hdu QueuingQueuing

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概述

Queuing

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 608 Accepted Submission(s): 278

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

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

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分析：开始研究4种状态fm,ff,mm,mf，TLE。接着研究6种状态 ffm,fmm,mfm,mmm,mff,mmf，继续超时。只好从队末加入往前分析，若新来的s[n]=m，那么n前面只要合法这个就合法，即f[n]+=f[n-1]。若新来一个f，那么必须s[n-2]=m。此时，若f[n-1]=m，那么n-2前面合法这个就合法，即f[n]+=f[n-3]；若f[n-1]=f，那么必须f[n-3]=m，那么只要n-3前面合法这个就合法，即f[n]+=f[n-4]。

#include <stdio.h>
#define N 1000001
#define M 30

char node[N][M];

void init()
{
    int i,j;
    for(i=1;i<=M;i++)
    {
        node[1][i]=2%i;
        node[2][i]=4%i;
        node[3][i]=6%i;
        node[4][i]=9%i;
        for (j=5;j<=N;j++)
        {
            node[j][i]=(node[j-1][i]+node[j-3][i]+node[j-4][i])%i;
        }
    }
}

int main()
{
    init();
    int n,m;
    while (~scanf("%d%d",&n,&m))
    {
        printf("%dn",node[n][m]);
    }
    return 0;
}