看我手写分析
题目大意：有3个整数 x[1], a, b 满足递推式x[i]=(a*x[i-1]+b)mod 10001。由这个递推式计算出了长度为2T的数列，现在要求输入x[1],x[3],......x[2T-1], 输出x[2],x[4]......x[2T]. T<=100,0<=x<=10000. 如果有多种可能的输出，任意输出一个结果即可。
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//拓展欧几里德用法详见紫书314页
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
using namespace std;
typedef long long LL;
const int mod = 10001;
void ex_gcd(LL a, LL b, LL &d, LL &x, LL &y) {
if (b == 0) d = a, x = 1, y = 0;
else {
ex_gcd(b, a % b, d, y, x);
y -= x *(a / b);
}
}
int main()
{
int n, x[210];
while (cin >> n) {
for (int i = 1, j = 1; i <= n; ++i, j += 2)
cin >> x[j];
for (LL a = 0; a <= 10000; ++a) { //暴力枚举a
LL d, X, Y, c;
ex_gcd(a + 1, mod, d, X, Y);
c = x[3] - a * a * x[1];
if (c % d == 0) { //若 非c|d，无整数解
bool bingo = true;
LL b = X * (c / d) % mod;
for (int i = 1, j = 3; i < n; ++i, j += 2) {
if (x[j] != (a * (a * x[j - 2] + b) + b) % mod) {
bingo = false;
break;
}
}
if (bingo) {
for (int i = 1, j = 1; i <= n; ++i, j += 2)
cout << (a * x[j] + b) % mod << endl;
}
}
}
}
return 0;
}
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