Problem Description
In the new year party, everybody will get a "special present".Now it's your turn to get your special present, a lot of presents now putting on the desk, and only one of them will be yours.Each present has a card number on it, and your present's card number will be the one that different from all the others, and you can assume that only one number appear odd times.For example, there are 5 present, and their card numbers are 1, 2, 3, 2, 1.so your present will be the one with the card number of 3, because 3 is the number that different from all the others.
 
Input
The input file will consist of several cases.  Each case will be presented by an integer n (1<=n<1000000, and n is odd) at first. Following that, n positive integers will be given in a line, all integers will smaller than 2^31. These numbers indicate the card numbers of the presents.n = 0 ends the input.
 
Output
            For each case, output an integer in a line, which is the card number of your present.
 
Sample Input
5
1 1 3 2 2
3
1 2 1
0
 
Sample Output
3
2
Hint

Hint

use scanf to avoid Time Limit Exceeded
 
Author
8600
 
Source
HDU 2007-Spring Programming Contest - Warm Up (1)
 
Recommend
8600
 

题目内存限制:1024K,所以不能简单地用数组存然后再处理。

这题有一个好的方法—位异或。

位异或的运算法则吧:

1、a^b = b^a。

2、(a^b)^c = a^(b^c)。

3、a^b^a = b。

对于一个任意一个数n,它有几个特殊的性质:

1、0^n = n。

2、n^n = 0。

所以可以通过每次异或运算,最后剩下的值就是出现奇数次的那个数字。

#include <stdio.h> 
int main() 

    int n,x,ans; 
    while(scanf("%d",&n),n) 
    { 
        ans = 0; 
        while(n--) 
        { 
            scanf("%d",&x); 
            ans ^= x; 
        } 
        printf("%d\n",ans); 
    } 
    return 0; 
}

Find your present (2) (位异或)的更多相关文章

  1. HDU 2095 find your present (2)

    HDU 2095 find your present (2) 解法一:使用set 利用set,由于只有一个为奇数次,对一个数进行查询,不在集合中则插入,在集合中则删除,最后剩下的就是结果 /* HDU ...

  2. HDOJ(HDU) 1563 Find your present!(异或)

    Problem Description In the new year party, everybody will get a "special present".Now it's ...

  3. find your present (2) 2095

    Problem Description In the new year party, everybody will get a "special present".Now it's ...

  4. 跳转时候提示Attempt to present on while a presentation is in progress

    出现这种情况,例如:我在获取相册图片后,直接present到另一个页面,但是上一个页面可能还未dismiss,所以,要在获取相册图片的dismiss方法的complete的block里面写获取图片及跳 ...

  5. find your present (感叹一下位运算的神奇)

    find your present (2) Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Ot ...

  6. HTTP Status 400 - Required String parameter 'userName' is not present 错误

    HTTP Status 400 - Required String parameter 'userName' is not present 错误 先mark  有时间详细写 参考链接: https:/ ...

  7. Linux 克隆虚拟机引起的“Device eth0 does not seem to be present, delaying initialization”

    虚拟机Vmware上克隆了一个Red Hat Enterprise Linx启动时发现找不到网卡,如下所示,如果你在命令窗口启动网络服务就会遇到"Device eth0 does not s ...

  8. required string parameter XXX is not present

    @RequestParam jQuery调用方式: deleteFile: function(filePath) { return ajax({ method: 'POST', url: '/cm/s ...

  9. 启动网卡报:Device eth0 does not seem to be present”解决办法

    Device eth0 does not seem to be present”解决办法 : 用ifconfig查看发现缺少eth0,只有lo:用ifconfig -a查看发现多出了eth1的信息. ...

随机推荐

  1. PHP开发Apache服务器配置

    照此配置流程,绝对一路畅通,可保无虞. 昨天弄了个PHP小程序,想在本地跑一下测试,可是工作电脑没有安装环境,于是下载了一个wamp,一路畅通,Apache.Mysql.PHP就 全有了.启动wamp ...

  2. 开源的连接池技术DBCP和C3P0

    概述: Sun公司约定: 如果是连接池技术,需要实现一个接口! javax.sql.DataSource;   相关jar包和资料下载 1.1  DBCP连接池: l  DBCP 是 Apache 软 ...

  3. 【转】YUV420P的格式以及转换为RGB565的代码(Android摄像头的输出一般为YUV420P)

    http://blog.csdn.net/daisyhd/article/details/38866809 static void cvt_420p_to_rgb565(int width, int  ...

  4. Java-struts2的问题 java.lang.NoClassDefFoundError: org/apache/commons/lang3/StringUtils

    缺commons-lang3-3.1.jar,添加之后就可以了

  5. 启动MYSQL密码审计插件

    http://www.innomysql.com/article/25717.html [root@server-mysql plugin]# pwd /usr/local/mysql56/lib/p ...

  6. careercup-链表 2.6

    2.6 给定一个有环链表,实现一个算法返回环路的开头结点. 类似leetcode中 Linked List Cycle II C++实现代码: #include<iostream> #in ...

  7. Servlet配置文件

    <url-pattern>/servlet/demo</url-pattern> 1.以 / 开头, /代表工程路径:(必须要加 / ) 2.以 * 开头,必须加后缀名 /* ...

  8. 一个类搞定UIScrollView那些事

    前言 UIScrollView可以说是我们在日常编程中使用频率最多.扩展性最好的一个类,根据不同的需求和设计,我们都能玩出花来,当然有一些需求是大部分应用通用的,今天就聊一下以下需求,在一个categ ...

  9. angularjs 更新局部作用域

    前几天项目需要,做了一个背景遮罩的弹出框,html采用js动态添加进去的,结果发现angularjs绑定在这里面不起作用,搜索下解决了,记录下: var smallApplyParent = docu ...

  10. 11.12 noip模拟试题

    题目名称 加密 冒泡排序图 重建可执行文件名 encrypt bubble rebuild输入文件名 encrypt.in bubble.in rebuild.in输出文件名 encrypt.in b ...