HDU 5650 so easy（数学找规律）

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本文介绍了一个关于集合中所有子集进行异或运算的问题，并提供了一种简洁高效的解决思路及代码实现。

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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

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

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;
}