Emirp numbers | Program to print Emirp numbers less than n ; Function to find reverse of any number ; Sieve method used for generating emirp number ( use of sieve of Eratosthenes ) ; Create a boolean array " prime [ 0 . . n ] " and initialize all entries it as true . A value in prime [ i ] will finally be false if i is Not a prime , else true . ; If prime [ p ] is not changed , then it is a prime ; Update all multiples of p ; Traverse all prime numbers ; Find reverse a number ; A number is emrip if it is not a palindrome number and its reverse is also prime . ; Mark reverse prime as false so that it 's  not printed again ; Driver program
#include <bits/stdc++.h> NEW_LINE using namespace std ; int reverse ( int x ) { int rev = 0 ; while ( x > 0 ) { rev = ( rev * 10 ) + x % 10 ; x = x / 10 ; } return rev ; } void printEmirp ( int n ) { bool prime [ n + 1 ] ; memset ( prime , true , sizeof ( prime ) ) ; for ( int p = 2 ; p * p <= n ; p ++ ) { if ( prime [ p ] == true ) { for ( int i = p * 2 ; i <= n ; i += p ) prime [ i ] = false ; } } for ( int p = 2 ; p <= n ; p ++ ) { if ( prime [ p ] ) { int rev = reverse ( p ) ; if ( p != rev && rev <= n && prime [ rev ] ) { cout << p << " ▁ " << rev << " ▁ " ; prime [ rev ] = false ; } } } } int main ( ) { int n = 40 ; printEmirp ( n ) ; return 0 ; }