Given an array with
n![技术分享]()
integers, assume
f(S)![技术分享]()
as the result of executing xor operation among all the elements of set
S![技术分享]()
.
e.g. if
S={1,2,3}![技术分享]()
then
f(S)=0![技术分享]()
.
your task is: calculate xor of all
f(s)![技术分享]()
,
here
s?S![技术分享]()
.
This problem has multi test cases. First line contains a single integer
T(T≤20)![技术分享]()
which represents the number of test cases.
For each test case, the first line contains a single integer number
n(1≤n≤1,000)![技术分享]()
that represents the size of the given set. then the following line consists of
n![技术分享]()
different integer numbers indicate elements(
≤10
9![技术分享]()
![技术分享]()
)
of the given set.
For each test case, print a single integer as the answer.
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}
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做这道题需要知道的是异或运算满足交换律(a^b)^(c^d)=a^b^c^d ,并且a^a=0; a^0=a;只有N为1时,集合中的元素只出现奇数次,其他情况集合中的元素都是出现偶数次,所以此时结果都为0;
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
int t,i,j,k,l,m,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
scanf("%d",&m);
for(i=1;i<n;i++)
scanf("%d",&k);
if(n==1)
printf("%d\n",m);
else
printf("0\n");
}
return 0;
}