如何使Haskell使用另一个列表，取决于Haskell中的以前的答案(How to make Haskell use another list depending on the previous a

如何使Haskell使用另一个列表，取决于Haskell中的以前的答案(How to make Haskell use another list depending on the previous answer in Haskell)

我正在做项目欧拉问题63，在那里我必须找到存在的数字的数量：

x^(n) == y

其中n是y的长度。

很快就会发现，这种情况的结果在奇数和偶数之间交替出现，所以我在Haskell中提出了这个问题：

prob63 = [ n | n <- nums n , i <-[1..10], i^(length $ show n) == n] nums n | odd (n) == True = filter odd [n..] | otherwise = filter even [n..]

如果n < - [1 ..]，prob63会产生如下所示的流：

[1,2,3,4,5,6,7,8,9,16,25,36,49,64,81,125,216,343,512,729,1296,2401,4096,6561,16807,32768,59049,117649,262144,531441]

但是这很慢，我之后想到的不起作用。 我需要的是，根据以前的答案，它将开始测试n的奇数或偶数。 我如何从现有的东西中解脱出来？

I am doing project euler question 63 where I must find the amount of numbers that exist where:

x^(n) == y

Where n is the length of y.

It soon emerges that the results for this condition alternate between odd and even and so I came up with this in Haskell:

prob63 = [ n | n <- nums n , i <-[1..10], i^(length $ show n) == n] nums n | odd (n) == True = filter odd [n..] | otherwise = filter even [n..]

If n <- [1..], prob63 produces a stream that looks like this:

[1,2,3,4,5,6,7,8,9,16,25,36,49,64,81,125,216,343,512,729,1296,2401,4096,6561,16807,32768,59049,117649,262144,531441]

But this is slow and what I came up with after doesn't work. What I need is, depending on the previous answer, it will start testing the odd or even integers from n. How would I go about this from what I already have?

最满意答案

再看看输出。 元素不会在奇数和偶数之间交替出现！

1,2,3,4,5,6,7,8,9,16,25,36,49,64,81,125,216,343,512,729,1296,2401,4096,6561,16807,32768,59049,117649,262144,531441]

---

在实现您要求的代码后，我注意到了这一点。 它的输出与您的代码的输出不匹配。 无论如何，我会在这里发布代码：

prob63 :: [Integer] prob63 = test 1 where test s = next : test (next + 1) where next = head [n | n <- [s,s+2..], i <-[1..10], i^(length $ show n) == n]

Look again at the output. The elements do not alternate between odd and even!

1,2,3,4,5,6,7,8,9,16,25,36,49,64,81,125,216,343,512,729,1296,2401,4096,6561,16807,32768,59049,117649,262144,531441]

---

I noticed this after implementing the code that you ask for. Its output does not match the output of your code. I'll post the code here anyway:

prob63 :: [Integer] prob63 = test 1 where test s = next : test (next + 1) where next = head [n | n <- [s,s+2..], i <-[1..10], i^(length $ show n) == n]