标签：数学 codeforces
题目链接：http://codeforces.com/contest/469/problem/A
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).
4 3 1 2 3 2 2 4
I become the guy.
4 3 1 2 3 2 2 3
Oh, my keyboard!
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.
代码如下：
#include <cstdio>
#include <cstring>
int main()
{
int n;
int vis[1117];
while(~scanf("%d",&n))
{
int x, y;
memset(vis,0,sizeof(vis));
int tt;
scanf("%d",&x);
for(int i = 0; i < x; i++)
{
scanf("%d",&tt);
vis[tt] = 1;
}
scanf("%d",&y);
for(int i = 0; i < y; i++)
{
scanf("%d",&tt);
vis[tt] = 1;
}
int flag = 0;
for(int i = 1; i <= n; i++)
{
if(vis[i] == 0)
{
flag = 1;
break;
}
}
if(flag)
{
printf("Oh, my keyboard!\n");
}
else
printf("I become the guy.\n");
}
return 0;
}CodeForces 469A. I Wanna Be the Guy（数学）
标签：数学 codeforces
原文地址：http://blog.csdn.net/u012860063/article/details/39453765
