斐波那契数列Ⅳ【矩阵乘法】

>Link

ssl 1531

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>Description

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

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>解题思路

这道题仍然是矩阵乘法例题，思路同斐波那契数列Ⅱ和斐波拉契数列Ⅲ

打题解打得不想说话了╥﹏╥
$A$矩阵为 $\left[\begin{array}{cccc}1& 1& 3& 1\end{array}\right]$

$B$矩阵为 $\left[\begin{array}{cccc}0& 1& 0& 0\\ 1& 1& 0& 0\\ 0& 1& 1& 0\\ 0& 1& 1& 1\end{array}\right]$

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>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;
	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 >> 1);
	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 = 4;
	A.a[1][1] = A.a[1][3] = A.a[1][4] = A.a[2][3] = 0;
	A.a[2][4] = A.a[3][1] = A.a[3][4] = A.a[4][1] = 0;
	A.a[1][2] = A.a[2][1] = A.a[2][2] = A.a[3][2] = 1;
	A.a[3][3] = A.a[4][2] = A.a[4][3] = A.a[4][4] = 1;
	power (n - 1);
	ans.x = 1, ans.y = 4;
	ans.a[1][1] = ans.a[1][2] = ans.a[1][4] = 1;
	ans.a[1][3] = 3;
	ans = ans * B;
	printf ("%lld", ans.a[1][1]);
	return 0;
}