acm寒假特辑1月20日 CodeForces

C - 3 CodeForces - 469A

There is a game called “I Wanna Be the Guy”, consisting of n levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.

Little X can pass only p levels of the game. And Little Y can pass only q levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?

Input
The first line contains a single integer n (1 ≤  n ≤ 100).

The next line contains an integer p (0 ≤ p ≤ n) at first, then follows p distinct integers a1, a2, …, ap (1 ≤ ai ≤ n). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It’s assumed that levels are numbered from 1 to n.

Output
If they can pass all the levels, print “I become the guy.”. If it’s impossible, print “Oh, my keyboard!” (without the quotes).

Examples
Input
4
3 1 2 3
2 2 4
Output
I become the guy.
Input
4
3 1 2 3
2 2 3
Output
Oh, my keyboard!
Note
In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.

In the second sample, no one can pass level 4.

大概题意：两人玩游戏，都有自己不会的关，两人合作能打到n关吗？？
数据量不大，直接模拟。

//miku saiko
#include<iostream>
using namespace std;
int main()
{
	int n, x, a[105] = { 0 }, y, t1, oj = false;
	a[0] = 1;
	cin >> n >> x;
	for (int i = 0; i < x; i++)
	{cin >> t1; a[t1] = 1;}
	cin >> y;
	for (int i = 0; i < y; i++)
	{cin >> t1; a[t1] = 1;}
	for (int i = 1; i <= n; i++)//寻找
	{
		if (!a[i])
		{
			oj = true; break;
		}
	}
	if(oj)
		cout<< "Oh, my keyboard!" << endl;
	else
		cout<< "I become the guy." << endl;
    return 0;
}