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; 
}