Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
Problem 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：
给定一个走路规则，如果你当前在(x,y)那么你接下来可以走到(x+lcm(x,y),y)或者(x,y+lcm(x,y))
lcm(x,y)是x,y的最小公倍数。
然后给你终点，要求所有合法的起点的个数
题解：
我们来假设gcd(x,y)=k
则x=nk,y=mk,lcm(x,y)=nmk
然后(nk,mk)的下一步就是(n(m+1)k,mk)或者(nk,m(n+1)k)
那么对于每个(x,y)来说，都是逆推一次的事情。要么是x/=(y+1)【x>y】要么是y/=(x+1)【y>x】
路径是唯一的0.0
#include<bits/stdc++.h>
#define ll long long
using namespace std;
int x,y;
int w33ha(int CASE){
scanf("%d%d",&x,&y);
int cnt=1;
if(x<y)swap(x,y);
int t=__gcd(x,y);
while(x%(y+t)==0){
++cnt;
x=x/(y/t+1);
if(x<y)swap(x,y);
}
printf("Case #%d: %d\n",CASE,cnt);
return 0;
}
int main(){
int T;scanf("%d",&T);
for(int i=1;i<=T;i++)w33ha(i);
return 0;
}