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Haskell Quiz/PP Pascal/Solution Kelan

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< Haskell Quiz | PP Pascal(Difference between revisions)
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(+cat)
Current revision (21:15, 14 December 2009) (edit) (undo)
m
 
Line 4: Line 4:
<haskell>
<haskell>
import System.Environment
import System.Environment
 +
import Control.Monad (forM_)
nCr n r =
nCr n r =
Line 22: Line 23:
digits = 1 + ( truncate . logBase 10 . fromIntegral . trimax $ rows )
digits = 1 + ( truncate . logBase 10 . fromIntegral . trimax $ rows )
pad s = ( replicate ( digits - length s ) ' ' ) ++ s
pad s = ( replicate ( digits - length s ) ' ' ) ++ s
-
for_ = flip mapM_
 
-
for_ [ 0 .. ( rows - 1 ) ] $ \ n -> do
+
forM_ [ 0 .. ( rows - 1 ) ] $ \ n -> do
putStr . concat . replicate ( rows - n - 1 ) . pad $ ""
putStr . concat . replicate ( rows - n - 1 ) . pad $ ""
-
for_ [ 0 .. n ] $ \ r -> do
+
forM_ [ 0 .. n ] $ \ r -> do
putStr . pad . show . nCr n $ r
putStr . pad . show . nCr n $ r
putStr . pad $ ""
putStr . pad $ ""
putStrLn ""
putStrLn ""
</haskell>
</haskell>

Current revision

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 ""