1041 Be Unique (20 分)

Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1,10​4​​]. The first one who bets on a unique number wins. For example, if there are 7 people betting on { 5 31 5 88 67 88 17 }, then the second one who bets on 31 wins.

Input Specification:

Each input file contains one test case. Each case contains a line which begins with a positive integer N (≤10​5​​) and then followed by N bets. The numbers are separated by a space.

Output Specification:

For each test case, print the winning number in a line. If there is no winner, print None instead.

Sample Input 1:

7 5 31 5 88 67 88 17

Sample Output 1:

31

Sample Input 2:

5 888 666 666 888 888

Sample Output 2:

None
 #include<iostream>
#include<map> using namespace std; int main()
{
int N, temp, index = , min = ;
map<int, int> m; cin >> N; for (int i = ; i<N; ++i)
{
cin >> temp; if (m.find(temp) == m.end())
m.insert(pair<int, int>(temp, index++));
else
m[temp] = ;
} map<int, int>::iterator begin = m.begin(), end = m.end(), i, minIndex; for (i = begin; i != end; ++i)
{
if (i->second> && i->second<min)
{
min = i->second;
minIndex = i;
}
} if (min == )
cout << "None" << endl;
else
cout << minIndex->first << endl;
}

PAT 甲级 1041 Be Unique (20 分)的更多相关文章

  1. PAT 甲级 1041 Be Unique (20 分)(简单,一遍过)

    1041 Be Unique (20 分)   Being unique is so important to people on Mars that even their lottery is de ...

  2. PAT 甲级 1041. Be Unique (20) 【STL】

    题目链接 https://www.patest.cn/contests/pat-a-practise/1041 思路 可以用 map 标记 每个数字的出现次数 然后最后再 遍历一遍 找到那个 第一个 ...

  3. PAT Advanced 1041 Be Unique (20 分)

    Being unique is so important to people on Mars that even their lottery is designed in a unique way. ...

  4. PAT (Advanced Level) Practice 1041 Be Unique (20 分) 凌宸1642

    PAT (Advanced Level) Practice 1041 Be Unique (20 分) 凌宸1642 题目描述: Being unique is so important to peo ...

  5. PAT甲 1041. Be Unique (20) 2016-09-09 23:14 33人阅读 评论(0) 收藏

    1041. Be Unique (20) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Being uniqu ...

  6. PAT 甲级 1035 Password (20 分)

    1035 Password (20 分) To prepare for PAT, the judge sometimes has to generate random passwords for th ...

  7. PAT 甲级 1073 Scientific Notation (20 分) (根据科学计数法写出数)

    1073 Scientific Notation (20 分)   Scientific notation is the way that scientists easily handle very ...

  8. PAT 甲级 1050 String Subtraction (20 分) (简单送分,getline(cin,s)的使用)

    1050 String Subtraction (20 分)   Given two strings S​1​​ and S​2​​, S=S​1​​−S​2​​ is defined to be t ...

  9. PAT 甲级 1046 Shortest Distance (20 分)(前缀和,想了一会儿)

    1046 Shortest Distance (20 分)   The task is really simple: given N exits on a highway which forms a ...

随机推荐

  1. 『Python』socket网络编程

    Python3网络编程 '''无论是str2bytes或者是bytes2str其编码方式都是utf-8 str( ,encoding='utf-8') bytes( ,encoding='utf-8' ...

  2. Eclipse+Spring boot开发教程

    一.安装 其实spring boot官方已经提供了用于开发spring boot的定制版eclipse(STS,Spring Tool Suite)直接下载使用即可,但考虑到可能有些小伙伴不想又多装个 ...

  3. java变量的作用域和基本数据类型转换

    1.变量的作用域 赋值运算符 变量名 = 表达式 列: a = (b+3)+(b-1) 表达式就是符号(如:加号,减号)与操作数(如:b,3)的组合 自动类型转换(隐式类型转换):从小类型到大类型可以 ...

  4. java课堂笔记3

  5. day051 Django创建

    Django的下载安装 下载Django: pip3 install django==1.11.14 创建Django project(项目) 步骤1: 步骤2: 步骤3: 配置settings属性 ...

  6. 阶段01Java基础day17集合框架03

    17.01_集合框架(HashSet存储字符串并遍历) A:Set集合概述及特点 通过API查看即可 B:案例演示 HashSet存储字符串并遍历 HashSet<String> hs = ...

  7. Linux的相关概念

    1 Linux的相关概念 1.1 什么是操作系统? 操作系统(英语:operating system,缩写:OS)是管理计算机硬件与软件资源的计算机程序,同时也是计算机系统的内核与基石.操作系统需要处 ...

  8. git一些有用的命令

    更改本地和远程分支的名称 git branch -m old_branch new_branch # Rename branch locally 本地分支改名 git push origin :old ...

  9. echarts双y轴折线图柱状图混合实时更新图

    先看下效果,自己用ps做了张gif图,发现很好玩啊..不喜勿喷 自己下载个echarts.min.js 直接上代码: <!DOCTYPE html><html><head ...

  10. ecmall 后台添加新菜单

    所谓的开发新菜单,其实是和开发模块相对比的,之前说的开发模块,是在应对较大的,或者较为复杂,又相对独立于其他功能的项目需求. 而开发菜单,就是简单的在后台增加一个一级菜单以及其子菜单,或者直接在现有的 ...