uva729 The Hamming Distance Problem（全排列）

最新推荐文章于 2019-05-04 00:13:33 发布

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本文介绍如何计算两个等长比特串之间的汉明距离，并提供了一种生成所有与初始全零比特串具有特定汉明距离的比特串的方法。通过使用C++实现的程序示例，展示了这些比特串是如何按照字典升序排列生成的。

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 distanceH from 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

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

Print a blank line between datasets.

1

4 2

Sample Output

全排列的题。但是别忘了，要从最小的开始全排列。例如00001111，000111.#include<iostream>
#include<stdio.h>
#include<string>
#include<algorithm>
using namespace std;

int main()
{
   int t;
   scanf("%d",&t);
   while(t--)
   {
       int m,n;
       int a[20];
       scanf("%d%d",&m,&n);

       for(int i=0;i<m;i++)
       {
           if(i<m-n)
            a[i]=0;
           else
            a[i]=1;
       }
       do
       {
           for(int i=0;i<m;i++)
            printf("%d",a[i]);
            printf("\n");
       }while(next_permutation(a,a+m));
       if(t)
        printf("\n");
   }
    return 0;
}