tourist's blog

Codeforces Round #740 Editorial
By tourist, 3 years ago, In English

Hope you enjoyed the round!

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3 years ago, # |
  Vote: I like it +22 Vote: I do not like it

Thanks for the fast editorial :)

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

Blogewoosh #8 when?

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3 years ago, # |
← Rev. 2   Vote: I like it +41 Vote: I do not like it

Thanks for the interesting problems. Overall an awesome contest.Orz. Will make sure I don't miss your future rounds :)

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3 years ago, # |
  Vote: I like it +8 Vote: I do not like it

Thanks for an amazing round and fast editorial:)

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

Awesome contest. Nice challenges! Thank you!

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

Can anyone explain how to do D1? I am not able to understand

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

Was O(N √N log(N) ) intended to fail on DIV2 D1?

I believe my submission: 126895459 had that complexity yet it tled on D1 or am I missing something? :(

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

    Nope, but I couldn't find an error in your solution (sorry =(, I am very tired). But you can check mine which has the same complexity and maybe figure out the problem.

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

      I think constant factor of my code was greater than that of yours (around 5 times).

      It required heavy optimizations to pass(barely):); But I still don't understand why the constant factor my code was so high compared to yours.

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

    I think it was. I did the exact same thing as you and it tled. I mean it takes 2*10^5*447*14=1.25e9 operations, which is improbable to finish in 6 seconds.

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3 years ago, # |
← Rev. 2   Vote: I like it -8 Vote: I do not like it

Thank you for the editorial. In problem A (Simply Strange Sort), why do we have to start swapping odd indices first? Can't we start swapping even indices first? For instance, for this case, why should the answer be equal to 7?

7

6 4 5 3 2 7 1

if we start swapping odd initially. The answer is 7.

  • 4 6 3 5 2 7 1
  • 4 3 6 2 5 1 7
  • 3 4 2 6 1 5 7
  • 3 2 4 1 6 5 7
  • 2 3 1 4 5 6 7
  • 2 1 3 4 5 6 7
  • 1 2 3 4 5 6 7

but if we start from even indices, the answer is 6.

  • 6 4 5 2 3 1 7
  • 4 6 2 5 1 3 7
  • 4 2 6 1 5 3 7
  • 2 4 1 6 3 5 7
  • 2 1 4 3 6 5 7
  • 1 2 3 4 5 6 7

Please, help. What am I missing?

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    3 years ago, # ^ |
      Vote: I like it +16 Vote: I do not like it

    We don't choose the start. We perform operations on the basis of iteration parity and iterations start from 1 . We are not trying to minimize the iterations. Just performing the iterations till the array becomes sorted. :)

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

    Well, there is a sentence in paragraph 4 says "On the i-th iteration, the algorithm does the following:" which means your first iteration must be odd (your counting must start with number one)

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3 years ago, # |
  Vote: I like it +99 Vote: I do not like it

I was surprised to see the $$$m \leq 2000$$$ requirement in E, as I have a very simple $$$\mathcal{O}(n^2 \log \max a_i)$$$ solution.

In every step, we try to find the path that

  • starts from $$$0$$$
  • loops back to itself
  • doesn't contain only already beaten caves

that requires the minimum initial strength. Then, we merge nodes on that path with 0, and update the values of minimum initial strength and strength gained over initial strength. We repeat this until all nodes have been processed.

This is clearly correct, since taking the path looping back to itself with minimum required strength cannot hurt you: you end up in a cycle, so you can still go anywhere afterwards, and any solution to the problem has to have you travel a cycle.

To find this loop, we binary search the initial strength required to beat it, then do a DFS (or BFS, the order doesn't matter) of caves we can get to with that initial strength. We maintain for every cave the previous cave, and don't allow going back to it. Once we have two paths to some cave, we have found a cycle, as we could take the one that ends with greater strength in the forward-direction, and the other going back.

