JaySharma1048576's blog

Codeforces Round #785 (Div. 2)
By JaySharma1048576, 3 years ago, In English

Hello Codeforces,

I invite you to participate in Codeforces Round 785 (Div. 2) which will be held on Apr/30/2022 17:35 (Moscow time). The round will be rated for participants of Division 2 with rating strictly less than 2100. But as always, the participants of Division 1 are welcome to take part out of competition.

You will be given 6 problems and 2 hours to solve them. Moreover, there will be atleast $$$1$$$ and atmost $$$1048576$$$ interactive problems. So, you are advised to refer to the guide on interactive problems in case you are not familiar with them.

I would like to thank the following people for their contribution to the round:

This will be my second Codeforces round and based on the feedback by participants on Round #730, I have implemented the following changes:

  • The round will have no theme. I have tried to keep the problem statements as short and to the point as possible.
  • The pretests have been made stronger (specifically, more pretests have been included).
  • I have tried to keep the samples and their explanation as useful as possible.
  • No floating-point precision issues this time.
  • The I/O constraints for the interactive problem(s) have been kept low to provide a fair ground to the languages with slower I/O.

The score distribution will be released strictly before the round and the editorial will be released strictly after the round.

Good luck and have fun!

Disclaimer

UPD: The score distribution is $$$500-750-1500-2000-2750-3250$$$.

UPD: Editorial

UPD: Congratulations to the winners

Div. 1 + Div. 2 -

  1. ksun48
  2. Kirill22
  3. JiBaDa
  4. Geothermal
  5. errorgorn

Div. 2 -

  1. JiBaDa
  2. A-SOUL_Ava
  3. Beduardo
  4. happy-potato
  5. Moomer

First solves -

Problem Participant Time
A A_G 0:02
B Geothermal 0:05
C quailty 0:03
D BalintR 0:16
E zsxdcfv 0:27
F ksun48 0:11
  • Vote: I like it
  • +369
  • Vote: I do not like it

Comments Comments (123)
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3 years ago, # |
  Vote: I like it +64 Vote: I do not like it

As a taster.. I want to assure the round is well balanced, sweet and spicy

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    3 years ago, # ^ |
      Vote: I like it +68 Vote: I do not like it

    By looking at your pfp, i can confirm that you are a taster :P

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      3 years ago, # ^ |
        Vote: I like it +44 Vote: I do not like it

      Yes I do.. "easly morning dream" is always right!

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3 years ago, # |
  Vote: I like it +84 Vote: I do not like it

As a tester, problem statements are short and all have expected rating less than 1048576.

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

    Do you mean delta <= 1048576 but > 0?

    :)

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3 years ago, # |
← Rev. 2   Vote: I like it +60 Vote: I do not like it

As a tester still waiting for jay to set a round without an interactive problem

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3 years ago, # |
  Vote: I like it +43 Vote: I do not like it

As a tester, I agree that no powers of 2 were harmed.

JaySharma1048576 orz

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3 years ago, # |
  Vote: I like it +88 Vote: I do not like it

As a tester, I must say that it was quite hard to solve those 1048576 interactive problems.

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3 years ago, # |
  Vote: I like it +98 Vote: I do not like it

As a tester, I solved at most $$$1048576$$$ interactive problems.

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3 years ago, # |
  Vote: I like it +33 Vote: I do not like it

I care about tags :)

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3 years ago, # |
← Rev. 2   Vote: I like it -139 Vote: I do not like it

Oh no, mathforces again. Btw how did Mike let the "cheater" set a round again?

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3 years ago, # |
  Vote: I like it +57 Vote: I do not like it

As a tester, good luck.

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3 years ago, # |
  Vote: I like it +61 Vote: I do not like it

No powers of 2 were harmed but testers were harmed with the powers of 2.

