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Tutorial of Codeforces Round #599 (Div. 2)

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Isn't div1 B a weaker version of 920E - Connected Components??

This one has a more elaborate editorial :)

190E

653E (a little longer)

920E

1242B

all about components of complement graph :)

Thank you for sharing...Solved.

It seems that the data is so weak.

I used

`std::map<int,int>`

in Div1. C but it passed the system test.I used

`int`

at first too, and I realized the error after the first submission.But this doesn't seem matter according to the data.

I suspect 7500 in div1 C editorial is a typo of 75000.

Also, $$$\varnothing$$$ has been rendered wrongly as $$$varnothing$$$.

gen please look into this and the above comment.

Hi, my implementation for Div2 B2 is O(N^2).

This is how my algorithm works: If s[i] != t[i], I used an inner loop (with index j=i+1) to find s[i] == s[j] or s[i] == s[j].

However, the editorial states that the complexity is O(N). Is my analysis for the complexity wrong? Or is there a more efficient way of implementation?

Thanks!

I think you are right . :)

I suggest you simulation it by

vector,vec[i] present the position occurence insort, which complexity may beO(n),butn<=50,So,there is unnecessary。No. complexity is O(n) for div2 B2. here link to O(n) solution https://codeforces.com/contest/1243/submission/64410431

Your code also seems to have O(n^2) complexity. Is there anything that I am missing?

I think your solution is $$$O(n^2)$$$

i have tried to write the code in o(nlogn) using set hovever if you use unordered set in my code the complexity would become o(n). https://codeforces.com/contest/1243/submission/64422994

In my experience with Java/Kotlin it's not good to use unordered HashSets if you want to repeatedly get a single element from them (e.g. using Kotlin's extension

`.first()`

function; presumably the most idiomatic Java equivalent is`.iterator().next()`

) as it potentially might need to step through empty buckets in $$$O(n)$$$ time. (A LinkedHashSet is better for this use case.) I'm unsure if this is true for C++ unordered sets as well, but it probably works in a similar mannerThe system test for 1C is too weak.

sysjuruo hacked lots of submissions after contest.

I hacked them by my hand :(

It spent me an hour.

Just imagine how many hack you can do in 3.5 years

I think the editorial of div 2 B2 is incomplete.

There is no explanation on how to efficiently find characters in the strings. My implementation (64439067) used TreeSets, which pushes the complexity up to $$$O(n \log n)$$$. I think it can be done in $$$O(n)$$$ using arrays/vectors and pointers but the implementation is non-trivial. (Edit: Apparently it doesn't matter too much though cause $$$n \leq 50$$$)

Disproved; see belowI found I needed to check both for $$$s_i = t_j$$$ and $$$t_i = s_j$$$, but I'm not exactly sure how to construct the counterexample. All I know is that in this submission: 64438904 where I only checked for $$$t_i = t_j$$$ and $$$t_i = s_j$$$, it failed in case 49 of test set 2. Is there a test for the case where the strings are swapped?

no its complete. if total number of each character is even you can always find a solution within 2*n moves.check this https://codeforces.com/contest/1243/submission/64410431

I give up. I'm not sure how leaving out the parity check can cause a false

negative. Unless there's a difference in implementation that someone can enlighten me to.edit: Found my error. Apparently "find $$$t_j = t_i$$$ else find $$$s_j = t_i$$$" fails, but "find $$$s_j = s_i$$$ else find $$$s_j = t_i$$$" passes.

Can you tell me its sufficiency? The above seems to prove only the necessity. Thank you！

Instead of an explicit parity check, I simply try to make $$$s_i = t_i$$$ for each $$$i$$$. If the algorithm fails, it means it can't find a match for $$$t_i$$$, which only can happen if $$$cnt(t_i)$$$ is odd.

thx

Can someone please explain the time complexity in Div2C Editorial

You try all the numbers smaller than sqrt(n) to see if it's a divisor of n, so the complexity if O(sqrt(n)).

Some more duplicate problems on div1B.

It is not similar to this problem, during contest i also thought it is similar to this problem but after making required change it kept giving me wrong answer on pretest then again after some modification i passed pretest but it fail on main test.

And the reason behind that in Div1B is we can have the edge in its own vertex set as well that differ the complete approach. In 1228-D it is strict that we won't have edge in own vertex set but that is not the case here

for problem C in div1，15·5000=7500???

Can someone please explain the case of n=pq for 1242A Tile Painting

Try n = 21 = 3*7, p=3, q=7 and gcd(p,q)=1

Choose say i=2 and j=4

Then 2,2+3,2+3+3,2+3+3+3, ... will have same color

And 4,4+7,4+7+7, ... will have the same color

By Chinese remainder theorem, x=11 will satisfy:

11 = 2 (mod 3)

11 = 4 (mod 7)

So 2 and 11 (= 2+3+3+3) have same color

So 4 and 11 (= 4+7) have same color

So 2 & 4 have same color

According to Chinese remainder theorem because gcd(3,7)=1 we could have chosen any i,j and found such an x.

