JZYZOJ1371 青蛙的约会 扩展欧几里得 GTMD数论

Posted 2020-10-13 鲸头鹳

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

 1 #include<iostream>  
 2 #include<cstdio>  
 3 #include<cstring>  
 4 #include<algorithm>  
 5 #include<cmath>
 6 using namespace std;
 7 const int maxn=100010;
 8 void exgcd(int a,int b,long long &x,long long &y){
 9     if(!b){
10         x=1;y=0;
11         return;
12     }
13     exgcd(b,a%b,x,y);
14     long long w=x;x=y;
15     y=w-y*(a/b);
16 }
17 long long gcd(long long x,long long y){
18     while(y){
19         int w=y;y=x%y;x=w;
20     }
21     return x;
22 }
23 int main(){
24     long long x,y,m,n,l;
25     scanf("%I64d%I64d%I64d%I64d%I64d",&x,&y,&m,&n,&l);
26     long long t,z=x-y,zz=n-m,k,d=gcd(n-m,l);
27     if(z%d) printf("Impossible\\n");
28     else{
29         z/=d;
30         exgcd(zz/d,l/d,t,k);
31         t=z*t-z*t/l*l;
32         if(t<0)t+=l;
33         printf("%I64d\\n",t);
34     }
35     return 0;
36 }

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