In article
<67930447-88f9-4fdc-bd1a-8ca096d441bf@[EMAIL PROTECTED]
>,
William James <w_a_x_man@[EMAIL PROTECTED]
> wrote:
>On May 5, 11:49 am, an...@[EMAIL PROTECTED]
(Anton Ertl)
>wrote:
>> Still small enough to solve in a few lines:
>>
>> http://www.complang.tuwien.ac.at/forth/programs/euler/63.fs
>>
>> \ Problem:
>> \ The 5-digit number, 16807=75, is also a fifth power. Similarly, the
>> \ 9-digit number, 134217728=89, is a ninth power.
>>
>> \ How many n-digit positive integers exist which are also an nth power?
>> \ They don't count 0^1 (0 is not deemed to be positive)
>>
>> \ Solution:
>> \ 10^n produces a n+1-digit integer, so we need only look at bases 0-9.
>>
>> : n-digits { n -- n2 }
>> \ number of n-th powers with n digits
>> \ for n=1, it does not count 0 and 1 (I correct for the 1 below)
>> \ this works by just seeing what the lowest x is that fits in
>> \ 10^(n-1), and then rounding x up to an integer
>> n 1- 0 d>f n 0 d>f f/ falog f>d drop 9 swap - ;
>
>Clever, but the stack makes it hard to follow the code.
>
>>
>> : solve ( -- )
>> 1 1 begin ( n sum )
>> over n-digits dup 0> while
>> + swap 1+ swap repeat
>> drop nip ;
>
>Again, the stack makes it very difficult to follow.
>When one sees the +, he can't easily tell what two
>items are being added together. This code is hard
>to understand and to modify.
>
>Ruby:
>
>def n_digits n
> 9 - ( 10 ** ((n-1.0) / n) ).to_i
>end
>
>sum = 1
>(1 .. 9999).each{|i|
> n = n_digits(i)
> break if n < 1
> sum += n }
>p sum
Please don't publish spoilers without the words spoiler in
the subject line.
Groetjes Albert
--
--
Albert van der Horst, UTRECHT,THE NETHERLANDS
Economic growth -- like all pyramid schemes -- ultimately falters.
albert@[EMAIL PROTECTED]
&=n http://home.hccnet.nl/a.w.m.van.der.horst
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