UVA729 UVALive5366 The Hamming Distance Problem【置换】

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 H 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
N! / (N − H)!H!
    This number can be very large. The program should work for 1 ≤ H ≤ N ≤ 16.
    Print a blank line between datasets.
Sample Input
1

4 2
Sample Output
0011
0101
0110
1001
1010
1100

Regionals 1991 >> North America - East Central NA

问题链接：UVA729 UVALive5366 The Hamming Distance Problem
问题简述：（略）
问题分析：
    两个等长字符串之间的海明距离(Hamming distance)是两个字符串对应位置的不同字符的个数。换句话说，它就是将一个字符串变换成另外一个字符串所需要替换的字符个数。
    该问题用置换函数来解决，需要注意字符串数组的初始化。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++程序如下：

/* UVA729 UVALive5366 The Hamming Distance Problem */

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int t, n, h;
    cin >> t;
    while(t--) {
        scanf("%d%d", &n, &h);

        string s;
        for(int i = 1; i <= n - h; i++)
            s += '0';
        for(int i = n - h + 1; i <= n; i++)
            s += '1';
        do {
            cout << s << endl;
        } while(next_permutation(s.begin(), s.end()));
        if(t) cout << endl;

    }

    return 0;
}