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
|
