SPatrik's blog

By SPatrik, history, 2 months ago, In English,

Hello, Codeforces!

I am glad to invite you to take part in Codeforces Round #577 (Div 2), which will be held on Sunday, 4 August at 16:35 UTC.

You will be given 5 problems, one of them will have 2 subtasks and 2 hours to solve them.

The round will be rated for the second division.

Huge thanks to _kun_ and KAN for helping to prepare the round.

I would like to thank to 300iq, isaf27, V--gLaSsH0ldEr593--V, pllk, mohammedehab2002, tractor74.ru, Rox, opukittpceno_hhr for testing the round.

And thanks to MikeMirzayanov for the great codeforces and polygon platforms.

This is my first Codeforces round. Hope you will enjoy it.

Good luck and have fun!

UPD: The scoring distribution is: 500 — 1000 — 1500 — 2000 — (2000 + 1000)

UPD2: Editorial

UPD3: Congratulations to the winners:

Div2:

1: wwaynetuu

2: jerome_mei

3: satu0king

4: Yazmau

5: Tlatoani, 2-SAD, boba5551

Unofficial Div1:

1: neal

2: uwi

3: scott_wu

4: I_love_Tanya_Romanova

5: ecnerwala

 
 
 
 
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2 months ago, # |
  Vote: I like it +135 Vote: I do not like it

All the " Author's First Codeforces round " are always carefully designed. Can't wait attending this contest.

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2 months ago, # |
  Vote: I like it +41 Vote: I do not like it

Wish you the best of luck for organising your first contest

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2 months ago, # |
  Vote: I like it +13 Vote: I do not like it

Which question has two subparts? Please share the points of all problems.

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2 months ago, # |
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The time is not friendly for Chinese

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2 months ago, # |
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I hope it's the first of many.

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2 months ago, # |
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We expect problems would be interesting and this round will be remarkable for PARTICIPANTS & JUDGES.

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

Which problem has two subtasks?

And the what's the markes for each problem?

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2 months ago, # |
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i have a question... current my email "not visibled" in codeforces, how to fix???

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    2 months ago, # ^ |
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    It means other users can't see your email.If you want to show it,you can change it in "settings".

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2 months ago, # |
  Vote: I like it -35 Vote: I do not like it

Hopefully, I will become spec.

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2 months ago, # |
  Vote: I like it +31 Vote: I do not like it

Hope C1 and C2 I want rate inflation

Spoiler
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2 months ago, # |
  Vote: I like it +41 Vote: I do not like it

coding sucks

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2 months ago, # |
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Hope this round goes well.

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2 months ago, # |
Rev. 2   Vote: I like it -8 Vote: I do not like it

As always, I find out everything at the last moment( Good luck to you to!

Как всегда, узнаю всё в последний момент( Удачи и Вам!

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

It's interesting that writing of contest problems is a free creativity. So many authors use imagination and fantasy almost without limits, sometimes even insert jokes.

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    2 months ago, # ^ |
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    i love good jokes

    here is one:

    what did the microscopic cow say?

    Spoiler
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      2 months ago, # ^ |
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      This is anime cow

      Meow -> nya

      Moo -> μ

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      2 months ago, # ^ |
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      It's a most funny joke that I see in problems

      B. Tavas and SaDDas

      Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."

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2 months ago, # |
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This is my first time! gogogo~!

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2 months ago, # |
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Guess I'll be master after this round.

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2 months ago, # |
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There are fewer people take part in this contest.

Maybe it is too late,but I'll still enjoy it!

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2 months ago, # |
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I hope the problem statements are short and clear.

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2 months ago, # |
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Please don't be stuck, server

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2 months ago, # |
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When codeforces adds another speedforces.

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2 months ago, # |
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WHY CANT I HACK PROBLEMS?

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2 months ago, # |
  Vote: I like it +68 Vote: I do not like it

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    2 months ago, # ^ |
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    me too =))

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    2 months ago, # ^ |
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    And doing it really slow with tons of WA... Rip rating

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    2 months ago, # ^ |
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    relatable

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    2 months ago, # ^ |
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    Maybe you shouldn't post meme which everyone see thousands of times?

