Thanks!

You get a big ugly warning message that you will be registering "out of competition" (i.e. unrated), but you should still be able to click past the agreement and then solve the problems during contest time

solve question ...you will be registered unofficially !!

That's what divisions are all about.

it will be hilariously sad if you get only two points

i hope you will get 2 points for this contest ^^

It seems that you get 2 points.emmmmmmmmmm That's so interesting

QWQ

Походу да :(

My worst contest ever. :/

I should have solved all, but did fiddling on B for like two hours. Really loosing the fun of this.

no it was, you think like that only beacuse you did just two problems. Problems were really good!

If you swap D and C it's actually relatively balanced, maybe the difficulty between A and B is a bit too wide though.

No, in div3 and educational contests tests aren't divided into pretests and tests and there are no hacks during the contest time, so your code runs on all the tests during the contest. After that there's a 12hr hacking phase before the final results are announced.

k can be up to 10^12

If the strings don't have the same letters, answer NO. If some letter appears at least twice, you can swap those two letters to be adjacent, then swap them in the first string and make an arbitrary swap in the second string, therefore we can answer YES. Now assume duplicate letters don't appear, and consider the strings as permutations.

You can do any even number of swaps of adjacent elements to one string without changing the other. Also, every inversion of some intervals either doesn't change the parity of the permutations or changes the parity of both permutations. If the parities are equal, we can make the strings equal with an even amount of swaps, therefore we answer YES. If they are not equal, they cannot be made equal, and therefore we answer NO.

Code: 64228071

Just get the indexing right, again and again and again. Boring and frustrating.

Problem A? Implementation is not trivial ?

You might be doing something that's undefined behavior

Yep, undefined behavior

Clang++17 Diagnostics says: p71.cpp:12:7: runtime error: index -1 out of bounds for type 'int [1001]' SUMMARY: UndefinedBehaviorSanitizer: undefined-behavior p71.cpp:12:7

In a[qan - 1] when qan is 0

Ternary search if you want

It can be solved in O(1) time. 64209894

Binary search on the number of coins of value n. So lower_limit = 0, upper_limit = a

if mid*n > total then upper_limit = mid — 1

else find the number of coins of value 1 (total-mid*n). If this value is <= b then "YES" else lower_limit = mid + 1

Code : 64218027

Me too.

No. Not for Div 3s and Educational Rounds where it is open hacking. But best hackers for opening hacking will be featured in the blog post.

B to D, greedy. Not that standard. D could be harder.

use lightweight sites (m1.codeforces.com) (m1/m2/m3)

yeah. I faced this for half an hour.

Yeah, my fellow Indian contestants and I were also not able to access the contest, but it was accessible using m2.codeforces.com

I had same problem and I thought this contest would be unrated. I tried to access another web site but unfortunately I forget my password. Zzz

Okay, My approach is , first of all sort the array. Now we have to split the array into such sub-segments that each segment is consists of at least 3 numbers. Now, Let our answer be ans. At first we can say that ans = arr[n-1]-arr[0] . Now for each valid splits in position i, we can improve our answer by arr[i]-arr[i+1] . we have to find a set of such i, that summation of all arr[i]-arr[i+1] is as lowest as possible. Let it be Sum . So , our final answer will be , ans= arr[n-1] — arr[0] + Sum . How do we find such Sum ? Just used dp on every i where i>=2 && n-i<=3

I did it like this-
First sort the array
Now the obervation is that each segment would of atleast length 3 and atmost length 5. It makes sense because if we have a segment of length 6 then we can split it into two segments of 3 and we are sure to get the smaller sum because the array is sorted.

now we can use dp , here dp(i,j) is the min cost if we have a segment of length j ending on ith index. Of course only 3 values of j are possible.
Since you could not get idea, I am not writing the complete equations and mentioning only the state. You can look at my solution if you want.

ps: I hope this passes the sys tests too.

First, Sort array $$$a$$$ in increasing order. Each team contains at least 3 members and you want to minimize the total diversity of all teams. So, there is no reason to build a team more than 5 members because you always can split the teams (>5 members) into smaller teams and reduce the diversity. Now, you can solve problem with DP technique in linear time.
Let $$$dp(i)$$$ be the minimal total diversity of the school which consists of first $$$i$$$ members.
Transition:

Base case:

• $dp[1] = dp[2] = Infinity$
• $$$dp[3] = a[3] - a[1]$$$
• $$$dp[4] = a[4] - a[1]$$$
• $$$dp[5] = a[5] - a[1]$$$

Run transition to calculate $$$dp[i]$$$ with $$$6 \leq i \leq n$$$.
The answer is $$$dp[n]$$$. You can construct the configuration by yourself.

you can construct the configuration after getting the answer, by checking all the state from dp[n] to dp[1],details can be referred to my code.

me too,,,, but it doesn't seem to be system testing.

Yeah, really bad pretests in this contest !

It's not going to be done today. haha

Congrats

Well, if string has 2 or more equal letters in it, then the answer is YES. In other case the length of string is not more then 26

Yes bro , I have experienced same issue. Can anyone please explain the above with a small test case ?

consider the case: 4 4 2 1 3