HDU 3802 矩阵快速幂 化简递推式子 加一点点二次剩余知识

求$G(a,b,n,p) = (a^{frac {p-1}{2}}+1)(b^{frac{p-1}{2}}+1)[(sqrt{a} + sqrt{b})^{2F_n} + (sqrt{a} - sqrt{b})^{2F_n}] (mod p)$

左边可以看出是欧拉判定准则，那么只有当a,b其中一个满足是模p下的非二次剩余时G()为0。

右边的式子可以先把平方放进去，发现这个已经是通项公式了，那么$a+b+sqrt{ab}$和$a+b-sqrt{ab}$就是它的特征根了，反代回二阶特征方程，得到递推式的两个系数$2(a+b)$，$-(a-b)^2$，然后可以使用矩阵快速幂了，注意其项数是$F(n)$，指数循环节，欧拉降幂，其矩阵快速幂的过程中模p-1就行了。

/** @Date    : 2017-10-11 22:30:33
  * @FileName: HDU 3802 斐波那契特征方程解递推式 矩阵快速幂.cpp
  * @Platform: Windows
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;

LL mod;

LL fpow(LL a, LL n)
{
	LL res = 1LL;
	while(n)
	{
		if(n & 1LL) res = (res * a % mod + mod) % mod;
		a = (a * a % mod + mod) % mod;
		n >>= 1LL;
	}
	return res;
}

struct matrix
{
	LL mat[2][2];
	matrix(){MMF(mat);}
	void id(){
		mat[0][0] = mat[1][1] = 1LL;
	}
	matrix operator *(const matrix &b){
		matrix c;
		for(int i = 0; i < 2; i++)
			for(int j = 0; j < 2; j++)
				for(int k = 0; k < 2; k++)
				{
					c.mat[i][j] = (c.mat[i][j] + mat[i][k] * b.mat[k][j] % mod) % mod;
				}
		return c;
	}
	matrix operator ^(LL n){
		matrix res;
		matrix a = *this;
		res.id();
		while(n)
		{
			if(n & 1LL)
				res = res * a;
			a = a * a;
			n >>= 1LL;
		}
		return res;
	}
};

LL fib(LL n)
{
	mod--;
	matrix T;
	T.mat[0][0] = 1LL, T.mat[0][1] = 1LL;
	T.mat[1][0] = 1LL, T.mat[1][1] = 0LL;
	matrix tmp = (T^(n-1));
	LL ans = ((tmp.mat[0][0] + tmp.mat[1][0])%mod + mod) % mod;
	mod++;
	return ans; 
}

LL fun(LL a, LL b, LL n)
{
	LL ans = (fpow(a, (mod - 1)>>1) + 1LL) * (fpow(b, (mod-1)>>1) + 1LL) % mod;//注意这里也要取模...
	if(ans == 0)
		return 0;
	if(n < 2)
		return (a + b) * 2LL % mod * ans % mod;
	LL e = fib(n);
	//cout << e << endl;
	if(e == 0)
		e = mod - 1;
	matrix A;
	A.mat[0][0] = (a + b) * 2LL % mod, 	   A.mat[0][1] = 1LL;
	A.mat[1][0] = (b - a) * (a - b) % mod, A.mat[1][1] = 0LL;
	A = A^(e - 1);
	ans = (ans * ((a + b) * 2LL % mod * A.mat[0][0] + A.mat[1][0] * 2LL % mod) % mod) % mod;
	while(ans < 0)
		ans += mod;
	return ans;
}
int main()
{
	int T;
	cin >> T;
	while(T--)
	{
		LL a, b, n;
		scanf("%lld%lld%lld%lld", &a, &b, &n, &mod);
		LL ans = fun(a, b, n);
		printf("%lld
", ans);
	}
    return 0;
}