Queuing

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3141    Accepted Submission(s): 1431

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

 

//矩阵快速幂解递推式：f(n)=f(n-1)+f(n-2)+f(n-3)
#include<cstdio>
#include<cstring>
using namespace std;
struct node{
    int mat[4][4];
};
int L,M;
const int n=4;
node mat_multi(node a,node b)
{
    node c;
    int i,j,k;
    memset(c.mat,0,sizeof(c.mat));
    for(i=0;i<n;i++)
        for(j=0;j<n;j++)
            for(k=0;k<n;k++)
            {
                c.mat[i][j]+=a.mat[i][k]*b.mat[k][j];
                c.mat[i][j]%=M;
            }
            return c;
}
node pow_mod(node a,int t)
{
    node c;
    int i;
    memset(c.mat,0,sizeof(c.mat));
    for(i=0;i<n;i++)
    c.mat[i][i]=1;
    for(;t>0;t>>=1)
    {
        if(t&1) c=mat_multi(a,c);
        a=mat_multi(a,a);
    }
    return c;
}
int main()
{
    int i,t;
    int b[4]={9,6,4,2};
    node c;
    while(~scanf("%d%d",&L,&M))
    {
    c.mat[0][0]=1;
    c.mat[0][1]=0;
    c.mat[0][2]=1;
    c.mat[0][3]=1;

    c.mat[1][0]=1;
    c.mat[1][1]=0;
    c.mat[1][2]=0;
    c.mat[1][3]=0;

    c.mat[2][0]=0;
    c.mat[2][1]=1;
    c.mat[2][2]=0;
    c.mat[2][3]=0;

    c.mat[3][0]=0;
    c.mat[3][1]=0;
    c.mat[3][2]=1;
    c.mat[3][3]=0;
        int ans=0;
        if(L==0) ans=0;
        else if(L==1) ans=2;
        else if(L==2) ans=4;
        else if(L==3) ans=6;
        else if(L==4) ans=9;
        else{
        t=L-4;
        c=pow_mod(c,t);

        for(i=0;i<4;i++)//构造矩阵的（L-4）次幂后 再乘以前4项就是结果
        ans+=c.mat[0][i]*b[i];
        }
        printf("%d
",ans%M);
    }
}

　　公式是从网上找的。。。。