Блог пользователя TheScrasse

Автор TheScrasse, история, 4 года назад, По-английски

1485A - Add and Divide

Author: TheScrasse
Preparation: MyK_00L

Hint 1
Hint 2
Hint 3
Solution

Official solution: 107232596

1485B - Replace and Keep Sorted

Author: TheScrasse
Preparation: Kaey

Hint 1
Hint 2
Hint 3
Hint 4
Solution

Official solution: 107232462

1485C - Floor and Mod

Authors: isaf27, TheScrasse
Preparation: Kaey

Hint 1
Hint 2
Solution

Official solution: 107232416

1485D - Multiples and Power Differences

Author: TheScrasse
Preparation: MyK_00L

Hint 1
Hint 2
Hint 3
Solution

Official solution: 107232359

1485E - Move and Swap

Author: TheScrasse
Preparation: TheScrasse

Hint 1
Hint 2
Hint 3
Solution

Official solution: 107232216

1485F - Copy or Prefix Sum

Author: TheScrasse
Preparation: TheScrasse

Hint 1
Hint 2
Hint 3
Solution

Official solution: 107232144

Разбор задач Codeforces Round 701 (Div. 2)
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4 года назад, # |
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Love editorials with hints! Also a good but not great contest!!

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4 года назад, # |
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Video editorial for C Floor and Mod: https://youtu.be/ap-q_wQn7BE Solution in O(sqrt(x)).

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4 года назад, # |
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Can anyone explain the hint A problem to me

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    4 года назад, # ^ |
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    The idea is to first increase b as much as you want then perform operation 1. The reason is :- Lets say you divide first and then increase b, our a will be floor(a/b) and b will be b+1. If we increase b first then finally a will be floor(a/(b+1)) and b will be b+1. You can similarly prove for more number of operations. So once you know that we have to increment b first, the question is by how much. You will see that if we start with b and increment b till it reaches b+k ( k steps ), you will now need to integer divide a by b+k until a != 0. Lets call the number of steps to do so to be 'l'. So total number of steps will be l+k. I observed that we dont need to go very far away as that would increase k by very much but l wont be affected that much. So I started with b and went until b+100 and took the minimum value of l+k for every b in this range. P.S : Sorry I dont know how to use mathematical symbols

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      4 года назад, # ^ |
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      Sorry ,but i dont get,how do u know to increase b only to a 100????

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        4 года назад, # ^ |
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        Its pure intuition,as you can see in the editorial, it turns out that we dont even need to increase above 30. To get b->b+k we need k steps and to get a->0 we need O(log(a)) steps where base is b+k, we can observe that as b+k increases, the function log(a) doesnt decrease very fast and thats why I thought we dont need to increase b by a lot as that would increase total number of steps more than what log(a) would decrease and the net effect would be that we would require more steps.

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          4 года назад, # ^ |
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          I also didnt get how 30 comes...and how should I even realize that...:( Bad Problem

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            4 года назад, # ^ |
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            I approached in this way: Let's not keep track of count of type2 operation before doing type1 operation.

            Code
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            4 года назад, # ^ |
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            You cant understand solution --> bad problem.

            Amazing logic

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              4 года назад, # ^ |
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              XD...was waiting for this comment to appear...no not understanding the editorial is not the reason to consider it a bad problem...

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          4 года назад, # ^ |
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          we need almost 30 operations when it s 2 I agree. but how is it related to incrementing b by only 30 times and checking the minimum possibility scenario of each incremented b. why not more than 30 increments of b?

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            4 года назад, # ^ |
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            Because if you use 30 steps for incrementing then you need one more step for dividing, making it totally 31 steps. Instead of that you could have.simply used 30 steps all in dividing, which would have brought a down to 0 already.

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        4 года назад, # ^ |
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        We can rewrite the question as min(x-b+(ln(a)/(ln(x))) for all x>=b. Of course, there is some rounding off business, which we do at last. So consider this fnc, if we differentiate this function, we get 1-(ln(a)/x), so we get a minimum at x=ln(a), but it may be also possible that ln(a) is less than b, so if we take the worst possible ln(a) it won't cross ln(10^9), which comes around 25. So we can iterate x from b to ln(a)+5, and obtain a minimum value. Well, this analysis is rough. So b we can increase up to up to 25 or so, but the editorial gives a better limit.