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    2 months ago, # ^ |
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    When you wait the end of system tests and the change of rating

    [Same picture]

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    2 months ago, # ^ |
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    I wish I were you :)

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    2 months ago, # ^ |
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    SO TRUE!! i did the first three so fast but after that i was just staring at my moniter thinking

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2 months ago, # |
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D is too hard for D.

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    2 months ago, # ^ |
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    And E is too hard for E as well

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    2 months ago, # ^ |
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    D is not too hard. It is too shitty.

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      2 months ago, # ^ |
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      What is the solution? I feel fucking stupid. :'(

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        2 months ago, # ^ |
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        think it's just some dp on leftmost/rightmost element in each row. lots of typing

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        2 months ago, # ^ |
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        DP. Iterate over rows and for each treasure holding how fast you can reach that treasure with all treasures in the same row be gotten too.

        That's just my thinking. Late for submiting. RIP rate:)

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        2 months ago, # ^ |
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        I think modifiable BFS or something similar.

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        2 months ago, # ^ |
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        $$$dp_{row, lasty, lastcol}$$$, where $$$row$$$ is the number of row (we need to consider only rows with at last one treasure but in fact it does not matter), $$$lasty$$$ is either $$$0$$$ or $$$1$$$ and means that the last $$$y$$$ on the current row we visited is the leftmost $$$y$$$ of some treasure on the current row or the rightmost such $$$y$$$. And $$$lastcol$$$ is either $$$0$$$ or $$$1$$$ too and means that after visiting the leftmost/rightmost $$$y$$$ with the treasyre we go to the left safety column or to the right safety column from this $$$y$$$. Transitions are very easy to understand but hard do code.

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          2 months ago, # ^ |
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          Can you elaborate more ? I am not able to understand :(

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            2 months ago, # ^ |
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            Since you can go only up, never down, you have to collect all treasures of one level at once. This end in standing on the leftmost treasure of one level or the rightmost one.

            From these two positions there are again four paths to the left/right treasure of the next level.

            These paths are simple. You can allway either go left, then up, then to the treasure, or go right, then up, then to the treasure.

            So, you need to write that down in terms of dp recurrence, which is no rocket science, but nasty index fiddling.

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          2 months ago, # ^ |
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          Said Vovuh and get WA)

          You lost chanse to become a master on this contest((

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            2 months ago, # ^ |
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            You can see the bug in diff between WA42 and AC. It is so stupid. Just like me.

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      2 months ago, # ^ |
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      Exactly, it is just tons of typing. And E seemed like this as well :(

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2 months ago, # |
Rev. 2   Vote: I like it +9 Vote: I do not like it

TypingForces. Also D is unusually tough.

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2 months ago, # |
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I can't see B solution meh feels stupid

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2 months ago, # |
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How to solve B?

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    2 months ago, # ^ |
      Vote: I like it +23 Vote: I do not like it

    You can always make everything to zero if and only if :

    1. Sum of all given numbers is even.
    2. No element of array exceeds half of the sum.
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      2 months ago, # ^ |
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      Try 3 3 3 5

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        2 months ago, # ^ |
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        ???

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        2 months ago, # ^ |
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        I suppose you mean 4 numbers, (3, 3, 3, 5).

        (1, 4), (1, 4), (2, 4), (2, 4), (3, 4), (1, 3), (2, 3).

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          2 months ago, # ^ |
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          Or if you meant 3 numbers (3, 3, 5), answer is NO since the sum is odd.

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        2 months ago, # ^ |
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        • sum=14
        • 5<=7
        • 3 3 2 4

        • 3 2 2 3
        • .....
        • 0 2 2 0
        • .....
        • 0 0 0 0
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        2 months ago, # ^ |
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        Use the last element with the first three, then it's 2,2,2,2 and it's easy.

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      2 months ago, # ^ |
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      Could you explain the second point please?

