【Loj10222】佳佳的Fibonacci

题面

题解

可以发现(T(n))无法用递推式表示。

于是我们做如下变形：

[T(n) = sum _ {i = 1} ^ n i imes f_i \ S(n) = sum _ {i = 1} ^ n f_i \ herefore nS(n) - T(n) = sum _ {i = 1} ^ {n - 1} (n - i)f_i \ ]

令(p(n) = nS(n) - T(n))

则(p(n + 1) = p(n) + S(n))

用矩阵乘法即可。

#include<cstdio>
#include<cstring>
#include<algorithm>
#define RG register
#define file(x) freopen(#x".in", "r", stdin);freopen(#x".out", "w", stdout);

inline int read()
{
	int data = 0, w = 1;
	char ch = getchar();
	while(ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
	if(ch == '-') w = -1, ch = getchar();
	while(ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = getchar();
	return data * w;
}

int n, Mod;
inline void Add(int &x, const int &y) { x += y; if(x >= Mod) x -= Mod; }
inline int Mul(const int &x, const int &y) { return 1ll * x * y % Mod; }

template<int N, int M>
struct Matrix
{
	int a[4][4];
	Matrix() { memset(a, 0, sizeof(a)); }
	inline int *operator [] (const int &id) { return a[id]; }

	template<int K> inline Matrix<N, K> operator * (const Matrix<M, K> &b) const
	{
		Matrix<N, K> c;
		for(RG int i = 0; i < N; i++)
			for(RG int j = 0; j < M; j++)
				for(RG int k = 0; k < K; k++)
					Add(c[i][k], 1ll * a[i][j] * b.a[j][k] % Mod);
		return c;
	}
};

Matrix<1, 4> S;
Matrix<4, 4> T;

int main()
{
	n = read(); Mod = read();
	int Tmp = n;

	S[0][1] = T[0][0] = T[0][1] = T[0][2] = T[1][0] = T[1][2] = T[2][2] = T[2][3] = T[3][3] = 1;
	while(Tmp)
	{
		if(Tmp & 1) S = S * T;
		T = T * T, Tmp >>= 1;
	}

	printf("%d
", (Mul(n, S[0][2]) - S[0][3] + Mod) % Mod);
	return 0;
}