ACM-ICP 4000 C国际大学生程序设计竞赛北京赛区(2017)网络赛 Bounce

题目7 : Bounce

时间限制:1000ms
单点时限:1000ms
内存限制: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?

输入

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)

输出

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).

样例输入

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

样例输出

2
3
5
5
7
8
7
9
11
39

首先计算球经过的路径长度 为L= lcm(m-1,n-1)+1;再减去两倍的走过两遍的点的个数： 为2*( (L/(m-1)-1)*(L/(n-1)-1)/2） = (L/(m-1))*(L/(n-1));

#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iomanip>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>
using namespace std;
long long m,n;
long long gcd(long long a,long long b)
{
if(b==0)return a;
else return gcd(b,a%b);
}
long long lcm(long long a,long long b)
{
return a/gcd(a,b)*b;
}
int main()
{
while(cin>>m>>n)
{
long long l=lcm(m-1,n-1);
long long a=l/(m-1),b=l/(n-1);
cout<<(l-a*b+a+b)<<endl;
}
return 0;
}