Uva 12169

http://blog.youkuaiyun.com/u013451221/article/details/38497029

感谢原博主提供的思想。

带扩展二字的算法都可以出奇迹啊！

#include<iostream>
using namespace std;
typedef unsigned long long LL;
LL num[300];
void Gcd(LL a,LL b,LL &gcd,LL &x,LL &y)
{
	if(!b) { gcd =a ; x= 1; y = 0;}
	else {
		Gcd(b,a%b,gcd,y,x); y -= x*(a/b);
	}
}

int main() {
	int T;
	while(cin >> T) 
	{
		for(int i = 1; i < T*2; i+=2)
		cin >> num[i];
		for(int a = 0; a <= 20000; ++a) 
		{
			LL temp = num[3] - a*a*num[1]; //推导式子：  num2 = (num1 * a + b) %10001; 
			LL gcd, x, y;  				   //            num3 = (num2 * a + b) % 10001;
			Gcd(10001, a+1, gcd, y, x);		//           num3 - a*a*num1 = (a+1) * b + 10001 * k;
			if(temp % gcd) continue;          //            思路就是枚举a，求满足式子b；在推导num数组，看是否符合。 
			x = x * temp / gcd;       // 解为（x*t/gcd,y*t/gcd) 一组解 
			bool jug = true;
			for(int i = 2; i <= T *2; ++i) 
			{
				if(i & 1) {
					if(num[i] != (num[i-1] * a + x) % 10001)
					{ jug = false; break;}
				}
				else num[i] = (num[i-1] * a + x) % 10001;
			}
			if(jug) break;
		}
		for(int i = 2; i <= T*2; i+=2)
		cout << num[i] << endl;
	}
	return 0;
}