#!/usr/bin/perl
use v5.35.0;

# Machine setup requires solving paired equations of the form
#
# AX * a + BX * b = PX
# AY * a + BY * b = PY
# 
# Solutions are
# 
# a = (PX * BY - BX * PY) / (AX * BY - BX * AY)
# b = (AX * PY - PX * AY) / (AX * BY - BX * AY)
# 
# and are required to be integers because they represent numbers of
# button presses, so they only exist when the numerators are multiples
# of the (common) denominator.
# 
# Solutions are fully determined so there is no need to search for
# optimal A and B button press combinations per machine; there is no
# "cheapest way" to win the prize, they're either winnable or not.

my ($ax, $ay, $bx, $by, $px, $py, $cost);
my $adjustment = 10000000000000;
my $patience = 'inf';
while (<>) {
	print;
	($ax, $ay) = ($1, $2) if (/Button A: X\+(\d+), Y\+(\d+)/);
	($bx, $by) = ($1, $2) if (/Button B: X\+(\d+), Y\+(\d+)/);
	if (/Prize: X=(\d+), Y=(\d+)/) {
		($px, $py) = ($1 + $adjustment, $2 + $adjustment);
		say "$ax $ay $bx $by $px $py";
		my $denom = $ax * $by - $bx * $ay;
		my $num_a = $px * $by - $bx * $py;
		my $num_b = $ax * $py - $px * $ay;
		if ($num_a % $denom || $num_b % $denom) {
			say "Not winnable";
		}
		else {
			my $a = $num_a / $denom;
			my $b = $num_b / $denom;
			say "Button A x $a, button B x $b";
			if ($a > $patience || $b > $patience) {
				say "Too slow";
			}
			else {
				my $tokens = 3 * $a + 1 * $b;
				say "Spent $tokens tokens";
				$cost += $tokens;
			}
		}
	}
}
say "\nTotal spent: $cost"