hdu 2769 uva 12169 Disgruntled Judge 拓展欧几里德

//数据是有多水 连 10^10的枚举都能过

关于拓展欧几里德：大概就是x1=y2,y1=x2-[a/b]y2,按这个规律递归到gcd(a,0)的形式，此时公因数为a，方程也变为a*x+0*y=gcd(a,0)的形式,显然解为x=1,y=0,然后再递归回去就能得到解（a*x+b*y=gcd(a,b)的解）

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<queue>
 #include<vector>
 #include<map>
 #include<stack>
 #include<string>

 using namespace std;

 long long T;
 long long x[];

 void exGED(long long a,long long b,long long &d,long long &x,long long &y){
     if (b==){
             x=;
             y=;
             d=a;
     }
     else{
             exGED(b,a%b,d,x,y);
             long long tmp=x;
             x=y;
             y=tmp-(a/b)*y;
     }
 }

 bool solve(long long a){
     long long d,b,k;
     long long tmp=x[]-a*a*x[];
     exGED(a+,,d,b,k);
     if (tmp%d!=) return false;
     b=b*(tmp/d);
     for (long long i=;i<=*T;i++){
             if (i%==){
                     x[i]=(x[i-]*a+b)%;
             }
             else{
                     if (x[i]!=((x[i-]*a+b)%)){
                             return false;
                     }
             }
     }
     for (long long i=;i<=*T;i+=){
             printf("%I64d\n",x[i]);
     }
     return true;
 }

 int main(){
     scanf("%I64d",&T);
     for (long long i=;i<*T;i+=){
             scanf("%I64d",&x[i]);
     }
     //solve(1096);
     for (long long a=;a<=;a++){
             if (solve(a)) break;
     }
     return ;
 }
 /*
 3
 17
 822
 3014
 */