如何为 Haskell 中的函数提供特定范围的输入

Question

我是 Haskell 的新手，我想知道是否可以提供前 51 个素数的范围作为输入而不是键入素数，请参阅第 2 行？

如果是，请修改代码和post解决方案。

  1 perfect_nums n = (2^(n-1)) * ((2^n)-1)
  2 fperfect_nums = [perfect_nums x | x <- [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233]]
  3 main = print $ fperfect_nums

Answer 1

您可以编写生成素数的函数或使用库。为此，我喜欢使用 arithmoi：

import Math.NumberTheory.Primes (nextPrime, unPrime)

perfectNums :: Integer -> Integer
perfectNums n = (2^(n-1)) * ((2^n)-1)

fperfectNums :: [Integer]
fperfectNums = [perfectNums (unPrime x) | x <- take 51 [nextPrime 0 ..]]

main :: IO ()
main = print fperfectNums

Answer 2

你已经有了

fperfect_nums = [perfect_nums x | x <- [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233]]
              = [perfect_nums x | x <- primes51]

primes51 = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233]

也就是

 = take 51 $ [2 .. 288] `minus`
      [4, 6 .. 288] `minus`       -- all the multiples of 2 in the range
       [9, 12 .. 288] `minus`      -- all the multiples of 3 in the range
        [25, 30 .. 288] `minus`     -- and of 5, 7, 11, ...
         [49, 56 .. 288] `minus`     -- until we see that
          [121, 132 .. 288] `minus`   -- 13*13 = 169  whereas
           [169, 182 .. 288]           -- 17*17 = 289   which is above 233

 = take 51 $ [2 .. 288] `minus`
      (let {p=2} in [p*p, p*p+p .. 288]) `minus` 
       (let {p=3} in [p*p, p*p+p .. 288]) `minus` 
        (let {p=5} in [p*p, p*p+p .. 288]) `minus` 
         (let {p=7} in [p*p, p*p+p .. 288]) `minus` 
          (let {p=11} in [p*p, p*p+p .. 288]) `minus`  -- 13*13 = 169
           (let {p=13} in [p*p, p*p+p .. 288])          -- 17*17 = 289

 = take 51 $ [2 .. 288] `minus` (
      (let {p=2} in [p*p, p*p+p .. 288]) `union` 
       (let {p=3} in [p*p, p*p+p .. 288]) `union` 
        (let {p=5} in [p*p, p*p+p .. 288]) `union` 
         (let {p=7} in [p*p, p*p+p .. 288]) `union` 
          (let {p=11} in [p*p, p*p+p .. 288]) `union`
           (let {p=13} in [p*p, p*p+p .. 288]) )

 = take 51 $ [2 .. 288] `minus` ( foldr1 union
      [[p*p, p*p+p .. 288] | p <- [2,3,5,7,11,13]] )

minus xs ys = [ x | x <- xs, not (elem x ys)]
union xs ys = xs ++ ys

所以是的，确实可以提供前 51 个质数的范围作为输入，而不是自己键入所有 51 个质数。为此，您只需要自己输入前 6 个质数。

你实际上有 61 个，

primesTo288 = 2 : [3 .. 288] `minus` ( foldr1 union
      [[p*p, p*p+p .. 288] | p <- [2,3,5,7,11,13]] )
 = 2 : 3 : [5,7 .. 288] `minus` ( foldr1 union
      [[p*p, p*p+2*p .. 288] | p <- [3,5,7,11,13]] )

现在让它无界....

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