By MikeMirzayanov, 7 years ago, In English

Codeforces Round 440 will start on October 15 (Sunday), 08:05 (UTC). It will be based on Technocup 2018 Elimination Round 2. So, if you are a Russian-speaking high-school student, please take part in the Technocup 2018.

Codeforces Round 440 is open and rated for everyone.

Wish you good luck and bugless code.

Editorial is posted.

Congratulations to winners!

Technocup official round:

  1. Mikhniuk
  2. scanhex
  3. Krisha
  4. lis
  5. FalseMirror

Div. 1:

  1. khadaev
  2. Errichto
  3. eddy1021
  4. FizzyDavid
  5. fateice

Div. 2:

  1. wdmmsyf
  2. Legilimens2023
  3. destinydrifter
  4. oscar114514
  5. OMS
  • Vote: I like it
  • +188
  • Vote: I do not like it

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7 years ago, # |
  Vote: I like it +39 Vote: I do not like it

Mandatory comment- "Hope the problem statements are as short as this blog".

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7 years ago, # |
  Vote: I like it +110 Vote: I do not like it

Clash with OpenCup. ;(

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7 years ago, # |
  Vote: I like it +75 Vote: I do not like it

Is it me or is the contest title is longer than the announcement :3

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7 years ago, # |
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I guess it's the first time that we don't see any "thanks" or even any username in an announcement :D

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7 years ago, # |
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This is like the shortest announcement ever of all time.

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7 years ago, # |
  Vote: I like it +25 Vote: I do not like it

A good time for Chinese students!

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7 years ago, # |
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I have an onsite contest after 15mins for 5 hours, and then I can take part in this cf round. A good time for me.

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7 years ago, # |
Rev. 2   Vote: I like it -32 Vote: I do not like it

I have doubt.

When I opened 'contests' page, it showed me two separate contests(div1 and div2), but in the announcement it is given (Div1 + Div2). Which one it is going to be ? MikeMirzayanov

UPD : when I posted this comment, there were only 2 contests. (contests 440Div1 , 440Div2).

That's why I got confused. Right now contests page is fully updated. NH_Apu

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7 years ago, # |
  Vote: I like it +14 Vote: I do not like it

I Wish everyone positive rating!

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

why do timing varies a lot recently ??

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7 years ago, # |
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I hope I won't get a WA on problem A, the way it happened in the previous contest :(

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7 years ago, # |
  Vote: I like it +26 Vote: I do not like it

It will be at 3 AM here in Colombia. Guess is a good way to start a Sunday ¯_(ツ)_/¯ .

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7 years ago, # |
  Vote: I like it -12 Vote: I do not like it

2 PM in Bangladesh..Surely the best time except for missing lunch

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7 years ago, # |
  Vote: I like it +12 Vote: I do not like it

The Terms of agreement when registering is in Russian "Вы подтверждаете, что будете работать самостоятельно во всех этапах Технокубка и не будете нарушать какие-либо другие правила." So I'm wondering whether an English version of problem statements will be given during the real contest. Can anyone explain about this a little bit? Thanks a lot.

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7 years ago, # |
  Vote: I like it -8 Vote: I do not like it

gl hf

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7 years ago, # |
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number of problems and the difficulty is about the 1st round ? because that round was easier than ususal

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7 years ago, # |
Rev. 2   Vote: I like it +10 Vote: I do not like it

What's the "Practice Round 2"?

EDIT: never mind, seems to be for Technocup official participants only

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

Hope there will be no geometry problems

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7 years ago, # |
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May be it's a silly question.

but what does "00:04" like things below every submission means??

is it time of submission or what

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    7 years ago, # ^ |
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    You mean the time that is shown below the submission during the contest.

    Its the time when you've submitted the solution after the start of the contest. '00:04' Would mean that you've submitted the solution after 4 minutes

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7 years ago, # |
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How long will the contest last?

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7 years ago, # |
  Vote: I like it +32 Vote: I do not like it

have no time to receive it :D

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

Why can I register for both Codeforces Round 440 (Div 1) and Technocup 2018 round 2?

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7 years ago, # |
  Vote: I like it +35 Vote: I do not like it

How many problems?

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

YouHaveShort Blog.

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7 years ago, # |
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Registration button didn't work for me...

