Vladosiya's blog

By Vladosiya, history, 2 months ago, translation, In English

1931A - Recovering a Small String

Idea: myav Prepared by: myav

Tutorial
Solution

1931B - Make Equal

Idea: MikeMirzayanov Prepared by: Vladosiya

Tutorial
Solution

1931C - Make Equal Again

Idea: senjougaharin Prepared by: senjougaharin

Tutorial
Solution

1931D - Divisible Pairs

Idea: MikeMirzayanov Prepared by: Vladosiya

Tutorial
Solution

1931E - Anna and the Valentine's Day Gift

Idea: Gornak40 Prepared by: Gornak40

Tutorial
Solution

1931F - Chat Screenshots

Idea: senjougaharin Prepared by: senjougaharin

Tutorial
Solution

1931G - One-Dimensional Puzzle

Idea: senjougaharin Prepared by: senjougaharin

Tutorial
Solution
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2 months ago, # |
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Can anyone explain why for Problem. F this submission gets TLE.

246345877

and this submission gets accepted?

246347175

I don't feel like there's any difference in logic, just the way of implementation is different.

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    2 months ago, # ^ |
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    In the second submission there is an optimization of the DFS function: do not run a search if the vertex has already been visited

    My submissions:

    246247300

    246250491

    You can see the difference more clearly here

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      2 months ago, # ^ |
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      are you talking about the external dfs optimization from main function or internal dfs optimization?

      I mean the lines

      le(i, 1, n){
      			if(!vis[i]){
      				if(dfs(i)){
      					cout << "NO" << endl;
      					flg = 1;
      					break;
      				}
      			}
      		}
      

      or

      	for(auto c: graph[v]){
      		if(dfs(c)){
      			return true;
      		}
      	}
      
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        2 months ago, # ^ |
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        le(i, 1, n){ if(!vis[i]){ if(dfs(i)){ cout << "NO" << endl; flg = 1; break; } } } I'm talking about this

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    2 months ago, # ^ |
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    In the TLE submission your first loop (to clear graph / vis / dfsStack) will run 10000(t) * 200005(NxM) operations. The accepted one only clears the memory needed for each test case.

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      2 months ago, # ^ |
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      so I don't need to initialize it with size NxM? and resize it to n+1 in every test case while clearing it in the end of each test case?

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    2 months ago, # ^ |
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    The solution in which you are getting TLE. Coz:

    You are unnecessary initializing array of maxN again and again, even when the Value of N be small.

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    2 months ago, # ^ |
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    In the second submission the graph is a array of sets, so when you insert a repeated edge it doesn't get duplicated. In the first submission you are duplicating a lot of edges (because you are pushing them in a vector), and going through all of them during the dfs.

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      2 months ago, # ^ |
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      I will try it a little while. Hope it works.

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      2 months ago, # ^ |
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      I'm wrong, this still would not cause TLE (the total amount of edges over all tests is less than 2*10⁵), robostac gave the right answer.

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2 months ago, # |
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A different way to do F:

  • Each element (after the 0th element in the kth list) can only be it's current position or its current position+1

  • Store both possibilities in a vector

  • Remove possibilities for elements when they're no longer possible (i.e. the current element's position in the kth list conflicts with the elements position in a previous list)

  • The answer is no if an element has no possibilities, or if its only possibility conflicts with another elements only possibility

  • Otherwise the answer is yes

Code: https://codeforces.com/contest/1931/submission/246250979

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2 months ago, # |
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How was my solution hacked? https://codeforces.com/contest/1931/submission/246212609

I see it was probably due to collisions in the map but does that mean I need a better hashing function for pairs? If so can someone suggest one?

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2 months ago, # |
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can somebody help me understand why this solution got TLE Problem D

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    2 months ago, # ^ |
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    probably, for multiset's count function, if I ain't mistaken, it works in O(logn) time

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    2 months ago, # ^ |
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    I think it is because of s.count( {cx,cy} ), that is not O(1) but O(count_of({cx,cy}) + log(N)), so worst case in an array with all ones and x=2, you get O(N^2) time complexity. I think in C++ you should be going for a unordered_map<pair<int,int>, int> in which the values are the count so you can do the addition in O(1).

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2 months ago, # |
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Good contest but again I will most likely stay with newbie.

