[Luogu_P3390] [Template] Matrix fast power

[Template] Matrix fast power

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Topic link: [Template] Matrix fast power

Problem solving ideas

This is a template question, the question is very clear that this is a quick exponent .
In fact, it is similar to fast power, but there is one thing to pay attention to
fast power $ans\; is$ generally assigned as$1.\; The$ purpose is to make$ans$ can get the value directly in the first operation.
But what should we use to achieve the goal in matrix multiplication?
The easiest way is to performans ans for thefirst time$In$ the calculation of$a$$n$$s$ , it is judged and then assigned.
But there is another idea: after
a matrix is ​​multiplied by another matrix, the original matrix can be obtained. What will this matrix be?
Such as:
$$\left|\begin{array}{cc}1& 2\\ 3& 4\end{array}\right|$$
We go back and get:
$$\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|$$
Verify that it is correct.

In this way, quick power is enough

code

#include<iostream>
#include<cstdio>
#include<cstring>
#define lzh using
#define ak namespace
#define ioi std
lzh ak ioi;

const int mod=9973;

int n;
int a[4][4];
int ans[4][4];
int t[4][4];

void add()
{
    
    
	memset(t,0,sizeof(t));
	for(int i=1;i<=3;i++)
		for(int j=1;j<=3;j++)
			for(int k=1;k<=3;k++)
				t[i][j]=(t[i][j]+ans[i][k]*a[k][j])%mod;
	for(int i=1;i<=3;i++)
		for(int j=1;j<=3;j++)
			ans[i][j]=t[i][j];
}

void cf()
{
    
    
	memset(t,0,sizeof(t));
	for(int i=1;i<=3;i++)
		for(int j=1;j<=3;j++)
			for(int k=1;k<=3;k++)
				t[i][j]=(t[i][j]+a[i][k]*a[k][j])%mod;
	for(int i=1;i<=3;i++)
		for(int j=1;j<=3;j++)
			a[i][j]=t[i][j];
}

void ksm(int b)
{
    
    
	for(int i=1;i<=3;i++)
		ans[i][i]=1;
	while(b)
	{
    
    
		if(b&1)
			add();
		cf();
		b>>=1;
	}
}

int main()
{
    
    
	cin>>n;
	if(n<=2)
	{
    
    
		cout<<1<<endl;
		return 0;
	}
	a[1][1]=0,a[1][2]=1,a[1][3]=0;
	a[2][1]=1,a[2][2]=1,a[2][3]=0;
	a[3][1]=0,a[3][2]=1,a[3][3]=1;
	ksm(n-2);
	cout<<(ans[1][2]+ans[2][2]+ans[3][2])%mod<<endl;
}