【矩阵乘法】裴波拉契数列III

Description

求数列f[n]=f[n-1]+f[n-2]+1的第N项.f[1]=1,f[2]=1.

Input

n(1＜n＜２^31-1)

Output

第Ｎ项的结果 mod 9973

Sample Input

12345

Sample Output

8932

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解题思路

详情参见裴波拉契数列II

+1其实很好处理
把A矩阵改为

B矩阵改为

$A=$ { $0，1，0，1，1，1，0，0，1$｝
$B=$ { $f[n-2],f[n-1],1$}
$A\ast B$
$=${ $f[n-2]\ast 0+f[n-1]\ast 1+1\ast 0,f[n-2]\ast 1+f[n-1]\ast 1+1\ast 1,f[n-2]\ast 0+f[n-1]\ast 0+1\ast 1$}
$=${ $f[n-1],f[n],1$}

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Code

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

const int Mod = 9973;
long long n;

struct DT{
    
    
	int n, m;
	int aed[5][5];
}A, B, Ac;

DT operator *(DT a, DT b){
    
    //矩阵乘法
	DT c;
	c.n = a.n, c.m = b.m;
	memset (c.aed, 0, sizeof (c.aed));
	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.aed[i][j] = (c.aed[i][j] + a.aed[i][k] * b.aed[k][j] % Mod) % Mod;
	return c;
}

void power (long long n){
    
    //快速幂
	if (n == 1)
	{
    
    
		Ac = A;
		return;
	}
	power (n / 2);
	Ac = Ac * Ac;
	if (n % 2) Ac = Ac * A; 
}

int main(){
    
    
	A.n = A.m = 3;
	A.aed[1][2] = A.aed[2][1] = A.aed[2][2] = A.aed[3][2] = A.aed[3][3] = 1;
	B.n = 1, B.m = 3;
	B.aed[1][1] = B.aed[1][2] = B.aed[1][3] = 1;
	//初始化矩阵
	scanf ("%lld", &n);
	power (n - 1);
	B = B * Ac;
	printf ("%d", B.aed[1][1]);
}