Perfect Number Problem & Solution

A perfect number is a positive integer that is equal to the sum of its positive divisors, excluding the number itself. A divisor of an integer x is an integer that can divide x evenly.

Given an integer n, return true if n is a perfect number, otherwise return false.

See the perfect number problem on LeetCode.

C++ Solution

#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")

static const int _=[](){ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();

class Solution {
  bool checkPerfectNumber(int num) {
    if (num == 1) {
      return false;

    int sum = 0;

    // To avoid exceeding the time limit, go up to `sqrt(num)`.
    for (int i = 2; i <= sqrt(num); ++i) {
      if (num % i == 0) {

        // Adds up both the divisor and the result at the same time.
        sum += i + num / i;

    return sum + 1 == num;