A simple algorithm follows from this reasoning:

• You can build at most as many figures of type 1 as your budget allows;
• Once you've chosen the number n1 of figures of type 1, you can only buy a number of figures of type 2 going from 0 to the maximum allowed by your budget (minus what you spent on type 1);
• Once you've chosen the number n1 and n2 of figures of type 1 and 2, you can only buy a number of figures of type 3 going from 0 to the maximum allowed by your budget (minus what you spent on type 1 and 2);
• Same as above for type 4 to the last

As a simple sloppy example, if you have only three types:

$p1 = ... # Price of figures of type 1
$p2 = ... # Price of figures of type 2
$p3 = ... # Price of figures of type 3
$budget = ... # Total budget

for (my $i1=0; $i1 * $p1 <= $budget; $i1++) {
    for (my $i2=0; $i2 * $p2 <= $budget - $i1 * $p1; $i2++) {
        for (my $i3=0; $i3 * $p3 <= $budget - $i1 * $p1 - $i2 * $p2; $
+i3++) {
            print "You can buy $i1 type 1 + $i2 type 2 + $i3 type 3\n"
+;
        }
    }
}
[download]

Of course you should never code it this way and should instead be using arrays and an iteration which lends itself better to extensions. Anyway, I hope the algorithm is clear.

Cheers

Antonio

Update: I have taken for granted that you could use more than one figure of each type. If that assumption is not valid, please disregard the solution or add the additional condition that all indices ($i1, $i2 ...) should not exceed 1.

The stupider the astronaut, the easier it is to win the trip to Vega - A. Tucket