codeforces 469A - I Wanna Be the Guy set做法（适合set萌新）

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
inputCopy
4
3 1 2 3
2 2 4
outputCopy
I become the guy.
inputCopy
4
3 1 2 3
2 2 3
outputCopy
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.
链接：https://codeforces.ml/problemset/problem/469/A
题意：两个人的数字合起来起来可以达到层数的话 就能过关。
思路：想到set中无重复的数字，最后直接用 s.size()==t 来判断两个人是否能达到层数。
ac代码如下：

#include
#include
using namespace std;
int main(){
	set s;
	int t,n,a;
	cin>>t;
	cin>>n;
	while(n--){
		cin>>a;
		if(s.find(a)==s.end()){
			s.insert(a);
		}
	}
	cin>>n;
	while(n--){
		cin>>a;
		if(s.find(a)==s.end()){
			s.insert(a);
		}
	}
	if(s.size()==t){
		cout<<"I become the guy."<