chokudai's blog

By chokudai, history, 4 years ago, In English

We will hold AtCoder Beginner Contest 178.

The point values will be 100-200-300-400-500-600.

We are looking forward to your participation!

  • Vote: I like it
  • +106
  • Vote: I do not like it

| Write comment?
»
4 years ago, # |
  Vote: I like it +8 Vote: I do not like it

Reminder: Contest starts in 10 mins!

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

Are there any other contest on atcoder for beginner? ABC are very good but aren't that frequeant and i tried AGC and i shouldn't have... so any other recommendation

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

    The ARC's should be doable, they're gonna be frequent again soon.

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

deleted

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

    Inclusion Exclusion. Number of ways that we can pick any value from 0 to 9 = 10^N . Number of ways that 0 doesnt occur = number of ways to pick any value from 1 to 9 = 9^N. By same logic, number of ways that 9 doesn't occur anywhere = 9^N Number of ways that both 0 or 9 don't occur = 8^N Hence answer is 10^N — 9^N — 9^N + 8^N . Use mods wherever required

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

      I tried (nC2)*2*10^(n-2). nC2 to choose any two places for 0 and 9. 2 because {0,9} has two permutations. 10^(n-2) to fill remaining places with 0-9. I am still not sure why this approach is wrong. Could someone point out?

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

        You are counting duplicates

        for example,

        when 0,9 is fixed like this 0_9 and like this 09_ in both cases you are counting 099

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

    Why so many downvotes he is just asking the approach didn't say anything wrong or asked anything wrong U can just use Inclusion exclusion

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

      He asked during contest which is not allowed

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

        I was just try to be first so i can get reply so i can get reply quickly without having intention to get solve yes i can give these immediate after contest but i have a online class during the contest for that i gave that during the contest to be honest i have no intention to get a solve from here during contest

        • »
          »
          »
          »
          »
          4 years ago, # ^ |
            Vote: I like it +31 Vote: I do not like it

          The issue is that even if you weren't going to cheat yourself, if someone answered you during contest, anyone else could have seen the answer too, which wouldn't have been fair.

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

    In this question you need to find all sequences so:

    1. There is some i such that Ai = 0 (count of such i may be > 1)
    2. There is some i such that Ai = 9 (count of such i may be > 1)

    (both conditions must be met)

    Let's iterate k — number of positions i such that Ai = 0/9

    Number of ways to create "good sequences" with 0/9 is 2^k — 2 (on every position you have 2 ways to put 0/9) (-2 because you don't need to create sequences 00..000 and 99..999)

    Let's count C(n, k) — the number of ways to select a set of i positions from n positions, where we want to put 0/9

    Finally we want to fill the remaining positions with number 1..8. It is equal to 8^(n — i)

    And final number of ways for some k equal to C(n, k) * (2^k — 2) * (8^(n — i))

    P.S. more details about Binomial Coefficients you can read here Binomial Coefficients

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

Any ideas on how to do F? If the frequency of any number in first array is greater than n — (frequency of that element in the second array) than answer is "No" else answer exists.

I made a logic that the maximum frequent number in the first array will get the least frequent number from the second array. If there are multiple numbers with the same frequency then the smallest number with the smallest frequency goes to the largest number with the largest frequency. I was not quite able to implement it.

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

    I understand, I thought there's just 1 minute left which won't really make any difference. One cannot implement it that quick. Sorry though I should have been a bit more patient

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

Solved E but coudnt solve C and D :(.

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

    approach for e

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

      There are only two cases to consider, if we only consider results such that Xi <= Xj.

      1. If Yi <= Yj, then the distance is (Xj + Yj) — (Xi + Yi)
      2. Otherwise, the distance is (Xj — Yj) — (Xi — Yi)

      By breaking it down into these cases, I have gotten rid of the absolute value function making it much easier to reason about the distances.

      So, we simply pick points with minimum and maximum x+y, and compute the distance.
      Then pick points with minimum and maximum x-y, and compute the distance.
      One of those two distances is your maximum.

      This can be done in O(n).

      My submission

      • »
        »
        »
        »
        4 years ago, # ^ |
        Rev. 2   Vote: I like it +25 Vote: I do not like it

        Copy-pasted from: stackoverflow Why can't you provide the link directly instead of copy pasting the text XD

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

          its simple as you prob. dont want to jump from one site to another for a simple reason ,so posting a simple basic explanantion is not that big of issue

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

      The manhattan distance of two points is the distance of the parallel diagonals of these points.

