HDU 2064 Queuing

原题传送门：http://acm.hdu.edu.cn/showproblem.php?pid=2604

 

Queuing

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2538    Accepted Submission(s): 1194

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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F(n)表示为L=n,时候E序列的数目；

分析：如果长度n的序列中 1）如果最后一个m，那么只要长度为n-1的序列中所有E序列的最后加上m，就可以构成长度为n的E序列(+=F(n-1))。2）如果最后一个是f，那么长度n-1的序列没有能保证如果n-1长度序列为E，再加上f还是E序列的，也不能保证加上f一定是O序列，所以我接着深入！2.1 如果长度n序列中最后两个为mf、ff、貌似都不能保证一定是E或者O序列，我们继续深入！如果长度n序列中最后3个为mmf、fmf、mff、fff，显然fff、fmf一定是O序列！！而mff不能保证一定是E或者O序列，而mmf如果长度n-3的序列是E序列，那么序列后加上mmf也一定是E序列(+=F(n-3))； 对于mff序列，我们继续深入，长度n序列中最后4个为，mmff、fmff 对于fmff一定是O序列，而mmff一定是E序列，那么很显然了(+=F(n-4))

因此可以推出递推公式F(n) = F(n-1) + F(n-3) + F(n-4);

那么很容易想到循环递推;

#include<cstdio>
#define MAXN 1000010
int L,M;
int F[MAXN];
int main()
{
    F[1] = 2; F[2] = 4; F[3] = 6; F[4] = 9;

    while(~scanf("%d%d",&L,&M))
    {
        for(int i=5;i<=1000000;i++)
            F[i] = (F[i-1]+F[i-3]+F[i-4])%M;
        printf("%d\n",F[L]%M);

    }

}

我会告诉你OJ上的结果，

Run ID

Submit Time

Judge Status

Pro.ID

Exe.Time

Exe.Memory

Code Len.

Language

11164180

2014-07-24 10:45:23

Time Limit Exceeded

2604

5000MS

4104K

293 B

G++

 其实对于这种超时，很容易想到用矩阵快速幂

构造矩阵

/*
1,0,1,1
1,0,0,0
0,1,0,0
0,0,1,0*/

#include<cstdio>
#include<cstring>
int L,M,ans;
int init[4] = {2,4,6,9};
struct Matrix{
    int mat[4][4];
    Matrix operator * (const Matrix rhs) const{
        Matrix temp;
        for(int i=0;i<4;i++){
             for(int j=0;j<4;j++){
                temp.mat[i][j] = 0;
                for(int k=0;k<4;k++)
                {
                    temp.mat[i][j] += this->mat[i][k]*rhs.mat[k][j]%M;
                    temp.mat[i][j] %= M;
                }
            }

        }

        return temp;
    }
};
Matrix mt,ret;

int main()
{
    while(~scanf("%d%d",&L,&M))
    {
       memset(mt.mat,0,sizeof(mt.mat));
       memset(ret.mat,0,sizeof(ret.mat));
       for(int i=0;i<4;i++)mt.mat[0][i] = i==1?0:1;
       for(int i=0;i<3;i++)mt.mat[i+1][i] = 1;
       for(int i=0;i<4;i++)ret.mat[i][i] = 1;

       ans = 0;
       if(L<=4){
           printf("%d\n",L==0?0:init[L-1]%M);
       }else{
            L = L-4;
            for(int i=0;i<4;i++) ret.mat[i][i] = 1;
            while(L)
            {
                if(L&1)ret = ret*mt;
                mt = mt*mt;
                L>>=1;
            }
            for(int i =0;i<4;i++){
                ans = (ans+ret.mat[0][i]*init[3-i]%M)%M;
            }

            printf("%d\n",ans);
       }
    }
    return 0;
}

//超时代码
/*
#include<cstdio>
#define MAXN 1000010
int L,M;
int F[MAXN];
int main()
{
    F[1] = 2; F[2] = 4; F[3] = 6; F[4] = 9;

    while(~scanf("%d%d",&L,&M))
    {
        for(int i=5;i<=1000000;i++)
            F[i] = (F[i-1]+F[i-3]+F[i-4])%M;
        printf("%d\n",F[L]%M);

    }

}*/