haskell foldr和变量

Question

我怎样才能创建一个与此类似的函数,但n作为一个变量,其数字1以xs的长度值开头并以xs的长度值结尾？

例如,我想[1,2,3]返回结果3*1 + 2*2 + 1*3.

function :: [Int] -> Int
function xs = foldr (\x acc -> (x*n) + acc) 0 xs

Answer 1

惯用的Haskell答案可以是:

function = sum . zipWith (*) [1 ..] . reverse

从右到左阅读:你反转列表(reverse),这样做你不需要向后计数(从n到1)而是向前计数(从1到n)...然后用索引用*(zipWith (*) [1..])来压缩它最后总结了(sum).

Answer 2

使用zipWith:

import Data.List

function :: [Int] -> Int
function xs = sum $ zipWith (*) xs lst
              where
                lst = unfoldr (\x -> Just (x-1,x-1)) $ (length xs) + 1

main = putStr $ show $ function [40,50,60]