如下的题：
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}
其实求symmetric集我自己有几个办法：
1.并集-交集
2.(p-q)V(q-p) 也就是两个的差集的并集
可是这两个方法的实现都很麻烦，有没有更简单的实现呢？
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我们以前讨论过类似题的，楼主看这个帖子
http://bbs.bc-cn.net/viewthread.php?tid=160933&star=at#
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谢谢，我看看
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好像没有什么最终的解决方案？
而且和我的问题有差距啊，我是问：除了我说的两种方法，还有没有更简单的方法？
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归并会不会
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会啊。
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最后一个人的代码你看了吗？？？
不能解决你的问题么？？
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哪里啊？呵呵，我不太明白，你能不能说得明白些啊？谢谢
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不能啊，怎么解决，请教高手，大家都来讨论讨论啊，共同进步啊！
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没试了一下，好像是对的吧。
#include<iostream>
using namespace std;
#include<algorithm>
int k,a[20],b[20],c[20],d[20];
int BinSearch(int R[],int K)
{
int low=0,high=k-1,mid;
while(low<=high){
mid=(low+high)/2;
if(R[mid]==K) return mid;
if(R[mid]>K)
high=mid-1;
else
low=mid+1;
}
return -1;
}
int main()
{
int n,m,cas,i,p,l;
cin>>cas;
while(cas--)
{ k=l=p=0;
memset(d,0,sizeof(d));
while(cin>>a[l]&&a[l])l++;
while(cin>>b[k]&&b[k])k++;
sort(&b[0],&b[k]);
for(i=0;i<l;i++)
if(BinSearch(b,a[i])!=-1)d[BinSearch(b,a[i])]=1;
else c[p++]=a[i];
for(i=0;i<k;i++)
if(d[i]==0) c[p++]=b[i];
if(p==0)cout<<"{}\n";
else {
sort(&c[0],&c[p]);
cout<<"{"<<c[0];
for(i=1;i<p;i++)
cout<<","<<c[i];
cout<<"}\n";
}
}
return 0;
}
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