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PONumber of combinations with LEGO plastic bricks C++
primarykey
Id
10396058
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10396656
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2
ClosedDate
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12
CommunityOwnedDate
CreationDate
2012-05-01T09:49:47.217
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2
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2012-09-10T22:39:24.363
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2012-09-10T22:39:24.363
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1
Score
6
ViewCount
955
LastEditorDisplayName
text
Body
You have some LEGO plastic bricks, all bricks are 1x1x1. Also you have one tile, 1xN (N <= 80), on which you should place the LEGO bricks. You can order them in sequences (one sequence is correct if it has 3 or more consecutive bricks), also you must have at least 1 empty space between 2 sequences. You should calculate the number of different combination you can place the bricks on the tiles. Here's an example: If the tile is 1x7, there are 17 different combinations. input : 7 output : 17 <a href="http://mendo.mk/task_files/kocki.gif" rel="nofollow noreferrer">pic of the example http://mendo.mk/task_files/kocki.gif</a> Also if you have no bricks it's counted as 1 combination. I have worked on this problem and i found the way to calculate possible combinations of if the max length of the tile is 14 (3 sequences). I found it using for loops. My biggest problem is the huge number of for loops that i need to run. For example, for 1 sequence i use 1 for loops, for 2 sequences 2 loops + 1 for 1 sequnce...so if i use all 80 bricks, i can create 20 sequence and i will have to use over 210 for loops which is huge number. So it'll be good if i can nest them in one. I tried it and it get messy and it doesn't give correct answers. New code: <pre><code>#include <iostream> using namespace std; int main() { long long int places, combinations = 1; cin >> places; long long int f[80], g[80]; f[0] = 0; f[1] = 0; f[2] = 0; g[0] = 1; g[1] = 1; g[2] = 1; for(int i = 3; i<=places; i++) { f[i] = f[i-1] + g[i-3]; g[i] = f[i-1] + g[i-1]; } combinations = f[places] + g[places]; cout << combinations; return 0; } </code></pre>
Tags
<c++><algorithm><loops><combinations><lego>
Title
Number of combinations with LEGO plastic bricks C++
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