We can sort edges from vertices in increasing order of the strength of the monster on the cave on the other end, and break once the monster is too strong for the strength coming from our current cave. If we ever end up in the same cave multiple times, we are done, so every vertex gets handled at most once every step. Thus, we take $$$\mathcal{O}(n)$$$ time every step.

We might need to add $$$\mathcal{O}(n)$$$ loops, and do $$$\mathcal{O}(\log \max a_i)$$$ DFS's every time, each taking $$$\mathcal{O}(n)$$$ time, so the total time complexity is $$$\mathcal{O}(n^2 \log \max a_i)$$$.

Code: 126910619

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    3 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    complexity of your DFS should be O(n + m)?

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

      No. Think about using DFS to find a cycle in a undirected graph. While we haven't found a cycle, we have handled at most one in-edge to every vertex (as following the backpointers from a vertex with two in-edges gives a cycle) -- thus we handle $$$\mathcal{O}(n)$$$ edges in the worst case, and the DFS runs in $$$\mathcal{O}(n)$$$ time.

      Here the DFS is a bit different, but we similarly return once we have two in-edges to a vertex, so the same proof works.

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

I like how the same problem is the easiest and toughest in this round, just different constraints.

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

https://mirror.codeforces.com/contest/1561/submission/126865692 ---> TLE one https://mirror.codeforces.com/contest/1561/submission/126911328 ---> accepted (same code) my submissions of today A question both codes were same one at the time of contest this gave me tle on test case 11 in system testing but now after the system testing i submitted the same code again by removig some comments it got accepted .... with 31ms

can anyone explain why it happened......

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

    This is very strange. I think it might actually be a problem with the judge. One thing to note though is that creating arrays with variable length is not allowed in c++ standard. Most people create them with constant sizes. However, I doubt that's the problem because it apparently works in gnu c++.

    Maybe MikeMirzayanov or someone who knows better can check...

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

    It is matter of chance that it got accepted the second time, you are checking the last element also in the odd loop. a[i] > a[i+1] gives undefined behavior for i == n. Maybe your loop ran infinite times in systests.

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3 years ago, # |
  Vote: I like it +166 Vote: I do not like it

I was quite right with my prediction...

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

Thanks for the editorial and the contest. It was great.

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3 years ago, # |
  Vote: I like it +53 Vote: I do not like it

I think the editorial of 1561C — Deep Down Below meant to say $$$b_i=\max(\ldots)$$$ instead of $$$\min$$$, likewise for $$$p$$$.

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

What was the role of less space(128 mb) in Div2 D1. And how many iterations can be run in 6 sec time limit?

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

    2e8(cpp)*6==?

    and the first question is explained in the editorial

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

What's the reason behind stopping $$$O(N \log^2 N)$$$ solutions to pass in Div.2 D2? All the solutions utilizing factors rather than multiples already struggle with $$$10^6$$$ inputs and exceed the given 6 seconds, whereas $$$O(N \log^2 N)$$$ fenwick tree solutions take < 2 seconds on those tests.

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

    I thought of fenwick tree approach but only towards the end. Did you find any?

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

      The best I managed is the one described in the last line of editorial (it TLEs on test 7 because of the extra log factor). So fenwick simply doesn't work because it could be avoided by simple prefix/suffix sums or difference arrays, but I got stuck on it. Here are my two submissions for the sake of comparison (in the AC one I traverse from the front, but it is essentially the same thing).