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        2 months ago, # ^ |
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        Try to think number $$$x$$$ as $$$x$$$ balls of one color. We can match a ball with another ball with another color (pair with same balls is not allowed). Then, we note that if there are odd balls, we can't complete matching. This is first point. If we have too many balls of one color (by "too many", I mean more than half of all balls), by the pigeonhole principle (although it is pretty straightforward), at least one pair must contain two balls of that color.

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      2 months ago, # ^ |
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      What about 2 2 2

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      2 months ago, # ^ |
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      please is there a proof for point 2?

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        2 months ago, # ^ |
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        If a color has cardinality over n/2, there is no solution because you will need to pair up two elements of a color at least once. Otherwise, we construct a solution as follows: make each move by choosing of the two largest groups of colors. It is intuitively clear to me that making this kind of move will ensure that, in subsequent moves, no color gets to cardinality over (total number of balls)/2. And if we always preserve this property, there will always be at least 2 piles.

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      2 months ago, # ^ |
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      I also thought first one but the second was too away from me :'( how did you find this solution? (I mean the second approach)

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        2 months ago, # ^ |
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        It seems quite awkward for me to explain how I approached this problem since I struggled twice (If you take a look at my submission, you'll notice that I tried 2 completely different logic which both were not working).

        However, I think for this problem some caseworks can help. I basically noticed second approach when I thought of following case :
        2 numbers, (1, 7). Sum is even, but we can't do the task.

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          2 months ago, # ^ |
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          I also noticed that (1, 7) case but I couldn't bring more :(

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            2 months ago, # ^ |
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            Maybe more casework would have been helpful. Restricting to 2 or 3 numbers (which is hand-doable)? Such as (1, 3), (2, 2), (2, 3), ... and thinking with more data.

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      2 months ago, # ^ |
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      How do I prove it?

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      2 months ago, # ^ |
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      Try 12 3 9 4

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      2 months ago, # ^ |
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      how do you know that these are all the necessary conditions? How do you know there aren't more? I thought of these in the contest but didn't think it was that simple

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      2 months ago, # ^ |
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      can you prove it?

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    2 months ago, # ^ |
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    sum all elements. Than if the sum is odd then return "NO" else check if the largest element is bigger than the sum/2, if it is than "NO" else "YES"

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    2 months ago, # ^ |
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    Just think nothing!

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2 months ago, # |
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How to solve D?

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2 months ago, # |
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Unfortunately, there were no server issues today as well :(

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2 months ago, # |
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Is N^4 the intended complexity for E1 ?

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2 months ago, # |
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What is pretest 5 in B ?

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Logic behind B?

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    2 months ago, # ^ |
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    sum(1,n) % 2 =0 and sum(1,n-1) >=a[n] with array a sorted -_- -_-

    bad at english sorry -_-

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    2 months ago, # ^ |
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    Think about this test case: 6 6 4. I think you will understand the logic behind solution of B.

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    2 months ago, # ^ |
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    If the biggest element is not larger than the sum of the rest, then it is possible. If the sum of all elements is even, of course.

    Basically, if all this holds, then you can take pair "smallest element — largest element" and decrease them until the smallest one becomes 0. Then you just reduced the number of elements, but the two main conditions still hold. You can show by induction that you cannot end up in a situation when you have only one number left -- otherwise the first condition did not hold from the very beginning.

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      2 months ago, # ^ |
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      wow!

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      2 months ago, # ^ |
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      Consider the following array: 4, 17, 19. Sum % 2 == 0 and 19 < 17 + 4. So it should work.

      If I apply your method of taking smallest — largest pair and decreasing the following happens:

      First step, take 4 and 19 and reduce and you get 0, 17, 15. Next, take 17 and 15 and you are left with 0, 2, 0.

      Did I misunderstand anything?

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        2 months ago, # ^ |
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        You don't have to reduce 4 all the way to 0 because when we have 1 17 16 then we will decrease 1 and 17 because they are the biggest and smallest right now. This will retain our claim

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          2 months ago, # ^ |
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          Ah I see, thanks for clearing it up.

