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 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.

 #include <iostream>
#include <cstdio>
#include <cstring> using namespace std; int main()
{
int n, cnt1, cnt2;
string src, line;
while(cin>>n)
{
cnt1 = ;
cnt2 = ;
cin >> src;
for(int i = ; i < n; i++)
{
cin>>line;
if(src[] == line[] && src[] == line[] && src[] == line[]) cnt1++;
if(src[] == line[] && src[] == line[] && src[] == line[]) cnt2++;
}
if(cnt1==cnt2) cout<<"home"<<endl;
else cout<<"contest"<<endl;
}
return ;
}

765A 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 - Neverending competitions

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

  3. 【codeforces 765A】Neverending competitions

    [题目链接]:http://codeforces.com/contest/765/problem/A [题意] 给你一个人的n个行程 行程都是从家到某个地方或从某个地方到家; 且是无序的,且如果到了非 ...

  4. Codeforces Round #397 by Kaspersky Lab and Barcelona Bootcamp (Div. 1 + Div. 2 combined)

    运气好,分到的房里我最先开始Hack C题,Hack了12个,听说F题沙雕莫队但我不会,最后剩不到15分钟想出E题做法打了一波结果挂了,最后虽然上分了但总有点不甘心. 最后A掉ABCD Hack+12 ...

  5. CodeForces765A

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

  6. 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 ...

  7. Codeforces Round #397 题解

    Problem A. Neverending competitions 题目大意 一个团队有多个比赛,每次去比赛都会先订机票去比赛地点,然后再订机票返回.给出\(n\)个含有起止地点的购票记录(不按时 ...

  8. Improving your submission -- Kaggle Competitions

    1: Improving Our Features In the last mission, we made our first submission to Titanic: Machine Lear ...

  9. Getting started with Kaggle -- Kaggle Competitions

    1: The Competition We'll be learning how to generate a submission for a Kaggle competition. Kaggle i ...

随机推荐

  1. c++官方文档-动态内存

    #include<iostream> #include <new> using namespace std; int main() { /** * 动态内存 * url: ht ...

  2. UVA327

    模拟 这个问题的任务是求解一组c语言里的表达式,但是你不需要知道c语言是怎么解决这个问题!每一行一个表达式,每个表达式的组成不会超过110个字符.待求解的表达式只包含一个int类型变量和一个组有限的操 ...

  3. bootstrapValidator针对设置赋值进行验证

    bootstrapValidator在提交的时候可以进行验证,但是对于点击输入框进行赋值的时候验证失效. 解决方法: 然后在设置change方法方可解决.

  4. JAVA 读取配置文件 xxx.properties

    package config_demo; import java.io.InputStream; import java.util.Properties; public class UrlDemo { ...

  5. HTML5 Canvas 小例子 旋转的图片

    <一>CSS部分 @charset "utf-8"; *{ padding:; margin:; outline: none; } #canvas{ position: ...

  6. leetcode13

    public class Solution { private int ChangeToInt(char c) { ; string s = c.ToString(); switch (s) { ca ...

  7. 15.Result配置详解

    转自:https://wenku.baidu.com/view/84fa86ae360cba1aa911da02.html 说明:在前面的许多案例中我们所用到的Action基本都继承自ActionSu ...

  8. 用yield 实现协程 (包子模型)

    协程是一种轻量级的线程 无需线程上下级的开销, 所有的协程都在一个线程内执行 import time def consumer(name): print('%s is start to eat bao ...

  9. Spring MVC 视图及视图解析器

    org.springframework.web.servlet.view.InternalResoureceViewResolve 把逻辑视图改为物理视图 可混用多种视图 不进过Handler直接进入 ...

  10. JS 判断鼠标滚轮的上下滚动

    JS 判断鼠标滚轮的上下滚动   <script type="text/javascript"> var scrollFunc = function (e) { e = ...