feecIe6418's blog

By feecIe6418, 2 years ago, In English

Thanks for participation!

If your solution to D involves any data structures and is not $$$O(n)$$$ -- please read the "solution 1". I believe it is very interesting, but to make the difficulty suitable for D we allowed not $$$O(n)$$$ solutions.

1696A - NIT orz!

Hint 1
Solution

1696B - NIT Destroys the Universe

Hint 1
Hint 2
Solution

1696C - Fishingprince Plays With Array

Hint 1
Hint 2
Solution

1696D - Permutation Graph

This problem has two different solutions. The first one is more beautiful, but less straight-forward.

Hint 1 for solution 1
Hint 2 for solution 1
Hint 3 for solution 1
Solution 1
Hint 1 for solution 2
Hint 2 for solution 2
Solution 2

1696E - Placing Jinas

Hint 1
Hint 2
Hint 3
Solution

1696F - Tree Recovery

Hint 1
Hint 2
Solution

1696G - Fishingprince Plays With Array Again

Hint 1
Hint 2
Hint 3
Solution

1696H - Maximum Product?

Hint 1
Hint 2
Solution
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2 years ago, # |
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Thank you for fast editorial!

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2 years ago, # |
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fast editorial, good problems

great round!

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2 years ago, # |
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What a speed!

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Editorials with hints. Appreciable efforts..Thanks!

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2 years ago, # |
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I hope I'm not the only one who got WA on question B for not realizing that the answer is at most 2... XD

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2 years ago, # |
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Best way to present editorial is with hints Thanks for fast and crisp editorial

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Thank you for the fast editorial feecIe6418 Congrats on your first Global Round. It's hard to imagine that you single-handedly created all except for one problems for this round, that's amazing! Great work!

Feedback
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2 years ago, # |
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Thanks for the fast editorial. By the way,E<D.

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2 years ago, # |
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Feeling sorry for Tourist. Dude is facing a lot of negative deltas in recent times. I remember reading a blog few days back, when someone predicted Tourist's fall. It's becoming real now :(

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    2 years ago, # ^ |
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    Just out of curiosity, could you link that blogpost? I'd like to read it too

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    2 years ago, # ^ |
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    Brother Legends never dies Tourist is Legendary No one takes his place

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      2 years ago, # ^ |
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      There is nothing called legendary. If you consistently put in the effort for over a decade(He has been into mathematics and programming since age 10), you could become like him(may be better than him).

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        2 years ago, # ^ |
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        And you're still newbie after 3 years.

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    2 years ago, # ^ |
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    A Lion takes its one step back before making its bigger attack

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    2 years ago, # ^ |
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    Normal competition when everyone is too good.

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    2 years ago, # ^ |
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    Share the blog . I want to see what happen to our king !!.

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    2 years ago, # ^ |
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    i Dont think its a downfall if you get -56 for global rank 7.

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2 years ago, # |
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Thank you for Editrorial!

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2 years ago, # |
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Solution for E without using the identity given in the editorial: Let $$$D_i$$$ be the $$$i$$$'th diagonal, the set of $$$(x, y)$$$ such that $$$x+y=i$$$. Calculate the answer by summing each diagonal. When moving from $$$D_i$$$ to $$$D_{i+1}$$$, the current number of moves is simply multiplied by $$$2$$$, except in the case where some cell in $$$D_i$$$ is a "boundary" (the cell down or to the right is black). But there are only $$$O(n)$$$ such boundary cells, so we can simply subtract their contribution as we go along.

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2 years ago, # |
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Plz show codes for the questions

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2 years ago, # |
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Are you sure memory limit will not exceed if we expand them? If not why did it happen to me?

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    2 years ago, # ^ |
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    You shouldn't make the whole array just you can store the occurences

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      2 years ago, # ^ |
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      can you pls explain how to store the occurrences such that they appear in the correct order as well?

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        2 years ago, # ^ |
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        I did it just storing a pair of values: the factors and its number of ocurrences. For example, for M = 3 and A = {6 3 4} -> {[2,3] [1,3] [4,1]} Did it for both vectors and compare them

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        2 years ago, # ^ |
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        161770389 You can check here. I've just stored it as pair like if A={4,2,6} and M=2 then it will be {{2,3},{3,2}}.

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      2 years ago, # ^ |
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      This is how I rewrote my solution after hitting the memory limit.

      https://codeforces.me/contest/1696/submission/161775684

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2 years ago, # |
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The problem F is great !

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2 years ago, # |
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Easier solution for F:
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2 years ago, # |
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D was really great, I have solved it with stack, segment tree, and bfs :).

