最新文章专题视频专题问答1问答10问答100问答1000问答2000关键字专题1关键字专题50关键字专题500关键字专题1500TAG最新视频文章推荐1 推荐3 推荐5 推荐7 推荐9 推荐11 推荐13 推荐15 推荐17 推荐19 推荐21 推荐23 推荐25 推荐27 推荐29 推荐31 推荐33 推荐35 推荐37视频文章20视频文章30视频文章40视频文章50视频文章60 视频文章70视频文章80视频文章90视频文章100视频文章120视频文章140 视频2关键字专题关键字专题tag2tag3文章专题文章专题2文章索引1文章索引2文章索引3文章索引4文章索引5123456789101112131415文章专题3
当前位置: 首页 - 科技 - 知识百科 - 正文

hdu2095findyourpresent(2)找到只出现一次的数字

来源:动视网 责编:小采 时间:2020-11-09 08:31:59
文档

hdu2095findyourpresent(2)找到只出现一次的数字

hdu2095findyourpresent(2)找到只出现一次的数字:find your present (2) Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/1024 K (Java/Others) Total Submission(s): 15349Accepted Submission(s): 5821 Problem Description In the new year party, everybody will get a special present.Now
推荐度:
导读hdu2095findyourpresent(2)找到只出现一次的数字:find your present (2) Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/1024 K (Java/Others) Total Submission(s): 15349Accepted Submission(s): 5821 Problem Description In the new year party, everybody will get a special present.Now


find your present (2) Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/1024 K (Java/Others) Total Submission(s): 15349Accepted Submission(s): 5821 Problem Description In the new year party, everybody will get a special present.Now

find your present (2)

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/1024 K (Java/Others)
Total Submission(s): 15349 Accepted Submission(s): 5821


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


数据有点水, 不符合要求的都是成对出现,所以有人用位运算^给A了

我是用map做, 之前还想排序下搞,发现内存限制,,,, 1024*1024 / 4=262144 无法满足要存的10^6

所以,只能map了.

#include 
#include 
#include 
using namespace std;

int main()
{
	int n,i,a;
	mapmy;
	map::iterator it;
	while(scanf("%d",&n),n)
	{
	for(i=0;isecond==1)
	{
	printf("%d\n",it->first);
	break;
	}
	}
	my.clear();
	}
	return 0;
}

文档

hdu2095findyourpresent(2)找到只出现一次的数字

hdu2095findyourpresent(2)找到只出现一次的数字:find your present (2) Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/1024 K (Java/Others) Total Submission(s): 15349Accepted Submission(s): 5821 Problem Description In the new year party, everybody will get a special present.Now
推荐度:
标签: 找到 的数字 your
  • 热门焦点

最新推荐

猜你喜欢

热门推荐

专题
Top