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    7 years ago, # ^ |
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    Still can't register in div.2 ((

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      7 years ago, # ^ |
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      Russian-speaking high-school students have to register in this contest.[contest:http://codeforces.net/contest/870]

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7 years ago, # |
  Vote: I like it +29 Vote: I do not like it

Mandatory comment- "Thankyou for the short problem statements".

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7 years ago, # |
  Vote: I like it +41 Vote: I do not like it

Disclaimer: My argument may be wrong, and if it is, I don't mean to offend anyone, so apologies in advance.
The contestant who is currently second (at the time of commenting) in the leaderboard looks mighty suspicious. He has solves solves for 3 different problems in the space of 6 minutes, including solving D in 2 minutes. Check out his submissions.

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

    How is it possible to do D problem in 2 minutes :O.Is it a joke? :|

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    7 years ago, # ^ |
      Vote: I like it -10 Vote: I do not like it

    Same case in hackerrank. People have multiple accounts, view problems from one account and once they got the solutions, they submit all at once.

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    7 years ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    Looks like someone else submitted his C first and later he resubmitted. Look at the programs. There is no template in the first submission of C.

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    7 years ago, # ^ |
      Vote: I like it +13 Vote: I do not like it

    I think destinydrifter can explain better

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      7 years ago, # ^ |
        Vote: I like it +31 Vote: I do not like it

      It's extremely sad, given nothing has been done about this. The contestant was not kicked out of the contest, and even received a significantly large rating increase (+281). Please look into this MikeMirzayanov, KAN.

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7 years ago, # |
  Vote: I like it +65 Vote: I do not like it

In Div-1 B, just ask n queries of the form (i, 0) and n queries of the form (0, i).
Now, if you know p[i], you can get p[j] using p[i] ^ query(i, 0) ^ query(j, 0).
And if you know b[i], you can get b[j] using b[i] ^ query(0, i) ^ query(0, j).

Now, one by one, fix the the value of p[0] from 0 to n-1. Then you can recover the whole permutation p[].
You know that b[p[0]] = 0, so you can recover the whole inverse permutation b[] as well.

Check if both are valid. All elements of p[] are distinct and 0 <= p[i] < n && b[p[i]] = i.
It this is true, this is a valid answer. Simply count the total and print any one of these.

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    7 years ago, # ^ |
      Vote: I like it +10 Vote: I do not like it

    I only stuck for checking is it valid permutation. Simply I do not see why it means that all n2 pairs will produce correct answer as valid permutation.

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      7 years ago, # ^ |
        Vote: I like it +41 Vote: I do not like it

      If you do these 2n queries all other queries are redundant.
      Q(i, j) = Q(i, 0) ^ Q(0, 0) ^ Q(0, j)

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    7 years ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    Yeah so querying these is enough.

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    7 years ago, # ^ |
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    I did not check that p[] are distinct, but I don't think that there could be 2 the same elements.

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

    All elements of p are always distinct in all potential permutations.

    Say pi = pj. But then xor(pi, pj) = 0. Hence pi = pj in all potential permutations and there is no answer.

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Besides, you need to apply this procedure only for even n. For odd n, p0 is uniquely determined, hence only one possible permutation.

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    7 years ago, # ^ |
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    My solution:

    I first find the value of a[0]^b[i] for all 0<=i<n (by query(0,i)) let v[i]=a[0]^b[i] for all 0<=i<n Next, I fix the value a[0]=j, 0<=j<n Now, its not hard to produce b and the corresponding a array using the above definitions. Then, just check the validity of this solution.(All this is being done for a fixed value of j) What's wrong with this approach? I am getting WA at test 7

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Since you haven't done queries of the form query(i, 0), the information you have is not equivalent to all the information you would have if you made all n2 queries.

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7 years ago, # |
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Ok, so what are the hacks on B? Thanks for short statements, by the way.

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    7 years ago, # ^ |
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    3 2 1 8 1

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    7 years ago, # ^ |
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    Depends on the solution. I used the following hack a lot for the ones who used sorting:
    3 2
    1 2 1

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7 years ago, # |
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literally 1 sec from submitting D and the contest ends :/ .

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

How to solve Div-1 C?

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    7 years ago, # ^ |
    Rev. 2   Vote: I like it +76 Vote: I do not like it

    Let's say that two points are connected if they lie on the same horizontal or vertical line. Find connected components. For each of them, let L denote the number of possible different lines you can draw, and let P denote the number of points.