The number 1048576 still haunts me :P

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3 years ago, # |
  Vote: I like it +14 Vote: I do not like it

Debugging interactive can be hard. Need this feature(google code jam has this) : https://mirror.codeforces.com/blog/entry/102256

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

    This most probably will not be implemented on codeforces as their team is small and this feature is not very important. Better to setup your own environment. -is-this-fft- blog will be helpful for this.

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3 years ago, # |
  Vote: I like it +14 Vote: I do not like it

It seems to be interesting contest

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3 years ago, # |
  Vote: I like it +36 Vote: I do not like it

I really liked all your interactive problems. Looking forward to this round as well.

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

Indian Round Let's Go!...

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3 years ago, # |
  Vote: I like it +27 Vote: I do not like it

Hey Jay! Good to see you back again with your interesting rounds!

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3 years ago, # |
← Rev. 2   Vote: I like it +124 Vote: I do not like it

No powers of 2 were harmed during the making of this round.

Not even $$$1$$$?

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

Why is 1048576 occursed frequently?

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

Excited for an Indian Round !

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

Hoping to retrieve my lost rating :)

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3 years ago, # |
  Vote: I like it -28 Vote: I do not like it

Hopefully I reach specialist today

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3 years ago, # |
← Rev. 2   Vote: I like it -31 Vote: I do not like it

.

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

Super excited for this round !!

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

good luck everyone

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3 years ago, # |
  Vote: I like it +33 Vote: I do not like it

Problem D IS SUCH A PAIN

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3 years ago, # |
  Vote: I like it +154 Vote: I do not like it

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

I really dislike Problem B. Problem B took me more than 1 hour, while I solved Problem C in less than 20 minutes. Proving that you algorithm works for Problem B is not straightforward, and while I got the main idea after a few minutes, I spent the entire hour trying to prove that it is true. I was extremely tempted to try proof by "accepted/try and see" vs mathematically proving a solution actually works, especially looking at the number of accepted submissions. At least I feel this shouldn't be encouraged for div2AB.

One might argue that I just suck/am slow at proving things, which is fair, but I honestly don't think I'm that much worse than an average Div2 contestant in my proof skills.

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

    Tf were you proving?

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

      Let $$$prev(i)$$$ denote the maximum index $$$j<i$$$ such that $$$s[j]==s[i]$$$. Then $$$s$$$ is balanced if and only if $$$s[prev(i), ..., i]$$$ is balanced for all $$$i$$$.

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

    shit happens. i solved ab in 14 mins and couldnt handle with c lol

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

    You can just check for all pairs of letters that condition is satisfied with prefix sums in n * 26^2, and there is nothing to prove there.

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

      Did that pass? The time limit was $$$1$$$ second for $$$2*10^5*26*26 = 135200000$$$ which I thought wouldn't pass. I ended up with $$$O(26n)$$$ by noticing a counter example can always start and end with the same character.

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

      another way was calculating total number of distinct letters in the string (let it be) k, and for all 1 <= i, j <= n, s[i] == s[j], j > i you can check if j — i >= k. it must be true for all such pairs

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

    my solution was to count the number of distinct characters. Now let's say the string s has 4 distinct characters. Then the first 4 characters must be some permutation of these 4 characters, and every set of 4 consecutive characters after must be the same permutation. EX: abdcabdcabd

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

      That passed the pretests? I thought of that and then realized the counterexample, what if the string doesn't have a permutation of the $$$d$$$ distinct characters as a prefix of length $$$d$$$ and instead has something like $$$abbabcabcabc$$$.

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

        but for this case, the answer is NO, obviously, isn't it?

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

        It's no because of the substring 'bb'. Here the difference between the number of b's and a's is 2

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

    only need to check if distance of two adjacent same characters is the number of distinct characters in s.because each of them should appear exactly once in this substring.

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

      Thanks, I didn't think of that. I ended up checking adjacent same characters, but brute-forcing the count for characters in the range using presum table.

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

      How to implement that in O(n)?

