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.

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:-