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### yutaka1999's blog

By yutaka1999, history, 18 months ago,

Hello Everyone!

Japanese Olympiad in Informatics Spring Camp 2019 will be held from Mar. 19 to Mar. 25.

There will be four online mirror contests during the camp.

The contest duration is 5 hours and there are 3 or 4 problems in each day. Problem statements will be provided both in Japanese and English.

The contest site and registration form will be announced about 30 minutes before the contest.

Details are available in the contest information page.

We welcome all participants. Good luck and have fun!

UPD: The link to the contest page is added. Schedule is updated.

• +156

 » 18 months ago, # |   0 Auto comment: topic has been updated by yutaka1999 (previous revision, new revision, compare).
•  » » 18 months ago, # ^ |   0 Good Luck Coders ;)
 » 18 months ago, # |   +16 Will there be mirrors after contest? The timezone isn't good for European programmers :(
•  » » 18 months ago, # ^ |   -10 since koosaga will make virtual participation to day 1 or day 2 of JOI spring camp , so I think there will be a mirrors after the contest.
•  » » » 18 months ago, # ^ |   +10 I think it won't be a mirror but just an analysis mode of the contest with all problems available.
•  » » 18 months ago, # ^ |   +2 Unfortunately, there won't. But you can upsolve after the contests.
•  » » 18 months ago, # ^ |   +2 They traditionally opened judge servers, as far as my memory goes.And, as my name is called, I'll also advertise it here: https://codeforces.com/blog/entry/66014
 » 18 months ago, # |   0 Will the japanese team be selected based on these 4 contests?
•  » » 18 months ago, # ^ | ← Rev. 2 →   +32 Definitely! The team will be selected practically just with the spring camp results. I will participate in it and very looking forward to it :) P.S. I, of course, want to compete even with open contestants, so I am really looking forward to seeing many open contest participants and I definitely recommend to participate. It's really a high-quality contest.
•  » » » 18 months ago, # ^ |   +10 good luck!
•  » » » 18 months ago, # ^ |   +10 Good luck! And who are the other participants? :)
•  » » » » 18 months ago, # ^ |   +6 There are twenty-one participants in JOI Spring Camp. I don't list up because there are lots, but the list is available here in Japanese. Four of them are participants (Japan 1 or 2 Team) of IOI 2018. It includes Masataka Yoneda (E869120) who got gold medal in IOI, and Hirotaka Yoneda (square1001), Koichi Namekata, and Yasutaka Hiraki. But we still can't expect who will qualify... Final Round Winner and Runner-up was not IOI 2018 participants.
 » 18 months ago, # |   0 Auto comment: topic has been updated by yutaka1999 (previous revision, new revision, compare).
 » 18 months ago, # |   +23 where is site and registration form?
•  » » 18 months ago, # ^ | ← Rev. 2 →   +3 They will eventually be on https://contests.ioi-jp.org/joi-sp-2019/index-en.html.A link to the registration page will appear on this page about 30 minutes before the beginning of the contest.And the contest begins in about 35 minutes:  March 20, 2019 to March 23, 2019 10:00-15:00 +0900 (JST) 01:00-06:00 (UTC/GMT) Edit: this is going by what is on the contest site, there is discrepancy with the blog...
•  » » » 18 months ago, # ^ |   +3 But the time is 00:30 in the blog..
•  » » » » 18 months ago, # ^ |   +3 Sorry for the delay.Today's contest will start in 01:30 in UTC, one hour after the initial plan.
•  » » » » » 18 months ago, # ^ |   +8 still nothing
•  » » » » » 18 months ago, # ^ |   0 Why the contest site is still to be announced?
 » 18 months ago, # |   +16 Will the lecture materials be available?
 » 18 months ago, # |   0 I'm curious that how did people solve problem "Meetings"? I heard many (at least three types) solutions, so I expect that many people solved in different ways.
