flaviu2001's blog

Codeforces Round #676 (Div. 2)
By flaviu2001, history, 4 years ago, In English

Hi Codeforces!

stefdasca, koala_bear00 and I are very excited to announce our first contest Codeforces Round #676, which will take place Oct/18/2020 12:05 (Moscow time). The round will be rated for participants with rating up to 2099.

The tasks were written by me with help from stefdasca and koala_bear00 and you have to help some of the authors' favorite musical artists to solve the problems they're faced with.

We hope we compiled a very interesting contest with memorable tasks :)

Special thanks to:

You will be given 2 hours to solve 5 problems, good luck everyone and have fun!

UPD 1: After the round you can watch videos explaining the solutions to the tasks on stefdasca's Youtube channel.

UPD 2: The round was rescheduled, because of intersection with other scheduled contests.

UPD 3: The scoring distribution is standard 500 — 1000 — 1500 — 2000 — 2500.

UPD 4: The editorial was posted and you can check stefdasca's video solutions aswell.

UPD 5: The round is finished, we are glad everything went smooth and hope you enjoyed our tasks!

Div1 winners (unofficial):

  1. I_love_Tanya_Romanova
  2. LJC00118
  3. 244mhq
  4. LayCurse
  5. uwi

Div2 winners:

  1. RGB_ICPC2
  2. _ztyqwq
  3. AnakMbah1937
  4. SakuraiMomoka
  5. asdsasd

Congratulations to those above and to everyone for participating!

  • Vote: I like it
  • +564
  • Vote: I do not like it

Comments Comments (183)
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4 years ago, # |
  Vote: I like it +30 Vote: I do not like it

Auto comment: topic has been updated by flaviu2001 (previous revision, new revision, compare).

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

i lost the rated contest again :weary:

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    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it +47 Vote: I do not like it

    look on the bright side, you participated in another unrated contest ^o^

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4 years ago, # |
  Vote: I like it +224 Vote: I do not like it

As a tester, you know ... what I want :)

PS
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4 years ago, # |
  Vote: I like it +79 Vote: I do not like it

As a tester, I would like to say "Good Luck everyone. Have Fun!!"

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4 years ago, # |
  Vote: I like it +137 Vote: I do not like it

As a tester, I would like to say that I haven't seen the problem yet.

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4 years ago, # |
  Vote: I like it +17 Vote: I do not like it

Notice the unusual time !
Good luck for your maiden contest !!

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4 years ago, # |
← Rev. 2   Vote: I like it -235 Vote: I do not like it

The blog is written $$$4$$$ days before the contest -> bad contest

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4 years ago, # |
  Vote: I like it +17 Vote: I do not like it

As a tester, I would like to say, good luck and have fun!

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

What time is it? 18:35 or 21:05?

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

Auto comment: topic has been updated by flaviu2001 (previous revision, new revision, compare).

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

Auto comment: topic has been updated by flaviu2001 (previous revision, new revision, compare).

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

It seems the blog text does not use the "Contest time" feature of the blog editor, like [contest_time:1421]

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    4 years ago, # ^ |
      Vote: I like it +60 Vote: I do not like it

    Oops, let's say I'm worse at writing blogs than problems.

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

I expect that this contest can be held smoothly.

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4 years ago, # |
← Rev. 5   Vote: I like it +42 Vote: I do not like it

As a Romanian, I am excited to participate in a round made by Romanians!

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

Good luck to everyone

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

if someone knows why there is no announcement of the Codeforces Raif Round 1??

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

    Now there's an announcement of the Codeforces Raif Round 1.

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

What is "Raif" in Codeforces Raif Round 1 (Div. 1 + Div. 2)?

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

Clash with codechef cook-off, but do we even care?

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

    I do, cook-offs are decent and it's always better to have more contests to participate in.

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

What a great way to promote a youtube channel :D, JK,

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

stefdasca orz

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

I am a newbie so should i participate in div2 or div3 contests Please help. i want to increase my rating seriously

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

    Participate in both.