      126906387 126864992

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3 years ago, # |
← Rev. 2   Vote: I like it +19 Vote: I do not like it

The ordering for 1561C - Deep Down Below can be found with exchange argument too. Consider two sequence of monsters A(a1,a2,...,ap) and B(b1,b2,...,bq) now,

since from the editorial it is clear that, initial power level needs to be greater than max( a1 -0 , a2 - 1 , a3 -2 , ... , ap - (p-1) ) if we go through cave A we create two auxiliary sequence from the original sequence A_aux( ( a1 -0 , a2 - 1 , a3 -2 , ... , ap - (p-1) ) ) and B_aux( b1 - 0 , b2 - 1 , .... , bq - (q-1) )

if A is traversed before B the minimum initial power level in this case needs to be x1 = max( max(A_aux) , ( max(B_aux) - size(A_aux))) [ size(A_aux) is subtracted from the second part because before going to second cave power is incremented by size(A_aux) ]

if B is traversed before A the minimum initial power level in this case needs to be x2 = max( max(B_aux) , ( max(A_aux) - size(B_aux))) [ size(B_aux) is subtracted from the second part because before going to second cave power is incremented by size(B_aux) ]

now we need our initial power level to be small in case to A traversed before B , i.e x1 < x2 => max( max(A_aux) , ( max(B_aux) - size(A_aux))) < max( max(B_aux) , ( max(A_aux) - size(B_aux))) . this custom sort rules can be used to get the ordering of the caves.

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

problem D1 ,how to work sqrt(n) ,please anyone explain this...

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    3 years ago, # ^ |
      Vote: I like it +16 Vote: I do not like it

    If you consider $$$x = p^2 - k$$$, such that $$$p^2 - k > (p-1)^2$$$, then it is straightforward to show that $$$\left\lfloor\frac{x}{\lfloor\sqrt{x}\rfloor}\right\rfloor - \left\lfloor\frac{x}{\lfloor\sqrt{x}\rfloor + 1}\right\rfloor \ge 1$$$ (left-hand-side can only be either $$$1$$$ or $$$2$$$), which means that consecutive addenda up to this threshold point to different cells. Note: $$$\left\lfloor\sqrt{p^2-k}\right\rfloor = p-1$$$, for $$$k>0$$$.

    As to how pick up that threshold might be of form $$$\lfloor\sqrt{x}\rfloor$$$, one of the options would be to start from solution of the related continuous problem. Consider some generic function $$$f(x)$$$, and minimize functional $$$F[f] = -f(x)$$$ with respect to the (continuous) constraint $$$\frac{x}{f(x)} - \frac{x}{f(x)+1} = a, a \ge 1$$$. Or equivalently (using Lagrange multipliers): $$$F[f] = -f(x) + \lambda \left(a - \frac{x}{f(x)} + \frac{x}{f(x)+1}\right) \to f(x)=\sqrt{\frac{x}{a} + \frac{1}{4}}-\frac{1}{2}$$$.

    Then maximize $$$f(x)$$$ within the space of valid $$$a$$$'s: $$$(a=1) \to f(x) = \sqrt{x + \frac{1}{4}}-\frac{1}{2}$$$

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

In Problem C, I tried Binary Search on $$$x$$$ in my submission https://mirror.codeforces.com/contest/1561/submission/126887144, but was faced with the problem of which array to enter first. I tried implementing some sorting algorithm to sort the vectors according to the current power in the binary search, and if for two vectors $$$a$$$ and $$$b$$$, there exists $$$1 \le i \le min(a.\text{size}(), b.\text{size}())$$$ such that $$$\text{power} \le a[i]$$$ but $$$\text{power} > b[i]$$$, then $$$b$$$ has to come before $$$a$$$ in the order. (You can see other details in the cmp function) However, that solution doesn't pass. Any help?

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

    Lets say b={1,1,10} a={1,3} and power is 2 then according to your logic b should come before a(for i=3). however it cannot enter there.

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

      Yes, and that's OK. Since the Binary Search takes care of which power to choose. For example, in your case, $$$b$$$ comes before $$$a$$$, but anyways $$$\text{power} = 2$$$ doesn't work. Anyways, my submission outputs $$$7$$$ for the input you gave, which is correct.

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

      Anyways, I realized where the error is. Thanks everyone!

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

I love how Div 1F and Div 2A are the same problem. I just glanced at 1F to see if I understood the problem (I usually don't), and it was a pleasant surprise.

