Problem 38

Take the number 192 and multiply it by each of 1, 2, and 3:

192 x 1 = 192

192 x 2 = 384

192 x 3 = 576

By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)

The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).

What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, … , n) where n > 1?

特に工夫もせずに。

import Data.List
concatProduct m n
| length x > 9 =[]
| length x == 9 =x
| otherwise = concatProduct (m+1) n
where x = concatMap (show.(n*))$ [1..m]
isPandigital = ("123456789"==).sort
p038 = maximum[(read::String->Integer).concatProduct 2$ n|
n<-[1..9999],(isPandigital.concatProduct 1) n ]
More Reading
Newer// Problem 37
Older// Problem 39