AOJ2101 Perfect Number
#include<iostream> using namespace std; int main(){ int n, f; while(cin >> n, n != 0){ f = 0; for(int i = 1; i * i <= n; i++){ if(n % i ==0){ f += i + n / i; if(n == i * i){ f -= i; } } } if(f > n * 2)cout << "abundant number" << endl; else if(f < n * 2)cout << "deficient number" << endl; else cout << "perfect number" << endl; } return 0; }