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抱歉，如果问题不清楚，我想知道如何用 C++ 编写一个程序，使用公式输出一些句子的所有组合C=n!/(n-k)!。例如，这就是我想要打印的那种东西：

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...

另外，是否可以有多达 10 亿个组合，或者有一些限制？

编辑。

我尝试了以下程序，但我找不到改变上面公式中“k”变量的方法。

// 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;
}
4

2 回答 2

2

你可能的意思是你想要'n'字符串的所有可能组合。有n！可能的情况。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东西

do-while 循环将是：

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()) );

另外，是否可以有多达 10 亿个组合，或者有一些限制？

唯一的限制是内存。在此示例中，您只有 1 个存储所有字符串的向量。所以如果你有 10 个字符串，你将有 10 个！= 3,628,800 种不同的组合，但内存本身只是您的向量使用的内存，它有 10 个字符串。

于 2013-06-19T20:56:11.490 回答
1

您可以使用它next_permutation来执行此操作。

于 2013-06-19T20:19:04.763 回答