vovuh's blog

By vovuh, history, 7 months ago, In English

Notice the unusual start time.

Hello!

Codeforces Round #674 (Div. 3) will start at Sep/28/2020 11:05 (Moscow time). You will be offered 6 or 7 problems (or 8) with expected difficulties to compose an interesting competition for participants with ratings up to 1600. However, all of you who wish to take part and have rating 1600 or higher, can register for the round unofficially.

This round consists mostly of the problems of the first stage of All-Russian Olympiad of School Students in Saratov and will be hosted during the real competition time. The problems were invented and prepared by Ivan BledDest Androsov, Alexander fcspartakm Frolov and me.

The round will be hosted by rules of educational rounds (extended ACM-ICPC). Thus, during the round, solutions will be judged on preliminary tests, and after the round it will be a 12-hour phase of open hacks. I tried to make strong tests — just like you will be upset if many solutions fail after the contest is over.

You will be given 6 or 7 (or 8) problems and 2 hours to solve them.

Note that the penalty for the wrong submission in this round (and the following Div. 3 rounds) is 10 minutes.

Remember that only the trusted participants of the third division will be included in the official standings table. As it is written by link, this is a compulsory measure for combating unsporting behavior. To qualify as a trusted participants of the third division, you must:

  • take part in at least two rated rounds (and solve at least one problem in each of them),
  • do not have a point of 1900 or higher in the rating.

Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you.

Thanks to MikeMirzayanov for the platform, help with ideas for problems and for coordination of my work. Thanks to my good friends Daria nooinenoojno Stepanova, Mikhail awoo Piklyaev, Maksim Neon Mescheryakov and Ivan BledDest Androsov for help in round preparation and testing the round. Also thanks to Artem Rox Plotkin and Dmitrii _overrated_ Umnov for the discussion of ideas and testing the round!

Good luck!

UPD: Thanks to Ivan MrReDoX Ushakov, Ivan Ivan19981305 Georgiev and Dmitry nuipojaluista Kadomtsev for testing the round!

UPD2: Editorial is published!

 
 
 
 
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7 months ago, # |
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*Notice the unusual start time.

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7 months ago, # |
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I think vovuh should update the blog mentioning unusual start time.

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7 months ago, # |
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weird time tho

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7 months ago, # |
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Sorry cannot participate because of the practical class at the same time

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Me Also. xD :) (SAVE_20200927_1219521f486f74dbb24105.jpg)

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7 months ago, # |
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Why so unusual start time ?? :(

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    7 months ago, # ^ |
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    "This round consists mostly of the problems of the first stage of All-Russian Olympiad of School Students in Saratov and will be hosted during the real competition time."

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      7 months ago, # ^ |
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      I See, its going to be fun then!!!!!!!!!

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    7 months ago, # ^ |
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    yeah...i think all contests should be held usual time:)

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7 months ago, # |
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So unusual time. Its lunchtime here in India.

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7 months ago, # |
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All-Russian Olympiad alright imma skip this one

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7 months ago, # |
  Vote: I like it +42 Vote: I do not like it

Me after remembering the previous div2 contest based on russian Olympiad for school students: i am gonna skip this because of unusual time.

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    7 months ago, # ^ |
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    Its Div-3 bro, what do you got to lose, its unrated for you

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7 months ago, # |
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Bruh this contest is suit for me but i think i won't participate(because of the unusual start time)

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[]

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7 months ago, # |
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really wanted to participate but wont be able to due to my online classes.hope for some magic

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7 months ago, # |
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Really excited for this contest, but it's unusual time is not suitable for most of the college students in India as we have online lecture at that time.

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7 months ago, # |
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submit F in last 5 seconds and find my output is following:

YES
1 1 1 1
1 1 1 3
4 2 3 1
RDLL
DRDD
ULUU
NO

instead of

YES
1 1 1 1
1 1 1 3
4 2 3 1
R D L L
D R D D
U L U U
NO

Hope my solution is wrong XD

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    7 months ago, # ^ |
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    Round is not even started yet, but u already submitted, Ur coding speed is so enormous!