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2 months ago, # |
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Hey, I have submitted a java code for problem C but it gave me error on test case 3 whereas when i submitted the exact same code in c++ it passed all test cases why

java link: https://codeforces.com/contest/1931/submission/246230264

c++ link: https://codeforces.com/contest/1931/submission/246276255

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2 months ago, # |
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In G we don't need to calculate factorials to $$$4 \times 10^6$$$ since the biggest factorial we need in combinations is actually $$$2 * max(c_{i}) \leq 2 \times 10^6$$$

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2 months ago, # |
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i think something wrong in solution code of problem A ._.

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2 months ago, # |
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Thanks for good Round. Good F

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2 months ago, # |
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Why D's solution is correct? It uses a hash table (dict), so is it possible to hack this hash table?

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2 months ago, # |
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A lot of dict/unordered hacks this round. Although I still see quite a lot of solutions not hacked that use the same approach for problem D.

Nice round. Had no idea toposort is available in standard Python library.

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2 months ago, # |
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could any explain why in F , having cycle in graph means there is no logical order of numbers ?

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    2 months ago, # ^ |
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    consider the following case

    1 2 3 4 2 3 1 4 3 4 3 2

    the graph would be

    2 -> 3 -> 4 and 3 -> 1 -> 4 and 4 ->3 ->2 if you construct graph for these nodes then there's a cycle. now answer to why does it make sense: we have to follow the order no matter what. here 2 comes before 3 and 4, similarly 3 comes before 1 and 4 but the 3rd one, which creates the cycle, has that 4 comes before 3. it is possible to have 3 before 4 and 4 before 3 in the line simultaneously. so that's why having cycle means no ordering for the data.

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    2 months ago, # ^ |
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    consider these two statements,

    1 comes before 2

    2 comes before 1

    they can't be true at the same time,

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2 months ago, # |
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Was I the only one who had an unusually hard time in this round? T_T

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2 months ago, # |
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Nice problems , had fun solving them

I really hope more rounds like this exist

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2 months ago, # |
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why the code of problem A uses the string cur outside loop so I think this isn't true; please anyone can review it?

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    2 months ago, # ^ |
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    It's correct because the first triples that it finds is guaranteed to be the minimum, so everything after that affecting the cur variable doesn't matter

    (That also means it doesn't need to continue the loop to find the minimum here and can return immediately when it finds the first triples)

    But I agree that appending to the cur variable after that is definitely unnecessary and the logic is also wrong IF the minimum is not the first triples lol)

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2 months ago, # |
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I still don't understand anything in G, can someone please explain me

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    2 months ago, # ^ |
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    This was my thought process when upsolving G.

    The main observation needed is to notice that 1 + 3 = 1 and 2 + 4 = 2, so we can condense all 3 and 4 pieces. However, it is possible to start the sequence with a 3 or 4 piece, so for this reason we can imagine having an "extra" 1 or 2, respectively. Now there are a few cases:

    • No correct puzzle sequence can have consecutive 1's or 2's, so if $$$ |c_1 - c_2| > 1 $$$, the answer is 0.

    • If $$$ c_1 = c_2 = 0 $$$, the answer is 1 if $$$ c_3 = 0 $$$ and/or $$$ c_4 = 0 $$$ and 0 otherwise. This is true because 3 cannot mesh with 4.

    • If $$$ c_1 = c_2 $$$ and $$$ c_1, c_2 > 0 $$$, either 1 or 2 can be used to start the puzzle sequence, so we can fix the first number and calculate the rest of the sequence. Counting the # of ways to condense $$$ c_3 $$$ 3 pieces into $$$ c_1 + 1 $$$ 1 pieces is the same as counting the # of ways to put $$$ c_3 $$$ indistinguishable balls into $$$ c_1 + 1 $$$ boxes, which is $$$ \binom{c_1 + c_3}{c_1} $$$ (this can be visualized with the stars and bars technique). Note that 1 is added to $$$ c_1 $$$ for that "extra" piece, if we fix 1 to be the first puzzle. As such, the answer is $$$ = \binom{c_1 + c_3 - 1}{c_1 - 1} \cdot \binom{c_2 + c_4}{c_2} + \binom{c_1 + c_3}{c_1} \cdot \binom{c_2 + c_4 - 1}{c_2 - 1} $$$

    • If $$$ c_1 = c_2 + 1 $$$ or $$$ c_1 + 1 = c_2 $$$, then we are required to fix either 1 or 2 to be the first piece, respectively. Then it's the same exact idea as the previous case

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      2 months ago, # ^ |
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      your explanation is better than editorial, thanks...

