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
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
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
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;
}