nthperm(): Replace one %= by a -= remainder * divisor computed earlier

[davidavdav]

This is a tiny rewrite that makes nthperm!() a little faster, by about 15% or so. The speed results from replacing an effective k = k ÷ f by k = k - j * f, where j has been computed in an earlier %.

(Incidentally, this kind of code could benefit from a generic function divisor, remainder = integer_divide(num, denom))

Further some small cosmetic changes, and an unnecessary +1-1 removal for j.

[codecov-io]

Codecov Report

Merging #55 into master will not change coverage.
The diff coverage is 100%.

@@           Coverage Diff           @@
##           master      #55   +/-   ##
=======================================
  Coverage   96.96%   96.96%           
=======================================
  Files           6        6           
  Lines         594      594           
=======================================
  Hits          576      576           
  Misses         18       18

Impacted Files	Coverage Δ
src/permutations.jl	100% <100%> (ø)	⬆️

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[davidavdav]

(Incidentally, this kind of code could benefit from a generic function divisor, remainder = integer_divide(num, denom))

Just found out there is such a function, divrem(), but this is slower than just doing a ÷ and a * as in the current patch.

[mschauer]

Yes, by k - div(k, f) * f == k % f

[mschauer]

Not passing tests after resolving conflict

[mschauer]

The issue is:
54bd3ee#diff-0da9215c8a020f2964e085699848213d