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

      I only get notifications of div1 and div2 contests. till now i have not got even one div3 contest notification so that i can participate

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

        Frequency of Div3 contests is lesser than Div2 so you should try to participate in every Div2 and Educational Rounds. Educational Rounds are good too.

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

          oh thanks for the info, btw i saw some participants graph starting from 1200 and for some it was below 1200 , what is this method of starting and rating further

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

            Starting rating changed some time ago. You can learn more about it here -> https://mirror.codeforces.com/blog/entry/77890

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

              hey today i gave a div 2 contest , i was only able to do the first problem(this was my first contest ever). the soln got accepted but in my account ,contest section its showing nothing

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

                Rating changes happen some time after the contest (usually few hours). That's when your profile page section will be updated. If you want to know rating changes as soon as possible, there are some pretty accurate CF Predictors, that you can find just by googling them.

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

This contest clash with the cook-off.....They should prepone or postpone the contest whichever is possible

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

A lots of thanks to MikeMirzayanov and other involved organisers for rescheduling the round:)

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4 years ago, # |
  Vote: I like it +17 Vote: I do not like it

Why can't CodeChef reschedule for CodeForces?

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    4 years ago, # ^ |
      Vote: I like it +17 Vote: I do not like it

    Cause codechef scheduled the contest before codeforces

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

      In that case, CodeChef two monthly short contest are always prescheduled... So does that mean CodeForces will always have to keep that in mind.. Only a bit unfair, But this also shows how considerate Codeforces is for all of us..

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

Great another 2AM contest

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

As a tester, I can say that the problems are really interesting!

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

    Really interesting and a hard jump from C->D? Usually I have noticed Div-2 contests with five problems have a large jump in difficulties from C to D, hopefully this round is more balanced.

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

      You see the thing is not that simple, say the author wants to have a difficult problem D (say rating of atleast 2000), for an average participant it requires 40-60 min atleast, so if C is also medium difficult most of them won't have to time solve D or even read E.So they generally make C a bit easy and make D and E kinda difficult, so though there is a huge gap in difficulty but still you will be able to attempt it properly. Though no idea how this contest is gonna be

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4 years ago, # |
← Rev. 3   Vote: I like it +28 Vote: I do not like it

stefdasca Can you leave a dot in comment section? So that we can upvote :p

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

That's a good time for Chinese competitor!Thanks flaviu2001!

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

what will be the scoring distribution? or is there a fixed scoring distribution for div2 rounds and I'm just not aware of it?

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

    The scoring distribution is standard 500 — 1000 — 1500 — 2000 — 2500.

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

Whoah that's really early round, gives me no time to have dinner. Anyways glhf

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

15 min to be ready (:

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

Good time for a participant from India. Just after lunch full of Energy.

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

In question B what does "adjacent by side" means does that exclude diagonal cells?

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

all problems have nice music :) thanks <3

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

Well, cool round! Problem C was sickkkk

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

https://www.youtube.com/watch?v=ebgjlXyH9s8

Do watch the video editorial here

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4 years ago, # |
← Rev. 2   Vote: I like it -43 Vote: I do not like it

Sorry, but it was a very stupid problem B.

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

    don't comment during ongoing contest.

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

      Sorry. I edited the comment and left only the assessment.

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4 years ago, # |
← Rev. 2   Vote: I like it +21 Vote: I do not like it

The Kind of day i want to wake up to almost everyday during lockdown!!!

Afternoon — CF round

Evening — Kickstart

Night — CC round Cook Off

HAPPPYYYYYYYYYYYY!!!!!!!!!!!!!!!!!!!!!!!!!

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4 years ago, # |
  Vote: I like it +41 Vote: I do not like it

Gap between D and E is very Huggggggggge.

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

stefdasca, flaviu2001, koala_bear00 You guys surely have a great taste in music:) I had never imagined that I would be introduced to new music via a contest!

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

Am i the only person who find C much easier than B and regret for wasting too much time on B instead of solve C. Or i overcomplicated B?

