Geeksforgeeks Solution For " Sum of divisors "

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Problem :- Given a natural number, calculate sum of all its proper divisors. A proper divisor of a natural number is the divisor that is strictly less than the number.

For example, number 20 has 5 proper divisors: 1, 2, 4, 5, 10, and the divisor summation is: 1 + 2 + 4 + 5 + 10 = 22.

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Solution :- 

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,i,res=0;
        cin>>n;
        for(i=1;i<=n/2;i++)
        {
            if(n%i==0)
            {
                res+=i;
            }
        }
        cout<<res<<endl;
    }
    return 0;
}

Output:-