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          2 months ago, # ^ |
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          So, basically you have to sort the whole array after each turn to make sure the arr[0] and arr[n-1] are getting reduced by 1 where they are smallest and largest pair. Got it. Looks a lot simpler this way.

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    2 months ago, # ^ |
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    I solved without sorting. 'YES' if sum of all elements % 2 == 0 and there's no element which is more than all other values' sum, 'NO' otherwise

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2 months ago, # |
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any guesses on test case6 of question 3??

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Test 5 B == killer.

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Anybody know what pretest 7 in D was?

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2 months ago, # |
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Dafuq was B omfg

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Logic behind c?

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    2 months ago, # ^ |
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    binary search

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      2 months ago, # ^ |
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      More precise please

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        2 months ago, # ^ |
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        If you want the median to be at least X, you need at least (n / 2) + 1 elements to be greater or equal than X.

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        2 months ago, # ^ |
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        Then binary search the value of the median. Let's say the value you want to check is X. Then you have to check, whether you can increase all numbers with indexes from N/2 to N up to X, using only k moves. For this you need: a) prefix sums of array elements, b) to search the first element in the second half of the array which is already larger than X (also binary search).

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        2 months ago, # ^ |
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        You sort the array. Find out the middle element each time. Increment it by 1. But array must remain sorted throughout. Thus, a multiset might help. (I didn't implement it during contest, but I guess the logic is correct. Correct me if I am wrong)

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      2 months ago, # ^ |
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      binary search gave me TLE on 2 solution one n*log(n) & the other log(n)² , had to implement the O(n) solution

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        2 months ago, # ^ |
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        If you implement binary search correctly, it must be sth like ($$$n \log n + n \log A$$$) where $$$A = 2e9$$$ and $$$n = 2e5$$$. BTW I would really like to learn how $$$O(n)$$$ solution works...

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          2 months ago, # ^ |
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          sorry was wrong it's o(n)+n log(n) bcs of the sort at first, the idea is to start from middle and check if you can make all the numbers from middle to current index equal to the next element value on the sorted array, and substract the cost. sol

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        2 months ago, # ^ |
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        Your binary search was TLE because mid was overflowing. int mid = (l+r)/2; l = 10^9 and r = 2*10^9

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2 months ago, # |
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A, B, C are standard problems. D is a good problem. For this thank you author.

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i wanna to ask why my code wrong on pretest 2. https://ideone.com/rwIkPi. they say it is undefined behaviour, but i didn't figure out where it happens.

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2 months ago, # |
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B reminded me of AGC 036 C.

By the way, could somebody hack my solution for D? I solved in a dijkstra-like style. I checked 998244353 times but I couldn't find my bug.

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    2 months ago, # ^ |
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    My solution is working fine on custom testing on cf for all sample 3 test cases . But failing for 1st test case while submitting. can anyone plss check it

    https://codeforces.com/contest/1201/submission/58300351

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      2 months ago, # ^ |
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      So what about your mistake? At least, now custom invocation shows "10000000000000000", not the right answer.

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    2 months ago, # ^ |
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    OMG, edit distance was 1...

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      2 months ago, # ^ |
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      what is the mistake?

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

When you propose a contest, you is asked to indicate the features of the tasks that are given. To whom what today's tasks seemed unusual and interesting? What is their feature? Who, after solving the problem, got the joy that you came up with a cool solution and it came in? Personally, I did not like any of the ABC. Moreover, I didn’t have joy after the decision of the old C. In addition, D and E, in my opinion, are very difficult for D and E. Sorry from my poor English

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My solution is working fine on custom testing on cf for all sample 3 test cases . But failing for 1st test case while submitting. can anyone plss check it

https://codeforces.com/contest/1201/submission/58300351

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2 months ago, # |
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Hey guys, Could anyone please explain to me what is the hack means for contests?

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    2 months ago, # ^ |
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    If your solution's imperfect(even though you passed pretest, it's possible), the others who locked that problem can hack your solution with data which is counterexample.

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      2 months ago, # ^ |
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      Could you please more clear :D I am new for contests

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        2 months ago, # ^ |
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        During the contest, your submissions are only tested on a limited set of test (pretests). It's possible that an incorrect solution can pass pretests, so during the contest, other people in your room who have solved the same problem can choose to lock their problem and "hack" your solution (provide a test case where your soln will fail). Hackers are awarded +100 pts for succesfull hack, -50 for unsuccessful.

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        2 months ago, # ^ |
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        The pretest doesn't cover all of test data. So although you passed pretest, it can be a wrong answer. After pretest passed, you can lock the problem. It means you cannot change your answer and you can read other user's code. When you reading code, if someone's code have counterexample, you can enter counterexample and if the hacking attempt's successful, you can get 100point. but if it failed, you lose 50point. Can you understand? (sorry for my poor english)

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          2 months ago, # ^ |
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          Yes finally I understood :), Thanks a lot for your reply

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2 months ago, # |
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Wasn't the contest a bit unbalanced? Still, quite a good contest for a first timer setter. :)

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2 months ago, # |
  Vote: I like it +6 Vote: I do not like it

Problem D looks similar to a recent task from TCO, except that the one from topcoder restrict the number of safe columns to exactly two.

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

--deleted--

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    2 months ago, # ^ |
      Vote: I like it +13 Vote: I do not like it

    (le + ri) may not fit in signed int.

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    2 months ago, # ^ |
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    int le = arr[n / 2];
    int ri = arr[n / 2] + k + 2, mi;
    mi = (le + ri) / 2;
    

    Someone got overflow...

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2 months ago, # |
  Vote: I like it +21 Vote: I do not like it

A nice version of problem D: 1066F - Yet another 2D Walking.

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2 months ago, # |
  Vote: I like it +38 Vote: I do not like it

Okay, why is E1 worth 2000 and E2 worth 1000? E1 is nothing but dumb coding that doesn't even require much thinking from my perspective, while E2 is a beautiful (in my opinion) and hard problem.

Subtasks are getting a bit annoying on CF :(

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    2 months ago, # ^ |
    Rev. 2   Vote: I like it +30 Vote: I do not like it

    From perspective of Div2 users both of them are almost unsolvable)
    P.S. Sorry, not almost:D

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    2 months ago, # ^ |
    Rev. 2   Vote: I like it +11 Vote: I do not like it

    "Dumb coding"

    When even with the unoficials contestants, only 2 persons for 7000 contestants solved it, and that even high rated like scott wu, ecnerwala and a lot others red contestants couldn't solve it, I'm not sure the problem was so "dumb".

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    2 months ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    Do you see better scoring solution under assumption that subtasks are there?

    I agree that it sounds questionable when you look at it from this perspective, but it all kinda makes sense. Proper solution gives you E1+E2, for which 3000 is reasonable. During the contest I spent 25-30 minutes thinking on E2 before abandoning it — and the case work that I had at that point was more complicated and messy than the editorial.

    Yes, E1 is straightforward coding which requires even less thinking than D — but it is, by div2 standard, quite a lot of tedious coding. I don't think it would be fair to say that 58306483 is worth less than solving D. Is it really that wrong to look at it like "either you just code the trivial part and get 2000 for it, or you solve a full problem for 3000"?

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2 months ago, # |
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Even though I may not have solved most of the problems, I quite enjoyed the contest. The questions were challenging and intriguing at the same time.

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2 months ago, # |
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Can anyone recommend some problems to solve to develop skills to solve problems like B ?

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2 months ago, # |
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A, B and C are too easy and D was too tough I think this contest just focuses on speed.

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2 months ago, # |
Rev. 3   Vote: I like it +22 Vote: I do not like it

SPatrik, can you explain to me why did you need this statement that knights are allowed to capture each other? Does it affect anything except solutions that fall as mine because the case was forgotten when I need to capture an opponent`s knight? And why was not it covered by pretests?

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

    Without capturing, the problem would be really easy. Just select the knight that can reach it's target faster. With capturing, there is a situation where you have to move more tricky. First go to the opponent target, then to your target. Read the editorial for more details.

