### Nuu's blog

By Nuu, history, 4 years ago,

Hi guys, Today i sees a apparently easy math problem (but i can't solved):

Two persons walk simutanious around two circles with size A and B with positions [1,2,3,...,A] and [1,2,3,...,B].

Then follow Q querys. Each query give two positions x and y, and asks the number of step for person A arrive on position x and person B arrive on position y (on same time) using O(1).

Thanks guys, i'm very novice in math problems.

• -3

 » 4 years ago, # |   +3 It asks for a question like k mod A = x and k mod B = y. This is Chinese remainder theorem application.
 » 4 years ago, # | ← Rev. 2 →   0 Dear Nuu,The equivalent Chinese Remainder problem is: Find the smallest positive integer 0 <= n <= A*B - 1 such that n mod A = x - 1 and n mod B = y - 1.Best wishes,