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      7 months ago, # ^ |
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      I miss the golden time in which I can delete the misplaced comment. T_T

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    7 months ago, # ^ |
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    well looks like your wish came true xD

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7 months ago, # |
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4 AM for me then.

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Deleted for posting at a wrong place.

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7 months ago, # |
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Bruh ,you ain't getting me this time .

I still remember that russian 6th grade student's round . This one is for all russian student olympiad. Even more tougher.

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7 months ago, # |
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BledDest and vovuh are gem problemsetter

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    7 months ago, # ^ |
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    are ratings updated?

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      7 months ago, # ^ |
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      not yet !

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        7 months ago, # ^ |
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        oh fine.Normally how long does it take for updation? I am new here so I don't know much

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          7 months ago, # ^ |
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          it may very contest to contest , we can say anything that in that amount time it will update, usually in div 3 it take more time then div 2. hope it help you . all the best keep coading bro

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7 months ago, # |
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should i eat lunch or give this contest?

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7 months ago, # |
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I'll be eating and crying at the contest time as it's lunchtime and I suck at DIV 3://

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7 months ago, # |
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Gonna wake up at 4 am(my local time) for this contest! Very excited!

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7 months ago, # |
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Your text to link here...

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7 months ago, # |
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Time too bad. Have classes. Any after 3pm UTC works fine.

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really weird time !

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Every time I want to improve my CP skills, my college is letting me down and this is another example.

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7 months ago, # |
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It would be pretty good if CF were in this time rectangle in the future.

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7 months ago, # |
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Another suitable time for Chinese......

But "based on" makes me afraid!

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    7 months ago, # ^ |
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    I can finally have a CF competition and have not to sleep at 1:00 in China :)

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Oops, sorry It was the wrong post :((

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Spoiler
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vovuh score distribution plzz..

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    7 months ago, # ^ |
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    There are no score distributions in Div.3

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7 months ago, # |
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IN INDIA IT's OUR ONLINE CLASS TIME

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    7 months ago, # ^ |
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    It's impossible to conduct a contest such that time will be suitable for everyone across the world.

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7 months ago, # |
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This is the first time a div3 round is not rated for me!

idk i just want to share this

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7 months ago, # |
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No as a tester, I would say sth comment

Wierrrrrd!

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7 months ago, # |
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After a long time, a rated div3 for me :P.

glhf

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2olj4n.jpg

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haha .... i had to choose between this and online class :(

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7 months ago, # |
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After the contest ends, how to solve D?

How to not double count subarrays with zero sum?

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How to solve F?

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    7 months ago, # ^ |
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    we want to count the total number of abc, ab?, ?b?, ???, ??c, a?c, a??, ?bc subsequences occurring in all the strings, we can replace ? with the character that we need to make it abc. I've named variables for better understanding. In my code, k = count of ? so far submission

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    7 months ago, # ^ |
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    started from back of string and kept the count of subsequences of type abc,bc and c

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how to solve c?

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    7 months ago, # ^ |
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    binary search

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    7 months ago, # ^ |
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    For this, I got a pattern of the required number of moves like 0, 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, ......

    Hope it will be helpful.

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      7 months ago, # ^ |
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      Took time for figuring that out. Found editorialist solution more easy btw.

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    7 months ago, # ^ |
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    I found x*(x+1)>=n using binary search but it wasn't correct for last two sample test cases and then I found x*x>=n as well as x*(x+1)>=n separately and took minimum of both, it got accepted.

