Parliament

Time limit: 1.0 second
Memory limit: 64 MB
A
new parliament is elected in the state of MMMM. Each member of the
parliament gets his unique positive integer identification number during
the parliament registration. The numbers were given in a random order;
gaps in the sequence of numbers were also possible. The chairs in the
parliament were arranged resembling a tree-like structure. When members
of the parliament entered the auditorium they took seats in the
following order. The first of them took the chairman’s seat. Each of the
following delegates headed left if his number was less than the
chairman’s, or right, otherwise. After that he took the empty seat and
declared himself as a wing chairman. If the seat of the wing chairman
has been already taken then the seating algorithm continued in the same
way: the delegate headed left or right depending on the wing chairman’s
identification number.
The
figure below demonstrates an example of the seating of the members of
parliament if they entered the auditorium in the following order: 10, 5,
1, 7, 20, 25, 22, 21, 27.
During
its first session the parliament decided not to change the seats in the
future.
The speech order was also adopted. If the number of the session was odd
then the members of parliament spoke in the following order: the left
wing, the right wing and the chairman. If a wing had more than one
parliamentarian then their speech order was the same: the left wing, the
right wing, and the wing chairman. If the number of the session was
even, the speech order was different: the right wing, the left wing, and
the chairman. For a given example the speech order for odd sessions
will be 1, 7, 5, 21, 22, 27, 25, 20, 10; while for even sessions — 27,
21, 22, 25, 20, 7, 1, 5, 10.
Determine the speech order for an even session if the speech order for an odd session is given.

Input

The first line of the input contains N, the total number of parliamentarians. The following lines contain N integer numbers, the identification numbers of the members of parliament according to the speech order for an odd session.
The total number of the members of parliament does not exceed 3000. Identification numbers do not exceed 65535.

Output

The
output should contain the identification numbers of the members of
parliament in accordance with the speech order for an even session.

Sample

input output
9
1
7
5
21
22
27
25
20
10
27
21
22
25
20
7
1
5
10
Problem Source: Quarterfinal, Central region of Russia, Rybinsk, October 17-18 2001
 

timus 1136 Parliament(e)的更多相关文章

  1. timus 1136 Parliament(二叉树)

    Parliament Time limit: 1.0 secondMemory limit: 64 MB A new parliament is elected in the state of MMM ...

  2. ural 1136. Parliament

    题目链接:http://acm.timus.ru/problem.aspx?space=1&num=1136 题目描述:给定一个按照(左子树-右子树-根)(即先序)遍历序列的树,求其按照 右子 ...

  3. 1136. Parliament(二叉树)

    1136 先由后左 再父 建一个二叉树 #include <iostream> #include<cstdio> #include<cstring> #includ ...

  4. URAL 1136 Parliament 二叉树水题 BST后序遍历建树

    二叉树水题,特别是昨天刚做完二叉树用中序后序建树,现在来做这个很快的. 跟昨天那题差不多,BST后序遍历的特型,找到最后那个数就是根,向前找,比它小的那块就是他的左儿子,比它大的那块就是右儿子,然后递 ...

  5. URAL 1136 Parliament (DFS)

    题意 输入一棵树的后缀表达式(按左-右-中顺序访问),这棵树的每一个结点的数值都比它的左子树结点的数值大,而比它的右子树结点的数值小,要求输出其按右-左-中顺序访问的表达式.所有的数都为正整数,而且不 ...

  6. codeforces 644A Parliament of Berland

    A. Parliament of Berland time limit per test 1 second memory limit per test 256 megabytes input stan ...

  7. Timus OJ 1997 Those are not the droids you're looking for (二分匹配)

    题目链接:http://acm.timus.ru/problem.aspx?space=1&num=1997 这个星球上有两种人,一种进酒吧至少玩a小时,另一种进酒吧最多玩b小时. 下面n行是 ...

  8. cdoj 1136 邱老师玩游戏 树形背包

    邱老师玩游戏 Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.uestc.edu.cn/#/problem/show/1136 Desc ...

  9. [置顶] 【J2SE 】1136 容器之旅

    开篇引言 本篇文章我将要详细的介绍一下什么是容器?以及什么是1136?来系统全面的了解容器,以及容器的应用,下面就进入我们的容器之旅吧! 1.什么是容器? 用来存储和组织其他对象的对象.我们也可以这样 ...

随机推荐

  1. KStar ----BPM应用框架,K2 的新星

    “KStar”是基于K2 BPM搭建的应用框架产品,将K2最佳实践方案以产品的形式呈现给用户,该框架面向SOA服务,便于二次开发和扩展,流程设计.用户组织.业务表单.流程管理.系统集成等开发工作,都按 ...

  2. The Coco-Cola Store C(Contest #3 )

    Once upon a time, there is a special coco-cola store. If you return three empty bottles to the shop, ...

  3. 踏着前人的脚印学Hadoop——序列化,Writerable

    package org.apache.hadoop.io; import java.io.DataOutput;import java.io.DataInput;import java.io.IOEx ...

  4. Python单元测试——unittest

    unittest是python自带的一个模块 python344\Lib\unittest 官方参考文档: http://docs.python.org/2.7/library/unittest.ht ...

  5. hive 中 union all

    hive 中的union all是不能在sql语句的第一层使用的,否则会报 Top level UNION is not supported currently 错误: 例如如下的方式: select ...

  6. IOS socket开发基础

    摘要 详细介绍了iOS的socket开发,说明了tcp和udp的区别,简单说明了tcp的三次握手四次挥手,用c语言分别实现了TCPsocket和UDPsocket的客户端和服务端,本文的作用是让我们了 ...

  7. 查看ubuntu文件目录的大小和文件夹包含的文件数 zT

    查看ubuntu文件目录的大小和文件夹包含的文件数 查看linux文件目录的大小和文件夹包含的文件数 统计总数大小 du -sh xmldb/ du -sm * | sort -n //统计当前目录大 ...

  8. HDOJ-三部曲一(搜索、数学)-1012-Shredding Company

    Shredding Company Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other) T ...

  9. GridView自定义之前后效果对比

  10. Use XML in Windows Phone 8.0

    I have an XML file stored in DataModel folder, the structure is shown as below: <?xml version=&qu ...