【Matrix Multiplication】【POJ 3233】Matrix Power Series

topic

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

enter

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. The
input contains only 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 rows, each row contains n non-negative integers below 32768, and 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 is given. Output the elements of S modulo m in the same way as A is given
.

Sample

input
2 2 4
0 1
1 1

output
1 2
2 3

Problem-solving ideas

Simplify
the meaning of the question first.
It is easy to deduce that the answer matrix is {a ^n-1 , sn-2} and the
multiplication matrix is

Insert picture description here
Then convert back.
The multiplication matrix (1,1) becomes matrix A
(1,2) and (2,2) becomes the identity matrix (that is, the value on the diagonal is 1, and the other grids are 0)
~~Because the product of a matrix multiplied by the identity matrix is equal to this matrix~~
, the multiplication matrix of the sample is as long as this
Insert picture description here

Code

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
struct lzf{
    
    
	long long n,m,h[120][120];
}a,b,x;
long long n,k,mo;
lzf operator * (lzf l,lzf y)
{
    
    
	x.n=l.n,x.m=y.m;
	memset(x.h,0,sizeof(x.h));
	for (int i=1;i<=x.n;i++)
	    for (int k=1;k<=l.m;k++)
	        for (int j=1;j<=x.m;j++)
	            x.h[i][j]=(x.h[i][j]+l.h[i][k]*y.h[k][j]%mo)%mo;
	return x;
}
void power(long long n)
{
    
    
	 while (n)
	 {
    
     
	 	   if (n & 1) a=a*b;
	 	   b=b*b;
	 	   n>>=1;
	 }
}
int main()
{
    
    
	scanf("%lld%lld%lld",&n,&k,&mo);
	a.n=n,a.m=2*n;
	b.n=b.m=2*n;
	for (int i=1;i<=n;i++)
	    for (int j=1;j<=n;j++)
	    {
    
    
	        scanf("%lld",&a.h[i][j]);  //输入矩阵A
	        b.h[i][j]=a.h[i][j];  //相乘矩阵（1，1）是矩阵A
	    }
    for (int i=1;i<=n;i++)
        b.h[i][i+n]=b.h[i+n][i+n]=1;  //（1，2）（2，2）是正对角线为1
    power(k);  
    for (int i=1;i<=n;i++)
    {
    
    
        for (int j=1;j<=n;j++)
            printf("%lld ",a.h[i][j+n]);
        printf("\n");
    }  //输出最后的和
    return 0;
}