•  » » 18 months ago, # ^ | ← Rev. 2 →   0 Can you describe some solutions?I've solved this contest in analysis mode, and it was really hard to push my solution to get OK (because it makes ~$41\,000$ queries)
•  » » » 18 months ago, # ^ |   +5 I solved during contest, finally around 39100 queries (as I remember) in 17-ary tree worst case. First, if we ask $query(0, x, y)$, we get LCA of $x$ and $y$ when the root is $0$. The first sub-problem is that, assuming the graph is line graph and one edgepoint is $0$, find the graph in $O(V \log V)$. We can do it easily because we can simply sort by distance from zero because $LCA(x, y)$ will be $x$ iff $d(0, x) < d(0, y)$. Then, let's return to main problem. Supposing that we fix $x$ and we ask $query(0, x, i)$ for all $i$, the set $S$ of return value will be all vertices in path from $0$ to $x$. We can get the order of path by solving sub-problem. This routine takes $O(V + |S| \log |S|)$. Then, thinking about distinguishing return value of $query(0, x, i)$, we can separate into the same problem for sub-tree. It works well if we decide the value of $x$ randomly. I submitted this solution first and got 78 points with about 44000 queries, but doing somewhat optimization I got 100 points. E869120 has much more efficient algorithm (about 22000 queries, possibly more than expected solution), so I hope that he will write the solution there.
•  » » » 18 months ago, # ^ |   +18 My randomised solution which gets ~25k queries on worst case 17-ary tree: Pick a random vertex as root. We will solve the problem recursively. We will pick a random vertex and find all vertices that belong to the path from the root to this random vertex by simply iterating over all vertices in this recursive call and doing a $query(root, randv, w)$. If the result of this query is $w$ we know that $w$ belongs to the path. If that's not the case we will add $w$ to a temporary set $S_{result}$. After that we can find the order of the vertices on this path easily in expected $O(n \log n)$ with a "quick sort"-like approach (pick a random vertex and divide the path into to new paths — one to the left of the vertex and one to the right).Well finally we run the recursion with root equal to every vertex on the path and set of vertices equal to $S_v$, where $v$ is the vertex on the path.
•  » » » 18 months ago, # ^ | ← Rev. 2 →   +20 I solved it during online contest. My solution works in about ~22k queries on 17-ary tree(I'm not sure it is the worst case).At first, there are tree with two nodes, 0 and 1. and there is an edge between them. from 2 to n-1, we will add a node to tree. To add a node $u$, we can use centroid decomposition on current tree. Let the centroid of the tree is $m$. There are $k$ child of $m$, $v_1$, $v_2$, ..., $v_k$. Assume that the size of subtree is nonincreasing order, $|Subtree(v_1)|> ... > |Subtree(v_k)|$. Ask query $(u, v_1, v_2)$. If the answer is $u$, then $u$ is on the path from $v_1$ to $v_2$ and we can find the location of $u$ by one more query: $(u, v_1, m)$. If the answer is $v_1$, we can restrict the range to $Subtree(v_1)$, so we can do centroid decomposition on $Subtree(v_1)$ recursively. The case of $v_2$ is similar. If the answer is $z$, a totally unknown node, then make an edge between $u$ and $z$. $z$ is on the path from $v_1$ to $v_2$, so we can handle it similarly. If the answer of the query is $m$, then just move to query $(u, v_3, v_4)$, .. and so on. If the process is finished and can't find the location of $u$, then add an edge $(u, m)$.18*18*8>2000, so I think 9+9+4 query is enough to last insertion. Also, former insertions needed smaller number of query.
 » 18 months ago, # | ← Rev. 4 →   +14 After some hour struggling to solve Meetings problem, I suddenly realized that this problem is very similar to This Problem in CSAcademy, which I solved it before. The only difference between two problem is the constraints of input tree. (Meetings : max degree <= 18, CSAcademy problem : trees are generated with RNG) I copy-pasted my previous code and it gave me 100 points.I'm not blaming the problem setters, (There are way too many problems in the world, and it is very hard to avoid the coincidence) but I just feel sad that I became more stupid than last year...
