havaliza's blog

By havaliza, 6 years ago, In English,

Hi all! :)

I'm glad to invite you to participate in Codeforces Round #148 today. I (Hamed Valizadeh) am the author of this round.

I'd like to thank Gerald (Gerald Agapov) MikeMirzayanov (Mike Mirzayanov) Delinur (Maria Belova) and Saeed_Reza (SaeedReza Seddighin) who helped me in preparing the round.

Score distribution will be the standard 500-1000-1500-2000-2500 in both divisions.

Hope you find the problems interesting to solve.

Good luck and have fun ;)

Update. Contest is over. Congratulations to the winners of both divisions! :)

Div1:

  1. tourist
  2. cerealguy
  3. Dmitry_Egorov
  4. RAVEman
  5. UESTC_Nocturne

Div2:

  1. KotonohaKatsura
  2. goooooooopan
  3. jthread
  4. kolina
  5. xcodevn

And congrats to Endagorion who was the only one solving 238D - Tape Programming correctly during the contest.

BTW, I hope you didn't get sick of that boring problem! :-"

Update 2. The editorial is ready now, sorry for the delay. http://codeforces.com/blog/entry/5765

 
 
 
 
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6 years ago, # |
  Vote: I like it +49 Vote: I do not like it

Two consecutive contests with Iranian authors, very good :-)

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6 years ago, # |
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questions are going to be VERY hard :|

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    6 years ago, # ^ |
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    Why do you think so ?

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    6 years ago, # ^ |
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    fahmidan sualasham sakhte che berese be hale suala

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      6 years ago, # ^ |
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      may be the only reason why he's taken this low rate is that he's written it in phinglish(in farsi means writing persian in english).it's the translatoin: even understanding the questions is hard so how U wanna solve it??!

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6 years ago, # |
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Thankyou -- Good luck and have fun — -

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6 years ago, # |
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Codeforces Round #148 will lucky for div 2, :)

148 div 2 = 74

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6 years ago, # |
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be Sure that Question will be So hard ! :) Do not Put your Rating in Danger !!! =))) GL everyone !

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    6 years ago, # ^ |
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    why are they going to be hard?

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      6 years ago, # ^ |
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      Because it has had a Three month Preparing Process and its Tasks Have been prepared in This Months :D and its Authors Are Havaliza + Saeed_reza

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        6 years ago, # ^ |
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        Please, do not scare me!!! I hope get DIV1 today :(

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          6 years ago, # ^ |
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          Good and Hards Tasks... Now I am DIV 1 contestant!!!

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6 years ago, # |
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I think there's going to be a lot of math in this contest.

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6 years ago, # |
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When the lags will be fixed?

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    6 years ago, # ^ |
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    When the first contestants will give up ;) (seriously, I hope that the lag will not alter too much the contest :/ )

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6 years ago, # |
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Again it is — the crash!

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6 years ago, # |
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This is not the first contest which gives me "**Bad gateway**" errors (502 — 504) instead of a contest start. I understand that is not supposed to be so, however, I would want to hear what exactly goes wrong with the server. Moreover, I believe there are plenty of nice guys over here who could probably be very experienced and helpful in configuring network/servers. Why don't we post some CodeForces issues as a public discussion? This kind of a solution would allow CF users to help the platform somehow.

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6 years ago, # |
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Pressing the F5 Key all the time...

Hope this is getting fixed soon!

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    6 years ago, # ^ |
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    Maybe the servers work has worsened because of you and some other users pressing F5 all the time?)

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6 years ago, # |
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Isn't it better to start on the 5 PM than 4.55 PM ?

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6 years ago, # |
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Will it start soon !?!

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6 years ago, # |
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nooooooooooooooooooooooooooooooooooo

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6 years ago, # |
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I think Codeforces should use safe mode during contest time as they did few times in past. Server down before contest is ok but this may happen in contest time too. Topcoder also closes practise rooms during contest.

