#include <iostream>
using namespace std;

int main()
{

    int n;
    cin >> n ;

    while(n--)
    {
        int x = 0,sum = 0;
        cin >> x;
        for( int i = 1; i * i <= x; i++)
        {
            if( x%i == 0 ) 
            {
                if(i<x) sum += i;
                if(i != x/i && x/i < x) sum += x/i;
            }
        }
        if (sum == x) cout<< x <<" is perfect"<<endl;
        else cout<< x <<" is not perfect"<<endl;
    }


    return 0;
}

也可以先考虑1的值，因数是成对存在的,可以对x进行开方,求出前一半的因数i,那后一半的可以用x/i求出来。前一半中包含1,那么后一半中就包含他本身,所以sum求完以后要减去它本身,这样在与x进行比较.这样写的好处是不会超时。

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    int n,x;
    cin>>n;
    int s;
    int sum;
    while(n--)
    {
        cin>>x;
        sum=0;
        if(x==1)
            cout<<x<<" is not perfect"<<endl;
        else
        {
            for(int i=1;i<=sqrt(x);i++)
                if(x%i==0)
                {
                    sum+=i;
                    s=x/i;
                    //printf("%d %d\n",i,s);
                    sum+=s;
                }
            //printf("%d\n",sum);
            if(sum/2==x)
                cout<<x<<" is perfect"<<endl;
            else
                cout<<x<<" is not perfect"<<endl;
        }
    }
    return 0;
}