### tourist's blog

By tourist, 21 month(s) ago,

Hope you enjoyed the round!

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 » 21 month(s) ago, # |   +22 Thanks for the fast editorial :)
 » 21 month(s) ago, # |   +7 Blogewoosh #8 when?
 » 21 month(s) ago, # | ← Rev. 2 →   +41 Thanks for the interesting problems. Overall an awesome contest.Orz. Will make sure I don't miss your future rounds :)
 » 21 month(s) ago, # |   +8 Thanks for an amazing round and fast editorial:)
 » 21 month(s) ago, # |   +13 Awesome contest. Nice challenges! Thank you!
 » 21 month(s) ago, # |   +7 Can anyone explain how to do D1? I am not able to understand
 » 21 month(s) ago, # |   0 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? :(
•  » » 21 month(s) ago, # ^ |   0 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.
•  » » » 21 month(s) ago, # ^ | ← Rev. 2 →   0 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.
•  » » 21 month(s) ago, # ^ |   +1 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.
 » 21 month(s) ago, # | ← Rev. 2 →   -8 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?76 4 5 3 2 7 1if 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?
•  » » 21 month(s) ago, # ^ |   +16 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. :)
•  » » 21 month(s) ago, # ^ |   0 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)
 » 21 month(s) ago, # |   +99 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
•  » » 21 month(s) ago, # ^ |   +8 complexity of your DFS should be O(n + m)?
•  » » » 21 month(s) ago, # ^ |   +4 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.
 » 21 month(s) ago, # |   +31 I like how the same problem is the easiest and toughest in this round, just different constraints.
 » 21 month(s) ago, # | ← Rev. 2 →   +3 https://codeforces.com/contest/1561/submission/126865692 ---> TLE one https://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 31mscan anyone explain why it happened......
•  » » 21 month(s) ago, # ^ | ← Rev. 3 →   0 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...
•  » » 21 month(s) ago, # ^ |   0 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.
 » 21 month(s) ago, # |   +166 I was quite right with my prediction...
 » 21 month(s) ago, # |   +10 Thanks for the editorial and the contest. It was great.
 » 21 month(s) ago, # |   +53 I think the editorial of 1561C — Deep Down Below meant to say $b_i=\max(\ldots)$ instead of $\min$, likewise for $p$.
•  » » 21 month(s) ago, # ^ |   +58 Thanks, fixed.
 » 21 month(s) ago, # |   0 What was the role of less space(128 mb) in Div2 D1. And how many iterations can be run in 6 sec time limit?
•  » » 21 month(s) ago, # ^ |   0 2e8(cpp)*6==?and the first question is explained in the editorial
 » 21 month(s) ago, # | ← Rev. 2 →   +6 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.
•  » » 21 month(s) ago, # ^ |   0 I thought of fenwick tree approach but only towards the end. Did you find any?
•  » » » 21 month(s) ago, # ^ |   0 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).
 » 21 month(s) ago, # | ← Rev. 2 →   +19 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.
 » 21 month(s) ago, # |   0 problem D1 ,how to work sqrt(n) ,please anyone explain this...
•  » » 21 month(s) ago, # ^ |   +16 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: $(a=1) \to f(x) = \sqrt{x + \frac{1}{4}}-\frac{1}{2}$
 » 21 month(s) ago, # | ← Rev. 2 →   0 In Problem C, I tried Binary Search on $x$ in my submission https://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?
•  » » 21 month(s) ago, # ^ |   0 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.
•  » » » 21 month(s) ago, # ^ |   0 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.
•  » » » 21 month(s) ago, # ^ |   0 Anyways, I realized where the error is. Thanks everyone!
 » 21 month(s) ago, # |   +2 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.
 » 21 month(s) ago, # |   +38 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.
 » 21 month(s) ago, # |   0 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.
•  » » 21 month(s) ago, # ^ |   0 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.
•  » » 21 month(s) ago, # ^ |   +3 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).
 » 21 month(s) ago, # |   +9 Very Good Contest.
 » 21 month(s) ago, # |   0 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.
•  » » 21 month(s) ago, # ^ |   0 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 $O(n^2)$.
•  » » » 21 month(s) ago, # ^ |   0 What exactly do you mean by the O(n) transitions?
•  » » » » 21 month(s) ago, # ^ |   0 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.
 » 21 month(s) ago, # | ← Rev. 2 →   0 Can anyone share a way to find a "closed-form" solution in problem B. Charmed by the Game?
 » 21 month(s) ago, # |   +31 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.
 » 21 month(s) ago, # | ← Rev. 4 →   +8 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
•  » » 21 month(s) ago, # ^ | ← Rev. 2 →   +10 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
•  » » » 21 month(s) ago, # ^ |   +3 Wow, I didn't think that such small operations can affect performance greatly. Thanks for fixing my code and pointing out what's wrong!
»
21 month(s) ago, # |
Rev. 4   -8