I got stuck with the tutorial with at first because 11-2 = 9, and 9 does not divide n=21, so why would "x" and "i" have the same color? But you "step" by increments of p and each step has the same color. So, if (x-i) is divisible by p, then x and i must have same color.

p.s. how does one insert newlines in these posts without a whole blank line??

Can people who submitted "gcd of all divisors except 1" in div.2 C describe their way of thinking? I know that it works, you can prove it with editorial's proof, but to me it makes no sense and seems like a random guess.

Think about it in terms of Bezout's Identity.

Let us take two integers $$$x$$$ and $$$y$$$.

Suppose, $$$n = Ax + By$$$

Let $$$g = \gcd(x, y)$$$

$$$n = Agx' + Bgy' = g(Ax' + By')$$$

This means that we can write $$$n$$$ as a linear combination of $$$x$$$ and $$$y$$$ if $$$g$$$ divides $$$n$$$.

Conversely, if we find a combination such that $$$Ax + By = g$$$, we can reach any multiple of $$$g$$$ too.

Suppose, we want to reach $$$kg$$$, then $$$(kA)x + (kB)y = kg$$$

Now, let us take $$$N = p_1 ^{a_1} p_2^{a_2} ... p_k^{a_k}$$$

Now, gcd of any two primes is $$$1$$$. This means that if there are at least $$$2$$$ primes, we can reach all the integers as $$$A p_1 + B p_2$$$

I suspect that this was the idea behind writing gcd of all the divisors. Suppose $$$g$$$ is the gcd, then we can paint $$$[1, g]$$$ with different colours but after that we cannot use any more colours.

The editorial and the intended solution went one step further and noticed that we do not really need to know the GCD of all the divisors since $$$2$$$ primes are always coprime to each other and if there is only $$$1$$$ prime, then that prime number is the gcd.

Let us generalise the question. Suppose we were given a set $$$ S = (a_1, a_2, \dots , a_n)$$$ and told that if $$$(i - j) \in S$$$, then $$$i$$$ and $$$j$$$ have to be the same colour, the solution is the gcd of $$$S$$$. We can reach any integer that is a multiple of the gcd of $$$S$$$. So we can only choose the colours in $$$[1, g]$$$. Whenever we colour $$$x$$$, the colour of $$$x + g$$$ is decided.

However, this question has an additional step that $$$S$$$ is only the divisors of an integer so this allows us to make it sharper. :)

Thanks a lot, your answer was helpful for me!

My solution was to find the gcd too, but it only made intuitive sense during the contest so thanks for this proof!

DIV 2B. if both strings are already same then answer is yes.

I had a different, possibly simpler solution to div1B.

All nodes with more than $$$\frac{n}{2}$$$ edges of zero cost are in the same zero-cost component. Loop all nodes with more than n/2 edges with connected to each other, so connect them in the DSU in O(n log n). Loop all nodes with at least $$$\frac{n}{2}$$$ edges with cost 1, and connect all nodes they have zero-cost edges to in total time O(m log n). Total complexity is O((m + n) log n).

Code: 64382712 (I use $$$\frac{n}{2} - 10$$$ here because I didn't want to think about the constants)

If you use DSU with path compression, the $$$O(\log n)$$$ factor can be reduced to $$$O(\alpha(n))$$$, where $$$\alpha$$$ is the inverse Ackermann function, a function that grows so slowly that it can be considered $$$O(1)$$$ for any data set that can

physically fitin thisuniverse. But in my experience uploading DSU solutions with/without path compression, I can't seem to see much appreciable difference.This was really nice observation and pretty easy to implement as well

Thanks Mango , your codes are the best. :)

Div 1 B can be solved using Prim's also.

I keep a set of pairs,which are disconnected vertices , first entry will be number of edges from connected part to the vertex and select will be vertex number. and i keep a map from second entry to first entry of set.

Every time fetch the first element in set, if its degree equal no of connected nodes, increment answer by 1.then update map and set with edges of this node.

Here is my submission : https://codeforces.com/contest/1242/submission/64390575

THANKS

Could you give me a intuition explanation for Chinese Remainder Theorem?

Well, I did the same thing, except that I determined, for every vertex

`i`

, any other vertex`j`

that:a.

`i`

isn't connected to`j`

with a 1-weight edge,b.