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6 years ago, # |
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Friends, nothing is perfect!!!!

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6 years ago, # |
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nemishe hala ke khodet irani hasti ye noskhe farsi sualam bedi nashinim mani konim?

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6 years ago, # |
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The server needs to be upgraded.

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6 years ago, # |
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Div 2 — B is not well defined. When we erase a character at position n and cp points to the character at position n+1, does it still point to that character (now at n) or does it remain at n+1?

Why the negative votes and no replies? Is there any place where we can ask questions to the organizers?

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6 years ago, # |
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Very easy problems. Thx!

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    6 years ago, # ^ |
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    Easy for you?

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    6 years ago, # ^ |
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    I never felt as stupid as I did on this contest =D even in div2

    Teach me, master, I want them to be easy for me too.

    WTB editorial!

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    6 years ago, # ^ |
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    I think, you cannot make the conclusions becouse of first and second tasks.

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6 years ago, # |
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Looks like its going to take a while to reach division one :s

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6 years ago, # |
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In problem D, div 2: I think just care the case n==3, another case the anwser is "2 2 2 2 2 ...". Is it right?

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    6 years ago, # ^ |
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    no but there are only two options: two smallest in one subsequence or each in a different one. there is no way to decrease the maximum value of F so we have to increase the minimum of F.

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    6 years ago, # ^ |
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    No, the answer is without doubt "2 2 2 2 2 ..." only when n==2. In other cases we should check more conditions.

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it looks problem E,DIV 2 was easier than D...:D

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    6 years ago, # ^ |
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    and it was the opposite in div 1 !!!!

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    6 years ago, # ^ |
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    I recall a similar problem, at least on the surface, which also asked the minimum number of edges which direction you had to change. Maybe some people solved that problem and chose this instead of D, since it looked very familiar.

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    6 years ago, # ^ |
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    You can't erase your comment, prepare to receive minus by the non-contestants who will not get it now that the system test phase is ended ;)

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      6 years ago, # ^ |
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      cool!!!

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        6 years ago, # ^ |
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        for E,DIV2 my submission passed in pretests but failed in system test 7...can anyone help here..

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          6 years ago, # ^ |
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          could have gained 200 more ratings(Expert!!) insted of gaining just 42, if final tests would have been passed for E!!!

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6 years ago, # |
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Are the pretests of С (div1) strong enough? I wrote solution 5 minutes before the end and realized that sometimes it can't find solution. So I just put line: if (best == INF) cout << 2. And it passed pretests! I'm 99% sure my solution is wrong but maybe someone can prove it? :)

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    6 years ago, # ^ |
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    It depends on your algorithm ... Maybe you can always find a solution?

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      6 years ago, # ^ |
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      I think it doesn't work on bamboo graphs (e.g. O-O-O-O). Idea is the following: Choose one vertex i = 1..n and run DFS on tree to mark for every vertex: is the parent edge forward or backward. If this edge (to vertex j) is backward, we claim pair (i, j) to be good. If both pair (i, j) and pair (j, i) is good, we can relax answer with number of backward edges on the way from the root minus one.

      P.S.: Ouch, looks like I deleted some lines and now it relaxes just if pair (i, j) is good :)

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        6 years ago, # ^ |
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        Aw .. I'm a little confused .. Hope you pass it!

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I liked the contest. At the beginning problems confused me a lot, especially problem A. But when I thought a bit about them, they turned out to be solvable.

Unfortunately my Div 1 C failed...

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    6 years ago, # ^ |
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    I can't agree more ~ The problems are all interesting and skillful.

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ahhh... at last system test started. hope it won't take long time to finish.

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Judging in not working!

Edit: Now working

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6 years ago, # |
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I missed a hack on problem B :( Someone in my room outputs "1 1" if n == 2...

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B Div 2 много у кого упала ...