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2 years ago, # |
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Was it really a good idea to give n+1 number in array for task E ? I didn't solve it only because of this. The tasks mustn't be with such a potential trap I think.

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2 years ago, # |
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D is really cool, I solved it with mono stack and binary search

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    2 years ago, # ^ |
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    I cant understand the second approach given in editorial . Need help

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could someone please find the mistake in my submission for C?

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2 years ago, # |
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thanks for the editrorial!

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2 years ago, # |
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pls give some proofs for math formulas you are using in E solution otherwise it doesnt tutorial, because your solution based only on this math fact so you have to prove it

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    2 years ago, # ^ |
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    It's just hockey stick identity. Nothing more, nothing less. I don't think one has to prove a well-known, simple identity in combinatorics in their editorial

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    2 years ago, # ^ |
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    I'll leave a link to the solution of last problem from last Starters since it also uses this identity and gives a 'proof' based on another (maybe more well-known) identity.

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    2 years ago, # ^ |
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    For row i : the summation is iCi + (i+1)Ci + (i+2)Ci ... + (i+v[i]-1)Ci.

    Now if we know iCi = 1 = (i+1)C(i+1)

    Replacing this in the above formula we get the summation for row i as :

    (i+1)C(i+1) + (i+1)Ci + (i+2)Ci ... + (i+v[i]-1)Ci.

    Combine the first 2 terms : (i+1)C(i+1) + (i+1)Ci = (i+2)C(i+1) [ Using the simple property nCr = (n-1)C(r-1) + (n-1)Cr ]

    Replacing this in the equation we get :

    (i+2)C(i+1) + (i+2)Ci ... + (i+v[i]-1)Ci

    Again now, we can combine the next 2 terms and so on. So finally we get (i+v[i])C(i+1)

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      2 years ago, # ^ |
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      Thinking on how to come up with a visual approach, I have found the following one. It does nothing different than what 123thunderbuddie did, but supports the idea visually.

      We want to find the sum of the number of ways to reach every cell in the $$$i^{th}$$$ row. Assuming there are $$$c$$$ cells, the sum is equal to the following (just what 123thunderbuddie wrote):

      $$$\displaystyle \binom{i}{i} + \binom{i+1}{i} + \cdots + \binom{i+c-1}{i}$$$

      Suppose there is a row below that has the same number of cells $$$c$$$ (not the $$$(i+1)^{th}$$$ row but a temporary one). This sum is actually equal to number of ways to reach the last cell in that row, which is $$$\binom{i+c}{i+1}$$$ (number of ways to reach cell $$$(i+1, c-1)$$$ from $$$(0, 0)$$$). Here is a visualization:

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Why is this submission with long long giving WA, but this isnt.

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    2 years ago, # ^ |
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    The variable is z_cnt is not initialized in both of them resulting in UB. You got lucky with the second solution.

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Kudos to the authors for an amazing round. Finally reached Purple!

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2 years ago, # |
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I've noticed that in G there are no small tests before the first big one, maybe that's the case in other problems too, keep in mind that it's bad.

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I am not sure of the time complexity of this solution for D can someone help out https://codeforces.me/contest/1696/submission/161799451

I feel it should be $$$O(n)$$$ because any next greater or next smaller chain is travelled once.

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2 years ago, # |
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We can do D in O(N). By maintaining premax,premin,sufmax,sufmin.

Spoiler
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    2 years ago, # ^ |
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    thanks for the code

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    2 years ago, # ^ |
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    Doesn't this code will add multiple edges between two nodes? Though it won't affect our result.

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I solved problem D using greedy approach :) 161785869 here it is

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Can someone explain C please

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    2 years ago, # ^ |
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    Just maintain vector of pairs where pairs.first = actual value and pair.second = frequency . Make vector of both a and b . And check at last a == b or not. Since we can divide every no m times even in the worst case it's log(n), which means atmost 30times for each element. which is still O(N). Code--

    Spoiler
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2 years ago, # |
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Another solution for F: Let $$$S(u)$$$ be a set of pairs $$$(x, y)$$$ such that $$$d(x, u) = d(y, u)$$$. Notice that $$$u$$$ with minimal number of elements in $$$S(u)$$$ is a leaf and the vertex connected to it is a vertex $$$v$$$ such that $$$S(v)$$$ contains $$$S(u)$$$ and for every pair $$$(x, y)$$$ such that $$$(x, y)$$$ belongs to $$$S(v)$$$ and doesn't belong to $$$S(u)$$$ either $$$x = u$$$ or $$$y = u$$$. We can remove that leaf and continue the process.