 » 18 months ago, # |   +1 i woke up 4:00 am because of this contest and the contest started after 1.5 hour. So can someone tell me when is the second contest exactly ?
•  » » 18 months ago, # ^ |   0 I don't know but I expect that it is 90 minutes after contest (10:30 — 15:30 UTC+9), because such information is written in contest information page.
 » 18 months ago, # |   0 How to solve day 1 naan?I solve the small subtasks by a greedy approach but the numbers grow quickly and exceed the limit for the last subtask with the same solution.
•  » » 18 months ago, # ^ | ← Rev. 2 →   -14 I've used some __int128 + hack that if you want to make a fraction (A, B) with $B > 10^9$, just convert it to fraction ((A * 10^9 + B — 1) / B, 10^9).
•  » » 18 months ago, # ^ | ← Rev. 3 →   +30 Break the naan evenly(by happiness) into N pieces for each person.Call p(i,j) as the position of the jth cut for the ith person The B of p(i,j) should be N*v<=2e8Repeat j from 1 to N, take the smallest p(i,j) out of the remaining people and stop considering i afterwards It is easy to see that it works (its just a slight modification to the greedy)
 » 18 months ago, # |   0 Has Day2 started yet? Or is it delayed
•  » » 18 months ago, # ^ |   0 it starts in 13 minutes
 » 18 months ago, # |   +6 how to solve day2 first problem antennas?
 » 18 months ago, # |   +20 If it's not hard, could you please add all previous day's problems in the analysis session? yutaka1999
•  » » 18 months ago, # ^ |   +20 It should be OK now.
•  » » 18 months ago, # ^ |   0 Could you please tell me what or in a better way, where is the analysis session you mentioned? :D
•  » » » 18 months ago, # ^ |   +23
•  » » » » 18 months ago, # ^ |   0 Ahaa, Thanks!
•  » » » » 18 months ago, # ^ |   +10 Does this link works? (connects too long on my browser)
 » 18 months ago, # |   +10 How to solve d2p1? To me, it seems the hardest problem in day 2..BTW, p2 and p3 were interesting, too. Great contest!
•  » » 18 months ago, # ^ | ← Rev. 2 →   +26 It was hard but I guess not the hardest. Though hard to explain.For each query $[L, R)$ I find $i, j$ $(L \leq i < j < R)$ which both can see each other and $H_i - H_j$ is maximum. (To solve the whole problem reverse the array and run this again)In this way $i$-th antenna can communicate with antennas in $[i+A_i, i+B_i+1)$ so make it two event. Add this antenna at $i + A_i$ and remove it at $i+B_i+1$. Set $c_i$ to $h_i$ if this antenna is present now and $-inf$ otherwise. Set $d_i$ to currently maximum difference value $H_j - H_i$. Now sweep from left to right and process events. After processing events till $j$, the answer for every queries $[l,j)$ is $min(d_l, d_{l+1}, .., d_{j-1})$. Also for the $j$-th antenna and every antenna $k$ which $j - B_j \le k \le j - A_j$ update $d_k$ to $max(d_k, c_k - H_j)$ (notice that if $k$-th antenna is not present then $c_k$ is $-inf$ and $d_k$ doesn't change).We can implement changes to array $c$ and $d$ using segment tree with lazy propagation. So it's done in $O(nlgn)$ time complexity.
•  » » » 18 months ago, # ^ |   0 At "antenna k which j−Aj≤k≤j−Bj", shouldn't it be j−Bj≤k≤j−Aj ?
•  » » » » 18 months ago, # ^ |   0 Yup. My mistake, thanks!
•  » » » 3 months ago, # ^ |   0 Hey, can you explain the segment tree part? How are you modifying dk to max(dk,ck-hj)
 » 18 months ago, # |   +7 Will there be editorials in English after the camp has ended?
 » 18 months ago, # |   +11 How to solve dishes from day 2?