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

        Check out my solution, if you want. I had the same approach. 155410182

        UPD: It is very bad implementation probably, but idc :D

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

          I like your implementation. Here is another way to do the last part of your code.

          while (sz(ans) < sz(s)) ans += ns;   
if (ans.substr(0,sz(s)) != s) cout << "NO";
else cout << "YES";
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            3 years ago, # ^ |
              Vote: I like it -6 Vote: I do not like it

            Oh yeah... This looks much more easier, thanks :D

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

    Probably, you choose a harder approach. My approach:

    Let's say the string has k distinct characters. Then the first k characters of a must contain all distinct characters, otherwise frequency of at least one letter in first k characters of string will be >= 2, making it imbalanced. Then, this substring of prefix must repeat, if it doesn't you can take [2...k+1], it will be imbalanced.

    Code of above approach
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3 years ago, # |
  Vote: I like it +2 Vote: I do not like it

How can I solve problem D?

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

How do you solve problem C. I tried to do a knapsack variationy sort of thing, but it counted the solutions 1 3 1 and 3 1 1 as different.

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    3 years ago, # ^ |
      Vote: I like it +9 Vote: I do not like it

    you can probably google it by searching "unbounded knapsack"

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

    similar problem: CSES Coin Combinations II

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    3 years ago, # ^ |
      Vote: I like it +9 Vote: I do not like it

    This is simply solvable. Assume $$$p$$$ is a set of weights. Then usual knapsack algorithm is

    for (int i = 0; i < n; i++) {
    for (int j : p) {
        if (i + j < n) ...
    }
}

    Do instead

    for (int j : p) {
    for (int i = 0; i < n; i++) {
        if (i + j < n) ...
    }
}
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      3 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Could you please explain why this works though?

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

        The idea is that once you pick a weight w[i], you make sure that you pick all possible amounts of it now so that you don't pick the w[i] again in the future. That way you wont have a case like 1,3,1.

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

      Also instead of for (int i = 0; i < n; i++) if (i + j < n) { ... }

      you can do for (int i = 0; i+j < n; i++) { ... }

      (not important for correctness or speed)

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

    Backpack Counting.

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3 years ago, # |
← Rev. 2   Vote: I like it +17 Vote: I do not like it

Perhaps I overcomplicated the casework in D but come on... This problem was just painful. On another note, does anyone know why mixing puts and cout results in a wrong answer?

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

    Probably, you did not use flush in $$$cout$$$. Then they alternate flushing and produce mess in output. For $$$cout$$$ it is $$$cout.flush()$$$ and $$$cout « endl$$$. $$$puts$$$ always flushes.

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

      That's probably it (since I'm not using endl). I'll be more careful next time (thank god I was able to suspect that puts was the problem).

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

    always use same output method.

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

    You are using ios_base::sync_with_stdio(false), which does not preserve the order of output streams.

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

I literally submitted problem D 2 seconds before the contest ends but it said the contest is over... I forgot to take the answer mod 1e9 + 7. So sick of crap like this...

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

ModuloForces

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

In problem D, let say for a common difference d and first number c, how do you find smallest number in that sequence which exist in sequence B but not in sequence C?

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3 years ago, # |
  Vote: I like it +36 Vote: I do not like it

Trash Indian Round. Terrible problems : C , D . Maybe F is a good one , but generally a trash round.

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

how to solve E and F? E: I know from one index, at most 20 power could be extended. But don't get how to deal with K.

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    3 years ago, # ^ |
      Vote: I like it +26 Vote: I do not like it

    F: The goal is to find a good way to assign 0~1023 to all the cells, and edge connecting u and v is u^v. Gray code will do.

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

      $$$2^5$$$ x $$$2^5$$$ gray code is yielding a sum of approximately 84k

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

        You should use the 0, 2, 4, 6, 8-th bit for one dimension and 1, 3, 5, 7, 9 for the other dimension.

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

    For problem E, consider for a certain $$$(i, j), i \le j$$$, how many times does $$$A_i ^ {A_{i + 1} \cdots A_j}$$$ contribute to the final answer, when putting XOR both between $$$A_{i - 1}$$$ and $$$A_i$$$, and between $$$A_j$$$ and $$$A_{j + 1}$$$.

