http://acm.hdu.edu.cn/showproblem.php?pid=2604


 

Problem Description

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.

题意：长度为n的队列，每一位是f或m，求n长度不含fmf和fff的队列可能个数。

思路：自己开始时考虑2^n-含那两种的个数，但是含的比不含要难求。含的要考虑前含&&后不含，前不含&&后含，前含&&后含，而不含只有前不含&&后不含一种情况。

考虑不含的情况，设f（n）为前n位不含的组合数。若最后一位是m，转化为f（n-1），

若最后一位是f,那么最后三位：fff,fmf,mmf,mff，

fff和fmf是不满足的，mmf转化为f（n-3），mff的话，倒数第四位不能是f只能是m，即mmff，转化为f（n-4）

故f(n)=f(n-1)+f(n-3)+f(n-4)。

转化为矩阵即为：  
1 0 1 1       F(N-1) =F(N)     
1 0 0 0  *    F(N-2) = F(N-1)   
0 1 0 0       F(N-3) =F(N-2)
0 0 1 0       F(N-4) =F(N-3)

#include<bits/stdc++.h>
using namespace std;

int n,mod;
struct mat{
	int a[4][4];
};

mat mul(mat x,mat y)
{
	mat res={0};
	for(int i=0;i<4;i++)
		for(int j=0;j<4;j++)
			for(int k=0;k<4;k++)
				res.a[i][j]=(res.a[i][j]+x.a[i][k]*y.a[k][j])%mod;
	return res;
}

int pow(int n)
{
	mat c={0},res={0};
	c.a[0][0]=1;
	c.a[0][2]=1;
	c.a[0][3]=1;
	c.a[1][0]=1;
	c.a[2][1]=1;
	c.a[3][2]=1;
	for(int i=0;i<4;i++)res.a[i][i]=1;
	while(n)
	{
		if(n&1)res=mul(res,c);
		c=mul(c,c);
		n=(n>>1);
	}
	return (9*res.a[0][0]+6*res.a[0][1]+4*res.a[0][2]+2*res.a[0][3])%mod;
}

int main()
{
	//freopen("input.in","r",stdin);
	while(cin>>n>>mod)
	{
		if(n==0)cout<<0<<endl;
		else if(n==1)cout<<2%mod<<endl;
		else if(n==2)cout<<4%mod<<endl;
		else if(n==3)cout<<6%mod<<endl;
		else if(n==4)cout<<9%mod<<endl;
		else cout<<pow(n-4)<<endl;
	}
	return 0;
}