every task is worth 1 point

What about getting Grey color.. (jk) ;)

are you blessing me or cursing me?

++ (•_•)

profile pic checks out

But u r already purple (magic)

But you are already CM (of course, magic)

You are welcome to joining any content if you are interested,and in Div.3,one problem equal 1 points,and there is time penatly,in other word,the few time you use to solve the problem,the penatly lower,your rank will become higher. You can use whatever lang you like,like C/C++,python and otherelse

You can compute for all pairs of of snakes $$$i$$$ and $$$j$$$, the minimum separation needed between the two snakes assuming snake $$$j$$$ will be placed after $$$i$$$ in the strip. You can then run a TSP-like DP.

Yeah, just sort and binary search on the left and right bounds for the interval to pair with the current element (when iterating through from left to right). be sure to pair with only elements that come after the current one.

I did E without any binary search. I used a priority queue to simulate the "events" than happen when reaching a number, and maintain 3 counts (Positives = 0, Negatives = 0, Unprocessed = n).

• At a[i]: Positives++, Unprocessed--
• At a[i] + 1: Positives--, Negatives++
• At b[i]: Counts don't change, but this is where you can also take it as the cost of the tree to sell.
• At b[i] + 1: Negatives--

After each "timestamp" is done, check if Negatives <= k then the current sum can be calculate as cost * (Positives + Negatives + Unprocessed) (The value at the moment can be taken as the cost).

Problem E: you can do it in 2 ways, 1st — Prove that the answer lies in the unique pairs of (ai, bi) always, can be proved using forming the loss function as we do in ternary search and observing that its convex, 2nd — use generic binary search with low =0, high = max(bi) and then check for each mid if we are well within the limits of <= k negative reviews if we are then find how many are willing to pay * mid, and find the max accordingly

yes

That's the same test case where my code fails ..35th token from tc 2.. if u get where exactly your approach fails , pls post here

yes, its because the hashmap would take up space of 10^9 ints in the worst case which is around 4gb of space whereas the memory limit is only 256MB.

I think G was pretty standard problem. If you precompute how much distance between 2 snake should be at the start, then you can just apply bitmask dp. Total complexity would be (n^2*q+2^n*n^2).

The customer will buy its not your choice so if you keep it 9 then there are 2 penalties which is not allowed. Cost me 7 more minutes, resulted in only 4 problems solved :')

Your isValid() is not suitable for binary search because it produces "ok" values for small prices, then it may produce some -1 for larger prices, and then again "ok" for even larger prices.

You can use the lower_bound.

You can use pbds ordered_set data structure, but you have to manually import it though.

you meant lower_bound function on sorted array?

Bro no disrespect or anything in any way but you're 1903 and never heard of ordered_set?

Try this input.

1
2 1
5 5
100 100

Your answer gives 100, but 100 should cause 2 negative reviews, which is not allowed.

The vector pair approach treats each event as independent, but in fact you should only update the answer after all xs which are the same have finished processing.

You can refer to CLIST (which hasn't finished estimation yet). The official estimation will be days later.

It behaves like a piecewise constant function rather than unimodal, as the function only changes values at different values of (ai, bi). So the answer must lie in the different unique values of ai and bi, then you can use trivial binary search or do like I did, here

You can find that while the price is higher than the max price, there will be no customers buying, and at the same time,there will be no customers giving negative reviews. So your point 1 is wrong, because the number of negative reviews may decrease when you're higher the price.

there will be a price at which the number of negative reviews will be greater than K

This assumption is wrong, for a quick example.

3 2
1 3 5
2 4 6

Update: even top 4 on the scoreboard (amhdaimm) was using it, but he participated unofficially

If you got an approach, you can save the unnecessary plag attacks by using classes. Whats the big deal to cry here buddy. Listen dude, try to get the solutions from gpt for the problems. Seriously bro ? Dont blabber for no reason

I guess you didn't get the meaning of competitive programming. It's all about the approach and not the syntax. If it is, we wouldn't even have stls for cpp. I hope you got it lol

which problems you mean?