Queuing

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

Total Submission(s): 36 Accepted Submission(s): 29

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

思路：矩阵乘法，转移就行了

用 f->1 m->0

an 表示 mm

bn 表示 mf

cn 表示 fm

dn 表示 ff

那么很明显 an=cn-1+an-1

bn=an-1 因为不能由 cn-1 fm->fmf

cn=bn-1+dn

dn=bn;

构造出矩阵即可！

  1 /*
  2 Author:wuhuajun
  3 */
  4 #include <cmath>
  5 #include <cstdio>
  6 #include <algorithm>
  7 #include <cstring>
  8 #include <string>
  9 #include <cstdlib>
 10 using namespace std;
 11 
 12 typedef long long ll;
 13 typedef double dd;
 14 const int maxn=210;
 15 
 16 struct matrix
 17 {
 18     ll m[5][5];
 19 } c,base,ans;
 20 ll ans1,mod,r,t;
 21 int n,l;
 22 
 23 void close()
 24 {
 25 exit(0);
 26 }
 27 
 28 matrix mul(matrix a,matrix b)
 29 {
 30     memset(c.m,0,sizeof(c.m));
 31     for (int i=1;i<=4;i++)
 32         for (int j=1;j<=4;j++)
 33             for (int k=1;k<=4;k++)
 34             {
 35                 c.m[i][j]+=a.m[i][k]*b.m[k][j];
 36                 c.m[i][j] %= mod;
 37             }
 38     return c;
 39 }
 40 /*
 41 void print(matrix a)
 42 {
 43     for (int i=1;i<=4;i++)
 44     {
 45         for (int j=1;j<=4;j++)
 46             printf("%lld ",a.m[i][j]);
 47         puts("");
 48     }
 49 }
 50 */
 51 void ass(int x,int y,matrix &base)
 52 {
 53     base.m[x][y]=1;
 54 }
 55 
 56 void work()
 57 {
 58     memset(base.m,0,sizeof(base.m));
 59     ass(1,2,base);ass(1,3,base);
 60     ass(2,4,base);
 61     ass(3,2,base);
 62     ass(4,1,base);ass(4,4,base);
 63     memset(ans.m,0,sizeof(ans.m));
 64     for (int i=1;i<=4;i++)
 65         ans.m[i][i]=1;
 66     n=l; n-=2;
 67     while (n!=0)
 68     {
 69         if (n & 1)
 70             ans=mul(base,ans);
 71         n/=2;
 72         base=mul(base,base);
 73     }
 74     for (int i=1;i<=4;i++)
 75         for (int j=1;j<=4;j++)
 76             ans1+=ans.m[i][j];
 77     /*
 78     t=0;
 79     for (int i=1;i<=4;i++)
 80         t+=ans.m[6][i];
 81     t %= mod;
 82     ans1-=t;
 83     for (int i=1;i<=5;i++)
 84         ans1+=ans.m[5][i];
 85         */
 86 }
 87 
 88 void init()
 89 {
 90     while (scanf("%d %I64d",&l,&mod)!=EOF)
 91     {
 92         ans1=0;
 93         if (n==1 && n==2)
 94             ans1=0;
 95         /*
 96         else
 97         {
 98             while (n!=0)
 99             {
100                 if (n & 1)
101                     ans1=(ans1 * r) % mod;
102                 n/=2;
103                 r=(r * r) % mod;
104             }
105             work();
106         }
107         */
108         work();
109         ans1+=mod;
110         ans1 %= mod;
111         printf("%I64d\n",ans1);
112     }
113 }
114 
115 int main ()
116 {
117     init();
118     close();
119     return 0;
120 }

posted on 2013-09-23 17:09 cssystem 阅读(281) 评论(0) 编辑收藏举报