矩阵快速幂和矩阵加速

矩阵快速幂

先贴一下快速幂的板子

void ksm(int a, int k) { // a^k
	int res;
	while (k) {
		if (k & 1) res *= a;
		k >>= 1;
		a *= a;
	}
	return res;
} 

---

众所周知，矩阵乘法的公式是

$C_{i,j} = \sum_{k=1}^n b_{i,k} \times c_{k,j}$

所以写个矩阵的结构体，和乘法的函数，套进快速幂就行了。时间复杂度 $O(n^3 \log k)$

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 100 + 5, INF = 0x3f3f3f3f, mod = 1e9 + 7;
inline int read() {
	int x = 0, f = 0; char ch = 0;
	while (!isdigit(ch)) f |= ch == '-', ch = getchar();
	while (isdigit(ch)) x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
	return f ? -x : x;
}
int n;
struct mat {
	int m[N][N];
	mat() {
		memset(m, 0, sizeof(m));
		for (int i = 0; i < N; i++) m[i][i] = 1;
	}
};


mat mul(mat a, mat b) {
	mat c;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++) {
			c.m[i][j] = 0; //Mention! 
			for (int k = 1; k <= n; k++) 
				c.m[i][j] += a.m[i][k] * b.m[k][j] % mod;
			c.m[i][j] %= mod;	
		}
	return c;
}
mat ksm(mat a, int k) {
	mat res;
	while (k) {
		if (k & 1) res = mul(res, a);
		k >>= 1;
		a = mul(a, a);
	}
	return res;
}
signed main() {
	n = read(); int k = read();
	
	mat a; 
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
			a.m[i][j] = read();
	mat ans = ksm(a, k);
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++)
			printf("%lld ", ans.m[i][j]);
		puts("");
	}		
	return 0;
}

【模板】矩阵加速

已知一个数列 $a$，它满足
$a_{x}=\left\{\begin{array}{ll} 1 & x \in\{1,2,3\} \\ a_{x-1}+a_{x-3} & x \geq 4 \end{array}\right.$
求 $a$ 数列的第 $n$ 项对 $10^{9}+7$ 取余的值。

已知 $f_{i-1},f_{i-2},f_{i-3}$ 发现可以推出 $f_{i}, f_{i+1}, f_{i+2}$

那么有

$\begin{bmatrix} f_i &f_{i+1} & f_{i+2} \end{bmatrix}= \begin{bmatrix} f_{i-1} &f_{i-2} & f_{i-3} \end{bmatrix} \times \begin{bmatrix} ? & ? & ? \\ ? & ? & ? \\ ? & ? & ? \end{bmatrix}$

根据递推式

$\begin{cases} f_{i} = 1 \times f_{i-1} + 0 \times f_{i-2} + 1 \times f_{i-3} \\ f_{i+1} = f_{i} + f_{i-2} = 1 \times f_{i-1} + 1 \times f_{i-2} + 1 \times f_{i-3} \\ f_{i+2} = f_{i+1} + f_{i-1} = 2 \times f_{i-1} + 1 \times f_{i-2} + 1 \times f_{i-3} \end{cases}$

竖着写，用来加速的矩阵为

$\begin{bmatrix} 1 & 1 & 2\\ 0 & 1 & 1\\ 1 & 1 & 1\\ \end{bmatrix}$