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
•  » » 21 month(s) ago, # ^ | ← Rev. 2 →   +1 You can use spoiler to show your code: Code#include using namespace std; void s() { int n; cin>>n; vector< pair> >cave; for(int i=1;i<=n;i++) { int k; cin>>k; vectortemp; for(int j=1;j<=k;j++) { int armor; cin>>armor; temp.push_back(armor); } cave.push_back({k,temp}); } sort(cave.begin(),cave.end()); int ans = 100009; int l = 0 , r = 100005; while(l<=r) { int mid = l + (r-l)/2; bool possible=true; int power = mid; for(int i=0;ipower) { cout<>t; while(t) { s(); t-=1; cout<
 » 21 month(s) ago, # |   -16 I was in awe for DIV2-D1, time limit was 6s and O(n^2) was not passing.
•  » » 21 month(s) ago, # ^ |   +18 Not even $O(N\sqrt{N}\log{N})$ will pass, so a TLE for $O(N^2)$ is kinda obvious.
 » 21 month(s) ago, # |   +3 Can anyone explain more clearly problem D2div2/Bdiv1
•  » » 21 month(s) ago, # ^ |   0 My solution for D1 covers a part of the idea for D2. Hint for D2:Use prefix sums instead of BIT and compute recurrence in reverse order
 » 21 month(s) ago, # |   +8
 » 21 month(s) ago, # |   0 can anyone explain how to generate all divisors of n with it's prime factors ,div2/d2
 » 21 month(s) ago, # |   +2 in problem D, if i is divisor of x+1 why we need to replace i-1 to i ?
 » 21 month(s) ago, # |   +9 My solution to div2. D1/D2 or div1. BLets 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://codeforces.com/contest/1561/submission/126890430
 » 21 month(s) ago, # | ← Rev. 2 →   +5 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 .
•  » » 21 month(s) ago, # ^ |   0 can you explain and share your solution :) ?
•  » » » 21 month(s) ago, # ^ |   0 The solution is added.
 » 21 month(s) ago, # |   0 Why "Tutorial is not available" for 1558F
 » 21 month(s) ago, # |   +11 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!
 » 21 month(s) ago, # |   +3 Thanks for the amazing problems as alwayscheers \(^o^)/
 » 21 month(s) ago, # |   +10 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?
 » 21 month(s) ago, # |   +3 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?
•  » » 21 month(s) ago, # ^ |   +6 See "The harmonic lemma" in https://codeforces.com/blog/entry/53925.
 » 21 month(s) ago, # |   0 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.e19 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?
•  » » 21 month(s) ago, # ^ |   0 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
•  » » 4 months ago, # ^ |   0 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))
 » 21 month(s) ago, # |   +2 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.
 » 21 month(s) ago, # |   0 Why is there only a problem A code?
 » 21 month(s) ago, # |   +10
•  » » 21 month(s) ago, # ^ |   0 Thanks for you sharing!
 » 21 month(s) ago, # |   +2 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.
•  » » 21 month(s) ago, # ^ |   +18 I will try to tell how i came up with my solution. We are moving from n towards 1. 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. 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 $\sum n/i$.
 » 21 month(s) ago, # |   0 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.
 » 21 month(s) ago, # |   +52 When will the editorial for Div1F be available? Meanwhile, here is a sketch of my solution. Hint 1Consider solving the problem when the given sequence, instead of being a permutation, consists of only 0 and 1. Hint 2The number of steps necessary to sort the permutation is equal to the maximum of these values: the number of steps needed to make $n$ the rightmost element of the sequence the number of steps needed to make $n-1$ and $n$ the two rightmost elements of the sequence (not necessarily in this order) the number of steps needed to make $n-2$, $n-1$ and $n$ the three rightmost elements of the sequence (not necessarily in this order) and so on. Finding any one of these values is equivalent to the 0-1 sequence problem mentioned earlier. Thus, the original problem is reduced to $n$ copies of the new problem, with one 0 changed to an 1 each time. Hint 3To solve the 0-1 sequence problem, compute, from right to left and for each 1, after how many steps will that 1 have no zeros to the right of it. Call this function $f$. Hint 4By considering whether each 1 will "collide" with the 1 immediately to the right of it after some steps, turns out there is a not-too-complicated formula for $f$ that depends on the $f$ value of the 1 immediately after the current 1. Hint 5A segment tree works for quickly computing this function as the 0-1 sequence is updated.
•  » » 21 month(s) ago, # ^ |   +62 Soon
 » 21 month(s) ago, # |   0 When can tourist update the Tutorial for Problem F?
 » 21 month(s) ago, # |   0 Is there a more clever (sub $N^2$) solution for the first problem?
 » 11 months ago, # |   0 Great round
 » 10 months ago, # |   0 how can we solve problem C(div2) with binary search ?
 » 10 months ago, # |   0 "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)