[SSL 1530] Fibonacci Sequence III [Matrix Multiplication]

Fibonacci Sequence III

$Time$ $Limit:10000MS$ $Memory$ $Limit:65536K$
$Case$ $Time$ $Limit:1000MS$

Description

Find the sequence $f[n]=f[n-1]+f[n-2]+1$ The first of$N$ items.$f[1]=1,f[2]=1$.

Input

$n(1＜n＜{\mathrm{２}}^{31-1})$

Output

The first $N$ item result$mod$ $9973$

Sample Input

12345

Sample Output

8932

analysis:

Matrix multiplication and this template is similar to
the same considerations matrix $[f_{n-1},f_{n-2},1]$
Multiply a certain matrix to get
$$[f_{n-1},f_n,1]=[f_{n-1},f_{n-2}+f_{n-1},1]\; It\; is$$
easy to get the matrix as:
$$0,1,0$$

$$1,1,0$$

$$0,1,1$$
Thenjustdomatrix multiplicationdirectly

CODE：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int mod=9973;
long long n;
struct matrix{
    
    
	long long n,m;
	long long G[11][11];
}base,A,B;
matrix operator *(matrix a,matrix b)
{
    
    
	matrix C;
	C.n=a.n;C.m=b.m;
	for(int i=1;i<=C.n;i++)
		for(int j=1;j<=C.m;j++)
			C.G[i][j]=0;
	for(int k=1;k<=a.m;k++)
		for(int i=1;i<=a.n;i++)
			for(int j=1;j<=b.m;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(long long x){
    
    
	if(x==1){
    
    
		A=base;
		return;
	}
	ksm(x/2);
	A=A*A;  //快速幂
	if(x&1) A=A*base;
}
int main(){
    
    
	scanf("%lld",&n);
	base.n=3;base.m=3;
	base.G[1][1]=0;base.G[1][2]=1;base.G[1][3]=0;
	base.G[2][1]=1;base.G[2][2]=1;base.G[2][3]=0;  //乘的矩阵
	base.G[3][1]=0;base.G[3][2]=1;base.G[3][3]=1;
	if(n<=2){
    
    
		printf("1");
		return 0;
	}
	else{
    
    
		B.m=3;B.n=1;
		B.G[1][1]=1;B.G[1][2]=1;B.G[1][3]=1;  //设的矩阵
		ksm(n-1);
		B=B*A;
		printf("%lld",B.G[1][1]);
	}
	
	return 0;
}