不定期更新一些Haskell刷题记录

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Last digit of a large number

题目要求输入两个整数a, b, 输出 a^b 的最后一位数

题解:

lastDigit :: Integer -> Integer -> Integer lastDigit _ 0 = 1 lastDigit a b = case mod a 10 of 0 -> 0 1 -> 1 2 -> [6,2,4,8] !! fromInteger (mod b 4) 3 -> [1,3,9,7] !! fromInteger (mod b 4) 4 -> if odd b then 4 else 6 5 -> 5 6 -> 6 7 -> [1,7,9,3] !! fromInteger (mod b 4) 8 -> [6,8,4,2] !! fromInteger (mod b 4) 9 -> if odd b then 9 else 1

Integer和Int要转换

这题的意思应该都能看懂吧不想赘述…最直接的解法了

当然每次刷hasekell的题目就是为了看别人的一行流解法:

lastDigit :: Integer -> Integer -> Integer lastDigit a b = ((a `rem` 10) ^ ((b - 1) `rem` 4 + 1)) `rem` 10

不会真的有人不知道rem是商吧， 不会吧不会吧

拿这题作为记录的第一题好像不太行

Primes in numbers

原题:

Given a positive number n > 1 find the prime factor decomposition of n. The result will be a string with the following form : "(p1**n1)(p2**n2)...(pk**nk)" where a ** b means a to the power of b with the p(i) in increasing order and n(i) empty if n(i) is 1. Example: n = 86240 should return "(2**5)(5)(7**2)(11)"

题意就是因数分解然后格式化输出

题解:

import Data.List prime_factors :: Integer -> String prime_factors = concatMap (\x -> if length x == 1 then "(" ++ show (head x) ++ ")" else "(" ++ show (head x) ++ "**" ++ show (length x) ++ ")" ) . group . prime_factors' 2 . fromInteger prime_factors' :: Int -> Int -> [Int] prime_factors' y x | y ^ 2 > x = [x] | x `mod` y == 0 = y : prime_factors' y (x `div` y) | otherwise = prime_factors' (y + 1) x Integer类型转化为Int类型prime_factors’函数得到整数的因数列表group给列表中相同内容分组，刷题常用函数concatMap按照要求转化为格式化字符串. 挺好用的，柯里化