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Educational Codeforces Round 50 Editorial
By pikmike, history, 2 years ago, translation, In English

1036A - Function Height

Tutorial
Solution (Vovuh)

1036B - Diagonal Walking v.2

Tutorial
Solution (Vovuh)

1036C - Classy Numbers

Tutorial
Solution with combinatorics (PikMike)
Solution with precalc (PikMike)

1036D - Vasya and Arrays

Tutorial
Solution (Ajosteen)

1036E - Covered Points

Tutorial
Solution (PikMike)

1036F - Relatively Prime Powers

Tutorial
Solution (PikMike)

1036G - Sources and Sinks

Tutorial
Solution (BledDest)
Tutorial of SRound #001
 
 
 
 
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2 years ago, # |
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I found a strange thing in F. Two ACs codes output different result. Detail in https://codeforces.com/blog/entry/61728. Can anyone help me?

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2 years ago, # |
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Can someone prove the author's solution for B?

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    2 years ago, # ^ |
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    i thought about this problem lot after the contest.

    what i came up with is that there is 8 cases for m,n,k: odd/odd/odd, odd/odd/even....even/even/even. all cases reduce down to the odd/odd/odd case or the even/even/even case. we can see that it is possible to solve the odd/odd/odd or even/even/even case using all diagonal moves as long as m,n is reachable in k moves.

    for an example of "reduction", the even/odd/odd case reduces to the even/even/even case because you can "spend" one non-diagonal move in the n-direction. [1]

    the author's solution is basically "reducing" down all cases to o/o/o or e/e/e cases with mod. see my coded sol here: 42690270

    [1] in fact, the o/o/o case reduces down to the e/e/e case by spending 1 diagonal move

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    2 years ago, # ^ |
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    We can change the problem to this: you should use 1/-1/0 k times totally to get m and n respectively.(remember that m and n are nothing to do with) And according to the problem and greedy algorithm the num of 1/-1 should be as much as possible.

    for m we should at least use 1 m times and if k<m then it is impossible to get to (m,n) if k==m then for m we are done if k>m then we can add the combination of (1 and -1), then the left (m-k)%2(1 or none) should be 0. for n we do the same.

    so now for m and n, the num of 0 should be at least (m-k)%2 and (n-k)%2 respectively. And for the 0 in m, we must use a 1/-1 in n to construct a valid move (0,1) or (0,-1), the same for n. at last there are (m-k)%2+(n-k)%2 moves not diagonal, so the answer is k-(m-k)%2-(n-k)%2.

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2 years ago, # |
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Randomized solution for problem G: here

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    2 years ago, # ^ |
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    cool.

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    2 years ago, # ^ |
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    I will try to explain why this solution works: obviously (unless I missed some bug of course, but from conceptual standpoint the solution seems pretty sound), the solution in question can make only one type of mistake: answer "YES" when the actual answer is "NO".

    When the answer is "NO"? When there is at least one "bad" permutation of numbers from 1 to k. I will show that in this cases there are actually a lot more bad permutations than just one. Indeed, consider the set X from the editorial.

    To explain the main idea lets consider the simplest case when |X| = |f(X)|. Then any permutation that matches elements of X only with elements of f(X) and matches elements {1, 2, ..., k} - X with elements {1, 2, ..., k} - f(X) is a bad permutation: when doing a graph walk starting with some source in X, we will never leave reach any source not from X. Therefore there is at least |X|!(k - |X|)! ≥ [k / 2]![(k + 1) / 2]! bad permutations.

    To deal with the case |f(X)| < |X|, we will take some subset Y of X with size |f(X)|, then and we can similarly show that all permutations that match sources from Y to sinks in f(X) and sources not from Y to sinks not from f(X) are bad permutations. Therefore there is at least |Y|!(k - |Y|)! ≥ [k / 2]![(k + 1) / 2]! bad permutations.

    All in all, if there is one bad permutation, there are at least [k / 2]![(k + 1) / 2]! of them. So the probability of finding a bad permutation randomly is at least in the worst case of k = 20. From the other hand, on my computer, the code in question checks random permutations before timing out. In the end, the probability of failure is something of the order  ≈ 10 - 8, which is good enough.

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    2 years ago, # ^ |
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    cool

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2 years ago, # |
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Wasted 1 hour trying to solve a harder version of Problem C. I thought classy numbers were those number which had the number of distinct non zero digits <= 3. And when TC 2 and 3 were not passing, I tried so hard debugging.

After successfully wasting 1 hour, read the question again and BAM!!

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    2 years ago, # ^ |
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    what it mean prefix in the tutorial for C??

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2 years ago, # |
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The data for G maybe too weak? I can easily hack my code.

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2 years ago, # |
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The problem A is so easy, but I solved it late because of my poor English and my poor wi-fi.

