mohammedehab2002's blog

Codeforces Round #563 (Div. 2)
By mohammedehab2002, 5 years ago, In English

Hi!

I'm back with not one, not two, but three contests, although I have no promises about when to expect them....

The first of them, codeforces round #563, will take place on Jun/03/2019 17:05 (Moscow time). It's rated for the second division, but, as usual, first division participants can take part out of competition.

I'm the problemsetter of the round. I'd like to thank KAN for coordinating the round (and his patience .. try coordinating ~20 problems), arsijo for helping with the preparation, Um_nik, _overrated_, Aleks5d, wiwitrifai, pllk, Bedge, Ivan19981305, and PrianishnikovaRina for testing the round, and MikeMirzayanov for the great codeforces and polygon platforms.

In this round, you'll be given 6 problems and 2 hours to solve them.

UPD: I decided to drop the 3 seconds rule. The scoring distribution is 500-1000-1500-1750-2500-2500. That means you should probably read both E and F :D

Good luck & Have fun!

UPD: here's the editorial.

UPD: congratulations to the winners!

Div.1+Div.2:-

  1. tribute_to_Ukraine_2022
  2. E869120
  3. 800iq
  4. cerberus97
  5. Anadi

Div.2:-

  1. 800iq
  2. Alex18mai
  3. Mikaeel
  4. prick
  5. wasyl

See you in the second round :D

  • Vote: I like it
  • +466
  • Vote: I do not like it

Comments Comments (146)
| Write comment?
»
5 years ago, # |
← Rev. 2   Vote: I like it +24 Vote: I do not like it

Trying to make enough money to afford MIT. I see you. Only a few ten thousand dollars left np

→ Reply
»
5 years ago, # |
  Vote: I like it +90 Vote: I do not like it

By tradition, the scoring distribution will be announced 3 seconds before the contest.

So accurate!

→ Reply
»
5 years ago, # |
  Vote: I like it -13 Vote: I do not like it

Imagine next rounds you would actually use high precision clocks to measure the interval of announcing score distributions... ;)

→ Reply
»
5 years ago, # |
  Vote: I like it -7 Vote: I do not like it

The time is good for Chinese.

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

i m very excited about the new contest

→ Reply
»
5 years ago, # |
← Rev. 2   Vote: I like it -23 Vote: I do not like it

4 contests in the next 7 days, great.

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

Looking forward to more fun XOR questions

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

Upvoted for creating 3 contests simultaneously (Great job!).

Downvoted for not capitalize "Codeforces" and "Polygon".

→ Reply
»
5 years ago, # |
  Vote: I like it +10 Vote: I do not like it

It seems that another competition season has come

So many contests these days :)

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

mohammedehab2002 previous rounds all problem names start with "Mahmoud"(round #396), "Ehab"(round #525) and "Mahmoud and Ehab"(round #473). Guess which start is next ??

→ Reply
  • »
    »
    5 years ago, # ^ |
      Vote: I like it +13 Vote: I do not like it

    To get all the combination, Ehab and Mahmoud is the last one to start with :)

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

    the next start is an empty string

    → Reply
  • »
    »
    5 years ago, # ^ |
      Vote: I like it -14 Vote: I do not like it

    or maybe concatenated version MahmoudEhab:)

    → Reply
»
5 years ago, # |
  Vote: I like it -22 Vote: I do not like it

up to which rating second division?

→ Reply
  • »
    »
    5 years ago, # ^ |
    ← Rev. 3   Vote: I like it +8 Vote: I do not like it

    Upto 2099, but if div1 and div2 rounds are held simultaneously people who are rated greater than equal to 1900 must participate in div1

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

Wait, sorry guys I am new here. Do you have to start at that exact time? Or is there a contest window where we can choose to do it?

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

    You need to register before the contest starts. Registration's usually open till 5 minutes prior to the start of a contest. Some contests support extra reg which opens 10 minutes after the contest start and continues for 20 minutes.

    → Reply
  • »
    »
    5 years ago, # ^ |
      Vote: I like it +41 Vote: I do not like it

    There's no window, the contest starts at the same time for everyone.

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

Let's get ready to rumblllllllleeeeeeeeeeeeeeeeeeee!!!!!!!!!!!!!!

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

Hope this round not unrated and no accident

→ Reply
»
5 years ago, # |
  Vote: I like it -6 Vote: I do not like it

Do we have a hacking phase?

