Knowledge required: number theory, binary search

First of all, it is important to recognize the relationship between ~N!~ and its number of trailing zeros. A zero is formed with a pair of a factor of ~2~ and a factor of ~5~. In fact, it is not necessary to account for the count of both, since the number of zeros is dependent on the lesser of the two, and the number of ~5~ will always be less or equal than that of ~2~s.

To solve for the 50% subtask, we can simply brute-force the value of ~N~ such that ~N!~ will have a number of trailing zeros in the range of ~[a,b]~.

Time Complexity: ~\mathcal{O}(b)~

To solve for the rest of the cases, it is important to recognize that the number of trailing zeros of ~N!~ will always be non-decreasing in relationship to an increasing ~N~. Hence, we can use binary search to find:

1. The minimum ~N~ such that ~N!~ has at least ~a~ trailing zeros.
2. The maximum ~N~ such that ~N!~ does not have more than ~b~ trailing zeros.

The number of valid values of ~N~ can be calculated by subtracting the two and adding one.

Time Complexity: ~\mathcal{O}(\log_2 \max(b))~