【HDU】5584LCM Walk-（数学推导）

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概述

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2708 Accepted Submission(s): 1426

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

Source

2015ACM/ICPC亚洲区上海站-重现赛（感谢华东理工）

Recommend

wange2014

题目大意：有一只青蛙,它从起点(x,y)出发,每次它会走LCM(x,y)步[LCM：最小公倍数]到达点(x+LCM(x,y),y)或点(x,y+LCM(x,y)),最终,它会到达点(ex,ey),现给你终点(ex,ey),要你求出它的起点有多少种可能

思路：

可以证明：两个数的gcd一直是k，

我们发现当前位置是(x,y)时,如果x>y,那么当前位置一定是由(x1,y)走到的,如果x<y,当前位置一定是由(x,y1)走到的,由这点确定了路径的唯一性

对于(x,y),假设x>y,x=nk,y=mk,设先前点为（x1，y),则x1+gcd*x1*y=x,所以先前点为(x/(y/k+1),y)，如果x不再是(y+k)的倍数(即:(y/k+1)*k的倍数),则表示不能再逆推。
重要公式：gcd(X,Y) = gcd(X, Y - K*X) = gcd(X - K * Y, Y)

代码：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#define ll long long
using namespace std;
int gcd(int a,int b)
{
return b==0?gcd(b,a%b):a;
}
int main()
{
int x,y,T;
scanf("%d",&T);
int t=0;
while(T--)
{
scanf("%d%d",&x,&y);
if(x<y)
swap(x,y);
int k=gcd(x,y),cnt=1;
while(x%(y+k)==0)
{
cnt++;
x=x/(y/k+1);
if(x<y)
swap(x,y);
}
t++;
printf("Case #%d: %dn",t,cnt);
}
return 0;
}