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6 years ago, # |
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I did not spend much time on the problem E because of the following test:
7 8
1 2
2 4
1 3
3 4
2 5
5 7
3 6
6 7
3
1 4
2 7
3 7

The answer is 2 but if we consider only stops reachable in any case for each bus route we don't find a solution. I am wondering how many of those who submitted this problem know about this test?

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6 years ago, # |
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Very nice problems...But I'm lack in experience and my solution of B got FST...TAT

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6 years ago, # |
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I'm unluky :( ;(

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6 years ago, # |
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Pleeeease, don't make us wait, update the rating.

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6 years ago, # |
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Can smb explain to me how to solve problem C(Div2) or A(Div1)? Thanks in advance

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    6 years ago, # ^ |
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    hey, count how many number you can put in the place of a1, then how many number you can put in the place of a2 and so on .

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    6 years ago, # ^ |
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    2^m — 1 ways for the first number, 2^m — 2 for the second etc.

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    f(n) = f(n-1) * (2^m — (n-1) — 1), f(1)=2^m-1. Here 2^m — (n-1) -1 means you have this number of ways to append a number to the valid sequence with length n-1. "-1" since you cannot use 0.

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      6 years ago, # ^ |
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      I don't understand.

      Why can you simply add unused numbers (apart 0) to a sequence of length n-1?

      Look at this sequence (n=3,m=3): 5 (101), 2(010), 7(111)

      All numbers are different, but 5^2^7 = 0.

      So it is a wool sequence: l=1,r=3

      Why am i wrong?

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        6 years ago, # ^ |
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        At begining you have only "0" as foribbden element.

        After first addition, you have one more foribbden element (the added one).

        After second addition, you have next foribbden element (first ^ second). And so on.

        So, you must look at "accumulate xor" not (simply) values.

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          6 years ago, # ^ |
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          Thanks for your explanation.

          I had misunderstood the meaning of "n-1".

          And you can also be sure that all of the "n-1" accumulated XORed values will be different, so that there are exactly n-1 distinct numbers to be excluded:

          Else, it implies the previous n-1 elements would be a wool sequence:

          If XOR (1 -> x) = XOR (1 -> y), y>x would imply that XOR(x+1->y) = 0, implying it was a Wool sequence.

          Impressive how some people solved it within minutes =|

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6 years ago, # |
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I wish this contest were the Elimination Round for BAYAN ... Thank you for the good contest :)

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6 years ago, # |
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Finally "Expert"! :)

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    6 years ago, # ^ |
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    Multi account is bad. There is a lot of "Ashvili" account on Kutaisi, with one or two contests on the history, like if they create a new account when they are too low to gain a lot of rating. The codes present the same style (same freopen (why not, its common), but reduced indentation too), and you said "Finally Expert" like if it wasn't your first contest, but in your profile, you are a new contestant. Suspicion...

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      6 years ago, # ^ |
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      More: the header "cmath" is often more present than "math.h" for C++, but here, for all the bukhnikashvili, there is "math.h".

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      6 years ago, # ^ |
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      Anyway, he does not hide anything :D

      @Lashabuxo I'm hope that you remain blue after a few contests :)

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      6 years ago, # ^ |
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      Haha. Better to sit quiet if you're cheating! :P

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Nice Contest :D hope to see the Editorial soon..

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Can someone explain why in the sample test for div2 b: 7 4 1>3>22< 1 3 4 7 7 7 1 7 the answer for call l = 1 and r = 7 is 2 3 2 1 0 0 0 0 0 0.

