# 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 2L 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;

|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|

``` 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 }```

657 篇文章64 人订阅

0 条评论

## 相关文章

### 关于web.xml3.0启动报错

34770

Holding Bin-Laden Captive! Time Limit: 2000/1000 MS (Java/Others)    Memory Limi...

283110

28650

### HDUOJ---（4708）Herding

Herding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (J...

357100

19310

### HDUOJ---hello Kiki

Hello Kiki Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K...

28490

### hdu 1695 GCD（莫比乌斯反演）

GCD Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/...

41260

3.9K30

### POJ-1322 Chocolate（概率DP）

Chocolate Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 92...

33930

### HDU----(4291)A Short problem(快速矩阵幂)

A Short problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32...

35260