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. git命令行删除远程分支

    先查看远程分支 git branch -r 使用下面两条命令来删除远程分支 git branch -r -d origin/branch-name git push origin :branch-na ...

  2. 关于Javascript中声明变量、函数的笔记

    一.概念 1.变量声明 在JavaScript中,变量一般通过var关键字(隐式声明,let关键字声明除外)进行声明,如下通过var关键字声明a,b,c三个变量(并给其中的a赋值): var a=1, ...

  3. UITabBar 设置选中、未选中状态下title的字体颜色

    一.如果只是设置选中状态的字体颜色,使用 tintColor  就可以达到效果 self.tabBar.tintColor = [UIColor redColor]; 二.但如果要将未选中状态和选中状 ...

  4. html5--5-6 绘制圆/弧

    html5--5-6 绘制圆/弧 学习要点 掌握arc() 方法创建圆弧/曲线(用于创建圆或部分圆) 矩形的绘制方法 rect(x,y,w,h)创建一个矩形 strokeRect(x,y,w,hx,y ...

  5. 关于URL编码的一些结论

    转载自:http://www.ruanyifeng.com/blog/2010/02/url_encoding.html与http://www.ruanyifeng.com/blog/2007/10/ ...

  6. 记录ubuntu16.04版本安装过程中遇到的问题

    记录ubuntu16.04版本安装和使用过程中遇到的些问题,方便以后查看,主要内容有: 1. ubuntu源替换 2. windows与vmware ubuntu文件夹共享 3. putty连接ubu ...

  7. Tensorflow和Caffe 简介

    TensorFlow TensorFlow 是相对高阶的机器学习库,用户可以方便地用它设计神经网络结构,而不必为了追求高效率的实现亲自写 C++或 CUDA 代码.它和 Theano 一样都支持自动求 ...

  8. JAVA 中的堆和栈

    栈与堆都是Java用来在Ram中存放数据的地方.与C++不同,Java自动管理栈和堆,程序员不能直接地设置栈或堆.     Java的堆是一个运行时数据区,类的对象从中分配空间.这些对象通过new.n ...

  9. hihoCoser(#1149 : 回文字符序列)

    时间限制:2000ms 单点时限:1000ms 内存限制:256MB 描述 给定字符串,求它的回文子序列个数.回文子序列反转字符顺序后仍然与原序列相同.例如字符串aba中,回文子序列为"a& ...

  10. calico在docker上的部署及验证

    1. 背景 以下的部署以五台服务器环境为例: 服务器1: hostname为etcdnode1, IP为192.168.56.100 服务器2: hostname为etcdnode2, IP为192. ...