A - I Wanna Be the Guy
Problem description
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.
解题思路:标记一下1~n出现的数字,如果都出现了,则输出"I become the guy.",否则输出"Oh, my keyboard!",水过!
AC代码:
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,p,q,x;bool flag=false,used[];
memset(used,false,sizeof(used));
cin>>n>>p;
for(int i=;i<=p;++i){cin>>x;used[x]=true;}
cin>>q;
for(int i=;i<=q;++i){cin>>x;used[x]=true;}
for(int i=;i<=n;++i)
if(!used[i]){flag=true;break;}
if(flag)cout<<"Oh, my keyboard!"<<endl;
else cout<<"I become the guy."<<endl;
return ;
}
A - I Wanna Be the Guy的更多相关文章
- 题目1162:I Wanna Go Home(最短路径问题进阶dijkstra算法))
题目链接:http://ac.jobdu.com/problem.php?pid=1162 详解链接:https://github.com/zpfbuaa/JobduInCPlusPlus 参考代码: ...
- BNUOJ 52308 We don't wanna work! set模拟
题目链接: https://acm.bnu.edu.cn/v3/problem_show.php?pid=52308 We don't wanna work! Time Limit: 60000msM ...
- HDU 5308 I Wanna Become A 24-Point Master(2015多校第二场)
I Wanna Become A 24-Point Master Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 ...
- 2015多校联合训练赛 hdu 5308 I Wanna Become A 24-Point Master 2015 Multi-University Training Contest 2 构造题
I Wanna Become A 24-Point Master Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 ...
- HDU 5308 I Wanna Become A 24-Point Master
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5308 题面: I Wanna Become A 24-Point Master Time Limit ...
- 2015 Multi-University Training Contest 2 hdu 5308 I Wanna Become A 24-Point Master
I Wanna Become A 24-Point Master Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 ...
- When you want to succeed as bad as you wanna breathe, then you’ll be successful.
上周末登了次山,回来就各种矫情犯懒.今天周四一周又要完蛋,我发现自己真的是对时间流逝无可奈何.然后中午看了把小码哥网站还有MJ博客什么的,各种首期班大爆照,心中羞愧无比.年纪大也不能放弃自己啊,要不人 ...
- We don't wanna work!
We don't wanna work! [JAG Asia 2016] 两个set,一个代表工作的,一个代表不工作的 其实是一个很简单的模拟,但是我竟然排序之前标号.... 检查代码的时候要从头开始 ...
- 2016弱校联盟十一专场10.3---We don't wanna work!(STL--set的使用)
题目链接 https://acm.bnu.edu.cn/v3/contest_show.php?cid=8504#problem/C 代码如下: #include <iostream> # ...
随机推荐
- asp.net mvc学习入门
MVC是什么? M: Model就是我们获取的网页需要的数据 V: View就是我们的aspx页面,注意这是一个不包含后台代码文件的aspx页面.(其实带有.asp.cs文件也不会有编译错误,但是这样 ...
- 【sqli-labs】 less37 POST- Bypass MYSQL_real_escape_string (POST型绕过MYSQL_real_escape_string的注入)
POST版本的less36 uname=1&passwd=1%df' or 1#
- beetl模板入门例子
加入maven依赖 <dependency> <groupId>org.beetl</groupId> <artifactId>beetl-core&l ...
- monad - the Category hierachy
reading the "The Typeclassopedia" by Brent Yorgey in Monad.Reader#13 ,and found that " ...
- PAT_A1148#Werewolf - Simple Version
Source: PAT 1148 Werewolf - Simple Version (20 分) Description: Werewolf(狼人杀) is a game in which the ...
- PAT_A1140#Look-and-say Sequence
Source: PAT A1140 Look-and-say Sequence (20 分) Description: Look-and-say sequence is a sequence of i ...
- python PIL图像处理-生成图片验证码
生成效果如图: 代码 from PIL import Image,ImageDraw,ImageFont,ImageFilter import random # 打开一个jpg图像文件: im = I ...
- 【JavaScript】不使用正则表达式和字符串的方式来解析浏览器的URl地址信息
1.比如我们要获取的网站URl地址是:https://music.163.com/#/playlist?id=2384581760 一般我们能够想到的方式是直接使用正则表达式获取使用字符串直接解析的方 ...
- CentOS7.4上搭建rocketMQ集群
一.rocketMQ集群部署方案优缺点对比: 多Master模式(2m-noslave) : 一个集群无Slave,全是Master,例如2个Master或者3个Master 优点:配置简单,单个Ma ...
- 08.Web服务器-3.Web静态服务器
1.显示固定的页面 from socket import * from multiprocessing import * import os def handleClient(clientSocket ...