TJU1081

没错，看懂题目即可

#include<iostream>
#include<cmath>
using namespace std;

#define YES {cout<<"It's a MaoLaoDa number."<<endl;continue;}
#define NO {cout<<"It's not a MaoLaoDa number."<<endl;continue;}
int main()
{
    long n,test,m;
    int flag1,flag2,temp;
    while(cin>>n)
    {
        flag1=1;flag2=0;m=(long)sqrt((double)n);
        for(temp=2;temp<=m;temp++)
            if(n % temp == 0)
            {flag1=0;break;}
            if(flag1)
                NO
                test=n/temp;
            m = (long)sqrt((double)test);
            for(temp=2;temp<=m;temp++)
                if(test % temp == 0)
                {;flag2=1;break;}
                if(flag2)
                    NO
                else
                YES
    }
    return 0;
}