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;
}