507.Perfect Number

507.Perfect Number

难度:Easy

We define the Perfect Number is a positive integer that is equal to the sum of all its positive divisors except itself.

Now, given an integer n, write a function that returns true when it is a perfect number and false when it is not.
Example:
Input: 28
Output: True
Explanation: 28 = 1 + 2 + 4 + 7 + 14
Note: The input number n will not exceed 100,000,000. (1e8)

求出所有的约数，不过需要注意相加时候的平方根和1.

class Solution {
public:
    bool checkPerfectNumber(int num) {
        if(num<1 ) return false;
        int s=sqrt(num);
        bool flag= s*s==num;
        int sum =1 +flag*s;
        for(int i=2;i<s+!flag;i++)
            if(num%i==0) sum+=i+num/i ;

        return sum == num;
            
        
    }
};

执行用时 :4 ms, 在所有 C++ 提交中击败了84.67%的用户 内存消耗 :7.9 MB, 在所有 C++ 提交中击败了95.98%的用户