您将如何在纸上手动计数？您将检查最后一位数字。如果为0，则将其设置为1。如果已经为1，则将其设置回0，然后继续下一个数字。因此，这是一个递归过程。

以下程序通过更改序列来生成所有可能的组合：

#include <iostream>

template <typename Iter>
bool next(Iter begin, Iter end)
{
    if (begin == end)      // changed all digits
    {                      // so we are back to zero
        return false;      // that was the last number
    }
    --end;
    if ((*end & 1) == 0)   // even number is treated as zero
    {
        ++*end;            // increase to one
        return true;       // still more numbers to come
    }
    else                   // odd number is treated as one
    {
        --*end;            // decrease to zero
        return next(begin, end);   // RECURSE!
    }
}

int main()
{
    char test[] = "0000";
    do
    {
        std::cout << test << std::endl;
    } while (next(test + 0, test + 4));
}

该程序适用于任何类型的任何序列。如果您同时需要所有可能的组合，只需将它们放入集合中，而不是打印出来。当然，您需要其他元素类型，因为您不能将C数组放入向量中。让我们使用字符串向量：

#include <string>
#include <vector>

int main()
{
    std::vector<std::string> combinations;
    std::string test = "0000";
    do
    {
        combinations.push_back(test);
    } while (next(test.begin(), test.end()));
    // now the vector contains all pssible combinations
}

如果您不喜欢递归，这里有一个等效的迭代解决方案：

template <typename Iter>
bool next(Iter begin, Iter end)
{
    while (begin != end)       // we're not done yet
    {
        --end;
        if ((*end & 1) == 0)   // even number is treated as zero
        {
            ++*end;            // increase to one
            return true;       // still more numbers to come
        }
        else                   // odd number is treated as one
        {
            --*end;            // decrease to zero and loop
        }
    }
    return false;              // that was the last number
}