[Logo P3390] [Quick Matrix Power] [Template]

Title description

Given an n×n matrix A, find A ^k

Input format

Two integers n,k in the first line, n lines in the next, n integers in each line, the j-th number in the i-th line represents Ai,j

Output format

Output A ^k

There are a total of n rows, each with n numbers. The j-th number in the i-th row represents A ^k i,j, and each element is modulo 10 ⁹ +7.

Sample input and output

Enter #1

2 1
1 1
1 1
1
2
3

Output #1

1 1
1 1
1
2

analysis:

The
key to the template problem of matrix multiplication and matrix fast power is to redefine " $\ast$ ".
Let the multiplication sign be defined as matrix multiplication.
Then just call it directly.

Upload code

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;

ll n,k;
const int mod=1000000007;

struct matrix
{
    
    
	ll f[101][101];
}A,B;

matrix operator *(matrix a,matrix b)
{
    
    
	matrix C;
	for(int i=1;i<=n;i++)
	{
    
    
		for(int j=1;j<=n;j++)
		{
    
    
			C.f[i][j]=0;
		}
	}
	for(int k=1;k<=n;k++)
	{
    
    
		for(int i=1;i<=n;i++)
		{
    
    
			for(int j=1;j<=n;j++)
			{
    
    
				C.f[i][j]=(C.f[i][j]+a.f[i][k]*b.f[k][j]%mod)%mod;
			}
		}
	}
	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()
{
    
    
    cin>>n>>k; 
    for(int i=1;i<=n;i++)
	{
    
    
		for(int j=1;j<=n;j++)
		{
    
    
			scanf("%lld",&A.f[i][j]);
		}
	}
	ksm(k);
	for(int i=1;i<=n;i++)
	{
    
    
		for(int j=1;j<=n;j++)
		{
    
    
			cout<<B.f[i][j]<<' ';
		}
		cout<<endl;
	 } 
	return 0;
}