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

题目:

A. Neverending competitions
time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

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 3capital 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".

Examples
input
4
SVO
SVO->CDG
LHR->SVO
SVO->LHR
CDG->SVO
output
home
input
3
SVO
SVO->HKT
HKT->SVO
SVO->RAP
output
contest
Note

In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.

思路:因为输入的情况必定合法,所以可以用n的奇偶性判断是home还是contest

 #include <bits/stdc++.h>

 using namespace std;

 #define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-;
const double pi=acos(-1.0);
const int K=1e5+;
const int mod=1e9+; int n;
int main(void)
{
cin>>n;
if(n&)
printf("contest\n");
else
printf("home\n");
return ;
}

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

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

    A. Neverending competitions 题目连接: http://codeforces.com/contest/765/problem/A Description There are ...

  2. 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 ...

  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. tp三级联动

    <script type="text/javascript">$(document).ready(function(){  $("#province" ...

  2. 使用scp命令传输文件

    1. 从远端复制文件到本地: sudo scp root@192.168.0.1:remote_path/remote_file . 2. 从本地复制文件到远端: sudo scp local_fil ...

  3. 【BZOJ】3400: [Usaco2009 Mar]Cow Frisbee Team 奶牛沙盘队(dp)

    http://www.lydsy.com/JudgeOnline/problem.php?id=3400 既然是倍数我们转换成mod.. 设状态f[i][j]表示前i头牛modj的方案 那么答案显然是 ...

  4. ReSharper 配置及用法(ZHUANG)

    1:安装后,Resharper会用他自己的英文智能提示,替换掉 vs2010的智能提示,所以我们要换回到vs2010的智能提示 2:快捷键.是使用vs2010的快捷键还是使用 Resharper的快捷 ...

  5. django用户认证系统——登录4

    用户已经能够在我们的网站注册了,注册就是为了登录,接下来我们为用户提供登录功能.和注册不同的是,Django 已经为我们写好了登录功能的全部代码,我们不必像之前处理注册流程那样费劲了.只需几分钟的简单 ...

  6. <转>RestKit在iOS项目中的使用,包含xcode配置说明

    本文转载至 http://www.cnblogs.com/visen-0/archive/2012/05/03/2480693.html 最近在iPhone工程中添加RestKit并编译,但是由于之前 ...

  7. 面试之Java持久层(十)

    91,什么是ORM?         对象关系映射(Object-Relational Mapping,简称ORM)是一种为了解决程序的面向对象模型与数据库的关系模型互不匹配问题的技术: 简单的说,O ...

  8. angular_文本变化

    注意,在input中用ng-change的时候,一定要结合着ng-model用 开头,注意在这里添加了ng-app <!DOCTYPE html> <html lang=" ...

  9. 一起学 Java集合框架、数据结构、泛型

    一.Java 集合框架 集合框架是一个用来代表和操纵集合的统一架构.所有的集合框架都包含如下内容: 接口:是代表集合的抽象数据类型.接口允许集合独立操纵其代表的细节.在面向对象的语言,接口通常形成一个 ...

  10. 160701、理解 Promise 的工作原理

    Javascript 采用回调函数(callback)来处理异步编程.从同步编程到异步回调编程有一个适应的过程,但是如果出现多层回调嵌套,也就是我们常说的厄运的回调金字塔(Pyramid of Doo ...