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

    We just have to count the number of times $$$pow(l,l+1,\ldots,r)$$$ appears in the final expression. As you said, $$$r-l \leq 20$$$ holds since we take the answer modulo $$$2^{2^{20}}$$$, so we can brute force all pairs.

    The sequence will look something like $$$\ldots A_{l-2} ~?~ A_{l-1} \oplus A_l ~\hat{}~ \ldots ~\hat{}~ A_r \oplus A_{r+1} ~?~ A_{r+2} \ldots$$$. We just need to count the number of ways to fill in the $$$?$$$ such that we satisfy the number of $$$\hat{}$$$ being more than $$$k$$$. This is simple combinatorics and we get some expression of the form $$$\sum\limits_{k=0}^m \binom{n}{k}$$$.

    Remember we only need this expression under modulo $$$2$$$, also we will only require the value for about $$$20$$$ values of $$$n$$$ since $$$r-l \leq 20$$$, so just do prefix sums.

    Btw, you can find $$$\binom{n}{k}$$$ under modulo $$$2$$$ easily as $$$\binom{n}{k} \pmod 2= 1$$$ iif $$$(k \& n)=k$$$. We can show this using Kummer's theorem, Lucas's theorem or just staring at an image of a Sierpiński triangle

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

Problem C is similar to this problem

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

Tests for problem B is very weak. My submission https://mirror.codeforces.com/contest/1673/submission/155406728 fails on test

1
abaabba

of which the answer should be NO, but this code prints YES.

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

The pretests have been made stronger (specifically, more pretests have been included).Are you sure for problem B?:(

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3 years ago, # |
  Vote: I like it +39 Vote: I do not like it

So much 2 in problem E...

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

Can somebody please tell me what's wrong with my approach in C:

I made an array of all palindroms from 1 to 1e5 (v) (v[j] — j-th palindrome starting from 1) I also made a 2-dimensional array dp, when for each i<=4*1e4 and j that j-th palindrome (lets say v[j]) (starting from 1) is no more than 1e5, dp[i][j] = number of different sums of palindromes which are equal i and in each sum all the palindroms used are equal or less than v[j]. I filled this array for i from 1 to 40000

for each i I found certain l that v[l] (l-th palindrome) <=i, but v[l+1]>i;

Let's notice that dp[i][j] (v[j]<=i) = dp[i][j-1]+dp[i-v[j]][j], because for each sum of palindromes equal i, and each palindrome <=v[j], either in this sum is included v[j] as a palindrome, and all the other palindromes are <=v[j], (dp[i-v[j]][j] sums in total), or v[j] is not included => all the other palindromes are <=v[j-1] (dp[i][j-1] sums in total).

since dp[i][0] = 0, then for each 1 <= j <= l dp[i][j]=dp[i][j-1]+dp[i-v[j]][j]. if j>l, dp[i][j] = dp[i][l], because i < v[l+1] <= v[j] (therefore in each possible sum of palindromes equal i every palindrome is <=v[l], and dp[i][j] = amount of all possible sums of palindromes equal i = dp[i][l].)

After filling our array for each 1 <= i <= n, we can print dp[n][r], if n < v[r] <=1e5 and it will be our answer, (dp[n][r] = number of sums of palindromes equal n, and therefore each palindrome <= n < v[r], so dp[n][r] = amount of all possible sums).

I will be glad if someone can explain what's wrong with my approach. Or is it correct and my realisation is wrong?

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

    I looked at your solution. Your dp is correct, but you output the dp[n][i], such that it has the maximum value (since it's in modulo 1e9+7 that would be wrong). Just output dp[n][590] and that would pass the testcases

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

      Thank you so much, AC now)

      (OMG i can't believe I made this mistake, i'm such a moron. No words for that)

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3 years ago, # |
  Vote: I like it +14 Vote: I do not like it

Technical question: The API was unavailable while contest. I this something that will happen now more often intentionally, or a coincidence?