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3 years ago, # |
  Vote: I like it +38 Vote: I do not like it

I have a solution to div1E that does not involve binary searching the initial power and runs in $$$O(NM \log N)$$$. We start off with 1 power. We run Dijkstra on the graph several times, each time starting from the set of defeated caves so far, to find the shortest path to each unvisited cave that requires the least increase to the starting power. This requires keeping track of the power increase along the path. The dijkstra comparator breaks ties among two nodes $$$u$$$ and $$$v$$$ by preferring the one that grants the largest power up to that point.

Technically there are negative cycles, but we just ignore cycles altogether by not revisiting processed nodes. At the end we have a shortest path tree. We can find in this tree the "cheapest" (requiring least increase in starting power) node that has at least one non-tree edge adjacent to it, then "accept" the increase in starting power and visit the path + cycle from this non-tree edge. Each iteration takes $$$O(M \log N)$$$ or $$$O(N \log N + M)$$$ and there are at most $$$O(N)$$$ iterations until we visit all the caves, so $$$O(NM \log N)$$$ in total.

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

I had a slightly different approach for Div 2 D / Div 1 B.

Calculating the ways you can reach a number by subtracting was straightforward, but I used suffix sums to calculate the sum of ways you can reach a given number by dividing another number.

The code can be seen here.

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

    That's what I also did. It's seems way more intuitive.

    However I didn't think that you could also hold a suffix sum with the memory constraint so I used the dp array as the suffix sum. I think it's a pretty neat trick 126918785.

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    3 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    In your code dp[i] stands for the sum of ways we can go from n to i. But I have a slightly different way of thinking from yours too: instead of going from n to 1, we can go from 1 to n. dp[i] stands for the sum of ways we can go from i to 1. For every i, for every divisor j ranging from 2 to n / i, it is possible to go from [i * j, (i + 1) * j) to i dividing j. So simply add dp[i] to dp[i * j, (i + 1) * j) using a prefix sum. the time complexity is O(n + n/2 + ... + n/n)=O(n log n).

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3 years ago, # |
  Vote: I like it +9 Vote: I do not like it

Very Good Contest.

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

Can someone please explain the reccurence relation in 1558B- Up The Strip and why it’s time complexity is O(n^2) , Thank you in advance.

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

    The time complexity of a dp recurrence is $$$no. of$$$ $$$states$$$ $$$*$$$ $$$no. of$$$ $$$transitions$$$. Since there are $$$O(n)$$$ states it can visit and it has $$$O(n)$$$ transitions, the total time complexity is $$$O(n^2)$$$.

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

      What exactly do you mean by the O(n) transitions?

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

        Transition time basically means how much time it takes to compute a single dp state. Here, to compute $$$f(x)$$$, you would need to compute $$$\sum_{y=1}^{x-1} {f(x-y)}$$$ and $$$\sum_{z=2}^x {f(x/z)}$$$, both of which take $$$O(n)$$$ time each, therefore $$$O(n)$$$ time total.

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

Can anyone share a way to find a "closed-form" solution in problem B. Charmed by the Game?

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

My solution for D has $$$O(M \sqrt{M})$$$ complexity, but doesn't require to roll back any changes. It got AC in 592 ms. I process insertions in the same order as they are given in the input, and use linked list data structure. Each element of the list contains some continuous range of positions that have already had some element inserted right before them.

Any insertion from the input can increase the size of the list by up to $$$2$$$. Each time the size of the list becomes greater than some constant around $$$\sqrt{M}$$$ (I chose $$$520$$$) we need to squeeze all the elements of the list into one range, so our list has only one element in it after that.

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3 years ago, # |
← Rev. 4   Vote: I like it +8 Vote: I do not like it

Such a shame that I wasn't able to solve Div 2 D1 yesterday since I was so close to the official tutorial.

I wasn't able to solve the floor function inequality, so I used binary search to find the upper and lower bound of $$$z$$$ instead, which makes my solution $$$O(N\sqrt{N}\log{N})$$$. That turned out too slow for the problem constraints though.

