JZYZOJ1371 青蛙的约会 扩展欧几里得 GTMD数论

http://172.20.6.3/Problem_Show.asp?id=1371

题意是两个青蛙朝同一个方向跳

http://www.cnblogs.com/jackge/archive/2013/04/22/3034925.html详细的题解，大概是网上能看到的最简单易懂的扩展欧几里得讲解了

 

代码

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 const int maxn=;
 void exgcd(int a,int b,long long &x,long long &y){
     if(!b){
         x=;y=;
         return;
     }
     exgcd(b,a%b,x,y);
     long long w=x;x=y;
     y=w-y*(a/b);
 }
 long long gcd(long long x,long long y){
     while(y){
         int w=y;y=x%y;x=w;
     }
     return x;
 }
 int main(){
     long long x,y,m,n,l;
     scanf("%I64d%I64d%I64d%I64d%I64d",&x,&y,&m,&n,&l);
     long long t,z=x-y,zz=n-m,k,d=gcd(n-m,l);
     if(z%d) printf("Impossible\n");
     else{
         z/=d;
         exgcd(zz/d,l/d,t,k);
         t=z*t-z*t/l*l;
         if(t<)t+=l;
         printf("%I64d\n",t);
     }
     return ;
 }