Another Euler problem.

Find the greatest product of five consecutive digits in the 1000-digit number.73167176531330624919225119674426574742355349194934

96983520312774506326239578318016984801869478851843

85861560789112949495459501737958331952853208805511

12540698747158523863050715693290963295227443043557

66896648950445244523161731856403098711121722383113

62229893423380308135336276614282806444486645238749

30358907296290491560440772390713810515859307960866

70172427121883998797908792274921901699720888093776

65727333001053367881220235421809751254540594752243

52584907711670556013604839586446706324415722155397

53697817977846174064955149290862569321978468622482

83972241375657056057490261407972968652414535100474

82166370484403199890008895243450658541227588666881

16427171479924442928230863465674813919123162824586

17866458359124566529476545682848912883142607690042

24219022671055626321111109370544217506941658960408

07198403850962455444362981230987879927244284909188

84580156166097919133875499200524063689912560717606

05886116467109405077541002256983155200055935729725

71636269561882670428252483600823257530420752963450

Scanning through some of the many Project Euler problems, I could see that converting large integers into lists of digits was going to be a recurring theme, so I decided to handle it once and for all.

integer_to_digits(N) -> [ D - $0 || D <- integer_to_list(N)].

I could not find out whether Erlang has a line continuation convention, so I just concatenated the 1000 digits into one long integer inside Emacs. After that a simple recursion does the trick.

max(N,M) -> if N > M -> N; true -> M end. maxprod(List) -> maxprod(1, List). maxprod(Prod,[A,B,C,D,E] ) -> max(Prod, A*B*C*D*E); maxprod(Prod, [A,B,C,D,E|Tail]) -> maxprod(max(Prod, A*B*C*D*E), [B,C,D,E|Tail]).