Power of XOR.cpp

#include <bits/stdc++.h>
using namespace std;

/*  Given array a of n integers. All integers are present in
	even count except one. Find that one integer which has
	odd count in O(N) time complexity and O(1) space;
	N < 10^5
	a[i] < 10^5 */

int main(void){
	int n;
	cin>>n;
	
	int x;
	int ans=0;
	
	for(int i=0;i<n;++i){
		cin>>x;
		ans^=x;
	}
	
	cout<<ans<<endl;
}