POJ 1061

题意:

　　两只青蛙在同一条纬度上,它们各自朝西跳,问它们要跳多少步才能碰面(必须同时到达同一点).

分析:

　　假设它们跳了t步才相遇,青蛙a初始坐标为x,青蛙b初始坐标为y,则跳了t步相遇后a的坐标为 x+m*t-p1*l, b的坐标为 y+n*t-p2*l (p1,p2分别表示a,b跳

的圈数) x+m*t-p1*l = y+n*t-p2*l => x-y = (n-m)*t + (p1-p2)*l => x-y = (n-m)*t + p*l.

代码如下:

 //方法一

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <fstream>
 #include <ctime>
 #include <cmath>
 #include <cstdlib>
 #include <algorithm>
 #include <set>
 #include <map>
 #include <list>
 #include <stack>
 #include <queue>
 #include <iterator>
 #include <vector>

 using namespace std;

 #define LL long long
 #define INF 0x3f3f3f3f
 #define MOD 1000000007
 #define MAXN 10000010
 #define MAXM 1000010

 LL extend_gcd(LL a, LL b, LL &x, LL &y)
 {
     if(b == )
     {
         x = ;
         y = ;
         return a;
     }
     else
     {
         LL r = extend_gcd(b, a%b, y, x);
         y -= x*(a/b);
         return r;
     }
 }

 int main()
 {
     LL x, y, m, n, l;
     while(scanf("%lld%lld%lld%lld%lld", &x, &y, &m, &n, &l)==)
     {
         LL t, p;
         LL ans = extend_gcd(n-m, l, t, p);
         if((x-y)%ans)
             printf("Impossible\n");
         else
         {
             t = t*(x-y)/ans;
             t = (t%(l/ans)+(l/ans))%(l/ans);
             printf("%lld\n", t);
         }
     }

     return ;
 }

 //方法二

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <fstream>
 #include <ctime>
 #include <cmath>
 #include <cstdlib>
 #include <algorithm>
 #include <set>
 #include <map>
 #include <list>
 #include <stack>
 #include <queue>
 #include <iterator>
 #include <vector>

 using namespace std;

 #define LL long long
 #define INF 0x3f3f3f3f
 #define MOD 1000000007
 #define MAXN 10000010
 #define MAXM 1000010

 LL extend_gcd(LL a, LL b, LL &x, LL &y)
 {
     if(b == )
     {
         x = ;
         y = ;
         return a;
     }
     else
     {
         LL r = extend_gcd(b, a%b, y, x);
         y -= x*(a/b);
         return r;
     }
 }

 LL cal(LL a, LL b, LL c)
 {
     LL x, y;
     LL ans = extend_gcd(a, b, x, y);
     if(c%ans)
         return -;
     x *= c/ans;
     b /= ans;
     if(b < )
         b = b*(-);
 //    b =  abs(b);  //abs只能对int取绝对值
     ans = x%b;
     if(ans <= )
         ans += b;
     return ans;
 }

 int main()
 {
     LL x, y, m, n, l;
     while(scanf("%lld%lld%lld%lld%lld", &x, &y, &m, &n, &l)==)
     {
         LL sum = cal(n-m, l, x-y);
         if(sum == -)
             printf("Impossible\n");
         else
             printf("%lld\n", sum);
     }

     return ;
 }