uva 729 - The Hamming Distance Problem

最新推荐文章于 2019-09-25 10:52:57 发布

原创最新推荐文章于 2019-09-25 10:52:57 发布 · 511 阅读

0 ·

CC 4.0 BY-SA版权

uva 同时被 2 个专栏收录

26 篇文章

订阅专栏

ACM

19 篇文章

订阅专栏

本文介绍了汉明距离的概念及其计算方法，并提供了一种通过XOR运算来找出两个10位字符串之间的汉明距离的示例。同时，文章还讨论了如何生成所有与全0字符串具有特定汉明距离的字符串，并给出了相应的C语言实现代码。

The Hamming Distance Problem

The Hamming distance between two strings of bits (binary integers) is the number of corresponding bit positions that differ. This can be found by using XOR on corresponding bits or equivalently, by adding corresponding bits (base 2) without a carry. For example, in the two bit strings that follow:

                               A      0 1 0 0 1 0 1 0 0 0
                               B      1 1 0 1 0 1 0 1 0 0
                            A XOR B = 1 0 0 1 1 1 1 1 0 0

The Hamming distance (H) between these 10-bit strings is 6, the number of 1's in the XOR string.

Input

Input consists of several datasets. The first line of the input contains the number of datasets, and it's followed by a blank line. Each dataset contains N, the length of the bit strings and H, the Hamming distance, on the same line. There is a blank line between test cases.

Output

For each dataset print a list of all possible bit strings of length N that are Hamming distance Hfrom the bit string containing all 0's (origin). That is, all bit strings of length N with exactly H 1's printed in ascending lexicographical order.

The number of such bit strings is equal to the combinatorial symbol C(N,H). This is the number of possible combinations of N-H zeros and H ones. It is equal to

$\begin{displaymath}{N!} \over {(N-H)! H!}\end{displaymath}$

This number can be very large. The program should work for $1 \le H \le N\le 16$ .

Print a blank line between datasets.

Sample Input

1

4 2

Sample Output

方法 ：确定遍历上限和下限。

代码 ：

#include 
#define N 17
int num_one(int n)
{
    int num;
    num = 0;
    while(n){
	num++;
	n = (n-1) & n;
    }
    return num;
}
char* atob(char str[], int inter, int n)
{
    int i, t;
    for(i = 0; i < n; i++)
	str[i] = '0';
    str[n] = '\0';
    t = 1;
    for(i = 0; i < n; i++){
	if(inter&t){
	    str[n-i-1] = '1';
	}
	t <<= 1;
    }
    return str;
}
int main()
{
    char str[N];
    int n, h, i, inter, t, low, up;
    scanf("%d", &t);
    while(t-- > 0){
        scanf("%d %d", &n, &h);
        low = 1;
        up = 1;
        for(i = 0; i < h; i++)
            low <<= 1;
        low--;
        for(i = 0; i < n; i++)
            up <<= 1;
        up--;
        for(inter = low; inter <= up; inter++){
            if(num_one(inter) == h)
                printf("%s\n", atob(str, inter, n));
        }
        if(t)
            printf("\n");
    }
    return 0;
}