#1584 : Bounce

時間限制:1000ms

單點時限:1000ms

記憶體限制:256MB

描述

For Argo, it is very interesting watching a circle bouncing in a rectangle.

As shown in the figure below, the rectangle is divided into N×M grids, and the circle fits exactly one grid.

The bouncing rule is simple:

1. The circle always starts from the left upper corner and moves towards lower right.

2. If the circle touches any edge of the rectangle, it will bounce.

3. If the circle reaches any corner of the rectangle after starting, it will stop there.

Argo wants to know how many grids the circle will go through only once until it first reaches another corner. Can you help him?

輸入

The input consists of multiple test cases. (Up to 105)

For each test case:

One line contains two integers N and M, indicating the number of rows and columns of the rectangle. (2 ≤ N, M ≤ 109)

輸出

For each test case, output one line containing one integer, the number of grids that the circle will go through exactly once until it stops (the starting grid and the ending grid also count).

樣例輸入

2 2
2 3
3 4
3 5
4 5
4 6
4 7
5 6
5 7
9 15

樣例輸出

2
3
5
5
7
8
7
9
11
39

思路：從左上角開始，所以可以先去除左上角的行和列（這裡的行列所包含的只講過一次的方個數有（n-1）/g+(m-1)/g個），然後對剩下的（n-1）行和（m-1）列劃分，

如果令tp=GCD（n-1,m-1）劃分為（n-1）/g* （m-1）/g個單元，每個單元裡有（g-1）個只經過一次的方格。

所以總次數是：

ans=（n-1）/g + (m-1)/g + (g-1) * (n-1)/g* (m-1)/g;

參考文章：https://blog.csdn.net/Feynman1999/article/details/78073669

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long LL;
LL gcd(LL x,LL y)
{
	return y==0?x:gcd(y,x%y);
}
int main(void)
{
	LL n,m,tp,ans;
	while(~scanf("%lld%lld",&n,&m))
	{
		tp=n<m?gcd(m-1,n-1):gcd(n-1,m-1);
		ans=(n-1)/tp+(m-1)/tp+((n-1)/tp*(m-1)/tp)*(tp-1); 
		printf("%lld\n",ans);
	}
	return 0;
}