[

矩

阵

乘

法

]

裴

波

拉

契

数

列

I

V

[矩阵乘法]裴波拉契数列IV

[矩阵乘法]裴波拉契数列IV

Description

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

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Input

n(1＜n＜２³¹-1)

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Output

一个数为裴波拉契数列的第n项mod 9973;

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Sample Input

10000

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Sample Output

4399

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题目解析

对于为什么用矩阵乘法来做，详见博客斐波那契数列II

关于递推式略， 详见博客斐波那契数列III,并请独自尝试通过类比来推递推式。

然后可以构造出一个

4

∗

4

4 * 4

4∗4的矩阵

T

T

T

∣

0

1

0

0

1

1

0

0

0

1

1

0

0

1

1

1

∣

\begin{vmatrix} 0 & 1 & 0 & 0\\ 1 & 1 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 1 & 1 & 1\\ \end{vmatrix}

∣∣∣∣∣∣∣∣​0100​1111​0011​0001​∣∣∣∣∣∣∣∣​

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Code

#include <cmath>

#include <cstdio>

#include <iostream>

using namespace std;

int nt;

const int MOD = 9973;

struct matrix

{

	int n, m;

	int t[10][10];

}t1, t2, t3;

matrix operator *(matrix t, matrix r)

{

	matrix c;

	c.n = t.n, c.m = r.m;

	for (int i = 1; i <= c.n; ++ i)

	 for (int j = 1; j <= c.m; ++ j)

	  c.t[i][j]=0;

	for (int k = 1; k <= t.m; ++ k)

	 for (int i = 1; i <= t.n; ++ i)

	  for (int j = 1; j <= r.m; ++ j)

	   c.t[i][j] = (c.t[i][j] + t.t[i][k] * r.t[k][j] % MOD) % MOD;

	return c;

}

void rt (int k)

{

	if (k == 1)

	{

		t2 = t1;

		return;

	}

	rt (k / 2);

	t2 = t2 * t2;

	if (k & 1) t2 = t2 * t1;

}

int main()

{

	scanf ("%d", &nt);

	if (nt == 1)

	{

		printf ("1");

		return 0;

	}

	t3.n = 1;

	t1.n = t1.m = t3.m = 4;

	t1.t[1][1] = 0, t1.t[1][2] = 1, t1.t[1][3] = 0, t1.t[1][4] = 0;

	t1.t[2][1] = 1, t1.t[2][2] = 1, t1.t[2][3] = 0, t1.t[2][4] = 0;

	t1.t[3][1] = 0, t1.t[3][2] = 1, t1.t[3][3] = 1, t1.t[3][4] = 0;

	t1.t[4][1] = 0, t1.t[4][2] = 1, t1.t[4][3] = 1, t1.t[4][4] = 1;

	t3.t[1][1] = t3.t[1][2] = t3.t[1][4] = 1; t3.t[1][3] = 3;

	rt (nt - 1);

	t3 = t3 * t2;

	printf ("%d", t3.t[1][1]);

	return 0;

}