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    6 years ago, # ^ |
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    1>3>22<

    last 6 numbers are ignored for convenience

    1-> CP=1 DP=R MAP=0>3>22< X= 0 1 0 0

    2-> CP=2 DP=R MAP=0>3>22< X= 0 1 0 0

    2-> CP=3 DP=R MAP=0>2>22< X= 0 1 0 1

    2-> CP=4 DP=R MAP=0>2>22< X= 0 1 0 1

    2-> CP=5 DP=R MAP=0>2>12< X= 0 1 1 1

    2-> CP=6 DP=R MAP=0>2>11< X= 0 1 2 1

    2-> CP=7 DP=L MAP=0>2>11< X= 0 1 2 1

    2-> CP=6 DP=L MAP=0>2>10< X= 0 2 2 1

    2-> CP=5 DP=R MAP=0>2>00< X= 0 3 2 1

    2-> CP=5 DP=R MAP=0>2>0< X= 1 3 2 1

    2-> CP=5 DP=L MAP=0>2>< X= 2 3 2 1

    2-> CP=4 DP=R MAP=0>2> X= 2 3 2 1

    2-> CP=5 END

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If i can solve 3 problems in div2 & 2 problems in div1...Whose 1 will be better???

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    6 years ago, # ^ |
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    It may vary between man to man. But according to my opinion, solving 3 in div II is only better confirming that any 2 of these 3 also appear on div I(Like 2 from (C-E))

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6 years ago, # |
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plz someone tell the algorithm for solving Div 2 problem E ..........

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I see some ppls failed Problem B (div1) on test 6, giving the same wrong answer 190 (P.S. correct answer would be 200).

Is that a coincidence?

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    6 years ago, # ^ |
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    Many people have the same, good, algorihtm. Similarly, many people have the same, but incorrect algorithm :P Just coincidence :)

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      6 years ago, # ^ |
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      ;) But when you think carefully, there is no way you can come up with a combination which is even better than the best case, even with an "incorrect" algorithm.

      What you can do with an "incorrect" algorithm, is to get a worse answer. Am I right?

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    6 years ago, # ^ |
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    Oh no I got WA on #6 and my answer was 190...However,I don't know why it's wrong = =

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      6 years ago, # ^ |
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      I would suggest that you might have calculated the minimum value of the maximum value and the maximum of the minimum value separately while they might not be achieved at the same time.

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        6 years ago, # ^ |
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        eh..no..I have thought about this — -...It's so kind of you to view my solution 2500928 (cause my poor english I cannot describe it TAT)

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          6 years ago, # ^ |
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          int min1=a[0].first+a[1].first+(i==1?h:0);

          This give me accepted. thx

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            6 years ago, # ^ |
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            oh no I misunderstood the problem,I thought f(i,j)=Ai+Aj+h if at least one of i and j is in the first subsequence!!Ahhh terrible~~~~~

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Where else can i find the current contest questions being discussed????????

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Сложный раунд=(

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    6 years ago, # ^ |
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    if this was an answer to my question then sir plzzzzz give link coz i dont understand Russian............plzzzzzzz provide the link.......

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      6 years ago, # ^ |
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      You should use translator or simply skip comments in Russian ;)

      He wrote something about 'complex round =('

      And... here is right place to discuss this round's problems

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where can I find the solutions to the problems in contests?

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    6 years ago, # ^ |
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    Try to take a look at others' solutions but don't copy them.

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Is there any editorials?

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6 years ago, # |
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UPD Ok it's published now

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6 years ago, # |
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hey guys can anyone tell me how to solve Div 2 Problem C

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    6 years ago, # ^ |
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    try to make a new sequence of n numbers for each valid answer sequence that i'th element of new sequence would be equal to xor of first i elements of answer sequence . each element of new sequence will be different form others and it's between 1 and 2^m -1 ... so for every n and m number of answer sequences would be equal to this :

    (2^m -1)*(2^m -2)*...*(2^m -n)

    I hope this will help you :)

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      6 years ago, # ^ |
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      I can't get the first sentence at all...did you really mean the second word "sequence"? Each element of new sequence will be different from others? Are you sure? I thought (1,2,1,2) is valid since 1^2^1^2 == 0.

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        6 years ago, # ^ |
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        But the request is to count the sequences that are NOT a wool sequence...So (1,2,1,2) is invalid..

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6 years ago, # |
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When the next contest start?