Perfect Number

/**
  *Euclid proved that 2p−1(2p−1) is an even perfect number whenever 2p−1 is prime, where ppp is prime.
  *
  *For example, the first four perfect numbers are generated by the formula 2p−1(2p−1), with ppp a prime number, as follows:
  *
  *for p = 2:   21(22 − 1) = 6
  *for p = 3:   22(23 − 1) = 28
  *for p = 5:   24(25 − 1) = 496
  *for p = 7:   26(27 − 1) = 8128.
  *
  *Prime numbers of the form 2p−1 are known as Mersenne primes. For 2p−1 to be prime, it is necessary that ppp itself be prime.
  *However, not all numbers of the form 2p−1 with a prime ppp are prime; for example, 211−1=2047=23×89 is not a prime number.
  *
  *You can see that for small value of ppp, its related perfect number goes very high. So, we need to evaluate perfect numbers
  *for some primes (2,3,5,7,13,17,19,31) only, as for bigger prime its perfect number will not fit in 64 bits.
*/

class Solution {
private:
    int primeN(int prime){
        return (1 << (prime - 1)) * ((1 << prime) - 1);
    }
    
public:
    bool checkPerfectNumber(int num) {
        std::vector<int> primes={2, 3, 5, 7, 13, 17, 19, 31};
        for(unsigned int cpt=0; cpt<primes.size(); cpt++){
            if(primeN(primes[cpt])==num){
                return true;
            }
        }
        return false;
    }
};