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      7 months ago, # ^ |
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      but why you have to find x*(x+1) in the first place?

      thanks

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    7 months ago, # ^ |
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    Suppose we perform the +1 operation $$$x$$$ times and the duplicate operation $$$y$$$ times. Clearly we want to perform all the $$$x$$$ operations first, so that each subsequent $$$y$$$ operation adds as much as possible. Then the sum of the array will be $$$(1+x)+y(1+x)=(1+x)(1+y)$$$. We want $$$(1+x)(1+y) \ge n$$$, so each of $$$(1+x)$$$ and $$$(1+y)$$$ is either $$$\lfloor \sqrt(n) \rfloor$$$ or $$$\lfloor \sqrt(n) \rfloor+1$$$. Then we can just try the cases.

    94077437

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      7 months ago, # ^ |
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      so each of (1+x) and (1+y) is either ⌊(√n)⌋ or ⌊(√n)⌋+1. Then we can just try the cases.

      i didnt get the last part how this is possible?

      and why the solution does not go beyond the root n thanks

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        7 months ago, # ^ |
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        If we want to minimize $$$a+b$$$ and want $$$ab\ge n$$$, then we want $$$a$$$ and $$$b$$$ to be as close to each other as possible (just try a few values... it's very intuitive). That means each of $$$a,b$$$ must be around $$$\sqrt{n}$$$. In this problem, $$$a=1+x$$$ and $$$b=1+y$$$.

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          7 months ago, # ^ |
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          ohh ok got it thanks

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          7 months ago, # ^ |
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          We can apply AM>=GM 
          
          ((x+1)+(y+1))/2 >=sqrt((x+1)(y+1))
          
          ((x+1)+(y+1))/2 >=sqrt(n)
          
          ((x+1)+(y+1)) >=2*sqrt(n)
          x+y >= 2*sqrt(n)-2    -----for n not perfect square
          x+y >= 2*sqrt(n)-1    -----for n perfect square  
          
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    7 months ago, # ^ |
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    it is better to increase 1 up to some number let say x and then append this x up to when the array sum becomes equals or greater than target value n

    here x should be the floor(square root of n) or 1 plus to this value

    here is mine solution 94092529

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    7 months ago, # ^ |
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    7 months ago, # ^ |
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    I pre calculated maximum possible sum for i operations (1 <= i <= 63244). To maximise, wee can use floor(i/2) operations of type 1 and ceil(i/2) of type 2.

    When the values are precalculated, we can just use the lower_bound to find the index for a particular n.

    https://codeforces.com/contest/1426/submission/94156182

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7 months ago, # |
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what was the testcase no 10 of problem D??

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Can some one give me idea about B & D

thanks in advance

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    7 months ago, # ^ |
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    For B you just have to input 4 nums n times , let's say x,y,a,b and check (y==a && m%2==0) since it is not possible to build a 2X2 tile in odd m , and for symmetric if you have y==a you can just use only that tile many times to build a matrix.

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      7 months ago, # ^ |
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      OPS i was checked for (x==d&&y==a)

      that's why i got wrong verdict It's my bad luck

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    7 months ago, # ^ |
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    for B: if any 2x2 tile has diagonal element same then answer will be yes, keep in mind that tile length and width is 2 so the the resultant square will be multiple of two

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    7 months ago, # ^ |
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    In your B, a doesn't have to be equal to d, just b should be equal to c and m should be even.

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    7 months ago, # ^ |
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    For B,

    let say matrix is

    A B

    C D

    If m is odd, answer will be NO

    For even m If for any matrix B == C then answer will be YES else NO

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7 months ago, # |
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Problem F is exactly the same as ABC104 Problem D

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How to solve D neatly ? I can think of one way where we count the disjoint subarrays with zero sum.

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    7 months ago, # ^ |
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    Think of it as you have, say $$$k$$$ intervals (imagine as segments on the $$$x$$$-axis, from $$$interval[i][0]$$$ to $$$interval[i][1]-1$$$) and each interval has elements that sums to $$$0$$$. Now basically, you need to find the total number of vertical lines to make, such that every segment is cut at least once. Implementation

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How to solve D? went upto prefix sums and thought if prefixsum[i] is 0 or has already occurred then add 1 to the answer but it doesn't seem to work because segments can collide. Can anybody explain a neat approach?

