HDU 5650 so easy

so easy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 617    Accepted Submission(s): 413

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)

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:   5674  5673  5672  5671  5670 

题意：给出N个数的集合，这个集合的所有子集合的元素异或运算后，再将这些子集合的运算结果进行异或运算，问最后结果。

做这道题需要知道的是异或运算满足交换律（a^b)^(c^d)=a^b^c^d  ,并且a^a=0;  a^0=a;只有N为1时，集合中的元素只出现奇数次，其他情况集合中的元素都是出现偶数次，所以此时结果都为0；

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
	int t,i,j,k,l,m,n;
    scanf("%d",&t);
    while(t--)
    {
    	scanf("%d",&n);
    	scanf("%d",&m);
    	for(i=1;i<n;i++)
    	scanf("%d",&k);
    	if(n==1)
    	printf("%d\n",m);
    	else
    	printf("0\n");
	}
	return 0;
}