[Logo P3390] [Template] Matrix Fast Power [Matrix Multiplication]

Topic background

l i n k link l i n k
matrix fast power

Title description

Given n × nn×nn×matrixAA of nA , findA k A^kAk

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 A i, j A i,jAt i ,j

Output format

Output A k A^kAThere are
n rows in k , and each row has n numbers. The j-th number in the i-th row represents(A k) i, j (A^k)i,j(Ak)i,j , each element pair1 0 9 + 7 10^9+7109+7 Take the modulo.

Sample input and output

Enter #1

2 1
1 1
1 1

Output #1

1 1
1 1

analysis:

There are many ways to write matrix multiplication + matrix fast power template. Note n, kn, kn,k openlong longlong l o n g long l o n g
matrixalso needs to open the calculation process may explodeint inti n t
is thematrix multiplication infast power.Overload* * is sufficient.

CODE:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=105;
const int mod=1000000007;
typedef long long ll;
ll n,k;
struct matrix{
    
    
	ll G[N][N];
}A,B;
matrix operator *(matrix a,matrix b)  //重载乘号
{
    
    
	matrix C;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			C.G[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.G[i][j]=(C.G[i][j]+a.G[i][k]*b.G[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;  //快速幂
	if(x&1) B=B*A;
}
int main(){
    
    
	scanf("%lld%lld",&n,&k);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			scanf("%lld",&A.G[i][j]);
	ksm(k);
	for(int i=1;i<=n;i++)
	{
    
    
		for(int j=1;j<=n;j++)
			printf("%lld ",B.G[i][j]);
		printf("\n");
	}
	
	return 0;
} 

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Origin blog.csdn.net/dgssl_xhy/article/details/111058072