HDU 5584 LCM Walk 数学

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原文链接：http://www.cnblogs.com/qscqesze/p/5029555.html

文章标签：

#php #数据结构与算法

本文介绍了HDU OJ上LCMWalk问题的解决方法，该问题涉及数学理论中的最小公倍数概念。通过分析跳跃路径，文章给出了寻找可能起始网格数量的有效算法。

LCM Walk

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

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

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。

题解：

假设x = pt,y =qt

所以(pt,qt)，那么下一步就可以走到(pt(1+q),qt)或者走到(pt,(1+p)qt)

那么很显然我们走到了(x,y) 那么上一步就是 (x/(y+1),y)和(x,y/(x+1))

代码：

#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);
    }
}