hdu---(2604)Queuing(矩阵快速幂)-优快云博客

本文介绍了一种计算特定条件下E-队列数量的方法，并通过矩阵快速幂算法实现了高效的求解过程。对于长度为L的队列，排除含有特定子序列的队列，计算剩余队列的数量。

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

8
7
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：

 1 //#define LOCAL
 2 #include<cstdio>
 3 #include<cstring>
 4 using namespace std;
 5  //matrix --> ¾ØÕó
 6 int mat[4][4];
 7 int ans[4][4];
 8 int len,m;
 9 
10 void init()
11 {
12     int cc[4][4]={
13           {1,0,1,1},{1,0,0,0},
14           {0,1,0,0},{0,0,1,0}};
15 
16   for(int i=0;i<4;i++)
17   {
18       for(int j=0;j<4;j++)
19     {
20       mat[i][j]=cc[i][j];
21       if(i==j) ans[i][j]=1;
22       else ans[i][j]=0;
23     }
24   }
25 }
26 void Matrix(int a[][4],int b[][4])    //¾ØÕóÏà³Ë
27 {
28     int i,j,k;
29     int c[4][4]={0};
30     for(j=0;j<4;j++){
31       for(i=0;i<4;i++){
32           for(k=0;k<4;k++){
33           c[j][i]=(c[j][i]+a[j][k]*b[k][i])%m;
34         }
35       }
36     }
37 
38     for(j=0;j<4;j++)
39        for(i=0;i<4;i++)
40            a[j][i]=c[j][i];
41 
42 }
43 
44 void pow(int n)
45 {
46     while(n>0)
47     {
48       if(n&1) Matrix(ans,mat);
49       n>>=1;
50       if(n==0) break;
51       Matrix(mat,mat);
52     }
53 }
54 int main()
55 {
56   #ifdef LOCAL
57    freopen("test.in","r",stdin);
58   #endif
59   int f[4]={2,4,6,9};
60   while(scanf("%d%d",&len,&m)!=EOF)
61   {
62       if(len==0)printf("%d\n",0);
63       else if(len<=4)printf("%d\n",f[len-1]%m);
64       else{
65       init();
66      pow(len-4);
67      printf("%d\n",(ans[0][0]*f[3]+ans[0][1]*f[2]+ans[0][2]*f[1]+ans[0][3]*f[0])%m);
68       }
69   }
70  return 0;
71 }

View Code