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    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:

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 int main(){
     4     int n,p,q,x;bool flag=false,used[105];
     5     memset(used,false,sizeof(used));
     6     cin>>n>>p;
     7     for(int i=1;i<=p;++i){cin>>x;used[x]=true;}
     8     cin>>q;
     9     for(int i=1;i<=q;++i){cin>>x;used[x]=true;}
    10     for(int i=1;i<=n;++i)
    11         if(!used[i]){flag=true;break;}
    12     if(flag)cout<<"Oh, my keyboard!"<<endl;
    13     else cout<<"I become the guy."<<endl;
    14     return 0;
    15 }
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  • 原文地址:https://www.cnblogs.com/acgoto/p/9159193.html
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