Queuing（以前写的没整理）

Queuing

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

Total Submission(s): 62 Accepted Submission(s): 46

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

Recommend lcy

/*
以前做的没整理的
*/
#include<cstdio>
using namespace std;

int n,m;
int f[5]={0,2,4,6,9};
int a[4][4],b[4][4],c[4][4];

void multi(int (*x)[4],int (*y)[4])
{
  int i,j,k;
  for(i=0;i<4;i++)
    for(j=0;j<4;j++)
      for(c[i][j]=k=0;k<4;k++)
        c[i][j]+=x[i][k]*y[k][j];
  for(i=0;i<4;i++)
    for(j=0;j<4;j++)
      x[i][j]=c[i][j]%m;
}

int main()
{
  //freopen("1.in","r",stdin);  
  
  int i,j;
  while(scanf("%d%d",&n,&m)==2)
    {
      if(n<=4){printf("%d\n",f[n]%m);continue;}
      a[0][0]=a[0][2]=a[0][3]=1,a[0][1]=0;
      a[1][0]=1,a[1][1]=a[1][2]=a[1][3]=0;
      a[2][0]=a[2][2]=a[2][3]=0,a[2][1]=1;
      a[3][0]=a[3][1]=a[3][3]=0,a[3][2]=1;
      
      for(i=0;i<4;i++)
        for(j=0;j<4;j++)
          b[i][j]=(i==j);
      for(n-=4;n;multi(a,a),n>>=1)if(n&1)multi(b,a);
      for(j=0,i=0;i<4;i++)j+=b[0][i]*f[4-i];
      printf("%d\n",j%m);
    }
  return 0;
}