poj3233 Matrix Power Series

分治即可

当然也可以把矩阵看成元素用矩阵快速幂做

#include <iostream>
#include <cstdio>
using namespace std;
int n, k, m;
struct Matrix{
	int num[35][35];
	Matrix operator*(const Matrix &x)const{
		Matrix re;
		for(int i=1; i<=n; i++)
			for(int j=1; j<=n; j++){
				re.num[i][j] = 0;
				for(int k=1; k<=n; k++)
					re.num[i][j] = (re.num[i][j] + num[i][k] * x.num[k][j]) % m;
			}
		return re;
	}
	Matrix operator+(const Matrix &x)const{
		Matrix re;
		for(int i=1; i<=n; i++)
			for(int j=1; j<=n; j++)
				re.num[i][j] = (num[i][j] + x.num[i][j]) % m;
		return re;
	}
}a, dan;
Matrix ksm(Matrix aa, int b){
	Matrix re=dan;
	while(b){
		if(b&1)	re = re * aa;
		aa = aa * aa;
		b >>= 1;
	}
	return re;
}
Matrix sum(int x){
	if(x==1)	return a;
	Matrix tmp=sum(x/2);
	if((x&1)==0)
		return tmp+tmp*ksm(a, x/2);
	else
		return tmp+tmp*ksm(a, x/2)+ksm(a, x);
}

int main(){
	cin>>n>>k>>m;
	for(int i=1; i<=n; i++)
		for(int j=1; j<=n; j++)
			if(i==j)	dan.num[i][j] = 1;
			else	dan.num[i][j] = 0;
	for(int i=1; i<=n; i++)
		for(int j=1; j<=n; j++)
			scanf("%d", &a.num[i][j]);
	Matrix ans=sum(k);
	for(int i=1; i<=n; i++){
		for(int j=1; j<=n; j++)
			printf("%d ", ans.num[i][j]);
		printf("
");
	}
	return 0;
}