I use some tooling that uses the API to parse problem statements, tests etc.

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

    I met the same situation: cf-tool warns Cannot find csrf, and I can neither login nor parse problem statements.

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

What's an upper bound on the number of divisors of $$$n$$$?. I mean, is it something like $$$2$$$ to the power of [number of smallest prime numbers with product up to $$$n$$$]? For example if $$$n = 10^9$$$ I can take up to $$$2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29$$$ so is $$$2^{10}$$$ a safe upper bound? I was a little bit confused about iterating divisors in problem D, so I didn't even try, but I guess that's what I had to do.

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

Just asking: what is the point of making F interactive? I don't think there exist offline solutions since you can just ask all cells anyway.

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    3 years ago, # ^ |
      Vote: I like it +31 Vote: I do not like it

    Even if I ask the queries offline, the grader still needs the length of the roads to generate the queries which itself needs the problem to be interactive.

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

I found problem C very hard to solve. At first, I tried to formulate the number of ways to create the integer using only 1-9, but got stuck there. Can anyone please tell me how to approach this kind of problem and solve it?

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

    You have to realize that there are only ~500 palindromes less than 4e4. So you can simply do dp on this where dp[i] is the number of ways to get sum i. The dp method is mentioned somewhere above.

    My submission: 155444840

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

    My thought process (and I think pretty much everyone else's) was, 'This $$$n \le 4 \cdot 10^4$$$ constraint is a bit weird, surely there must not be too many palindromic numbers anyway', so I quickly generated the palindromes and saw there were around $$$500$$$, so the constraints were enough for a knapsack, and I just had to be careful to first iterate the palindromes in the knapsack to count only ordered sums.

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

    You can find all the palindromes less than 4e4 with a simple python script. There are less than 500. after that, Visit the land of CSES Problemset, and find the one named Coin Combinations II. It will guide you.

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3 years ago, # |
  Vote: I like it -20 Vote: I do not like it

Generating all palindromic numbers for C was available online on GFG https://www.geeksforgeeks.org/generate-palindromic-numbers-less-n/

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

    I'm pretty sure that you can use google, also it's not even that hard to generate palindromes

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

oh

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    3 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it
    Input
    Expected Output
    Your Output

    Although your submission outputs the correct answer, it has a non-zero exit code, with stack smashing errors.

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3 years ago, # |
  Vote: I like it +18 Vote: I do not like it

Ratings updated preliminarily. We will remove cheaters and update the ratings again soon!

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

Glad to be green again :D Thanks for this amazing contest!

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

If you prefer command line, checkout CF Doctor for more flexibility and faster feedback.

If you are/were getting a WA/RE verdict on problems from this contest, you can get the smallest possible counter example for your submission on cfstress.com. To do that, click on the relevant problem's link below, add your submission ID, and edit the table (or edit compressed parameters) to increase/decrease the constraints.

If you are not able to find a counter example even after changing the parameters, reply to this thread (only till the next 7 days), with links to your submission and ticket(s).

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3 years ago, # |
  Vote: I like it -30 Vote: I do not like it

The problems were so interesting and fun. Unlike other contests, AB was really good. I like them

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

When you find out that all div2 winners are alt accounts

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

Really? just 2 seconds? Humans can do that?

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

in this submittion: https://mirror.codeforces.com/contest/1673/submission/155426364. I write vector cnt(26,0); in line 23. then I have Runtime error on test 11 but I write int cnt[26]; for(int k=0; k<=25;k++) cnt[k]=0; then there is no error. I want to know WHY???

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

I am avinash204, I am writing about the C problem of the contest. It happens that my solution coincides with another solution. The C problem was similar to a problem that I have done before on CSES (https://usaco.guide/problems/cses-1636-coin-combinations-ii-ordered/solution) and I happen to use the some part of the solution which I submitted last time on CSES. The AI has detected my solution to be coinciding with another user. Please look into the matter.

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