•  » » 18 months ago, # ^ |   +30 I got 74 points in contest, and there was a little typo. I upsolved it.Let $X[i]$ be maximum $j$ that satisfies $A[1]+A[2]+...+A[i]+B[1]+B[2]+...+B[j] \le S[i]$. Likewise, $Y[i]$ be maximum $j$ that satisfies $A[1]+A[2]+...+A[j]+B[1]+B[2]+...+B[i] \le T[i]$. Put a blue point in $(i, X[i])$ with score $P[i]$ and red point in $(Y[j], j)$ with score $Q[i]$. Order of steps can be represented by a path from $(0,0)$ to $(n,m)$ in lattice graph. $(i,j)$ means you already have done $i$ steps of IOI Donburi and $j$ steps of JOI curry. For a path, the whole score of process is (sum of scores of blue points above or on the path) + (sum of scores of red points below or on the path).It can solved directly with segment tree and lazy propagation in $O((N+M) log (N+M))$ time, but there are something to handle, so I tried to make it easier.For each red point in $(x,y)$ with score $q$, make it blue and change its location to $(x+1,y-1)$ and change its score with $-q$. After this process, only blue point remains and the problem becomes easier. After calculate answer, you should add $q$ for each red point. Then you can get the same answer with original problem.Only blue-point problem can be solved by segment tree. Time complexity : $O((N+M) log (N+M))$ .
 » 18 months ago, # |   0 Is there a way to solve task "examination" from day 1 without 2D binary indexed tree?
•  » » 18 months ago, # ^ |   +10 use MO's algorithm
•  » » » 18 months ago, # ^ |   0 Can you elaborate a little bit? I don't really see how MO's algorithm is related to this problem.
•  » » » » 18 months ago, # ^ |   +10 you just put pointer 1 on A and pointer 2 on B, and maintain a BIT on a+b valuesThen you just do like how you do it on regular Mo's
•  » » 18 months ago, # ^ | ← Rev. 2 →   +8 It can be solved by (CDQ) divide-and-conquer.Sort all students and queries together by one of the 3 parameters. Recursively divide them into halves according to that parameter.For each sub-problem obtained from above, consider only students in one half, only queries in another half. Re-sort them together by one of the two remaining parameters. Then handle each of them by operations on a segment tree based on the last parameter.
•  » » 18 months ago, # ^ |   0 you can use ordered_sets in segment tree.
•  » » 18 months ago, # ^ |   0 Interesting solutions. Thanks.
•  » » 18 months ago, # ^ | ← Rev. 4 →   +5 Here is my method :Represent students as points $(x, y)$ where $x = S_i$ and $y = T_i$.In subtask 2, we want to know the number of such points in $Q$ infinite rectangles covering all points $(x, y)$ with $x \geq X_i$ and $y \geq Y_i$. Sort students/points by decreasing $S_i$. Sort rectangles by decreasing $X_i$ and process them in this order, adding step by step "new discovered points" in a 1D segment tree on ordinates.After adding all "new points discovered with $x \geq X_i$", do a Range Sum Query on $[Y_i ; +\infty[$ to know the answer for this rectangle. Final complexity is $O((N+Q+Y_{max}) \log Y_{max})$.Subtask 3 : Thinking on paper, we see that we're now querying the intersection of an infinite triangle and an infinite rectangle. I count the number of points in the "big triangle" and I substract the number of points in the "little triangle" below the rectangle. I use two segment trees, one on $T_i$ values, other one on $S_i + T_i$ values. It needs two different passes with different sorting. There are some rectangles which should be treated as in subtask 2 instead (when $Z_i$ is useless).Substask 4 : I used dynamic segment tree, it's slow but it pass the TL.
