Problem 8: Sum of primes

Q:Find the sum of all the primes below two million.

Sol:

The logic used to solve the problem was the same as in problem 7 of finding the nth prime.
Code for the problem is also very similar,the small difference being finding the sum.

#include<cstdio>
#include<cmath>
using namespace std;

int main()
{
    int n;
    scanf("%d",&n);
    int arr[500000];
    int i=2;
    arr[0]=2;
    arr[1]=3;
    int k=1,j;
    long long sum=5;
    int nextprime,flag=0,c=1;
    nextprime=5;
    while(nextprime<n)
    {
               flag=0;
               for(j=0;((j < i)&&(arr[j]<=sqrt((double)nextprime)));j++)
               {
                               if(nextprime%arr[j] == 0)
                               {
                                              flag=1;
                                              break;
                               }
               }
               if(flag==0)
               {
                          arr[i]=nextprime;
                          if (arr[i]< n)
                            sum=sum+(long long)(arr[i]);
                          i++;
               }
               if(c==-1)
                        k++;
               nextprime=6*k +c;
               c=-c;
    }
    printf("%lld\n",sum);

}

The program takes 2-3 seconds to execute for the value of n being two million.



Please post your comments and doubts.
Suggestions for better algorithms are always welcome :)