P3390 [Template] Matrix fast power (matrix multiplication + fast power)

P3390 [Template] Matrix fast power

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Problem solving ideas

This title is a matrix multiplication + fast power template
first talk about matrix multiplication
Baidu Encyclopedia explanation of

code implementation

	for(int o=1;o<=k;o++)//先枚举中间的数，更快  
	 for(int i=1;i<=n;i++)
	  for(int j=1;j<=m;j++)
	   c[i][j]+=a[i][o]*b[o][j];

Then there is the fast power
Baidu Encyclopedia
to facilitate the implementation of the power
code

node ksm(node x,long long k)
{
    
    
	node y=x;//初值
	while(k)
	{
    
    
		if(k%2)y=x*y;//奇数
		x=x*x;
		k/=2;//除2
	}
	return y;
}

Finally, this question is to combine these two algorithms

AC code

#include<cstdio>
using namespace std;
long long n,k;
struct node//结构体
{
    
    
	long long oo[105][105];
}a;
node operator*(node x,node y)//更改“*”的定义，变为矩阵乘法
{
    
    
	node z;
	for(long long i=1;i<=n;i++)
	 for(long long j=1;j<=n;j++)
	  z.oo[i][j]=0;
	for(long long o=1;o<=n;o++)  
	 for(long long i=1;i<=n;i++)
	  for(long long j=1;j<=n;j++)
	   z.oo[i][j]=(z.oo[i][j]+x.oo[i][o]*y.oo[o][j])%1000000007;
	return z;
}
node ksm(node x,long long k)//快速幂
{
    
    
	node y=x;
	while(k)
	{
    
    
		if(k%2)y=x*y;
		x=x*x;
		k/=2;
	}
	return y;
}
int main()
{
    
    
	scanf("%lld%lld",&n,&k);
	for(long long i=1;i<=n;i++)
	 for(long long j=1;j<=n;j++)
	  scanf("%lld",&a.oo[i][j]);
	a=ksm(a,k-1);//快速幂，记住是k-1，因为有初值
	for(long long i=1;i<=n;i++,printf("\n"))
	 for(long long j=1;j<=n;j++)
	  printf("%lld ",a.oo[i][j]);
	return 0;
}

Thank you