### tourist's blog

By tourist, 13 months ago,

Hope you enjoyed the round!

• +346

 » 13 months ago, # |   +22 Thanks for the fast editorial :)
 » 13 months ago, # |   +7 Blogewoosh #8 when?
 » 13 months ago, # | ← Rev. 2 →   +41 Thanks for the interesting problems. Overall an awesome contest.Orz. Will make sure I don't miss your future rounds :)
 » 13 months ago, # |   +8 Thanks for an amazing round and fast editorial:)
•  » » 13 months ago, # ^ |   0 will someone please help me telling what's wrong with my div2 B submission: #include using namespace std; int main(){ int t; cin>>t; while(t--){ int a, b; cin>>a>>b; //consider the case when alice serves for the for the first time set ans; int p = (a+b)/2; int q = p; if((a+b)&1){ ++p; } for(int x = 0; x<=p; x++){ // a = p - x + y => y = a - p + x; int y = a - p + x; int k = x + y; if(x>=0&&y>=0&&(b==q - y + x)&&((x+y)<=(a+b))) ans.insert(k); } swap(p, q); for(int x = 0; x<=p; x++){ // a = p - x + y => y = a - p + x; int y = a - p + x; int k = x + y; if(x>=0&&y>=0&&(b==q - y + x)&&((x+y)<=(a+b))) ans.insert(k); } cout<
•  » » » 13 months ago, # ^ |   0 You can have a look on this code.https://codeforces.com/contest/1558/submission/128401048
•  » » » 12 months ago, # ^ |   0 you are supposed to add a condition y<=q when you insert k
 » 13 months ago, # |   +13 Awesome contest. Nice challenges! Thank you!
 » 13 months ago, # |   +7 Can anyone explain how to do D1? I am not able to understand
 » 13 months 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? :(
•  » » 13 months 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.
•  » » » 13 months 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.
•  » » 13 months 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.
 » 13 months 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?
•  » » 13 months 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. :)
•  » » » 13 months ago, # ^ |   0 Thank you. I should have read the problem statement carefully. The algorithm consists of iterations, numbered with consecutive integers starting with 1.
•  » » 13 months 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)
 » 13 months 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
•  » » 13 months ago, # ^ |   +8 complexity of your DFS should be O(n + m)?
•  » » » 13 months 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.
 » 13 months ago, # |   +31 I like how the same problem is the easiest and toughest in this round, just different constraints.
 » 13 months 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......
•  » » 13 months 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...
•  » » » 13 months ago, # ^ |   0 I also thought so,when I actually saw it during sys tests.
•  » » » 13 months ago, # ^ |   0 felt very sad after this happened ..... i would be near to pupil , but after this now again its so far -120 in the contest due to this ....
•  » » 13 months 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.
 » 13 months ago, # |   +166 I was quite right with my prediction...
•  » » 13 months ago, # ^ |   0 Dr. Strange !
•  » » » 13 months ago, # ^ |   -11 What are instincts ? How much one can rely on their instincts ? When two kind instincts appear which one you should rely on? These kind of questions will always be asked by the mankind...and there will be no end to seeking
 » 13 months ago, # |   +10 Thanks for the editorial and the contest. It was great.
 » 13 months ago, # |   +53 I think the editorial of 1561C — Deep Down Below meant to say $b_i=\max(\ldots)$ instead of $\min$, likewise for $p$.
•  » » 13 months ago, # ^ |   +58 Thanks, fixed.
 » 13 months 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?
•  » » 13 months ago, # ^ |   0 2e8(cpp)*6==?and the first question is explained in the editorial
 » 13 months 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.
•  » » 13 months ago, # ^ |   0 I thought of fenwick tree approach but only towards the end. Did you find any?
•  » » » 13 months 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).
 » 13 months 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.
 » 13 months ago, # |   0 problem D1 ,how to work sqrt(n) ,please anyone explain this...
•  » » 13 months 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: $(a=1) \to f(x) = \sqrt{x + \frac{1}{4}}-\frac{1}{2}$
 » 13 months ago, # |   0 I think f(n) = n not f(1) = 1 for 1558 B Up The Strip
 » 13 months 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?
