-- 阶乘函数
fact m = product [1..m] 
 
-- 对于mXn的矩阵，获取所有可能的路径数量。
path_num m n = fact (m-1 + n-1) `div` ((fact (m-1)) * (fact (n-1)))
 
-- 创建由 (1,1) 到 (m, n) 的一个验证列表
create_mn m n = (m, n) : (m, n+1) : create_mn (m+1) (n+1)
 
-- 获取前20个验证结果 
top20mn = take 20 $ create_mn 1 1
 
main = do
        -- 输出验证的序列前20个
        print $ top20mn
        -- 输出对应的路径数量
        print $ map (uncurry path_num) top20mn

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