PAT甲级——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 <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
- 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
https://pintia.cn/problem-sets/994805342720868352/problems/994805444361437184 Being unique is so imp ...
- 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 (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 ...
- 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——A1041 Be Unique
Being unique is so important to people on Mars that even their lottery is designed in a unique way. ...
- 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
- CountDownLatch和CyclicBarrier和Semaphore最通俗形象解释
应该还有好多同学对这三个的区别比较模糊,网络上其他文章说的也比较专业化.所以我在这里举个例子说明这三个的区别. 我们假定有一场百米比赛,比赛包括十个运动员和一个裁判,每个运动员和每个裁判都是一个线程, ...
- Windbg 大改版,值得期待
早上从twitter上面看到一篇文章,看到windbg会提供一个Time Travel Debugging(TTD) 功能,该功能会在未来的版本引入. Time travel debugging: I ...
- Jshint 安装方法
首先在编辑器中搜索扩展程序 "Jshint" 并安装,安装成功后 打开Javascript文件会出现报错提示: "Failed to load jshint librar ...
- 牛牛的DRB迷宫(DP、二进制编码器)
牛牛的DRB迷宫I 链接:https://ac.nowcoder.com/acm/contest/3004/A来源:牛客网 题目描述 牛牛有一个n*m的迷宫,对于迷宫中的每个格子都为'R','D',' ...
- UVA 11584 入门DP
一开始把它当成暴力来做了,即,从终点开始,枚举其最长的回文串,一旦是最长的,马上就ans++,再计算另外的部分...结果WA了 事实证明就是一个简单DP,算出两个两个点组成的线段是否为回文,再用LCS ...
- HDU 1588 矩阵快速幂 嵌套矩阵
这个题目搞了我差不多一个下午,之前自己推出一个公式,即 f[n+k]=k*f[n]+f[n-1]结果发现根本不能用,无法降低复杂度. 后来又个博客的做法相当叼,就按他的做法来了 即 最终求得是 S(n ...
- Java并发分析—Lock
1.Lock 和 Condition 当使用synchronied进行同步时,可以在同步代码块中只用常用的wait和notify等方法,在使用显示锁的时候,将通过Condition对象与任意Lock实 ...
- kotlin黑马影音项目学习笔记
1.包布局 --------model--------presenter----------------impl----------------interf--------view--------ui ...
- Go-方法-接口-异常处理-错误处理
方法 什么是方法? 方法其实就是一个函数,在 func 这个关键字和方法名中间加入了一个特殊的接收器类型.接收器可以是结构体类型或者是非结构体类型.接收器是可以在方法的内部访问的. package m ...
- SEO初步学习之影响网站排名的因素
本文介绍一些比较明显的因素,一些隐藏较深的原因还有待发掘: 1.采集网站内容,即抄袭其他网站的内容. 2.新站上传后建议不要有大的改动. 3.标题频繁修改. 4.大量投放垃圾外链. 5.不做友链,交友 ...