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          4 года назад, # ^ |
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          You differentiated it incorrectly:

          $$$d_x \frac{ln(a)}{\ln(x)} = - \frac{\ln(a)}{x \ln^2(x)}$$$

          Therefore after differentiation you must get the equation

          $$$1 - \frac{\ln(a)}{x \ln^2(x)} = 0$$$

          The upper limit for $$$a$$$ is $$$10^9$$$. Therefore we arrive to

          $$$x \ln^2(x) = 9 \ln(10)$$$

          The left part is strictly less than $$$0$$$ for $$$x \leqslant 6$$$ and it becomes positive for $$$x \geqslant 7$$$, where $$$x \in N$$$.

          As the minimum of $$$f(x) = x - b + \frac{\ln(a)}{\ln(x)}$$$ is somewhere between $$$6$$$ and $$$7$$$ we suffice with at most $$$6$$$ iterations provided that we start from $$$b = 1$$$ (ending up at $$$7$$$).

          Otherwise we need to iterate upwards only until $$$b >= 7$$$. So that if you start with $$$b = 10$$$ you do zero iterations!

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      4 года назад, # ^ |
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      You can type $...$ to insert math, where "..." is latex math expression. Common examples:

      $a_{1} + a^{2}$ --> $$$a_{1} + a^{2}$$$

      $\frac{a}{b}$ --> $$$\frac{a}{b}$$$

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    4 года назад, # ^ |
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    It wants to say that we should increment $$$b$$$ first as much we want and then do the divide operations . Because that way number of division operations would be less.

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4 года назад, # |
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Why solution of problem E's complexity is $$$O(n\log n)$$$?

upd:now it's ok

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4 года назад, # |
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Good contest but I think first 4 problems were all heavily based on Maths (except B).Would have been great contest if atleast one was from another topic making problemset more balanced.

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    4 года назад, # ^ |
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    B,C,D involved math but some ideas outside math was also required . Like binary search in C (not necessary though) , prefix sum in B , chess coloring in D. Also though i didn't read E,F they are tagged as DP and data structures.

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4 года назад, # |
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Thanks for the interesting problems and the fast editorial!

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4 года назад, # |
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Very nice problems. Every problem had something to offer at least till D(didn't see E and F). I wish I could upvote more for the nice editorial with hints. Looking forward to more contests from the authors.

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4 года назад, # |
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I solved A but don't know why it worked , now i am reading editorial of A, lol

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4 года назад, # |
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Thanks for fast editorial

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4 года назад, # |
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I love math forces.

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4 года назад, # |
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Can someone please tell why this code for problem C is giving TLE, inspite of O(sqrt(x)) solution? Code: 107231229

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4 года назад, # |
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I see what you did there ;) oie-h-Rh1llj1l-GLh

Problem — Floor and Mod

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4 года назад, # |
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The contest was really enjoyable! Thanks for the fast editorials!

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4 года назад, # |
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was a good contest .. good work authors ...and editorial with hints is really a good idea :)

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4 года назад, # |
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I used the same logic in div2A. Can someone please help me where I went wrong?

Link to submission: https://codeforces.me/contest/1485/submission/107217438

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    4 года назад, # ^ |
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    I guess it's because double are devil. On case 5 test 3 your l should be 3 but its just à little bit less because of approximation your flag isn't activated, rewrite without double and log and it should work

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      4 года назад, # ^ |
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      Thanks, got it. I never knew this happened with log. Could you explain why exactly does this happen?

      PS. It's so annoying to get WA for such a silly mistake, could've been a specialist for the first time mayb :( .

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4 года назад, # |
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Why does first solution in F has complexity O(N^2*log N)? Don't we have O(N^2) elements in map in worst case (if almost all subsegments have different sums) and get complexity O(N^3*log N)?

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    4 года назад, # ^ |
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    The possible different sums for each length are $$$O(n)$$$, because they only depend on the rightmost $$$i$$$ such that $$$b_i = \sum_{k=1}^{i} a_k$$$.

