10235 - Simply Emirp

/*
若一數字為質數且起位元顛倒後與原先不同且依然為質數則為  EMIRP
若非但此數為質數則為  PRIME
再者為合數  NOT PRIME
解法:埃拉托斯特尼篩法
*/
#include <iostream>
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
using namespace std;
int p[1000000] = {0};


int reverse(int n){
	int r = 0;
	while(n>0){
		r*=10;
		r+=n%10;
		n/=10;
	}
	return r;
}

int main(){
	p[0] = 1;
	p[1] = 1;
	p[2] = 0;
	
	for(int i = 2 ; i < 1000000 ; i ++){
		if(p[i]==0)
		for(int j = i+i ; j < 1000000;j+=i){
			p[j]=1;
		
		}	
	
	}
	

	int n= 0;
	while(scanf("%d",&n)==1){
	
		int r = reverse(n);
		if(p[n]==0&&p[r]==0&&n!=r)cout << n << " is emirp."<<endl;
		else if(p[n]==0)cout << n << " is prime."<<endl;
		else if(p[n]==1)cout << n << " is not prime."<<endl;
	}
	return 0;
}