"sieve of eratosthenes c++" Code Answer's
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sieve of eratosthenes c++
Old-fashioned Okapi
on Apr 25, 2020
// 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;
}
prime of sieve
Frightened Falcon
on May 27, 2020
//sieve of eratosthenes or prime of sieve
#include<iostream>
#include<math.h>
using namespace std;
void primeofsieve(long long int n)
{
long long int arr[n]={};
for(int i=2;i<=sqrt(n);i++)
{
for(long long int j=i*i;j<=n;j+=i)
arr[j]=1;
}
for(long long int i=2;i<=n;i++)
{
if(arr[i]==0)
cout<<i<<" ";
}
}
int main()
{
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
long long int n;
cin>>n;
cout<<"PRIME NUMBERs ARE : ";
primeofsieve(n);
return 0;
}sieve of eratosthenes c++
tough life
on Nov 09, 2020
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;
}
}
Source: cp-algorithms.com
sieve of eratosthenes c++
tough life
on Nov 09, 2020
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;
}
}
Source: cp-algorithms.com
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