Solution for C is simply,

R n-1

L n

L 2

Which takes about 15 minutes to come with idea and solve.

For B my idea is

Let b1 and b2 two adjacent cell of F and c1,c2,c3 be adjacent cell of b1 and b2.

Then i checked condition for possible cases which takes 1 hour to implement.

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

    Actually it was reverse for me,It took 20 min for B and an hour for C.

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

    For me B was much much easier than C, I think it depends

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

    C wasn't obvious at all, at least not for me. I wasted 1 hour 45 minutes in it and no AC xD

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

    For B it will be easier if you check two adjacent cells of S and two adjacent cells of F.

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

    Just make both the two cells at one end 0 or both the two cells at another end 1 or vice versa. check this condition I was first tried to implements bfs and got confused with conditions then I got this one. Though ~50 minutes are wasted. Here is my submission.

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

    It took me more time to understand C than solve B lol, So many misunderstandings in C for me

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

ConstructiveForces...xD....nice problems. Here are my submissions..hope to pass ST. Still 20 short of Specialist tho :(

A
B
C

Any ideas on D?

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

    Think about an alternative way of going to a neighboring cell.

    Path C6 == Path C1 + Path C5

    So, the cost of C6 = minimum of the cost of C6 and (cost of C1 + cost of C5)

    A similar approach for all other paths.

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

      How did you got all possible ways of going from one cell to another?

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

        First, compute the optimal cost to all the directions, Then you need to do basic math to get the optimal answer.

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

          And how did you get that maximum only addition of two paths will give minimum cost? Why not three or more?

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

            I don't think there exists any other optimal way except those two. You should try to observe the figure given in the problem.

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

            For simplicity consider $$$C2$$$(pure right), only $$$C1,C2,C3$$$ can give some right component. If you include any of $$$C4,C5,C6$$$ then you must need extra $$$C1,C2,C3$$$. Say if you want to use $$$C6$$$ then your path will be sth like $$$C6,C2,C3 > C2$$$ or $$$C6,C1,C3,C3>C1+C3$$$.

            Therefore minimum cost of $$$C2$$$ will be $$$C2$$$ itself, or optimal $$$C1$$$ + optimal $$$C3$$$. In fact, the minimum cost of $$$C2$$$ will be $$$C2$$$ itself, or the original $$$C1$$$ + original $$$C3$$$.

            The proof based on no two consecutive directions are not optimal. Assume $$$C1,C2$$$ are not optimal, then $$$C1>C2+C6$$$ and $$$C2>C1+C3$$$. Adding two inequalities gives $$$C1+C2>C1+C2+C3+C6 \implies 0>C3+C6$$$. Contradiction. If minimum cost of $$$C2$$$ is optimal $$$C1$$$ + optimal $$$C3$$$, then optimal $$$C1$$$ must be itself, the original cost of $$$C1$$$.

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

    After applying Floyd Warshall's Algorithm, you may use Greedy approach to find the answer.

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

      Target cell can have huge coordinates, so how is a graph algorithm feasible here? Is there some trick to compress the graph or just some O(1) thing?

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

        I mean considering a graph with only seven vertices. Essentially to find the minimum cost to reach from $$$ (x,y) $$$ to $$$ (x+dx,y+dy) $$$.

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

    Optimal right move = min(c[2],c[3]+c[1]) and Optimal left move = min(c[5], c[4]+c[6])

    For positive x: You just gotta find where y lies w.r.t x. There will be 2 choices of movement if (y>x or y<0), and if 0<=y<=x there are 3 possible choices of movement, try to figure them out.

    for negative x: just put x=-x, y=-y and swap opposite Cs and do the same as positive x

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

ppl solved D with O(1)?? I keep getting wrong answer on testcase 3 too. Is O(1) method correct way? find enclosing axis then max three ways to reach point??

