Видно, ты очень его любишь)

thats what they all say

Btw what is red blue and green testing?

idk why, but it was always 1900

no

I guess you don't understand what that part means. First, you should probably read the last paragraph of this blog.

From the blog:

-- We will exclude from the official standings of Div. 3 rounds and put in a separate rooms all those who can’t be reliably called to be a real participant. Accounts that materially participated in less than 2 rating rounds (materially means solved at least one problem there) before the start of the Div. 3 rounds, and those who have ever gained 1900 or more rating units will not get into the official standings and will be assigned to separate rooms. However, this does not mean that there is no rating recalculation for them. Thus, the rating will be updated for all users whose rating is strictly less than 1600 at the time of the start of a round.--

In short, everyone who has $$$< 1600$$$ rating will receive rating change after the round, but only some of these participants will be shown in the "official standings". Namely,

• People who have participated in less than $$$5$$$ rated contests (originally $$$2$$$ as you can see) are excluded from the official standings.
• People who have reached $$$\ge 1900$$$ rating, and later dropped back to $$$< 1600$$$ rating are excluded from the official standings.

These rules are in place to try to make sure that everyone in the "official standings" is actually at Div.3 skill level and not someone much better just having low rating. I think it is more fair to make this rating upper bound at $$$1900$$$ than $$$1600$$$, since reaching $$$\ge 1600$$$ rating doesn't mean you are actually that good (you could've gotten lucky).

But your name isn't on blue testers within the blog ?

you will find vika in this contest too .. no worries

You are correct, it was moved.

XD

hi,i'm from vietnam too.probably i'm younger than you.can i chat with you about solutions,idea and ways to deal with problem?

I'm really grateful for the opportunity this contest gave me to reach the blue！

I dont understand what is the Vika Problems? Is it Thanos ?

Where is it written that there will be ?

Did you even read the announcement?

If your rating is less than 1600, then the round will be rated for you.

It will...... .......................................... .......................................... ............................ ............................ ............................ ...................... be green hh

Maybe.

he can't control that , even mike can't , it's not working as it is contest time and too much pressure on the judge

cout << "Hello, Codeforces!" << '\n';

<333

Your uzeli array is uninitialised so has random undefined contents on each run.

1，3，5，6 ? 2，3，5，6 ? I was totally confused during the contest :D

your template.

Yes solved the problem D with quadratic and got WA, and I know it time to use bi-search

Extremely short solution without binary search: https://mirror.codeforces.com/contest/1862/submission/220208197

Do you fully understand the statement, that's confuse. If you are, the problem is just maintaining a priority queue to keep track of the largest value, and maintain the size of queue to be less than or equal to m.

I think the problem is that d*i can give overflow , when you're calculating res

Yes it is a valid strategy. I solved D using same approach.

I did the same thing tho https://mirror.codeforces.com/contest/1862/submission/220247861

You don't need the shortest continuous subarray. The answer is maximum of: for each position sum of the greatest m positive elements to the left minus the position multiplied by d. If you get more than m positive elements to the left, you just erase the minimum of the set or the priority queue and update the sum.

Same

I used knapsack to solve F, gave me a very easy implementation!

another step back?

I had a similar solution, mine was segmentation error (I believe to be memory exceeded)

Hey Cheater, why have you inserted unnecessarily string/char etc. in your code of D.

Extremely sorry bro, I mistakenly thought you cheated. Sorry please forgive me. Sorry Again.

if((sum-dp[F]<=W)||(sum-dp[W]<=F))

You only need to check one of these. Suppose sum-dp[F] <= W is true:

dp[F] is the sum of a subset of monsters sum - dp[F] is the sum of the remaining set of monsters

Since sum - dp[F] <= W, we can kill the remaining set of monsters with W. In other words, sum - dp[F] <= dp[W].

Therefore, sum - dp[W] <= dp[F] <= F and the second statement is true as well.

The same argument applies if you want to prove the second statement implies the first.

i did the same

Lol i thought it was question F nvm mb

Have a good day!

this is the equivalent of

for i from 1 -> n:

for j from sum -> a[i]:

dp[j] |= dp[j — a[i]]

generates all subset sum which are possible.

OK, I think I got it!

For the interested:

dp[amount] should be $$$1$$$ when we can form amount by summing some elements of s together, and $$$0$$$ when we cannot.

Therefore dp should have enough size to store all possible different amounts. In this case, from $$$0$$$ to $$$10^6$$$.

Initially dp will have all values set to $$$0$$$ (we don't know yet which values we can form), except for the $$$0$$$-th bit (we always can sum up to $$$0$$$, just choose no elements from the set).

The line that sets the $$$0$$$-th bit to 1 is: dp.set(0).

Then we iterate over all elements of our set. Say that our set s is {3, 2}.

In the first iteration of the for, we know that we can form s[0] — that is, when we choose only the first element to form our sum — so we should set the s[0]-th bit of dp to $$$1$$$.

After the first iteration, the $$$3$$$-rd (and $$$0$$$-th) bit of the bitset will be $$$1$$$:

dp equals 00000 ... 0001001.

In the second iteration, when considering s[1], we know that for each value we already know we can form, we can also form that value + s[1]. Since each bit already set in dp represents a value we know we can form, by left shifting each bit $$$1$$$ of dp by s[1], we get all new values we can form.

In the example, since s[1] is $$$2$$$, by shifting all bits of dp by $$$2$$$, we have:

dp << 2 equals 00000 ... 0100100 (now we know we can form 2 and 5).

By using | between dp and dp << 2, we join all the values we already knew we could form and these new values we just learned we can form: dp |= dp << s[1].

dp |= dp << 2 equals 00000 ... 0101101 (we can form 0, 2, 3 and 5).

The same idea for s[1] is valid for any s[i] that comes after, and after the for, we know all values from $$$0$$$ to $$$10^6$$$ that can be formed, since they're set to $$$1$$$ in the bitset.

There are probably better and shorter explanations elsewhere, but since this is a Div3, I think it means no harm putting mine here.

F wasn't that hard to be honest. I needed around 30 minutes to solve it(spent 10/15 minutes on E after solving C). I immediately had the binary search plus knapsack idea. But wasn't really sure if it would pass(I was thinking about normal knapsack not the bitset one). Then I was able to reduce it to only knapsack without the binary search.

I participated virtually though!

I was also surprised

I copy the SortedList template from someone's submission in Atcoder contest，it is crazy fast，prefect to solve this problem.

yeah , I never use " " , I always use "\n" and never had a problem.

Yes. 220222897