Name: Anonymous 2011-02-05 13:59
factorial 1->1; n->n*(factorial n-1)compare to Haskell
factorial::Integer->Integer
factorial 0=1
factorial n=n*factorial(n-1)
factorial 1->1; n->n*(factorial n-1)
factorial::Integer->Integer
factorial 0=1
factorial n=n*factorial(n-1)
factorial 1->1; n->n*(factorial n-1)
product :: [Integer] -> Integer
product [] = 1
product (x:xs) = x * product xs
factorial n = procuct [1..n]