BOJ 6975 - Deficient, Perfect, and Abundant [C++]

문제

링크

풀이

#include <iostream>
using namespace std;
 
void solve(void) {
  int n; cin >> n;
 
  int ans = 0;
  for (int i=1; i<=n-1; i++) {
    if (!(n % i)) ans += i;
  }
 
  if (ans < n) cout << n << " is a deficient number.";
  else if (ans == n) cout << n << " is a perfect number.";
  else cout << n << " is an abundant number.";
  cout << "\n\n";
}
 
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
 
  int t; cin >> t;
  while (t--) solve();
  return 0;
}