<n;i++){
    	sum+=res[nums[i]];
    }


    return sum;
}
};

16000