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4 года назад, # |
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In the editorial of problem A, it is given that even if we check the value of b up to 6 it will work. can someone explain why ?

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    4 года назад, # ^ |
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    Not a formal proof, but here's an intuitive way to explain it: log(1e9)/log(b+1)+1<log(1e9)/log(b)

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      4 года назад, # ^ |
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      I did not understand how it proves that, can you explain it?

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        4 года назад, # ^ |
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        The number of moves needed without increasing $$$b$$$, is $$$\frac{log(a)}{log(b)}$$$

        So, if $$$\frac{log(a)}{log(b+1)}+1>\frac{log(a)}{log(b)}$$$, it means it's not convenient to increase $$$b$$$.

        For all $$$a$$$ up to $$$10^9$$$, this is true for every $$$b >= 6$$$, so it is never convenient to increase $$$b$$$ if it is $$$>=6$$$

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          4 года назад, # ^ |
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          For all $$$a$$$ up to $$$10^9$$$, this is true for every $$$b>=6$$$, so it is never convenient to increase $$$b$$$ if it is $$$>=6$$$

          We can say this only by verifying or there is some short way to get the above conclusion using the formula in your comment ?

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            4 года назад, # ^ |
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            Yeah, you can solve for $$$b$$$ in that inequality. Also, that inequality is just $$$log_{b + 1}a + 1 < log_ba$$$.

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              4 года назад, # ^ |
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              Ok, so i think we need to first put the value of $$$a$$$ and then solve for $$$b$$$ .But to justify for all $$$a<=1e9$$$ using this formula , we need to iterate for all values of $$$a$$$.

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    4 года назад, # ^ |
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    It's trivial to prove that the answer is <= 31 because you can solve it with +1 to make it 2 and divide it by 2 30 times. So test it up until 31 if that makes you happy.

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4 года назад, # |
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I understood the O(n^2logn) solution of problem F. But I don't get how to get rid of first layer of dp. Specifically in the editorial —

Let's try to get rid of the first layer of the dp. It turns out that the operations required are....

can anyone explain this ?

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4 года назад, # |
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Here are video solutions to all problems, a challenge version of E, and a lesson on how to misread problems (of course, the last two things are totally unrelated)

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4 года назад, # |
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An easier solution for A with dp if you don't know how to prove your greedy solution, here. Dp state is represented by $$$a$$$ & $$$b$$$. At every stage we have two choices:
1) divide $$$a$$$ by $$$b$$$.
2) increase $$$b$$$ by 1.
Now take the minimum of these two at every state. Also in this way we can see that $$$b$$$ can go till 1e9, but we can observe that $$$a$$$ will be zero if we divide it once with atmost 13 consecutive numbers in worst case. So we can stop if $$$b$$$ goes beyond.(for safety i put it 20 instead of 13 in code).

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4 года назад, # |
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I got WA in Problem A because I used a log() function :((

for (int i = b; i < b + 50; i++)
        ans = min(ans, floor(log(a) / log(i)) + i - b);

107195035

When values are for e.g. a = 18^7 and b = 18, log(a) / log (b) returns 6.9999999999994... something. Does anyone know a better way to use log() and floor() to bypass this issue?

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4 года назад, # |
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what is greedy solution in problem A and how to prove the last statement in editorial :

If we notice that it is never convenient to increase $$$b$$$ over $$$6$$$, we can also achieve a solution with better complexity.

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    4 года назад, # ^ |
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    $$$a = 10^9, b = 1$$$

    $$$2^{30} > a, steps = 31$$$
    $$$3^{19} > a, steps = 21$$$
    $$$4^{15} > a, steps = 18$$$
    $$$5^{13} > a, steps = 17$$$
    $$$6^{12} > a, steps = 17$$$
    $$$7^{11} > a, steps = 17$$$
    $$$8^{10} > a, steps = 17$$$
    $$$9^{10} > a, steps = 18$$$
    $$$10^{10} > a, steps = 19$$$

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      4 года назад, # ^ |
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      thanks , example can help to visualize . How to prove it formally ? Like for every $$$a,b <= 1e9$$$ ?

