Hdoj 2604

原题链接

描述

Queues and Priority Queues are data structures which are known to most computer scientists. The Queue occurs often in our daily life. There are many people lined up at the lunch time. Now we define that ‘f’ is short for female and ‘m’ is short for male. If the queue’s length is L, then there are 2L numbers of queues. For example, if L = 2, then they are ff, mm, fm, mf . If there exists a subqueue as fmf or fff, we call it O-queue else it is a E-queue.
Your task is to calculate the number of E-queues mod M with length L by writing a program.

输入

Input a length L (0 <= L <= 10e6) and M.

输出

Output K mod M(1 <= M <= 30) where K is the number of E-queues with length L.

样例输入

3 8
4 7
4 8

样例输出

6
2
1

思路

首先是数学上找出递推关系式$a_{n} = a_{n-1} + a_{n-3} + a_{n-4}, n > 4$
接下来当然可以去解通项，不过略麻烦而且不好算，故直接构建矩阵，用矩阵快速幂。

$$\left[ \begin{matrix} a_{n} & a_{n-1} & a_{n-2} & a_{n-3} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right] = \left[ \begin{matrix} a_{n-1} & a_{n-2} & a_{n-3} & a_{n-4} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} \right] \left[ \begin{matrix} 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{matrix} \right]$$

代码

#include <cstdio>
#include<cstring>
#define ll long long
#define maxn 4
using namespace std;

int k, mod;

struct Mat
{
	ll f[maxn][maxn];
	void cls(){memset(f, 0, sizeof(f));}//全部置为0 
	Mat() {cls();}
	friend Mat operator * (Mat a, Mat b)
	{
		Mat res;
		for(int i = 0; i < maxn; i++) for(int j = 0; j < maxn; j++)
			for(int k = 0; k < maxn; k++)
				(res.f[i][j] += a.f[i][k] * b.f[k][j]) %= mod;
		return res;
	}
};

Mat quick_pow(Mat a)  
{  
    Mat ans;
    for(int i = 0; i < maxn; i++) ans.f[i][i] = 1;
    int b = k;
    while(b != 0)  
    {
        if(b & 1) ans = ans * a;
        b >>= 1;
        a = a * a;
    }
    return ans;  
}

int main()
{
	Mat A, B;
	B.f[0][0] = 9; B.f[0][1] = 6; B.f[0][2] = 4; B.f[0][3] = 2;
	A.f[0][0] = A.f[0][1] = A.f[1][2] = A.f[2][0] = A.f[2][3] = A.f[3][0] = 1;
	while(~scanf("%d %d", &k, &mod))
	{
		Mat C;
		if(k <= 4) {printf("%d\n", B.f[0][4-k] % mod); continue;}
		k -= 4;
		C = quick_pow(A);
		C = B * C;
		printf("%d\n", C.f[0][0]);
	}
	return 0;
}