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.
It asks for a question like k mod A = x and k mod B = y. This is Chinese remainder theorem application.
Dear Nuu,
The equivalent Chinese Remainder problem is: Find the smallest positive integer
0 <= n <= A*B - 1
such thatn mod A = x - 1
andn mod B = y - 1
.Best wishes,