Matrix Power Series - 矩阵快速幂对分块矩阵加速

Posted 2021-02-19 emcikem

题目
其中(A)是一个(n imes n)的矩阵，(S_k = A + A^2 + A^3 + … + A^k)，求(S_k)
按照数论出现和，那么构造一个和数论一起递推的式子(S_k = S_{k - 1} + A^k)
那么假设(A)不是矩阵

[left[egin{array}{l} 1 & 1 & A end{array} ight] imes left[egin{array}{l} S_{k - 1}A^k end{array} ight] = left[egin{array}{l} S_kA^{k + 1} end{array} ight]]

用E代替1，用矩阵代替数字，转换一下就是

[left[egin{array}{l} E & E & A end{array} ight] imes left[egin{array}{l} S_{k - 1}A^k end{array} ight] = left[egin{array}{l} S_kA^{k + 1} end{array} ight] ]

构建了个分块矩阵，大小是(2n imes 2n)
加速矩阵(left[egin{array}{l} E & E & A end{array} ight]) 初始矩阵(left[egin{array}{l} S_1A^2 end{array} ight])

#include <iostream>
#include <cstdio>
#include <cstring>
#define ll long long
using namespace std;
const int N = 61;
int mod, k, n;
struct Matrix{//矩阵
    int n,m;
    int a[N][N];
    Matrix(int x,int y):n(x),m(y){memset(a,0,sizeof(a));}
    Matrix operator * (const Matrix &b){
        Matrix ans(n,b.m);
        for(int i = 0; i < n; i++){
            for(int j = 0; j < b.m; j++){
                for(int k = 0; k < m; k++){
                    ans.a[i][j] = (ans.a[i][j] + a[i][k] * b.a[k][j] % mod) % mod;
                }
            }
        }
        return ans;
    }
};
Matrix ksm(Matrix a, ll b){
	Matrix ans(a.n, a.m);
	for(int i = 0; i <= max(a.n, a.m); i++)
		ans.a[i][i] = 1;

	while(b){
		if(b & 1)ans = ans * a;
		a = a * a;
		b >>= 1;
	}
	return ans;
}
Matrix a(35, 35);
void solve(){
	Matrix base(k * 2, k * 2);
	for(int i = 0; i < k; i++)
		base.a[i][i] = base.a[i][i + k] = 1;

	for(int i = k; i < 2 * k; i++)
		for(int j = k; j < 2 * k; j++)
			base.a[i][j] = a.a[i - k][j - k];

	base = ksm(base, n - 1);
	

	Matrix ans(k * 2, k);
	for(int i = 0; i < k; i++)
		for(int j = 0; j < k; j++)
			ans.a[i][j] = a.a[i][j];

	a = a * a;
	for(int i = k; i < 2 * k; i++)
		for(int j = 0; j < k; j++)
			ans.a[i][j] = a.a[i - k][j];

	ans = base * ans;
	for(int i = 0; i < k; i++){
		for(int j = 0; j < k; j++)
			printf("%d ", ans.a[i][j]);
		putchar(‘
‘);
	}
}
int main(){
	scanf("%d%d%d", &k, &n, &mod);
	for(int i = 0; i < k; i++)
		for(int j = 0; j < k; j++)
			scanf("%d", &a.a[i][j]);
	solve();
	return 0;
}