•  » » 18 months ago, # ^ |   +5 I got a solution: Lets divide our queries into 2 types : a+b>=c and a+b<=c For type 1(a + b >= c) : we could only consider a and b, thus the answer could be easily found(use sweeping line and BIT or else). For type 2(a + b < c) : consider the image below \| \| \|"This area is what we want" \ |\ ----------------------------------------------- "limit A" |\ |\"limit C" "limit B" and this is equal to area X — area Y \ \ \"Area X" \ ----------------------------------------------- "limit A" \ \ \"limit C" \"Area Y"| \| \| \| | |\ |\"limit C" "limit B" And it's easy to find Area X and Area Y(same technique as you find the answer for type 1)
•  » » » 18 months ago, # ^ | ← Rev. 2 →   0 I did manage to approach the problem till this point but I don't know how to go further. So we need compressed 2-D BIT? Can you share your code so that I can learn from it?
•  » » » » 18 months ago, # ^ |   0 no just sweeping line and 1D-BIT( or treap or anything else)for example, to get the size of Area X, you put all points satisfying limit A into your data structure, and ask how many of them is above limit C.
 » 18 months ago, # |   +33 Someone asked me how to solve transportation. I'll leave it here. 53N+1 queriesNo algorithm other than Dijkstra solves the problem even in $B = 0$, so it should be related. Dijkstra algorithm is inductive, which starts from the knowledge of $dist[0] = 0$ (Nearest vertex of 0 is 0 itself) and find the second-nearest, third-nearest, and so on.If both of them know the distance for $k$-nearest vertex, how can they find which is $k+1$-nearest? Usually, they are adjacent to some found vertex, but you don't have full information so it's not guaranteed. However, it turns out that one of two has a correct answer, which means, If they know the $k$ nearest vertex correctly, and they just relaxed to their best knowledge, then one of them has the answer. This fact is not hard to prove.Now, you should communicate to know which is the $k$-nearest vertex. Both should compare their candidates, and pick the one which gives smaller distance. You can do like this: A sends the distance of best candidate. (20 bits) B compares it with his/her best. (0 bits) If B's solution is better, tell A that it is better, and the distance, and also its vertex name (1 bit + 20 bits + 11 bits) Otherwise, tell A that it is worse, ask A the vertex name (1 bit + 11 bits) Both have full information so they can relax. However, someone should call A to do this again until it found all distance. Call A after B relaxed everything. (1 bit) 20 + 0 + 1 + 20 + 11 + 1 = 53. Also, initial stage requires 1 bits to wake A up. 31N+1 queriesHere goes the key idea: You don't need to spend 20 bits for distance. The distance for the new candidate will either be infinite or less than $maxDist + 500$ (when $maxDist$ is the maximum overall currently found distances). Thus you can just send the delta of distance (511 if its infinity). A sends the distance of best candidate. (9 bits) B compares it with his/her best. (0 bits) If B's solution is better, tell A that it is better, and the distance, and also its vertex name (1 bit + 9 bits + 11 bits) Otherwise, tell A that it is worse, ask A the vertex name (1 bit + 11 bits) Both have full information so they can relax. However, someone should call A to do this again until it found all distance. Call A after B relaxed everything. (1 bit) 9 + 0 + 1 + 9 + 11 + 1 = 31. Initial stage still requires 1 bits to wake A up. 29N queriesSome constant optimization. You don't have to tell A it is worse or good. Just send the distance 511. Then A realizes something is wrong, so he/she will start sending out bits. Don't wake A up using precious bits. A should call itself if everything seems to be done. And you get WA cause you spend 58001 bits. Just prune somewhere. I pruned the first phase of distance sending.
•  » » 18 months ago, # ^ | ← Rev. 2 →   +10 Also, transportation is a cool innovation because now we can see, there can be a lot of problems like on distributed code jam at IOI format :)
•  » » 18 months ago, # ^ |   0 SpoilerWe can improve the solution further by randomization into $(9*1.5+1+11)N=25.5N$ bits.
•  » » » 18 months ago, # ^ |   0 How?
•  » » » » 18 months ago, # ^ |   +10 SpoilerEvery round, we flip a coin to determine A or B send first. If A is sent first, and it's already less than B, we don't need to send B anymore. The expectation is $9+1+9*0.5$.