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    2 years ago, # ^ |
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    Notice that u with minimal number of elements in S(u) is a leaf.

    Do you know how to prove this? It does make sense intuitively.

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Is it just me or are the codes not attached?

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Hope I'm not the only one who overcomplicated D with sparse tables, sets and binary searches.

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I was dazzled when I found the "more beautiful but less straightforward" solution to problem D. It was perfectly splendid! orz

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Any other guy who got wrong answer on problem B untill realising answer can be atmost 2 !!

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"EDITRORIAL" of Codeforces Global Round 21. Lol :D

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I have another solution for F. It was very painful to implement, though, so I couldn't finish it during the contest. Maybe some of you had a similar thought.

First of all, if x and y are even distance apart, then there must be a valid z that's equidistant from them (center of their path). Also, if x and y are an odd distance apart, then no valid z will occur, clearly. Therefore we can create a graph where there is an edge between x and y if there is a valid such z. This graph must be bipartite, so we will be able to split vertices into 2 categories.

Now we will solve for the adjacent vertices of every vertex. After this, it will be pretty easy to construct a solution (and check if it is false) as we can just add in all the apparent edges. Consider the vertex v. Now, for all other vertices, we can form equivalence groups by their distance to v. Then, for each of these groups, we can check if every 2 vertices of that group are equidistant. If so, then clearly every one of these vertices is in a different component of the "graph without v". Now the size of the largest of those groups must be the degree of v, as each outgoing edge of v signifies one connected component on the "graph without v". Of course, we still are not able to pinpoint which one of these groups contains the vertices distance 1 around v. Let's say there are m such groups (groups where every 2 vertices are equidistant from each other and every vertex is equidistant from v with the largest possible size). Then they must be "circles" around v from distance 1 to m. We are trying to find the group of distance 1. This is where the bipartite property comes in: consider any point p from a group k distance away from v. If p and v are in different bipartite categories, then they are at an odd distance away from each other. Otherwise, they are even distanced away, and there will be exactly one single vertex z that is equidistant from them both, which will be in the middle of the path between p and v. This z will clearly be in the equivalence group with distance k/2 away from v. let's call this operation (obtaining a new vertex with k/2 distance) splitting. Now consider the equivalence group that splits the most. It is clear that if $$$k=2^ba$$$( a odd), then b will be the times it splits. Then, it is obvious that the largest power of 2 less than or equal to k will satisfy this property. The important part is that a vertex with a power-of-2 distance will eventually split into a vertex of distance 1 from v, and so we can obtain all the other vertices of distance 1 around v, and we are done.

I believe this method as described takes $$$O(n^3\log{n})$$$ time, and it could be simplified to $$$O(n^3)$$$

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    16 months ago, # ^ |
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    I came up with this solution too, and it unfortunately is incorrect (or perhaps incomplete).

    Then they must be "circles" around v from distance 1 to m.

    This statement is false (I had based my solution around this assumption too).

    Consider the following tree (I have highlighted the maximum size levels of vertices which have all their pairwise lca's at root (which you call "circles") with red):

    Spoiler

    So we can easily construct trees in which circles are not at one contiguous range of distance, but rather present at disjoint ranges of distance (and this totally breaks our ability to find the circle at distance 1 by halving the number of valid of levels at each step).

    Implementation of this approach: link

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2 years ago, # |
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Here is my solution to problem D:

I have a recursive function that takes two-parameter $$$(l, r)$$$ positions of the starting and ending index in an array and calculates the optimal answer for this range.

For the optimal answer, I first find the index of the minimum element and maximum element in the range $$$(l, r)$$$ and store them in $$$s$$$, $$$e$$$. $$$(s < e)$$$. I am using a segment tree and a map to find the index of the minimum and maximum in the range $$$(l, r)$$$. It is optimal to have an edge between $$$(s, e)$$$, so we add one edge.

So, The answer for the range $$$(l, r)$$$ is $$$1$$$ + the answer for $$$(l, s)$$$ + the answer for $$$(e, r)$$$, and the base case is that if $$$l$$$ == $$$r$$$, there is no need to add an edge, so return $$$0$$$.

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2 years ago, # |
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Here is another solution for F:

Assume that for some two nodes $$$u$$$ and $$$v$$$, we know there exist an edge between them in the solution.

From this, we can find all vertices $$$w$$$ such that $$$dist(u, v) = dist(v, w)$$$, then the solution must also contain the edge $$$(v, w)$$$.

Repeating that process, we can find the solution (if exists) in a bfs-like way.

