UVa12169

/*UVa12169 - Disgruntled Judge
--a,b都在10000以内，这样可以枚举a的值，然后根据x1和x3计算出b的值，再遍历数组X判断他们是否合法
--很容易得到x3=(a*a*x1+(a+1)*b)%10001;变形一下可以得到：(a+1)*b+10001*k=x3-a*a*x1;直接利用扩展欧几里得算法
求解b在[0,10000]的唯一解。
*/
#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
using namespace std;
const int maxn =100 + 10;
const int mod = 10001;
typedef long long LL;
LL F[maxn * 2];

//扩展欧几里德算法，求解ax+by=gcd(a,b)系数
void extendEuclid(LL a, LL b, LL &d, LL &x, LL&y){
	if (!b){ d = a; x = 1; y = 0; }
	else{ extendEuclid(b, a%b, d, y, x); y -= x*(a / b); }
}
//求ax+by=n的一个解(x0,y0)
//所有解为(x0+t*b,y0-t*a),t是任意整数
LL solveLiner(LL a, LL b, LL n, LL&x, LL&y){
	LL d;
	extendEuclid(a, b, d, x, y);
	if (n%d)return 0;  //无解
	x = (n / d)*x; y = (n / d)*y;
	return 1;
}

inline LL answer(LL a, LL b, LL x){ return (a*x + b) % mod; }

int main(){
	int T,i;
	LL a, b,k;
	while (~scanf("%d", &T)){
		for (i = 1; i <= 2 * T - 1; i += 2)
			scanf("%d", &F[i]);
		for (a = 0; a < mod; a++){
			solveLiner(a + 1, mod, F[3] - a*a*F[1], b, k);
			b = (b%mod + mod) % mod; //计算出b在[0,10000]的解
			for (i = 2; i <= 2 * T; i++){
				LL temp = answer(a, b, F[i - 1]);
				if (i & 1 && temp != F[i])break;
				F[i] = temp;
			}
			if (i == 2 * T + 1)break;
		}
		for (i = 2; i <= 2 * T; i += 2)
			printf("%I64d
", F[i]);
	}
	return 0;
}