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        4 года назад, # ^ |
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        1). $$$a = 10^9$$$, $$$b \in [1, 10^9]$$$

        As you can see before, the distribution converges at $$$5$$$ and diverges from $$$9$$$ for $$$b = 1$$$ the same holds for any $$$b$$$ cuz the difference stays the same.

        2). $$$a \in [1, 10^9], b = 1$$$

        Likewise, $$$\exists x \mid log_2a + 1 > log_3a + 2 > \dots > log_xa + x - 1 < log_{x + 1}a + x < \dots$$$.

        And I can state the same from above, the distribution converges at $$$x$$$ and diverges from $$$x + 1$$$ for $$$b = 1$$$ the same holds for any $$$b$$$ cuz the difference stays the same.

        Thus, $$$6$$$ is the upper-bound for this problem since $$$a$$$ can be at most $$$10^9$$$.

        P.S. The upper-bound could vary accordingly to $$$a$$$.

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4 года назад, # |
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For F the editorial definition is slightly different from what I thought. Define dp[i][j] as the number of ways of using i numbers and getting j = CURRENT_SUM — sum[1....i]. Under this definition, taking a[i] = b[i] makes the next j be the same as j so if we do nothing we already do the transition. For the other transition, we make j = a[i] — sum[1...i] so we take the total number of ways we have so far and exclude the number of ways such that the sum is the same in this transition or simply override dp[i][j] with the total number of ways in i-1 as the official solution does. In the end the code is equal but this way of thinking seems slightly cleaner (in my opinion)

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4 года назад, # |
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Very nice problems!! Sad on missing out on Master due to FST in problem B :(

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4 года назад, # |
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Hi, I think that I might have a better solution for problem div2A. 107216266 I can't say what the complexity of my submission is with total certainty but I think that maybe it is better than that of the proposed solution. Basically, I try all possible divisors in increasing order while keeping track of the previous answer. When it gets to the point where the current solution is worse than the previous one, then it stops the loop and returns the previous answer.

I don't have a formal mathematical proof to state why this works, but it does. It would be great if someone could explain what the complexity of my code is :)

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    4 года назад, # ^ |
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    Your solution is pretty much the same as the official solution, the complexity is $$$O(log(n))$$$, but it can be done in $$$O(1)$$$ if you use the builtin log function (paying attention to floating point errors).

    The reason why in the editorial we say our complexity is $$$O(log^2(n))$$$ is because we did not prove that the number of increases needed is $$$O(1)$$$ (it's not, it's prob something like $$$O(log(log(n))$$$, 6 for these constrains, but if $$$b$$$ is greater than that, than you don't need to increase at all), for simplicity we just proved it is $$$O(log(n))$$$

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    4 года назад, # ^ |
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    A quick proof of correctness is to note that the number of steps required is the pointwise floor of a convex function.

    $$$ f(x) = \lfloor g(x) \rfloor \text{, where } g(x) = x-b + 1 + \frac{\ln a}{\ln x} $$$
    $$$ g'(x) = 1 - \frac{\ln a}{x \cdot (\ln x)^2} $$$

    $$$g'$$$ is obviously increasing, so that $$$f(x)$$$ is non-increasing for $$$x \cdot (\ln x)^2 \leq \ln a$$$ and non-decreasing for $$$x \cdot (\ln x)^2 \leq \ln a$$$. This also means that the optimal value of $$$x$$$ is of order $$$\frac{\log a}{(\log \log a)^2}$$$.

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4 года назад, # |
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Great Contest, 1st, 2nd, and 4th questions were purely math-based. They should include problems from other topics as well.

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4 года назад, # |
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all you have to do in D

just get the lcm for the numbers from 1 to 16

and get for every number from the 1 to 16 the multiple to it and call it x and the different between the lcm and x have 4'th sqrt (yeaaaaaaaaah all that just in 1 minute brute force in your local pc)

color the grid as a chess board

replace the black cells by the LCM

and replace the white cells by the answer that you got from the brute force

do you need a proof?

ok i don't give a sh....

i am just upset because I did not try in D in the contest :(

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    4 года назад, # ^ |
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    How do you write a brute force efficiently?

