一组的排列 [英] Permutations of a set
问题描述
我有一组X项，但想要长度为Y（X> Y）的排列。我知道< algorithm>中的排列函数，但是我不认为
相信这会做我想要的。有没有办法，或者通过
STL或其他一些库来做这个，还是我需要自己编写
代码？
I have a set of X items, but want permutations of length Y (X > Y). I
am aware of the permutation functions in <algorithm>, but I don''t
believe this will do what I want. Is there a way, either through the
STL or some other library to do this, or do I need to write my own
code?
推荐答案
John Trunek写道：
John Trunek wrote:
我有一组X项，但想要长度为Y（X> Y）的排列。我知道< algorithm>中的排列函数，但我不相信这会做我想要的。有没有办法，无论是通过STL还是其他库来做这件事，还是我需要自己编写代码？
I have a set of X items, but want permutations of length Y (X > Y). I
am aware of the permutation functions in <algorithm>, but I don''t
believe this will do what I want. Is there a way, either through the
STL or some other library to do this, or do I need to write my own
code?
你的术语是非标准的（我不是在谈论C ++
标准）。集合S的置换是从S到S的双射映射。
我的猜测是你想要一组Y-
子集的排列图像（其中S是你的X项集合，其中一个''Y子集''
的S是S的一个子集，它有Y个元素。）
你需要为此编写自己的函数。我认为最简单的方法就是使函数递归，至少在开始时是这样的。
在这里发布你的尝试，我相信你会得到帮助。
-
问候，
巴斯特。
Your terminology is non-standard (and I''m not talking about the C++
standard). A permutation of a set S is a bijective map from S to S.
My guess is that you want the set of images of permutations of Y-
subsets of S (where S is your set of X items, and where a ''Y-subset''
of S is a subset of S which has Y elements).
You need to write your own function for this. It will be easiest
to make the function recursive, at least to begin with, I think.
Post your attempt here and I''m sure you will get help.
--
Regards,
Buster.
John Trunek写道：
John Trunek wrote:
我有一组X项，但想要长度为Y（X> Y）的排列。我知道< algorithm>中的排列函数，但我不相信这会做我想要的。有没有办法，无论是通过STL还是其他库来做这件事，还是我需要自己编写代码？
I have a set of X items, but want permutations of length Y (X > Y). I
am aware of the permutation functions in <algorithm>, but I don''t
believe this will do what I want. Is there a way, either through the
STL or some other library to do this, or do I need to write my own
code?
你的术语是非标准的（我不是在谈论C ++
标准）。集合S的置换是从S到S的双射映射。
我的猜测是你想要一组Y-
子集的排列图像（其中S是你的X项集合，其中一个''Y子集''
的S是S的一个子集，它有Y个元素。）
你需要为此编写自己的函数。我认为最简单的方法就是使函数递归，至少在开始时是这样的。
在这里发布你的尝试，我相信你会得到帮助。
-
问候，
巴斯特。
Your terminology is non-standard (and I''m not talking about the C++
standard). A permutation of a set S is a bijective map from S to S.
My guess is that you want the set of images of permutations of Y-
subsets of S (where S is your set of X items, and where a ''Y-subset''
of S is a subset of S which has Y elements).
You need to write your own function for this. It will be easiest
to make the function recursive, at least to begin with, I think.
Post your attempt here and I''m sure you will get help.
--
Regards,
Buster.
巴斯特写道：
Buster wrote:
John Trunek写道：
John Trunek wrote:
我有一组X项，但想要长度为Y的排列（X> Y） 。我知道< algorithm>中的排列函数，但我不相信这会做我想要的。有没有办法，无论是通过STL还是其他库来做这件事，还是我需要编写自己的代码？
你的术语是非标准的（我是'我不是在谈论C ++
标准。集合S的排列是从S到S的双射映射。
我的猜测是你想要S的Y-
子集的排列图像集（其中S是你的X项集合） ，并且S的''Y子集''是S的子集，其中包含Y元素。）
I have a set of X items, but want permutations of length Y (X > Y). I
am aware of the permutation functions in <algorithm>, but I don''t
believe this will do what I want. Is there a way, either through the
STL or some other library to do this, or do I need to write my own
code?
Your terminology is non-standard (and I''m not talking about the C++
standard). A permutation of a set S is a bijective map from S to S.
My guess is that you want the set of images of permutations of Y-
subsets of S (where S is your set of X items, and where a ''Y-subset''
of S is a subset of S which has Y elements).
s /排列/排列的图像图像/
您需要为此编写自己的功能。我认为最简单的方法就是使函数递归，至少在开始之后。
在这里发布你的尝试，我相信你会得到帮助。
s/images of permutations/permutations of images/
You need to write your own function for this. It will be easiest
to make the function recursive, at least to begin with, I think.
Post your attempt here and I''m sure you will get help.
-
问候，
巴斯特。
--
Regards,
Buster.
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