HDU-5584-LCM Walk【2015上海赛区】【数论】

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HDU-5584-LCM Walk

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        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

题目链接：HDU-5584

题目大意：假设该青蛙起点位置为(x, y)，他每次能到达的位置是（x + lcm(x, y), y）或者（x,y + lcm(x, y)）。已知终点位置（a, b）问有多少种情况的起始位置能到达该点。

题目思路：推公式。

假设初始位置为（x , y）, g为gcd(x,y)，则，下一步为（x + lcm(x,y),y）或者（x,y + lcm(x,y)）。

假设m1为x/g,m2为y/g，则起始位置为（m1g,m2g）

又因为：lcm(x,y) = m1m2g.所以下一步为（m1g + m1m2g,m2g）或者（m1g,m2g +m1m2g）；

已知终点位置为（a, b）又因为 起始位置和终点位置gcd一定相同 :所以：

        x = a * g / (g + b);
        y = b;

(假设x为较大的值)

以下是代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
long long gcd(long long a, long long b)
{   
    return b == 0 ? a : gcd(b, a % b);   
}

int main(){
    int T;
    cin >> T;
    int cas = 1;
    while(T--)
    {
        long long a,b,ans = 1;
        scanf("%lld%lld",&a,&b);
        if (a < b) swap(a,b);
        long long g = gcd(a,b);

        while((a % (g + b)) == 0)
        {
            a = a * g / (g + b);
            ans++;
            if (a < b) swap(a,b);
        }
        printf("Case #%d: %lld\n",cas++,ans);
    } 
    return 0;
}