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    7 months ago, # ^ |
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    i counted all the segment that collides in one then also ans was coming wrong on testcase 10 anyone can explain

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      7 months ago, # ^ |
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      • If you find subarray with sum zero ending at index i then you insert very big number between index i-1 and i.
      • Now you don't need to care about elements to the left of index i.
      inline void solve()
      {
      	ll n, i, s = 0, ans = 0;
      	cin >> n;
       
      	ll a[n];
      	set<ll> st;
       
      	for (i = 0; i < n; i++)
      		cin >> a[i];
       
      	for (i = 0; i < n; i++)
      	{
      		s += a[i];
      		if (st.find(s) != st.end() || s == 0)  //there is case of zero subarray sum
      		{                                      //So we increase our ans and
      			ans++;                         //clear set and  make sum = a[i]
      			st.clear();                   
      			s = a[i];
      		}
      		st.insert(s);
      	}
       
      	cout << ans << endl;
       
      }
      
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    7 months ago, # ^ |
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    went up to prefix sums and thought if prefixsum[i] is 0 or has already occurred then add 1 to the answer — it's perfectly okay till now then you just need to clear your container and do everything that you need by assuming the current position of the array as the starting position. Hopefully, it will do your job.

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    7 months ago, # ^ |
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    Think of it as you have, say $$$k$$$ intervals (imagine as segments on the $$$x$$$-axis, from $$$interval[i][0]$$$ to $$$interval[i][1]-1$$$) and each interval has elements that sums to $$$0$$$. Now basically, you need to find the total number of vertical lines to make, such that every segment is cut at least once. Implementation

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The number of participant's are not so many for the unusual start time.

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Since the contest time is unusual,I bunked my class to participate in the contest.The Dashboard was quite interesting and I enjoyed the contest very much.

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thanks vovuh for fast editorial ^.^

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Wow,The editorial is ready!!!

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Just when I was submitting the D problem ,my electricity supply went off,gave the contest from phone...poor me,but problems were superb and am glad I participated..Thanks vovuh

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7 months ago, # |
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Can someone help with with this submission? I could not think of a testcase that fails. 94110488

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    7 months ago, # ^ |
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    your code will fail for when n=1 and X =1 Cuz (1-2)/1=-1 So else part will give -1+1=0

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Did codeforces recently changed the rating formula?

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    7 months ago, # ^ |
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    I dont know but maybe this might be the reason...the number of participants is less hence you may see less change in delta even if you have good rank.

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      7 months ago, # ^ |
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      Yeah, I thought that I was really hoping to touch expert but nevermind I try next round.

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        7 months ago, # ^ |
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        Yes and I wanted to be specialist..sed..next round.

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        7 months ago, # ^ |
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        I followed you, and I think if you are not hacked, you will turn blue.

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        7 months ago, # ^ |
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        i'm also very close to blue :(

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Excuse me! Why is E like A in terms of difficulty?

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    7 months ago, # ^ |
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    The fact that it is E has psychological effect hence the number of solves is almost 10 times less then A..despite the difficulty level.

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      7 months ago, # ^ |
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      Number of solves is 10 times less mainly because there are 3 more questions between A and E which people attempt, and mostly move ahead only when they are done with those.

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    7 months ago, # ^ |
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    There is a general thing. People can solve A because its A, people cant solve E even if its easy cause thats E

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There is a question: I did not participate in this round because of something. If I hack someone, will I be rated?

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I started the contest half an hour late... yet managed to solve A,B,C,D ! Today was a good day for me!

