Queuing

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3032    Accepted Submission(s): 1379


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 2^L 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.

Input Input a length L (0 <= L <= 10 ⁶) and M.
Output Output K mod M(1 <= M <= 30) where K is the number of E-queues with length L.
Sample Input 3 8 4 7 4 8
Sample Output 6 2 1
Author WhereIsHeroFrom FFF FMF

一.找遞推公式

考慮最後若直接加M 顯然可以成立 所以F[n]=F[n-1]+X

若要在後面加F 考慮N-1是沒辦法的

考慮N-2 FF MF 依舊沒辦法

考慮N-3 FFF FMF MMF MFF （FFF,FMF 已經不可能 不考慮）

顯然新增MMF 前面N-3無論是什麼都不可能會出現FFF,FMF了

所以F[n]=F[n-1]+F[n-3]+X

但是 若新增MFF 可能會出現FMFF

考慮N-4 FMFF MMFF 顯然MMFF可以取了

所以 F[n]=F[n-1]+F[n-3]+F[n-4]

二.設計矩陣

注：矩陣十分好設計  第一排即為遞推公式的係數

      以後幾排類似於單位矩陣

三.矩陣快速冪

類似非遞迴快速冪

node kuaisumi(node A,int N,int mod)
{
	node di=e;
	while(N>0)
	{
		if(N&1)
		{
			di=MatrixMult(di,A,mod);
		}
		A=MatrixMult(A,A,mod);
		N=N>>1;
	}
	return di;
}

最後程式碼如下：

#include <cstdio>  
#include <cstdlib>  
#include <cmath>  
#include <cstring>  
#include <ctime>  
#include <algorithm>  
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313   
using namespace std;
int L,M;
struct node
{
	int mat[5][5];
}a,e,ans;
int mat2[5];
void CSH()
{
	mat2[1]=9;
	mat2[2]=6;
	mat2[3]=4;
	mat2[4]=2;
	for(int i=1;i<=4;i++)
{
		e.mat[i][i]=1;
		mat2[i]=mat2[i]%M;
}
	memset(a.mat,0,sizeof(a.mat));
	a.mat[1][1]=a.mat[1][3]=a.mat[1][4]=1;
	a.mat[2][1]=a.mat[3][2]=a.mat[4][3]=1;
}
node MatrixMult(node A,node B,int mod)
{
	node p;
	memset(p.mat,0,sizeof(p.mat));
	for(int i=1;i<=4;i++)
	 for(int j=1;j<=4;j++)
	 {
	 	 for(int k=1;k<=4;k++)
	 	p.mat[i][j]+=A.mat[i][k]*B.mat[k][j];
	 	p.mat[i][j]=p.mat[i][j]%mod;
	 }
	 return p;
}
node kuaisumi(node A,int N,int mod)
{
	node di=e;
	while(N>0)
	{
		if(N&1)
		{
			di=MatrixMult(di,A,mod);
		}
		A=MatrixMult(A,A,mod);
		N=N>>1;
	}
	return di;
}
void solve()
{
	int ANS=0;
	for(int i=1;i<=4;i++)
	{
		ANS+=ans.mat[1][i]*mat2[i];
		ANS=ANS%M;
	}
	printf("%d\n",ANS);
}
int main()
{
	while(cin>>L>>M)
	{
		CSH();
		if(L>4)
		ans=kuaisumi(a,L-4,M);
		if(L>4)
		solve();
		else 
		printf("%d\n",mat2[L]);
	}
	return 0;
}