find your present (2)

最新推荐文章于 2021-04-11 12:36:27 发布

原创最新推荐文章于 2021-04-11 12:36:27 发布 · 569 阅读

0 ·

CC 4.0 BY-SA版权

HDOJ--ACMSteps 专栏收录该内容

18 篇文章

订阅专栏

本文介绍了一个通过位运算找出唯一奇数次出现的礼物编号的算法。该算法适用于一系列礼物编号中仅有一个编号出现奇数次的情况。输入为一组整数，输出即为所求的特殊礼物编号。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

find your present (2)

Time Limit: 1000/2000 MS (Java/Others) Memory Limit: 32768/1024 K (Java/Others)

Total Submission(s): 1835 Accepted Submission(s): 543

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

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

位运算的巧妙运用！

#include<stdio.h>
int main()
{
   int i,k,n,s;
   while(scanf("%d",&n),n)
   {
   s=0;
    for(i=0;i<n;i++)
{
scanf("%d",&k);
s^=k;
}
   printf("%d\n",s);
   }
}