Thanks vovuh for the awesome round :)

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    7 months ago, # ^ |
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    Why good day? You managed to get lot of penalties with better performance due to be late

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      7 months ago, # ^ |
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      Solving 4 problems in 1.5 hrs was a boost in confidence for me! I think that matters more than the penalties :)

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Today's Div. 3 (Round #674):

Spoiler alert for those who are thinking to give contest (virtually)
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When will rating will be changed

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    7 months ago, # ^ |
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    Rating for Div3 and Educational rounds(Div2) will be updated 1-2 hours after the open hacking phase is completed.

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The advantages of being a Chinese have been revealed :) The start time is 16:05 for me.

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Is there any benefit for successful hacking, in this round?

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Why don't we go beyond sqrt(n) in problem C ?

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7 months ago, # |
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is there any point for hacking in div3?

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Can anyone tell me on what test case my solution for B fails? It is hacked. Here is my submission.

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    7 months ago, # ^ |
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    1

    2 2

    3 5

    6 4

    3 6

    5 4

    Your code print YES. But the ans is NO

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Can someone explain how hacks work for ICPC-style rounds like this one? Is there any competitive reward for successful hacks?

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Hands down the best contest I have participated in so far!

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Can anyone explain how this works:

Note that the penalty for the wrong submission in this round (and the following Div. 3 rounds) is 10 minutes.

I see xin_chen, CupCapCup and midheight at places [3, 4 and 5] accordingly. But that doesn't seam right: Their places are increasing but their (finish time + #WA * penalty) is decreasing >> [ (55 + 4 * 10) == 95 , (70 + 2 * 10) = 90 , (62 + 10) = 72] Plus, their number of WA decreasing: [4, 2, 1] but Penalty is increasing: 188, 193, 199 — it could make sense if their finish time was increasing, but it's not the case.

How to make any sense of it? Thank you in advance.

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

    The penalty of xin_chen is: 2 + 9 + 13 + 27 + 42 + 55 (number of minutes since the start until the task is solved) + 4 * 10 (for WA) = 188

    Hope that will make it clearer

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      7 months ago, # ^ |
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      Wow, Super! Thank you, that made everything clear.

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7 months ago, # |
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Why my rating is still not updated?

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7 months ago, # |
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waiting for rating changes ...

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7 months ago, # |
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Will it not be rated for untrusted participants? As my ratings haven't changed yet, being < 1600 and already 5+ hours have passed after the end of the hacking phase.

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7 months ago, # |
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As I am new to Codeforces so can anyone tell me that in this contest will I get any rating or not because I have solved the 3 questions but and also got my rank but not any rating and even in my contest section it is not showing anything.

please help me out?

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7 months ago, # |
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Generally within 12 hours after completion of hack phase, rating change occurs

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7 months ago, # |
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when will the ratings change...how long does it generally take??.

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7 months ago, # |
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I solved 3 problems and I hope that soon change of my rating class come up to me from newbie to pupil!!

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7 months ago, # |
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When are the ratings going to update for this round?

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7 months ago, # |
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Why hasn't the rating updated after the contest? it's almost 24 hours now...

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

When the ratings will be updated?

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7 months ago, # |
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Was this Contest unrated ?

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7 months ago, # |
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Solved 3 for the first time and I get a 0 rating change

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    7 months ago, # ^ |
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    I also face the same issue, I had solved three problems in less time. I think the rating should be increased but the rating is decrease for the same. why?

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7 months ago, # |
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I had no rating change....when will it happen?

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7 months ago, # |
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For whatever reason, the final updated rank is significantly higher than what it was initially after the contest. Generally as far as I've seen, the rank drops, as the ones out of competition are taken off the list. Is this peculiar only to me or am I not seeing something?

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7 months ago, # |
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I'm asking again, please can someone tell why don't we proceed beyond sqrt(n)+5 in problem C. Is there any mathematical proof for it? Thanks in advance!

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7 months ago, # |
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Hi everyone! Can someone please help me with this solution from ecnerwala. How did he find max-flow in non winning edges in the code here — https://codeforces.com/contest/1426/submission/94072786

How did he arrive at that one liner solution? Just a little insight will be very helpful.

Thanks a lot