使用列表理解的素因分解

本文介绍了使用列表理解的素因分解的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 我想在Haskell中仅使用列表解析方法和/或。（函数组合运算符）查找给定数字的所有素数因子。 例如， pfactors 120 必须产生 [2,2,2,3,5] 输出。 我试过了： pfactors n = [p | p 但是当我调用 pfactors 120 时，结果是 [2,3,5] ，并非所有的主要因素。解决方案以下是我的做法： pfactors :: Integer - > [整数] pfactors n = [p | p ，[d | d ，_< - [p | i 它本质上就是您的解决方案，但不同之处在于，它在最后包含了一个额外的列表理解，其中包含与 p 因素一样多的元素， code> n 。 声明我真的不会这样做。 / p> 编辑我觉得上面写的脏，所以作为参考，这更接近我写的内容： pfactors':: Int - > [Int] pfactors'= unfoldr firstFactor where firstFactor n = listToMaybe [（f，n`div` f） | f ，n`mod` f == 0] > Data.List（unfoldr）， Data.Maybe（listToMaybe） I want to find all prime factors of a given number using only list comprehension method and/or . (function composition operator) in Haskell. I specifically want to avoid a recursive solution.For example, pfactors 120 must produce [2,2,2,3,5] output.I tried:pfactors n = [p | p <- [2..n], n `mod` p == 0, [d | d <- [1..p], p `mod` d == 0] == [1,p]]But when I call pfactors 120, the result is [2,3,5], not all prime factors. 解决方案 Here's how I'd do it:pfactors :: Integer -> [Integer]pfactors n = [ p | p <- [2..n] -- Possible factors , [d | d <- [1..p], p `mod` d == 0] == [1,p] -- Are prime , _ <- [ p | i <- [1..n], n `mod` p^i == 0] ] -- Divisible powersIt's essentially the solution you have, but the difference is that it has an extra list comprehension at the end which contains as many elements as p factors into n.Disclaimer I really wouldn't do it like this in reality.EDIT I felt dirty writing the above, so for reference, this is something closer to what I would write:pfactors' :: Int -> [Int]pfactors' = unfoldr firstFactor where firstFactor n = listToMaybe [(f, n `div` f) | f <- [2..n] , n `mod` f == 0]Dependencies: Data.List (unfoldr), Data.Maybe (listToMaybe) 这篇关于使用列表理解的素因分解的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！