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: 28Output: TrueExplanation: 28 = 1 + 2 + 4 + 7 + 14Note: 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