Therefore, for each $$$x$$$ from $$$2$$$ to $$$n$$$ we can suppose that the edge $$$(1, x)$$$ exists in the solution, try to find that solution, and see if it is valid.

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Great round. I especially liked that nothing between A and E is too implementation-heavy.

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Damn my $$$O(n \log^2n)$$$ solution for D managed to pass. :)

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Minor issue: blog title has a typo (Editorial not editrorial)

feecIe6418

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2 years ago, # |
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Here is $$$O(n^3)$$$ solution to F. First we can find the farthest distance from any vertex $$$x$$$ to other. So we know the diameter of the tree. If we root the tree at the center, we can know the depth of every vertex. So we can determine at least one edge in the tree (node with depth 1 and the root), then use the method in the editorial. It also shows there is at most one valid answer.

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2 years ago, # |
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That's quick.

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2 years ago, # |
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I think the contest has too much observation.

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    2 years ago, # ^ |
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    As is literally every codeforces round including div 3, div.2, educational, div1+div2 combined, div1, olympiad-based rounds?

    (yes I know that div.4 exists but I don't have a very fond experience in div.4)

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a bit of pity for D

pass pretest make me think it works anyway

but didn't come up with cases like 1 100000 2 99999 3 99998 4 99997 ...

then got fst

try to consider more next time. or maybe always need to try to achieve as low complexity as i can

it's actually a great problem though, and solution 1 is awesome.

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For D, what if the maximum element is 1 or n?

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Could someone please help me to find me the error in my code of problem D I am using approach 2 https://codeforces.me/contest/1696/submission/161834104

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In problem G, when I set

const double INF = 1.0 / 0.0;

It got TLE on test 76 (or Pretest Passed 5880ms, and TLE on test 7 if I unroll the loops in matrix multiplication myself) with GNU C++14 (32bit), but ~850ms with GNU C++20 (64bit, winlib).

However, when I set

const double INF = 1e18;

Its speed remains unchanged with GNU C++20 (64bit, winlib), but when compiled by GNU C++14 (32bit), it works in <500ms and becomes the fastest submission.

Is there any logic behind this (about inf in double)?

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    2 years ago, # ^ |
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    An explanation for the first (but not very informative) is that diving by zero (even with floating numbers) is undefined behavior, (according to [expr.mul.4], or discussions in StackOverflow), and anything can happen with undefined behavior.

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    2 years ago, # ^ |
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    a better suggestion might be using std::numeric_limits<double>::infinity() as the value for inf. this is not undefined behaviour and still compares bigger than the maximum value representable with double.

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

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

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can anyone please explain the proof for problem c with intution by taking an example? i couldnt understand editorial

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    2 years ago, # ^ |
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    Try expanding $$$a$$$ and $$$b$$$ into $$$a^\prime$$$ and $$$b^\prime$$$ in the example test cases.

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      2 years ago, # ^ |
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      i have expanded both the arrays upto maximum extent as possible but im not getting proof/intution.

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        2 years ago, # ^ |
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        Now notice that it is possible to obtain $$$a^\prime$$$ from $$$a$$$ after some operations, and it is possible to obtain $$$b$$$ from $$$b^\prime$$$ after some operations, so if $$$a^\prime = b^\prime$$$, we can obtain $$$b$$$ from $$$a$$$.

        We still need to prove that if $$$a^\prime \ne b^\prime$$$, it is impossible to obtain $$$b$$$ from $$$a$$$ after any number of operations.

        To do so, note that the fully expanded form of $$$a$$$ does not change when you perform one of the operations on it.

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shit contest. downvoted

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What does it mean by "z will always become a submask of itself after any number of operations, so ai will always be a submask of (ai or z) after any number of operations"? Please explain.

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    2 years ago, # ^ |
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    Basically, $$$z$$$ will only lose bits.

    So $$$(a_i\mathrm{\ or\ } z^\prime) \le (a_i\mathrm{\ or\ } z)$$$, where $$$z^\prime$$$ is $$$z$$$ after some operations.

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      2 years ago, # ^ |
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      Okay, but what is the meaning of submask here specifically. Is it like changing the set bits of z after any number of operations so the resulting bit representation i.e. bit mask obtained is called submask?

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        2 years ago, # ^ |
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        $$$z^\prime$$$ is a submask of the bitmask $$$z$$$, since every bit that is set in $$$z^\prime$$$ is also set in $$$z$$$, but not necessarily vice versa.

