Ace of Aces
Description
There is a mysterious organization called Time-Space Administrative Bureau (TSAB) in the deep universe that we humans have not discovered yet. This year, the TSAB decided to elect an outstanding member from its elite troops. The elected guy will be honored with the title of "Ace of Aces".
After voting, the TSAB received N valid tickets. On each ticket, there is a number Ai denoting the ID of a candidate. The candidate with the most tickets nominated will be elected as the "Ace of Aces". If there are two or more candidates have the same number of nominations, no one will win.
Please write program to help TSAB determine who will be the "Ace of Aces".
Input
There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:
The first line contains an integer N (1 <= N <= 1000). The next line contains N integers Ai (1 <= Ai <= 1000).
Output
For each test case, output the ID of the candidate who will be honored with "Ace of Aces". If no one win the election, output "Nobody" (without quotes) instead.
Sample Input
3
5
2 2 2 1 1
5
1 1 2 2 3
1
998
Sample Output
2
Nobody
998
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std; int a[];
int f[]; int main()
{
int t,n,x,flag;
int maxx;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
memset(f,,sizeof(f));
maxx=-;flag=;
for(int i=;i<n;i++)
{
scanf("%d",&a[i]);
f[a[i]]++;
}
for(int i=;i<;i++)
{
if(f[a[i]]>maxx)
{
maxx=f[a[i]];
x=a[i];
}
}
for(int i=;i<;i++)
if(f[a[i]]==maxx&&a[i]!=x)
flag++;
if(flag>)
printf("Nobody\n");
else
printf("%d\n",x);
}
return ;
}
Ace of Aces的更多相关文章
- zjuoj The 12th Zhejiang Provincial Collegiate Programming Contest Ace of Aces
http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5493 The 12th Zhejiang Provincial ...
- 第十二届浙江省大学生程序设计大赛-Ace of Aces 分类: 比赛 2015-06-26 14:25 12人阅读 评论(0) 收藏
Ace of Aces Time Limit: 2 Seconds Memory Limit: 65536 KB There is a mysterious organization called T ...
- ZOJ 3869 Ace of Aces
There is a mysterious organization called Time-Space Administrative Bureau (TSAB) in the deep univer ...
- 水题 ZOJ 3869 Ace of Aces
题目传送门 水题,找出出现次数最多的数字,若多个输出Nobody //#include <bits/stdc++.h> //using namespace std; #include &l ...
- 140 - The 12th Zhejiang Provincial Collegiate Programming Contest(浙江省赛2015)
Ace of Aces Time Limit: 2 Seconds Memory Limit: 65536 KB There is a mysterious organization c ...
- 【windows 访问控制】十一、C# 实操 对象 System.Security.AccessControl 命名空间
AccessControl 命名空间 结构图 解说: DirectorySecurity=目录ACLFileSecurity=文件ACLFileSystemAuditRule=目录和文件中SACL中的 ...
- ACE的构建(VC++6.0环境)
ACE的构建(VC++6.0环境)Windows下ACE的构建1. 将ACE-5.5.zip解压到所需的安装目录,此处以E:/为例,解压后形成ACE_wrappers文件夹,因此ACE将会存在于ACE ...
- Windows Server 2008及以上系统磁盘无法查看(About UAC and ACE)
在windows Server2008及以上系統,如果UAC Enabled,ACE列表中不會包含Administrators成員的SID,所以即使你是administrators的成員,也無法訪問D ...
- Microsoft ACE OLEDB 12.0 数据库连接字符串
Excel 97-2003 Provider=Microsoft.ACE.OLEDB.12.0;Data Source=c:\myFolder\myOldExcelFile.xls;Extended ...
随机推荐
- BZOJ 3391: [Usaco2004 Dec]Tree Cutting网络破坏( dfs )
因为是棵树 , 所以直接 dfs 就好了... ---------------------------------------------------------------------------- ...
- [javascript]在浏览器端应用cookie记住用户名
<!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...
- 如何用 new 来动态开辟一个二维数组
一般的做法是: int **p = new int*[m]; //m行n列型 for (i = 0; i < m; ++i) { p[i] = new int[n]; for (j = 0; j ...
- 深入JDK源码之Arrays类中的排序查找算法(转)
原文出处: 陶邦仁 binarySearch()方法 二分法查找算法,算法思想:当数据量很大适宜采用该方法.采用二分法查找时,数据需是排好序的. 基本思想:假设数据是按升序排序的,对于给定值x,从序列 ...
- AzureDev 社区活动获奖者公布
今天,我们高兴地宣布 AzureDev社区活动的获奖者,并向这 5 个非盈利技术教育组织发放 10 万美元奖金.在 2013 年的Build大会上宣布的 AzureDev 活动专注于通过代码改变世界, ...
- in window js 未定义和undifined的区别
浏览器报错:未定义和undifined不是同一概念,前者是没有申明,后者是没有赋值. 1: <html> <body> <script> ...
- struts2_4_为Action属性注入值
Struts2为Action中的属性提供了依赖注入功能,在struts2的配置文件里,能够为Action中的属性注入值,属性必须提供setter方法. 1)employeeAction类: publi ...
- java序列化对象 插入、查询、更新到数据库
java序列化对象 插入.查询.更新到数据库 : 实现代码例如以下: import java.io.ByteArrayInputStream; import java.io.ByteArrayOutp ...
- BZOJ 1475: 方格取数( 网络流 )
本来想写道水题....结果调了这么久!就是一个 define 里面少加了个括号 ! 二分图最大点权独立集...黑白染色一下 , 然后建图 : S -> black_node , white_no ...
- shell字符串替换