Zlobober's blog

By Zlobober, 5 years ago, translation, In English,

Hi everybody!

Today at 12:00 Moscow Time there will be Codeforces Round #279, dedicated for second division contestants. Round is based on the problems of the Saratov second round of All-Russian School Olympiad in Informatics 2014-2015 that is held at the same time in Saratov.

This round was brought to you by ikar, HolkinPV, IlyaLos, fcspartakm that are all members and ex-members of Saratov SU 2 team.

There will be 6 tasks for 2 hours 30 minutes. Scoring will be announced just before the round starts.

Round will be rated for second divison participants. First division contestants may participate out of competition.

UPD: Scoring: 500-1000-1500-2000-2000-2500.

Good luck!

 
 
 
 
  • Vote: I like it
  • +149
  • Vote: I do not like it

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

ranked!?

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

waiting! waiting!!! good luck to all participants!

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

wish high rating

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

When you visit codeforces.com

Can you look the message "connecting to fonts.googleapis.com ..." with 10 sec?

»
5 years ago, # |
  Vote: I like it -19 Vote: I do not like it

6 problems. Wow!!!!!!!!

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

10 minuets to start. Why yet not scoring have been published? Waiting for it.

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

Delay? Again

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    yep

  • »
    »
    5 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    When it was moved last time, servers were showing they are busy almost all the time, hopefully today it will be better...

    • »
      »
      »
      5 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      hope that also last time I asked to be unrated because of the same reason but they told me they didn't face any problem during contest :D now it seems that it's not mine only :D

      • »
        »
        »
        »
        5 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        There was a lot of contestants complaining, I didn't understand why it was rated, I left the contest earlier also...

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

Scoring have been published. Best luck to all participants.

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

Contest not start

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

Not again! It's becoming a tradition to delay the contest right before it starts!

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    This time, I am glad the contest got delayed. Forgot to register earlier.

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

again more 10 minutes waiting :D nice. P.S. good luck everybody

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

when will i be candidate master???

  • »
    »
    5 years ago, # ^ |
    Rev. 4   Vote: I like it 0 Vote: I do not like it

    same problem here :D

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Just when you will have the quality to acquire it. Best luck to you in this round. :D

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

Almost every contest gets delayed these days! Anyone knows what the problem is?

  • »
    »
    5 years ago, # ^ |
    Rev. 4   Vote: I like it -8 Vote: I do not like it

    Because the contestants in codeforces are increasing exponentially.

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

I wish will be candidate master today)

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

I arrived at home just now from school (it's 12:30 here)

I'm so stressful now... :(

Please pray a good contest for me... :(

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

It delays for 10 minutes! Am I right?

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

Good luck to all!

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

Good luck guys !

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

Its good that this contest has 6 problems. Better chance to improve ratings. Wishing high ratings to everyone :)

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Really? It depends on number of problems? I don't think so...

    • »
      »
      »
      5 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Its not always correct, but most of the times it is easier to solve 3 problems out of 6 than solving 3 out of 5. Of course, it includes the fact that you should solve the 3 problems fast.

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

Is there any way to help Codeforces to strengthen its servers ? I think it worth to pay for having a faster and stronger programming contest platform.

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

Can you unblock gym?

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

Hello, why all problem pages are blocked? is this because contest is running?

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

Does the interface always inform about hacks with delay?

I got a message about my stupid C solution being hacked only after about 20 minutes and hadn't enough time to correct it =(

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

Why all wrote n^2 to C. I earned 5 hacks because of that. (actually n is the length of string)

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    All my hacked solutions were exactly the same. My test was s = '9' 10^6 times, a = 9, b = 2

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

Awesome contest! Thanks for the problems. :)

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

Omg, after I locked my C I found it will fail with

1230
123 10

...I'm so unhappy :-(

BTW: where I can test hacking using generator? Because I wanted to try huge input for heu2013201410's solution in my room — 27...

