hdu 2604 Queuing （矩阵高速幂）

Queuing

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

Total Submission(s): 2773    Accepted Submission(s): 1275

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

记答案为f[n],则易得f[0]=0,f[1]=2,f[2]=4,f[3]=6;f[[4]=9;

当长度为N时，若最后一个字符为M。前N-1个字符没有限制，即为F(N-1);

当最后一个字符串为F的时候，就必须去除最后3个字符是fmf和fff的情况（倒数第二个字符为F、M均有可能会不满足情况），此时最后3个字符可能为mmf和mff。

当后3个字符为mmf时。前N-3个字符没有限制。即F(N-3);

可是当最后四个字符为mmff时，前N-4个字符无限制，即为F(N-1);

即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)      （即是f[n]=1*f[n-1]+0*f[n-2]+1*f[n-3]+1*f[n-4];）
   

1 0 0 0  *    F(N-2)  = F(N-1)   （以下为单位矩阵）

0 1 0 0       F(N-3)　　F(N-2)

0 0 1 0       F(N-4)　　F(N-3)

#include"iostream"
#include"stdio.h"
#include"string.h"
#include"algorithm"
#include"queue"
#include"vector"
using namespace std;
#define N 4
#define LL __int64
struct Mat
{
    LL mat[N][N];
};
int M,n=4;
int p[5]={0,2,4,6,9};
Mat operator *(Mat a,Mat b)
{
    int i,j,k;
    Mat c;
    memset(c.mat,0,sizeof(c.mat));
    for(i=0;i<n;i++)
    {
        for(j=0;j<n;j++)
        {
            c.mat[i][j]=0;
            for(k=0;k<n;k++)
            {
                c.mat[i][j]+=(a.mat[i][k]*b.mat[k][j])%M;
            }
            c.mat[i][j]%=M;
        }
    }
    return c;
}
int fun(Mat &a,int k)
{
    int i;
    Mat ans;
    memset(ans.mat,0,sizeof(ans.mat));
    for(i=0;i<n;i++)
        ans.mat[i][i]=1;
    while(k)
    {
        if(k&1)
            ans=ans*a;
        k>>=1;
        a=a*a;
    }
    LL s=0;
    for(i=0;i<n;i++)
    {
        s+=ans.mat[0][i]*p[n-i];
        s%=M;
    }
    return s;
}
int main()
{
    int i,l;
    Mat a;
    while(scanf("%d%d",&l,&M)!=-1)
    {
        if(l<=n)
        {
            printf("%d\n",p[l]%M);
            continue;
        }
        memset(a.mat,0,sizeof(a.mat));
        a.mat[0][0]=a.mat[0][2]=a.mat[0][3]=1;
        for(i=1;i<n;i++)
            a.mat[i][i-1]=1;
        printf("%d\n",fun(a,l-4));
    }
    return 0;
}