如何编写'n'字符串的所有组合？重复：C=n！/(n-k)！

How can I write all the combinations of 'n' strings? With Repetition: C=n!/(n-k)!

本文关键字：n-k 组合 字符串 何编写 重复      更新时间：2023-10-16

对不起，如果问题不清楚，我想知道如何在c++中编写一个程序，可以输出一些句子的所有组合，使用公式C=n!/(n-k)!。例如，这就是我想要打印的东西:

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
And so on...

还有，是否可能有多达10亿个组合，或者有一些限制?

编辑。

我试了下面的程序，但是我找不到一个方法来改变上面公式中的"k"变量。

// 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;
}

您的意思可能是需要'n'字符串的所有可能组合。有n个!可能的情况下。您可以使用std::next_permutation方法如下:

我想你所有的句子都是这样的std::string:

// 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;
}

这是一个简单的打印示例。如果你有超过4个字符串，在do-while循环中你会使用这样的语句

do-while循环应该是:

do {
  //Print all the sentences in my vector :
  for( auto i = myVectorOfStrings.begin(); i != myVectorOfStrings.end(); ++i)
    std::cout << *i << ' ';
  // Go to the next line
  std::cout << std::endl;
} while ( std::next_permutation(myVectorOfStrings.begin(),myVectorOfStrings.end()) );

还有，是否可能有多达10亿个组合呢有限制吗?

唯一的限制是内存。在这个例子中，你只有一个存储所有字符串的向量。所以如果你有10根弦，你就有10根!= 3,628,800种不同的组合，但内存本身只是你的向量使用的内存，有10个字符串

您可以使用next_permutation来完成此操作

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