Rendering_Engine's blog

By Rendering_Engine, history, 8 months ago, In English

Hi Everyone ! I am not able to understand how to break this problem into subproblems ?
Can someone help please ? Here is the problem link : Problem

Tags dp, cses
 
 
 
 
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8 months ago, # |
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Just Bumping it . In case someone wanna help .

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8 months ago, # |
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Think about the last row of the tower. You can either have both cells as a part of the same block, or both cells as parts of different blocks. Let's call towers of the first kind type-1 towers, and the towers of the second kind type-2 towers.

Let's call $$$a_n$$$ be the number of type-1 towers and $$$b_n$$$ be the number of type-2 towers.

For computing $$$a_n$$$, look at what remains when you remove the last row from the tower. Corresponding to a type-1 tower of height $$$n - 1$$$, there will be two such towers, and corresponding to a type-2 tower of height $$$n - 1$$$, there will be one such tower, so $$$a_n = 2a_{n - 1} + b_{n - 1}$$$. In a similar manner you have $$$b_n = 4b_{n - 1} + a_{n - 1}$$$.

Then you can either precompute the answers to all possible queries, or use matrix exponentiation to solve the recurrence. The answer is $$$a_n + b_n$$$.

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    8 months ago, # ^ |
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    Thanks a lot .

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    6 weeks ago, # ^ |
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    When computing bn, you consider the last row to build the type-2 tower right ? But I don't understand why it can be 4 * b(n-1) + a(n — 1) sir. Thanks you if you can explain to me!!

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      5 weeks ago, # ^ |
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      Because the lower n-1 level can have either one a[n-1] and one b[n-1] because of solid border, or 2 b[n-1] breaking any of the borders, and b[n-1] when breaking both the borders. Here border is the line between the nth and (n-1)th row.

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5 months ago, # |
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can someone tell me the rating of this problem acc to codeforces