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Re^2: How to change these small one-liners into Perl6 code?

by ABCXYZ (Novice)
on Dec 23, 2012 at 19:18 UTC ( #1010114=note: print w/replies, xml ) Need Help??

in reply to Re: How to change these small one-liners into Perl6 code?
in thread How to change these small one-liners into Perl6 code?

I was simply translating my Perl5 one-liners on ProjectEuler into Perl6's code.

These are all on its first problem:
If we list all the natural numbers below 10 that are multiples of 3 or + 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000. (I generalized this to: below 10**$n for code #1 and #2, or below $N for code #3)
This is my fastest solution in Perl5: (not in the code #1..4 above)
perl5 -e '$n=3;print 2,3x--$n,1 .6x$n+2'

The following is the origin of code #1, which works beyond the limit of 64 bit int, since it uses string op:
$n=3;print 2,(3x--$n.1 .6x--$n.8)=~s/^18/3/r

This is the origin of code #2:
$n=3;$_=2 .3x--$n.1 .6x$n;substr($_,-1)+=2;print

So code #3 is a more general solution for $N which is typically not an integer exponentiation of 10:
map{$s+=int$_*($i=abs int 999/$_)*++$i/2}(3,5,-15);print$s

They all give the same result: 233168 for $n=3, but work for other $n as well.

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Re^3: How to change these small one-liners into Perl6 code?
by raiph (Chaplain) on Dec 24, 2012 at 01:51 UTC should interest you.

    Five solutions have been added to this repo over the last 4 years. The latest one was added 18 days ago:

    say [+] grep * %% (3|5), ^1000;

    This means "say the sum of numbers that are divisible by 3 or 5 in the range zero up to (but not including) 1000".

      Nice code!

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