在haskell中以给定方式添加列表元素

Question

有些人帮我下面的功能，但是我需要它的反转意味着下面的输出将是我的函数的输入，下面的输入将是我函数的输出

ReturnAnInt

Answer 1

我认为您正在寻找此功能的反转

conv = concat . f
    where f [] = []
          f (x:xs) = map (x+) xs : f xs

> conv [3,10,5]
[13,8,15]

> conv [2,5,15,34]
[7,17,36,20,39,49]

我发现了这样的事情。请注意，对于长度小于3的情况，无法做出独特的解决方案，处理这种情况......

solve x = (reverse . (h :) . map (subtract h) . map head . split') rx
        where rx@(a:b:c:_) = reverse x
              h = (a+b-c) `div` 2

> solve [7,17,36,20,39,49]
[2,5,15,34]

> solve [13,8,15]
[3,10,5]

你可以查看是否

conv . solve = solve . conv = id

分裂＆＃39;在我的屏幕上向上滚动，它被定义为

split' = go 1
     where go _ [] = []
           go n x = take n x : go (n+1) (drop n x)