Find the Product of first N Prime Numbers | C ++ implementation of above solution ; 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 ; Set all multiples of p to non - prime ; find the product of 1 st N prime numbers ; count of prime numbers ; product of prime numbers ; if the number is prime add it ; increase the count ; get to next number ; Driver code ; create the sieve ; find the value of 1 st n prime numbers
#include " cstring " NEW_LINE #include <iostream> NEW_LINE using namespace std ; #define MAX  10000 NEW_LINE bool prime [ MAX + 1 ] ; void SieveOfEratosthenes ( ) { memset ( prime , true , sizeof ( prime ) ) ; prime [ 1 ] = false ; for ( int p = 2 ; p * p <= MAX ; p ++ ) { if ( prime [ p ] == true ) { for ( int i = p * 2 ; i <= MAX ; i += p ) prime [ i ] = false ; } } } int solve ( int n ) { int count = 0 , num = 1 ; long long int prod = 1 ; while ( count < n ) { if ( prime [ num ] ) { prod *= num ; count ++ ; } num ++ ; } return prod ; } int main ( ) { SieveOfEratosthenes ( ) ; int n = 5 ; cout << solve ( n ) ; return 0 ; }