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".

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.

题意:告诉我们飞机的航班流程,年初在家,(注意题目说从家里出发是会回来的)最后问我们,他是在home还是contest

解法:既然从家里出发会返回,那么我们只要比较出现家的次数奇偶性就行,偶数就是在家,奇数就是还没有回来

 #include<bits/stdc++.h>
using namespace std;
int n;
char s1[],s2[];
int main()
{
int ans=;
cin>>n;
scanf("%s",s1);
for(int i=;i<n;i++)
{
scanf("%s",s2);
if(strstr(s2,s1)!=NULL)
{
ans++;
}
}
if(ans%)
{
cout<<"contest";
}
else
{
cout<<"home";
}
return ;
}

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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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 ...

  7. 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 ...

  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 ...

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

    地址:http://codeforces.com/contest/765/problem/C 题目: C. Table Tennis Game 2 time limit per test 2 seco ...

随机推荐

  1. MT6737 Android N 平台 Audio系统学习----录音到播放录音流程分析

    http://blog.csdn.net/u014310046/article/details/54133688 本文将从主mic录音到播放流程来进行学习mtk audio系统架构.  在AudioF ...

  2. 解决Android Studio Fetching Android SDK component information失败问题【转】

    本文转载自:http://blog.csdn.net/isesar/article/details/41908089 Android Studio 安装完成后,如果直接启动,Android Studi ...

  3. C语言中的排序算法--冒泡排序,选择排序,希尔排序

    冒泡排序(Bubble Sort,台湾译为:泡沫排序或气泡排序)是一种简单的排序算法.它重复地走访过要排序的数列,一次比较两个元素,如果他们的顺序错误就把他们交换过来.走访数列的工作是重复地进行直到没 ...

  4. 详解linux中install命令和cp命令的区别

    基本上,在Makefile里会用到install,其他地方会用cp命令. 它们完成同样的任务——拷贝文件,它们之间的区别主要如下: .最重要的一点,如果目标文件存在,cp会先清空文件后往里写入新文件, ...

  5. vim的tab缩进及用空格设置

    编辑~/.vimrc文件,分别设置用空格而不是用tab,一个tab多少个空格,自动缩进多少宽度,显示行号. set expandtabset tabstop=4 set shiftwidth=4 se ...

  6. C++之remove和remove_if

    一.Remove()函数 remove(beg,end,const T& value) //移除区间{beg,end)中每一个“与value相等”的元素: remove只是通过迭代器的指针向前 ...

  7. 2.CSS 颜色代码大全

    确实使用,不用重复造轮子了!!! 摘自:http://www.cnblogs.com/circlebreak/p/6140602.html

  8. Win 7下破解Loadrunner 11(带中文版下载地址)

    空间管理您的位置: 51Testing软件测试网 » 测试是一种生活态度 » 日志 与您一起分享在测试过程中的快乐与辛酸... Win 7下破解Loadrunner 11(带中文版下载地址) 上一篇  ...

  9. 华为G700电脑版Root软件-Eroot

    华为G700电脑版Root软件--eroot.zip(10.5M) 查阅全文 ›

  10. OutputDebugString()输出调试的使用