•  » » » » » 18 months ago, # ^ |   +10 Ohh... that’s a nice trick (and a good alternative to constant optimization)
•  » » » » » » 18 months ago, # ^ | ← Rev. 3 →   -8 Поставь "Класс по братски"
 » 18 months ago, # |   -32 Hi everyone, let's raise my contribution please))
 » 18 months ago, # |   +10 what is solution for problem 1 or problem 2?
•  » » 18 months ago, # ^ |   +21 I assume you are talking about Day 3. Q1: Centroid Decomposition. Choose a centroid, and assume you pick nodes from subtress of at least 2 children from it, or you pick the node itself. Then you can do greedy, greedily pick the node that descreases the cost most. By some small-to-large, you can compute the costs in O(size*lg).You then update the answer of [1,size], so solving a subtree is O(size*lg), and overall complexity is O(N*lg^2)I think you need special handle for E=1, but shouldn't be trouble. For E>=2 we only pick leaves actually, but i don't see how to use that.Q2: Simple DP. Note that there exists a optimal solution that does all "set range" before "flip range". Also, all "set range" do not overlap itself, and same for all "flip range".Consider that you have x "set ranges" and you have to open/close range for y "flip ranges". Then the cost is x+ceil(y/2).Run DP from left to right, maintain dp[i][j], which is the minimum x+y you can do, for prefix i,j=0: no "set ranges" are openedj=1: You have opened a range for "set range to 0" and haven't closed itj=2: You have opened a range for "set range to 1" and haven't closed itThen you can do transition, as following: for(int i=1; i<=n ;i++){ for(int j=0; j<3 ;j++) dp[i][j]=1e9; for(int j=0; j<3 ;j++){ for(int k=0; k<3 ;k++){ int cur; if(k==0) cur=(a[i]-48)^(dp[i-1][j]&1); else cur=(k-1)^(dp[i-1][j]&1); dp[i][k]=min(dp[i][k],dp[i-1][j]+(k!=0 && k!=j)*2+(cur!=b[i]-48)); } } } I tried the best to explain :)
•  » » » 18 months ago, # ^ |   0 i got it, Thanks.
•  » » » 18 months ago, # ^ |   +13 I have came up with solution of "Designated Cities" similar to you (greedy, with centroid decomposition $O(n \log^2 n)$), but there was more better solution explained by maroonrk in the editorial time of onsite contest. It was a genius-like solution. First, think about the query of $E = 2$. It can be solved by DP in $O(n)$. Then, realize that the designated cities of optimal solution for $E \geq 3$ completely includes the optimal solution for $E = 2$. Therefore, it means that greedy will work after we determined two vertices when $E = 2$. The time complexity is $O(n \log n)$.
•  » » » » 18 months ago, # ^ |   -6 how u r able to come up with such complex solutions
 » 18 months ago, # |   +12 Problems from Day 3 are all available here: https://oj.uz/problems/source/378
 » 18 months ago, # |   +15 How to solve Day4 Q1?
•  » » 18 months ago, # ^ |   +18 Sort by $C$. Let $F(i, j)$ be a max. possible cost obtained from interval $i, j$ (paying exactly $2(C_j - C_i)$ for circular tour). Let $opt(i)$ be the maximum $j$ such that $F(i, j)$ becomes maximum. You can prove that $opt(i) \le opt(i+1)$, and you can calculate $F(i, j)$ in $O(\log N)$ using persistent segtree, thus you can use D&C optimization to solve the problem in $O(N\log^2 N)$.
•  » » » 18 months ago, # ^ |   -21 How to prove that opt(i) <= opt(i+1)?
•  » » » 18 months ago, # ^ |   +8 In fact you can solve it without persistent segtree. Using that fact, that if you will store two pointers at the current segement, and will move it during the D&C (and maintaining two sets, classic) it will be $O(n \log^2 n)$
•  » » » » 18 months ago, # ^ |   0 Yes, I solved just like that and it is similar to IOI 2014 — Holiday.
