A Fixed Remainder — Solution & Editorial

Back to the Problem

The qualifying x are B, B+A, B+2A, ... — count them within [1, N], remembering that x = 0 does not count when B = 0.

Complexity: O(1)

Watch out: B = 0 excludes x = 0 (not in [1, N]) and B ≥ A yields zero — two separate traps.