sieve of eratosthenes c

//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; }

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#include <stdio.h>
#include <math.h>
#include <stdbool.h>

int main()
{
    int n,i,j; //declaring required variables
    scanf("%d",&n); //scanning n(limit numbers)
    bool arr[n+1]; //declaring an array which contains true and false values
    arr[1]=false; //setting the first index to false (1 is not a prime number)
    
    for(i=2;i<=n;i++){
        arr[i]=true; //we are assuming that all the numbers are true which means prime
    }
    
    for(i=2;i<sqrt(n)+1;i++){ //running loop to the sqrt on n + 1
        if(arr[i]==true){ //if the number is true of prime 
            for(j=i+i;j<=n;j+=i){ //run the loop to n
                arr[j]=false; //assume that the divisble numbers are not prime
            }
        }
    }
    
    for(i=2;i<=n;i++){
        if(arr[i]==true){
            printf("%d ",i); //printing the numbers if that is true 
        }
    }
    
    

    return 0;
}

Posted by: Guest on September-27-2021

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