•  » » 13 months 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.
•  » » » 13 months 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.
•  » » » 13 months ago, # ^ |   0 Anyways, I realized where the error is. Thanks everyone!
 » 13 months 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.
 » 13 months 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.
 » 13 months ago, # | ← Rev. 2 →   0 The first one is Accept,the second one is WrongAnswer,Why? The only difference between them is ll ans = 0; for (int i = 1; i < n; i++) { if (temp[i - 1].first + temp[i - 1].second <= temp[i].first) { ans += temp[i].first - (temp[i - 1].first + temp[i - 1].second); } } and ll ans = 0; ll aa = temp[0].first; for (int i = 1; i < n; i++) { aa += temp[i - 1].second; if (aa <= temp[i].first) { ans += temp[i].first - aa; aa = temp[i].first; } } 
•  » » 13 months ago, # ^ |   0 They look like the same but their basically doing different things.While the second code uses var aa which takes temp.first and keeps adding temp.second till the if statement evaluates to true again, the first code uses only the previous values (i.e.: i-1 in temp.first and temp.second).
•  » » » 13 months ago, # ^ |   0 Thanks.
 » 13 months 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.
•  » » 13 months 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.
•  » » » 13 months ago, # ^ |   0 hey , why are you adding dp[i+1] again at the end of the outer loop?
•  » » 13 months 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).
 » 13 months ago, # |   +9 Very Good Contest.
 » 13 months 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.
•  » » 13 months 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 $O(n^2)$.
•  » » » 13 months ago, # ^ |   0 What exactly do you mean by the O(n) transitions?
•  » » » » 13 months 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.
•  » » » » » 13 months ago, # ^ |   0 Thank you very much.
•  » » » » » 13 months ago, # ^ |   0 Nice one
 » 13 months ago, # | ← Rev. 2 →   0 Can anyone share a way to find a "closed-form" solution in problem B. Charmed by the Game?
 » 13 months 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.
 » 13 months 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
•  » » 13 months 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
•  » » » 13 months 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!
»
13 months 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
•  » » 13 months 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<
•  » » » 11 months ago, # ^ |   0 I am getting wrong answer...but my code is somehow similar to your code @henry-tb ...can u help me out where am i doing wrong #include using namespace std; #define ll long long #define nln "\n" #define N 100005 bool check(vector&a, ll val) { ll v = val; for(int i=0; i a[i]) v++; else return false; } return true; } void solve() { int n; cin>>n; vector>a(n); for(int i=0; i>m; vectorv(m); for(int j=0; j>v[j]; a[i] = v; } vector>ar; for(int i=0; i>t; while(t--) solve(); } 
 » 13 months ago, # |   -16 I was in awe for DIV2-D1, time limit was 6s and O(n^2) was not passing.
•  » » 13 months ago, # ^ |   +18 Not even $O(N\sqrt{N}\log{N})$ will pass, so a TLE for $O(N^2)$ is kinda obvious.
 » 13 months ago, # |   +3 Can anyone explain more clearly problem D2div2/Bdiv1
•  » » 13 months 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
•  » » » 13 months ago, # ^ |   0 Can you please explain how exactly you would use prefix sums?
 » 13 months ago, # |   +8
 » 13 months ago, # |   0
 » 13 months ago, # |   0 can anyone explain how to generate all divisors of n with it's prime factors ,div2/d2
 » 13 months ago, # |   +2 in problem D, if i is divisor of x+1 why we need to replace i-1 to i ?
