zoukankan      html  css  js  c++  java

  • 9.3.2 Queuing

    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 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

    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
",ans1);
112     }
113 }
114 
115 int main ()
116 {
117     init();
118     close();
119     return 0;
120 }
  • 相关阅读:
    调试
    webpack output的path publicPath
    CSS实现单行、多行文本溢出显示省略号
    docker安装mysql
    构建docker镜像
    Tensorflow博文列表
    ML理论知识博文列表
    Python博文列表
    Opencv博文收藏列表
    Centos文章列表
  • 原文地址：https://www.cnblogs.com/cssystem/p/3335191.html
Copyright © 2011-2022 走看看