HDU 5584 LCM Walk 數學

LCM Walk

Time Limit: 20 Sec

Memory Limit: 256 MB

題目鏈接

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

HINT

 

題意

給你(x,y) 而後能夠走到(x+lcm(x,y),y)或者走到(x,y+lcm(x,y))

而後如今給你一個位置，問你起點有多少種。

題解：

假設x = pt,y =qt 

因此(pt,qt)，那麼下一步就能夠走到(pt(1+q),qt)或者走到(pt,(1+p)qt)

那麼很顯然咱們走到了(x,y) 那麼上一步就是 (x/(y+1),y)和(x,y/(x+1))

代碼：

#include<iostream>
#include<stdio.h>
using namespace std;

int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}
int ans = 0;
void solve(int x,int y)
{
    ans++;
    if(x<y)swap(x,y);
    if(x%(y+1)==0)solve(x/(y+1),y);
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int cas=1;cas<=t;cas++)
    {
        ans = 0;
        int x,y;
        scanf("%d%d",&x,&y);
        int p = gcd(x,y);
        x = x/p,y = y/p;
        solve(x,y);
        printf("Case #%d: %d\n",cas,ans);
    }
}