题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5584
Description
A frog has just learned some number theory, and can‘t wait to show his ability to his girlfriend.
Now the frog is sitting on a grid map of infinite rows and columns. Rows are numbered 1,2,? from the bottom, so are the columns. At first the frog is sitting at grid (sx,sy), and begins his journey.
To show his girlfriend his talents in math, he uses a special way of jump. If currently the frog is at the grid (x,y), first of all, he will find the minimum z that can be divided by both x and y, and jump exactly z steps to the up, or to the right. So the
next possible grid will be (x+z,y), or (x,y+z).
After a finite number of steps (perhaps zero), he finally finishes at grid (ex,ey). However, he is too tired and he forgets the position of his starting grid!
It will be too stupid to check each grid one by one, so please tell the frog the number of possible starting grids that can reach (ex,ey)!
Input
First line contains an integer T, which indicates the number of test cases.
Every test case contains two integers ex and ey, which is the destination grid.
? 1≤T≤1000.
? 1≤ex,ey≤109.
Output
For every test case, you should output "Case #x: y", where x indicates the case number and counts from 1 and y is the number of possible starting grids.
Sample Input
3
6 10
6 8
2 8
Sample Output
Case #1: 1
Case #2: 2
Case #3: 3
题意：
给你(x,y) 然后可以走到(x+lcm(x,y),y)或者走到(x,y+lcm(x,y))
然后现在给你一个位置，问你起点有多少种。
假设x = at,y =bt ；
所以(at,bt)，那么下一步就可以走到(at(1+b),bt)或者走到(at,(1+a)bt)
除一下gcd，那么我们走到了(x,y) 上一步就是 (x/(y+1),y)和(x,y/(x+1))。
//#include<bits/stdc++.h>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
#define MAXN 200005
long long mod = 1e9+7;
int get_gcd(int x,int y)
{
if(!x)return y;
return get_gcd(y%x,x);
}
int main()
{
int n,x,y;
scanf("%d",&n);
for(int i=1; i<=n; ++i)
{
int ans=0;
scanf("%d%d",&x,&y);
int c=get_gcd(x,y);
x/=c,y/=c;
if(x>y)swap(x,y);
while(y%(x+1)==0)
{
ans++;
y/=(x+1);
if(x>y)swap(x,y);
}
printf("Case #%d: %d\n",i,++ans);
}
return 0;
}