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      2 months ago, # ^ |
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      Could you clarify why we need to add one? I'm not entirely clear on what you mean by an "extra piece." Could you elaborate further on that point?

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        2 months ago, # ^ |
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        We assume that every 3 and 4 piece must be in a sequence of 1333... or 24444.... However it is also possible to place a sequence of 3's or 4's at the beginning of the sequence, without a corresponding 1 or 2 behind it. So adding an invisible 1/2 lets us strictly associate every 3/4 with a corresponding 1/2, which makes calculation easier.

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          2 months ago, # ^ |
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          Thank you! Your explanation was really helpful.

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2 months ago, # |
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Anyone can check my wrong in this code for problem E? It's call wrong in test 3, but i think i have same ideal with propers.. 246215307

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2 months ago, # |
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why my solution TLE on system test.
is unordered_map of unordered_map took significantly more time to find than unordered_map that pair as key?

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2 months ago, # |
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Can some one explain the approach for problem G in detail?

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2 months ago, # |
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Problem G:

Let the four types be denoted by $$$\color{purple}{b_1}$$$, $$$\color{blue}{b_2}$$$, $$$\color{olive}{b_3}$$$ and $$$\color{teal}{b_4}$$$ respectively.

Lemma 1: Between any two $$$\color{purple}{b_1}$$$, there must be at least one $$$\color{blue}{b_2}$$$. It is easy to see that one cannot fit only $$$\color{olive}{b_3}$$$ and/or $$$\color{teal}{b_4}$$$ between two $$$\color{purple}{b_1}$$$.

Lemma 2: Between any two $$$\color{blue}{b_2}$$$, there must be at least one $$$\color{purple}{b_1}$$$. It is easy to see that one cannot fit only $$$\color{olive}{b_3}$$$ and/or $$$\color{teal}{b_4}$$$ between two $$$\color{blue}{b_2}$$$.

It follows that between any two $$$\color{purple}{b_1}$$$, there must be exactly one $$$\color{blue}{b_2}$$$. Likewise, between any two $$$\color{blue}{b_2}$$$, there must be exactly one $$$\color{purple}{b_1}$$$. In other words, they must follow an alternating pattern $$$\color{purple}{b_1}, \color{blue}{b_2}, \color{purple}{b_1}, \color{blue}{b_2}, \ldots$$$ or $$$\color{blue}{b_2}, \color{purple}{b_1}, \color{blue}{b_2}, \color{purple}{b_1}, \ldots$$$, with possibly some $$$\color{olive}{b_3}$$$ and $$$\color{teal}{b_4}$$$ in between.

Lemma 3: If a block exists to the left of a $$$\color{olive}{b_3}$$$, it must either be another $$$\color{olive}{b_3}$$$ or a $$$\color{purple}{b_1}$$$. If a block exists to the right of a $$$\color{olive}{b_3}$$$, it must either be another $$$\color{olive}{b_3}$$$ or a $$$\color{blue}{b_2}$$$.

Lemma 4: If a block exists to the left of a $$$\color{teal}{b_4}$$$, it must either be another $$$\color{teal}{b_4}$$$ or a $$$\color{blue}{b_2}$$$. If a block exists to the right of a $$$\color{teal}{b_4}$$$, it must either be another $$$\color{teal}{b_4}$$$ or a $$$\color{purple}{b_1}$$$.

Putting everything together, any valid arrangement will be of the form

  • $$$\{\color{teal}{\{b_4, b_4, \ldots\}}, \color{purple}{b_1}, \color{olive}{\{b_3, b_3, \ldots\}}, \color{blue}{b_2}, \color{teal}{\{b_4, b_4, \ldots\}}, \color{purple}{b_1}, \color{olive}{\{b_3, b_3, \ldots\}}, \color{blue}{b_2}, \ldots\}$$$ or
  • $$$\{\color{olive}{\{b_3, b_3, \ldots\}}, \color{blue}{b_2}, \color{teal}{\{b_4, b_4, \ldots\}}, \color{purple}{b_1}, \color{olive}{\{b_3, b_3, \ldots\}}, \color{blue}{b_2}, \color{teal}{\{b_4, b_4, \ldots\}}, \color{purple}{b_1}, \ldots\}$$$