→ Reply
»
5 years ago, # |
  Vote: I like it +15 Vote: I do not like it

Scoring distribution shall be announced 3 seconds before contest

I decided to drop the 3 seconds rule. The scoring distribution is 500-1000-1500-1750-2500-2500.

Rules are made to be broken here. xD

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

You really like GCD and prime numbers ! (WOW!)

→ Reply
»
5 years ago, # |
← Rev. 2   Vote: I like it -12 Vote: I do not like it

Apparently, it wasn't.

→ Reply
  • »
    »
    5 years ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    if only query d at beginning and query s on LCA

    then will get FST

       /\
  /\ \
 /\ \ \
/\ \ \ \

    but I have no idea to solve it correctly now : /

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

How To solve problem D ?

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

    Think in terms of prefix xors.

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

    Instead of directly construct the array, just construct the prefix xor version of the array S1, S2,..., Sn. So the condition that no subseg has the xor equals to 0 or x is the same as the array S is distinct and there is no pair in S that xor of them is x.

    So iterate from 1 -> 2^n-1, if x^i also belongs to the given range [1, 2^n) then just pick one, or else pick both.

    To construct the answer array A from S, use : Ai = Si^S(i-1).

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

      What is prefix xor ? And why prefix xor array changes the condition?

      → Reply
  • »
    »
    5 years ago, # ^ |
    ← Rev. 2   Vote: I like it +16 Vote: I do not like it

    First observation, if there is no number x restricted. You can simply construct something like 1 2 1 4 1 2 1 8 1 2 1 4 1 2 1... This pattern is optimal because at every i-th number where i is 2^x, you cannot add any new number using numbers in range [1..2^x), any new addition will cause XOR to be 0. Hence, you will need to add 2^x every time you are at the 2^x-th number (1, 2, 4, 8, 16, ...).

    Now, what about the restricted number? Well you can simply take the biggest 2^x that forms x. e.g. if x is 5, restricted number will be 4. Now, you can simply skip the restricted number when you want to form the sequence. (Why biggest? To minimize the largest number in the sequence) e.g. x = 5 1 2 1 8 1 2 1 16 1 2 1 8 1 2 1, if at any time the number has exceed 2^n, just stop adding new sequences.

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

    What can be pretest 7?

    → Reply
    • »
      »
      »
      5 years ago, # ^ |
      ← Rev. 2   Vote: I like it 0 Vote: I do not like it

      I thought of something like 4 5. Answer should be longer than 1 2 1 8, ... But I initially only answered 1 2 1 which is a wrong answer

      → Reply
  • »
    »
    5 years ago, # ^ |
    ← Rev. 3   Vote: I like it +20 Vote: I do not like it

    I will explain the complete solution of D.

    Hint 1
    Hint 2
    Source Code
    → Reply
»
5 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I found B to be harder than C and D maybe because I was approaching it wrong way.

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

How to solve B? And why does it get TLE 55041556 ?

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

    To be honest, B is so simple. If we have both odd and even numbers, then the sorted array is the answer, or else just print the original array. Correct me if I'm wrong

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

      Maybe i dont understand you but what about this test: 3 2 3 2 1 (n = 5)

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

        It is 1 2 2 3 3. Because both odd and even exist, we can easily prove that using the allowed operation could sort the array, and obviously the lexicographically smallest array.

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

        "You can perform the following operation on it as many times as you want:". Of course i again solve another problem (( .

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

    if the input contains both odd and even numbers the result will always be a sorted array.

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

    You need to notice that, if there are only odd or only even numbers you can not make any swap. Otherwise you can sort the whole array.

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

    If we have at least one odd and another even number, we will be able to reorder the whole array (swap two odd numbers or two even numbers). For example, if you wanna swap two odd numbers o1 and o2, with an even number e, you can do:

    swap o1 and e, swap e and o2, swap e and o1.

    And the position of o1 and o2 was swapped as a result.

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

Hints for F please.

→ Reply
  • »
    »
    5 years ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    Firstly, ask the distance from 1. Now find the centroid of the current tree and check if it is an ancestor by checking if dist(root,centroid)+dist(x,centroid) == dist(root,x) , if that holds true, query the next node on the path and make it the root and now distance from root is dist(x,centroid)-1. otherwise, you can delete the subtree rooted at the centroid and everything else remains the same.

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

I've seen successful hacks made. Any hints what they were about?

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

    In my room there were submissions that checked if there is any odd numbers and sorted based on that (I found one more submission like that after the end of contest :@ ) .