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      4 года назад, # ^ |
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      all that will just to get the answer for every number and take it by your hands

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        4 года назад, # ^ |
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        Nvm, I misunderstood. I thought that you got the idea by bruteforcing the entire solution.

        By the way, you don't even need to brute force values on white cells, you can just put $$$720720 + a_{i, j}^4$$$.

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          4 года назад, # ^ |
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          yp, i was in brain lag

          edit: i was thinking that the next number is lcm * aij^4 instead of lcm + aij^4 :(

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4 года назад, # |
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Great problems+fast editorial with hints are the best combo.

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4 года назад, # |
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For C I have good solution it don't requires math just logical binary search. See think of the graph of quotients plotting on (b) axis. For calcuating the peak of quotients basic observation will be for random number x on b-axis having remainders [a>=(x+1)(x-1)]=>>(a=x^2-1) =>>a+1=x^2.This equation comes from X=(Y.r)+r is now x=(y+1)r. So min value of(sqrt(a+1),b). Cool I got peak ! This is digram.. Now iterate from right end using binary search locate the location where quotient is still the same. And subtract both the points multiplying by number of quotients. In this keep way iterating to peak ((sqrt(a+1),b).

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4 года назад, # |
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Can anyone tell me why this code for E gets WA on test 3? I followed the (alternate) strategy given in the editorial.

submission

EDIT : very dumb error

Here is the corrected implementation of the editorial approach, if it helps.

Corrected code

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4 года назад, # |
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I have a different explanation for problem F , though the final approach is almost the same.

At each point we have two choices :

1) Make $$$a[i] = b[i]$$$

2) Make $$$a[i] = b[i]-$$$(prefix_sum up to $$$i-1$$$ elements in a[]).

So answer could have been $$$2^n$$$ except when both choices gives the same array.

When both the choices give the same array , prefix_sum of $$$i-1$$$ elements should be 0. This way I got 2 tasks , (1) Find the number of distinct arrays for first i elements : dp1 , (2) Find the number of distinct arrays for first i elements having sum 0 : dp2.

Then we directly have, $$$ dp1_i = (2*dp1_{i-1}) - dp2_{i-1}$$$.

To calculate $$$dp2_i$$$, we have one more observation that, when we do operation of type (1), $$$b[i]$$$ adds to the previous prefix_sum and for type (2), prefix_sum becomes $$$b[i]$$$. This tells us that, prefix sum of i elements in a[] will be a subarray of b ending at i.

So $$$dp2_i$$$ considers the smallest subarray of b ending at $$$i-1$$$ and having sum 0 (This is a standard task using hashmap). Let's say $$$[x,i-1]$$$. Then we take a unique subarray of $$$x-1$$$ elements, do a type 2 operation for x and do type 1 operation for elements in range $$$(x+1,i-1)$$$. This gives $$$dp2_i = dp1_{x-1}$$$.

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4 года назад, # |
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In problem C, I'm not getting how we're getting the no. of special pairs just by counting the possible no. of b, because can't there be different a's for a fixed b and a fixed k? If so, then how are they being considered in our answer?

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    4 года назад, # ^ |
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    $$$a = kb + k$$$.

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      4 года назад, # ^ |
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      In EDITORIAL Therefore, for any fixed k>0, the number of special pairs (a≤x; b≤y) is max(0,min(y,x/k−1)−k). Why we have subtracted k from min(y,x/k−1) and taken maximum of both to count special pairs???

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        4 года назад, # ^ |
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        Let's look at the four inequalities required for solving this question:

        1. b > k (derived in the editorial)

        2. 1 <= b <= y (given in the question)

        3. 1 <= k(b + 1) <= x (derived in the editorial)
          Modifying equation 3, we get

        4. 1/k — 1 <= b <= x/k — 1

        5. 1 <= k < b (derived in the editorial)

        (Why k != 0? If k = 0, then a = k*(b + 1) = 0 but a >= 1, so the minimum value of k is 1)

        Now from equation 1, we know that the minimum value of b is greater than k and from equation 2 we know that the upper limit of b is y.