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2 years ago, # |
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Why this code is working fine for question A? for(int i = 0; i < a.size(); i++){ cin >> a[i]; a[i] = a[i] | z; z = a[i] & z; ans = max(ans, a[i]); } Here I am performing both the operations — bitwise or and bitwise and. Also why it is not needed to perform second operation i.e bitwise as given in tutorial, any proof?

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    2 years ago, # ^ |
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    You are doing:

    	x = (a[i] | z)
    	y = x & z
    	z = y
    

    Every bit in z that is 1 is also 1 in x, so (x & z) equals z. In other words, you aren't changing the value of z and the second operation is doing nothing.

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Have a solution different from the tutorial for problem D which does not use segment tree or stacks, very difficult for me to explain how it works but the following ideas might help-

$$$1.$$$ $$$Its$$$ $$$always$$$ $$$optimal$$$ $$$to$$$ $$$jump$$$ $$$to$$$ $$$the$$$ $$$farthest$$$ $$$index$$$ $$$from$$$ $$$current$$$ $$$index.$$$

$$$2.$$$ $$$We$$$ $$$always$$$ $$$jump$$$ $$$from$$$ $$$a$$$ $$$local$$$ $$$maxima$$$ $$$to$$$ $$$a$$$ $$$local$$$ $$$minima,$$$ $$$and$$$ $$$then$$$ $$$to$$$ $$$a$$$ $$$local$$$ $$$maxima$$$ $$$and$$$ $$$so$$$ $$$on.$$$

161975222

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2 years ago, # |
Rev. 2   Vote: I like it +18 Vote: I do not like it

For F, for each $$$i$$$, one can assume that edge $$$(1, i)$$$ is in the tree, and then try to construct the rest of the tree, as per hint 2. We can then check if the tree is valid.

At the start, for each node $$$v$$$, we can do a $$$O(n^2 \alpha(n))$$$ precomputation of the equivalence classes of nodes equidistant from $$$v$$$. Then, we can do each tree reconstruction and check in $$$O(n^2)$$$, for a total complexity of $$$O(n^3 \alpha(n))$$$.

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

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2 years ago, # |
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Can someone explain the equation given in Problem E?

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2 years ago, # |
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Hi, could someone take a look at why I'm getting WA on test case 3 of F? My submission is here.

Specifically, I create equivalence classes for each node; each equivalence class represents equal distance from said node. This is done using DSU's, which should make the whole thing have complexity $$$O(n^3)$$$. Thanks!

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2 years ago, # |
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I got a solution for F which solves it in $$$O(n^3)$$$

Well-known Theorem: there is atmost two centres in a tree (a center is a vertex with minimum value of maximum distance of a vertex from the given vertex: this value is called eccentricity).

Note that using the given information, we can find the eccentricity of a vertex: this is just the number of equivalence classes of vertices wrt any given vertex (two vertices belong to the same equivalence class wrt a given vertex if they have the same distance from that given vertex).

So we can easily find the centre(s) of the tree.

If there are two centers, there is a theorem that they must be connected, so we find an edge immediately and proceed to construct the tree.

If there is one center, then we can prove that all adjacent vertices have eccentricity of exactly one greater than the center, and all other non-adjacent vertices have eccentricity $$$\geq e+2$$$, where $$$e$$$ is eccentricity of the center. So in this case, we can find any node with $$$e^\prime = e+1$$$ and hence, these two must be connected, and we have found an edge.

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2 years ago, # |
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Can someone explain me why I am getting TLE on problem C in test 11 because according to me my program is running in O((n+k)logmV) (as given in the editorial) . Instead of storing frequency I am just using a count variable to check whether expansion of both array will be same. thanks https://codeforces.me/contest/1696/submission/173329854

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23 months ago, # |
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In problem E, if you don't know the equation : $$$\sum_{i = 0}^kC_{n+i}^n = C_{n+k+1}^{n+1}$$$, here's the proof:

first, you have to know $$$C_{a}^b = C_{a - 1}^b + C_{a - 1}^{b - 1}$$$

then,

$$$ \sum_{i = 0}^kC_{n+i}^n = C_n^n+C_{n+1}^n+C_{n + 2}^n + ...+C_{n+k}^n\\ =C_{n+1}^{n+1} + C_{n+1}^n + C_{n + 2}^n ... + C_{n + k}^n\\ =C_{n + 2}^{n + 1} + C_{n + 2}^n + ... + C_{n + k}^n\\ =C_{n + 3}^{n + 1} + ... + C_{n + k}^n\\ =...\\ =C_{n + k}^{n + 1} + C_{n + k}^n\\ =C_{n + k + 1}^{n + 1} $$$
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12 months ago, # |
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Thanks!