PAT-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, 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 <set>
using namespace std; int main(){
int n;
int i,t;
cin>>n;
set<int> s,s1;
int record[n];
cin>>t;
s.insert(t);
s1.insert(t);
record[] = t;
for(i = ; i < n; i++){
cin>>t;
record[i] = t;
if(s1.find(t) != s1.end() && s.find(t) != s.end()){
s.erase(t);
}
else if(s1.find(t) == s1.end()){
s.insert(t);
s1.insert(t);
} }
if(s.empty()){
cout<<"None";
}else{
i = ;
while(s.find(record[i]) == s.end()){
i++;
}
cout<<record[i]<<endl;
}
return ;
}
PAT-1041 Be Unique的更多相关文章
- PAT 1041 Be Unique[简单]
1041 Be Unique (20 分) Being unique is so important to people on Mars that even their lottery is desi ...
- pat 1041 Be Unique(20 分)
1041 Be Unique(20 分) Being unique is so important to people on Mars that even their lottery is desig ...
- PAT 1041 Be Unique (20分)利用数组找出只出现一次的数字
题目 Being unique is so important to people on Mars that even their lottery is designed in a unique wa ...
- 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 ...
- 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 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
1041 Be Unique Being unique is so important to people on Mars that even their lottery is designed in ...
- 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 ...
- 【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 ...
随机推荐
- C#验证码使用
1.C#创建验证码 1.1 创建获取验证码页面(ValidateCode.aspx) <html xmlns="http://www.w3.org/1999/xhtml"&g ...
- Volatile vs. Interlocked vs. lock
今天在stackoverflow上看到一个关于Volatile, Interlock, Lock的问题,发现回答的特别好,所以就想到把它翻译一下, 希望给那些对它们有疑惑的人提供点帮助 :假设有一个类 ...
- COJ 2110 Day7-例3
Day7-例3 难度级别:C: 运行时间限制:5000ms: 运行空间限制:256000KB: 代码长度限制:2000000B 试题描述 输入 输入的第一行包含整数n和k,其中n(2 ≤ n ≤100 ...
- DFS hdu 1016
http://acm.hdu.edu.cn/showproblem.php?pid=1016 #include <iostream> using namespace std; int a[ ...
- SRM 396(1-250pt)
DIV1 250pt 题意:对于一个字符串s,若对于每一个i = 0 to s.size()-p-1都有s[i] = s[i+p]则称字符串s是p循环的."CATCATC", &q ...
- Java 类加载机制详解
一.类加载器 类加载器(ClassLoader),顾名思义,即加载类的东西.在我们使用一个类之前,JVM需要先将该类的字节码文件(.class文件)从磁盘.网络或其他来源加载到内存中,并对字节码进行解 ...
- set和replace方法的区别
对已有值的元素处理上两者是相同的,但是对于一个不存在的元素,set的作用就和add相当,replace则是只能对已经存在的元素进行处理如:表中某个字段值是空(null),如果某个字段为空,则通过查询方 ...
- opentesty--luasocket 安装
转载请注明原文地址:http://www.cnblogs.com/dongxiao-yang/p/4878323.html luasocket安装过程中遇到不少坑,之前采用的是从公司以前服务器的里面找 ...
- Java学习的随笔(2)Java语言的三大特性
1.面向对象的三大特性 面向对象的三大特性主要包括:继承.封装.多态 (1)继承:就是指子类(导出类)获得了基类的全部功能(所有的域和方法). 注:在子类中,想要调用基类的方法可以使用“super. ...
- java中的String.format使用
format(String format, Objece... argues)函数相当于C语言中的printf函数,但是相对来说更灵活. 和C中的printf函数差不多,在fo ...