码迷,mamicode.com
首页 > 其他好文 > 详细

HDU 2604 Queuing

时间:2016-07-21 21:27:10      阅读:151      评论:0      收藏:0      [点我收藏+]

标签:

Queuing

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4638    Accepted Submission(s): 2045


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.
技术分享

  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

 

Source
 
Recommend
lcy
 
对于长度为n的队列q(n),做以下讨论。
技术分享
图中 ,".....m"表示以字符‘m‘结尾的queue,其它类似。E_q表示E_queue。
可得q(n)=q(n-1)+q(n-3)+q(n-4)
 
转换为矩阵,运用矩阵快速幂运算进行求解。
技术分享
即,
n>=4,
技术分享
 
 
 
 

HDU 2604 Queuing

标签:

原文地址:http://www.cnblogs.com/cumulonimbus/p/5692903.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!