PikMike's blog

By PikMike, history, 7 days ago, translation, In English,

990A - Commentary Boxes

Tutorial
Solution (PikMike)

990B - Micro-World

Tutorial
Solution (adedalic)

990C - Bracket Sequences Concatenation Problem

Tutorial
Solution (Ajosteen)

990D - Graph And Its Complement

Tutorial
Solution (Ajosteen)

990E - Post Lamps

Tutorial
Solution (PikMike)

990F - Flow Control

Tutorial
Solution (PikMike)

990G - GCD Counting

Tutorial
Solution with Möbius (Bleddest)
Solution with centroid (adedalic)
 
 
 
 
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6 days ago, # |
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Can someone explain a bit more about the complexity for the problem E "Post Lamps" ?

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    6 days ago, # ^ |
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    Essentially it uses the fact that the sum of reciprocials of numbers up to n is about ln(n) for large n. The proof outline is as follows. We start from the fact (1+1/n)^n<e<(1+1/n)^(n+1). Now by taking logarithm we get that 1/(1+n)<ln(1+1/n)<1/n. Summing over all n gives desired.

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6 days ago, # |
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I passed GCD Counting with . Was a solution with that complexity intended to pass?

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    5 days ago, # ^ |
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    Was your solution for each subtree, calculate number of paths into the current node with each gcd and then find the contribution of each pair of paths from different subtrees? I wrote such a solution but I believe the complexity can be bounded by 160 N log N. This is for the case of a k-ary tree where we have f(N) = k f(N/k) + 160 N. Other trees only seem to require less operations — a line graph is only 160N.

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6 days ago, # |
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In Problem G , solution with centroid's code , there's an extra " ' " in the countdown fifth line of function "solve()" .

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6 days ago, # |
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in problem E the answer of first example ,if I let the lamp has 1 power place in position 0 2 4 5 then segment【0;n】 can be covered ,the answer is 4 why?

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6 days ago, # |
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PikMike Can you explain the complexity of your " get " function for the problem E Post Lamps ?

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    5 days ago, # ^ |
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    get(i) takes about steps to finish. Then it's the same as in Eratosthenes sieve, for example. in total.

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      5 days ago, # ^ |
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      I think it takes more than n / i steps for every i since we are not iterating on the multiples of i. But I am not able to bound it to n / i . Can you give an explanation for the bounding part ?

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        5 days ago, # ^ |
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        It takes more than but it is . Try doing two steps. You put segment, find the next free position, put another segment. Either the next free position will be i + 1 away from the first one, or you will quit function immediately. Then you reach n in about steps with some constant factor.

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          5 days ago, # ^ |
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          May be, We can block some positions such that every i (1 to k) will take O(n) steps. Won't that make this algorithm O(n*k) which will TLE. Or can we always prove that complexity is O(n*lg(n)) ?

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            5 days ago, # ^ |
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            OMG, guys, can't you like draw it yourselves?

            Imagine the worst case for placing two segments. It's when there are just two free positions and then lots of blocked ones. So you put one to the first position, one to the second position and then you will go right to l + 1 from the first free position or there will be no free spaces.

            y = 0 is the axis (black is blocked, white is free). The other two are two placed segments.

            You can draw other cases with less blocked positions, with more free positions to the left and see that the case I drew is the worst of them.

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      2 days ago, # ^ |
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      any help with why below code gives Time limit exceeded on Test #12? I think I am doing in nlogn only. But could not get this one.

      https://codeforces.com/contest/990/submission/39269221

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6 days ago, # |
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In G; anyone care to explain how this relation is derived?

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    6 days ago, # ^ |
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    We can use inclusion-exclusion to get the following:

    , where S is some set of prime numbers (possibly an empty set), and p(S) is the product of its elements (and h(p(S)) denotes the number of paths where gcd is divisible by every prime from S).

    The number of such sets is infinite, but since for every p(S) > 2·105 h(p(S)) = 0, we can iterate on the numbers from 1 to 2·105 and check whether there exists some set such that p(S) is equal to this number.

    In fact, if some number x is divisible by some number q2 (where q is prime), then there is no set S such that p(S) = x (and μ(x) = 0). Otherwise, there is exaclty one set which is the factorization of x, and μ(x) =  - 1 if and only if the size of this set is an odd number.

    Is this explanation clear? Or should I clarify something?

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      6 days ago, # ^ |
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      Right. The first sentence was the missing link I needed, thanks a lot!

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6 days ago, # |
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Can anyone explain problem D. How are we calculating the results ?

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    6 days ago, # ^ |
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    By "calculating the results", I am guessing you mean obtaining the graph to be outputted.

    So, we need a components in the graph. An easy way to do this is mentioned in the editorial:

    We have a total of n vertices. An easy way to obtain a-1 vertices is to simply take a-1 vertices and leave them unconnected to any other vertices. Thus, we obtain a-1 components. Now, just join the remaining vertices, which together form a single connected component.

    Finally, we obtain a-1 + 1 = a vertices as required in the problem.

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6 days ago, # |
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    5 days ago, # ^ |
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    At line 41 Use :
    int cntop[maxn]={0}, cntcl[op] ={0};
    Instead :
    int cntop[maxn], cntcl[op] ;
    rep(i,maxn) cntop[i] = cntcl[i] = 0 ;

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      5 days ago, # ^ |
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      No that was not the problem. cntcl[op] -> cntcl[maxn]

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5 days ago, # |
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Can someone explain the solution with centroid decompression in the problem G?

