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BledDest's blog

By BledDest, history, 16 months ago, In English

1765A - Access Levels

Idea: BledDest, preparation: awoo

Tutorial
Solution (awoo)

1765B - Broken Keyboard

Idea: vovuh, preparation: vovuh

Tutorial
Solution (vovuh)

1765C - Card Guessing

Idea: DStepanenko, preparation: BledDest

Tutorial
Solution (awoo)

1765D - Watch the Videos

Idea: BledDest, preparation: DmitryKlenov

Tutorial
Solution (DmitryKlenov)

1765E - Exchange

Idea: BledDest, preparation: BledDest

Tutorial
Solution (BledDest)

1765F - Chemistry Lab

Idea: awoo, preparation: awoo

Tutorial
Solution (awoo)

1765G - Guess the String

Idea: BledDest, preparation: BledDest

Tutorial
Solution (BledDest)

1765H - Hospital Queue

Idea: Neon, preparation: Neon

Tutorial
Solution (Neon)

1765I - Infinite Chess

Idea: DmitryKlenov, preparation: dmitryme

Tutorial
Solution (awoo)

1765J - Hero to Zero

Idea: BledDest, preparation: BledDest

Tutorial
Solution (BledDest)

1765K - Torus Path

Idea: adedalic, preparation: adedalic

Tutorial
Solution (adedalic)

1765L - Project Manager

Idea: BledDest, preparation: awoo

Tutorial
Solution (awoo)

1765M - Minimum LCM

Idea: BledDest, preparation: Neon

Tutorial
Solution (Neon)

1765N - Number Reduction

Idea: Neon, preparation: Neon

Tutorial
Solution (Neon)
  • Vote: I like it
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16 months ago, # |
  Vote: I like it +5 Vote: I do not like it

Unfortunately, the editorials for two problems are not ready yet. They will appear here as soon as they're written.

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    5 weeks ago, # ^ |
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    it's been more than a year and the editorials are still not ready yet

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    5 weeks ago, # ^ |
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    it's been more than 2 years but the editorials are still not ready yet

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16 months ago, # |
  Vote: I like it +3 Vote: I do not like it

In problem Torus Path wording "Note that you can't visit all vertices on the antidiagonal (vertices .... at the same time." — is quite confusing.

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16 months ago, # |
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can problem N can be solved by using stack data structures ?

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    15 months ago, # ^ |
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    Yes indeed! You need to check for all 1 <= i <= n-1, if s[i] > s[i+1] then we can delete s[i] to obtain a smaller number. You will continue doing this till k is 0. One special case, when stack's first element is greater than 0 and all the remaining values are 0, when placing a value x > 0 and x < stack.top, if k >= stack size, delete all the elements from the stack and push the current one . P.S — You need vector to do this stack operation. Check my submission

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      11 months ago, # ^ |
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      Thanks :)

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      6 months ago, # ^ |
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      nice solution!

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      5 months ago, # ^ |
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      I liked your implementation! This code indeed requires a good sense of implementation! How did you come up with this?

      Have you done similar problems before?

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15 months ago, # |
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The problem K is easily solvable using DP. But the greedy method is quite tricky, at least for me.

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    7 months ago, # ^ |
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    In your code,using 'for loop', you have have handled the case of going from last to first within a row.But you have not done anything in case of going from last column to first column,whenever lvl==n if prev!=n-1 , you are just returning INT_MIN.why have you ignored that case??

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    2 months ago, # ^ |
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    Do you have some kind of proof why the answer would be moving y-axis wise and not x-axis wise. In ur code, the incrementing lvl parameter is a sifn of u are moving yaxis wise

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      2 months ago, # ^ |
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      The for loop iterates over x-axis.The modulo operation handles the case of going from last to first column.

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15 months ago, # |
  Vote: I like it 0 Vote: I do not like it

The problem D can be done in O(N) using just two pointers.

AC submission: https://codeforces.com/contest/1765/submission/186659367

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    15 months ago, # ^ |
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    Sorting does require O(NlogN) operations, you probably meant O(N) operations after sorting.

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    14 months ago, # ^ |
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    Your solution seems ease to understand. Thank you for sharing!. But TC is O(Nlog(N)) because of sorting.

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14 months ago, # |
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Neon Can u plss explain why the limit of for loop in problem M is g * g <= n??

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    10 months ago, # ^ |
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    It is for finding divisor of the number which you will get within root n

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14 months ago, # |
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Is there anyone to fail my code ? imo my logic is true and working every test cases that i used. Any help appreciated. 191443868

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13 months ago, # |
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How to approach problem K that at max n-1 anti-diagonal elements can be visited editorial proof is understandable but how to approach such a mathematical problem???

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    12 months ago, # ^ |
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    can you explain how at max n-1 anti diagonal elements can be visited. I was unable to understand this part of editorial

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8 months ago, # |
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Thank you BledDest for one of the most beautiful problems I've recently seen (J).

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8 months ago, # |
  Vote: I like it 0 Vote: I do not like it

In problem G, using the exact same idea from the editorial, and going backwards from the end instead of forwards, you end up with $$$1000/1.5 \approx 667$$$ steps on average. Even better, when you get a reply $$$x > 2$$$, you can find $$$x$$$ values in one move.

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3 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

The only observation I made in problem H is its a "DAG" in entire day.one could go to hospital and stand in queue and come back, even then I am trying to solve this problem lol and I found that N<=2000 after an entire day one more thing to make me laugh