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2 years ago, # |
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In problem E, Can anyone please give proof of why exactly GCD + 1 value will give integer co-ordinates?

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2 years ago, # |
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[Deleted]

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2 years ago, # |
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http://codeforces.com/contest/1036/submission/42706346 can someone pls point out the mistake i am making. upd :: got it

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2 years ago, # |
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In Problem G, it is possible to find for every sources all achievable sinks in a linear time. For this it is necessary to store in each visited vertex an integer (the set of bits of achievable sinks).

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    2 years ago, # ^ |
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    Can you explain the approach behind recursion in problem C?

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2 years ago, # |
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Can problem F be solved with some precalculations (generations of correct numbers), simular to problem C?

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2 years ago, # |
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can anyone explain classy numbers editorial ??

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2 years ago, # |
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there is a easy way to solve problem C with DP :) 42742195

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    2 years ago, # ^ |
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    It seems to be an interesting approach, but I don't really understand it. Could you please explain it briefly?

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      2 years ago, # ^ |
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      f(i,j) means The i th bit of a number(unit's digit,ten's digit...) and already has j non-zero digits.

      DP uses flag to control upper bound.If it is not the upper bound,we can remember the result of f(x,y).(That's why I initialize f to -1 and update f(x,y) when flag=0).f(x,y) will be accessed multiple times, so we can speed it up by remember it .

      My English is poor( I use the google translate QAQ ).

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    2 years ago, # ^ |
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    Thanks.I can't understand about how (!flag&&~f[x][y]) works,what's mean ?it's mean if f[x][y] has been calculated ,then return f[x][y]? to be honest i want to use pinyin.

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2 years ago, # |
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Why dont we use -C instead of C? I read in internet that the formula will use -C instead of C? I mean this fragment: long long dx = det(l1.C, l1.B, l2.C, l2.B); long long dy = det(l1.A, l1.C, l2.A, l2.C);

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    2 years ago, # ^ |
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    By the properties of determinant det(-a, b, -c, d) = det(a, -b, c, -d) = -det(a, b, c, d)

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      2 years ago, # ^ |
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      Ahh it is used then when he does: x = -dx / d; y = -dy / d; right?

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2 years ago, # |
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Why is Problem A tagged with FFT ? Is there a solution with FFT as well ?

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2 years ago, # |
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very very good~ !!

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2 years ago, # |
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E has a tag "fft". Can it be done using FFT too? If yes, then how?

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2 years ago, # |
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For problem 1036E — Covered Points , can anybody explain what's special with the test 14? I've seen some people failing this test.

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2 years ago, # |
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Could someone explain how mobius function could be used in problem F. All I understood is that we are trying to remove the bad elements from the answer. Basically, could someone explain me the formula?

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2 years ago, # |
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in problem G, how about a 4-vertex graph: 1->2, 3->2, 3->4 ??? It's obvious that 1 is a source but the set of sink it can reach is just {2}???

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2 years ago, # |
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is there a proof for problem b ?

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2 years ago, # |
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Can someone explain the brute force generation for C

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    21 month(s) ago, # ^ |
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    pos represents the number of digits,cnt represents count of non-zero digits and cur represents the value generated. first we take the generated number and multiply by 10 and increasing number of digits. if a number is classy number then its multiple of 10 will also be a classy number. then we check if the number has less than 3 non-zero digits or not.if it has less than 3 non-zero digits then we can put any number between 1 to 9 at units place and generate another classy number(multiply by 10 and adding the number). we also increase the count of non-zero digit.

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      5 months ago, # ^ |
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      I was really confused. thanks for the explanation.

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19 months ago, # |
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please can anyone explain the formula in problem C, i dont know why it return exactly the beautiful numbers on [1,X), why cur-- like that, that means for(int cur = 3; cur>=0; cur --) , i think i'll insert s[i] in a set so that its size is cur ... A lot of things i cant understand. I've read PikMike's code and the formula for hours. Help me please.

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10 months ago, # |
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Proof why GCD works in E

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10 months ago, # |
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The solution to G reminds me of Hall's marriage theorem, and I finally figured out how to solve it in polynomial time.

Take the graph of sources and reachable sinks. If no perfect matching exists, by Hall's marriage theorem, there is a nonempty proper subset of sources $$$X$$$ such that $$$|X|>|f(X)|$$$, so the answer is NO. Otherwise, we need to ensure that $$$|X|<|f(X)|$$$. This means $$$f(X)$$$ has sinks besides the ones matched to sources in $$$X$$$. Connect each sink to the source it is matched to in the perfect matching. The answer is YES if and only if the the resulting graph is strongly connected.

Code: 67053099

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9 months ago, # |
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Can anyone explain F with little more details ?

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