xor

#include <stdio.h>
#include <string.h>

#define NMAX 100000
#define KMAX 20
#define VMAX (1 << KMAX)
using namespace std;
int a[NMAX];
int count[VMAX];

int main()
{
	FILE* fin = fopen("xor.in", "r");
	FILE* fout = fopen("xor.out", "w");

	int n, m;

	fscanf(fin, "%d %d", &n, &m);

	int i, k, j;
	for (i = 0; i < n; i++)
	{
		fscanf(fin, "%d", &a[i]);
	}

	int comb = n * (n - 1) / 2;

	long long countXor0;
	int len;

	len = 2;

	int x = 0;
	for (k = KMAX - 1; k >= 0; k--)
	{
		memset(count, 0, sizeof(count));
		for (i = 0; i < n; i++)
		{
			count[a[i] >> k]++;
		}

		countXor0 = 0;
		x = x << 1;

		for (i = 0; i < len; i++)
		{
			//acum nr de posibilitati astfel incat sa avem (x0)
			j = i ^ x;
			if (i == j)
			{
				countXor0 += (long long)count[i] * (count[i] - 1) / 2;
			}
			if (i < j)
			{
				countXor0 += (long long)count[i] * count[j];
			}
		}

		if (countXor0 < m)
		{
			//alegem 1
			x++;
			m -= countXor0;
		}

		len = len * 2;
	}

	fprintf(fout, "%d", x);

	fclose(fin);
	fclose(fout);

	return 0;
}