Codevs 2606 Approximate and problem

To see this problem we first consider the linear sieve. However??? 2*10^9 knelt down directly.

Consider the contribution of each factor I to the answer for [N/i]*i. So? Chunking can be done.

#include <iostream>
#include <cstdio>
using namespace Std;
Long long x, y;
Long Long ans=0;
Long long work (Long long N)
{
ans=0;
Long Long i=1;
while (i<=n)
{
Long Long j=n/(n/i);
ans=ans+ (n/i) * ((J*j+j+i-i*i)/2);
i=j+1;
}
return ans;
}
int main ()
{
scanf ("%lld%lld", &x,&y);
printf ("%lld\n", Work (y)-work (x-1));
return 0;
}

Codevs 2606 Approximate and problem