PAT 甲级 1041 Be Unique (20 分)
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 [1,104]. 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 (≤105) 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
- 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 ...
- 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 ...
- 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. ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 S1 and S2, S=S1−S2 is defined to be t ...
- 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
- 『Python CoolBook』C扩展库_其三_简单数组操作
点击进入项目 这里的数组要点在于: 数组结构,array.array或者numpy.array 本篇的数组仅限一维,不过基础的C数组也是一维 一.分块讲解 源函数 /* Average values ...
- .NET:bin 与 obj,Debug 与 Release ,区别与选择
bin 与 obj bin 目录:用来存放编译的结果. ( bin是二进制binrary的英文缩写,因为最初C编译的程序文件都是二进制文件 ) 编译的结果,有 Debug 和 Release 两个版本 ...
- session和cookie的应用场景和区别
cookie存于客户端,session存于服务器端. cookie的作用是与服务器进行交互,作为http规范的一部分存在,而web storage是为了在本地存储“数据”而生. cookie应用场景: ...
- .net反编译的九款神器
本人搜集了下8款非常不错的.Net反编译利器: 1.Reflector Reflector是最为流行的.Net反编译工具.Reflector是由微软员工Lutz Roeder编写的免费程序.Refle ...
- python之路-----网络编程
一:客户端/服务端j架构(C/S架构) 在生活中,最常见的C/S架构即是APP和网页服务器这个架构.Client和Server之间利用网络来进行通信,客户端和服务器端的程序不同,用户的程序主要在客户端 ...
- Com 调用word和excel
using Microsoft.Office.Interop.Word;using System;using System.Collections.Generic;using System.Compo ...
- 201671010142 <java程序设计>初次学习心得与感悟
从开始对JDK的配置就遇到了问题,从这点就可以知道自己知识的薄弱.又知道了在控制台下一些常用命令的掌握.对知识的理解挺艰难,比如遇到一个新的问题就不知道该从哪里入手,有时候还不知道到底问题是啥.接受能 ...
- js解密
import base64 src_code = 'Ly93eDIuc2luYWltZy5jbi9tdzYwMC8wMDc2QlNTNWx5MWcxaWlpOHNybThqMzB1MDE5NWRyMS ...
- 自动化测试-8.selenium操作元素之键盘和鼠标事件
前言 在前面的几篇中重点介绍了一些元素的定位方法,定位到元素后,接下来就是需要操作元素了.本篇总结了web页面常用的一些操作元素方法,可以统称为行为事件 有些web界面的选项菜单需要鼠标悬停在某个元素 ...
- iOS动画学习
学习一下动画,感谢以下大神的文章: UIView:基础动画.关键帧动画.转场动画 Core Animation :基础动画,关键帧动画,动画组,转场动画,逐帧动画 CALayer :CALaye ...