I Wanna Be the Guy

文章目录

一、 I Wanna Be the Guy

总结

一、 I Wanna Be the Guy

本题链接： I Wanna Be the Guy

题目：

A. I Wanna Be the Guy

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

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.

本博客给出本题截图：

题意：如果两个人会的所有level包括了题目中要求的所有level，那就输出I become the guy.，否则输出Oh, my keyboard!

AC代码

#include <iostream>
#include <set>
using namespace std;
set <int> s;
int main()
{
    int n;
    cin >> n;
    int p;
    cin >> p;
    for (int i = 0; i < p; i ++ )
    {
        int a;
        cin >> a;
        s.insert(a);
    }
    int q;
    cin >> q;
    for (int i = 0; i < q; i ++ )
    {
        int a;
        cin >> a;
        s.insert(a);
    }
    if (s.size() == n)
        puts("I become the guy.");
    else puts("Oh, my keyboard!");
    return 0;
}

总结

水题，不解释