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

    I guess it is possible, but it seemed so troublesome to correctly consider all cases that I abandoned this idea

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

    I guess you need to consider also max cases

    say you need to get to (7, 5)

    You can go 7 up-right and 2 down and this can also be optimal

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

    Yup thats how I did it, but among those 3 ways to reach a point, left and right movement have 2 ways each, I missed that part initially

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

I just dont get why authors dont get it .. Giving such stories just confuses us ... It does not make any thing "cool about your set". It makes it worse.

Man on a serious note, I know coming up with a interesting plot is definetly tough but I dont want to know that what the hell your character wants to achieve then first grab the info of what is doing what .. and not to mention the names u come up on

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4 years ago, # |
← Rev. 5   Vote: I like it +8 Vote: I do not like it

I tried to hack the following solution for A by generating 1e4 test cases where 1e8<=a, b<=1e9. This solution should time out but I wasn't able to hack it. Can someone please help me why it didn't TLE ? 

#include<bits/stdc++.h>
using namespace std;
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin>>t;
    while(t--){
        int min=0;
        long long a,b,temp;
        cin>>a>>b;
        for(int i=0; i<=b; i++){
            temp = (a^i)+(b^i);
            if(temp>min){
                min=temp;
            }
        }
        cout<<temp<<endl;
    }
return 0;
}
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    4 years ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    Actually the valuable $$$min$$$ and the loop doesn't do anything. After the loop, $$$temp$$$ is always set to $$$a\oplus b + b\oplus b = a\oplus b$$$, so the compiler ignored the loop.

    Unluckily, $$$a\oplus b$$$ is just the correct answer.

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

      So the compiler after few test cases realized that temp is being set to a^b and hence ignored thousands of other test cases ? Can you please elaborate a bit more ?

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

        It's called as O2-optimization, you can google for more information.

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

          I got -150 for this ;-; . I will definitely check for -O2 optimization.

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

            Hello I did exactly same code but get TLE why?

                    cin>>a>>b;
	mi=inf;//9e18
	ans=0;
	lp1(i,0,b)//for(ll i=0;i<=b;++i)
	{
		mi=min(mi,((a^i)+(b^i)));
	}
	cout<<mi<<endl;

            Submission

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

              b can be upto 1e9. you can not traverse to 1e9. You need to optimise your code.

              You can read about O2 optimisation here

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

              This is not the same code. The code that gets optimized above is roughly equivalent to

              cin>>a>>b;
mi=inf;//9e18
ans=0;
lp1(i,0,b)//for(ll i=0;i<=b;++i)
{
    mi=(a^i)+(b^i);
}
cout<<mi<<endl;

              So you can see that in the code above, mi is always overwritten in every iteration. This makes it easy for the compiler to realize that only the final iteration of the loop needs to be performed. (The submitter got lucky that this apparently incorrect code always gives the correct answer.)

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4 years ago, # |
← Rev. 2   Vote: I like it +14 Vote: I do not like it

B was so stupid because it was all about typing, annoying question imo.

Unless there's a better way of solving it..

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

C was just solving this test case — "abc"

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

    How to convert this to a palindrome ?

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

      3 operations

      L 2 ==> babc

      R 2 ==> babcba

      R 5 ==> babcbab

      'babcbab' is a palindrome.

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

      abc

      Perform R 2

      abcb

      Perform L 3

      cbabcb

      Perform L 2

      bcbabcb

      Now just generalize this to a being the first character of the string, c being the last and b being all the ones in-between. This is actually how I solved it during contest as well.

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

Is it only me, who did B in 15 mins but struggled in C for an hour? C definitely was a good problem, but required some thinking, and took me an hour to come up with a 4 line solution :(

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

    You're not alone. And the sad thing is I didn't get it in contest but after it.

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

    I first did D, then looked again into C.

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

I missed C by one second. As soon as I submitted, the contest ended:(

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

we got speedforces again lol

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

    I coded D earlier than C

    So my rank is low (ノ￣ー￣)ノノ(º_ºノ)

    The 55-minute-long D makes me down. ╮(╯﹏╰）╭

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

Ohhhhhhh as a rock fan I looooove these discription! C and D were so cute！hard E,really. Like this contest!

