hdoj 5584 LCM Walk(数学，逆推)_hdoj 逆推-优快云博客

本文介绍了一道名为LCMWalk的问题解决方案。该问题要求计算从任意起点到给定终点的所有可能路径数量，通过特殊跳跃规则（跳至最近的公共倍数距离）。文章详细解析了算法思路，并给出了具体实现代码。

题意：

在位置(x, y)，可以走到(x+lcm(x, y), y) 或(x, y+lcm(x, y))或者不移动。给定终点位置，求可能的起点位置数。

思路：设当前在(x, y)处，下一步可以是(x+lcm(x, y), y)或(x, y+lcm(x, y)). 令gcd(x, y) = k, x = m*k,

y = n*k, 则(m*k+m*n*k, n*k)或(m*k, n*k+m*n*k) --> k*(m*(1+n), n)或k*(m, n*(1+m))

所以假设当前为(m, n)[m n互质且 m > n], 可以推得上一步为(m/(1+n), n) 当且仅当m%(1+n) == 0

代码：

#include<bits/stdc++.h>
using namespace std;
int main(void)
{
    int t, ca = 1;
    cin >> t;
    while(t--)
    {
        int ex, ey, ans = 1;
        scanf("%d%d", &ex, &ey);
        int k = __gcd(ex, ey);
        ex /= k, ey /= k;
        if(ex < ey) swap(ex, ey);
        while(ex % (1+ey) == 0)
        {
            ex /= (1+ey);
            ans++;
            if(ex < ey) swap(ex, ey);
        }
        printf("Case #%d: %d\n", ca++, ans);
    }
    return 0;
}

LCM Walk

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1350 Accepted Submission(s): 694

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