Euler problem 3.4

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primes = 2 : filter (null . tail . primeFactors) [3, 5 ..] primeFactors n = factor n primeswherefactor n (p : ps) | p * p > n = [n] | n `mod` p == 0 = p : factor (n `div` p) (p : ps) | otherwise = factor n ps f = last (primeFactors 600851475143)

— Haskell offical

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— Me@2022.09.28 11:44:20 AM

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2022.09.28 Wednesday (c) All rights reserved by ACHK

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