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4 years ago, # |
← Rev. 2   Vote: I like it -35 Vote: I do not like it

C is one of the worst Adhoc ever, you have to keep trying combinations until you get it, no clever obervation, no analysis, just have pattern aSb and keep trying.

B is equally annoying problem, so many if conditions, pure implementation

Really dissappointed

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

    There is a clever observation to make first and last characters same and then apply both the operations one at a time . I dont know if this is clever or not but I was not clever!

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

      Can you arrive at it without trying a zillion samples ?

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

        Yes? Clearly distinct characters will be the most difficult case and 3 is the smallest n, so try 3 random distinct characters. This gets an easily generalizable answer. As for solving this case, I just checked for a sample with a distinct character at the end (sample 2) and used it as the basis.

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

          I can propose solns that will work for abcde but not for abcdef

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

          Exactly what I did just by seeing the problem but I was trying to do it int three operations. I thought there is a way since "abc" can be done in just 3 operations. but bad luck went in wrong direction earlier!

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

        L2 R2 R 2*n-1 would be clever lol .Yes just by thinking

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

          It would have been, if you arrive at it with a proper algorithm,instead of noticing pattern in a few examples

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            4 years ago, # ^ |
            ← Rev. 2   Vote: I like it +6 Vote: I do not like it

            And what is a "proper algorithm" for solving a problem?

            In my experience, nearly all problems break down to analyzing it (possibly with some small cases or simplifications or maybe just writing a brute force), noticing interesting properties and then getting some intuition about how to proceed.

            Even most problems requiring formal algorithms or some data structures usually require you to realize that some properties hold which makes you think of that approach.

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

        Test case 1 makes you realize that you can always do it in two moves if the first and third characters are the same. The rest is thinking if you can obtain such a string in a few moves.

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

    Nah, C had a very nice observation. You don't need to try many samples, just take a string where all characters are distict. If that works, everything else will work. After that its just a bit of thinking and 3 lines of code

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

    Firstly, $$$type 1$$$ operation will make a prefix of the string a palindrome and $$$type 2$$$ operation will make a suffix of the string a palindrome. A palindrome can have at most $$$1$$$ element having frequency $$$1$$$. Here in the given string, you cannot change the frequency of the first and the last element at the first operation. But if the first and the last element are different, then you must have one of them in some operation. So, first, you do $$$R \ n-1$$$, and then to double the frequency of the last element of the given input string, you do $$$L \ n$$$. Now, it is easy to see that, doing another operation will result in a palindrome.

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

    I will give you best answer

    Consider a non palindrome of two parts X|Y

    where | is middle point ok?

    lets take for example initital string => X1.X2.X3|Y1.Y2.Y3

    This must be worst case right?

    Now do L,'2'

    -> X2 .X1.X2.X3|Y1.Y2.Y3

    Now Do, R,'2'

    -> X2.X1.X2.X3.Y1.Y2.Y3. Y2.Y1.X3.X2.X1 (length be sz)

    Now Do R, (sz-1) where** sz** is length of string reached above

    -> X2.X1.X2.X3.Y1.Y2.Y3. Y2.Y1.X3.X2.X1. X2

    This is palindrome. Just use induction for all lengths of X and Y

    ez clap

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

I did two bfs on B. smh

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

Thanks for Tzuyu's reference in Problem A

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

Is it just me or was this contest too much ad-hoc?? A,B,C,D all can be implemented in O(1), it was only about solving various test cases until you can see the pattern. I may be very wrong here but I think not much real programming or barely any Data structures and algos were used in this contest :/

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

Damn man, why all Constructive. Imma lose a 100 rating points, where gaining 14 would've put me over to CM. It was so hit or miss.

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4 years ago, # |
  Vote: I like it +55 Vote: I do not like it

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

In problem B, is diagonal movement allowed?

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

    no, adjacent by a side, not a corner

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

    No, as neighbors in the grid are those who share an edge.