      The points at a distance x from a given point p form a diamond.

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

How to do D?

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

    Simple DP. very similar to this

    Just coins here are {3, 4, 5 .. N}

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

      But using coin change dp method does not give us the different permutations right? Like for sum = 7. the answer = 3 ( {3,4} , {4,3} , {7} ). But dp gives us only answer = 2 ( {3,4} , {7} ). Please correct me if I'm wrong

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

        Imagine filling DP array for n = 7 by this code:

        Spoiler

        Hope you understood

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

          can you please describe how can we implement it by top-down dp approach?

          I have implemented this but it isn't correct.

          • »
            »
            »
            »
            »
            »
            3 years ago, # ^ |
            Rev. 3   Vote: I like it 0 Vote: I do not like it
            Code

            UPD: https://codeforces.com/blog/entry/82566?#comment-697474

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

    I did it using Combinatorics, Stars and bars. Divide the number into x sums, (1 to s/3) and for each, find all possible combinations, by keeping each of it exactly 3 initially.

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

      I also did the same,this approach is bit intuitive as this revises the P&C concepts :)

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

      Could you elaborate more on this approach?

      • »
        »
        »
        »
        4 years ago, # ^ |
        Rev. 2   Vote: I like it +11 Vote: I do not like it

        Sure.
        Firstly let's notice that we can only have sequences of size from 1 to s/3. (because each one should be at least 3)

        Then, let's call these "number of groups" as x,

        example: For s=9, we have,
        x=1, {3} ; (with 6 still remaining)
        x=2, {3,3} (with 3 still remaining)
        x=3, {3,3,3} (with 0 remaining)

        Now, we can distribute the remaining sum, for each x, among the x groups, in (n+r-1)C(r-1) ways,
        This is, distributing n identical objects among r people (stars and bars).
        So iterating over each x from 1 to s/3 by calculating the combination for each iteration, it's easy to find the answer.

        Submission

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

How to solve E? I tried to find smallest and largest points on x and y axis(total 8 points) and found maximum among them.

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

      Thanks! Completely forgot to break mod, went in the wrong direction. :(

  • »
    »
    4 years ago, # ^ |
    Rev. 3   Vote: I like it +1 Vote: I do not like it

    I removed the mod (absolute value) by considering 4 cases.

    case 1 : xi > xj and yi > yj

    we need to find now max(xi-xj + yi-yj) = max((xi+yi)-(xj+yj)) = max(xi+yi)-min(xj+yj)

    for other cases I multiplied coords by -1 and solved by case 1 again

    case 2 : multiply all x by -1

    case 3 : multiply all y by -1

    case 4 : multiply all x and y by -1

  • »
    »
    4 years ago, # ^ |
    Rev. 3   Vote: I like it +11 Vote: I do not like it

    Manhattan distance between two points is: |$$$x$$$a — $$$x$$$b| + |$$$y$$$a — $$$y$$$b|.

    Now if can be evaluated in 4 ways:
    $$$dist$$$ = ($$$x$$$a — $$$x$$$b) + ($$$y$$$a — $$$y$$$b)
    $$$dist$$$ = ($$$x$$$a — $$$x$$$b) — ($$$y$$$a — $$$y$$$b).
    $$$dist$$$ = -($$$x$$$a — $$$x$$$b) + ($$$y$$$a — $$$y$$$b).
    $$$dist$$$ = -($$$x$$$a — $$$x$$$b) — ($$$y$$$a — $$$y$$$b).

    Now if you simplify them a little then the distance can be one of the two below:
    1) $$$dist$$$ = ($$$x$$$a + $$$y$$$a) — ($$$x$$$b + $$$y$$$b)
    2) $$$dist$$$ = (-$$$x$$$a + $$$y$$$a) + ($$$x$$$b — $$$y$$$b)

    Now the distance will be maximum from above expressions.
    submission link

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

      Thanks. Here is shorter implementation I did with your help.

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

        Oh, i see. It's really concise. 80% of time i overkill (: problems.

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

    the approach to solve was same as this question.

    you were just given distance function instead of f(i, j)

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

    Alternatively, you can calculate the convex hull of the given points and brute force the vertices of the polygon.

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

      This will TLE if to much points are on the convex hull, it could be all.

