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LCM Walk
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 47 Accepted Submission(s): 31
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
Sample Output
Case #1: 1
Case #2: 2
Case #3: 3
Source
2015acm上海区域赛的第三道水题。。第一开始以为是推公式然后o（1）求出答案，然而貌似并不能，最后还是想了个暴力枚举公因子吧。。
容易得知，x,y里面肯定是较小的数不变，较大的那个数是从之前某个数变化来的，假设x>y，（x,y）是从（x1，y）变化来的，那么：
x = x1 + x1*y/gcd(x1,y)；则x1 = x/（1 + y/gcd（x1,y））；
那么就很好说了，枚举gcd（x1,y），即枚举y的因子，反求出x1,然后判断x1是否合理，合理的话就继续递归（x1,y），这里枚举因子有一个细节需要
注意，就是对于y是完全平方数的时候，枚举上界是sqrt（y-0.5），然后对于x = sqrt（y）的情况特判，因为忘了注意这点此贡献了一次WA。。
为什么要这样子呢。。因为O（根号n)枚举因子时，如果i是y的因子，那么y/i也是y的因子，这里要判断两个因子，但是i*i=y时，必须只判断一次
#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <cmath>
using namespace std;
int t;
int x,y;
int ans;
int gcd(int x, int y)
{
return x == 0?y : gcd(y%x,x);
}
void dfs(int x, int y)
{
ans++;
if(x < y) swap(x,y);
int p = sqrt(y - 0.5);
int i;
for(i = 1; i <= p; ++i)
{
if(y % i == 0)
{
if(x%(1+y/i) == 0&&gcd(x/(1+y/i),y) == i) dfs(x/(1+y/i),y);
if(x%(1+i) == 0&&gcd(x/(1+i),y) == y/i) dfs(x/(1+i),y);
}
}
if(i*i == y)
{
if(x%(1+i) == 0&&gcd(x/(1+i),y) == i) dfs(x/(1+i),y);
}
}
int main()
{
int cas = 1;
for(cin >> t; cas <= t; ++cas)
{
ans = 0;
scanf("%d%d",&x,&y);
dfs(x,y);
printf("Case #%d: %d\n",cas,ans);
}
}
HDU5584 LCM Walk 数论
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原文地址：http://www.cnblogs.com/Norlan/p/5004565.html
