Task 1: Disarium Numbers

Write a script to generate first 19 Disarium Numbers.

A disarium number is an integer where the sum of each digit raised to the power of its position in the number, is equal to the number.

For example,

518 is a disarium number as (5 ** 1) + (1 ** 2) + (8 ** 3) => 5 + 1 + 512 => 518


MY NOTES: ok, sounds pretty easy.  Once I'd solved it, as it seemed rather
slow, I profiled it with NYTProf and generated successive optimised versions
ch-1[a-g].pl - and ch-1g.pl runs about 7 times faster than the initial version.

GUEST LANGUAGE: As a bonus, I also had a go at translating ch-1.pl
into C (look in the C directory for that).


Task 2: Permutation Ranking

You are given a list of integers with no duplicates, e.g. [0, 1, 2].

Write two functions, permutation2rank() which will take the list and
determine its rank (starting at 0) in the set of possible permutations
arranged in lexicographic order, and rank2permutation() which will take
the list and a rank number and produce just that permutation.

Please checkout this post for more informations and algorithm:
  https://tryalgo.org/en/permutations/2016/09/05/permutation-rank/

Given the list [0, 1, 2] the ordered permutations are:

0: [0, 1, 2]
1: [0, 2, 1]
2: [1, 0, 2]
3: [1, 2, 0]
4: [2, 0, 1]
5: [2, 1, 0]

and therefore:

permutation2rank([1, 0, 2]) = 2

rank2permutation([0, 1, 2], 1) = [0, 2, 1]


MY NOTES: hmm.. I hate permutations, never managed to memorise the
algorithm in all my years of programming.  However, I've got a module
Perm.pm which permutes the digits in a number, so let's reuse it to solve
this pesky problem in the simplest possible way: Perms' "next_perm()"
function generates all the perms in order, so sticking that inside
various simple counting loops does the trick, albeit very inefficiently.
That'll do!

I didn't have time to translate this (and the Perms module) to C, although it
should be pretty straightforward.