Question

我正在尝试这样的事情：

combination no 1: sentence1 sentence2 sentence3 sentence4

combination no 2: sentence1 sentence2 sentence4 sentence3

combination no 3: sentence1 sentence3 sentence2 sentence4

combination no 4: sentence1 sentence3 sentence4 sentence2

combination no 5: sentence1 sentence4 sentence3 sentence2

combination no 6: sentence1 sentence4 sentence2 sentence3

等等......

现在，使用以下代码，我如何处理公式中的“k”变量？有什么遗失？再次，它是关于重复的组合所以我认为公式是C = n！/（n-k）！

// next_permutation example
#include <iostream>     // std::cout
#include <algorithm>    // std::next_permutation, std::sort
#include <string>       // std::string
#include <vector>       // std::vector

int main () {
  std::string sentence1 = " A Sentence number one ";
  std::string sentence2 = " B Sentence number two ";
  std::string sentence3 = " C Sentence number three ";
  std::string sentence4 = " D Sentence number four ";

  // Store all the elements in a container ( here a std::vector)
  std::vector<std::string> myVectorOfStrings;      
  // In the vector we add all the sentences.
  // Note : It is possible to do myVectorOfStrings.push_back("Some sentence");
  myVectorOfStrings.push_back(sentence1);
  myVectorOfStrings.push_back(sentence2);
  myVectorOfStrings.push_back(sentence3);
  myVectorOfStrings.push_back(sentence4);

  // The elements must be sorted to output all the combinations
  std::sort (myVectorOfStrings.begin(),myVectorOfStrings.end());


  std::cout << "The 4! possible permutations with 4 elements:\n";
  do {
    //This printing can be improved to handle any number of sentences, not only four.
    std::cout << myVectorOfStrings[0] << ' ' << myVectorOfStrings[1] << ' ' << myVectorOfStrings[2] << ' ' << myVectorOfStrings[3] << '\n';
  } while ( std::next_permutation(myVectorOfStrings.begin(),myVectorOfStrings.end()) );

  std::cout << "After loop: "  << myVectorOfStrings[0] << ' ' << myVectorOfStrings[1] << ' ' << myVectorOfStrings[2] << ' ' << myVectorOfStrings[3] << '\n';

  return 0;
}

Answer 1

您所说的k会自动处理。假设，如果您在向量中插入两个相似的句子，则说有4个句子，其中两个是相同的。然后，排列总数为4!/2!=12。因此，该函数仅打印12个排列（不是24）。使用类似的句子here查看修改后代码的输出：

“n”个字符串的所有可能组合，重复：C = n！/（n-k）！

1 个答案: