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. 数组和List以指定的方式拼接成字符串类型

    /// <summary> /// list转换成格式的字符串 /// </summary> /// <param name="param">拼 ...

  2. sevrlet进行用户名密码校验

    在eclipse中建立了web项目,实现注册和登录还有在个人中心显示用户名密码 注册功能 源码如下 package com.sevlet.demo; import java.io.IOExceptio ...

  3. SAP发布REST/HTTP接口

    1.SE24新建类:ZCL_REST_QUERY 激活,然后添加interface:IF_HTTP_EXTENSION并激活. 2.实现IF_HTTP_EXTENSION~HANDLE_REQUEST ...

  4. 92. Reverse Linked List II 反转链表 II

    网址:https://leetcode.com/problems/reverse-linked-list-ii/ 核心部分:通过a.b.c三个变量之间的相互更新,不断反转部分链表 然后将反转部分左右两 ...

  5. vim 插件 -- NERDTree

    介绍 NERDTree 插件就是使vim编辑器有目录效果. 所谓无图无真相,所以直接看这个插件的效果图吧. 下载 https://www.vim.org/scripts/script.php?scri ...

  6. Vuejs的$nextTick原理

    本质: nextTick,本质上是一个异步API,表示当前同步流程执行完成后再调用传入的函数. 根据环境不同,异步API的实现可以分别通过: setTimeout(0), new Promise(), ...

  7. LeetCode 257 二叉树的所有路径

    题目: 给定一个二叉树,返回所有从根节点到叶子节点的路径. 说明: 叶子节点是指没有子节点的节点. 示例: 输入: 1 / \ 2 3 \ 5 输出: ["1->2->5&quo ...

  8. centos7下zabbix4.0配置磁盘IO监控

    一:准备 1.1:安装sysstat yum -y install sysstat 1.2:安装zabbix-get yum install -y zabbix-get.x86_64 1.3:iost ...

  9. 创建学生类 有姓名 学校 和年龄 覆盖toString() 1放到集合ArrayList 然后 2在第2个位置插入1个学生信息 3判断 刘德华这个学生是否存在 存在就打出来, 4输出全部学生信息 直接打印对象

    学生类 package com.lanxi.demo1; public class Student { //创建属性 姓名,学校,年龄 private String name; private Str ...

  10. Jsの练习-数组常用方法

    1. join() 方法: <!DOCTYPE html> <html lang="en"> <head> <meta charset=& ...