プログラミング・パラダイム（2019 年度）テスト解答例（一部）

Ⅰ
(1)

foo :: [(Integer,Integer)] -> Integer
foo []         = 0
foo ((x,y):xs) = if x > y then 1 + foo xs else foo xs

(2)

bar :: Integer -> [(Integer,Integer)]
bar n = [(i,j) | i <- [1..n], j <- [1..i], (i + j) `mod` 3 /= 0]

Ⅱ
(1) 9
(2) [(1,6),(2,3),(2,6),(3,4),(3,6),(4,6),(5,6),(6,6)]