Moral of the story: Math is important. Here's my submission

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

    Hello there, I tried modifying your code slightly to avoid the TLE and it worked. Because the % and / operations are time-consuming, I tried to use them as little as possible. Link

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      3 years ago, # ^ |
        Vote: I like it +3 Vote: I do not like it

      Wow, I didn't think that such small operations can affect performance greatly. Thanks for fixing my code and pointing out what's wrong!

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3 years ago, # |
← Rev. 4   Vote: I like it -8 Vote: I do not like it

This is my code for Question C, it's giving wrong answer and I can't understand where is the issue.Somebody please have a look and let me know what's wrong with my code.(I am using binary search approach here )

code
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    3 years ago, # ^ |
    ← Rev. 2   Vote: I like it +1 Vote: I do not like it

    You can use spoiler to show your code:

    Code

    upd: Thanks :)

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3 years ago, # |
  Vote: I like it -16 Vote: I do not like it

I was in awe for DIV2-D1, time limit was 6s and O(n^2) was not passing.

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    3 years ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    Not even $$$O(N\sqrt{N}\log{N})$$$ will pass, so a TLE for $$$O(N^2)$$$ is kinda obvious.

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3 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Can anyone explain more clearly problem D2div2/Bdiv1

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

    My solution for D1 covers a part of the idea for D2.

    Hint for D2:
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3 years ago, # |
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Video Editorial for problem D1 and idea for D2

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3 years ago, # |
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can anyone explain how to generate all divisors of n with it's prime factors ,div2/d2

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

in problem D, if i is divisor of x+1 why we need to replace i-1 to i ?

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3 years ago, # |
  Vote: I like it +9 Vote: I do not like it

My solution to div2. D1/D2 or div1. B

Lets define dp[I] as number of ways to reach I from N, therefore dp[N] = 1.

We can calculate the value of dp[I] using the formula (iterating I from N — 1 to 1) :-

dp[I] = sum(dp[I + 1], dp[N]) + sum(dp[J * I], dp[min(J * I + J - 1, N)]);

where J is [2, N / I]

The sum can be calculated with prefix sums in O(1) giving a time complexity of (N log N) where log N represent sum of harmonic series.

My submission: https://mirror.codeforces.com/contest/1561/submission/126890430

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3 years ago, # |
← Rev. 2   Vote: I like it +5 Vote: I do not like it

Why can't use greedy in Div. 2 C (I use greedy and I get AC.)

upd. Solution:

For each cave $$$i$$$, calculate

$$$\large p_i = \sum_{j=1}^{k_i}(a_{i,j}-j+1)$$$

and then select the cave in the order of $$$p_i$$$ from small to large .

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

Why "Tutorial is not available" for 1558F

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

Excellent contest. Thanks very much. I ran my algorithm exactly the same as the tutorial of div2d1 with case 200000 on my own machine and got exactly 6s runtime... so I submitted but got 3.35s runtime... CF's CPU is so fast!

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3 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Thanks for the amazing problems as always

cheers 

\(^o^)/
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3 years ago, # |
  Vote: I like it +10 Vote: I do not like it

Is there a reason why the constraint of div2D1 had to be 200000 instead of 100000?

I keep getting TLE with python even though 150000 would have worked.

Did anyone manage to implement the sqrt complexity editorial solution for D1 with python?

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3 years ago, # |
  Vote: I like it +3 Vote: I do not like it

In problem D1, can somebody explain why there is at most sqrt(n) different values of floor(n/x) for all x in [2,n]? is it just a well known thing?

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

For Problem D:

Can someone please elaborate a bit more on why will sqrt(n) we have no repetititon of the quotient.

For example for 19 If we print out 19/1,19/2 and so on i.e

19 9 6 4 3 3 2 2 2 1 1 1 1 1 1 1 1 1 1.