For such an arrangement to exist, it must hold that $$$\mathrm{abs}(c_1 - c_2) \leq 1$$$. Then, the number of valid arrangements is the number of ways to arrange $$$\color{olive}{b_3}$$$ and $$$\color{teal}{b_4}$$$ into the 'slots' between alternating $$$\color{purple}{b_1}$$$ and $$$\color{blue}{b_2}$$$. This can be done using the stars and bars technique.

Submission: 246362383

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2 months ago, # |
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Another idea for Problem F (indeed, what first came into my mind is not the solution in tutorial):

The first screenshot show the main order of users except the one (say, i) to which the screenshot belong. We can use the remaining screenshots to narrow the range of the i's place: if there's only one possible place, good; if there are multiple possible places, just choose one; if there's contradiction, then answer is "NO" directly. Adding i's place to the main order gives the whole order, compare it to remaining screenshots gives final answer (if contradictions exist, then "NO", otherwise, "YES"). Time complexity is OK.

https://codeforces.com/contest/1931/submission/246229823

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2 months ago, # |
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what are the prequesites for G . I didn't understand anything, and why am i still green and i hit the 1400 mark

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2 months ago, # |
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For the problem F, I do understand why existence of topological sort (and therefore absence of cycles) of such graph is necessary condition for existence of the order. However, why is it sufficient? I.e. What is exact proof that absence of cycles guarantee that such order of chat participants exists?

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    2 months ago, # ^ |
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    You have a set of statements where $$$a_i < a_{i+1}$$$ that you can get by ignoring $$$a_0$$$.

    That's an ordering. You cannot have $$$a_1 < a_2$$$, $$$a_2 < a_3$$$ and then another $$$a_3 < a_1$$$ due to transitivity of the ordering on integers. If $$$a_1 < a_2$$$ and $$$a_2 < a_3$$$ then $$$a_1 < a_3$$$. That's what cycle check is for. It finds the case where transitivity is broken if that case exists.

    If you're asking about missing statements to have an ordering where every $$$a_i$$$ is at its true position, then you can easily see that if transitivity holds, you can invent your own statements where it can still hold and create a full order.

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    2 months ago, # ^ |
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    If there are no cycles in the directed graph that means that you can run the topological sort algorithm on it.

    The correct sequence will be the sequence after we topologically sort it as it will be satisfying all the conditions ie all the edges in the graph are not forming a cycle.

    Therefore it is enough to no if we can topologically sort the graph or no ie check for any directed cycles.

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2 months ago, # |
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Can anyone explain for me what wrong in my details explanation of problem D: D_practice_Modular
I don't understand why a[i] mod x = (x - a[j] mod x) mod x instead of a[i] mod x = x - a[j] mod x

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    2 months ago, # ^ |
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    The value of a[i] mod x should always be in range [0, x - 1] (should be less than x). In the equation a[i] mod x = x - a[j] mod x this property will not always be true as if a[j] mod x becomes zero then a[i] mod x = x which is contradicting the property. For example

    a[j] = 10 , x = 5

    a[i] mod 5 = 5 - 10 mod 5

    a[i] mod 5 = 5

    thus to prevent this and keep the value of a[i] mod x in range we use the equation a[i] mod x = (x - a[j] mod x) mod x

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      2 months ago, # ^ |
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      I don't quite understand, one more than is, in range [0, 2x) we have two multiple of x are 0 and x. Why we only choose the case x in the expression instead of both cases? When a[i] mod x + a[j] mod x = 0 ==> a[i] mod x = a[j] mod x = 0

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        2 months ago, # ^ |
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        That case a[i] mod x = a[j] mod x = 0 actually gets considered on its own when we use a[i] mod x = (x - a[j] mod x) mod x.

        Let's say a[i] mod x = a[j] mod x = 0 thus a[i] mod x = (x - a[j] mod x) mod x LHS becomes 0 and RHS becomes (x - 0) mod x which is also 0. Hence LHS = RHS. So both cases are being considered in this single equation.

