1041. Be Unique (20)
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 <stdio.h>
int lottery(int *,int);
int main(void)
{
//freopen("in.txt","r",stdin);
int n;
int num[];
int i; scanf("%d",&n);
for(i=;i<n;i++)
{
scanf("%d",&num[i]);
}
lottery(num,n);
return ;
}
int lottery (int *num,int n)
{
int i,j;
int flag;
int count[]={}; for(i=;i<n;i++)
{
count[num[i]]++;
} for(i=;i<n;i++)
{
if(count[num[i]]==)
{
printf("%d\n",num[i]);
return ;
}
} printf("None\n");
return ;
}
1041. Be Unique (20)的更多相关文章
- 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 (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】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 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 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 ...
- 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 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 ...
随机推荐
- [c language] getopt 其参数optind 及其main(int argc, char **argv) 参数解释
getopt被用来解析命令行选项参数.#include <unistd.h> extern char *optarg; //选项的参数指针extern int optind, //下一次调 ...
- PHP中__autoload()的不解之处,求高手指点
一整段代码: 运行结果: 使用__autoload(),分为两页代码: 第一段代码: ACMEManager.php,代码如下: 运行结果:
- 对于C++中const & T operator= 的一点思考
一个正常的assignment操作符的声明是这样的. const elmentType & elmentType::operator=(const elmentType &rhs) 这 ...
- Poj 2187 Beauty Contest_旋转凸包卡壳
题意:给你n个坐标,求最远的两点距离 思路:用凸包算法求处,各个定点,再用旋转凸包卡壳 #include <iostream> #include <cstdio> #inclu ...
- GridBagLayout练习
摘自http://blog.csdn.net/qq_18989901/article/details/52403737 GridBagLayout的用法 GridBagLayout是面板设计中最复杂 ...
- CSS3初步
一.CSS与CSS3的区别 非常简单,CSS代表"Casading Style Sheets",就是样式表,是一种替代并为网站添加样式的标记性语言.现在所使用的CSS基本是在199 ...
- ios像素点颜色取样
一.像素点颜色取样 + (UIColor*) getPixelColorAtLocation:(CGPoint)point inImage:(UIImage *)image { UIColor* co ...
- Java并发编程学习笔记 深入理解volatile关键字的作用
引言:以前只是看过介绍volatile的文章,对其的理解也只是停留在理论的层面上,由于最近在项目当中用到了关于并发方面的技术,所以下定决心深入研究一下java并发方面的知识.网上关于volatile的 ...
- uglifyjs入门接触
一.背景 今天在看<锋利的jQuery>文时,突然看到Uglifyjs压缩工具,感觉值得一试(玩),所以网上稍微搜了一下资料,简单的运用了一下,发现入门非常简单,当然网上有很多在线压缩工具 ...
- Jquery如何获取控件ID
l 1.#id 用法: $(”#myDiv”); 返回值 单个元素的组成的集合 说明: 这个就是直接选择html中的id=”myDiv” l 2.Element 用法: ...