<blockquote>
<h1 id="heading">题目</h1>
<h2 id="description">Description</h2>
<blockquote>
Given an array with 
n 
n 
integers, assume f(S)f(S) as the result of executing xor operation among all the elements of set SS. e.g. if S={1,2,3}S={1,2,3} then f(S)=0f(S)=0.
your task is: calculate xor of all f(s)f(s), here s⊆Ss⊆S.
</blockquote>
 
<h2 id="input">Input</h2>
<blockquote>
This problem has multi test cases. First line contains a single integer T(T≤20)T(T≤20) which represents the number of test cases. 
For each test case, the first line contains a single integer number n(1≤n≤1,000)n(1≤n≤1,000) that represents the size of the given set. then the following line consists of nn different integer numbers indicate elements(≤109≤109) of the given set.
</blockquote>
<h2 id="output">Output</h2>
<blockquote>
For each test case, print a single integer as the answer.
</blockquote>
<h2 id="sample-input">Sample Input</h2>
<blockquote>
1 3 1 2 3
</blockquote>
<h2 id="sample-output">Sample Output</h2>
<blockquote>
0
</blockquote>
<h2 id="hint">Hint</h2>
<blockquote>
In the sample，S={1,2,3}S={1,2,3}, subsets of SS are: ∅∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}
</blockquote>
</blockquote>
<h1 id="heading-1">题解</h1>
非常简单的一道题，但是理解题意得难度要大于做出来……
其题意就是要求出集合{S}的所有子集{s}元素异或的异或
同一个数自己对自己异或为0
0对任何书异或都是该数本身
<code>a^b^a=b</code>
同时，异或操作满足交换律的结合律
因此，只要一个元素出现偶数次，就可以不再考虑该数
可以发现，除非只有一个元素，否则每个元素在最后总的计算中必定出现偶数次，因此，当n=0时，答案为该数本身，否则，答案为0
<h1 id="heading-2">代码</h1>
<pre style="background-color:#fff">/*
By:OhYee
Github:OhYee
Email:oyohyee@oyohyee.com
Blog:http://www.cnblogs.com/ohyee/

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/
#include &lt;cstdio&gt;
#include &lt;algorithm&gt;
#include &lt;cstring&gt;
#include &lt;cmath&gt;
#include &lt;string&gt;
#include &lt;iostream&gt;
#include &lt;vector&gt;
#include &lt;list&gt;
#include &lt;queue&gt;
#include &lt;stack&gt;
using namespace std;

//DEBUG MODE
#define debug 0

//循环
#define REP(n) for(int o=0;o&lt;n;o++)

const int maxn = 10000;
int n, m;
int Map[maxn][maxn];

void Do() {
 int n;
 scanf(&#34;%d&#34;, &amp;n);
 int temp, ans;
 if (n==1) {
 REP(n) {
 if (o == 0) 
 scanf(&#34;%d&#34;, &amp;ans);
 
 else 
 scanf(&#34;%*d&#34;, &amp;temp);
 }
 }
 else {
 REP(n)
 scanf(&#34;%*d&#34;);
 ans = 0;
 }
 
 printf(&#34;%d\n&#34;, ans);
}


int main() {
 int T;
 scanf(&#34;%d&#34;, &amp;T);
 while (T--) {
 Do();
 }
 return 0;
}
</pre>

HDU 5650.so easy