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

Well... Instead of CM i I failed system test for problem B.
Granted, I did implement it in a disgusting way, bit it passes pretests :/

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

Solution for C was so easy :(

crying

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

Ad-hoc Missiles. I'm not complaining though :))

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

Problem B: Consider the grid as:

A B C ...
D E F ...
G H I ...

A = Start Case 1. if B=D=0 and C=E=G=1 then STUCK. Case 2. if B=D=1 and C=E=G=0 then STUCK.

We have to make anyone of the two cases in atmost two steps.

Is this logic correct?

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    4 years ago, # ^ |
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    Logic only depends on, BD and HF.

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      4 years ago, # ^ |
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      Yes, this is also mentioned in editorial and I understood it. But why the above logic did not work? Am I missing something? Any idea.

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        4 years ago, # ^ |
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        I don't see any counter example for this strategy, i guess this should work well !

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

    Yeah, this works, I have implemented something similar.

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

    Your logic is correct. If $$$C=E=G$$$ then at most two steps. Otherwise there will be one of 0/1 appear two times, let's denote if by $$$x$$$. If at least one of $$$B,D$$$ is $$$y$$$ then set $$$C=E=G=x$$$ and $$$B=D=y$$$ which requires at most two steps. Else both $$$B,D$$$ are $$$x$$$, set the two $$$C,E,G$$$ with x to y.

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

I think the Problem BCD are boring

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

Great contest, great songs. However, I should not have listened to them during the contest.

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

This was a total disappointment since every question was a case work, I came here to solve the problems hoping to use my implementation and coding skills and not using my case work skills. I don't know about problem E but other 4 questions were total case work. Were you guys expecting us to write only if-else statements rather than using some great concepts out there?

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4 years ago, # |
  Vote: I like it -22 Vote: I do not like it

Was this contest unrated??

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

Problem E is so hard, so interesting, so mysterious. Thanks for the author's ideas and efforts!

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

As soon as I read the first line of problem A, I was like — Aaaah... Imma kill this round... sad not all the problems had "twice" background.

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4 years ago, # |
← Rev. 2   Vote: I like it -19 Vote: I do not like it

Note for C/C++ users: I lost problem D because I was doing negative % positive in the indexes of the c[] array. We know that (-1) % (6) should give 5 but in C/C++ when you do that you will get -1. So I wrote my own modulo to overcome this and I said it would be good to share it with you people:

#define mmod(a,b) ( (a >= 0) ? ((a%b)%b) : ( ( ( a + ( ((abs(a)+b)/b) * b ) ) % b ) % b ) )

Hope it help you and myself get out of Gold Nova! I mean Expert!

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

Problems B, C, and D were just if-else...Not so great round, Should rather call it speedforces, implementationforces, etc.

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

I have a Rating of 369 and I had solved 1 problem in this contest. Will my rating change as this is a Div 2 contest?

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

hello there, in the first question of Codeforces Round #676 (Div. 2) some test cases are wrong for the sake of understanding ..in one of the examples given in that question ..if we take a=28 and b=14 the output according to your code is 18 but

minimum possible answer is 10 if we take x as 4 28^4+14^4=10

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

    28^4+14^4==34 not 10 don't write like this in code because priority of '+' is greater than '^' so 28^4+14^4 will become 28^18^4 which is equal to 10 so write in code like this (28^4)+(14^4)

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

I was the first contestant to solve 1421C - Палиндромайзер, why didn't you put in the announcement people who were the first to solve each problem?

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

I found this because of your contest. Thank you :)

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

I gave the contest and used ideone.com to write my code. Someone stole my code, because of which I was marked a violater. But I haven't done any type of cheating. How can I get my rating back??

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

    Sorry we can't do anything about it, but in the future please don't use ideone or at least try to find a way to mark your source codes private.

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      4 years ago, # ^ |
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      Thanks for the reply. Will take care in future.

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4 years ago, # |
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Of course, A little easy.

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

Is no one going to speak about how incredibly cool the band references and music links were

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

And yes! We are celebrating TWICE's 5th anniversary today.

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