The symmetric difference of two sets is the set of elements belonging to one but not both of the two sets. For example, if we have two sets A = {1,2,3,4,5} and B = {3,4,5,6,7,8}, then the symmetric difference of A and B is the set {1,2,6,7,8}.

Given two sets of positive integers, display their symmetric difference.



Input

A positive integer will denote the number of cases. Both sets will be input from a single line. A zero (0) marks the end of each set. There will be no more than 20 numbers in each set, and no number within a set will be repeated. Each number in a set is a positive integer less than 65535.



Output

The set of integers making up the symmetric difference of the two sets. The numbers within a set may be sorted from smaller to bigger.



Sample Input
3
1 2 3 4 5 0 3 4 5 6 7 8 0
1 2 3 0 1 2 3 0
129 34 5 6 7 0 129 0
Sample Output
{1,2,6,7,8}
{}
{5,6,7,34}

开个数组筛选-,-

能具体说下思路么？本人是初学者谢谢

好多做法。
把两个集合一起读到一个数组里，排序，然后遍历，如果一个元素不等于它前面的也不等于它后面的就输出。

那怎么样输出三个呢？

有没有高人附一个完整代码？小弟在此等候

#include<cstdio>
#include<algorithm>
int hash[65536];
int main()
{
    int n,max_;
    scanf("%d",&n);
    for(int i=0;max_=-1&&i<n;++i)
    {  
        int v;
        memset(hash,0,sizeof(hash));
        for(int j=0;j<2;++j)
        while(scanf("%d",&v),v!=0)
        {
          if(!hash[v]) hash[v]=1;
          else hash[v]=0;
          max_=max_<v?v:max_;
        }
        putchar('{');
        for(int i=1;i<=max_;++i) if(hash[i]) printf("%d ",i);
        putchar('}');putchar('\n');
    }
    return 0;
}

谢谢你 虽然看不太懂

这已经不叫hash了，为了几十个数，申请几十万字节的空间，相当浪费。。。

那你觉得应该怎么样做呢？