Print all safe primes below N in C++

In this problem, we are given an integer N and we have to print all safe prime number whose values are less than N.

A safe prime number is a prime number which can be represented as [(2*p)- 1] where p is also a prime number.

Examples − 5[(2*2) +1] , 7[(2*3)+1].

Let’s take a few examples to understand the problem better −

Input: N = 12
Output: 5 7 11.

To solve this problem, we will find all the prime numbers less than N(for this we will use Sieve of Eratosthenes). And check if the prime number is a safe prime number or not and print if it is a safe prime number.

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
void printPrime(int n){
   cout<<n<<"\t";
}
void generateSafePrimes(int n){
   int prime[n + 1];
   for (int i = 2; i <= n; i++)
      prime[i] = 1;
   prime[0] = prime[1] = 0;
   for (int p = 2; p * p <= n; p++) {
      if (prime[p] == 1) {
         for (int i = p * 2; i <= n; i += p)
            prime[i] = 0;
      }
   }
   for (int i = 2; i <= n; i++) {
      if (prime[i] != 0) {
         int temp = (2 * i) + 1;
         if (temp <= n && prime[temp] != 0)
            prime[temp] = 2;
      }
   }
   for (int i = 5; i <= n; i++)
      if (prime[i] == 2)
         printPrime(i);
}
// Driver code
int main(){
   int n = 34;
   cout<<"safe Prime numbers less than "<<n<<" are :\n";
   generateSafePrimes(n);
   return 0;
}