`i`

and`j`

lie in different connected components.I do that by initializing

`j`

to 0 and looping over the vertices incident to the current`i`

, incrementing`j`

if necessary.However, this gives WA 13... :(

In div2 C, my implementation is: if n==1, ans is 1. Otherwise get the divisors of n. Get gcd of all the divisors except 1. Answer is the gcd. Here is my JAVA code: 64462563

Instead of divisors can find prime factors. Since gcd of prime factors is always 1, thus if there are more than 1 prime factor then answer will be 1 and if there is only 1 prime factor then answer will be that prime factor.

Please, someone, explain 0-1 MST problem's solution by using an example.

My solution for div1B:

Let G1 be the graph where only edges with cost 1 are considered. Pick any vertex (S) with minimum degree.

Let X be the set of vertices which have 0 cost edge with S and Y be the set of vertices which has 1 cost edge with S. Note that the size(Y) <= m/n.

Build a new graph as follows:

For each vertex in X add an edge to S (O(n))

For each vertex in Y add an edge to all its 0 cost neighbours (O((m/n)*n))

Now, the answer is Number of connected components of this new graph -1.

My Submission

Got you! Did you find this somewhere?

Let n be the number of vertices, m be the number of edges

sum(degree) = 2*m

n*min_degree <= 2*m

min_degree <= 2*m/n

size(Y) = min_degree <= 2*m/n

I dint find this anywhere. During the contest I was trying to find an equivalent sparse graph which would have the same number of connected components as the graph formed by 0 cost edges and eventually found this.

Proof of this please: "Now, the answer is Number of connected components of this new graph -1."

The "new graph" is equivalent to the graph mentioned in the editorial in terms of the number of connected components, just that the connected component which contains vertices in X is sparse.

I read your code and understood it. But can you explain the complexity since the worst time complexity seems O(n^2). Also why to find the vertex with min degree, I could just merge(union) all the vertex which have 0 weight edge between them in O(n^2)?

In the problem we need to find minimum degree vertex.As if you take any vertex randomly , then it might have degree about 10^5.(eg. In worst case, all 99999 vertices might be connected to a single vertex and if you select that vertex then it would have o(99999) * o(100000)[o(100000) is for zero weight components traverse]. But if you select minimum degree vertex then in worst case, minimum degree would be 10^(2.5) because in worst case a vertex can be connected to every other vertex which can only happen if n<=10(2.5) then it would have about 10^5 edges. Therefore, minimum degree must be smaller than equal to 10^(2.5). Hope you understood why are we minimum degree vertex :)

Why to take minimum? Edit: I figured it out.

In your code, you find value of cc before building the new graph and if this cc is greater than 1, "0" is printed, can u explain what's the reason behind finding cc here?

Can anybody explain Div2D in a more intuitive way? Like what is the key observation over here? With an example? I have read some of the comments explaining the solution but not how they got there.

can someone explain why i'm getting tle.. link — https://codeforces.com/contest/1243/submission/64400718

I solved Div2 D using the approach mentioned in the editorial,solution. However when I applied the same approach to this similar question 190E-Counter Attack, it is giving TLE although the time complexity is still the same,. Any help would be appreciated!

Thank you! Your code made me understand how to solve the problem.

Time limit is strict in this problem(190E - Counter Attack). TLE verdict is because of high constant factor in the search complexity of C++ std::set. Replacing set by sorted vector passes all test cases(64609809).

Thank you. Nice observation!

I found the typo of editorial of Div1D. It will be fixed soon.

On the editorial says "Every interval has exactly one non-self number", it means that the non-self numbers are periodic $$$($$$if n is non-self the $$$n$$$ + $$$k^{2}$$$ + $$$1$$$ is also non-self $$$)$$$; however, with $$$k$$$ = $$$2$$$, the numbers are $$$3, 9, 13, 18,...$$$ and clearly there is not a period. Thanks in advance.

$$$3$$$ is in first interval $$$[1, 5]$$$, $$$9$$$ is in second interval $$$[6, 10]$$$, $$$13$$$ is in third interval $$$[11, 15]$$$, etc. All non-self numbers can be in each interval without having exactly same period.

Ohh I understand thnks :)

Anyone can explain to me the div 1 — C / div 2 -E

I am not able to get the intuition and approach both.

I solved div1B using the algorithm provided int the editorial. But I still wonder why the complexity is O(nlongn + m), as we iterate every vertice from 1 ~ n and every time we iterate every zero weight component to try to unite them, it seems that the complexity should be n^2. Are there anyone explain it for me? Thanks a lot!

I will try to give you some intuition: Yes, you are right, for every vertex we are going over all the zero-weight-components, and you are right there can be N of them. But the complexity is not N^2, here's why: The number of zero-weight-components you can have at any given time is actually bounded by M/(N — 1), the number of edges. Because in order to leave a vertex unmerged, it needs to have a degree equal to N — 1. In other words, it needs to have an edge between it and every other node. This can happen at most M/(N — 1) times. Since M is actually min(10^5, N*(N — 1)/2), you can see that M/(N — 1) cannot exceed 10^3. And now, every node after that will have all of it's edges with weight 0, because we have already used all our weight 1 edges. So every subsequent node we consider will be merged.

