Agnimandur's blog

By Agnimandur, 3 months ago, In English

All the division 2 problems were created by Agnimandur. 1548E - Gregor and the Two Painters was created by Benq. I hope that this hint-based editorial helps you, no matter what your rating is! Solution code is provided in both C++, Java, and Kotlin when available.

Solution Code Repository


1549A - Gregor and Cryptography

Hint 1
Solution

1549B - Gregor and the Pawn Game

Solution 1

Hint 1
Hint 2
Solution

Solution 2

Hint 1
Hint 2
Hint 3
Hint 4
Solution

1549C - Web of Lies

Hint 1
Hint 2
Solution

1549D - Integers Have Friends

Hint 1
Hint 2
Hint 3
Solution

1549E - The Three Little Pigs

Solution 1

Hint 1
Hint 2
Hint 3
Solution

Solution 2

Hint 1
Hint 2
Hint 3
Hint 4
Solution

1549F1 - Gregor and the Odd Cows (Easy)

Hint 1
Hint 2
Hint 3
Hint 4
Solution

1549F2 - Gregor and the Odd Cows (Hard)

Hint 1
Hint 2
Hint 3
Hint 4
Hint 5
Solution

1548E - Gregor and the Two Painters

Hint 1
Hint 2
Solution
 
 
 
 
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3 months ago, # |
Rev. 2   Vote: I like it +65 Vote: I do not like it

Sorry about having to create a new editorial!

The old one passed some character limit built into Codeforces. Whenever I attempted to edit the old one, I was re-directed to the bug page, and the edit was not saved!

Thus, the editorial went offline for a while. Ultimately, a new and shorter editorial, was the only viable solution.

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    3 months ago, # ^ |
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    Could you post the old editorial page without the edits? There were several helpful (at least for me) discussions about GCD and solution of 1549D - Integers Have Friends without segment tree and sparse table.

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      3 months ago, # ^ |
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      Sorry, it's already been deleted.

      Ask the questions again and I will respond! .

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3 months ago, # |
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Can anyone please point out what's wrong with my code for 2D? 125022309 Getting Runtime error on test 3

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3 months ago, # |
Rev. 3   Vote: I like it 0 Vote: I do not like it

Can anyone please help me debug my solution for div 2 problem D. I am getting WA on test case 5.

Basically what I am doing is implementing binary search on the answer and if I find that lets say answer k is possible then I am searching for answer > k, and if answer k is not possible then I am searching for < k.

WA Solution

If my logic is wrong you are most welcome to correct it :)

UPDATE: Got it friends, there was a little problem in binary search implementation. AC Solution

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    3 months ago, # ^ |
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    Can you explain your check function?

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      3 months ago, # ^ |
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      So the param that I am passing to the function is the answer that I want to check whether it is possible or not. For this I am forming 2 arrays where array pre will store GCD of all the elements from index i to index j and suf will store GCD of all the elements from index j to i+m-1.

      And if you analyze it you can see that if(__gcd(suf[i], pre[i+m-1])!=1)return true; this piece of code will return true if GCD of m elements starting from i is > 1. If it is true then we can say that our array b has at least one subarray of size m who's GCD>1.

      (The main motive behind forming pre and suf arrays is to find GCD of any subarray of length m in O(1) time)

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        3 months ago, # ^ |
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        I think something is wrong with your binary search implementation.

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3 months ago, # |
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I can't understand how getting the value of p(k) can help in 1549E? How to expand it? What is polynomial long division? How to get the sum of coefficients of k^x? Also, a implementation with sufficient comment for this type of editorial would be good for Pupil like me.

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3 months ago, # |
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thanks for the editorial

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3 months ago, # |
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Solution for D div2 (B div1) without segment trees, sparse table:

After building the difference array $$$D$$$, for each $$$r$$$ we will maintain a list of uniques gcds that we can get if we start at position $$$r$$$ going backwards, i.e. all the possible values of $$$gcd_{l \dots r}=gcd(D_l, D_{l+1}, \dots, D_r)$$$ for all $$$1 \leq l \leq r$$$. Also for each value $$$g$$$ of this list we will also keep the maximum $$$length$$$ such that $$$gcd_{r-lenght+1 \dots r}=g$$$ so that we can update the answer for the problem. We can simulate all this process with a map.

Key observation: The size of the list at each iteration (that is, the number of unique gcds) is $$$O(\log MX)$$$, where $$$MX$$$ is the maximum value of $$$D_i$$$. That's because $$$gcd_{i \dots r}$$$ is a divisor of $$$gcd_{i + 1 \dots r}$$$ for each $$$1 \leq i < r$$$. So if $$$gcd_{i \dots r}$$$ is different from $$$gcd_{i + 1 \dots r}$$$, then $$$gcd_{i \dots r} \leq gcd_{i + 1 \dots r} / 2$$$.

Based on that, total complexity will be $$$O(n \cdot \log^2 MX)$$$ (the second $$$log$$$ is for the $$$gcd$$$ complexity). But maybe using the analysis of this blog (also cited in the editorial) the complexity can go down to $$$O(n \cdot \log MX + \log^2 MX)$$$ or $$$O(n \cdot \log MX \cdot \log \log MX + \log^2 MX)$$$ if you take into account the map complexity.

Link to my submission

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    2 months ago, # ^ |
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    I have another approach to get around the log factor from segtree / sparse table.

    I used a two pointer approach to find the subarray and represented the elements with a queue that always knows the gcd of contained elements. The technique is better known as a min-queue but same applies for gcd. It can be implemented with linked lists or like I did with two stacks.

    Here is my submission.

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3 months ago, # |
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in problem E,could give the code by way in the solution 2? please,i really want to know,it's so exciting.

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2 months ago, # |
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Here is a solution without the segment tree, just using prefix and suffix array for problem D solution

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2 months ago, # |
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Has anyone solved Div2D with Segment tree?. I am getting TLE with two pointers{ increase right pointer till gcd >1 in the difference array} and take the maximum size of all valid subarrays.

Here is my submission : 126619908

Is there any scope for improvement here?

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7 weeks ago, # |
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F2:

After I got struggled for hours with why "B is always even", I found "the area of the fence is an integer" in the problem statement.

QwQ.

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5 weeks ago, # |
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Agnimandur proof for B please