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[201];

void exGED(long long a,long long b,long long &d,long long &x,long long &y){
    if (b==0){
            x=1;
            y=0;
            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[3]-a*a*x[1];
    exGED(a+1,10001,d,b,k);
    if (tmp%d!=0) return false;
    b=b*(tmp/d);
    for (long long i=2;i<=2*T;i++){
            if (i%2==0){
                    x[i]=(x[i-1]*a+b)%10001;
            }
            else{
                    if (x[i]!=((x[i-1]*a+b)%10001)){
                            return false;
                    }
            }
    }
    for (long long i=2;i<=2*T;i+=2){
            printf("%I64d
",x[i]);
    }
    return true;
}

int main(){
    scanf("%I64d",&T);
    for (long long i=1;i<2*T;i+=2){
            scanf("%I64d",&x[i]);
    }
    //solve(1096);
    for (long long a=0;a<=10000;a++){
            if (solve(a)) break;
    }
    return 0;
}
/*
3
17
822
3014
*/