题目描述

时间超了 两个循环的原因吧

样例

#include <iostream>
#include <cstdio>
using namespace std;
int main(){
    int i,n,j,num,sum;
    cin>>n;
    for(i=0;i<n;i++){
        cin>>num;
        sum=0;
        for(j=1;j<num;j++){
            if(num%j==0)sum+=j;
        }if(num==sum)printf("%d is perfect\n",num);
        else printf("%d is not perfect\n",num);
    }

    return 0;
}

算法1

(暴力枚举) $O(n^2)$

blablabla

涉及质数的一定记得用sqrt优化下

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int main(){
    int i,n,j,num,sum;
    cin>>n;
    for(i=0;i<n;i++){
        cin>>num;
        sum=1;
        for(j=2;j<=sqrt(num);j++){
            if(num%j==0&&j!=sqrt(num))sum+=j+num/j;
            if(num%j==0&&j==sqrt(num))sum+=j;
        }if(num==sum&&num!=1)printf("%d is perfect\n",num);
        else printf("%d is not perfect\n",num);
    }

    return 0;
}

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla