xyz2606's blog

Contest 2050 and Codeforces Round #718 (Div.1 + Div.2)
By xyz2606, 4 years ago, In English

Hello Codeforces!

Contest 2050 and Codeforces Round 718 (Div. 1 + Div. 2) will take place on Apr/23/2021 17:35 (Moscow time). It's rated for everyone! The score distribution is 500--1250--1500--1750--2500--3000--3000--4000. You will have 2 hours and 45 minutes to solve 8 problems.

Contest 2050 is initiated by volunteers of the 2050 Conference. You can find out what is 2050 at the end of this blog.

The contest is prepared by

and we'd like to thank

See you on Friday and good luck!

We have the following introductory message by the 2050 volunteers:

2050 aims to equip young people to take action and to become volunteers. Every 'last weekend' of Apr, together with the youth, we celebrate the technology at Yunqi Town, China. 2050 is a volunteer-only, not-for-profit unconference.

"Most of the conferences in China are held for those who are already successful in their careers, but none targets young people. 2050 wants to fill the gap, giving Chinese youth a stage to make their own voices heard and maximize their energy and talent," said Dr. Wang Jian.

What to expect at 2050?

At 2050, we call it containers. There are 10 containers: 2050 Youth; Migratory Bird Program; Game On; Mindnet; Youth Stage; Starry Night Camping; Everything Grows; Explorer Space; New Gen Forum; Youth Reunion.

Special arrangement of this year

This year, because of the pandemic and travel restriction, many friends from overseas cannot join us in Hangzhou. This however, cannot stop us from connecting with each other. 2050 will be connecting with 100 cities, to dial in volunteers who cannot join us onsite.

The road to championship @ new gen forum

Algorithms and programming are their pleasures of life. With the belief of victory, they defeated the most powerful opponents. They worked hard to explore the boundaries of human intelligence on the road to the ultimate. We invite the top competitive programming players around the world to share their journeys on Saturday at 23:00~24:00 (UTC+8). If you are interested, please message littlelittlehorse.

Find out more at https://2050.org.cn/en/.

UPD. Score distribution 500--1250--1500--1750--2500--3000--3000--4000

UPD Congratulations to the winners

  1. Benq
  2. Um_nik
  3. mnbvmar
  4. Radewoosh
  5. -..

Editorial

Editorial

  • Vote: I like it
  • +532
  • Vote: I do not like it

Comments Comments (185)
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4 years ago, # |
  Vote: I like it +157 Vote: I do not like it

Should we expect 10 problems and there name mentioned above as containers? Also nice to see Retired_MiFaFaOvO after so many days

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4 years ago, # |
  Vote: I like it +2 Vote: I do not like it

Looking forward to become a volunteer at 2050 conference virtually : )

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4 years ago, # |
  Vote: I like it +48 Vote: I do not like it

Enjoying the alternate day contests!

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    4 years ago, # ^ |
      Vote: I like it +2 Vote: I do not like it

    The next contest is in a week. Sad

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      4 years ago, # ^ |
        Vote: I like it +4 Vote: I do not like it

      I think there will be a third division contest before educational round. It's usually been like this.

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        4 years ago, # ^ |
          Vote: I like it +3 Vote: I do not like it

        We have already had a "Div-3" contest on 10th April. So, the possibility is quite negligible.I mean, 2 "Div-3" contest in one month-it will not happen so far I think.

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      4 years ago, # ^ |
        Vote: I like it -37 Vote: I do not like it

      sad plus one

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4 years ago, # |
  Vote: I like it +43 Vote: I do not like it

it's really great when such organizations pay attention to codeforces, and the codeforces community pays attention to them accordingly!

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4 years ago, # |
← Rev. 2   Vote: I like it +5 Vote: I do not like it

honored to participate in this round! have fun!

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4 years ago, # |
← Rev. 2   Vote: I like it +55 Vote: I do not like it

OMG! Chinese Round, I'm so excited!

Edit: Why the downvotes?

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4 years ago, # |
  Vote: I like it -7 Vote: I do not like it

I heard Chinese rounds are generally difficult Is that true?? please correct me if i am wrong

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    4 years ago, # ^ |
      Vote: I like it +57 Vote: I do not like it

    Don't know about difficulty, but they can teach you a lot of stuff : ) Chinese informatics culture is really advanced and highly competitive.....at least I feel so.

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4 years ago, # |
  Vote: I like it -74 Vote: I do not like it

Notice the USUAL Timing !!!

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4 years ago, # |
  Vote: I like it -29 Vote: I do not like it

Length of the contest is not mentioned

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4 years ago, # |
  Vote: I like it -23 Vote: I do not like it

Good luck guys :) , hope every one is blessed with AC's and good rating (positive) changes

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    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it +34 Vote: I do not like it

    Do you know that's not possible (that everyone gets a positive rating change)?

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      4 years ago, # ^ |
      ← Rev. 2   Vote: I like it +29 Vote: I do not like it

      Instead of finding math in this, wish you luck for the contest bro Be Positive : )

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4 years ago, # |
  Vote: I like it +254 Vote: I do not like it

As a tester, I have to say it's a great round with interesting problems. Good luck and have fun!

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4 years ago, # |
  Vote: I like it -7 Vote: I do not like it

Good Luck everyone!!!

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4 years ago, # |
  Vote: I like it +37 Vote: I do not like it

"Migratory Bird Program"

Geese!?!?! You brought Waterloo geese to Hangzhou?!? (last I heard, geese are still a pest in the UK, after a bunch were shipped over as gifts for the Queen's Golden Jubilee...)

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4 years ago, # |
  Vote: I like it +56 Vote: I do not like it

What's up with the Friday Div 1s lately?

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4 years ago, # |
  Vote: I like it +4 Vote: I do not like it

can we expect that the starting problems will be of div 2 level? thanks!

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4 years ago, # |
  Vote: I like it -26 Vote: I do not like it

WLS!! DLS!!

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4 years ago, # |
  Vote: I like it +5 Vote: I do not like it

I am going to face Div1 + Div2 Round first time. Can somebody please send link of past contest of this type? This will help me in understanding difficulty of problems in it and practicing them.

Thankyou!

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4 years ago, # |
  Vote: I like it -96 Vote: I do not like it

As a non-tester I can't confirm that the problem statements are interesting because I haven't seen them yet.

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4 years ago, # |
  Vote: I like it +10 Vote: I do not like it

A new chinese round! I am so excited

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4 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

The length is 2:45, that means 15 minutes shorter than usual global round. It seems it will be hard!

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

a new chinese round!

I hope it will be successfully.

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4 years ago, # |
  Vote: I like it +33 Vote: I do not like it

what about score distribution ?

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4 years ago, # |
  Vote: I like it -250 Vote: I do not like it

Are Chinese Round Pretests as strong as their viruses??!

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4 years ago, # |
← Rev. 2   Vote: I like it +8 Vote: I do not like it

Hope the contest will be held successfully!

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4 years ago, # |
  Vote: I like it -10 Vote: I do not like it

How jqdai0815 not in top list rated users?!!

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4 years ago, # |
  Vote: I like it +4 Vote: I do not like it

Scoring distribution?

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4 years ago, # |
  Vote: I like it +19 Vote: I do not like it

There is a noticable difference between problem "A"(500) and "B"(1250).It will create a huge effect on the standings.

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    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it -50 Vote: I do not like it

    There is. A you will solve 15 mins before contest ends with 30 negative attempts and B you won't lol

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4 years ago, # |
  Vote: I like it +9 Vote: I do not like it

Good luck everyone!! Hope I will reach expert today :)

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4 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Hope this contest bring +ve rating change. Best of Luck to everyone.

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4 years ago, # |
← Rev. 2   Vote: I like it +8 Vote: I do not like it

i feel there could be 1 more problem between 1000- 1500 in this contest.

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Why is 2050 called 2050?

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Rating has taken a major dip. Hopefully, I will be able to recover this round.

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    4 years ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    good luck bro!!!

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      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      your submission heatmap is quite an inspiration. even I used to solve a lot of problems last year in quarantine but it is nowhere close to yours.

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

reminding everyone to drink water during the contest and stay hydrated : )

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    i remember a similar page on insta, reminding everyone to stop slouching while reading the comments of a post xd.

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4 years ago, # |
  Vote: I like it +5 Vote: I do not like it

My Only Dream In Life Is To Reach Pupil :)

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4 years ago, # |
  Vote: I like it +62 Vote: I do not like it

Guess people can have an early sleep after solving D.

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4 years ago, # |
  Vote: I like it +34 Vote: I do not like it

GridForces

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4 years ago, # |
← Rev. 2   Vote: I like it -7 Vote: I do not like it

Nice round!

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4 years ago, # |
  Vote: I like it +9 Vote: I do not like it

DEBUGS !!! kill me

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4 years ago, # |
  Vote: I like it +33 Vote: I do not like it

Implementationforces

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    4 years ago, # ^ |
      Vote: I like it +7 Vote: I do not like it

    I also felt so after wasting a lot of time on B but realised later That you can implement it without much efforts.

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4 years ago, # |
  Vote: I like it +29 Vote: I do not like it

I think very poor description for Problem C-Fillomino 2.

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4 years ago, # |
  Vote: I like it -7 Vote: I do not like it

Passed all pretests of D. Only 3 Minutes Left !!!

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4 years ago, # |
  Vote: I like it +5 Vote: I do not like it

Explanation of C and Input of D is Tricky to Understand.

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Please check this submission of C
https://codeforces.me/contest/1517/submission/114039804
How to use DP or optimize it ?

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    4 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    I did it using a version of DFS. Start visiting nodes from diagonal element. Continue visiting to the left node and push it to stack. When left is not feasible start visiting all the down elements

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      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Oh Thanks Feeling really bad that couldn't get this greedy approach during the contest :(

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

what was wrong in these solutions can anybody tell me
113994955 and 113989915

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

LongReadForces... Why would write such confusing statements???? Did too much math and don't know how to speak like human anymore?

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4 years ago, # |
  Vote: I like it +25 Vote: I do not like it

How to solve E?

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4 years ago, # |
  Vote: I like it +2 Vote: I do not like it

How to solve D

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    4 years ago, # ^ |
    ← Rev. 9   Vote: I like it +12 Vote: I do not like it

    First of all k needs to be even, else you can't come back to the original position. Consider a 3D dp -> $$$dp[iter][i][j]$$$ , where $$$dp[iter][i][j]$$$ represents the minimum cost to reach $$$(i,j)$$$ after $$$iter$$$ steps. Now, it can be built bottom up.

    1. Base case: $$$dp[0][i][j]=0$$$ .
    2. Now, minimum cost to reach $$$(i,j)$$$ after iter steps, is the same as the cost to reach one of it's neighbours in $$$iter-1$$$ steps and then going to $$$(i,j)$$$. $$$dp[iter][i][j]$$$ is minimum of
    • $$$dp[iter-1][i][j+1]$$$ + cost of going left
    • $$$dp[iter-1][i][j-1]$$$ + cost of going right
    • $$$dp[iter-1][i+1][j]$$$ + cost of going down
    • $$$dp[iter-1][i-1][j]$$$ + cost of going up


    You can perform this for $$$k/2$$$ steps (The path can be going on this $$$k/2$$$ step path twice). and cost for each would be $$$2*dp[k/2][i][j]$$$ .

    114025728

    In the above solution, you can optimize

    • the space by using two 2D dps.
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4 years ago, # |
  Vote: I like it +11 Vote: I do not like it

I think the English problem statements were not very clear.

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4 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Does anyone know a test case for problem C which will give answer -1

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

In F, can any of the volunteers be the centre?

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Holy shit I forgot to mod the answer in E....I feel so stupid rn ugh. The answer is around O(N^2) at most anyways, right? Why was mod even required?

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4 years ago, # |
  Vote: I like it +35 Vote: I do not like it

The matrix inputs are a bit confusing :/

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I would like to report the user deepanshupathak03 . During the contest he uploaded a video of the solution to the first problem . Link. In the video in the upper right corner you can clearly see his user name

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    4 years ago, # ^ |
      Vote: I like it +2 Vote: I do not like it

    How did you find it? Did you search for the problem during contest? ;)

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      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      No after the contest I was searching for video solutions.That's when I found it. Also you can see thatI submitted the solution to A before the video was released.

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4 years ago, # |
  Vote: I like it +81 Vote: I do not like it

Annoying Forces

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4 years ago, # |
  Vote: I like it +26 Vote: I do not like it

Solved 4 problems, logic was easy, but were difficult to implement. I wasn't expecting this from 2050 :(

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4 years ago, # |
  Vote: I like it +108 Vote: I do not like it

Problem C be like

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    4 years ago, # ^ |
      Vote: I like it +10 Vote: I do not like it

    Tell that to the guys that kept down unless overtaking :P

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4 years ago, # |
  Vote: I like it +3 Vote: I do not like it

We somehow survived 2020, and you think about 2050, isn't it too early to make a guess

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Is it possible to solve D using priority queue ??

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    4 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Without dp the complexity would be exponential afaik.

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

GridForces :))

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4 years ago, # |
← Rev. 2   Vote: I like it +31 Vote: I do not like it

-83 hack attempt -_- Maybe it's the maximum failed attempt in one contest.

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    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it +1 Vote: I do not like it

    Not to state the obvious or anything but have you seen the attempted ‘hacks’? The guy was very clearly doing it on purpose to sabotage his own rating. Don’t ask me why something would do that — I have no idea — but that’s what he was doing.

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I have no idea how my C passed, just kept greedily filling down left down....maybe it wont survive the system tests.

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

What a wonderful contest!! Btw how to solve D? I had some $$$O(n^2*k^2)$$$ idea but couldn't implement due to less time.

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    4 years ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    DP, Let dp[i][j][k] be the min-cost to travel exact $$$k$$$ step from point $$$(i, j)$$$. time complex: $$$O(n^2 * k)$$$ since you have at most four directions to go.

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      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Well,... how to calculate dp[i][j][k]?

      I mean, it is min of all dp[i][j][k-1] cells plus the costs of entering the adjacent cells of it. How to maintain them?

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        4 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        You can see my implement 114022406, I use std::vector<std::tuple<int, int, int>> dir[N][N]; and dir[i][j][d] = {x, xi, xj} means cell $$$(i, j)$$$ is connected with cell $$$(xi, xj)$$$ with cost x.

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        4 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Ah...I think I got it. It is not the min of all cells whos start k-1 dist away and ending in i,j.

        It is instead the paths of length k-1 from the 4 adjacent cells.

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It would probably have TLEd. You can do it with $$$O(nmk)$$$ DP.

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Well the logic I got - Vertices at distance x from (i,j) gives a diamond shaped closed path. For vertex at distance x, you can use dp from the neighbor vertices at distance x-1. Giving a solution of O(k^2). But couldn't implement due to less time

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4 years ago, # |
  Vote: I like it +5 Vote: I do not like it

I know how to deal E in $$$O(n)$$$, but my poor implement said no.

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4 years ago, # |
  Vote: I like it +21 Vote: I do not like it

the statement was difficult and annoying

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I just spent a lot of time trying to find where the error was for my submission of B , (I should have gone to C a lot earlier lesson learnt.) please help find it submission. My logic was to find the minimum lengths of all paths and distribute the top m to m athletes , (in the row where they are initially present) ,the remaining need not be changed.

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

can someone give hint for B

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    4 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    can minimum m values of all the n*m values be tiredness of m runners ?

    we know for sure that if above is possible, there is no better case.

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

I solved D but two bugs ruined my time. I learned two things:
1st : When you copy a code that is to be duplicated for multiple times for some identical cases, check twice whether you are missing any change that is done to the copied code.
2nd : When you are going to set a variable for a holding a minimum value, initialize it to the (maximum_value + 1) that it can take.
I know this are very basic and natural things, but this things ruined me over. Always basic and natural things are not visible to eyes.
Anyways good contest. Kudos to the team.

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4 years ago, # |
  Vote: I like it +2 Vote: I do not like it

How does the dp work in D? I was not able to find it.

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    4 years ago, # ^ |
    ← Rev. 3   Vote: I like it 0 Vote: I do not like it

    Judging from the other solutions, it seems that I overcomplicated it. But here was my thought process:

    The entire grid = -1 if k is odd (because for every step forward we take from a cell, in order to eventually come back we need a corresponding step backwards, and so this forces k to be even in order to be able to do this).

    Otherwise, You can recognize that the answer is always some "cycle" from the starting position (AKA, we start from a cell (i,j) to some other cell (i', j'), and we come backwards. We repeat this until we get K steps exactly). I think this is optimal because the path towards any cell and back has to be the same path, otherwise there was a better one and we could've taken that one forward and back. This gives the implication that we only care about the min cost of "cycles" of lengths that are divisors of k / 2 (apparently it is sufficient to check for k / 2, but i didn't make this observation EDIT: I am pretty sure this is the case since if any cycle of a smaller length was better, then it would have been included in paths of length k / 2, since they must divide it).

    So the answer for any cell is min(dp[i][j][d] * (k / d)) where d is some divisor of k / 2 and the dp[i][j][d] is the min cost of a path that is d away from (i, j). You can compute this by the transition that we either go left, down, right, up from any cell and this is a state forward.

    pseudo transition: dp[go in some direction][steps + 1] = min(that state, dp[here] + weight of edge) code: https://codeforces.me/contest/1517/submission/114033176

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4 years ago, # |
  Vote: I like it +13 Vote: I do not like it

This contest made me remember that I am bad at implementation :/

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4 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Editorials please

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Appeal : Start showing tests before system testing ends.

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4 years ago, # |
  Vote: I like it +11 Vote: I do not like it

Thanks for the contest! B was nice! So was D, but apparently I have the big stupid.

I also spent way too much time verifying I put in the right amount of zeroes in A. Really should've written 2050 * (ll) 1e15 instead...

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4 years ago, # |
← Rev. 3   Vote: I like it +9 Vote: I do not like it

ReadingForces
ImplementationForces

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4 years ago, # |
  Vote: I like it +6 Vote: I do not like it

finally pupil :'(

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4 years ago, # |
  Vote: I like it +15 Vote: I do not like it

BYE BYE PURPLE)))))))

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4 years ago, # |
  Vote: I like it -16 Vote: I do not like it

I can't figure out E... too many situations and so complex to implement

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4 years ago, # |
  Vote: I like it +10 Vote: I do not like it

Round is cool!

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can we not just calculate all possible path of length fc, where fc is a factor of k, from a cell (i, j) and then consider the min cost path? Wouldn't that give an optimum answer? Submission — https://codeforces.me/contest/1517/submission/114040509

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Waiting for Editorial...

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4 years ago, # |
  Vote: I like it +146 Vote: I do not like it

Let me leave some complaints about constraints/TLs.

D: I saw/heard many $$$O(n m k^3)$$$ passed or was on the border. Assuming the intended solution is $$$O(n m k)$$$, I think this is very bad (why such small $$$k$$$). If $$$O(n m k^3)$$$ is intended, then it is also bad as it was then a constant-factor-optimization contest.

E: I was so lucky that mine passed, but the problem has clearly a messy cases-analysis solution, so I don't welcome 1 sec for this. I personally think the interesting part of this problem lies in the observation part and $$$O(n^2)$$$ -> $$$O(n \log n)$$$ part is boring...

G: Assuming the intended solution uses network flow (I haven't got the details yet, but looks interesting), I am very curious about the time complexity and how you determined the constraints.

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    In Problem E can you help me figure out why CPPCP is not a valid case for sample 1.

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      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      You have 1st and 2nd occurences of $$$P$$$ at distance of $$$1$$$ and 2nd and 3rd occurence of $$$P$$$ have distance $$$2$$$, which is invalid.

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4 years ago, # |
  Vote: I like it +45 Vote: I do not like it

I found the problems very interesting, thank you for the round!

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

It really sucks not being able to solve Div2 D’s consistently even after practicising a lot.

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    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it +8 Vote: I do not like it 

    Hindi
असफलता एक चुनौती है, स्वीकार करो
क्या कमी रह गई, देखो और सुधार करो
जब तक न सफल हो, नींद चैन को त्यागो तुम
संघर्ष का मैदान छोड़ मत भागो तुम
कुछ किए बिना ही जय जय कार नहीं होती
कोशिश करने वालों की हार नहीं होती

English Translation

Failure is a challenge accept it
What is missing, look and improve
Until you succeed, give up sleep
Do not run away from the battlefield
You don't earn praises without doing anything
Those who try never fail.
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4 years ago, # |
  Vote: I like it -8 Vote: I do not like it

Failed D in system test because I put an n instead of m... now that's a dumb systest fail. Oh well, very easy ABCD.

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4 years ago, # |
  Vote: I like it +21 Vote: I do not like it

Congrats Benq for earning the Top Spot in Rating Rankings in advance... Wooooooah!!!

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4 years ago, # |
  Vote: I like it +48 Vote: I do not like it

To not keep you waiting, the ratings updated preliminarily. In a few hours, I will remove cheaters and update the ratings again!

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    4 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    Please do this always, first update then remove cheaters.

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    4 years ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    Do this in Educational rounds also :)

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Would it make sense to have an additional "Pending cheaters removal" contest leaderboard status before labelling it as "Final standings"?

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      4 years ago, # ^ |
        Vote: I like it +3 Vote: I do not like it

      I think it is a very good idea to have some temporary status for that

      Something like "Preliminary results"

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4 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Implementationforces? Problems were nice, but it took me a lot of time to implement.

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can we solve D using Dijkstra? https://codeforces.me/contest/1517/submission/114047555 (What am I doing wrong here?)

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Why am I getting "wrong answer You answer is larger than expected. (test case 1)".I have printed n*m values only. Submission

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4 years ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

multiset<pair<int,pair<int,int>>> s............ suppose I want to erase or find {1,{2,6}} and it is present in the set.........how can I erase or find a particular nested pairs in set.......

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    auto it = st.find(the object) and then you can erase it if its not st.end()

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Unrelated but why not use tuple<int, int, int> instead of pair<int, pair<int, int>>

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4 years ago, # |
  Vote: I like it +86 Vote: I do not like it

Weak pretest for E:every $$$a_i$$$ in pretests is a big number.

So if you read the condition wrong like this:

$$$\sum\limits_{x\in C}a_x\leq\sum\limits_{y\in P}a_y$$$

it passed the pretest and fst.So sad......

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Yesterday I recieved notification of recieving 379 rating and now my profile is showing I'm unrated. Why?????

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    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It will show up again, the rating is being recalculated after cheaters removal. :)

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4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Ratings return when?

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4 years ago, # |
  Vote: I like it +9 Vote: I do not like it

kingmanas stop messaging me again and again!! Saale gaandu ab aur msg mt kar mujhe tere chutiye msgs dekh dekhkr pakk gaya hun pareshan karna chod de Main ladki nahi hun chutiye jo ladki smjhkr peeche pda hua hai tu mere!! Ladka hun main aur tu nahi jaanta mujhe main kaun hun tere batch ka bhi nahi hun! Jab tereko maalum bhi hai main kaun hun chutiye toh kyun mujhe apni bakchodi sunaye ja raha hai itne mahine se?? Tere baap ka account ka hai kya jo baar baar msg kr raha hai mujhe ki main use na karun? Apna kaam kar na gaandu

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