    → Reply
  • »
    »
    5 years ago, # ^ |
    ← Rev. 2   Vote: I like it +3 Vote: I do not like it

    One of the mistakes on problem A was dividing sum of the numbers by 2 for checking the answer, while the sum is odd.

    Sorry for bad English. :)

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

      Your English is fine :D I understood

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

        Thanks :) I tried too much to write understandable.

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

How to solve F?

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

god damn it when I started to feel I'm halavin I failed on the fourth hack

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

    wow 3 hacks, how did you get them? was there a corner test case or something?

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

      Who did not solve A by sorting the array for some reason they were swapping the first different element from the first half with the other half and then print the array

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

        nice

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

        Isn't there an ambiguity in Problem A? In the end, it says "They must form a reordering of a. You are allowed to not change the order". What does that mean? As per my understanding, it meant that we're allowed rotate the array(so that ordering remains same) but, can't sort it completely.

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

          you are allowed to not change the order means you can leave it as it is and they must form a reordering of a means you can swap any element as many times as you want

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

Hints for F? Can it be solved using LCA?

→ Reply
»
5 years ago, # |
  Vote: I like it +30 Vote: I do not like it

→ Reply
  • »
    »
    5 years ago, # ^ |
    ← Rev. 2   Vote: I like it +14 Vote: I do not like it

    After solving (A,B,C)^zzzzzzzzzzzz
    01:59:29 Successful hacking attempt

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

How to think and approach problem D?

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

    Every number has only one number that makes (A^B == X) true. The problem asks for sub segment so it's another way to tell you to think of prefix xor. The answer should be clear after that.

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

fml, my bug on F was that I'm doing return the answer if there is only one edge from the current node, which should be the back edge to the parent, but I forgot the root!

Nice round, I really liked it :D

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

Any ideas for E? i see most solution use dp. can anyone explain!

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

    For E, $$$a_1=p_1^{e_1}p_2^{e_2}$$$. then max is $$$1+e_1+e_2$$$. that's exp step down. so $$$a_1$$$ must be $$$2^k$$$ or $$$2^{k-1}3$$$ (if $$$\leq n$$$). then remain is combination count. but no time to code.. the annoying $$$3$$$.

    Am I right? or is there a better solution? wait for tomorrow morning:) sleep now

    → Reply
  • »
    »
    5 years ago, # ^ |
    ← Rev. 3   Vote: I like it +64 Vote: I do not like it

    Actually, I didn't use DP. I used only multiplying. Seriously, I didn't use DP.
    My solution of problem E is as follows:

    ---

    Step 1: What is the maximal possible value of $$$f(p)$$$?
    Actually, the maximal possible value of $$$f(p)$$$ is as follows:

    • $$$1 \leq N \leq 1$$$ : $$$max=1$$$
    • $$$2 \leq N \leq 3$$$ : $$$max=2$$$
    • $$$4 \leq N \leq 7$$$ : $$$max=3$$$
    • $$$8 \leq N \leq 15$$$ : $$$max=4$$$
    • $$$16 \leq N \leq 31$$$ : $$$max=5$$$

    But why? Think about the sequence of {$$$g_1,g_2,g_3,...,g_k$$$} if you assume $$$g_1>g_2>g_3>...>g_n$$$. For example, if $$$A=$$$ {$$$4,6,2,1,5,3$$$}, the prefix gcd will be {$$$4,2,2,1,1,1$$$}, so $$$g$$$ will be {$$$4,2,1$$$}.

    Because the optimal sequence of $$$g$$$ is {$$$2^{k−1},2^{k−2},2^{k−3},...,4,2,1$$$}. Suppose $$$N=11$$$. For the sequence {$$$8,4,2,1,11,10,9,7,6,5,3$$$}, it is easy to find that $$$g$$$ is {$$$8,4,2,1$$$}.

    ---

    Step 2: How many ways are the maximal answer?
    Let's think about writing the value of $$$a_i$$$ which changed the prefix gcd. For example, if $$$g$$$ is {$$$8,4,6,5,2,3,1,7$$$}, the values that you will write will be {$$$8, 4, 6, 5$$$}.

    So what is the possible sequence of values which you will write? Suppose $$$N=8$$$.
    • 1st value: $$$8$$$. Only $$$1$$$ way.
    • 2nd value: $$$4$$$. Only $$$1$$$ way.
    • 3rd value: $$$2, 6$$$. $$$2$$$ ways are possible.
    • 4th value: $$$1, 3, 5, 7$$$. $$$4$$$ ways are possible.

