me too :>> From VNOI with love <3

orz

If it is not a binary-search problem :)

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Wish so :)

Trust me, by that time Vietnam would have the game in their bag anyway, so please join our contest! :)

I feel you, I can't stop supporting my national team. So I will write the contest later. =)))

A character smarter than the cat in your profile picture doesn't exist.

Actually the 9999 upper-bound is modifiable from the contest panel. Can change it to 1e9+7

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That means one day tourist will be affraid of being banned from div1

but tourist is not registered yet

And it comes true now

Register now for extra rregistration

I just submited from my phone and got accepted xdddd. Still can you please make it unrated for me. I dont know how to prove to you that i dont have any electricity now.

Sorry, we are all not perfect. This time I suggested such a change for the problem and it leads to a much more harder problem. This problem is really minor, all participants solved (thinking that they are solved) harder version actually solved the simplified version. I'll give a chance to exclude yourself from the rating if it really affects you. But 99% (or more) of users weren't really affected.

Actually, I saw what diligence and efforts were put by both the writers and the coordinator. They really tried to do their best. All the other problems are great.

Sorry again for the issue. I feel it mostly my mistake (I didn't ask writers to stress it with naive solution).

I hope you liked the problems!

Div1A reminds me of this CF995A — Tesla when people go derp mode real hard.

Contestants tried heuristic solutions, dijkstra, weird heaps, while you only need loops to solve it.

Find the next point to first using second type of queries (n-2 queries used).

Now we have 2 adjacent points. Now find areas of all the others with these 2.(n-2 queries).

Now the take highest area point (lets say h). Now using 1,h,i we can decide side of i to h. And with information of area we can sort points in each side.

1. we save the relevant number (1 to n)
   ex) (1,2,3) -> 1 save 2,3 and 2 save 1,3 and 3 save 1,2
   i used set supervising the array

1. for (1 to n) we find the number that has two sizes of set
   then, that number is what we start to solve the problem

1. then, keep track of that number
   also erase the already using number from the other set
   the size of set of that number is three , then that is not proper number to keep going

sorry, i am not good at english.. so i can't write any more

First of all, lock the first and second point. We'll consider other points to be on either side based on those two points.

We can see that the query $$$(2, 1, i, j)$$$ can tell that in a permutation starting from $$$1$$$, which point will come first: if the response is $$$1$$$, $$$i$$$ comes before $$$j$$$, and the other way around if the response being $$$-1$$$.

Loop through point from $$$3$$$ to $$$n$$$ to find out if each of them comes before or after $$$2$$$ in the permutation, using the query $$$(2, 1, 2, k)$$$. Here we can see that the points have been separated into two halves.

For each half, we can see that, following the permutation's order (i.e. the points' counter-clockwise order), the area of the triangle with point $$$1$$$, point $$$2$$$ and point $$$k$$$ forms a bitonic sequence: the area increases until the maxima value, then it decreases. From this knowledge, we may do the following:

• Use the query $$$(1, 1, 2, k)$$$ for each half to get all the required areas. Pick the point with the maximum area as maxima (if there are many points having the same maximum, pick an arbitrary one).
• Traverse through the not-chosen-as-maxima points in the area-increasing order, check if it comes before or after the maxima using the 2nd type of query (similar to the aforementioned part). If it comes before the maxima, add it to the left side of the current half, otherwise add it to the right side.

Total number of queries used will not surpass $$$3n-6$$$.

Why?

the same for me , and problem B changed during the contest was very confusing to me, !Rated.

when n == 2 && m == 2, then answer is -1

n = 2

case where you have 3 fridges and 2 chains, your code will fail. answer should be -1.

Solution:

Felt like C in difficulty to me, both implementation and solution

first i did the same, wa at 8. check all factors(no need prime) and pp

In array 1 1 1 0 0 1 1 1 The best solution would be 4 (divisible by 3)

When you realize you passed pretests with a blatantly incorrect solution ...

If a number X appears just 1 time in all triplets, it must be the first/last one in the original permutation. Save number->triplet mapping and start finding with X. It's an O(n) solution.

I'm pretty sure O(n^1.5) wouldn't need a large time limit, especially not for such low constraints. I've got $$$O(Q \sqrt{N \log N})$$$ where the log factor comes only from binary search. It still got TLE, but on pretests 29 and then 34, so it must be pretty close to passing.

I am the author of this problem. My solution runs in $$$O(QT\log{N} + QN/T)$$$ with some predefined constant $$$T$$$. I divide all queries into blocks of $$$T$$$. Choosing every $$$T$$$ between $$$175$$$ and $$$350$$$ should give you AC, unless your implementation has huge constant factor.

Can you explain the problem statement

I can't even understand it