poj 3233 Matrix Power Series（矩阵k次幂前缀和）

ACM题集：https://blog.csdn.net/weixin_39778570/article/details/83187443
题目链接：http://poj.org/problem?id=3233

Description

Given a n × n matrix A and a positive integer k, find the sum $S=A+A^2+A^3+\dots +A^k$.

Input

The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 10⁹) and m (m < 10⁴). Then follow n lines each containing n nonnegative integers below 32,768, giving A’s elements in row-major order.

Output

Output the elements of S modulo m in the same way as A is given.

求a的前n次幂缀和矩阵可以这么构造
| a^1, 1 |^(n)                         | a^n, a^0+a^1+a^2+...+a^(n-1) |
| 0 ,  1 |           =======>          | 0  , 1                       |
同理，求矩阵A的前n次幂的前缀和也可以使用相同的方法构造
| A^1, A^0 |^(n)                       | A^n, A^0+A^1+A^2+...+A^(n-1) |
| 0  , A^0 |         =======>          | 0  , A^0                     |	
这里的A^0是单位矩阵

code

#include
#include
#include
#include
#define fo(i,j,n) for(register int i=j; i<=n; ++i)
using namespace std;
const int N=62; // 注意要开两倍大
int n,k,mod;
struct mat{
	int mp[N][N],len;
	mat(int x){
		len = x;memset(mp, 0, sizeof(mp));
	}
	mat friend operator *(mat a, mat b){
		mat ret(a.len);
		// k,i,j比ijk更快! 
		fo(k,1,ret.len)
			fo(i,1,ret.len)
				fo(j,1,ret.len)
					ret.mp[i][j] = (ret.mp[i][j] + a.mp[i][k]*b.mp[k][j]) % mod;
		return ret;
	}
	mat friend operator ^(mat a, int n){
		mat ret(a.len);
		for(int i=1; i<=ret.len; i++)ret.mp[i][i] = 1;
		while(n){
			if(n&1)ret = ret*a;
			a = a*a;
			n>>=1;
		}
		return ret;
	}
	void debug(){
		for(int i=1; i<=len; i++){
			for(int j=1; j<=len; j++){
				printf("%d%c",mp[i][j],j==len?'\n':' ');
			}
		} 
	}
}; 
int main(){
	scanf("%d%d%d",&n,&k,&mod);
	// 构造矩阵 
	mat A(n+n);
	fo(i,1,n)fo(j,1,n)scanf("%d",&A.mp[i][j]);
	fo(i,1,n)A.mp[i][i+n] = A.mp[i+n][i+n] = 1; 
	// 计算 
	A = A^(k+1);
	fo(i,1,n)A.mp[i][i+n]=(A.mp[i][i+n]-1+mod)%mod; // 减去A^0
	// 输出 
	fo(i,1,n)fo(j,1,n)
		printf("%d%c",A.mp[i][j+n],j==n?'\n':' ');
	return 0;
}