•  » » » 16 months ago, # ^ |   0 I realized that calculating $F(i, j)$ is just finding the sum of M-2 maximum elements in range $[i+1, j-1]$, how can we solve this in $O(\log N)$ per query?
•  » » » » 16 months ago, # ^ |   +3 Construct persistent segment tree for every prefix, and for each query walk down the tree like binary search.
 » 18 months ago, # |   +3 How to solve day 4 Q3? My divide-and-conquer solution only gave me 40 points.
•  » » 18 months ago, # ^ |   0 Erasing an element also provides your some information. So you don't need to erase everything in the set when you solved $[l, r]$ and want to solve $[l, mid]$ and $[mid+1, r]$. You want to split some set into two halves. If one of them is full, you don't need to make a query to know which set it belongs to. To avoid hacks, random shuffle the set before making queries. Do not set $mid = (l+r) / 2$, try something like this: $mid = l * ratio + r * (1 - ratio)$. My code
•  » » » 18 months ago, # ^ |   0 What is the reason not using 0.5 as $ratio$?
•  » » » » 18 months ago, # ^ | ← Rev. 2 →   0 The real reason is $T(n) = T(x) + T(n-x) + n + \min(x, n-x)$.
•  » » 18 months ago, # ^ |   0 With some careful implementations and dirty optimizations, we can make it into about $9.705 * 10^5$ queries for $n=43000$ and $9.952 * 10^5$ queries for $n=44000$.My code
 » 18 months ago, # |   +3 I'm not sure what the cause is, but when I download https://cms.ioi-jp.org/dataset/day4-data.tar.gz and extract the file, the following error pops out: gzip: stdin: unexpected end of file tar: Unexpected EOF in archive tar: Unexpected EOF in archive tar: Error is not recoverable: exiting now 
•  » » » 18 months ago, # ^ |   0 It works now, probably the files were fixed
•  » » » » 18 months ago, # ^ |   +4 Please add them quickly. I am waiting for you to complete so that I can update OI Checklist :D
•  » » » » » 18 months ago, # ^ |   0 It would take some time to complete, because there are too many unusual problems this year and they cannot be automatically uploaded. I expect 2-4 weeks.
•  » » » » » » 18 months ago, # ^ |   0 Its Ok. I just needed the links to update the list. Thanks. :)
•  » » » » » » » 18 months ago, # ^ |   0 Links are finalized, you may use them :)
•  » » » » » » » 16 months ago, # ^ |   0 Maybe I should have posted here: link to the comment
 » 18 months ago, # | ← Rev. 2 →   +10 Will there be editorial for all the problems?
 » 18 months ago, # |   +8 What is the team that will go to IOI?
 » 16 months ago, # |   +37 You can solve every problem of JOI Spring Camp 2019 here: https://oj.uz/problems/source/375However there are two probable issues: Checker of Naan: I implemented using __int128 believing that the happiness of the fractional part of both ends can be represented within 128-bit numerator/denominator, but maybe my calculations were wrong. This is the code, hope there are no bugs. Transportations — this is the first Communication task, so maybe there could be some unpredicted bugs :)
 » 15 months ago, # |   0 Where can I make a submit for problem Road Service (day 2 B)?
•  » » 15 months ago, # ^ |   0 That problem is from Spring Camp 2018. Anyway, I don’t know :/
•  » » 12 months ago, # ^ | ← Rev. 2 →   0 RockyB U can submit here btw: https://dunjudge.me/analysis/problems/1555/
 » 13 months ago, # | ← Rev. 2 →   +8 Hi, I was trying to upload mergers from day 4 to my country's online judge, but when i downloaded the input there were a few input files like "0123.txt", does that mean that I should put this input file in multiple subtasks or something? Can someone help pls? Thanks :)[Edit: Solved]
 » 13 months ago, # |   +8 Hi, can anyone who have solved d3p3("Bitaro who leaps through time") please describe ur solution? Thank u.
 » 7 weeks ago, # |   0 Can somebody explain solution for Day 3 Problem 3 — Bitaro, Who Leaps Through Time? Thank you!