    For example, for a single point L = 2 and P = 1. For two points with the same y-coordinate it would be L = 3 and P = 2. For the sample test with four points it would be L = 4 and P = 4.

    It turns out that you can draw a subset of L possible lines if and only if its size doesn't exceed P. So the answer for this connected component is .

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      7 years ago, # ^ |
        Vote: I like it +11 Vote: I do not like it

      What is X?

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        7 years ago, # ^ |
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        sry, it was supposed to be P. corrected now.

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          7 years ago, # ^ |
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          Thanks! Great solution really. I was taking quite a complicated route only to arrive at the wrong formula.

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        7 years ago, # ^ |
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        I think he meant P.

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      7 years ago, # ^ |
        Vote: I like it +77 Vote: I do not like it

      You can also note that P ≥ L - 1, so the answer is either 2L or 2L - 1.

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    7 years ago, # ^ |
    Rev. 2   Vote: I like it +23 Vote: I do not like it

    Make a bipartite graph like this :

    In one of the parts put a node for every x. And in the other part put a node for every y. Now a point in the problem at coordinate (xi, yi) is an edge in the graph that is between node xi in th first part and the node yi in the other part .

    Now consider a connected component , consider that it has k nodes.

    If this component has a cycle , then you can make all 2k states for horizontal or vertical lines in this component .

    Otherwise when the component is a tree you can make all of the states minus the state that all the nodes (vertical and horizontal lines) appeared , so it has 2k - 1 states!

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    7 years ago, # ^ |
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    I see Div1 C smth likes ARC83 F: https://atcoder.jp/img/arc083/editorial.pdf.

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7 years ago, # |
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Hints for Div2 E?

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

how to do C?? i submitted in last 3 min..

if(x&1 || x<4) ans=-1;

if(x%4) ans=x/4

else ans=[(x-6)/4]+1

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    7 years ago, # ^ |
      Vote: I like it +11 Vote: I do not like it
        if (n % 4 == 0) return n / 4;
        else if (n % 4 == 1) return (n - 9) / 4 + 1;
        else if (n % 4 == 2) return (n - 6) / 4 + 1;
        else if (n % 4 == 3) return (n - 15) / 4 + 2;
    
    
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    7 years ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    "if(x&1 || x<4) ans=-1;"

    This is wrong. What if x = 9?

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    7 years ago, # ^ |
    Rev. 6   Vote: I like it 0 Vote: I do not like it

    My logic was:

    if N is Even:

           If N < 4: -1;

           Else If N >= 4: N / 4;

    else N is Odd:

           If N >= 9 and N != 11: 1 + (N — 9) / 4;

           Else : -1;

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    7 years ago, # ^ |
    Rev. 13   Vote: I like it 0 Vote: I do not like it

    every number 11 > can be expressed as sum of composite numbers

    so if n<=11: you can precalculate that that

    else if (n & 1) n-9,ans++

    if (n % 4 != 0) n-6,ans++

    ans += n/4

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    7 years ago, # ^ |
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    public static int solve(int n) {
    		int c=0;
    		if(n%2==1) {
    			n-=9;
    			c++;
    		}
    		if(n%4==2) {
    			n-=6;
    			c++;
    		}
    		if(n<0) {
    			return -1;
    		}
    		return c+n/4;
    	}
    

    31347433

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7 years ago, # |
Rev. 3   Vote: I like it 0 Vote: I do not like it

wait whoops i forgot to flush output for final answer for B and i still passed pretests even though the problem warns against this, i hope i pass systest...

EDIT @below: Nope, I used '\n' to end my line, I really hope I'm okay...

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7 years ago, # |
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Hacks for C?

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

    I don't know which ones where in pretests, so I'll write some special cases:

    5
    5
    9
    11
    13
    15

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7 years ago, # |
Rev. 4   Vote: I like it +59 Vote: I do not like it

Fun solution for Div.1 D:

The main observation is that the distance for every pair is either 0, 1, 2 or 3.

You can easily find the number of pairs with distance 1 (they have gcd ≠ 1 and this is a well known problem).

For the number of pairs with distance 2 the following condition should be true:

lp(X) * lp(Y) <  = n, where {X, Y} is the pair (lp(x) here is the least prime divisor of x).

That's true, because we can have a path like X -> (lp[x] * lp(y)) -> Y.

