One algorithm of the day –leetcode 507– perfect number –C++

【 topic description 】

【 原 文 】

All the factors of an integer are required.

I * I <=num; I * I <=num;

2. The initial value of sum is 1, and the initial value of I is 2. This avoids adding num to the loop.

【 Source code 】

class Solution {
public:
    bool checkPerfectNumber(int num) {
        if(num==0||num==1) return false;
        int sum=1;
        for(int i=2; i*i<=num; i++){if(num%i==0) sum=sum+i+num/i;
            if(i*i==num) sum-=i;
            if(sum>num) return false;
        }
        return sum==num?true:false; }};Copy the code