        So we get the following inequality if we merge equation 1, 2 and 4

        k < b <= min(y, x/k — 1)

        Now we got an open interval at the lower limit and closed interval at the upper limit, so the number of elements in the interval is Upper Limit — Lower Limt = min(y, x/k — 1) — k.

        Now the value of k can exceed the value of min(y, x/k — 1) since 1 <= k <= sqrt(x) and in this case, we will get negative number of elements in our interval which is not possible. So if we get negative number of elements, in such cases it means we have 0 elements. That is why we must take max(0, upper limit — lower limit).

        Example: If y = 2, x = 36, k = 6

        min(2, 36/6 — 1) — 6 = min(2, 5) — 6 = 2 — 6 = —4.

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          4 года назад, # ^ |
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          .

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            4 года назад, # ^ |
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            a mod b = k, that is, k is the remainder when we divide a by b.

            Now applying Euclid's Division Lemma we can write

            a = bq + k, where q = $$$\lfloor{\frac{a}{b}} \rfloor$$$ = a mod b = k (Given in question)

            Since k is the remainder, we 0 <= k < b by Euclid's Division Lemma or b > k.

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              3 года назад, # ^ |
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              Can you please explain why k is always less than or equal to sqrt(x)

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                3 года назад, # ^ |
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                $$$0$$$ $$$<=$$$ $$$k$$$ $$$<$$$ $$$b$$$ (By Euclid's Division Lemma)

                Multiply both sides by $$$k$$$.

                $$$k*k$$$ $$$<$$$ $$$k*b$$$

                $$$=>$$$ $$$k^2$$$ $$$<$$$ $$$k*b + k$$$ (Since $$$k$$$ is non — negative, the RHS sum can remain same or increase but never decrease)

                But $$$k*b + k$$$ $$$=$$$ $$$a$$$ $$$<=$$$ $$$x$$$ (Given in the question).

                Hence $$$k^2$$$ $$$<$$$ $$$x$$$ or $$$k$$$ $$$<$$$ $$$\sqrt{x}$$$.

                Now here we need to take equality in case $$$x$$$ is not a perfect square else you can drop the equality. This is because in case of non square number, we will be missing 1 number.

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          4 года назад, # ^ |
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          Thank You so much bro!

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          4 года назад, # ^ |
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          By that logic we can use equations 2 and 4 to get :
          elements in the interval is Upper Limit — Lower Limit = min(y,x/k-1) — max(1/k-1,1)) that should work right ?

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4 года назад, # |
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I tried to calculate $$$Dp_i$$$ from bottom to top in problem E, but got wa on test 3. It works like this:

bool cmp(int k,int b){return a[k]<a[b];}
...
sort(stage[i].begin(),stage[i].end(),cmp);
ll Max=-Inf;
rep(j,0,stage[i].size()-1){
    int k=stage[i][j];
    f[fa[k]]=max(f[k]-a[k]+maxx,f[k]+a[k]-minn);
    f[fa[k]]=max(f[fa[k]],a[k]+Max);
    Max=max(Max,f[k]-a[k]);
}
Max=-Inf;
per(j,stage[i].size()-1,0){
    int k=stage[i][j];
    f[fa[k]]=max(f[fa[k]],Max-a[k]);
    Max=max(Max,f[k]+a[k]);
}

Where $$$i$$$ is the depth. The nodes in $$$stage[i]$$$ has the same $$$dis$$$ which is $$$i$$$.

$$$minn$$$ is the minimum of $$$a_i$$$ at the same stage, $$$maxx$$$ is similar to $$$minn$$$.

Could anyone explain that why it is wrong? The submission : 107250435

Thanks!

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4 года назад, # |
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In Problem C, in the final expression, max(0,min(y,x/k−1)−k), Why are we subtracting k from min(y, x/k-1). Can someone please explain?

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    4 года назад, # ^ |
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    x/k gives a count of numbers which can be our b but we need a remainder equal to k, that's why we decrease it by 1 (think?), if it is greater than y, we can take any number up to y as our b, but wait.... we can't take any b which is less than or equal to our remainder, therefore we again subtract k from it.

