Difference between revisions of "Haskell Quiz/PP Pascal/Solution Kelan"
< Haskell Quiz | PP Pascal
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(+cat) |
m |
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<haskell> |
<haskell> |
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import System.Environment |
import System.Environment |
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+ | import Control.Monad (forM_) |
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nCr n r = |
nCr n r = |
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digits = 1 + ( truncate . logBase 10 . fromIntegral . trimax $ rows ) |
digits = 1 + ( truncate . logBase 10 . fromIntegral . trimax $ rows ) |
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pad s = ( replicate ( digits - length s ) ' ' ) ++ s |
pad s = ( replicate ( digits - length s ) ' ' ) ++ s |
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− | for_ = flip mapM_ |
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− | + | forM_ [ 0 .. ( rows - 1 ) ] $ \ n -> do |
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putStr . concat . replicate ( rows - n - 1 ) . pad $ "" |
putStr . concat . replicate ( rows - n - 1 ) . pad $ "" |
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− | + | forM_ [ 0 .. n ] $ \ r -> do |
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putStr . pad . show . nCr n $ r |
putStr . pad . show . nCr n $ r |
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putStr . pad $ "" |
putStr . pad $ "" |
Latest revision as of 21:15, 14 December 2009
Pretty basic solution. Find the biggest number the triangle will contain given the number of rows, use that number to determine column widths. Then just print out each row with the correct columns and spacing.
import System.Environment
import Control.Monad (forM_)
nCr n r =
( product [ ( s + 1 ) .. n ] ) `div` ( product [ 2 .. t ] )
where
s = max r ( n - r )
t = min r ( n - r )
-- maximum number in a pascal triangle
-- with given number of rows
trimax rows =
nCr ( rows - 1 ) ( ( rows - 1 ) `div` 2 )
main = do
args <- getArgs
let rows = read . head $ args
digits = 1 + ( truncate . logBase 10 . fromIntegral . trimax $ rows )
pad s = ( replicate ( digits - length s ) ' ' ) ++ s
forM_ [ 0 .. ( rows - 1 ) ] $ \ n -> do
putStr . concat . replicate ( rows - n - 1 ) . pad $ ""
forM_ [ 0 .. n ] $ \ r -> do
putStr . pad . show . nCr n $ r
putStr . pad $ ""
putStrLn ""