Matrix Power Series（矩阵快速幂+求幂和）_matrix矩阵的幂-优快云博客

Matrix Power Series

Time Limit: 3000MS		Memory Limit: 131072K
Total Submissions: 17862		Accepted: 7546

Description

Given a n × n matrix A and a positive integer k, find the sum S = A + A² + A³ + … + 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.

Sample Input

2 2 4
0 1
1 1

Sample Output

1 2
2 3

Source

POJ Monthly--2007.06.03, Huang, Jinsong

AC code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#define LL long long
#define MAXN 1000010
using namespace std;
const int INF=0x3f3f3f3f;
//----以下为矩阵快速幂模板-----// 
//const int mod=1000;//模3，故这里改为3即可 
int mod=1000;
const int NUM=32;//定义矩阵能表示的最大维数 
int N;//N表示矩阵的维数，以下的矩阵加法、乘法、快速幂都是按N维矩阵运算的 
struct Mat{//矩阵的类
	int a[NUM][NUM];
	Mat(){memset(a,0,sizeof(a));}  
	void init()//将其初始化为单位矩阵  
	{
		memset(a,0,sizeof(a));
		for(int i=0;i<NUM;i++)
		{
			a[i][i]=1;
		}
	}
};
Mat add(Mat a,Mat b)//(a+b)%mod  矩阵加法  
{
	Mat ans;
	for(int i=0;i<N;i++)
	{
		for(int j=0;j<N;j++)
		{
			ans.a[i][j]=(a.a[i][j]%mod)+(b.a[i][j]%mod);
			ans.a[i][j]%=mod;
		}
	}
	return ans;
}
Mat mul(Mat a,Mat b) //(a*b)%mod  矩阵乘法  
{
	Mat ans;
	for(int i=0;i<N;i++)
	{
		for(int j=0;j<N;j++)
		{
			ans.a[i][j]=0;
			for(int k=0;k<N;k++)
			{
				ans.a[i][j]=(ans.a[i][j]%mod)+(a.a[i][k]%mod)*(b.a[k][j]%mod);
			}
			ans.a[i][j]%=mod;
		}
	}
	return ans;
}
Mat power(Mat a,int num)//(a^n)%mod  矩阵快速幂 
{
	Mat ans;
	ans.init();
	while(num)
	{
		if(num&1)
		{
			ans=mul(ans,a);
		}
		num>>=1;
		a=mul(a,a);
	}
	return ans;
}
Mat pow_sum(Mat a,int num)//(a+a^2+a^3....+a^n)%mod 矩阵的幂和
{
	int m;
	Mat ans,pre;
	if(num==1)
		return a;
	m=num/2;
	pre=pow_sum(a,m);
	ans=add(pre,mul(pre,power(a,m)));
	if(num&1)
		ans=add(ans,power(a,num));
	return ans;
}
void output(Mat a)//输出矩阵 
{
	for(int i=0;i<N;i++)
	{
		for(int j=0;j<N;j++)
		{
			printf("%d%c",a.a[i][j],j==N-1?'\n':' ');
		}
	}
}
//----以上为矩阵快速幂模板-----// 
int main()
{
	//freopen("D:\in.txt","r",stdin);
	int n,m,i,j,T,k;
	scanf("%d%d%d",&n,&k,&m); 
	Mat A,S;
	N=n;
	mod=m;
	for(i=0;i<N;i++)
	{
		for(j=0;j<N;j++)
		{
			scanf("%d",&A.a[i][j]);
		}
	}
	S=pow_sum(A,k);
	output(S);

	return 0;
}