在Haskell中制作除数列表 [英] Making a list of divisors in Haskell

问题描述

我在euler项目中做了第21题。
一部分需要找到一个数字的正确因数列表。即剩余 n 和少于 n 的数字。所以我做了这个哈斯克尔，但GHCI对我很生气。

  divisors n = [n | n < -  [1 ..（n-1）]，n`rem` [1 ..（n-1）] == 0] 
  

问题是我不知道如何制作：

  n `rem` [1..（n-1）] 
  

比 n 平均分配 n 。

解决方案

您只需要一个单独的变量。

  Prelude>让除数n = [x | x < -  [1 ...（n-1）]，n`rem` x == 0] 
 Prelude>除数20 
 [1,2,4,5,10] 
前奏>除数30 
 [1,2,3,5,6,10,15] 
  

现在，如果你想提高效率，我们已经知道除数不会超过 n 的一半，我们知道1是除数一切。而且，让我们继续，让Haskell-y多一点，以避免列表理解：

让除数n = 1：filter（（== 0）。rem n）[2 .. n`div` 2]
Prelude>除数20
[1,2,4,5,10]
前奏>除数30
[1,2,3,5,6,10,15]
前奏>除数31
[1]

I am doing problem 21 in eulerproject. One part requires finding the list of proper divisors of a number. i.e. where there is remainder of n and some number of less than n. So I made this Haskell, but GHCI gets angry at me.

divisors n =[ n | n <- [1..(n-1)], n `rem` [1..(n-1)] ==0 ]

The problem is that I don't know how to make:

n `rem` [1..(n-1)]

so that it only returns the number less than n that divide evenly into n.

解决方案

You just need a separate variable.

Prelude> let divisors n = [x | x <- [1..(n-1)], n `rem` x == 0]
Prelude> divisors 20
[1,2,4,5,10]
Prelude> divisors 30
[1,2,3,5,6,10,15]

Now, if you wanted to make it a bit more efficient, we already know that a divisor won't be more than half of n, and we know 1 is a divisor of everything. And, let's go ahead and make it a bit more Haskell-y to boot, avoiding a list comprehension:

Prelude> let divisors n = 1 : filter ((==0) . rem n) [2 .. n `div` 2]
Prelude> divisors 20
[1,2,4,5,10]
Prelude> divisors 30
[1,2,3,5,6,10,15]
Prelude> divisors 31
[1]

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