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| -rw-r--r-- | challenge-149/james-smith/README.md | 21 |
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diff --git a/challenge-149/james-smith/README.md b/challenge-149/james-smith/README.md index a755db373e..9ebbcb0a39 100644 --- a/challenge-149/james-smith/README.md +++ b/challenge-149/james-smith/README.md @@ -20,6 +20,27 @@ https://github.com/drbaggy/perlweeklychallenge-club/tree/master/challenge-149/ja ## The solution +```perl +for( my($n,$ds,$i,$fa,$fb,%fib)=(@ARGV?$ARGV[0]:20,0,0,1,1,0,1,1,1); + $n; $i++,$ds=0 ) { ## 1 + $ds+=$_ foreach split //,$i; ## 2 + ($fib{$fa+$fb},$fa,$fb)=(1,$fb,$fa+$fb) if $ds > $fb; ## 3 + (say $i)**$n-- if exists $fib{$ds}; ## 4 +} +``` + +**Notes:** + + * Line 1 - We initialise everything inside the for loop + * `$n` is the number to print (and is based on what is passed at the command line) + * `$ds` is the digit sum (Note we reset it everytime through the loop in the incremement part of the loop + * `$i` current value being considered + * `$fa` & `$fb` - the highest two fibonacci numbers + * `%fib` hash whose keys are fibonacci numbers + * Line 2 - Computes the digit sum by splitting number on `//` I split into 1 character blocks + * Line 3 - Expand the fibonacci hash by 1 if the digit sum is greater than the highest fibonnaci number {we don't need to loop this as the digit sum of `$n+1` can only be at most 1 higher than that for `$n`. Note we just update $fb and $fa in this line + * Line 4 - Check to see if the digit sum exists, print and decrement counter - and return to the start of the loop. + # Challenge 2 - Largest Square ***Given a number base, derive the largest perfect square with no repeated digits and return it as a string. (For base>10, use ‘A’..‘Z’.)*** |