  • »
    »
    5 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    The answer should be "No"? As your test doesn't differ from

    120 12 1

    • »
      »
      »
      5 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Answer is Yes I think — 123 0

      • »
        »
        »
        »
        5 years ago, # ^ |
          Vote: I like it +19 Vote: I do not like it

        I think it's "NO". Both parts should be positive integers that have no leading zeros.

      • »
        »
        »
        »
        5 years ago, # ^ |
          Vote: I like it +1 Vote: I do not like it

        Both the resulting numbers should be positive integers. So the numbers can not be 123 0.

        • »
          »
          »
          »
          »
          5 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Thats good message, thanks and now its clear, why my hacking failed. Stil better -100 then -1000...

      • »
        »
        »
        »
        5 years ago, # ^ |
          Vote: I like it +1 Vote: I do not like it

        Actually I think answer should be NO according to 3rd test ?!

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Wanted to hack solution C by TL with test.

    1...1 (1 repeats 1e6 times)
    5 1
    

    But this test file has size (1e6 + 5) Bytes! And this is more then 256KB. Why does system not allow to upload big tests???

    Lost my hack because of this.

    • »
      »
      »
      5 years ago, # ^ |
        Vote: I like it +1 Vote: I do not like it

      How about generator?

    • »
      »
      »
      5 years ago, # ^ |
        Vote: I like it +1 Vote: I do not like it

      You can create generator — program, that generates the input. I wanted to try it first time today also. Motivation is clear, if everyone uploads 1MB big file it is a lot of network traffic...

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    A generator is a program which outputs the hack input in the correct format. I don't think that you can test it without a contest. Just write one and check if the generated input is ok.

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    The two parts must be positive integers
    is zero considered to be a positive integer ?

    • »
      »
      »
      5 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      No, if 0 would be considered into solution, the problem statement should say "non-negative integers"

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

Div.2 C. Couldn't understand why hack attempts were unsuccessful until realized that me forgot expand generated test changing magnitude(10^) from 1 to 6 :( .

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

The problem statement of F today was so bad and unclear that I had to read the statement over and over again and rewrite my code about 4 times from scratch because of not understanding what was actually I needed to do. I had to code this problem mostly on assumption. Glad it was only a div-2 contest. I hope the authors will provide easily understandable statements from next time.

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

why b=1000000007 is invalid test for hacking of C?

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    The number is too large, look at constraints. Has to be less than 10^8.

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Because b ≤ 108

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Because b must be in range [1..10^8] It clearly said in problem statement

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

this code is still wrong, but for this tc, it gives right answer and yet it got WA http://codeforces.com/contest/490/submission/8820977

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

New contest new color...

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

Can anyone tell me why am I getting a wrong answer on pretest 4 ? I'm new here! http://codeforces.com/contest/490/submission/8821196 (Ignore the biginteger part,coding begins at the function compute() ) thanks!

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Is it wrong because I messed up the concept of leading zeros / trailing zeros?! wrong condition?

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Because there is the answer for string "604": "60 4" where both of numbers are divisible by 6 and 4 respectively

    • »
      »
      »
      5 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      What should be the condition to check if there aren't any leading 0s? Is my implementation wrong?

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

Why can't I see the wrong test case when submitting problem E in practice?

Edit: It's fixed now

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

Surprisingly fast system test. I thought problem C can be solved using O(n2) algorithm. Too bad I didn't check the constraint for n. Anyway, thanks for the contest!

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

I think that codeforces should check the IP when a new user registers because i can bet that there are lots of div1 members who make clone accounts in order to participate in div2 rounds with rating and because of that our chances for a lower rating grow. Or we get a lower improvement than we should. Just my opinion.

  • »
    »
    5 years ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    Some nations do not have static IP.

    • »
      »
      »
      5 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I know that it cant be that accurate, but at least.. you block few users. With a bit luck, more users :D

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    IP based check is bad idea. But I have same feelings... Mbaybe solution could be that when there is no Div 1 competition, they will have same problems in and they can compete in speed typing... Or separate ranking can be created for such contests...