Why till 4 i.e sqrt(19) we dont have any repetition?

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    3 years ago, # ^ |
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    You need $$$\frac{n}{i} -\frac{n}{i+1} < 1$$$ or $$$i\cdot(i+1) > n$$$ and $$$i = \sqrt n$$$ is first integer that satisfie so anything larger that that will have consecutive difference with last less then 1 so repeat

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    22 months ago, # ^ |
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    Cause for x, numbers less than sqrt(x) on dividing will give say sqrt(x) quotient greater than sqrt(x), and for all numbers greater than sqrt(x) quotient will be less equal to sqrt(x) which is sqrt(x) at most. Giving O(2*sqrt(x)) = O(sqrt(x))

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

In editorial for Div2 D1,

Find the sum over all $$$z < sqrt(x)$$$ directly
is not correct ig. Let's say $$$x = 5$$$ & $$$z = 2$$$, then $$$z < sqrt(x)$$$ but $$$5 / 2 = 2$$$ which is less than $$$sqrt(5)$$$ and hence will be covered in loop for $$$c$$$ (refer the editorial).

so loop for $$$z$$$ should be $$$(x / z) > sqrt(x)$$$ or $$$z < floor(sqrt(x))$$$

correct me if I am wrong.

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    10 months ago, # ^ |
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    You are right. Here is my submission which implements the solution as per the editorial for D1: 243412291

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3 years ago, # |
← Rev. 4   Vote: I like it 0 Vote: I do not like it

In problem 1558A — Charmed by the Game,In the 0<=x<=p and 0<=y<=q.I submitted the same solution but with relaxed constraints i.e 0<=x<=a and 0<=y<=b 126977218.I don't understand why this solution is also giving AC.We can clearly see the constraints we are imposing on x and y in the case of my solution isn't same.Your help is appreciated.

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3 years ago, # |
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Why is there only a problem A code?

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3 years ago, # |
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Video Editorial for Div 2 Problem E / Div 1 Problem C

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3 years ago, # |
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Sorry for asking this, but can anyone please explain Div2-D2 in more detail? I looked at the others' solutions, but found it slightly hard to find the intuition behind their approach.

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    3 years ago, # ^ |
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    I will try to tell how i came up with my solution.

    1. We are moving from n towards 1.
    2. Using moves of kind 1, what are possibles positions from where you can reach i(current index)? it will (i+1,i+2......n). So we will add the ways of reaching these value to the ways of reaching i th position.
    3. Using moves of kind 2, what are possibles positions from where you can reach i(current index)? here things get a bit trick, see if at some place you used z=2 and reached i what were the possible x's? (2i to 2i+1 right? because 2i+2 onwards, on division by 2 will yeild $$$> $$$ i ), similarly by using z = 3 (3i to 3i+2) ,z = 4 (4i to 4i+3), and ya this will be O(nlogn) because it's $$$\sum n/i $$$.
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3 years ago, # |
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Can someone explain me how to do this part in the solution of 1558B?

We can achieve that by preparing a sieve of Eratosthenes, factorizing x and generating all its divisors.

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3 years ago, # |
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When will the editorial for Div1F be available? Meanwhile, here is a sketch of my solution.

Hint 1
Hint 2
Hint 3
Hint 4
Hint 5
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3 years ago, # |
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When can tourist update the Tutorial for Problem F?

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3 years ago, # |
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Is there a more clever (sub $$$N^2$$$) solution for the first problem?

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2 years ago, # |
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"For each i>1 that is a divisor of x+1, S(x+1) contains an occurrence of i that replaces an occurrence of i−1."

Could someone explain why this is true? (it is from the editorial of 1561D2)

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3 weeks ago, # |
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In problem B: Up the Strip,

By definition of the floor function, c≤xz<c+1 . Solving this inequality, we get z∈[⌊xc+1⌋+1;⌊xc⌋]

Can anyone explain exactly how to deduce the range?

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