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

        Yes, you're right. However, I got an Accepted verdict. I guess the test cases are not strong enough.

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

Pure Mathematics Contest!!

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

Maybe E is just a too well known problem, but F solution was more obvious to me than E solution.

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

    how to solve F ? It seems easy at first.

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

    how to solve F ? Think about 70 mins and more but still fail to figure

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

    What was the approach for F?

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

    Can you please explain your solution to E, I don't know which one test case was failing, I got 17 AC and 1 WA

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

      This might help https://atcoder.jp/contests/abc178/submissions/16705345 I sorted the points considering points to be equal if they lie in the same circle centered at the origin

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

      Put three comparators. one comapring x1-y1 other with x1+y1 one with y1-x1 Take the max of difference of the end terms and max of three would be the answer.

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

      Here is my take on E,

      ans=(x2-x1)+abs(y2-y1) ,x2>x1

      Now we have two options for answer

      ans1=(x2+y2)-(x1+y1) ==== max(x+y) -min(x+y) for optimal answer

      ans2=(x2-y2)-(x1-y1) ==== max(x-y) -min(x-y) for optimal answer.

      Final answer would be the max of two.

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

        Hey can you please suggest which one test case is failing for this solution

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

        I have a doubt. The two cases are coming from assumptions, like case 1 comes from y2 >= y1. So how is it that when we calcuate the answer, we don't check for these restrictions and still get the correct answer. Can't it be that for i and j which gives max(xi + yi) — min(xj + yj), yi is actually less than yj which makes that pair invalid?

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

          Actually I have this same doubt and I have used the same assumptions that invalid pairs wont affect the best answer for the problem and got a AC, I was really trying to find a intuitive or mathematical proof for the same but havent got one, I would really appreciate if someone comes up with a proof for the same.

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

          Here we are only concerned about the largest value

          Lets get our definitions clear first

          option1 = (xj-xi) + (yj-yi) for yj>=yi

          option2 = (xj-xi) + (yi-yj) for yi>=yj

          Now for yj>=yi we can never have option2>option1 and vice versa.

          So though we are considering some invalid pairs our answer is never going to be affected

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

    Can you check why I am getting WA on F. here is submission

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

    More than being well known, E is pretty easily searchable I think.

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

MathCoder literally :V

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

What are the solutions for problems E and F?

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

    e is well-known, u can google it

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

      I googled couldn't find the same problem, can you please provide a link where solution of this problem is explained. like gfg or something?

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

Hi, this is my code for F problem

I am failing in 3 test cases and I couldn't figure out those TCs,

you can find my code in this link please help me

https://ideone.com/7PaZ9k

My submission link https://atcoder.jp/contests/abc178/submissions/16710181

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

    can I get any test case for which my code is failing?

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

      there is dropbox link to all atcoder test cases in one of the rng_58's blog

»
4 years ago, # |
  Vote: I like it +118 Vote: I do not like it

F solution: Reverse array B, then A will be in ascending order and B will be in descending order. If we now look at positions where A[i] == B[i], they will form a segment [l, r], and will all have equal element A[i] == B[i] == C. Then you just need to swap all positions i from this segment with such positions j that A[j] != C and B[j] != C. And if there are not enough such positions j then I think it's not hard to prove that reordering is not possible at all.

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

    I think reordering is not possible only in the case where lets say x occurs m times in A and k times in B and both k>=(n+1)/2 and m>=(n+1)/2. By pigeonhole principle!

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

      No consider A = [1 1 1 1 2] B = [1 1 2 3 4], maybe we can say that frequency of some element must obey $$$f_A + f_B <= n$$$ ...

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

        Ya I missed it thanks , Amazing F!

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

        How to prove that answer is always Yes when every f_A + f_B <= n

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

          That is shown from answer of madn. When A is ascending and B is descending, and lets say $$$f_A + f_B = N$$$ for boundary condition, maximum intersection is possible if both A and B have near equal frequencies, intersection will be of length $$$\lfloor N/2 \rfloor$$$. So we have adequate numbers ($$$\lceil N/2\rceil$$$) to put in this positions of overlap, so answer is always possible

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

            oh,actually its very obvious...what am i thinking...Thanks!

  • »
    »
    4 years ago, # ^ |
      Vote: I like it +25 Vote: I do not like it
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +15 Vote: I do not like it

    This is the best solution I have ever read!!

