HDU 5650 so easy（数学找规律）

so easy

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 195 Accepted Submission(s): 157



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 ndifferent
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)

大体题意：
给你一个集合包含n个元素，f（s）表示s集合内所有元素异或的结果，求所有集合异或的结果。
思路：
看样例分析可看出来，因为没有重复元素，每一个数出现的次数都是2^(n-1)当n不是1的时候肯定是偶数，那么异或结果肯定是0，否则n是1的时候，相当与和0异或，结果是自身

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main(){
int T;
scanf("%d",&T);
while(T--){
int n,k;
scanf("%d",&n);
for (int i = 0; i < n; ++i)scanf("%d",&k);
if (n > 1)printf("0\n");
else printf("%d\n",k);
}
return 0;
}