P1349 广义斐波那契数列(矩阵乘法)

题目

P1349 广义斐波那契数列

解析

把普通的矩阵乘法求斐波那契数列改一改，随便一推就出来了

\[\begin{bmatrix}f_2\\f_1
\end{bmatrix}\begin{bmatrix}
p&q\\
1&0\\
\end{bmatrix}^{n-2}=\begin{bmatrix}f_n\\f_{n-1}
\end{bmatrix}\]

水题

代码

#include <bits/stdc++.h>
#define int long long

using namespace std;

const int N = 100;

int n, m, a1, a2, p, q;

struct matrix {
	int a[N][N];
	matrix() {
		memset(a, 0, sizeof a);
	}

	void InitMatrix() {
		a[1][1] = p, a[1][2] = q;
		a[2][1] = 1;
	}

	matrix operator * (const matrix &oth) const {
		matrix ans;
		for (int k = 1; k <= 2; ++k)
			for (int i = 1; i <= 2; ++i)
				for (int j = 1; j <= 2; ++j)
					ans.a[i][j] = (ans.a[i][j] + (a[i][k] * oth.a[k][j]) % m) % m;
		return ans;
	}

} init;

matrix qpow(matrix a, int b) {
	matrix ans = init;
	while (b) {
		if (b & 1) ans = ans * a;
		b >>= 1, a = a * a;
	}
	return ans;
}

template<class T>inline void read(T &x) {
	x = 0; int f = 0; char ch = getchar();
	for ( ; !isdigit(ch); ch = getchar()) f |= (ch == '-');
	for ( ; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
	x = f ? -x : x;
	return;
}

signed main() {
	read(p), read(q), read(a1), read(a2), read(n), read(m);
	if (n <= 2) {
		printf("%lld", n == 1 ? a1 : a2);
		return 0;
	}
	init.InitMatrix();
	init = qpow(init, n - 3);
	printf("%lld\n", (a2 * init.a[1][1] + a1 * init.a[1][2]) % m);
	return 0;
}