【SSL_1530】Pepperach series III

Fibonacci Sequence III

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Topic link: Peppolaci Sequence III

Problem solving ideas

设$\left|\begin{array}{ccc}f[n-2]& f[n-1]& 1\end{array}\right|$ 和 $\left|\begin{array}{ccc}0& 1& 0\\ 1& 1& 0\\ 0& 1& 1\end{array}\right|$
Others are almost the same as the Peppolac sequence II

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;
}