A proper divisor, or proper factor, of a natural number, is a factor of a number other than 1 and the number itself. For example, the proper divisors of 6 are 1, 2, and 3.
Given a positive integer, find its sum of proper divisors.
Sample Inputs/Outputs
Sample Input(s) | Sample Output(s) |
---|---|
The input consists of only one line containing a positive integer N (1 ≤ N ≤ 65535). | Output the sum of the proper divisors of N. |
6 | 6 |
12 | 16 |
Thought Process
Use a for loop from 1 to N - 1 (excluding N itself). Since the number is not large, you can directly iterate through all the divisors of N and sum them up. Finally, output the sum of the divisors.
Sample Code-ZeroJudge I025: Sum of Proper Divisor
#include <iostream>
using namespace std;
int main() {
cin.sync_with_stdio(0);
cin.tie(0);
int N;
cin >> N;
int ans = 0;
for (int i = 1; i<N; i++)
{
if (N % i == 0) ans += i;
}
cout << ans << "\n";
}
//Z.O.J. I025
//Dr. SeanXD