It's not very rigorous I know but I hope I could provide some intuition

I see, thanks a lot for your explaination!

Did anyone solved div2D exactly as explaind in editorial.

I tried to stay as faithful as possible to the explanation in the editorial, you can check my code here: https://codeforces.com/contest/1243/submission/64548383

Oh... It helped me a lot. Thanks for replying. I was stuck for 1 day.

Sorry to bother you again, but can you tell why i > v[i][j]?

It's just because we want to process the nodes whose label is less than the current one. This is because when we are iterating over node i, the sets for nodes whose index is greater than i have not been created yet, as we are only creating the set after finishing with i.

You could probably do an implementation where you create all the sets in the beginning, I don't see why it would not work. But as I told you, I tried to stay as close as possible to the editorial and this is what the editorial suggested.

oh, that means if we remove i>v[i][j] , then it would increase time complexity for visiting adjacent vertices by factor of 2, which is sufficient for passing the time limit. [I tried :-)]. Thanks again for help.

Could anybody help me with my code of Div1.C? It did output the correct answer when running with the first example input on my own computer, but I don't know why it just printed

`0`

when submitted.Here's the local output:

Here's the online output:

I'm really confused about this. And here's my submission 64595444. Thanks a lot!

Type of

`a`

is long long int, that is why your second`%d`

buffer doesn't work. Change`a`

to just`int`

or make`%lld`

for`a`

in your printf.That's what's the matter! Thank you very much! :)

Can someone explain 0-1 MST solution with DSU? I understood the general idea that we should count the number of zero-weight components. The part that I couldn't get starts from "For each of the zero weight components, we count the number of edges from this component to v."

I did not understand the editorial for D. I understood the claim but did not quite understand the proof neither how to come up with such claim out of blue. Did anyone come up with the solution in some other way? Or any better explanation?

I also think my sentence is bit messy.

Suppose you have $$$k^2$$$ self numbers and $$$1$$$ non-self number in $$$k^2+1$$$ size interval. Then each $$$k$$$ numbers of $$$k^2$$$ self numbers will be appended to $$$s$$$ at the same time, and they will generate new $$$k$$$ non-self numbers in total. Each non-self numbers will be included in next interval($$$k \cdot i, k \cdot i + 1, \ldots , k \cdot i + k - 1$$$ th intervals), because each summation will be fit in corresponding interval as I described as formula in editorial.

Now let's see this formula:

$$$k$$$ numbers of $$$t$$$-th subinterval(size $$$k$$$) of $$$i$$$-th subinterval(size $$$k^2 + 1$$$) will be $$$(i \cdot (k^2 + 1) + t \cdot k + j) + r[j]$$$ for $$$j = 1, 2, \ldots, k$$$ where $$$r[j]$$$ is $$$0$$$ or $$$1$$$, because if non-self number is inserted at the left(or inside) of the $$$t$$$-th subinterval, then part(or whole) of the subinterval will be shifted, so that $$$r[j]$$$ is offset consideration, and $$$\sum_{j} r[j] = offset$$$.

Since $$$0 \le t \lt k$$$ and $$$0 \le offset \le k$$$, this is always true: $$$1 \le \frac{k \cdot (k+1)}{2} - t + offset \le k^2 + 1$$$. This means generated sum of $$$t$$$-th subinterval will always be in $$$(k \cdot i + t)$$$-th interval. ($$$i$$$-th interval represents $$$[i \cdot (k^2+1) + 1, (i+1) \cdot (k^2+1)]$$$.)

Thank you.

I can not understand the solution of div2E-div1C until i read the problem statement again and realized that i did not read statement carefully before. I did not notice on the important condition "All of the integers are distinct".

for 1242B if we find the number of connected components of the given 1 weighted graph complement the problem is solved!

we can find the vertex v with minimum degree of the given 1 weighted graph.

Merge all vertices except ones adjacent to v into one component and make the root of them to v in the dsu.

the number of remaining vertices (adjacent to the v) can't be large!!

Let degree of v be d then d<n & n*d<=2m so d<=sqrt(m).

Now brute force for all zero edges between neighborhoods of v and the remaining vertices and complete the dsu.

complexity is o(n*sqrt(m)).

Div2E/Div1C: can we get more details about the graph we should construct?

Is it TRUE in 1242A Tile Painting that if all characters occur even number of times then their must exist answer with less than 2n swaps ?

Can someone please help me out in finding the reason for getting

TLEfor problem 1243D - 0-1 MST ??Here's my submission 64726030 ...According to me it's O(m)...hello, could someone help me with the explanation of the problem D div2? Please.