Theoretically yes, but as we didn't reach out to an Azerbaijan official, we could make an exception. Talk to whoever is in charge with the national selection in your country to let them know they should expect an invitation email regarding the contest, and give me (via PM) the contact information of that person. We'll send him/her an official invitation email and he/she will have to sign up the team. Afterwards, we'll send ids and passwords to the whole team. Consider, though, that we must move quickly as the contest is in 3 days and all this communication takes time.

Andrei Popa has been to IOI twice — he couldn't make it till there last year (was the first one not to qualify for IOI). Alex Tatomir was in 2016 at equality with 4th position and some tie-breaker criteria was used. And Cristea Theodor won the NOI for 10th grade last year (he's 11th grade now). This is like a summary to tell more about that and confirm your information.

The competition for IOI is intense in Romania. We have the NOI which shouldn't be a really difficult contest but it does have its own difficulties (some of them related to the format and to the fact that you don't have any interface). Anyway, that is also a grade-specific contest. After you're ranked in the first half for your grade, you can participate to a selection contest 0 (which, IMO, is the most risky one for good competitors), where again, you have no feedback, but you have IOI format (3 problems 5 hours) and top 20 students among all high school grades are chosen. From here, these 20 students go through 6 more selection contests and a final ranking is formed. These 6 contests are taken in 2 separate camps. Sometimes, because the second camp is too late they are forced to consider the first one for some selections as well (for example in 2016, we had to send the BOI team by the end of the first selection camp so the most recent ranking — the one after 3 contests — was used). Aand, now from year to year the selection criteria for BOI, CEOI differ, although, it is made sure that always the better ranked has more options to choose from. The top 4 students go to IOI every year. I'll tell you what happened last year when we didn't have to choose a team earlier (as in BOI 2016): We had the final rankings, first 4 went to IOI. Then, first 8 would go to BOI or CEOI (but not both, in order to give the chance to compete and train to more students): participants were asked in order of the rankings which one they prefer (and so, in order, I chose CEOI, Constantin-Buliga Stefan chose BOI, Constantinescu Andrei-Costin chose CEOI, Tamio Nakajima chose CEOI, Andrei Popa chose CEOI, and the rest didn't have a choice (the following two were Alex Tatomir and Chichirim George with the same score, who also got first position inf BOI)). Unfortunately, being really good in Romania is not enough to assure an IOI place, although I believe that the selection is objective. The biggest risk a good contender has to miss his deserved slots to a international competition is missing that 20 people team which would mean end of the story, and because of the no feedback thing (like not even on example, you're not sure they've saved on the flash drive the correct codes) the risk is amplified. Fortunately, this was not the case for Andrei Popa or Alex Tatomir (that is why you still got to see them in some international competition representing Romania).

Actually there will be an online mirror which starts soon, so I suppose discussing the solutions will happen after the end of the mirror.

We will publish the editorials after the end of the online mirror. Please do not discuss the problems until then.

Although we haven't published the editorials yet, you can find a sketch of the intended solution in this comment of atatomir.

We will publish the editorials after the end of the online mirror. Please do not discuss the problems until then.

We will publish the editorials after the end of the online mirror. Please do not discuss the problems until then.

Unfortunately not:( We had decided some upper bounds of how many medals we'll give when there were ~180 participants (in the meantime about 3 more unexpected teams appeared). We also chose the threshold for each medal so that the scores would slightly make sense (a 4 point distance from gold to silver, 6 point from silver to bronze, and 2 from bronze to nothing). If we gave a bronze for 110, we should give it for 109 and 108 as well, and it would lead us to 9 extra medals which already exceeds half of the number of contestants. The truth is that some of those registered didn't participate either, so it makes sense not to have really 1/2 of the number of registered users. However, I understand you, and I wish we could give more. Hopefully, next year we'll order the medals after the final rankings:)).

Also, don't forget that the medal you receive doesn't mean much: the position and score are those that matter towards your performance (after all the last gold may be closer in value to the first silver than to the first gold (not implying this is the case now), so a medal doesn't represent that much).

We're sorry for that. We still hope you enjoyed the contest.

Sorry for that — you were the only one who had trouble logging in. Yes, the prizes will be actual medals this year as well

Can you please explain in more details your optimisation on third problem? I don't get how you shrink the length to at most .

UPD. Got it in the comment below.

Hello dacin21 ! First of all, congratulations on your result ! Thank you for your sketches. Your solutions are very similar to ours. Regarding the proof of [N / 2] + 1, I will give a fast explanation. We try to find the position of every 1 in the hidden sequence.

Let's say that you want to find the position of a '1'. It has a zeros and b ones to the left. It also has c zeros and d ones to the right. We want to be able to do one of the following final queries:

• a zeroes; one; d ones -> length a + d + 1
• b ones; one; c zeros -> length b + c + 1

We also know that a + b + c + d equals N — 1. Hence, at least one of the queries is at most [N / 2] + 1. We add zeroes progressively until the sequence does not exist or we exceeded the limit. If we exceeded the limit in both parts, then both queries are final and any additional zero would lead to a 'false' answer.