And now we are left with pairs with length 3. The following conditions should be true:

lp(X) * lp(Y) > n and max(lp(X), lp(Y)) * 2 <  = n, where {X, Y} is the pair.

That's true, because we can have a path like X -> (2 * lp[x]) -> (2 * lp(y)) -> Y.

Well if we have found the number of pairs of distance 2 and we count the number of pairs for which max(lp(X), lp(Y)) * 2 <  = n is true and then remove the number of pairs with distance 2 we will get the number of pairs with distance 3.

Well now the only part is how to count the number of pairs with the given two conditions. We can do a fenwick tree on the values. Here is the code.

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    7 years ago, # ^ |
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    Why is distance being further than 3 impossible?

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      7 years ago, # ^ |
        Vote: I like it +20 Vote: I do not like it

      Because for every pair {X, Y} if there is a path, then we can go like X -> (2 * lp[x]) -> (2 * lp(y)) -> Y.

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    7 years ago, # ^ |
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    Could you explain how you count these pairs with a fenwick tree a little more? Maybe it's less complex than I'm making it out to be, but I got the conditions in-contest but couldn't figure out how to count them easily.

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

      First we do n - 1 point updates of form add 1 to lp(i).

      Now we start iterating all numbers from 2 to n. Let our current position/number be i.

      • We remove 1 from the fenwick tree at position lp(i). Now the number of pairs with lp equal to X is the value at position X in the fenwick tree.

      • To count the number of positions lp(X) * lp(Y) <  = n we can simply query the range [2;n / lp(i)].

      • For the second condition (lp(X) * lp(Y) > n and max(lp(X), lp(Y)) * 2 <  = n) we will have query in range [2;n / 2] — query in range [2;n / lp(i)].

      • And we only add the second part to the answer (the pairs with distance 3) if 2 * lp(i) <  = n.

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        7 years ago, # ^ |
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        What does cnt[i] calculate in your code?

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          7 years ago, # ^ |
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          Initially the number of numbers divisible by i. Then the number of pairs of numbers with gcd equal to i.

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            7 years ago, # ^ |
              Vote: I like it +44 Vote: I do not like it

            Ok I finally got the key part I've been missing , namely the fact that if x and y are distinct, not coprime and at most n then the product of the smallest prime factor of both numbers must be at most n.

            Proof : Let g be the gcd, then if g does not exceed sqrt(n) we're done. Otherwise, let x = ga, y = gb, and wlog b > 1. Then, the product of smallest prime divisors is at most gb which is at most n.

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    7 years ago, # ^ |
    Rev. 4   Vote: I like it +10 Vote: I do not like it

    I don't think we need BIT here.

    To find the number of pairs for which lp(i) * lp(j) <= n, just store the prefix frequencies of the lp array. Then for every i, add pref[n / lp[i]]. Remove the number of i, such that lp(i) * lp(i) <= n, and divide by 2.

    Link

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    7 years ago, # ^ |
      Vote: I like it +13 Vote: I do not like it

    Thank you for this solution. I would like to add that it can be adapted to O(n) by using two pointers instead of Fenwick trees. Code : 31374591

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7 years ago, # |
Rev. 3   Vote: I like it 0 Vote: I do not like it

Does any one solved problem div2 C using DFS / BFS or Recursion or dp ?

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

    I used greedy: keep subtracting 4 from the number.
    If n % 4 == 1: Use 9 as your composite number.
    If n % 4 == 2: Use 6 as your composite number.
    If n % 4 == 3: Use 15 as your composite number.
    This is the rough idea behind the problem.

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      7 years ago, # ^ |
        Vote: I like it +2 Vote: I do not like it

      you mean use 6 and 9 as composite numbers if n%4==3

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        7 years ago, # ^ |
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        Yes, my bad. :P
        I did assign a count of 2 to 15, so my final answer is fine.

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      7 years ago, # ^ |
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      I tried solving it using recursion , got TLE @test case 7

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        7 years ago, # ^ |
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        The number can be up to 10^9. What did your recursion do, exactly?

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        7 years ago, # ^ |
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        i solved using dp. Greedy till certain point ( ~5000), dp in the end.

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7 years ago, # |
  Vote: I like it +5 Vote: I do not like it

May be rename to Mathforces?

