1041 Be Unique

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 <string>
using namespace std;

int main(void)
{
int a[100001] = {0};
int b[100001] = {0};//用来区别该数字是否重复出现
int n,flag=0;
cin >> n;
for (int i = 0; i < n; i++)
{
cin >> a[i];
b[a[i]]++;
}
for (int i = 0; i < n; i++)
{
if (b[a[i]] == 1)
{
cout << a[i] << endl;
return 0;
}
}
if (flag == 0)
cout << "None" << endl;
return 0;
}

PAT甲级——1041 Be Unique的更多相关文章

  1. 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 desi ...

  2. 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 ...

  3. PAT 甲级 1041 Be Unique

    https://pintia.cn/problem-sets/994805342720868352/problems/994805444361437184 Being unique is so imp ...

  4. 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 ...

  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】1041. Be Unique (20)

    题目链接:http://pat.zju.edu.cn/contests/pat-a-practise/1041 题目描述: Being unique is so important to people ...

  7. 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. ...

  8. PAT甲级——A1041 Be Unique

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

  9. PAT Advanced 1041 Be Unique (20) [Hash散列]

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

随机推荐

  1. 判断苹果和安卓端或者wp端

    window.onload = function() { var u = navigator.userAgent; if(u.indexOf('Android') > -1 || u.index ...

  2. Html 常见标签,css基础样式,css的浮动和清流,浏览器的兼容

    1.html模板<!DOCTYPE html><html><head> <meta charset="UTF-8"> <tit ...

  3. 阿里云云服务器测试uwgis的基本流程

    基本背景 uWSGI是一个Web服务器,它实现了WSGI协议.uwsgi.http等协议.Nginx中HttpUwsgiModule的作用是与uWSGI服务器进行交换. 要注意 WSGI / uwsg ...

  4. (转)Java中的容器详细总结

    Java中的容器详细总结(编辑中) 原文链接:http://anxpp.com/index.php/archives/656/ 注:本文基于 Jdk1.8 编写 通常程序总是根据运行时才知道的某些条件 ...

  5. OSS 图片处理流程

    1.步骤一 2.步骤二 3.步骤三 4.步骤四 5.步骤五(步骤4完成会自动添加cname用户解析,不需要自己去加,只需要点击进来看下是否添加成功即可) 通过以上步骤就可以实现了图片服务的配置

  6. qt使用了qstackedwidget里面放置了widget后对该子widget设置的样式无效

    关键字:子窗口样式无效 QStackedwidget 问题: 我有一个对话框,里面放了一个qstackedwidget,qstackedwidget放了N个子窗口,使用addwidget添加上去了: ...

  7. Dijkstra与Floyd算法

    1. Dijkstra算法 1.1 定义概览 Dijkstra(迪杰斯特拉)算法是典型的单源最短路径算法,用于计算一个节点到其他所有节点的最短路径.主要特点是以起始点为中心向外层层扩展,直到扩展到终点 ...

  8. 爬虫防止浏览器防止debug处理

    方式一(基于你会前端,我比较喜欢这种方式) #复制html页面 #复制其中的js,css(css可有可无,如果加css和不加css情况不一样,网页可能做了css反爬处理) #全局搜索debug or ...

  9. iTOP-4418开发板TF卡烧写-引导uboot

    基于迅为iTOP-4418开发板 将 TF 卡接入开发板,将拨码开关设置为 TF 卡启动,进入 uboot 模式,如下图所示. 如下图所示,使用命令“fastboot”,接着就可以通过 OTG 给 e ...

  10. 201403-2 窗口 Java

    要想到定义一个窗口类,判断点在不在矩形里好判断 需要一个数组,存放结果 import java.util.ArrayList; import java.util.List; import java.u ...