BISHOP2 - Chessboard Billiard
Let's imagine: there is a chess piece billiard ball. Its movements resemble the ones of a bishop chess piece. The only difference is that when a billiard ball hits the board's border, it can reflect from it and continue moving.
More formally, first one of four diagonal directions is chosen and the billiard ball moves in that direction. When it reaches the square located on the board's edge, the billiard ball reflects from it; it changes the direction of its movement by 90 degrees and continues moving. Specifically, having reached a corner square, the billiard ball is reflected twice and starts to move the opposite way. While it moves, the billiard ball can make an infinite number of reflections. At any square of its trajectory the billiard ball can stop and on that the move is considered completed.
It is considered that one billiard ball a beats another billiard ball b if a can reach a point where b is located.
You are suggested to find the maximal number of billiard balls, that pairwise do not beat each other and that can be positioned on a chessboard n × m in size.
Input
The first line of input contains two integers n and m (2 ≤ n, m ≤ 106).
Output
Print a single line for each input line containing a single number, the maximum possible number of billiard balls that do not pairwise beat each other.
Example
Input: 3 4 2 2 5 5 Output: 2 2 5
hide comments
Santiago Palacio:
2013-01-27 23:01:06
There are inputs where n or m = 1, i have tested that. Nice problem. Last edit: 2013-01-28 04:43:47 |
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blashyrkh:
2011-09-23 12:14:51
Please explain result for n,m=3,4. Why 2? Where these billiard balls can be placed?
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Efim:
2011-06-30 19:41:18
how many lines are in input? |
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Alex Anderson:
2011-06-16 13:25:05
The example input of 1 1 and 1 2 is contradictory to the specifications. |
Added by: | Bidhan |
Date: | 2011-06-15 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 FSHARP FORTRAN HASK PYTHON PYTHON3 |
Resource: | Codeforces Beta Round |