a similar solution that avoids making an intermediate list:
>./ 13 */@:". \ n
regarding avoiding those items that contain a 0, for the problem at hand, I
would guess it isn't necessary since a 0 ensures that the product cannot be
maximum, and I would guess it slows things down. But if you must, then your
original approach seems right.
(#~ ( 13 = '0' i.~ ])"1) 13,\ 40{. n
7316717653133
6249192251196
2491922511967
4919225119674
9192251196744
1922511967442
9225119674426
2251196744265
2511967442657
5119674426574
1196744265747
1967442657474
9674426574742
6744265747423
7442657474235
>./ */@:"."1 (#~ ( 13 = '0' i.~ ])"1) 13,\ n
timespacex ' >./ 13 */@:". \ n'
0.00128096 31872
timespacex ' >./ */@:"."1 (#~ ( 13 = ''0'' i.~ ])"1) 13,\ n'
0.00065632 33792
turns out I was wrong and it is faster to filter out the 0s!
----- Original Message -----
From: mvillarino <[email protected]>
To: [email protected]
Cc:
Sent: Thursday, July 3, 2014 6:18:10 PM
Subject: [Jprogramming] Project Euler Problem 8
I came to this solution, which happens to be a transliteration of the
problem's formulation:
>./ */"1 ".items"1(13,\number)
where «numero» is the string with the number's digits (one thousand).
Please note I'm no programmer, and I'm learning a little J with no
other intent than just for curiosity.
Before proceeding into another Problem, I would like to modify the
solution above in order to avoid multiplying groups that contain
noughts. With iteration it would be something like instead of
".items"1 (13,\number), build the table ignoring _increasing some
counter that refers to position_ those subarrays that contains a zero
and the next few rows, where few is 13 minus the position of the last
zero counted, until the end of number is reached.
Is something alike that expressible with J? Can you give me some hints?
Best regards,
Marce
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