A. Neverending competitions

题目连接:

http://codeforces.com/contest/765/problem/A

Description

There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.

Jinotega's best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:

this list contains all Jinotega's flights in this year (in arbitrary order),

Jinotega has only flown from his hometown to a snooker contest and back,

after each competition Jinotega flies back home (though they may attend a competition in one place several times),

and finally, at the beginning of the year Jinotega was at home.

Please help them to determine Jinotega's location!

Input

In the first line of input there is a single integer n: the number of Jinotega's flights (1 ≤ n ≤ 100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega's home airport. In the next n lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega's home airport.

It is guaranteed that flights information is consistent with the knowledge of Jinotega's friends, which is described in the main part of the statement.

Output

If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".

Sample Input

4

SVO

SVO->CDG

LHR->SVO

SVO->LHR

CDG->SVO

Sample Output

home

Hint

题意

有个人要去参加比赛,他有n张飞机票,飞机票写着从A->B。

保证他只会两种旅行,home->xxx,xxx->home。

而且一开始在home

但是他的飞机票顺序是乱的,问你现在他在xxx,还是在home

题解:

其实飞机票都是误导你的。

如果飞机票数是偶数,那么就在家。

否则就在xxx

显然嘛。

代码

#include<bits/stdc++.h>
using namespace std; int main()
{
int n;
scanf("%d",&n);
if(n%2==0)cout<<"home"<<endl;
else cout<<"contest"<<endl;
}

Codeforces Round #397 by Kaspersky Lab and Barcelona Bootcamp (Div. 1 + Div. 2 combined) A. Neverending competitions 水题的更多相关文章

  1. Codeforces Round #397 by Kaspersky Lab and Barcelona Bootcamp (Div. 1 + Div. 2 combined) A B C D 水 模拟 构造

    A. Neverending competitions time limit per test 2 seconds memory limit per test 512 megabytes input ...

  2. Codeforces Round #397 by Kaspersky Lab and Barcelona Bootcamp (Div. 1 + Div. 2 combined) A - Neverending competitions

    地址:http://codeforces.com/contest/765/problem/A 题目: A. Neverending competitions time limit per test 2 ...

  3. Codeforces Round #397 by Kaspersky Lab and Barcelona Bootcamp (Div. 1 + Div. 2 combined) F. Souvenirs 线段树套set

    F. Souvenirs 题目连接: http://codeforces.com/contest/765/problem/F Description Artsem is on vacation and ...

  4. Codeforces Round #397 by Kaspersky Lab and Barcelona Bootcamp (Div. 1 + Div. 2 combined) E. Tree Folding 拓扑排序

    E. Tree Folding 题目连接: http://codeforces.com/contest/765/problem/E Description Vanya wants to minimiz ...

  5. Codeforces Round #397 by Kaspersky Lab and Barcelona Bootcamp (Div. 1 + Div. 2 combined) D. Artsem and Saunders 数学 构造

    D. Artsem and Saunders 题目连接: http://codeforces.com/contest/765/problem/D Description Artsem has a fr ...

  6. Codeforces Round #397 by Kaspersky Lab and Barcelona Bootcamp (Div. 1 + Div. 2 combined) C. Table Tennis Game 2 水题

    C. Table Tennis Game 2 题目连接: http://codeforces.com/contest/765/problem/C Description Misha and Vanya ...

  7. Codeforces Round #397 by Kaspersky Lab and Barcelona Bootcamp (Div. 1 + Div. 2 combined) B. Code obfuscation 水题

    B. Code obfuscation 题目连接: http://codeforces.com/contest/765/problem/B Description Kostya likes Codef ...

  8. Codeforces Round #397 by Kaspersky Lab and Barcelona Bootcamp (Div. 1 + Div. 2 combined) E. Tree Folding

    地址:http://codeforces.com/contest/765/problem/E 题目: E. Tree Folding time limit per test 2 seconds mem ...

  9. Codeforces Round #397 by Kaspersky Lab and Barcelona Bootcamp (Div. 1 + Div. 2 combined) D. Artsem and Saunders

    地址:http://codeforces.com/contest/765/problem/D 题目: D. Artsem and Saunders time limit per test 2 seco ...

随机推荐

  1. bzoj千题计划221:bzoj1500: [NOI2005]维修数列(fhq treap)

    http://www.lydsy.com/JudgeOnline/problem.php?id=1500 1.覆盖标记用INF表示无覆盖标记,要求可能用0覆盖 2.代表空节点的0号节点和首尾的两个虚拟 ...

  2. HDU 3511 圆扫描线

    找最深的圆,输出层数 类似POJ 2932的做法 圆扫描线即可.这里要记录各个圆的层数,所以多加一个维护编号的就行了. /** @Date : 2017-10-18 18:16:52 * @FileN ...

  3. Matrix67|自由职业者,数学爱好者

    Matrix67|自由职业者,数学爱好者 介绍一下你自己和所做的工作. 我叫顾森,网名 Matrix67,长住北京的重庆人,目前没有固定的职业.一会儿当当码农,一会儿做做编辑,一会儿教教数学,一会儿写 ...

  4. (64位)本体学习程序(ontoEnrich)系统使用说明文档

    系统运行:文件夹system下,可执行文件ontoEnrichment 概念学习 --------------------------------------------------------1.简 ...

  5. [转载]AngularJS视图

    http://www.yiibai.com/angularjs/angularjs_views.html <html> <head> <title>Angular ...

  6. Spring Mvc + Maven + yuicompressor 使用 profile 来压缩 javascript ,css 文件; (十)

    profile相关知识点: 在开发项目时,设想有以下场景: 你的Maven项目存放在一个远程代码库中(比如github),该项目需要访问数据库,你有两台电脑,一台是Linux,一台是Mac OS X, ...

  7. Guava HashMultiset(MultiSet)

    multiset:多重集合,和set唯一的不同是 set 集合中一个值只能出现一次,而multiset多重集合中一个值可以出现多次.一个典型的应用就是统计单词出现次数 举例: public class ...

  8. c++刷题(33/100)笔试题1

    笔试总共2小时,三道题,时间挺充裕的,但是最后只做了一道,原因在于自己很浮躁,不审题,不仔细思考.没过的两道都是稍微改一下代码就能过,但是没过就是没过,要引以为戒 题目1: 小W有一个电子时钟用于显示 ...

  9. Docker01 CentOS配置Docker

    Docker 是一个开源的应用容器引擎,让开发者可以打包他们的应用以及依赖包到一个可移植的容器中,然后发布到任何流行的 Linux 机器上,也可以实现虚拟化.容器是完全使用沙箱机制,相互之间不会有任何 ...

  10. Aho-Corasick 多模式匹配算法、AC自动机详解

    Aho-Corasick算法是多模式匹配中的经典算法,目前在实际应用中较多. Aho-Corasick算法对应的数据结构是Aho-Corasick自动机,简称AC自动机. 搞编程的一般都应该知道自动机 ...