BestCoder Round #77 (div.2)

so easy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 288    Accepted Submission(s): 215

思路：多余两个数的集合，在他的所有子集中每个元素都会重复出现偶数次。异或满足结合律。故为0；
一个数时则输出该数；
Problem Description

Given an array with

n

integers, assume f(S) as the result of executing xor operation among all the elements of set S . e.g. if S={1,2,3} then f(S)=0 .

your task is: calculate xor of all f(s) , here s⊆S .

 

Input

This problem has multi test cases. First line contains a single integer 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) that represents the size of the given set. then the following line consists of n different integer numbers indicate elements( ≤109 ) of the given set.

 

Output

For each test case, print a single integer as the answer.

 

Sample Input

   
   
   
   

    
    
    
    1
3
1  2  3
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0

In the sample，$S = \{1, 2, 3\}$, subsets of $S$ are: $\varnothing$, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}
   
   
   
   

 

Source

BestCoder Round #77 (div.2)

 

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#include <iostream>
#include<cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <cstring>
using namespace std;

int main()
{
   int t;
   int n;
   int x;
   cin>>t;
   while(t--)
   {
       cin>>n;
       for(int i=0;i<n;i++)
      scanf("%d",&x);
      if(n==1)
      cout<<x<<endl;
      else
        cout<<0<<endl;

   }
    return 0;
}