The Hamming Distance Problem UVA729

最新推荐文章于 2019-01-31 19:03:39 发布

最新推荐文章于 2019-01-31 19:03:39 发布 · 219 阅读

本文介绍了一种通过位操作计算两个二进制字符串之间的汉明距离，并利用全排列生成所有与指定汉明距离相对应的二进制字符串的方法。详细解释了汉明距离的概念，提供了实现代码，适用于计算机科学与编程领域。

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 containsN, the length of the bit strings andH, 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 lengthNthat are Hamming distanceHfrom the bit string containing all 0's (origin). That is, all bit strings of lengthNwith exactlyH1's printed in ascending lexicographical order.

The number of such bit strings is equal to the combinatorial symbolC(N,H). This is the number of possible combinations ofN-Hzeros andHones. 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

给出两个数，一个是总位数，另一个是1的位数，求全排列

#include<iostream>
#include<algorithm>
#include<string>

using namespace std;

int main()
{
    int n,k;
    cin>>n;
    for(k=1;k<=n;k++)
    {
        int r,t,i,j;
        cin>>r>>t;
        string str;
        for(i=1;i<=r-t;i++)
            str=str+'0';
        for(i;i<=r;i++)
            str=str+'1';
        cout<<str<<endl;
        while(next_permutation(str.begin(), str.end()))
        {
            cout<<str<<endl;
        }
        if(k!=n) cout<<endl;
    }
    return 0;
}