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      2 months ago, # ^ |
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      Of course, I meant that it would have been greater if knights hadn`t been able to move under attack at all, not allowing moving under attack and being immediately captured afterward.

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        2 months ago, # ^ |
          Vote: I like it +8 Vote: I do not like it

        Yeah, that could be done. That would remove the possibility of those minor error. I didn't think that way. There wouldn't be any other differences, capturing just makes the game more realistic.

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          2 months ago, # ^ |
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          If you like this so, this case should be in pretests at least :). But as I would think in your shoes, the statement that doesn`t affect should be removed. Anyway, thanks a lot, I like this problem very much despite that mistake!

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2 months ago, # |
  Vote: I like it +6 Vote: I do not like it

Are you kidding? A B C are too easy and D is so hard. it is not suitable as D.

For E & F , are you sure you were not make a kidding to us?

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2 months ago, # |
Rev. 3   Vote: I like it +27 Vote: I do not like it

Thanks to SPatrik for problem C. The same problem was given to the participants of Lipetsk Municipal Olympiad and some participants who wanted to qualify for Sirius (December 2018) about one year ago. So, I had the same problem prepared in Polygon and you helped me realize that my testset wasn't full and some authors' solutions weren't right. Thank you very much :)

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

for god sake WA on C because of int overflow, Shame ._.

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2 months ago, # |
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WOW fast system testing

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2 months ago, # |
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Can someone help me find out why is this submission getting MLE? 58305730

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    2 months ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    How deep can your recursion be? If it is too deep, it keeps all variables in the stack.

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      2 months ago, # ^ |
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      In the worst case the recursion's depth will be equal to the number of states which is N^4 * 2 which is ~ 5 million. Is it really enough to fill up 256 MB?

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        2 months ago, # ^ |
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        I do not really know, what exactly each call stores. For sure it is your method variables (passed + 2 internal), it is below 15 B though. But there should be something else, I guess.

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2 months ago, # |
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I dont get why a lot of people failed to solve B.

can somebody explain

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    2 months ago, # ^ |
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    Can you explain how you find it to be easy ? Also can you recommend similar problems ?

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      2 months ago, # ^ |
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      the theoretical problems about invariance would help

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        2 months ago, # ^ |
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        Thanks, Any source you aware of ?

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          2 months ago, # ^ |
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          i didn't know any english source.

          but i found THIS LINK for you. and the problems are good.

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            2 months ago, # ^ |
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            Is there an answer key for that link?

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              2 months ago, # ^ |
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              i didn't find any. you can ask here.

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    2 months ago, # ^ |
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    How to solve it ?

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      2 months ago, # ^ |
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      the sum of all elements must be even.(because we decrease the whole sum by 2 after each operation)

      also the biggest element should not be larger than half of the whole sum.(because for every operation that we decrease the biggest element there should be another element bigger than 0 to decrease)

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    2 months ago, # ^ |
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    i forgot that i have to reduce by 1 instead was trying to solve it by reducing full number from other

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    2 months ago, # ^ |
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    They didn't guess that one of elements maybe greater than all another. In this case also there is impossible do it

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    2 months ago, # ^ |
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    First I thought that you can simply take two biggest numbers, and reduce them as much as possible — and then repeat this procedure until all become 0. But this logic does not always work, for example, for the case 99 100 101.

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      2 months ago, # ^ |
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      did this soulution pass the pretest?

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        2 months ago, # ^ |
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        No. And it took me some time to figure it out.

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2 months ago, # |
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one should never overthink B otherwise, it cost him his rating

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    2 months ago, # ^ |
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    True, At the first glance I thought about a partitioning problem and this would have got TLE easily lol

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2 months ago, # |
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This is my last Div2 Round. But I was late in submitting E1 about 10 seconds so I missed the last chance to win in Div2. QAQ so sad....

