HDU 5650 异或

Posted 2020-06-29 半根毛线code

so easy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 235    Accepted Submission(s): 180

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： 给你一个集合S    f（s）代表其子集中的元素的异或值

        输出所有子集的异或值

题解：设集合有n个数，则包含x的子集个数有2^(n-1)个。 那么当n > 1时，x出现了偶数次，所以其对答案的贡献就是0；当 n = 1时，其对答案的贡献是 x。

    （两个相同的数的异或值为0）

5（10）=101（2）

5^5=0

  101

  101

=000

 

 

 1 #include<iostream>
 2 #include<cstring> 
 3 #include<cstdio>
 4 #define ll __int64
 5 using namespace std;
 6 int t;
 7 int n;
 8 ll exm;
 9 int main()
10 {
11   int t;
12   scanf("%d",&t);
13   for(int i=1;i<=t;i++)
14   {
15       scanf("%d",&n);
16       for(int j=1;j<=n;j++)
17         scanf("%I64d",&exm);
18       if(n==1)
19        printf("%I64d\n",exm);
20        else
21       cout<<"0"<<endl;
22   } 
23     return 0;
24 }

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