[ 矩 阵 乘 法 ] 裴 波 拉 契 数 列 I I I [矩阵乘法]裴波拉契数列III [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.

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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

12345

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Sample Output

8932

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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

我们考虑矩阵 ⊏ f [ n − 2 ] , f [ n − 1 ] , 1 ⊐ \sqsubset f[n - 2] , f[n - 1] , 1\sqsupset ⊏f[n−2],f[n−1],1⊐并利用斐波那契数列的递推关系来得到式子 ⊏ f [ n ] , f [ n − 1 ] ⊐ = ⊏ f [ n − 2 ] + f [ n − 1 ] , f [ n − 1 ] ⊐ \sqsubset f[n] , f[n - 1]\sqsupset = \sqsubset f[n - 2] + f[n - 1] , f[n - 1]\sqsupset ⊏f[n],f[n−1]⊐=⊏f[n−2]+f[n−1],f[n−1]⊐

然后可以构造出一个 3 ∗ 3 3 * 3 3∗3的矩阵 T T T
∣ 0 1 0 1 1 0 0 1 1 ∣ \begin{vmatrix} 0 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ \end{vmatrix} ∣∣∣∣∣∣​010​111​001​∣∣∣∣∣∣​
然后可以通过 f [ 1 ] , f [ 2 ] , 1 ∗ T = f [ 2 ] , f [ 3 ] , 1 f[1] , f[2] , 1* T = f[2] , f[3], 1 f[1],f[2],1∗T=f[2],f[3],1来实现代码了

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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 = 3;
	t1.t[1][1] = 0, t1.t[1][2] = 1, t1.t[1][3] = 0;
	t1.t[2][1] = 1, t1.t[2][2] = 1, t1.t[2][3] = 0;
	t1.t[3][1] = 0, t1.t[3][2] = 1, t1.t[3][3] = 1;
	t3.t[1][1] = t3.t[1][2] = t3.t[1][3] = 1;
	rt (nt - 1);
	t3 = t3 * t2;
	printf ("%d", t3.t[1][1]);
	return 0;
}