 » 13 months ago, # |   0 Will anyone please explain the idea of Problem B? Thanks in advance
•  » » 13 months ago, # ^ | ← Rev. 2 →   0 Without reducing the generality a >= b. First analyse the case (a+b) is even. First player has (a) won games. Lets distribute it between held and broken serves. Total quantity of serves for the first player is h = (a+b)/2. Minimum breaks quantity is a-h when first player holds all h his serves. Maximum breaks quantity is h+b, when the first player breaks all h serves of the opponent and the opponent breaks b serves of the first player. When we move one game from holds to breaks the total count of breaks increases by 2. So for even (a+b) answer sequence is a-h, a-h+2, ..., h+b When (a+b) is odd we may add to even case one break or one hold. Adding one break add 1 to the answer sequence. The answer sequence for odd (a+b) is: (a-b)/2, (a-b)/2+1, ..., h+b+1
•  » » » 13 months ago, # ^ |   0 can you explain why min breaks are a — h why maximum breaks h+b
•  » » » » 13 months ago, # ^ |   0 Minimum breaks we get when the first player holds all his serves — h games. From the rest games first player wins a-h games. All these a-h games will be breaks. Maximum breaks we get when the first player wins all h serves of opponent. Plus opponent wins b serves of the first player.
 » 13 months 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
 » 13 months 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 .
•  » » 13 months ago, # ^ |   0 can you explain and share your solution :) ?
•  » » » 13 months ago, # ^ |   0 The solution is added.
 » 13 months ago, # |   0 Why "Tutorial is not available" for 1558F
 » 13 months 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!
 » 13 months ago, # |   0 Even the first question was confusing yesterday the level of questions blow my mind
 » 13 months ago, # |   +3 Thanks for the amazing problems as alwayscheers \(^o^)/
 » 13 months 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?
 » 13 months 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?
•  » » 13 months ago, # ^ |   +6 See "The harmonic lemma" in https://codeforces.com/blog/entry/53925.
 » 13 months ago, # | ← Rev. 2 →   0 Hello , I am pretty sure that there is a mistake in engine check , please check those 2 solutions for problem C — Deep Down Below: this is my last submission in the contest : http://codeforces.com/contest/1561/submission/126904661and this is the exact code except that i removed the compare function passed to sort , which actually does exactly the same as sort except when pair.first == pair.second , i choose the cave with more monsters , and this should not affect the solution : http://codeforces.com/contest/1561/submission/126961415
 » 13 months 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?
•  » » 13 months 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
•  » » » 13 months ago, # ^ |   0 Thank You So Much!
 » 13 months ago, # |   0 tourist Editorial for 1558A - Charmed by the Game says "Analyzing the formulas further, we can find a "closed-form" solution". Can you explain how the equations of a and b got reduced to the final 3 "closed" forms ?
 » 13 months 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.
 » 13 months ago, # | ← Rev. 2 →   0 Hello guys, in problem 1561A — Simply Strange Sorting. Why can't we just look at a the sorted array and find the maximum difference? Example: 3 2 1 2 3 1 ==> i = 1 2 1 3 ==> i = 2 1 2 3 ==> i = 3 So the maximum distance is moved by 3 and 1 i.e. 2 and answer = 2 + 1 = 3. Using same logic we can solve for other test cases. Can anyone suggest why will this approach fail?
 » 13 months ago, # |   0 In the problem D2 ,when I write int dp[N]={0,1,2}; outside the main() I got a CE --Compiled file is too large [34420807 bytes], but maximal allowed size is 33554432 bytes-- But when I write dp[1]=1; dp[2]=2; in the main(). It will not have this problem. why?
 » 13 months ago, # | ← Rev. 4 →   0 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.
 » 13 months ago, # | ← Rev. 2 →   0 can someone help me with problem C, my code is giving right answers for some cases but giving ans + 1 for other. i stored the minimum powers to enter each cave and the power while exiting that cave in a vector pair and sorted according to the entry level powers and checked whether the exit level powers heres the code. 126983320
 » 13 months ago, # |   0 Why is there only a problem A code?
 » 13 months ago, # |   +10
•  » » 13 months ago, # ^ |   0 Thanks for you sharing!
 » 13 months 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.
•  » » 13 months 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 $\sum n/i$.
 » 13 months 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.
