[PKU 3233] Matrix Power Series [variation of matrix multiplication]

$Time$ $Limit:$ $3000MS$
$Memory$ $Limit:$ $131072K$

Description

$link$
Given a $n\times n$ 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.

Sample Input

2 2 4
0 1
1 1

Sample Output

1 2
2 3

analysis:

Direct calculation will overtime consider the matrix $[A^{i-1},S_{i-2}]$
$[A^i,S_{i-1}]=[A^{i-1},S_{i-2}]\ast F$
$F$ is thetransformation matrix,then$F$ :
$$A,E$$

$$O,E$$
其中$E$ represents theidentity matrix $O$ is all$0$ matrix.
Note that theinitial matrixconstruction should beexpanded toconstruct
$S$ is in theinitial matrix$[1][2]$

CODE：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
ll n,mod,k;
struct matrix{
    
    
	ll n,m;
	ll F[101][101];
}A,B,C;
matrix operator *(matrix a,matrix b)
{
    
    
	matrix c;
	c.n=a.n;c.m=a.m;
	for(int i=1;i<=c.n;i++)
		for(int j=1;j<=c.m;j++)
			c.F[i][j]=0;
	for(int k=1;k<=a.m;k++)
		for(int i=1;i<=c.n;i++)
			for(int j=1;j<=c.m;j++)
				c.F[i][j]=(c.F[i][j]+a.F[i][k]*b.F[k][j]%mod)%mod;  //矩阵乘法
	return c;
}
void ksm(ll x)
{
    
    
	if(x==1){
    
    
		B=A;
		return;
	}
	ksm(x/2);
	B=B*B;
	if(x&1) B=B*A;
}
int main(){
    
    
	scanf("%lld%lld%lld",&n,&k,&mod);
	C.n=n;C.m=2*n;A.n=n*2;A.m=n*2;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			scanf("%lld",&C.F[i][j]);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			A.F[i][j]=C.F[i][j];
	for(int i=1;i<=n;i++)
		for(int j=n+1;j<=n+n;j++)
			if(i+n==j)  //赋单位矩阵
				A.F[i][j]=1;  //相当于初始矩阵[1][2]的初值
	for(int i=n+1;i<=n+n;i++)
		for(int j=n+1;j<=n+n;j++)
			if(i==j)  //同单位矩阵
				A.F[i][j]=1;  //初始矩阵[2][2]的初值
	ksm(k);
	C=C*B;
	for(int i=1;i<=n;i++)
	{
    
    
		for(int j=n+1;j<=n+n;j++)
			printf("%lld ",C.F[i][j]);  //输出[1][2]
		printf("\n");
	}
	
	return 0;
}