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7 years ago, # |
Rev. 3   Vote: I like it +5 Vote: I do not like it

For some reason, for Div 2 C I thought that the queries themselves had to be prime to be valid, and I would need to print out -1 if the queried number wasn't prime.

I wrote up the entire code of getting all the primes less than sqrt(1 billion) by checking for primality with a sieve. Then I checked for each query number if it was not prime by seeing if it could be divided by any of these prime numbers < sqrt(1 billion) I found through the seive, and then I realized I had interpreted the problem wrong and had added more requirements than necessary!

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

    Anyway, how did you check primality of numbers up to 1e9 using sieve, not miller-rabin?

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      7 years ago, # ^ |
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      I found all the prime numbers less than sqrt(1e9) using a sieve. There were about 5,000 or so I think.

      I stored all these primes in a vector.

      And to check whether each queried number N was prime, I just iterated through the primes and checked if any of them could divide N.

      This would come out to about 5 * 1e8 time.

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        7 years ago, # ^ |
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        Thanks. An offline method, interesting :)

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7 years ago, # |
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waiting for system test.

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7 years ago, # |
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When will the system test start?

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7 years ago, # |
Rev. 2   Vote: I like it +434 Vote: I do not like it

Come on, khadaev. We agreed to submit on 1:30 -_-

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7 years ago, # |
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How to solve div.2 D? It's it about circles of permutation?

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    7 years ago, # ^ |
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    It's about n^2 brute-force. You can query (0..n-1, 0) and (0,0..n-1) and try every position of b0, say fix b0=i.Then reconstruct p from pi=b0^(saved query(i, 0) value). Finally, check whether it is a valid permutation and consistent with the queries.

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      7 years ago, # ^ |
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      Er...set b0 to i?Are you mean that assume b0 is i,then Query(0,0-n-1) and Query(0-n-1,0)...?

      It seems good...But how to construct all the p values or determine the value of the true b0...?

      Sorry for my late reply and my low intelligence...

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        7 years ago, # ^ |
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        Let Q0, Q1 be arrays of length N and Query(X, Y) be query value — i.e. P[X] xor B[Y].

        We can save Q0[i] = Query(i, 0) and Q1[i] = Query(0, i) for all 0<=i<N.

        Then the brute force begins.

        1. We fix b[0] = k

        2. We can reconstruct P by setting P[i] = Q0[i] xor k.

        3. Now we check whether P is a valid permutation.

        4. If it is valid, then we check P against our saved queries, namely Q1.

        5 Repeat from 1 to 4 for all 0<=k<N.

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7 years ago, # |
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I think div.2E is a problem of combination number,but I can't get the solution.

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Div.2D: I wrote all my queries to stdout and flushed, then read the all of the answers one by one, and my submission got an ILE? Somebody wrote one query to stdout and then read the answer at once, then he passed the pretest? Is it my fault? My Submissions:

cstdio version

iostream version

Practice submission(query and then read at once):

Practice version

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7 years ago, # |
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del

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    7 years ago, # ^ |
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    На прошлом раунде взяли 100 человек, так что еще неизвестно )

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I submitted solution to B after hack to my solution which got accepted. To recheck it i submitted my previous solution again and it showed the wrong answer. Now my solution (correct one) is not getting evaluated. Will they not consider my correct solution. :( EDIT : They evaluated it :)

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NOOOOOOOOO!!!!!!!In Div2 D During the contest I read the answer of 2*n times together and I got an ILE…… After contest I tried read every answer after questioning I got AC …… Oh my goodness……

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    7 years ago, # ^ |
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    Same thing happened to me, which wasted me 30 minutes and 7 incorrect submissions = almost 500 points.

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How to solve Div.1 E?

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    7 years ago, # ^ |
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    I think the main problem is constructing compressed tree for that k vertices. I remember it is to solve this: https://open.kattis.com/problems/optimistan.

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    7 years ago, # ^ |
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    Pick a pair of special vertexes U,V that are as far from each other as possible. Construct a chain that connects them (these are the points with d(U,W)+d(W,V)=d(U,V) ordered by distance from U). Consider subtrees rooted at each vertex from this chain. We can easily split all vertexes into these subtrees. The distances to each vertex X from X's root Y equals to d(U,X)-d(U,Y). So we may treat roots as special vertexes. Most of subtrees will have only one special vertex — the root. They may be processed in O(N) in total. For the other subtrees we recursively call the same procedure.