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2 months ago, # |
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Can someone please figure out what the mistake with my submission is? Any help is appreciated. https://codeforces.com/contest/1931/submission/246419122

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    7 weeks ago, # ^ |
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    	    sort(trailnums.begin(), trailnums.end(), [&](pair<long long int, long long int> a, pair<long long int, long long int> b) {
    	        return to_string(a.first).length() < to_string(b.first).length();
    	    });
    

    I think this part is the issue. You are sorting the numbers with trailing zeros by their absolute lengths in increasing order (instead of their numbers of trailing zeros in decreasing order). Seems like test #2 was only of integers from 1 to 10, thus this solution will be automatically be correct for just that case (but not when the integers are higher).

    For example: 4000000 should be prioritized more than 711100 since it has 6 trailing zeros (compared to 2 of 711100), but your sorting will make 711100 stand first.

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      6 weeks ago, # ^ |
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      Hi AkiLotus, I also tried that approach before using the number of trailing zeros in decreasing order:

      Submission:https://codeforces.com/contest/1931/submission/246418992

      But even that does not seem to work.

      Appreciate your effort in looking into my code.

      Thanks

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        6 weeks ago, # ^ |
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        I see now. The remtrail function isn't correct either, specifically this part.

            string ans = "";
            for(int i=0;i<n;i++) {
                if(s[i] == '0') break;
                else {
                    ans += s[i];
                }
            }
        

        Take a guess of what the output of remtrail(170011000) would be with this code. Correct answer is supposed to be 170011.

        Spoiler (open only when you figured out the bug)
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          6 weeks ago, # ^ |
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          Ahhh. I see!!! understood the mistake. Thanks a lot for the help

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2 months ago, # |
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From tutorial D: Since (ai + aj) mod x = 0, it follows that (ai mod x + aj mod x) mod x = 0. Since (ai − aj) mod y = 0, it follows that ai mod y − aj mod y = 0.

Quite not clear, how it follows, and why it's not (ai mod y − aj mod y) mod y = 0 for 2nd case then.

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    2 months ago, # ^ |
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    $$$

    0\ \le\ ai\ mod\ y\ \le\ y-1 \newline

    0\ \le\ aj\ mod\ y\ \le\ y-1 \newline

    -(y-1)\ \le\ (ai\ mod\ y − aj\ mod\ y)\ \le y-1 $$$

    So if $$$\ (ai\ mod\ y − aj\ mod\ y)\ mod\ y = 0$$$

    Then $$$\ (ai\ mod\ y − aj\ mod\ y) = 0$$$

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2 months ago, # |
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Solution for D without hash:

We need to count number of pairs equal with respect to $$$(mod\ y)$$$ and sum of which equal to $$$0$$$ with respect to $$$(mod\ x)$$$.

For each $$$a[i]$$$ store pair $$$ \{ a[i] (mod\ y), a[i] (mod\ x) \} $$$. Sort array of pairs.

Now pairs can be divided into continuous blocks of numbers equal with respect to $$$(mod\ y)$$$. And inside each block elements are sorted by their $$$(mod\ x)$$$.

So inside each block we can count number of pairs which add up to $$$0$$$ with respect to $$$(mod\ x)$$$ using 2 pointers in $$$O(n)$$$.

Complexity of solution $$$O(n * log(n))$$$. (Because of sorting)

My submission 246421822

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2 months ago, # |
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Could someone tell me why my code for Question D always returns compilation error? My submission: 246426188

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2 months ago, # |
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Could Someone explain why the following code fails in problem F. My logic for the problem is completely different compared to what the tutorial talks about, but i feel it should work.

I have first constructed a pattern using the first screenshot, using every element except the first, in the first screenshot.

Then I check, if all the other screenshots after this first screenshot follow that pattern or not.

https://codeforces.com/contest/1931/submission/246430531

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2 months ago, # |
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.

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2 months ago, # |
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In Problem D,when i use unordered_map,i got TLE on test 30.But if use map,this test only need 70ms.Is this intentional design by the author?

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    2 months ago, # ^ |
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    The unordered map uses a certain hashing algorithm and the author has put in testcases that cause collisions to it.

    You can read this for more information.

    So answering your question, yes it is intentional design by the author as the testcases are set in such a way.

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why tle? D link

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    2 months ago, # ^ |
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    int arr [][]= new int [ x ][ y ];

    Constraints: $$$1 \le x, y \le 10^9$$$

    You can't just create such a big matrix.

