我如何获得XPY的所有排列？ [英] How do I get all permutations of xPy?

问题描述

我想计算尺寸的Y一组尺寸X的就是说，如果我有（1,2,3）的所有排列和大小要2的所有排列，3P2，这将是（1， 2）（1,3）（2,1）（2,3）（3,1）（3,2）

I'd like to calculate all the permutations of size Y of a set of size X. That is if I had (1,2,3) and want all permutations of size 2, 3P2, it would be (1,2) (1,3) (2,1) (2,3) (3,1) (3,2).

无论是GSL和C ++ STL只提供XPX，我可以看到。可能有人点我在C / C ++库，能做到这一点，或勾勒出一个快速，高效的存储算法？

Both the GSL and C++ STL only provide xPx that I can see. Could someone point me at a C/C++ library which can do this or spell out a fast and memory efficient algorithm?

我试图解决一个很短的密文。我已经想通了两封信，并已决定做一个强力攻击。我有ouglg ouyakl和我检查对一个非常好的字典每种排列。我已经消除了2个英文字母所以其24P7或1744364160可能性，这是没有那么糟糕。我现在已经在运行一个Perl程序，所以这将是编程时间+运行时间，总有效率的一个有趣的测试。 ：）

I'm trying to solve a very short cryptogram. I've figured out two letters and have decided to do an brute force attack. I have "ouglg ouyakl" and am checking every permutation against a very good dictionary. I've eliminated 2 letters so its 24P7 or 1,744,364,160 possibilities which isn't so bad. I have a Perl program running now, so this will be an interesting test of the total efficiency of programming time + run time. :)

（不，我不只是要答案的密文。）

(No, I do not just want the answer to the cryptogram.)

推荐答案

我用的在此之前库（注意：这是C ++）在code，需要做同样的事情。它有排列组合，有和没有重复。对于您的问题，这应该足够了（未经测试...）：

I've used this library before (note it is C++) in code that needed to do something similar. It has permutations and combinations, with and without repetition. For your problem, this should suffice (untested...):

std::vector<int> v;
v.push_back(1);
v.push_back(2);
v.push_back(3);

std::vector<int>::iterator first = v.begin(), middle = v.begin() + 2, last = v.end();

do {
    // do stuff with elements in range first...middle (but dont change them)
} while(next_partial_permutation(first, middle, last));

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