Regarding "Balanced Tree", there are only two 2 contestants who solved it during the online mirror. Radewoosh solved it in the 5-hours period while you up-solved it. Congrats ! I am very glad that you found the same solution as us. I also invite anybody with different approaches (even for the chain/path subtask) to share their solution with us.

An official editorial will be posted on the website in the following days. I hope that all of you found at least one interesting problem during this contest. Feel free to give any relevant feedback so that we will improve the quality next year.

Also, as a completion of atatomir's comment, I wanted to add that in balanced tree, in the assessment part, you need at least a +2 (we have tests that require that, but only when the search interval is [1, 3] instead of [1, 2]). We could only prove +3 upper bound.

As for the dp part, unfortunately your solution was not the intended one (ours doesn't use dp in any way) and I'm not sure whether it's correct or not.

Judging by the ellipsis mark in your last sentence, I feel that we've let you down:( Still, we only wanted 2 of the problems to require dp (second and forth) as for the first one we considered partial minimum/maximum not to be a dp and third problem seemed impossible to approach by using dp (I'm still stunned that you did so and succeed). The second one also had an alternative greedy from what I could see in people's solutions, but the intended solution was dynamic programming as it was easier to think and to code. I genuinely think and hope that this year's edition was better than last one and that we're in a continuous improvement. Hope you've still enjoyed the contest and we hope to see you as a participant next year as well:)

I have a greedy solution for Del13

First, define d[i] = p[i] - p[i - 1] - 1 (consider there's a city #0 and a city #n+1). The problem asks to do one of these operations: choose some i and subtract one from d[i] and d[i - 1] or subtract 2 from d[i] if d[i] > 2. The problem is that d[i] must be "fixed" which means 1 or 2 should be subtracted from it using the first operation depending on its parity (because after reducing it by 2 several time using the second operation it will end up as 2 or 1 and you must go with the first operation).

Let's loop over d[i]. If it's odd, the only way to "fix" it is to subtract 1 from it and from d[i + 1]. If after the loop there's still an odd d[i], no solution. Now we're left with some even numbers and some of them are fixed. Let's define a connected component as a subarray [l;r] with non-zero elements such that d[l - 1] and d[r + 1] are zero (or out of bounds). We'll fix every connected component. If its size is even, fix every pair. Otherwise you must find an odd index with d[i] > 2 and fix it twice or d[i] is already fixed and skip it. If you can't, no solution. Now as every element is fixed you can just reduce all positive elements by 2s (put these operations first).

With great help from the organizers, you can submit ALL problems from the international round right now: https://oj.uz/problems/source/321

Where can we find problems of National Round?

For scoica you will use two dp's. To do that, you will run a dfs on the graph and you will practically go through the p nodes in reverse order. Let's note the list of the p nodes with S. dp[node]= the index in S of the last node (in reverse order) from S from the path from N to node. I.e. if the node is 5, N is 9, the path from 5 to 9 is 5 8 6 9 and the p nodes from S are 2 3 8 6, dp[5] will be 3 (value 8) because the last node (in reverse order) from the path from 5 to 9 is 8 (6 is the first one in reverse order). Also let nr[node]=the number of paths from node to N for which dp[node] has that value.

For each son of node you will have the following relations:

if (dp[son]<dp[node]) dp[node]=dp[son], nr[node]=1; else if (dp[son]==dp[node]) nr[node]+=nr[son];

Also, afterwards, if node is equal with S[dp[node]-1], decrease dp[node].

Finally, you will print nr[1].

For Cambridge, firstly, for an interval [p,q], you should sort the problems by their D. Why? Try to swap two of them after you sort all of them and you will see that if you cannot satisfy the condition when they are sorted, you also cannot satisfy the condition when you break their order by swapping two of them. So optimally, you should sort them by their D. To do this fast, you should use two pointers: i and j (i<=j). You go through the problems increasing the j while you can. By this, I mean that you should increase j as much as possible while the condition is satisfied for [i,j]. When you cannot increase j anymore, increase i until the condition is satisfied again. This has O(n) complexity and you will now that way for each j how much you can go to its left so that the condition is satisfied and answer the queries in O(1) each query. Now, how to check if the condition is satisfied? Use a segement tree. If values for D were below 10^6 each time you increase the j, you should update the segment tree with -T[j] on the interval [D[j],10^6]. When you increase the i, you should update the segment tree with T[i] on the interval [D[i],10^6]. Also as soon as you meet one D[j] on your way, the first thing you have to do is update the interval [D[j],D[j]] with D[j]. (Because later you will decrease this value with those Ts). If one value from the segment tree becomes negative at one moment, then the condition is not satisfied anymore. Now because Ds are below 10^9, you can either use normalisation, or a dynamic allocated segment tree.