
Ah sorry, I didn't notice that you were doing that. The effectiveness of
the trick really only comes into play though if you use an analytic
solution for finding the sum of the multiples of 3, etc.
I haven't tested this code in a while, but here's what I wrote some time
ago:
sum2 :: Integer -> Integer -> Integer -> Integer
sum2 a b ceiling = aX + bX - abX
where
aX = sum1 a ceiling
bX = sum1 b ceiling
abX = sum1 (a * b) ceiling
sum1 :: Integer -> Integer -> Integer
sum1 x ceiling = sum1' (even times) times x
where
times = ceiling `div` x
sum1' :: Bool -> Integer -> Integer -> Integer
sum1' True times x = area
where
area = (times + 1) * (times * x) `div` 2
sum1' False times x = max + area'
where
max = times * x
area' = sum1' True (times - 1) x
Please excuse the poor Haskell style as it is quite possibly the first
Haskell program I ever wrote.
On Wed, Jan 28, 2015 at 11:15 AM, Jean Lopes
Thats actually what I did...
2015-01-27 22:11 GMT-02:00 Lyndon Maydwell
: I remember that when I had a look at Euler 1 I found that there's a fun
solution that should run in "constant" time.
You can find the sum of the multiples of 3, add the multiples of 5, and then subtract the multiples of 3*5.
Is that the kind of thing you're looking for?
- Lyndon
On Wed, Jan 28, 2015 at 10:57 AM, Jean Lopes
wrote: I'm not really good at math, maybe I am missing something obvious ? Maybe some pointers as of where to start studying math in order to avoid this brute force attempts, at least to help in this particular problem
2015-01-27 21:38 GMT-02:00 Brandon Allbery
: On Tue, Jan 27, 2015 at 6:29 PM, Jean Lopes
wrote: The problem to be solved: https://www.hackerrank.com/contests/projecteuler/challenges/euler001
It's worth remembering that the Euler problems are all about math understanding; often they are designed such that brute force solutions will time out or otherwise fail.
-- brandon s allbery kf8nh sine nomine associates allbery.b@gmail.com ballbery@sinenomine.net unix, openafs, kerberos, infrastructure, xmonad http://sinenomine.net
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