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?

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.

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

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.

题意：给你一个数 N 意味着有 1 至 N 个关卡，然后 每行第一个数 p 代表后面有 p 个整数跟随。 每个整数代表 其 能过的关卡的编号。

   而现在有两组编号，题目就是要求我们把两组编号合起来，看里面是否能有 1 至 N 所有的数。

    如果有，则输出“I become the guy.” 否则输出Oh, my keyboard!

懂题意后这题就简单了，水掉就是！

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <vector>
#include <ostream>
#include <string>
#include <cstdlib>
#include <cmath>
#define PI 3.141592653
using namespace std;

int n;
int p,q;
int a[1000]; //用以标记

int main( )
{
    cin>>n;
    for(int i = 0;i <= n; i++)
    {
        a[i]=1; //初始化
    }
    cin>>p;
    int x;
    while(p--)
    {
        scanf("%d",&x);
        a[x]=0;
    }
    cin>>q;
    while(q--)
    {
        scanf("%d",&x);
        a[x]=0;
    }
    int judge=0;
    for(int i = 1;i <= n; ++i)
    {
        judge=judge+a[i];
    }
    if(judge)
        cout<<"Oh, my keyboard!"<<endl;
    else
        cout<<"I become the guy."<<endl;
    return 0;
}

1 1 0 4
2 3
0 1

3

2 3 0 20
15 17
23 26
1 4
7 11
15 17

20

所以。。。。暴力即可。。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <vector>
#include <ostream>
#include <string>
#include <cstdlib>
#include <cmath>
#define PI 3.141592653
using namespace std;

int p,q,l,r;
int vis[55][2];

int main( )
{
    cin>>p>>q>>l>>r;
    int a,b;
    int kis[1100];
    memset(kis,0,sizeof(kis));
    for(int i = 0;i < p; i++)
    {
        scanf("%d%d",&a,&b);
        vis[i][0]=a;
        vis[i][1]=b;
    }
    int judge=0;
    while(q--)
    {
        scanf("%d%d",&a,&b);
        for(int i = l;i <= r; i++)
        {
            if(kis[i]==0)
            for(int j = 0;j < p; j++)
            {
                if(a+i==vis[j][1]||a+i==vis[j][0]||b+i==vis[j][1]||b+i==vis[j][0])
                {
                    kis[i]=1;
                    break;
                }
                else if(b+i>vis[j][1]&&a+i<vis[j][1])
                {
                    kis[i]=1;
                    break;
                }
                else if(b+i>vis[j][0]&&a+i<vis[j][0])
                {
                    kis[i]=1;
                    break;
                }
                else if(b+i<vis[j][1]&&a+i>vis[j][0])
                {
                    kis[i]=1;
                    break;
                }
            }
        }
    }
    for(int i = l; i <= r; i++)
    {
        judge+=kis[i];
    }
    cout<<judge<<endl;
    return 0;
}

1

NO

8

YES
8 * 7 = 56
6 * 5 = 30
3 - 4 = -1
1 - 2 = -1
30 - -1 = 31
56 - 31 = 25
25 + -1 = 24

题意：从 1 至 N 个数取两个数，用“ + ”，“ - ”，“ * ”进行运算，并把得到的结果加到原来的数中。如此进行N-1次操作，凑出24即可。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <vector>
#include <ostream>
#include <string>
#include <cstdlib>
#include <cmath>
#define PI 3.141592653
using namespace std;

int n;

int main( )
{
    cin>>n;
    if(n<4)
    {
        cout<<"NO"<<endl;
    }
    else
    {
        cout<<"YES"<<endl;
        if(n%2==0)
        {
            printf("2 * 3 = 6\n");
            printf("6 * 4 = 24\n");
            printf("1 * 24 = 24\n");
            for(int i = 5;i < n; i+=2)
            {
                printf("%d - %d = 1\n",i+1,i);
                printf("1 * 24 = 24\n");
            }
        }
        else
        {
            printf("4 * 5 = 20\n");
            printf("3 + 20 = 23\n");
            printf("23 + 2 = 25\n");
            printf("25 - 1 = 24\n");
            for(int i = 6; i< n; i+=2)
            {
                printf("%d - %d = 1\n",i+1,i);
                printf("1 * 24 = 24\n");
            }
        }
    }
    return 0;
}

如有BUG，欢迎指出！