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

      Yeah I find it very simple and easy to see correctness. In my contest I was just wondering why they gave A, B in sorted order, maybe it was hint toward this solution

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

      Yes, key insight is that only a[i] = b[i] = c will be the only value, and form an interval

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

        Can you prove it ?

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

          if there are two different value that a[i] = b[i], let a[i] = b[i] = c, a[i+1] = b[i+1] = d, and c != d

          a[i] < a[i+1], so d > c

          b[i] > b[i+1], so d < c

          contradiction

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

    I am not claiming following is correct but I also cannot see why it is wrong. It must be wrong because it gives runtime error. The process was:

    1. Find net frequency of each element in the two arrays. If it is greater than n, then it is not possible to rearrange.
    2. Otherwise we use a max-heap to store pair of {"frequency in B", "number"}. Now we loop A from 1 to N and simply pop from heap and check it with A[i]. If not equal then we place it in our final array. Otherwise we pop another value and place it in final array. We push back the changed frequency.

    I think it runs into dead end where it no longer finds a suitable pair, but why it happens? Thanks!

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

    i think thats the easiest code to read and understand.

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

Man problem C ruined the round for me!:) Can anyone please explain the approach for it?

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

    10^n-2*9^n+8^n Inclusion exclusion

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

      Wow, thanks! Very interesting!

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

      Can you please help me to understand what wrong with this approach.

      Take two numbers and assign 0 and 9 to them , ways=n*(n-1)

      remaining (n-2) numbers can be arranged in 10^(n-2) ways making

      ans=n*(n-1)*10^(n-2)

      Thanks.

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

        The 10 ^(n-2) replicates in many cases when n>=3 see lets say you place 0,9 at 1 2 so possible values are 0 9 1, .....0 9 9

        SImilarly suppose in another permutation you fixed 0 9 at position 1 3 and other 10 values accordingly at remaining position .Observe both the vectors carefully.

        0 1 9........0 9 9.

        You can clearly see 0 9 9 appearing common in both thus WA!

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

        There will be repeation, like in your solution you will count (2 1 9 2) twice.

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

      Intuition behind this:

      There are $$$x$$$ ways of choosing which of $$$x$$$ numbers to place in a spot, each choice is independent of each other, so for $$$n$$$ places there are $$$x^{n}$$$ ways.

      Number of ways of placing numbers in the range $$$[0, 9]$$$ -> $$$10^{n}$$$.

      Number of ways of placing numbers in the range $$$[0, 8]$$$ or $$$[1, 9]$$$, that is the number of ways of generating an array that is guaranteed to be missing either a $$$0$$$ or an $$$8$$$ -> $$$9^{n}$$$ each.

      But note that we have double counted the number of ways to place $$$[1, 8]$$$ (both missing) while removing the bad array count, so add back the number of ways of doing that which is $$$8^{n}$$$.

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

    You can use DP for accounting for double counting

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

      Thanks! I tried to do a dp approach but i didn't manage to solve it during the contest! In my opinion, D was easier than C.

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

        Exactly! I was stumped by C as well. Here's my DP solution to C

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

    Answer = (total sequence with length n and all element <= 9) — (total sequences such that it contains neither 0 nor 9)

    first term above is simply pow(10,n) for the second term I used inclusion exclusion:

    A = Set of sequences that doesn't contain 0

    B = Set of sequences that doesn't contain 9

    n(A union B) = n(A) + n(B) - n(A intersection B)

    n(A) = n(B) = pow(9,n)

    n(A intersection B) = pow(8,n)

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

what's wrong in my solution of F? getting wrong answers on 6/50 test cases?

https://atcoder.jp/contests/abc178/submissions/16721049

UPD: I did some changes in my above code but still it's giving wrong answer on 3 test-cases :( https://atcoder.jp/contests/abc178/submissions/16724818

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

Anyways to do C or D without DP ?

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

    C is simple inclusion — exclusion. My one line code :

    cout << sub(add(power(10, n), power(8, n)), add(power(9, n), power(9, n)));

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

    D could be solve by brute force the number of elements in the sequence and the do a star-and-bar-like counting trick.

