sieve of eratosthenes c

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21 Apr 2016
1//sieve of eratosthenes or prime of sieve
4using namespace std;
5void primeofsieve(long long int n)
7	long long int arr[n]={};
8	for(int i=2;i<=sqrt(n);i++)
9	{
10		for(long long int j=i*i;j<=n;j+=i)
11			arr[j]=1;
12	}
13	for(long long int i=2;i<=n;i++)
14	{
15	    if(arr[i]==0)
16	    	cout<<i<<" ";
17	}
21int main()
24	#ifdef _DEBUG
25	freopen("input.txt", "r", stdin);
26	freopen("output.txt", "w", stdout);
27    #endif
28	long long int n;
29	cin>>n;
30	cout<<"PRIME NUMBERs ARE : ";
31	primeofsieve(n);
32	return 0;
08 May 2017
3                            Online C Compiler.
4                Code, Compile, Run and Debug C program online.
5Write your code in this editor and press "Run" button to compile and execute it.
9#include <stdio.h>
10#include <math.h>
11#include <stdbool.h>
13int main()
15    int n,i,j; //declaring required variables
16    scanf("%d",&n); //scanning n(limit numbers)
17    bool arr[n+1]; //declaring an array which contains true and false values
18    arr[1]=false; //setting the first index to false (1 is not a prime number)
20    for(i=2;i<=n;i++){
21        arr[i]=true; //we are assuming that all the numbers are true which means prime
22    }
24    for(i=2;i<sqrt(n)+1;i++){ //running loop to the sqrt on n + 1
25        if(arr[i]==true){ //if the number is true of prime 
26            for(j=i+i;j<=n;j+=i){ //run the loop to n
27                arr[j]=false; //assume that the divisble numbers are not prime
28            }
29        }
30    }
32    for(i=2;i<=n;i++){
33        if(arr[i]==true){
34            printf("%d ",i); //printing the numbers if that is true 
35        }
36    }
40    return 0;
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