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

When will ratings be updated ???

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

Anyone on how to approach Problem C. I thought about writing own function to check if the two numbers formed are divisible by the given number : eg. 64010 64 10 check for each value of i 6|4010, 64|010, 640|10 ... for divisibility Will this approach run in time?

  • »
    »
    5 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    try to check forward and backward (will be almost the same like forward )

    • »
      »
      »
      5 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Yeah I got it. Have to use exponentiation as well.

      Thanks

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    This is my code..have a look..just used the wrong condition to implement trailing 0s condition. Any feedbacks are welcome. http://codeforces.com/contest/490/submission/8821196

  • »
    »
    5 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    the idea starts from the observation:

    lets say you have the number 78541

    78541%x = ( (7*10^4)%x + (8*10^3)%x + (5*10^2)%x + (4*10^1)%x + (1*10^0)%x )%x

    if you dont get the further idea for the solution i will add the next part :D

    • »
      »
      »
      5 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      thanks... i got the idea. Couldn't expand the number during the contest, but I got the answer now. Thanks.

    • »
      »
      »
      5 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      cool, thanks!

»
5 years ago, # |
  Vote: I like it -8 Vote: I do not like it

For B I allocated int[] map = new int[1000000] instead of int[] map = new int[1000001] and I got a WA on Test 55 because of that. Cruel!

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

    Do people like downvoting stuff just for fun? This community should be a little more mature.

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

I got TLE in 'C' with O(n) solution. :/

  • »
    »
    5 years ago, # ^ |
      Vote: I like it +9 Vote: I do not like it

    don't use strlen(s), it has O(n) complexity, where n is length of s.

  • »
    »
    5 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    strlen() has a complexity of O(n), not O(1).

    So your overall complexity is O(n^2).

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    you repeatedly called strlen, which has n complexity;

»
5 years ago, # |
Rev. 2   Vote: I like it -11 Vote: I do not like it

OK

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

How I can solve problem C?

  • »
    »
    5 years ago, # ^ |
    Rev. 10   Vote: I like it 0 Vote: I do not like it

    Traverse a string in both directions and keep two arrays ad, bd.

    For ad[i] we traverse the string from the left to the right and find the reminder s[1: i]%a. There is a formula for this: ad[i] = (ad[i - 1] * 10 + s[i] - '0')%a.

    For the bd[i] we traverse string from the right to the left and find the reminder s[i + 1: n]%b. Also we keep the variable bteni which is equal to 10i %b. So there is a formula for bd: bd[i] = (bd[i + 1] + (s[i] - '0') * bteni)%b.

    Finally we traverse the arrays ad, bd and search for such index i, that ad[i] = bd[i + 1] = 0.

    • »
      »
      »
      5 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      thanks~

    • »
      »
      »
      5 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Could you explain why these formulas work? Or point to a source where this theorems are explained? Thanks

»
5 years ago, # |
  Vote: I like it +12 Vote: I do not like it

Rating....so late.

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

Can someone Please explain the idea behind problem C ? in detail preferably. Thank You. :)

  • »
    »
    5 years ago, # ^ |
    Rev. 4   Vote: I like it +6 Vote: I do not like it

    If we divide the sequnence into two parts: 1, 2, ..., i and i + 1, ..., n, then we need to check s(1, i) % a and s(i + 1, n) % b. Since s(1, i) % a = (s(1, i — 1) * 10 + s[i]) % a, s(i, n) % b = (s(i + 1, n) + 10^(n — i) * s[i]) % b, we can compute every s(1, i) and s(i, n) within O(n) time firstly. Now, we can check s(1, i) % a and s(i + 1, n) % b within O(1) time.

