A. MUH and Sticks
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Two polar bears Menshykov and Uslada from the St.Petersburg zoo and elephant Horace from the Kiev zoo got six sticks to play with and assess the animals' creativity. Menshykov, Uslada and Horace decided to make either an elephant or a bear from those sticks. They can make an animal from sticks in the following way:

  • Four sticks represent the animal's legs, these sticks should have the same length.
  • Two remaining sticks represent the animal's head and body. The bear's head stick must be shorter than the body stick. The elephant, however, has a long trunk, so his head stick must be as long as the body stick. Note that there are no limits on the relations between the leg sticks and the head and body sticks.

Your task is to find out which animal can be made from the given stick set. The zoo keeper wants the sticks back after the game, so they must never be broken, even bears understand it.

Input

The single line contains six space-separated integers li (1 ≤ li ≤ 9) — the lengths of the six sticks. It is guaranteed that the input is such that you cannot make both animals from the sticks.

Output

If you can make a bear from the given set, print string "Bear" (without the quotes). If you can make an elephant, print string "Elephant" (wıthout the quotes). If you can make neither a bear nor an elephant, print string "Alien" (without the quotes).

Sample test(s)
input
4 2 5 4 4 4
output
Bear
input
4 4 5 4 4 5
output
Elephant
input
1 2 3 4 5 6
output
Alien
Note

If you're out of creative ideas, see instructions below which show how to make a bear and an elephant in the first two samples. The stick of length 2 is in red, the sticks of length 4 are in green, the sticks of length 5 are in blue.

sb模拟题

唯一的难点是……cf太卡交都交不上去

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<ctime>
#include<set>
#define LL long long
using namespace std;
inline LL read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
LL a[10];
LL n,aa,bb;
LL s,mx,saves=-1;
int main()
{
for (int i=1;i<=6;i++)a[i]=read();
sort(a+1,a+7);
for (int i=1;i<=6;i++)
{
if(a[i]==a[i-1]||i==1)s++;
else s=1;
if(s==4) saves=a[i];
mx=max(mx,s);
}
if (mx<4)
{
printf("Alien\n");
return 0;
}
if (mx==6)
{
printf("Elephant\n");
return 0;
}
if (mx==5)
{
printf("Bear\n");
return 0;
}
for(int i=1;i<=6;i++)
if (a[i]!=saves)
{
if (!aa) aa=a[i];
else bb=a[i];
}
if (aa!=bb)
{printf("Bear\n");
}else printf("Elephant\n");
}

  

cf471A MUH and Sticks的更多相关文章

  1. 【CodeForces 471A】MUH and Sticks

    题 题意 给你六根木棍的长度,熊需要头比身体短,大象需要头和身体一样,四肢要一样长,否则就是外星人.请你判断能组成哪一个. 分析 暴力,循环看一下每根有几根相同的,然后如果有四根都是有四根相同的&am ...

  2. [CodeForces 471A] MUH and Sticks

    题目链接:http://codeforces.com/problemset/problem/471/A 题目数据规模1 - 9,可以用一个数组进行计数,减掉出现四次的数,看看还有几个是非零数,有一个就 ...

  3. HDOJ 1051. Wooden Sticks 贪心 结构体排序

    Wooden Sticks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) To ...

  4. POJ 2653 Pick-up sticks (线段相交)

    题意:给你n条线段依次放到二维平面上,问最后有哪些没与前面的线段相交,即它是顶上的线段 题解:数据弱,正向纯模拟可过 但是有一个陷阱:如果我们从后面向前枚举,找与前面哪些相交,再删除前面那些相交的线段 ...

  5. hduoj 1455 && uva 243 E - Sticks

    http://acm.hdu.edu.cn/showproblem.php?pid=1455 http://uva.onlinejudge.org/index.php?option=com_onlin ...

  6. POJ 2653 Pick-up sticks【线段相交】

    题意:n根木棍随意摆放在一个平面上,问放在最上面的木棍是哪些. 思路:线段相交,因为题目说最多有1000根在最上面.所以从后往前处理,直到木棍没了或者最上面的木棍的总数大于1000. #include ...

  7. POJ1065Wooden Sticks[DP LIS]

    Wooden Sticks Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 21902   Accepted: 9353 De ...

  8. 【POJ 2653】Pick-up sticks 判断线段相交

    一定要注意位运算的优先级!!!我被这个卡了好久 判断线段相交模板题. 叉积,点积,规范相交,非规范相交的简单模板 用了“链表”优化之后还是$O(n^2)$的暴力,可是为什么能过$10^5$的数据? # ...

  9. CF451A Game With Sticks 水题

    Codeforces Round #258 (Div. 2) Game With Sticks A. Game With Sticks time limit per test 1 second mem ...

随机推荐

  1. c++ 12

    一.模板与继承 1.从模板类派生模板子类 2.为模板子类提供基类 二.容器和迭代器 以链表为例. 三.STL概览 1.十大容器 1)向量(vector):连续内存,后端压弹,插删低效 2)列表(lis ...

  2. HDU1841——KMP算法

    这个题..需要对KMP的模板理解的比较透彻,以前我也只是会套模板..后来才知道..之会套模板是不行的..如果不能把握模板的每一个细节`,至少能搞清楚模板的每一个模块大体是什么意思.. 题意是给出两个串 ...

  3. 2016"百度之星" - 资格赛(Astar Round1) 1001

    思路:第一个做法就是:每读入起始位置i和结束位置j,就从这位置i到位置j计算,可是TLE了,后面我想想要是我输入一个最长的字符串,且以最大次数计算开始位置1到结束位置100000,那么这计算量是很大的 ...

  4. JavaScript 基础一

    内部: <Script Language="JavaScript" type="text/javascript"> JavaScript代码 < ...

  5. iOS 面试常见问题最全梳理

    序言 目前形势,参加到iOS队伍的人是越来越多,甚至已经到供过于求了.今年,找过工作人可能会更深刻地体会到今年的就业形势不容乐观,加之,培训机构一火车地向用人单位输送iOS开发人员,打破了生态圈的动态 ...

  6. uva 310 L--system(隐式图搜索+字符串处理)

     L-system  A D0L (Deterministic Lindenmayer system without interaction) system consists of a finite ...

  7. Weblogic的Admin server进程将CPU消耗尽问题解决

    1.serverCPU被耗尽,持续100% 以下附nmon图 2.两个weblogicadmin server进程将CPU耗尽 问题:24298进程,占用百分之四千多的CPU资源 23529进程,占用 ...

  8. LinkedList的分析(转)

    一.源码解析 1. LinkedList类定义. public class LinkedList<E> extends AbstractSequentialList<E> im ...

  9. java.util.concurrent.ExecutionException

    java.util.concurrent.ExecutionException: org.apache.catalina.LifecycleException: Failed to start com ...

  10. BackgroundWorker 后台进程控制窗体label、richtextbook内容刷新

    昨天写了一个从文章中提取关键词的程序,写完处理的逻辑后又花了好几个小时在用户友好性上.加了几个progressBar,有显示总进度的.有显示分布进度的..因为程序要跑好几个小时才能执行好,只加个总进度 ...