UVA 10235 - Simply Emirp Solution

UVA 10235 - Simply Emirp Solution

/*
It is a simple and easy problem. But has critical test case.
Critical case: 
If a number is prime and reverse number is equal to the given number then the output is "prime" not "emirp".
For example:
Given number 11.
11 number is prime. But reverse number is also 11.So, it is not "emirp". It is prime.
*/

Code:

#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#define MAX 1000001
using namespace std;

bool prime[MAX];
int reverseInt(int num)
{
    string str=to_string(num);
    reverse(str.begin(),str.end());
    num=stoi(str);
    return num;
}

int main()
{
    int i,j,x;
    memset(prime,1,sizeof(prime));
    for(i=4;i<MAX;i+=2)
        prime[i]=0;
    for(i=3;i<1001;i+=2)
    {
        if(prime[i])
        {
            for(j=i*i;j<MAX;j+=i)
                prime[j]=0;
        }
    }
   
    while(cin>>x)
    {
        if(prime[x])
        {
            i=reverseInt(x);
            if(prime[i] && x!=i)
                cout<<x<<" is emirp."<<endl;
            else
                cout<<x<<" is prime."<<endl;
        }
        else
            cout<<x<<" is not prime."<<endl;
    }
    return 0;
}

Input:

7
11
33
59
24
3
17

Output:
7 is prime.
11 is prime.
33 is not prime.
59 is prime.
24 is not prime.
3 is prime.
17 is emirp.