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    5 days ago, # ^ |
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    Key idea of this solution is next: let's look at the subtree in decomposition with its centroid c. You need to calculate all paths which go throw c, so its gcd will be divisors of a[c] and there will be no more than K different divisors of a[c] (K ≤ 160). So if you process each subtree (subtrees are formed by childs of c) not with but with (checking each gcd still takes O(K) operations) complexity will reduce from to something like .

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      4 days ago, # ^ |
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      Apparently a centroid decomposition involves recursively finding the centroid of each subtree, so this is not actually a centroid decomposition since the centroid is found only once. I believe this is only a constant-factor speedup over the naive O(NK log N) solution I described here, which was fast enough, runs in about 1 second.

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        4 days ago, # ^ |
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        I could miss something, but my proof was next: Let's look at all subtrees, which appeared in decomposition.

        Proposition: there O(N / K) subtrees with size at least K (let's name them big). To prove it, let's find all big subtrees which don't contain any other big subtree. There is no more than N / K such subtree. Let it will be 0-level. Let's look at subtrees, which contains 0-level subtrees. Name them 1-level. Since it's centroid decomposition size of each subtree of 1-level at least twice of size of 0-level. So there are no more than N / 2K such subtrees. And so on. So there no more than N / K + N / 2K + N / 4K... = O(N / K). Each big subtree is processes with O(K2) time. So total complexity for big subtrees is O(N / K·K2) = O(NK).

        Similarly, There are O(N) small subtrees, but each small subtree of size x is processed with O(xK) time. But there are no more than levels of small subtrees since after that they become big and each level is processed with O(NK) in total. Here we came to final complexity. Of course there is also to build decomposition and or to calculate gcd-s.

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          4 days ago, # ^ |
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          Ah, I think I misunderstood your last comment. It seems you are using a legit centroid decomposition, so the complexity makes sense. At first I didn't understand how centroids would work, since you calculate paths to the middle, but then for the parent calculation you need a different set of paths. But I guess you could find most of those by taking the gcd of the path to the centroid and the path from the centroid to its parent.

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5 days ago, # |
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I think that tests for problem E are weak, because my (I think!) wrong solution passed. I'll try to explain my logic:

Firstly, note that F(i) (editorial explanation) is decreasing with an increase i.

Second, assign every a[i] to min(a[j]), i <= j (obviously???)

a[i] is increasing with an increase i.

So we have two functions F(i) and a[i], answer for fixed i is F(i) * a[i].

Now every a[i] change to  - a[i].

So we have two decreasing functions.

Now launch ternary search xD

Answer is maximum F(i) * a[i] multiplied by -1 (because we changed a[i] to  - a[i]).

Please hack me, because no one was trying to do it (maybe, I'm a lucky one)

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    5 days ago, # ^ |
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    I found a hack to your solution.

    720720 0 100
    100 199 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 100000 100000 100000 100000 100000 100000 100000...
    

    Your program outputs 720720000 while the correct answer should be 199*(720720/2)=71711640

    I chose 720720 because it is the LCM of 1..16 so the first numbers all divide evenly into N. Then, you make the first few numbers cost equal to its size, so they all cost the same, so ternary search can't find the one that costs slightly below it's size. I padded on the 100000s so that it won't run into your brute force when the range is too small.

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5 days ago, # |
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Can anyone explain E? I struggling going through it. And why he has chosen this approach and its proof? Help!

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    5 days ago, # ^ |
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    It goes like this. You come up with straightforward O(n3) dp solution. Then you check the constraints and decide that there should be a greedy solution, not a dp one. After that you check which states are actually useful when updating the next states. Finally, you code it greedy using some precalc values to achieve the right complexity.

    The proof can be found in earlier comments of this blog.

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4 days ago, # |
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can anyone please explain problem C? I am not able to get the intuition to it from the editorial. Thanks in advance !

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4 days ago, # |
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Concerns about Problem: E Post Lamps.

Why are all the lamps being considered? Should not only ONE lamp with power >= 'some value L' and having the least cost be the optimal choice?

e.g: Suppose the maximum length of consecutive forbidden positions is 'L'. Then we must use a lamp with a power of at least 'L+1'. (We may put this lamp just before the start of that forbidden segment).

Now isn't it better to use a lamp with a power greater than 'L+1', but having a smaller cost? But this is not giving the optimal answer for some reason. I can not seem to find any example where a greater powered less cost lamp is not giving better answer.

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    4 days ago, # ^ |
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    Let array of costs be increasing array . For each L there are no ''a greater powered less cost lamp'', but optimal answer can be any of L.

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      2 days ago, # ^ |
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      Thank you for your response.

      I am getting it now.

      Here is the case that I was missing:

      pattern: 0XXXXX000

      lamp1: power: 6 , cost: 10 , ans: 2*10 = 20 ( 2 piece needed )

      lamp2: power: 8 , cost: 11 , ans: 1*11 = 11 ( 1 piece needed )

      Optimal answer is obtained by lamp2.

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3 days ago, # |
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I finally managed to pass tests with my solution for G. It is stupid dp on trees technique dp[v] = sack with pairs <gcd number, amount of paths>. Dp counted bottom-up then top-down. I think it can be hacked by ML verdict with careful test. Submission

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3 days ago, # |
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for E, "f(i) the minimal number of post lamps to cover segment [0; i] is clearly monotonous, thus you want to update states as early as possible"

^ i get the function f is monotonic but don't quite see the connection between f being monotonic and wanting to update states as early as possible" someone care to elaborate?

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3 days ago, # |
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This was really a nice contest.

In problem F what would happen if constraint fi to be under some c such that
 - c  <  =  fi  <  =  c

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3 days ago, # |
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Can somebody to help me to figure out whiy this is giving tle?