我想生成比特重presentation所有可能的组合（不重复）。我不能使用像升压或STL :: next_combination任何库 - 它必须是我自己的code（计算时间是非常重要的）

I'd like to generate all possible combination (without repetitions) in bit representation. I can't use any library like boost or stl::next_combination - it has to be my own code (computation time is very important).

下面是我的code（从那些计算器用户修改）：

Here's my code (modified from ones StackOverflow user):

    int combination  = (1 << k) - 1;
    int new_combination = 0;
    int change = 0;

    while (true)
    {
        // return next combination
        cout << combination << endl;

        // find first index to update
        int indexToUpdate = k;
        while (indexToUpdate > 0 && GetBitPositionByNr(combination, indexToUpdate)>= n - k + indexToUpdate)
            indexToUpdate--;

        if (indexToUpdate == 1) change = 1; // move all bites to the left by one position
        if (indexToUpdate <= 0) break; // done

         // update combination indices
        new_combination = 0;
        for (int combIndex = GetBitPositionByNr(combination, indexToUpdate) - 1; indexToUpdate <= k; indexToUpdate++, combIndex++)
        {
            if(change)
            {
                new_combination |= (1 << (combIndex + 1));
            }
            else
            {
                combination = combination & (~(1 << combIndex));
                combination |= (1 << (combIndex + 1));
            }
        }
        if(change) combination = new_combination;
        change = 0;
    }

其中， N - 所有元素， K - 组合元素的数量。 GetBitPositionByNr - 第k位的复位位置。 GetBitPositionByNr（13,2）= 3 13的原因是1101和第二位是在第三的位置。

where n - all elements, k - number of elements in combination. GetBitPositionByNr - return position of k-th bit. GetBitPositionByNr(13,2) = 3 cause 13 is 1101 and second bit is on third position.

这给了我正确的输出 N = 4，K = 2 是：

It gives me correct output for n=4, k=2 which is:

0011 (3 - decimal representation - printed value)
0101 (5)
1001 (9)
0110 (6)
1010 (10)
1100 (12)

此外，它给了我正确的输出 K = 1 和 K = 4 ，但是给了我错了outpu为 K = 3 是：

Also it gives me correct output for k=1 and k=4, but gives me wrong outpu for k=3 which is:

0111 (7)
1011 (11)
1011 (9) - wrong, should be 13
1110 (14)

我想这个问题是在内部，而条件（第二），但我不知道如何解决这个问题。

I guess the problem is in inner while condition (second) but I don't know how to fix this.

也许你们当中有些人知道越好（快）算法做要我要达到什么目的？它不能使用更多的内存（阵列）。

Maybe some of you know better (faster) algorithm to do want I want to achieve? It can't use additional memory (arrays).

下面是code对ideone运行： IDEONE

Here is code to run on ideone: IDEONE

推荐答案

如果有疑问，用蛮力。唉，生成所有的的变化与重复，的再过滤掉不必要的模式：

When in doubt, use brute force. Alas, generate all variations with repetition, then filter out the unnecessary patterns:

unsigned bit_count(unsigned n)
{
    unsigned i = 0;

    while (n) {
        i += n & 1;
        n >>= 1;
    }

    return i;
}

int main()
{
    std::vector<unsigned> combs;
    const unsigned N = 4;
    const unsigned K = 3;

    for (int i = 0; i < (1 << N); i++) {
        if (bit_count(i) == K) {
            combs.push_back(i);
        }
    }

    // and print 'combs' here
}

编辑：其他人已经指出，没有过滤和蛮力解决办法，但我还是想给大家介绍一下这个算法有一些提示：

Someone else already pointed out a solution without filtering and brute force, but I'm still going to give you a few hints about this algorithm:

大多数编译器提供某种形式的内在的人口数的功能。我知道，海湾合作委员会，并锵具有 __ builtin_popcount（）。使用这种内在的功能，我能加倍的code的速度。

most compilers offer some sort of intrinsic population count function. I know of GCC and Clang which have __builtin_popcount(). Using this intrinsic function, I was able to double the speed of the code.

既然你似乎是工作在GPU上，你可以并行化code。我已经做到了用C ++ 11的标准线程设施，并且我已经成功地计算所有32位重复进行任意选择的popcounts 1，16和7.1秒19日我的8核英特尔机。

Since you seem to be working on GPUs, you can parallelize the code. I have done it using C++11's standard threading facilities, and I've managed to compute all 32-bit repetitions for arbitrarily-chosen popcounts 1, 16 and 19 in 7.1 seconds on my 8-core Intel machine.

下面是最终的code我已经写了：

Here's the final code I've written:

#include <vector>
#include <cstdio>
#include <thread>
#include <utility>
#include <future>


unsigned popcount_range(unsigned popcount, unsigned long min, unsigned long max)
{
    unsigned n = 0;

    for (unsigned long i = min; i < max; i++) {
        n += __builtin_popcount(i) == popcount;
    }

    return n;
}

int main()
{
    const unsigned N = 32;
    const unsigned K = 16;

    const unsigned N_cores = 8;
    const unsigned long Max = 1ul << N;
    const unsigned long N_per_core = Max / N_cores;

    std::vector<std::future<unsigned>> v;

    for (unsigned core = 0; core < N_cores; core++) {
        unsigned long core_min = N_per_core * core;
        unsigned long core_max = core_min + N_per_core;

        auto fut = std::async(
            std::launch::async,
            popcount_range,
            K,
            core_min,
            core_max
        );

        v.push_back(std::move(fut));
    }

    unsigned final_count = 0;
    for (auto &fut : v) {
        final_count += fut.get();
    }

    printf("%u\n", final_count);

    return 0;
}