<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) {
        int cnt=0;
        for(int i=0;i<nums.size();i++){
            int c=0;
            int t=0;
            int s=1+nums[i];
            for(int j=2;j<nums[i];j++){
                if(nums[i]%j==0){
                    if(c<2){
                        s+=j;
                        c+=1;
                        }
                    else{
                        t=1;
                        break;
                    }
                }
            }
            if(t==0 and c==2){
                cnt+=s;
                cout<<nums[i]<<" ";
            }
        }
        return cnt;
    }
};
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