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in G, why do we need this condition? It is guaranteed that the sum of ci for all test cases does not exceed 4⋅1000000.

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https://codeforces.com/contest/1931/submission/246604571

Please help, I am getting TLE for problem D test 5. My code works by iterating k to find (a[i]+yk) % x == 0, a[i]+yk<=a[n-1]. Is it possible to make this solution faster?

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2 months ago, # |
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What does 'tout' mean in problem F's answer? Can't understand

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2 months ago, # |
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Getting RUNTIME_ERROR (Exit code is -1073741819) on large inputs for this solution https://codeforces.com/contest/1931/submission/246676323 for Problem E. Does anyone know what am I doing wrong here?

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    2 months ago, # ^ |
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    I tested your code and I think error occured when you sort vector b.

    This is an AC submission based on your code. 247704863

    I don't know what wrong on your code too (maybe Segment fault?).You can make some further study.

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2 months ago, # |
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My solution is giving the correct result on my external compiler but the codeforces judge is getting wrong answer. I am frustated looking for the problem. Can someone plz show me the problem with my code?

https://codeforces.com/contest/1931/submission/246796190

246796190

import java.util.*;

public class Main { public static void main(String[] args) { Scanner sc=new Scanner(System.in); int t=sc.nextInt(); while(t-->0){ int n=sc.nextInt(); int x=sc.nextInt(); int y=sc.nextInt(); int[] arr=new int[n]; for(int i=0;i<n;i++) arr[i]=sc.nextInt(); int count=0; HashMap<Integer, HashMap<Integer, Integer>> hm=new HashMap<>(); for(int j=0;j<n;j++){ int a=(x-arr[j]%x)%x; if(hm.containsKey(a)){ HashMap<Integer, Integer> hy=hm.get(a); if(hy.containsKey(arr[j]%y)){ count+=hy.get(arr[j]%y); } } if(!hm.containsKey(arr[j]%x)){ hm.put(arr[j]%x, new HashMap<>()); } HashMap<Integer, Integer> hy=hm.get(arr[j]%x); if(!hy.containsKey(arr[j]%y)){ hy.put(arr[j]%y, 1); } else hy.put(arr[j]%y, hy.get(arr[j]%y)+1); } System.out.println(count); } } }

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2 months ago, # |
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thanks for the problem E! really enjoyed solving it :)

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2 months ago, # |
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Can G use dp?

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2 months ago, # |
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#include<bits/stdc++.h> using namespace std; void solve(){ int n; cin>>n; vectorv(n); for(auto &i:v) { cin>>i; } if(n==1) { cout<<"yes\n"; return; }

long long sum=0;
for(auto i:v)
{
    sum+=i;
}
int req=sum/n;
int flag=0;
long long exc=0;
for(int i=0;i<n;i++)
{
    if(v[i]>=req)
    {
        exc+=(v[i]-req);

    }
    else{
        int req_am=req-v[i];
        if(exc>=req_am){
            exc-=req_am;
        }
        else{
            cout<<"no"<<endl;
        }
    }
}
cout<<"yes\n";

}

int main() { int t; cin>>t; while(t--) { solve(); } }

not run all test case

problem name = B.Make Equal

help me for run it

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2 months ago, # |
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In E can somebody explain why have we used ans -1 instead of ans.

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    2 months ago, # ^ |
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    ans : the amount of digits that we have in final number.

    but how many digits does $$$10^m$$$ have ? It's $$$m+1$$$.

    So we need to check if ans >= m+1 ,ans it's the same as ans-1 >= m.

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2 months ago, # |
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1931C can anyone tell me there is i>=1 that means first index of the array is not changeable in order to make the whole array equal . all the other elements of array needs to be equal as first index isn't it ?

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    2 months ago, # ^ |
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    select three integers i, j, x(1≤i≤j≤n)

    I think array index range from 1 to n , not 0 to n-1 , and it's ok. first element's index is 1.

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2 months ago, # |
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Can somebody pls help to identify the cause of runtime err in my soln : https://codeforces.com/contest/1931/submission/248125344

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2 months ago, # |
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In tutorial for D., why is the condition for y not the same as the condition for x? Since it a[i] - a[j] is either equal to y or 0 as well.