  • »
    »
    4 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it
    My approach of D (Without dp ) :
    First of all fix the length of the sequence .
    Suppose length is 5 ;
    so a1 + a2 + a3 + a4 + a5 = S ; now for all i , a[i]>=3 ;
    using stars and bars technique  ,
    no of solution  of that equation is ncr(s-3*length +length -1 , length-1);here length = 5 ;
    final answer = sum of solutions for each fixed length .
    
    Code
  • »
    »
    4 years ago, # ^ |
    Rev. 2   Vote: I like it 0 Vote: I do not like it

    I did D using combinatorics. It is the modified version of distributing n balls into r urns such that each urn contains at least 1 ball.

    Solution of standard problem: (n-1)C(r-1).

    Here problem can be reformulated as if you can use any number of urns (because the length of sequence can vary) and each urn should contain at least 3 balls.

    Solution: summation of (n-1-2*i)C(i-1) for all i, where i is the number of urns and n is the sum we want. The term -2*i because we first put 2 balls in every urn so that problem is now transformed into a standard one containing at least 1 balls, for each i.

    https://atcoder.jp/contests/abc178/submissions/16705277

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

For problem C: I'm thinking like this, place 0 and 9 in any two places (nC2) then other n-2 places will have 10 possibilities hence (10^(n-2)). So overall answer is nC2 *(10^(n-2)).

Can anyone say what fact am i assuming wrong/missing. Thanks :)

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

    nC2 * 10^(n-2) will duplicate

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

      can you give correct combination formula for this problem ?

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

        I used dp https://atcoder.jp/contests/abc178/submissions/16709001 Here,

        • dp[n][0][0] means sequence of length n and 0 zeros and 0 nines,

        • dp[n][0][1] means sequence of length n and 0 zeros and atleast 1 nine,

        • dp[n][1][0] means sequence of length n and atleast 1 zero and 0 nines,

        • dp[n][1][1] means sequence of length n and atleast 1 zero and atleast 1 nines

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

      so the 0's and 9's are getting duplicated?

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

        In cases when you put 9, 0 like this: ab90 09cd sequences may coincide, when a = 0, b = 9, c = 9, d = 0: 0990 0990

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

    There are repetitions in your formula. For n = 4 the answer is 974. Check with your submission

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

atdynamicprogrammer

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

lose again because of my speed

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

Was trying to do D , using DP but gave TLE... https://atcoder.jp/contests/abc178/submissions/16721075 Can someone pls help what is wrong

»
4 years ago, # |
Rev. 3   Vote: I like it +43 Vote: I do not like it

Editorial:

A
B
C
D
E
F

All codes:https://atcoder.jp/contests/abc178/submissions?f.User=Gary

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

    In $$$C$$$ constraints are smaller, so if someone has weak combinatorics(like me) then dp is more intuitive imo. here is my sol using dp:

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

    Hey, in problem D, I think you also need to add 1 to dp[i].

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

Is there a case to be handled for mod of negative numbers in C ?

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

    if a<0 then use mod-abs(a) will work.

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

    Don't forget to add mod when you are subtracting two numbers under mod.

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

    we can do this : ans = (a-b+MOD)%MOD;

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

      How does it work?Can you please explain or give me any reference?

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

        I'll leave that for you to figure out :)

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

And.. AtCoder becomes MathCoder.

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

c explanation ?

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

    Total Number: 10^n

    Without digit '9': 9^n

    Without digit '0': 9^n

    Without both: 8^n

    Final Result: 10^n — 9^n — 9^n + 8^n

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

    Total numbers you can make using all digits (10^n) — numbers with all digits but 9 (9^n) — all but 0 digits (9^n) + all but 0 and 9 (8^n) (because they were deleted twice);

»
4 years ago, # |
  Vote: I like it +54 Vote: I do not like it

Maybe this is an off-topic.

I think Atcoder should show counts of solve for each problem in dashboard/tasklist(like CF). I know it is in the standings page, but viewing standings itself some number of times is well-disgusting.

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

I have tried to solve F using bipartite graph matching but it failed just in test 16 :)))

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

I found my solution for F is realy complex(Though it is right)!!

Can you share your idea for F?

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

    Resulting ordering of B array is some cyclic shift of itself. (If no shifting is possible, no solution).

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

      Thank you for your reply:)

      But can you prove it?And how to achieve it?

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

        I did not scratch a full proof yet.

        For each element in B, we find what shifts are not possible. This can be done using set/segment tree.

