What is wrong with this approach :
I am trying to find the count of number of divisors in a-b by using seive , if i found that i is a divisor of j i just update the count of number of divisors of j by 2 because if k if div of n then n/k is also a div of n , and to avoid doing it twice for k and n/k i am using the check operator.

Can someone please explain it in a bit detail ? the methods by which this problem can be solved along with the complexity of the solution SPOJ.com - Problem NDIV