    So the number of possible sequence of values which you will write will be $$$1 \times 1 \times 2 \times 4 = 8$$$.
    But, how about the other values? If sequence is {$$$8,4,6,5$$$}, you should insert other values: {$$$1,3,7,2$$$}.

    You should think about inserting in order of $$$1, 3, 7, 2$$$. (Insert from values which is not divisible of $$$2$$$, next not divisible for $$$4$$$, next not divisible for $$$8, 16, ...$$$

    • There are $$$1$$$ ways to insert $$$1$$$: Just after $$$5$$$.
    • There are $$$2$$$ ways to insert $$$3$$$: Just after $$$5$$$, or just after $$$1$$$.
    • There are $$$3$$$ ways to insert $$$7$$$: Just after $$$5$$$, just after $$$1$$$, or just after $$$3$$$.
    • There are $$$5$$$ ways to insert $$$2$$$: Just after $$$6, 5, 1, 3$$$, or $$$7$$$.
    So for each sequence you wrote, there are $$$1 \times 2 \times 3 \times 5 = 30$$$ ways of insertion. It means there are $$$8 \times 30 = 240$$$ permutations of $$$N=8$$$ which is maximal.

    So, let's move on another example. Suppose $$$N=11$$$:
    • Ways to write: $$$1 \times 1 \times 3 \times 6 = 18$$$ ways
    • Ways to insert the other values: $$$1 \times 2 \times 3 \times 4 \times 5 \times 7 \times 8 = 6720$$$ ways
    • So, the answer will be $$$18 \times 6720 = 120960$$$.

    Did you get it? :)

    ---

    Step 3: Thinking about other cases, not only $$$g =$$$ {$$$2^{k−1},2^{k−2},...,4,2,1$$$}
    If $$$N=12$$$, not only {$$$8,4,2,1$$$} is maximal, but also {$$$12,4,2,1$$$}, {$$$12,6,2,1$$$}, {$$$12,6,3,1$$$} is maximal.
    So you should brute force sequence $$$g$$$ (There are up to $$$O(\log N)$$$ ways). If you determined the sequence $$$g$$$, you can do like step 2: first calculate the number of sequence that you will write, second calculate the number of ways of insertions and finally mutiplies.

    Since the complexity for each $$$g$$$ is $$$O(N)$$$, so the total complexity is $$$O(N \log N)$$$. If you precount factorial and inversions, you can also solve with $$$O(N+\log^2 N)$$$.

    Sorry for my poor English.

    Code (Uploaded at 2:01 JST, 56 minutes after the contest)
    → Reply
»
5 years ago, # |
  Vote: I like it 0 Vote: I do not like it

seems that problem B shares the same trick with problem C in Global Round 3..

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

    what i felt and is true for most cases : if u don't know proof just input garbage and get gold.

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

    Yeah exactly same over here too!

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

I like problem E.

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

Thanks for mathforces (no)

→ Reply
»
5 years ago, # |
← Rev. 4   Vote: I like it +19 Vote: I do not like it

61741833-1039620329569556-2416932415634145280-n

X is 29.

Apparently this tc was not there in PreTest of F.

Naive Solns asking only type 2 queries will take atleast 8 queries.

Extending this tc — for $$$n*(n+1)/2$$$ vertices will take $$$n$$$ queries.

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

In the problem A, it was mentioned in the problem that order of the array should not be changed! What did that line mean and how do we reorder it without changing the order? Please Help!

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

    "You are allowed to not change the order", not "You must to not change the order"

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

    Please read it carefully : 

    You are allowed to not change the order.

    I guess u miss read it as:

    You are not allowed to change the order.
    → Reply
»
5 years ago, # |
  Vote: I like it +1 Vote: I do not like it

$$$A, B$$$ and $$$C$$$ felt like typing test. The situation drastically changed afterward :(

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

Very very nice contest, 10/10. Tl63 just because of using included sort. I hate this test -_-. But in general all it was good.

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

Can someone explain how to solve this example in F?

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

    You can use a method named "Heavy-Light Decomposition".

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

wow I got TLE on test 63 and my rating is gone

→ Reply
  • »
    »
    5 years ago, # ^ |
      Vote: I like it +15 Vote: I do not like it

    The test 63 is the hack I used during the round, so I want to clarify a couple of points:

    First that seems the only test with anti Quick sort test, I can see that 150+ solutions failed this test and if it wasn't added all these solution may have been passed, I've also always see Quick sort solution pass sys test in previous rounds, so I think such test should be always added to any problem that require sorting.

