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Count the coins: Difference between revisions
→{{header|Haskell}}: more efficient method
(/* {{header|Haskell}}: simple method, not for big numbers) |
(→{{header|Haskell}}: more efficient method) |
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main = print (count 100 [1, 5, 10, 25])</lang>
Much faster, probably harder to read, is to update results from bottom up:
<lang haskell>count = foldr addCoin (1:repeat 0)
where addCoin c oldlist = newlist
where newlist = (take c oldlist) ++ zipWith (+) newlist (drop c oldlist)
main = do
print (count [25,10,5,1] !! 100)
print (count [100,50,25,10,5,1] !! 100000)</lang>
=={{header|Icon}} and {{header|Unicon}}==
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