hdu1563Find your present



     这道题就是找出来独特的数。举个例子，给出一组数，若一个数出现的次数只有一次，而其它数都出现两次，那么这个数就是你要找的数啦。先排个序，然后把它输出就行了。

刚开始看到题的时候我就没想起来能用排序来实现，看了别人的代码突然觉得这么整挺简单的。借鉴了别人的思路，编了个简单易懂的代码。

Find your present!

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2691    Accepted Submission(s): 1786

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.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<=200, and n is odd) at first. Following that, n positive integers will be given in a line. 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

51 1 3 2 231 2 10

 

Sample Output

32
ac代码：
#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;int main(){    int n,a[205],i;    while(scanf("%d",&n)&&n!=0)    {         memset(a,0,sizeof(a));         for(i=0;i<n;i++)           scanf("%d",&a[i]);         sort(a,a+n);         for(i=0;i<n;i++)  //不能漏掉这个 for语句，前边的 for语句对下边的 if语句不起作用             if(a[i]!=a[i-1]&&a[i]!=a[i+1])            printf("%d\n",a[i]);     }    //while(1);    return 0;}//简单易懂，