Matrix Power Series【Matrix Multiplication】

Time Limit: 3000MS Memory Limit: 131072K
Total Submissions: 35006 Accepted: 14029

---

Description

Given a $n\times n$ matrix A and a positive integer k, find the sum $S=A+A2+A3+\dots +Ak$ .

---

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

---

Topic translation

Given a $n\times n$ matrix a and a positive integer k, sum$S=a+A2+A3+\dots +Ak$。

The input
contains only one test case. The first line of input contains three positive integers $n（n\le 30）、k（k\le 10^9\left)andm\right(m<10^4)$. Then follow n lines, each line contains$For$ n non-negative integers below$3$$2$$7$$6$$8$ , the elements of A are given in the main order of the row.

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

---

Problem solving ideas

Matrix multiplication

Consider $1\times$Matrix of$2$$【A^n-1,S[n-2]】$ , note that the 2 elements of this 1×2 matrix are all square matrices of order r! We hope to obtain a 1×2 matrix[A n, S [n − 1]] = [A n − 1 ∗ A, A n − 1 + S [n − 2]by multiplying by a certain 2×2 matrix M$【A^n,S[n-1]】＝【A^n-1\ast A,An-1+S[n-2]]\; It\; is$
easy to construct this matrix M as:

where 4 elements are all square matrices of order r, O means a matrix of all 0s of order r, and E means an identity matrix (1 on the main diagonal, and all 0 on the other).
problem solved. Our complexity is:$\left(2r\right)3\ast logn$

In fact, this matrix is ​​used in that popular method. Here we use the ideas of the previous questions to easily construct this $2r\ast 2r$ matrix. Therefore: This idea is efficient, general, unified and harmonious!

---

Code

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
long long nn,k,mm;
struct c{
    
    
	long long n,m;
	long long a[100][100];
}A,B,CC;
c operator *(c A,c B){
    
    
	c C;
	C.n=A.n,C.m=B.m;
	for(int i=1;i<=C.n;i++)
		for(int j=1;j<=C.m;j++)
			C.a[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.a[i][j]=(C.a[i][j]+(A.a[i][k]*B.a[k][j])%mm)%mm;
	}	
	return C;
}
void poww(long long x){
    
    
	if(x==1)
	{
    
    
		B=A;
		return; 
	}
	poww(x>>1);
	B=B*B;
	if(x&1)
		B=B*A;
	
}
int main(){
    
    
	scanf("%lld%lld%lld",&nn,&k,&mm);
	CC.n=nn,CC.m=2*nn;
	A.n=2*nn,A.m=2*nn;
	for(int i=1;i<=nn;i++)
		for(int j=1;j<=nn;j++)
		{
    
    
			scanf("%lld",&CC.a[i][j]);
			CC.a[i][j]=CC.a[i][j]%mm;
			A.a[i][j]=CC.a[i][j];
		}	
	for(int i=1;i<=nn;i++)
		for(int j=nn+1;j<=2*nn;j++)
			if(j-nn==i)
				A.a[i][j]=1;	
	for(int i=nn+1;i<=2*nn;i++)
		for(int j=nn+1;j<=2*nn;j++)
			if(j==i)
				A.a[i][j]=1;
	poww(k);
	CC=CC*B;
	
	for(int i=1;i<=nn;i++)
	{
    
    
		for(int j=nn+1;j<=2*nn;j++)
			printf("%lld ",CC.a[i][j]);
		printf("\n");
	}		
}