HDU 5650：so easy【数学】

so easy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 292    Accepted Submission(s): 218

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)

题意：给你一个由n个数组成的数列S，f(s)函数为数列s中的每个元素相异或的结果，让你求出f(S的子集)相异或的结果。如样例：就是让你求f(0)^f({1})^f({2})^f({3})^f({1, 2})^f({1, 3})^f({2, 3})^f({1, 2, 3});

思路：x^x=0;而由n个数组成的数列，其每个元素出现的次数为2的n-1次方。即当n大于1时均为偶数，值为0；当n=1时，答案就是那唯一的一个数喽~

AC-code：

#include<cstdio>
#include<cmath>
int main()
{
	int T,n,i;
	__int64 ans,a;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		for(i=0;i<n;i++) 
			scanf("%I64d",&ans);
		if(n>1)
		{
			printf("0\n");
			continue;
		}
		else
			printf("%I64d\n",ans);
	}
	return 0;
}