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 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 <algorithm>
#include <cstdio>
#include <cstring>
#include <map>
#include <stack>
#include <vector>
#include <queue>
#include <set>
#define LL long long
using namespace std;
const int MAX = 1e5 + ; map <int, int> mp;
map <int, int> :: iterator iter;
int n, a, A[MAX]; int main()
{
// freopen("Date1.txt", "r", stdin);
scanf("%d", &n);
for (int i = ; i < n; ++ i)
{
scanf("%d", &a);
A[i] = a;
mp[a] ++;
} for (int i = ; i < n; ++ i)
{
if (mp[A[i]] == )
{
printf("%d\n", A[i]);
return ;
}
}
printf("None\n");
return ;
}
pat 1041 Be Unique(20 分)的更多相关文章
- 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)(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 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甲级】1041 Be Unique (20 分)(多重集)
题意: 输入一个正整数N(<=1e5),接下来输入N个正整数.输出第一个独特的数(N个数中没有第二个和他相等的),如果没有这样的数就输出"None". AAAAAccepte ...
- 1041 Be Unique (20分)(水)
Being unique is so important to people on Mars that even their lottery is designed in a unique way. ...
- 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 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 1035 Password(20 分)
1035 Password(20 分) To prepare for PAT, the judge sometimes has to generate random passwords for the ...
随机推荐
- [USACO10NOV]奶牛的图片Cow Photographs
题目描述 Farmer John希望给他的N(1<=N<=100,000)只奶牛拍照片,这样他就可以向他的朋友炫耀他的奶牛. 这N只奶牛被标号为1..N. 在照相的那一天,奶牛们排成了一排 ...
- libevent::bufferevent
#include <cstdio> #include <netinet/in.h> #include <sys/socket.h> #include <fcn ...
- 共轭梯度法求解协同过滤中的 ALS
协同过滤是一类基于用户行为数据的推荐方法,主要是利用已有用户群体过去的行为或意见来预测当前用户的偏好,进而为其产生推荐.能用于协同过滤的算法很多,大致可分为:基于最近邻推荐和基于模型的推荐.其中基于最 ...
- 后缀数组(SA)
学习了LRJ神犇的代码.orz. 首先真心建议了解下基数排序!!且要有一定的c++程序经验,否则程序很难看懂. 然后对着下面的程序调试(假装你已经会了算法思想) 弄个一个礼拜一下午就能学会了. 该算法 ...
- 百万年薪python之路 -- 迭代器
3.1 可迭代对象 3.1.1 可迭代对象定义 **在python中,但凡内部含有 _ _ iter_ _方法的对象,都是可迭代对象**. 3.1.2 查看对象内部方法 该对象内部含有什么方法除了看源 ...
- MacOs mysql 安装
1. 去官网下载mysql镜像:https://dev.mysql.com/downloads/file/?id=475582 2. 双击镜像文件 - > 双击.pkg文件 -> 出现 ...
- Codeblocks 等软件 修改源代码后 不能立即执行的解决办法||exe文件删除慢
不懈地奋斗了两天,终于找到原因了. 记录如下 症状: Codeblocks .Visual Studio 都出现此问题:修改源代码 无法立即执行 ,就是:cannot open output file ...
- MySQL基础篇(2)数据类型
MySQL提供了多种数据类型,主要包括数值型.字符串类型.日期和时间类型. 1.数值类型 整数类型:TINYINT(1字节).SMALLINT(2字节).MEDIUMINT(3字节).INT(INTE ...
- 如何在python文件中测试sql语句
在manage.py的同级目录下新建一个run.py import os if __name__ == '__main__': #加载Django项目的配置信息 os.environ.setdefau ...
- rem辅助式响应布局
<!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title> ...