This will allocate the books to schools in alphabetic order, skipping over any that need more than are left once their position is reached. How you would deal with any left unallocated at the end depends upon your needs.
If however, you want to try and skip an earlier large need because two later smaller ones would allow you to allocated the 400 exactly (or more exactly) then you are into what I believe is called a 'NP-complete problem'. Which is to say, you would probably need to try all the possible combinations (or is that permutations? I can never remember) in order to come to a descision as to which of them is the "Best fit". If this is the case, then say so, and we can help you
#! perl -slw use strict; my %table; my $n=0; for my $c ('A' .. 'Z') { $table{ +sprintf '%s%03d', $c, ++$n } = int(10+rand 100); } my $allocated = 0; for my $school (sort keys %table) { my $needed = $table{$school}; next if $allocated + $needed > 400; print "$school:$needed"; last if ($allocated += $needed) >= 400; } print 'Total allocated: ', $allocated; __DATA__ c:\test>226210 A001:107 B002:72 C003:15 D004:59 E005:106 O015:34 Total allocated: 393
Examine what is said, not who speaks.
The 7th Rule of perl club is -- pearl clubs are easily damaged. Use a diamond club instead.
In reply to Re: The Best fit by capacity of Box
by BrowserUk
in thread The Best fit by capacity of Box
by Gerryjun
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