我怎样才能写出'n'字符串的所有组合?重复:C=n!/(n-k)! [关闭]
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【中文标题】我怎样才能写出\'n\'字符串的所有组合?重复:C=n!/(n-k)! [关闭]【英文标题】:How can I write all the combinations of 'n' strings? With Repetition: C=n!/(n-k)! [closed]我怎样才能写出'n'字符串的所有组合?重复:C=n!/(n-k)! [关闭] 【发布时间】:2013-06-16 12:48:52 【问题描述】:对不起,如果问题不清楚,我想知道如何用 C++ 编写一个可以输出所有句子组合的程序,使用公式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 亿个组合或有一些限制?
编辑。
我尝试了以下程序,但找不到更改上述公式中“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;
【问题讨论】:
std::next_permutation
提示:“请为我编写整个程序”的请求不被看好。
好的,我记下了。
【参考方案1】:
您可能的意思是您想要所有可能的 'n' 字符串组合。 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 循环中你会使用类似this
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,是否有可能拥有多达 10 亿个组合? 有什么限制吗?
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,但内存本身只是包含 10 个字符串的向量使用的内存。
【讨论】:
如何更改公式中 k 的值?【参考方案2】:您可以使用next_permutation
来执行此操作。
【讨论】:
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