Answers for "sieve of eratosthenes c++ code"

C++
0

sieve of eratosthenes c++

int n;
vector<bool> is_prime(n+1, true);
is_prime[0] = is_prime[1] = false;
for (int i = 2; i <= n; i++) {
    if (is_prime[i] && (long long)i * i <= n) {
        for (int j = i * i; j <= n; j += i)
            is_prime[j] = false;
    }
}
Posted by: Guest on November-09-2020
0

sieve of eratosthenes c++

// 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; 
}
Posted by: Guest on March-13-2020
0

sieve of eratosthenes c++

int n;
vector<char> is_prime(n+1, true);
is_prime[0] = is_prime[1] = false;
for (int i = 2; i <= n; i++) {
    if (is_prime[i] && (long long)i * i <= n) {
        for (int j = i * i; j <= n; j += i)
            is_prime[j] = false;
    }
}
Posted by: Guest on November-09-2020
0

sieve of eratosthenes c++ code

#include <iostream>
const int len = 30;
int main() {
   int arr[30] = {0};
   for (int i = 2; i < 30; i++) {
      for (int j = i * i; j < 30; j+=i) {
         arr[j - 1] = 1;
      }
   }
   for (int i = 1; i < 30; i++) {
      if (arr[i - 1] == 0)
         std::cout << i << "\t";
   }
}
Posted by: Guest on July-05-2021

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