    Also I'm not very proud with that kind of hack, because all I did is check if a java solution that uses sort for primitive array, so I want to raise attention against this.

    For those who still want to use primitives you can build your own sort function like the radixSort function that I use in my solutions, wich I copied from uwi code ( I hope he doesn't mind )

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

JAVA submissions have been judged on the C++ time limit for the B question. Getting a TLE for an O(nlogn) approach. However I didn't use a StringBuilder which is bad on my part but still a correct solution should pass.

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

    Problem is not time limit. Problem is the inbuilt sort in Java have worst case complexity of n^2. Hence the time limit. There is blog somewhere on the codeforces discussing this SORTING IN JAVA.

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

Hope that the rankings would be revised. Otherwise rating would take a great hit.

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

55036867 why using StringBuildfer was wrong and a get TLE? java

UPD: 55052153 problem was oin primitive types, only in that, i am so disappointed=(

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

    Arrays.sort() for primitive data types in java uses quick sort O(n^2) while Collections.sort() uses merge sort . Quick sort can have O(n^2) complexity in worst case

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

      So what we are not supposed to use Arrays.sort() in future contests?

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

        in this particular case, we can use Integer insted of int, looks like its using another, type of sort, need to check source of class Arrays

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

      Also judgement was based on 1s time limit. For java it should be 2s i guess.

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

        Does the TLE differ for different languages?

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

          No I was mistaken. Checked previous contests and found that time limit is same for all languages. However many other platforms have extended limits for Java and Python.

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

    I also used BufferedReader, Arrays.sort(), and StringBuilder but still got a TLE :(

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

Lmao, I was so pissed when my naive solution for F had bugs and didn't get accepted, while so many people solved it.
Now I see anyway my naive solution would have got verdict Wrong answer on test 99xD.

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

Wrong answer on test 99. Such pain in F

→ Reply
»
5 years ago, # |
← Rev. 2   Vote: I like it +50 Vote: I do not like it

Problem F
qkbjv

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

I passed F in last 5 minutes. Exciting!

Thanks for the round very much!

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

Sorry, does the #23 testcase from F shows "wrong answer query limit exceeded"?

I made a submission with assertion but couldn't see that I was exceeding the limit. 55048680

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

    May be it's because you have a query in your main but you didn't count that ?

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

https://codeforces.me/problemset/problem/1054/D a question similar to today's D

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

Nice Contest.

Btw a simiar (a̶n̶d̶ ̶a̶n̶ ̶e̶a̶s̶y̶ ̶v̶e̶r̶s̶i̶o̶n̶?̶) of F recently appeared in Codechef Cook Off MYS00T

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

Problem F, a very nice tree and interactive problem! Thanks for writers.

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

Well I'm stupid and haven't debugged my F during contest, but here is my solution that should works but it got AC:

First of all let's know a height of x by first type of query with vertex 1, let it be H.

Let root be the lowest vertex that we know that has X in it's subtree.

So let choose a type of query randomly!

If we chose first type then just choose some vertex with height H in root subtree randomly and then change root!

If we chose second type then just change a root!

→ Reply
»
5 years ago, # |
  Vote: I like it +15 Vote: I do not like it

Was a randomized solution intended to pass for F?

Instead of using any heavy-light/centroid decomposition, I just picked a leaf at random for the current subtree, and then asked similar queries as given in the editorial.

https://codeforces.me/contest/1174/submission/55047296

→ Reply
  • »
    »
    5 years ago, # ^ |
    ← Rev. 3   Vote: I like it 0 Vote: I do not like it

    I dont think randomized was intended. But sometimes its difficult to break randomised. Until and unless probability of failing it is high.
    I did saw some randomised solns and one of my friend also used randomised in F.

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

Since editorial is not linked — EDITORIAL

→ Reply
»
5 years ago, # |
  Vote: I like it -16 Vote: I do not like it

^ Click here to go up! ^

→ Reply
»
5 years ago, # |
  Vote: I like it -15 Vote: I do not like it

how. to. calculate. the. contribution. ?why. the. contribution. for. me. is. negative. ? i. do. nothing. God

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

Very neat and clear problem descriptions. Thanks to the writer.

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

I think there is a time limit problem in problem 2 for java. Same logic gets accepted for C++ but TLE for java..

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

How to solve the problem D ??

→ Reply