题目链接:LCM Walk
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 491 Accepted Submission(s): 254
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)!
Every test case contains two integers ex and ey, which is the destination grid.
⋅ 1≤T≤1000.
⋅ 1≤ex,ey≤109.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
ll gcd(ll a,ll b)
{
if(b==)return a;
return gcd(b,a%b);
}
int main()
{
int t,cnt=;
ll x,y;
scanf("%d",&t);
while(t--)
{
ll ans=;
cin>>x>>y;
while()
{
if(x<y)
{
ll fx=x;
x=y;
y=fx;
}
ll fy=y/gcd(x,y);
if(x%(fy+)!=)break;
else
{
ll ax=x/(fy+);
if(ax+ax*y/gcd(ax,y)!=x)break;
else ans++,x=ax;
}
}
cout<<"Case #"<<cnt<<": "<<ans<<"\n";
cnt++;
} return ;
}