Odd Man Out

Given an odd list of couples of numbers (each number appears twice), find the number that occurs only once.
The solution is a mere xor on all the numbers as:


Also, xor is commutative:

so if we have a sample input A, B, C, A, B:

C++ implementation:

#include <iostream>
#include <fstream>
#include <iomanip>
using namespace std;

#define N 1024
typedef unsigned long long int64;

int n, g;
int64 card, cardXor;
ofstream ofile;
ifstream ifile;

int main() {
	int i, j;
	ifile.open("in.txt");
	ofile.open("out.txt");
	ifile >> n;
	for (i = 0; i < n; ++i) {
		// init
		cardXor = 0;
		// in
		ifile >> g;
		for (j = 0; j < g; cardXor ^= card, ++j)
			ifile >> card;
		// out
		ofile << "Case #" << i + 1 << ": " << cardXor << endl;
	}
	return 0;
}