 » 13 months ago, # | ← Rev. 2 →   0 I tried to implement 1561D1 - Up the Strip (simplified version) according to the editorial, but it's failing for large values can anyone tell what's wrong in my code ? solve functionint solve(int n ,int m ) { vector dp(n+1 , 0 ) ; dp[1] =1 ; int pref_sum = dp[1] ; for(int x =2 ; x <= n ; x++ ) { dp[x] += pref_sum ; dp[x] %= m ; for(int z = 2 ; z*x < x ; z++) { dp[x] += dp[x/z] ; dp[x] %= m ; } for(int c = 1 ; c*c <= x ; c++) { int range_left = (int)x/(c+1) + 1 , range_right = (int)x/c ; dp[x] += (( range_right - range_left + 1)*dp[c])%m ; dp[x] %= m ; } pref_sum += dp[x] ; } return dp[n] ; } 127054036
•  » » 13 months ago, # ^ |   0 In the second for loop you're condition is z*x
•  » » » 13 months ago, # ^ |   0 sober_phoenix, gjain369, you add the same cell in both loops as early as $x=5$: $c=2$ and $z=2\to \lfloor x/z \rfloor=2$. The reason for that comes from variability in bound of $c$. Given $z \in \left\{1, \dots, \lfloor\sqrt{x}\rfloor\right\}$, then $c \in \left\{1, \dots, \left\lfloor \frac{x}{\lfloor\sqrt{x}\rfloor + 1} \right\rfloor \right\}$. Starting from definition of the floor function: $\lfloor\sqrt{x}\rfloor \le \sqrt{x} < \lfloor\sqrt{x}\rfloor + 1$, one can show that $\lfloor\sqrt{x}\rfloor - 1 \le \left\lfloor \frac{x}{\lfloor\sqrt{x}\rfloor + 1} \right\rfloor \le \lfloor\sqrt{x}\rfloor$. e.g. $x=5$ vs $x=7$.But since value of $\lfloor \sqrt{x}\rfloor$ is known from $z$-loop, boundary on $c$ can be computed directly. changes to the codeint z = 2; for(; z*z <= x; z++) ... for(int c = 1, c_cut = x/z; c <= c_cut; c++) ... 
•  » » » » 13 months ago, # ^ |   0 Yes I knew there was some problem near $\sqrt{x}$. Thanks to your critical explaination.
•  » » » » 12 months ago, # ^ |   0 yeah , thanks for your explanation.
 » 13 months ago, # |   0 Did we need anykind of algorithm to solve problem B(div2)?if yes please tell me which one.
 » 13 months ago, # |   0 In 1561C, I didn't get the binary search approach. Can someone please help?
 » 13 months ago, # |   0 How can one solve 1561C - Deep Down Below using binary search?
 » 13 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. this line is written as a part of the second approach for the editorial of Div1B Up the Strip. I am unable to understand how we can arrive at this conclusion. Really appreciate some intuition on this. Thanks in advance
 » 13 months ago, # | ← Rev. 2 →   0 will someone please help me telling what's wrong with my div2 B submission: `#include using namespace std; int main(){ int t; cin>>t; while(t--){ int a, b; cin>>a>>b; //consider the case when alice serves for the for the first time set ans; int p = (a+b)/2; int q = p; if((a+b)&1){ ++p; } for(int x = 0; x<=p; x++){ // a = p - x + y => y = a - p + x; int y = a - p + x; int k = x + y; if(x>=0&&y>=0&&(b==q - y + x)&&((x+y)<=(a+b))) ans.insert(k); } swap(p, q); for(int x = 0; x<=p; x++){ // a = p - x + y => y = a - p + x; int y = a - p + x; int k = x + y; if(x>=0&&y>=0&&(b==q - y + x)&&((x+y)<=(a+b))) ans.insert(k); } cout<
 » 13 months 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.
•  » » 13 months ago, # ^ |   +62 Soon
 » 13 months ago, # |   0 When can tourist update the Tutorial for Problem F?
 » 13 months ago, # |   0 Sorry, but this is not a newbie friendly editorial.
 » 13 months ago, # |   0 Thanks for good editorial !!!
 » 12 months ago, # |   0 Is there a more clever (sub $N^2$) solution for the first problem?
 » 12 months ago, # |   0 What a marvellous solution Thanks MR
 » 2 months ago, # |   0 Great round
 » 7 weeks ago, # |   0 how can we solve problem C(div2) with binary search ?
 » 6 weeks 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)