    This process seems to terminate quickly. On every step each subtree will lose 2 special vertex, but it may possibly generate some new. The more subtrees there are, the faster the special vertexes are lost. It seems all vertexes will be processed in O(K) iterations and problem is solved in O(KN).

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Good, lucky problems, good contest as well as author!

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7 years ago, # |
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My screencast

Hope you find it interesting!

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7 years ago, # |
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For problem B

for k==2 how is it max(a[0],a[n-1])

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    7 years ago, # ^ |
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    1. Seeing, that you can achieve it, is pretty easy. You can take that one element as one segment, and the rest as the other.

    2. Let's assume we achieve an answer x such that x > max(a0, an - 1). If this element is in the first segment, then the first segment's value should be x, but since x > a0 the score will be at most a0. Same goes for the second segment, so x can be in neither segment, which is contradiction.

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    7 years ago, # ^ |
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    Splitting of array a of n elements into 2 subarray,it must be a prefix and a suffix. for the minimum to be larger, we make the 2 subsegments [0,n-2] and [n-1, n-1] or [0, 0] and [1,n-1]

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      7 years ago, # ^ |
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      subsegment can be:

      [0,n-3][n-2,n-1] like that...

      why these two??

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        7 years ago, # ^ |
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        2 subsegment : [0, x], [x+1, n-1] (0 <= x < n-1)

        the answer is max(mininum in [0, x], mininum in [x+1, n-1])

        so :

        mininum in [0, 0](it is a[0]) is larger than mininum in [0, x]

        mininum in [n-1, n-1](it is a[n-1]) is larger than mininum in [x+1, n-1]

        when a[0]>a[n-1] a[0] is larger than mininum in [0, x] and mininum in [x+1, n-1](for each x : 0 <= x < n-1)

        when a[0]<a[n-1] a[n-1] is larger than mininum in [0, x] and mininum in [x+1, n-1](for each x : 0 <= x < n-1)

        so for the minimum to be larger, we make the 2 subsegments [0,n-2] and [n-1, n-1] or [0, 0] and [1,n-1]

        sorry, my English is not very good, but you can try to understand it. :)

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Oh..There are so many hacks today! You see There are about one thousand and seven hundred people solved ABC during the contest. So do I. But my solution for B got a FST... Because I defined the mininum by -1e9+7 instead of -(1e9+7).. But I still got rank 180..Because I hacked eleven people :P

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Will there be an editorial?

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http://codeforces.net/contest/872/submission/31353320

check this solution for Div2D.

When something likes "5000 xor 4191" happens, will it get Runtime Error ? I think it will

but why it didn't happen ?

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    7 years ago, # ^ |
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    I think you don't need to do this. I'm just lucky. :)

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Is the editorial published ?

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How you guys came up with Div2-C . I mean is it writing bruteforce than generalizing it Or there is some other way to approach for solution?

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    7 years ago, # ^ |
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    I will try to explain a possible approach.

    First, if you want to represent n as a sum with lots of summands, those summands have to be small. Therefore you should ask yourself, what is the smallest suitable summand? It is 4. Can you use only 4? Only if n % 4 == 0. What about the other remainders modulo 4? You are obliged to use something else apart from 4.

    The next composite number is 6. You can see that if n % 4 == 2, using one 6 and all 4 is optimal. But you cannot obtain any odd n using only 4 and 6.

    Hence you need to use one more number. Maybe 8? No 8 is not useful as 4+4=8. Maybe 9? Yes! Using one 9 and all 4 you can obtain all n % 4 == 1.

    The only remaining case is n % 4 == 3. Can you get it using only 4, 6, 9? The answer is yes as 6 + 9 % 4 == 3. The issue is that this is no more trivially optimal and you have to check the optimality. However, checking it is super-easy, as the smallest useful number (i.e. it is composite and it cannot be generated as a sum of 4, 6, 9) is bigger than 6+9 = 15 and therefore using 6+9 is optimal if n % 4 == 3.

    p.s. I have ignored all small cases. As a byproduct of the explanations I have given ,those are all impossible (1, 2, 3, 5, 7, 11).

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And how many participants went to the final?

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Does anybody know why I receive "Denial of judgement" for sources 31428657 and 31428722 ? The error on test 88 is : Can't compile file: Compilation process returned error: execution failed because of unknown invocation error