Q:

# how to get the prime number in c++ where time complexity is 0(log n)

``````// C++ program to print all primes smaller than or equal to
// n using Sieve of Eratosthenes
#include <bits/stdc++.h>
using namespace std;

void SieveOfEratosthenes(int n)
{
// 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.
bool prime[n+1];
memset(prime, true, sizeof(prime));

for (int p=2; p*p<=n; p++)
{
// If prime[p] is not changed, then it is a prime
if (prime[p] == true)
{
// Update all multiples of p greater than or
// equal to the square of it
// numbers which are multiple of p and are
// less than p^2 are already been marked.
for (int i=p*p; i<=n; i += p)
prime[i] = false;
}
}

// Print all prime numbers
for (int p=2; p<=n; p++)
if (prime[p])
cout << p << " ";
}

// Driver Program to test above function
int main()
{
int n = 30;
cout << "Following are the prime numbers smaller "
<< " than or equal to " << n << endl;
SieveOfEratosthenes(n);
return 0;
}
``````
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