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

        my F solution

        Whenever there exists index i where A[i] is equal to B[i], shift array B one time. Loop until there are no such indices in one interation. I use two pointers to maintain the shift

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

          Can you prove it?

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

            Sorry, I didn't prove it during contest. I generate all possible cases for small N and it seemed work.

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

              alvinvaja , I really like the quality of your profile picture. May I know which camera was used?

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

                It wasn't mine. Got it from pinterest

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

            I think mine is similar where I take a number, x, in both a and b and if the start of b's interval of x isn't over the end of a's interval of x, I calculate how many shifts it would take to make b's start over a's end of x's interval (if b's start isn't already over a's end) and take the maximum shift for all numbers that are both in a and b. Maybe there's a counterexample but I couldn't find one.

            After the greatest shift needed is done, for that number x in b that needed the largest shift, all the numbers < x in b should be assigned to a number bigger than it in a and all the numbers >= x in b are shifted up to match numbers bigger than it in a until n, and then they match to lower numbers that are <= x because the shift circles it to the beginning of a. Unless the array is impossible I don't see how any numbers can conflict yet.

      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it +12 Vote: I do not like it

        I used the same approach, but I also can’t prove it. But I know how to achieve it. Each position of B can be expressed as "not shifting in range [l, r]" which is equivalent to

        ~[x >= l and x <= r] = (~(x >= l) or (x >= r + 1)).

        Now the requirement is just AND product of all boolean clauses which is nothing, but 2-SAT. I thought that they put some problems to practice Atcoder library, turned out that I overcomplicated things up. lol (They didn’t add library yet, so I got compile error.) code

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

    you can see this solution I just tried to solve F using this and got AC. Logic is simple as well as code. Here is my code for this approach

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

Can someone check why I'm getting WA on C . here is my submission

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

    Result of a1 — a2 + b1 may be negative, so just add MOD to it before taking remainder

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

    "ll ans = (a1 — a2 + b1)%M" should be ll ans = (a1 — a2 + b1 + M)%M;

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

    ll ans = (a1 - a2 + b1 + MOD)%M;

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

Just 2 min short to get AC in problem F. Nice F anyway

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

    What was your idea behind F? Was it greedy?

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

      The idea is tough to explain for me. I try. 1.A is in ascending order and B is in ascending order. So, just try to pick up elements like two pointer from start to end. Then, pick the unpicked elements too. Now, check the answer. If its OK then print.
      2. Otherwise, reverse B once again and go like 1. Check the answer. If OK then print it, else there is no answer.

»
4 years ago, # |
Rev. 2   Vote: I like it +7 Vote: I do not like it

A Combinatorics solution for D — Redistribution

Recall that, by the famous Stars and Bars Theorem, the number of non-negative integral solutions of

$$$ x_1 + x_2 + \dots + x_r = n $$$

is

$$$ {n + r - 1} \choose {r - 1} $$$

Let us denote this value as $$$stars(n, r)$$$. Clearly, the value is 0 if $$$n$$$ is negative.

Moreover, if we add a constraint that $$$x_i \geq t$$$, then, we can replace $$$x_i$$$ by $$$y_i$$$, where

$$$ x_i = y_i + t $$$

This transforms the problem to finding the non-negative integral solutions of

$$$ y_1 + y_2 + \dots + y_r = n - r \cdot t $$$

Clearly, the number of valid solutions is $$$stars(n - r \cdot t, \ r)$$$

So, for each $$$r$$$ from 1 to $$$n$$$, we sum up the values of $$$stars(n - 3 \cdot r, \ r)$$$.

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

DP Solution for Problem C — Ubiquity

Code
»
4 years ago, # |
Rev. 2   Vote: I like it +7 Vote: I do not like it

A great contest.

Unfortunately,I nealy AK(AC 5 problems),but I'm in trouble with the sixth problem.

All of them are Math problems(.(xiao xue ao shu in Chinese)

And there's Chinese in English task:配点。

Hope the abc will be better!

:)

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

So I did a randomized solution for F. I try three ways:

  1. Reverse $$$B$$$.

  2. Random shuffle $$$B$$$ for $$$T$$$ times.

  3. Rotate $$$B$$$ randomly for $$$T$$$ times.

$$$T = 20$$$ is sufficient to get AC, and it runs in 61 ms. Of course, I have no proof (well not even an intuition) for this. :D

So, I'll be glad if someone can hack this solution or give some proof/intuition for it.

