Find the sum of the first n terms of the Fibonacci sequence [matrix multiplication]

Description

数列$f[n]=f[n-1]+f[n-2]，f[1]=f[2]=1$ before$sum\; of\; n$ terms$s[n]$

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Input

N(１＜Ｎ＜2³¹)

Output

Nth result

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Sample Input

12345

Sample Output

8995

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Source

elba

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Problem solving ideas

Matrix multiplication

Method 1: Although we have $S[n]=F[n+2]-1$ , but this article does not consider this method, we want to get a more general method.

Method 2: Following the previous ideas, consider $1\times$Matrix of$3$$【f[n-2],f[n-1],s[n-2]】$ , we hope to multiply by a$3\times$A matrix of$3$ ,$1\times 3$的 矩阵 ：
$【f[n-1],f[n],s[n-1]】=【f[n-1],f[n-1]+f[n-2],s[n-2]+f[n-1]]$
Easy to get this$3\times$The matrix of$3$ is:

Then... It is easy to find that the matrix scale of this method is$(r+1)\ast (r+1)$ , which is much better than the previous popular method (method one).

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Code

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
const int INF=9973;
long long n;
using namespace std;
struct c{
    
    
	int n,m;
	int a[10][10];
}A,B,CC;
c operator *(c A,c B){
    
    
	c 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.a[i][j]=0;
	for(int k=1;k<=A.m;k++)
	{
    
    
		for(int i=1;i<=C.n;i++)
			for(int j=1;j<=C.m;j++)
				C.a[i][j]=(C.a[i][j]+(A.a[i][k]*B.a[k][j])%INF)%INF;
	}	
	return C;
}
void poww(long long x){
    
    
	if(x==1)
	{
    
    
		B=A;
		return; 
	}
	poww(x>>1);
	B=B*B;
	if(x&1)
		B=B*A;
	
}
int main(){
    
    
	scanf("%lld",&n);
	if(n==1){
    
    
		printf("1");
		return 0;
	}
	A.n=3,A.m=3;
	A.a[1][1]=0,A.a[1][2]=1,A.a[1][3]=0;
	A.a[2][1]=1,A.a[2][2]=1,A.a[2][3]=1;
	A.a[3][1]=0,A.a[3][2]=0,A.a[3][3]=1;
	poww(n-1);
	CC.n=1,CC.m=3;
	CC.a[1][1]=1,CC.a[1][2]=1,CC.a[1][3]=1;
	CC=CC*B;
	printf("%d %d\n",CC.a[1][1],CC.a[1][3]);
	
}