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    2 months ago, # ^ |
    Rev. 2   Vote: I like it +22 Vote: I do not like it

    I was not sure that my solution is correct and my implementation is correct. So I inserted some assertions in my code. But I accidentally did a wrong assertion. So you can see that I got some REs in my contest submissions. And then I replaced those assertions with exit(0) so I got WA. Then I realized that my assertions was wrong. As soon as I finished to erase them, the contest ended.

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    2 months ago, # ^ |
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    Maybe it's not your last Div2 round :)

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2 months ago, # |
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My rating did not update. Why?

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2 months ago, # |
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How this case could be "YES"? Problem B.

Test case 5 :

5

1000000000 1000000000 1000000000 1000000000 1000000000

Answer YES

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    2 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Just because i dont want do write zeroes, here is explanation for 5 10 10 10 10 10

    9 9 10 10 10 9 9 9 9 10 9 9 9 8 9 9 9 8 8 8 8 8 8 8 8 (and repeat until all are 0)

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    2 months ago, # ^ |
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    take half of 1000000000 and half of the next 1000000000 ,they cancel out one 1000000000 now we left with: 500000000 500000000 1000000000 1000000000 now it's trivial...

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    2 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    make it

    1000000000 1000000000 1000000000 0 0

    then you can make them

    1000000000 999999999 999999999 0 0

    999999999 999999998 999999999 0 0

    999999998 999999998 999999998 0 0

    as you can see ,you can reduce all of them by 2 after 3 operation. so you can make all of them zero as well

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    2 months ago, # ^ |
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    A similar example is {2,2,2,2,2} you can take operations like {1,1,2,2,2} {0,1,1,2,2} {0,1,1,1,1} {0,0,0,1,1} {0,0,0,0,0}

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    2 months ago, # ^ |
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    • 500000000 500000000 1000000000 1000000000 1000000000
    • 500000000 500000000 500000000 500000000 1000000000
    • 500000000 and 1000000000 give 500000000
    • so we get : 500000000 500000000 500000000 500000000
    • now you can conclude the result
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2 months ago, # |
  Vote: I like it -8 Vote: I do not like it

can anyone help me finding error in this code

int main()
  {   
      ios_base::sync_with_stdio(false);
      cin.tie(0);
      cout.tie(0);            
      //std::setprecision(20);    
      int tests=1;
        //freopen("input.txt", "r", stdin);
     //  cin>>tests;
       
     while(tests--) 
      {   
        ll n,k;
        cin>>n>>k;
        ll a[n+1];
        for(int i = 1;i<=n;i++)
           cin>>a[i];
         a[0]  = 0;
        sort(a,a+n+1);
        ll i = (n+1)/2,j;
        j = i;
        ll median = a[i];
        for(j = i;j<n;j++){
          if(a[j+1] > a[j])
          {
            if(k >= (a[j+1]-a[i])*(j-i+1)){
                k = k-(j-i+1)*(a[j+1]-a[j]);
                //cout<<a[j]<<" ";
                median = a[j+1];
            }
            else{
                break;
            }
          }
        }
        if(k){
         // cout<<k;
          ll len = (j-i+1);
          median = median + k/len;
        }
        cout<<median;
 
      }
}
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2 months ago, # |
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What wrong with me? It's second consecutive time I passed pretests and got wrong answer in system tests. It's not wrong approach, it is overflow integer. So sad for my rating.

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2 months ago, # |
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How to solve E? Is it SG function or something else?

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2 months ago, # |
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The code of neal’s problem D has a problem.In its last loop,safe_index might equal to safe.size(),and safe_col has random value.It might lead to “wrong answer”.(Sorry for my poor English)

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

Problem D:Could anybody tell me why I always fail on test 13? After reading the editorial,I found that my idea is correct.Maybe some details!

code:58360651

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

    Check this case: 2 2 1 1 2 1 2 (Ans:3), here the first row does not contain any treasure.

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2 months ago, # |
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In problem div2 C i think like the position of median is not changing right. so sort the array take sum from mid to n add k and divide it with number of element from median position to end that will give the max median . //{sum of(a[position of median to n]+k) divide (n-mid)} but its not working for all case. somebody have any idea