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4 года назад, # |
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Can anyone tell why this binary search solution for A failed?

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    4 года назад, # ^ |
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    Are you sure that the f is a decreasing function ? I think that it is a first decreasing and then increasing function.

    Ternary Search works for this kind of functions in O(log(n)).

    Submission

    But I can't prove that f is is a unimodal function !

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      4 года назад, # ^ |
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      I got it my solution is wrong because f is monotonic.

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        4 года назад, # ^ |
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        You mean not monotonic ?

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          4 года назад, # ^ |
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          No monotonic, i.e it is nondecreasing first then nonincreasing. If f(mid) = f(mid + 1) before the minima then I would get the wrong ans.

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        4 года назад, # ^ |
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          4 года назад, # ^ |
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          That's ternary search, not binary search, because you're looking at the slopes of adjacent elements. I'm not sure how to formally prove that the function is unimodal, but it's pretty easy to see intuitively.

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            4 года назад, # ^ |
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            Yep, but I split the search space into halves at each iteration. Also, the function is not unimodal.

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              4 года назад, # ^ |
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              It's still called ternary search, even though you're splitting the search space in half. See cp-algorithms for more detail. That's actually how I code ternary search if I'm not working with doubles (I actually did this solution in contest, so you can check my submission). Binary search only works with monotonic functions, but ternary search works with any unimodal function.

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                4 года назад, # ^ |
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                It is Binary Search. Yep, F is monotonic.

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                  4 года назад, # ^ |
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                  That example, is different, as you're not really searching on a function. You can read about monotonicity on boolean functions (functions used in binary search) here. If f was monotonic, it would mean that looking at a single value, it would be possible to determine whether to go to the right or to the left. Ternary search can essentially be viewed as binary search on the slopes of the function, instead of the values itself. That's what differentiates them. If you look at the slopes of unimodal function, and define $$$g(x) = f(x)-f(x-1)$$$, you can see that some prefix of a unimodal function will have some sign, and the suffix will have the opposite sign, which makes $$$g(x)$$$ monotonic, but $$$f(x)$$$ unimodal.

                  Btw, a function is unimodal if it first increases, then decreases (or vice versa). See here for more details.

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                  4 года назад, # ^ |
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                  A common definition is as follows: a function f(x) is a unimodal function if for some value m, it is monotonically increasing for x ≤ m and monotonically decreasing for x ≥ m. In that case, the maximum value of f(x) is f(m) and there are no other local maxima.

                  Yeah, I said f is not unimodal since, you can see four local minima here.

                  Arguing if my approach is binary search or ternary search is self-contradictory. I should've mentioned it Divide and conquer in general.

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                  4 года назад, # ^ |
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                  Those local minima you mentioned are also the global minima, so it still satisfies unimodality (it's decreasing instead of strictly decreasing for that segment). Because they are in the same contiguous segment, I think general consensus is that they are still global minima. At this point, we're really arguing about if it's convention to call a function which is decreasing but not strictly decreasing monotonically decreasing, and I guess it doesn't really matter. In general for ternary search/binary search purposes, I would call it unimodal/monotonic because it's still possible to binary search/ternary search over the functions.

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4 года назад, # |
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Shouldnt k <= root(x) for problem C be k < root(x)? Since if k^2 = x implies b=k-1 which is less than k

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4 года назад, # |
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I have a question: In problem C's editorial, once "b" and "k" is fixed, is the value of "a" determined and unique?

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4 года назад, # |
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How to prove the complexity for the O(n^2logn)-algorithm for prob F?

Thanks

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4 года назад, # |
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In problem 1485A - Add and Divide, BFS was fast and memory-efficient enough to get accepted.

107307717

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4 года назад, # |
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Can someone explain problem B editorial when x is in-between al and ar, this exactly "There are (ar−al+1)−(r−l+1) such values of x."

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    4 года назад, # ^ |
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    There are $$$a_r - a_l + 1$$$ integers between $$$a_l$$$ and $$$a_r$$$. Since $$$x \neq a_i$$$, you have to subtract the $$$r-l+1$$$ integers already in the array, in the range $$$[l, r]$$$.

