Given N, B, and D: Find a set of N codewords (1 <= N <= 64), each of length B bits (1 <= B <= 8), such that each of the codewords is at least Hamming distance of D (1 <= D <= 7) away from each of the other codewords. The Hamming distance between a pair of codewords is the number of binary bits that differ in their binary notation. Consider the two codewords 0x554 and 0x234 and their differences (0x554 means the hexadecimal number with hex digits 5, 5, and 4):

        0x554 = 0101 0101 0100
0x234 = 0010 0011 0100
Bit differences: xxx  xx

Since five bits were different, the Hamming distance is 5.

PROGRAM NAME: hamming

INPUT FORMAT

N, B, D on a single line

SAMPLE INPUT (file hamming.in)

16 7 3

OUTPUT FORMAT

N codewords, sorted, in decimal, ten per line. In the case of multiple solutions, your program should output the solution which, if interpreted as a base 2^B integer, would have the least value.

SAMPLE OUTPUT (file hamming.out)

0 7 25 30 42 45 51 52 75 76
82 85 97 102 120 127

题目要求：
求出N个数，是这些数种任意两个的汉明码>= D。（汉明码：两个数的二进制相同位上不同数字的个数）
代码如下：

/*
LANG: C
TASK: hamming
*/
#include<stdio.h>
int len, n, b, d, path[100], largest;
int Dis(int x)
{
    int z, num, i;
    for (i = 0; i < len; i++)
    {
        //对两个数字异或，得出的数的二进制含1的个数就是原来两个数在二进制下相同位上不同数字的个数 
        z = x ^ path[i];
        num = 0;//计算个数 
        while (z)//求两个数的二进制对应位上不同数字的个数 
        {
            z &= z - 1;
            num++;
        }
        if (num < d)
        {
            return 0;
        }
    }
    return 1;
}
void Dfs(int th)
{    
    if (len == n)
    {
        return;
    }
    int i, cost;
    for (i = th + 1; i < largest; i++)
    {
        if (Dis(i))
        {
            path[len++] = i;
            Dfs(i);
            break;
        }
    }
}
int main()
{
    freopen("hamming.in", "r", stdin);
    freopen("hamming.out", "w", stdout);
    int i;
    scanf("%d%d%d", &n, &b, &d);
    largest = 1 << b;//范围 
    path[0] = 0;
    len = 1;
    Dfs(0);
    for (i = 0; i < len; i++)
    {
        if ((i + 1) % 10 == 0)
        {
            printf("%d\n", path[i]);
        }
        else if (i + 1 != len)
        {
            printf("%d ", path[i]);
        }
        else
        {
            printf("%d\n", path[i]);
        }
    }
    fclose(stdin); 
    fclose(stdout);
    return 0;
}