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

    Assume odd n, and n/2 '1' in a[], and n-n/2 '1' in b. Then we need to rotate exactly n/2 positions.

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

      Reversing B works in this case.

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

        Ok, let the sympobls be '2', not '1'. And a '1' in first position of both arrays. Then the reverse of b[] does not work, but still we need to rotate exactly n/2 positions.

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

          Yeah it prints No for this case.

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

How to prove that on F, if a solution exists, one of the cyclic shifts of b is also a solution?

I wrote my solution based on it and it passed.

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

My Unofficial tutorial (A-F) for this contest.

And the Chinese version.

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

Just a thought: How to solve F if the elements are unsorted in both the arrays?

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

    You can look at my solution, which does not depend on whether the arrays are sorted or not.

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

    You can always sort them and use your solution for the sorted input...

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

Can somebody please help me find the only test case that is failing for my solution of problem E?

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

    Changing the initial value of your variable ans1 to LLONG_MIN should solve your problem, because it's not necessary that x+y or x-y is always greater than -1.

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

What's wrong with my solution of D?

I used dp, $$$dp_{ij}$$$ means the sequence‘s length is $$$i$$$ and its sum is $$$j$$$.

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

    The elements of the sequence can be arbitrary positive integers rather than digits.

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

      Yes, I get it! Thank you so much!

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

    Can be simplified using a 1D dp array. Starting from i=4, dp[i]=dp[i-1]+dp[i-3] and then take the mod.

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

Why do you have the problem whose solution is available in CPH? The most popular book on CP, talking about problem E.

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

Sry for a newbie question, how to we see the test cases at atcoder???

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

How to solve E??

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

Can someone plz tell me why my code for task C is failing for 5 test cases?

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mod 1000000007

ll help(ll a,ll b){
    if(b==0)return 1;
    if(b==1)return a;
    ll ans = 1;
    if(b&1){
        ans = help(a,b-1)*a;
    }
    else {
        ans  = help(a,b/2);
        ans = ans * ans;
    }
    return ans%mod;
}

int main(){
    ll n;
    cin>>n;
    ll x = help(10,n);
    ll y = help(9,n);
    ll z = help(8,n);
    ll res = x-2*y+z;      // error in this line
    cout<<res;
    return 0;    
}

UPD : Error found The error is, it should be replaced with ll res = (x- 2*y + z + 2*mod ) % mod;

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

    I guess you forgot to take mod from the res at the end

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

      I have changed it after you said, but still those same 5 test cases are failing. I'll update the latest code in my comment. Can you plz look into it again. Thanks!

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

        Since it is c++, mod can also give negative result, you need to handle this

»
4 years ago, # |
Rev. 2   Vote: I like it +25 Vote: I do not like it

F can be solved by maintaining an invariant as follows:

Hint1
Hint2
Hint3

Thanks to Yousef_Salama for showing me this idea, you can check his submission.

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

I have written unofficial English editorial, you can find it at: editorial

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

For F, I am reversing B and then checking if B[i]=A[i]. If so at index=p, I am reversing the array from B[0] to B[p] and from B[p+1] to B[n-1]. This works as for the reversed B, all numbers left of B[p] will be >= B[p] and to right of B[p] will be <= B[p]. Whereas in A, all numbers left of A[p] will be <= A[p] and to right of A[p] will be >= A[p]. Theoretically it should work but I am getting wrong answer in 3 of the 60 test cases and I cant figure out why. Can anyone point out the fault in my algo? Submission: https://atcoder.jp/contests/abc178/submissions/16733148

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

Problem F is EXACTLY the same with Perm Matrix (just with smaller constraints).

I've seen this problem,so I got an unexpected high rank.

My atcoder account: sjcsjcsjc

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

    It's nothing new at atcoder, many time I got the exact questions which I had previously solved or seen somewhere. Problem E was also copied in the same contest.

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

      Problem E is just a standard problem of changing Manhattan distance to Chebyshev distance.

      Problems using this technique are everywhere.

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

Hi can someone please tell why my submission for problem C is working in atcoder c++ compiler but not in my local compiler.

I have tried recursive dp solution for this problem and it is working well . But in some large input example for n = 869121 it gives segmentation fault in local compiler as well as in online cpp compiler but works well upon submitting can some please tell why this happening ? Can this be due to memory limit or recursion depth ? please help