RogerBW's Blog

Perl Weekly Challenge 72: factorial zeroes and line ranges 05 August 2020

I’ve been doing the Perl Weekly Challenges. The latest involved finding the trailing zeroes of a factorial and printing a range of lines from a file.

You are given a positive integer $N (<= 10).

Write a script to print number of trailing zeroes in $N!.

There's an easy way and a hard way to do this. Let's do the (computationally) hard way first: actually calculate the factorial (in a library where precision will be retained), and count the zeroes at the end.

use Math::GMP;

sub zerofact {
  my $n=shift;
  $n=Math::GMP->new($n);
  my $f=$n->bfac->get_str_gmp(10);
  my $l=0;
  $f =~ /(0+)$/;
  if (defined $1) {
    $l=length($1);
  }
  return $l;
}

The easier way, which I used in Raku, is not to calculate the factorial at all, but rather to consider that a trailing zero is the result of a pair of (2,5) among the prime factors. And there will always be at least as many twos as fives, so we just need to count the fives. So 0..4 has a result of 0, 5..9 a result of 1, 10..14 2, etc. A slight complication is that at n=25 we add two factors of five, at n=125 three, etc.

sub zerofact($n) {
  my $k=5;
  my $t;
  my $a=0;
  repeat {
    $a=floor($n/$k);
    $t+=$a;
    $k*=5;
  } while ($a>0);
  return $t;
}

(It's still slower than Perl though.)

You are given a text file name $file and range $A - $B where $A <= $B.

Write a script to display lines range $A and $B in the given file.

Well, that's very straightforward. The only slight cleverness is that we can jump to the end when we hit line number $B.

sub range {
  my $fn=shift;
  my $a=shift;
  my $b=shift;
  my $n=0;
  open I,'<',$fn;
  while (<I>) {
    $n++;
    if ($n > $b) {
      last;
    }
    if ($n>=$a) {
      print $_;
    }
  }
  close I;
}

and Raku is basically identical.

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