POJ 1061

 1 #include <stdio.h>
 2 #include <iostream>
 3 using namespace std;
 4 
 5 long long x,y,m,n,L,dist;
 6 //辗转相除求最大公约数
 7 int gcd(int a,int b)
 8 {
 9     if(b==0)return a;
10     return gcd(b,a%b);
11 }
12 // 把跳的慢的视为相对静止
13 int main()
14 {
15     while(cin>>x>>y>>m>>n>>L)
16     {
17         //计算出两只青蛙的距离差
18         //先比较快慢
19         if(m>n)
20             dist=(y-x+L)%L;
21         else
22         {
23             dist=(x-y+L)%L;
24             int temp=m;
25             m=n;
26             n=temp;
27         }
28 
29         //现在证明判断条件
30         //等式为： (i*L+dist)%(m-n)==0
31         //若有解 let t = gcd（L,m-n)
32         //let L = ct;m-n = bt;
33         //ict +dist = bkt
34         //dist = (bk - ic)t
35         //dist % t = 0; 
36         if(dist%gcd(m-n,L)!=0)
37         {
38             cout<<"Impossible"<<endl;
39             continue;
40         }
41         long long k=0;
42         int i =0;
43         for(int i=0;; i++)//暴力来试出结果
44         {
45             if((i*L+dist)%(m-n)==0)
46             {
47                 k=(i*L+dist)/(m-n);
48                 break;
49             }
50         }
51         cout<<k<<endl;
52     }
53     return 0;
54 }