  • »
    »
    5 years ago, # ^ |
    Rev. 3   Vote: I like it +5 Vote: I do not like it

    I Will explain it using an example:

    consider the number 116401024(call it N)

    we can split it as

    (1*10^8)+(1*10^7)+(6*10^6)+(4*10^5)+(0*10^4)+(1*10^3)+(0*10^2)+(2*10^1)+(4*10^0)
    
       a8       a7       a6       a5       a4       a3       a2       a1       a0 
    

    If this number is divisible by a number M then the modulus must be 0(i.e. N%M==0), to check this we can do:

    rem = 0;
    rep i from 0 to 8:
        rem = ( rem%M + ai%M )%M;
    if(rem == 0)
        the number is divisible by M
    

    Coming to the actual problem

    The possible ways we can split the number into two parts are :

    1 16401024
    11 6401024
    116 401024
    1164 01024
    11640 1024
    116401 024
    1164010 24
    11640102 4
    

    now for each of this splits we want to check if the first part is divisible by "a" and second part is divisible by "b".

    Now we can modify the above for loop by storing for each index i, whether the part of the number from 0 to i is divisible by a, like this:

    n = Number of digits in N;
    rem = N[0]%a;
    rep i from 1 to n-1:   --------------------------------> O(N)
        rem = ((rem*10)%a + (N[i]%a))%a
        if(rem == 0)
            divisible_by_a[i] = true;
        else
            divisible_by_a[i] = false;
    

    Similarly, we can do for b starting from least significant digit of the number.

    and once we have divisible_by_a and divisible_by_b, we can easily check if we can divide the number into two parts for each index i by the logic:

    (divisible_by_a[i] == true) && (divisible_by_b[i+1] == true)  && (N[i+1] != 0)    -------> O(1)
    
  • »
    »
    5 years ago, # ^ |
    Rev. 2   Vote: I like it 0 Vote: I do not like it

    Call the large number s and let n be its length.

    For every prefix, compute x[i] = s[0..i] % a. For every suffix, compute y[i] = s[i..n-1] % b. Now scan s from left to right and check if x[i] == 0 and y[i+1] == 0 and s[i + 1] != '0'. If this condition holds, we have an answer.

»
5 years ago, # |
  Vote: I like it +23 Vote: I do not like it
»
5 years ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

Can anyone please tell me, why my solution of C always takes 0 or 15 ms and suddenly on test 36 i got TIME_LIMIT_EXCEEDED? There are no cycles or anything and I'm getting really clueless... http://codeforces.com/contest/490/submission/8823963

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    There's also the weird thing, that it shows my output (which I think isn't supposed to be there, because I shouldn't have time to it), but there is no Answer...

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Don't use ansistring. Try to use array of char instead.

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

where is the editorial?

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

Can someone explain to me their approach for problem E? I got it accepted in practice but barely under the time limit so I guess there are faster approaches.

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

What's wrong with checker of problem D ?

Test #12

2160 3240
7200 384

Вывод
5
1280 1080
3600 384

Ответ
5
640 2160
3600 384

Протокол тестирования
wrong answer reported answer can be reached in minimum 1000000002 operations, but given result is 5 operations

Seems like my solution gave a valid answer, but checker haven't noticed that!

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

What is the approach for Problem B?

  • »
    »
    5 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    My approach:

    1. If we determine the first two elements, we can determine the rest uniquely. -> Make an array X such that X[A[i]]=B[i] for every i. Then answer[i+2]=X[answer[i]] for any i.

    2. The student ID of the second person is X[0]. The first person in the line has a = 0, b = (second person's ID).

    3. The student ID of the first person is A[i] which does not appear in array B. Let id = the first person's ID. There's no student such that b_i = id, because there's no one in front of him! And conversely, the first person is the only student with such property.

    We can uniquely determine the first two persons, which determine the rest uniquely.

    The total run time is O(n).

    My submission

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

please provide the editorial section ..please ASAP

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

Can someone please explain their approach for problem E? I got it accepted in practice but barely under the time limit so I guess there are faster approaches.

