斐波拉契数列Ⅲ【矩阵乘法】

>Link

ssl 1530

---

>Description

求序列 $f_n=f_{n-1}+f_{n-2}+1$ 中的第$N$项，其中 $f_1=f_2=1$

---

>解题思路

思路同斐波拉契数列Ⅱ

$f_n=f_{n-1}+f_{n-2}+1$，转移思路为下：
$\left[\begin{array}{ccc}{f}_{n-2}& f_{n-1}& 1\end{array}\right]\to \left[\begin{array}{ccc}{f}_{n-1}& f_n(f_{n-1}+f_{n-2}+1)& 1\end{array}\right]$

$\therefore$我们把$A$矩阵设为$\left[\begin{array}{ccc}1& 1& 1\end{array}\right]$
$B$矩阵设为$\left[\begin{array}{ccc}0& 1& 0\\ 1& 1& 0\\ 0& 1& 1\end{array}\right]$

---

>代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define LL long long
using namespace std;

const LL p = 9973;
struct matrix
{
    
    
	int x, y;
	LL a[10][10];
} A, B, ans;
LL n;

matrix operator *(matrix a, matrix b)
{
    
    
	matrix c;
	c.x = a.x, c.y = b.y;
	for (int i = 1; i <= c.x; i++)
	  for (int j = 1; j <= c.y; j++) c.a[i][j] = 0; //注意要清零
	for (int k = 1; k <= a.y; k++)
	  for (int i = 1; i <= a.x; i++)
	    for (int j = 1; j <= b.y; j++)
	      c.a[i][j] = (c.a[i][j] + a.a[i][k] * b.a[k][j] % p) % p;
	return c;
}
void power (LL k)
{
    
    
	if (k == 1) {
    
    B = A; return;}
	power (k / 2);
	B = B * B;
	if (k & 1) B = B * A;
}

int main()
{
    
    
	scanf ("%lld", &n);
	if (n == 1 || n == 2) {
    
    printf ("1"); return 0;}
	A.x = A.y = 3;
	A.a[1][1] = A.a[1][3] = A.a[2][3] = A.a[3][1] = 0;
	A.a[1][2] = A.a[2][1] = A.a[2][2] = A.a[3][2] = A.a[3][3] = 1;
	power (n - 1);
	ans.x = 1, ans.y = 3;
	ans.a[1][1] = ans.a[1][2] = ans.a[1][3] = 1;
	ans = ans * B;
	printf ("%lld", ans.a[1][1]);
	return 0;
}