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4 года назад, # |
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4 года назад, # |
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Hello. I've been trying to get "accepted" on problem C in Python3 but I get TLE while implementing the solution suggested here.

Here is the link to my submission: 107347518

I googled a bit and it turns out that the time limits don't seem to be language-specific, i.e. the same algo implementation might be passing in C but not in Python. link: https://codeforces.me/blog/entry/45228

Is this the case? Are some problems not meant to be solved as long as you don't write in the language intended or is something wrong with the submission I posted?

Thanks

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    4 года назад, # ^ |
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    Check the following update to your code.

    107361634

    The same update ran in PyPy 3.7 more than an order of magnitude faster than it did in Python 3.9.

    107361941

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      4 года назад, # ^ |
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      Thanks for the update!

      Yep, the code passes now at 1965ms and we have whole 35ms to spare!

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        4 года назад, # ^ |
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        Yes, the speed difference between the Python interpreter and the PyPy compiler in running the same code is obvious.

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4 года назад, # |
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There is a O(N) solution for F. Note that this solution uses a map. The solution is as follows -
Let f[i] denote the number of hybrid arrays of length i. Let g[i] denote the number of hybrid arrays of length i and sum of the array is 0. Given that we have a[0], a[1], a[2] . . . a[i-1] we have two options for the ith position — 1) Place b[i] 2) Place b[i] — (a[0] + a[1] + a[2] + . . . . a[i-1]) Now there will be over-counting when b[i] = b[i] — (a[0] + a[1] + a[2] + . . . . a[i-1]), or a[0] + a[1] + a[2] + . . . . a[i-1] = 0. Thus f[i] = f[i-1]*2 — g[i-1].

Now how to find g[i]? Observations — 1) The sum of the hyprid array 'a' of length i will always be a suffix of array 'b' ending at i. 2) Let j <= i be the maximum index where sum(b[j] . . . b[i]) = 0. Then for the sum of array 'a' to be equal to sum(b[j] . . . b[i]) = 0, we need two things — 2.1) a[j] = b[j] — sum(a[0] . . .a[j-1]) 2.2) For all j < k <= i, a[k] = b[k]. The array a[0]. . . a[j-1] can be anything we dont care. From 1 and 2 we can say g[i] = f[j-1]. 'j' can be found for every 'i' by maintaining map of prefix sums. 107307490

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    4 года назад, # ^ |
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    This is exactly the official solution, and it's $$$O(n\log n)$$$ because of the map.

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4 года назад, # |
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Hello, can anybody explain me div2-C, I'm not able to understand how is k<=sqrt(x)...Thanks in advance :)

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4 года назад, # |
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in problem C (floor and mod) max(0, min(0, x/k — 1) — k). Why do we subtract k from min(0, x/k — 1)?

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4 года назад, # |
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Can someone please explain me why my submission are getting TLE? I already try to optimize it but didn't work.

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4 года назад, # |
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I'm a bit confused with where the factor of 2 comes from in B. Does someone have an explanation for this? Specifically, why it's $$$2((a_r-a_l+1) - (r-l+1))$$$. Thanks!

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    4 года назад, # ^ |
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    I think I just understood it. I'll comment it here just in case someone else had the same question:

    consider the array [1, 2, 4, 5] for k = 5. Let $$$x = 3$$$. This mean that we could either replace 2 or 4. Similarly, $$$\forall x$$$ such that $$$a_l < x < a_r$$$ and $$$x \not= a_i$$$, it is the case that there will be two options for us to choose.

    The expression $$$(a_r-a_l + 1 - (r-l+1))$$$ counts the number of such $$$x$$$s. We just have to multiply that by two since for each such $$$x$$$, there are two options.

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12 месяцев назад, # |
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Yet again another chess board Yet another Problem to feel dumb I only once recognised this pattern once because I tried with few examples and it was convinent to throw examples but here there was one more step of disguise and it is even harder to check example and more paths to deviate from intended solution, yet I think these kind of constructive problem we need to think of some function of indexes f(i,j)