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

    Just find the smallest feasible number every time. I'm not sure what is the best algorithm to find this, but my solution is: First, compare the length of the string with the one of last string, and there are three cases: 1. s[i].size()<s[i-1].size(): impossible, print "NO" 2. s[i].size()>s[i-1].size(): set all '?' to 0 (If it's the first number, set it to '1') 3. s[i].size()==s[i-1].size(): Find the first digit j where s[i] and s[i-1] is different. There are two cases: 1. j==s[i].size(Can't find) or s[i][j]<s[i-1][j]: Find the rightest '?' before digit j whose corresponding digit 'k' in s[i-1] is not '9'(because no digit is larger than 9). If you can't find it, also print "NO". Otherwise, set the '?' to 'k+1'. Then set all '?' before this digit exactly the same as their corresponding digit in s[i-1], and all '?' after this digit to '0'. 2.s[i][j]>s[i-1][j]: Similar to the case 1, but don't need to do the first step because this digit is already larger.

    • »
      »
      »
      5 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Thank you very much. It's very similar to my solution but I used a recursive approach and I guess that's what's taking my algorithm longer.

  • »
    »
    5 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    I used binary search to solve E.
    Here is my submission.

    • »
      »
      »
      5 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Could you please explain how you used binary search? I didn't understand very well by reading the code.

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

When I was trying to hack a solution by the output generator, I got a message "FAIL Line [name=public_key] equals to "#include <b...spond to pattern "[1-9][0-9]{0,999999}" (stdin) [validator val.exe returns exit code 3]".

I didn't understand the meaning of it. Please anyone help me how to hack in problem C if anyone use only 10^5 array where I need 10^6.

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

Do you guys know when the editorial will be published? I am really interested what is the idea for D and how one can figure it out.

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Every time Polycapus eats the chocolate, he set one of the number to 2/3 or 1/2. So what you should consider is, whether these two products is the same after dividing all 2 and 3. If it is, you can set the power of 3 to the same number by performing several 2/3 cuts to the one which has more 3, and then set the power of 2 to the same number by 1/2 cuts.

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

My C problem reached TL on test 42. How to solve this problem correctly?

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    you are using substr and strlen functions and they both have o(n) so your code complexity have o(n^2) runtime. find a way to get ride of them both.

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

    Use bigmod. I think your solution is N^2

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

The contest is useless. I don't gain any thing throng the contest. The problems are so silly!!!

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    throng -> through

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

    That's really offensive thing to say. They spent a lot of time to write/test problems and put them together.

    • »
      »
      »
      5 years ago, # ^ |
      Rev. 2   Vote: I like it +1 Vote: I do not like it

      Sorry for my rude talk.Forgive me please.

      But I do think the problems need to be considered.....

  • »
    »
    5 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    Can you be more specific? Can you describe what was silly for each problem?

    • »
      »
      »
      5 years ago, # ^ |
      Rev. 3   Vote: I like it +3 Vote: I do not like it

      Eh.. All the problem just add some details on the common problems. Or just problems for details.

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

How to solve B?

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

Is there any tutorial?

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

The B has a error in your problem description.To give an example, how do you solve in this data? 4 0 2 1 3 3 5 4 0

It's easy to infer that the correct answer is 1 2 3 4 5.. but many programe are not give me this answer but also accept it! please view.

  • »
    »
    5 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    In your example you have n=4, but 5 different ids in the list. Change n to 5 and add a line with "2 4" — then you will get 1 2 3 4 5.

    • »
      »
      »
      5 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Must change to 5? Oh.. Maybe i have not read this problem carefully..

  • »
    »
    5 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    Your test is wrong not enough information while n == 4 if n == 4 you test data be like this 4 0 2 1 3 2 4 3 0 if n == 5 4 0 2 1 3 2 4 3 5 4 0 try again

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

In case anyone is looking for the Editorial as I was, it has been posted here.

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

Hello, When the next contest will be hold?

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

could you help me please to figure out why WA on test 20 ? http://ideone.com/0AzRks

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

After reading the tutorial, I realized how unnecessary this solution by me is: http://codeforces.com/contest/490/submission/8829276