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
5
1 1 3 2 2
3
1 2 1
0
Sample Output
3
2
题意：输入一串数，找出其中与其他数不同的数并输出。
输入：先输入一个数字n，表示需要输入的个数，当n为0的时候结束
输出：输出其中不同于其他数的数：
算法：采用暴力查找，当找到，break，未找到，就继续查找，其中主要代码如下：
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(j==i)
continue;
if(arr[i]==arr[j])
break;
}
if(j==n)
break;
}
第一个代表查找的次数，共有n个，第二个循环代表与所有的元素相比较，若没有找到相等的，j=n，代表当前a[i]是特殊的数，直接break，输出当前a[i].
全部代码如下：
#include<stdio.h>
int arr[210];
int main()
{
int n;
int i,j;
while(scanf("%d",&n),n)
{
for(i=0;i<n;i++)
scanf("%d",&arr[i]);
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(j==i)
continue;
if(arr[i]==arr[j])
break;
}
if(j==n)
break;
}
printf("%d\n",arr[i]);
}
return 0;
}