yujong_lee has asked for the wisdom of the Perl Monks concerning the following question:
I'm solving basic dynamic programming question. It's simple. input is single integer. if input is 3, all possible cases is 3,1+2,2+1,1+1+1 so out put is 4. if input is 4, output is 7 because 4 = 1+1+1+1 1+1+2 1+2+1 2+1+1 2+2 1+3 3+1. code above is my code to solve this.
use strict; use warnings; our @cache; sub find { my $num = shift; if(defined $cache[$num]) {return $cache[$num];} if($num == 0) {return 1;} if($num < 0) {return 0;} $cache[$num] = &find($num -1) + &find($num - 2) + &find($num - 3); return $cache[$num]; } my $test_case = <>+0; for(1..$test_case) { my $num = <>+0; print &find($num),"\n"; }
but the output is wrong. and I found that output of 2 is 3, instead of 2. also output of 3 is 5 instead of 4. to fix this, I write @cache1..3 = (1,2,4); and of course, It is fixed. But, I want to remove @cache1..3 = (1,2,4); and want to know why my original code is not working properly. when call solution(2), isn't it obvious that the output is solution(1) + solution(0) + solution(-1)? and when I call solution(0), solution(1), solution(-1) individually, the output is 1,1,0 so sum is 2. not 3. Thank you for your help.
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Re: problem while solving basic dynamic programming question
by haukex (Archbishop) on Mar 15, 2020 at 08:42 UTC | |
by yujong_lee (Novice) on Mar 22, 2020 at 05:56 UTC | |
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Re: problem while solving basic dynamic programming question
by vr (Curate) on Mar 15, 2020 at 11:11 UTC | |
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Re: problem while solving basic dynamic programming question
by tybalt89 (Monsignor) on Mar 15, 2020 at 15:11 UTC | |
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Re: problem while solving basic dynamic programming question
by jo37 (Curate) on Mar 15, 2020 at 08:17 UTC | |
by haukex (Archbishop) on Mar 15, 2020 at 13:51 UTC |