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}
