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:
    openfire 部署后报错: java.lang.IllegalArgumentException: interface xx is not visible from class loader
    Calendar
    list 移除值
    fastjson 返回json字符串,JSON.parse 报错
    tomcat 跨域
    spring boot 笔记
    Mybatis处理列名—字段名映射— 驼峰式命名映射
    hadoop的NullWritable
    CentOS中用Nexus搭建maven私服,为Hadoop编译提供本地镜像
    CentOS中配置xrdp,通过微软远程桌面访问CentOS桌面
  • 原文地址:https://www.cnblogs.com/cssystem/p/3335191.html
Copyright © 2011-2022 走看看