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

題意：輸入一串數，找出其中與其他數不同的數並輸出。

輸入：先輸入一個數字n，表示需要輸入的個數，當n為0的時候結束

輸出：輸出其中不同於其他數的數：

演算法：採用暴力查詢，當找到，break，未找到，就繼續查詢，其中主要程式碼如下：

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;
}
 

第一個代表查詢的次數，共有n個，第二個迴圈代表與所有的元素相比較，若沒有找到相等的，j=n，代表當前a[i]是特殊的數，直接break，輸出當前a[i].

全部程式碼如下：

#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;

}