杭电 2095

find your present (2)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/1024 K (Java/Others)
Total Submission(s): 15295    Accepted Submission(s): 5797

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）

 

 

这道题乍一看是用数组，可是要是用数组的话，由于题上的数非常大 编译器会崩溃，想我一開始就是用的数组就是编辑器崩溃了，后来正好看到了位运算，想到也许能够用位运算优化，要是用异或的话，那么同样的数运算运算就是0，那么那个剩下的唯一的奇数，（实际上在这里就不用在推断是不是奇数了），就是要求的，于是又一下代码：

<span style="font-size:18px;">#include<stdio.h>
int main()
{
	int n,x,m;
	while(~scanf("%d",&n),n)
	{
		x=0;
		while(n--)
		{
			scanf("%d",&m);
			x^=m;
		}
		printf("%d
",x);
	}
	return 0;
}</span>

<span style="font-size:18px;"></span> 

<span style="font-size:18px;">以下是之前我的错误的代码：</span>

<span style="font-size:18px;"></span><p>#include<stdio.h>
#include<string.h>
int a[1000001];
int s[1000001];
int b[1000001];
int main()
{
 int n,i;</p><p> while(~scanf("%d",&n),n)
 {
  memset(s,0,sizeof(s));
  for(i=1;i<=n;i++)
  scanf("%d",&a[i]);
  for(i=1;i<=n;i++)
  {
   s[a[i]]++;
   b[i]=i;  
  }
  for(i=1;i<=n;i++)
  {
   if(s[a[i]]&1)
   {
    printf("%d
",b[i]);
   }
  }
 }
 return 0;
}
</p>

<span style="font-size:18px;">//可以执行，可是由于数组设的过大，导致编译器崩溃没法通过。</span>