hdu---(2604)Queuing(矩阵快速幂)

Queuing

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2796    Accepted Submission(s): 1282

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

 

 

   首先我们不考虑去模的问题：

      l = 0                            0 种

      l = 1      e的数目有 f,m： 2 种

      l = 2        ...........ff,mm,fm,mf    4种

      l = 3                                         6

      l = 4                                        9

      l =  5                                       15

      l  =  6                                      25

      f5=f4+f3+f1;

      f6=f5+f4+f2;

  ------->  fn={   fn-1+fn-3+fn-4  n>4;

由齐次方程构造矩阵.....

|fn   |    |1,0,1,1|  |fn-1|

|fn-1|    |1,0,0,0|  |fn-2|

|fn-2| = |0,1,0,0|*|fn-3|

|fn-3|    |0,0,1,0|   |fn-4|

代码：

//#define LOCAL
#include<cstdio>
#include<cstring>
using namespace std;
 //matrix --> ¾ØÕó
int mat[4][4];
int ans[4][4];
int len,m;

void init()
{
    int cc[4][4]={
          {1,0,1,1},{1,0,0,0},
          {0,1,0,0},{0,0,1,0}};

  for(int i=0;i<4;i++)
  {
      for(int j=0;j<4;j++)
    {
      mat[i][j]=cc[i][j];
      if(i==j) ans[i][j]=1;
      else ans[i][j]=0;
    }
  }
}
void Matrix(int a[][4],int b[][4])    //¾ØÕóÏà³Ë
{
    int i,j,k;
    int c[4][4]={0};
    for(j=0;j<4;j++){
      for(i=0;i<4;i++){
          for(k=0;k<4;k++){
          c[j][i]=(c[j][i]+a[j][k]*b[k][i])%m;
        }
      }
    }

    for(j=0;j<4;j++)
       for(i=0;i<4;i++)
           a[j][i]=c[j][i];

}

void pow(int n)
{
    while(n>0)
    {
      if(n&1) Matrix(ans,mat);
      n>>=1;
      if(n==0) break;
      Matrix(mat,mat);
    }
}
int main()
{
  #ifdef LOCAL
   freopen("test.in","r",stdin);
  #endif
  int f[4]={2,4,6,9};
  while(scanf("%d%d",&len,&m)!=EOF)
  {
      if(len==0)printf("%d
",0);
      else if(len<=4)printf("%d
",f[len-1]%m);
      else{
      init();
     pow(len-4);
     printf("%d
",(ans[0][0]*f[3]+ans[0][1]*f[2]+ans[0][2]*f[1]+ans[0][3]*f[0])%m);
      }
  }
 return 0;
}