Euler problem 3.4

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primes = 2 : filter (null . tail . primeFactors) [3, 5 ..]

primeFactors n = factor n primes
  where
    factor 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