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In other words $$$\sum a \lt = L - (b_r - b_l) \lt = L - (b_j - b_i)$$$ where $$$l \lt = i \lt = j \lt = r$$$. Since we are iterating all possible $$$(l, r)$$$ we are also going over all $$$(i, j)$$$

No tougher A please, people are simply skipping contest after seeing A.

I was solving quadratic equation for A without seeing n <= 100 !!!

I iterated over each tea and checked till which person would the tea last until it's completed using binary search on prefix sum of persons capacities, all these persons in between will drink tea with there full capacities and last person will drink remaining, so i just stored and updated multiples of their capacities in segment tree and maintained an additional array that stores remaining tea that was drank by last person.

Hopefully it wont get hacked

Can someone explain why my solution 193899252 in contest for D gave wrong answer? It is something related to modulus

if n is even then "NO". else sum of first pair should be (3*(n+1)/2). proof: S1 + (S1+1) + (S1+2) + ....... + (s1+n-1) = 2n*(2n+1)/2

Is Div 2 D just a case work or is there a elegant method?

Think it this way: '-' sign can just travel anywhere along array, bcoz if we do operation on 1 positive and 1 negative number signs of them will interchange, in this way we can make '-' sign travel along array.

Even I wasted all time on this. But we can change last 'a' to 'b' because changing 'a' doesn't increase set size

if ai >= s: (ai-s, s), else (0, ai), but I couldn't figure out which is x and y.

you only need to split array into 2, If you get some gcd 'x' by splitting more than once, then you can club all untill there are 2 subarrays because each subarray is multiple of 'x' and sum of them will also be multiple of 'x'.

I think even setters must have thought greedy was right

As n <= 100, at max we can make 50 pairs right?, so isn't 51 enough. Just couldn't wrap my head around this.

Edit: It got Accepted, now someone should confirm whether my assumption is True or Tests are weak

maximum possible xor is all bits of n becoming 1. 2n will have one extra bit than n which will be greater than all bits of n becoming 1.

Oh Yeah!

Sum of numbers from 1 to number of attacks goes to max health within time limit. Don't know what I'm thinking during contest :(

*tourist after 28 mins

For B I wasted 30 mins thinking I can't simulate this as health of monster <= 10^9. Later I realized k goes to zero in at most 10^5 attacks

Interesting sponsor title, will it have math questions? (=^・^=)

It took you 2.4 sec in C++, I'm sure it won't work in Python, until they come up with solution that works with all language this is a really bad problem :(

Will this pass in python as well?

Same, I thought operations should be in order of 10^6 for a second time limit

How do you estimate time complexity of this? I thought of same thing but I assumed no. of operations as 3*10^3 * 10^5 (correct me if I'm wrong here). I thought that's not the number of intended operations for given Time Limit forcing me to think other ways. I tried my luck by submitting some bullshit at last moment which of-course failed

Yeah, just realised

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If anyone is new to codeforces today this type of contests happens once in a while and is referred as speedforces. Don't judge by this contest and leave codeforces :(

Can anyone look and tell why is my solution for G is giving RTE for testcase 3? 166727901

B is not a graph problem I guess. Hopefully my solution is right and I don't get FST :\

Waiting for tourist Vs jiangly showdown. Will no.1 change in this Chinese round?

Maybe most difficult one idk about that, but A, B and C's are not far from expected div2 difficulty

I got TLE by doing that but may be inefficiently. I stored all possible values for a number in a 2-D array and iterated 1 to 3000 fixing it as maximum and got maximum possible number of each element less than fixed maximum using binary search!

How did you do that?

God, I felt so useless when I took that much time to solve A with solid proof, until I saw comments. Feeling so idle in this contest :)

Congratulations, I think you'll be cyan today, I lost it with time and penalties:(

Pupil Tag is not leaving me so easily :(

F can be done with binary search

I don't think we have a chance to become specialists untill we solve all including G, so i would say don't feel bad :)

Can someone say how to solve C without hashing, bcoz my solution using dictionaries got hacked. Or is it fate of Python Users :(

Specialists are least worried of their rating drop today, bcoz they hav div 4 coming soon to rescue in that case.

Yes, its possible. Look at my code that I submitted after contest.

Yeah, I thought of that, but I'm struck in getting largest and smallest duration after every simulation. How do we efficiently do that?

Is greedily giving tasks to workers with no task right approach for C?

God, how much more time will it take to reach Cyan?

because mex of subarray (0,3,2,1) of a = 4 and same subarray in [3, 0, 4, 2, 1] is (0, 4, 2, 1) for which mex = 3 since mex is different for same subarray, this is not similar permutation

WHY???!!!

For A why is it always first element? Although I used brute force after long time by seeing the constraints

I don't know C++ but it looks like he's applying Brute force only if size of array is less than 100 or else he is printing NO ig

Are you sure memory limit will not exceed if we expand them? If not why did it happen to me?

This runs in O(2^n), simply this is not intended way of doing.

I'll suffer from PTSD from now on for string problems :(

Yes, if that contest of more ranked participant is still less than or equal to 6

Congrats for your Color Change :)

Intially I expected that I'll be Cyan for my rank and number of participants but realised It doesn't happen in Div4 and now you are seeding doubt for even green colour change

I think this is the longest time I waited for rating change

You actually did with O(N^2) complexity. Try doing it with O(N).

go through each character one by one.

1. if s[i] == t[i] ---> it's fine, leave it and go to next character.

1. else, string can be changed only if (s[i] == 'a'and t[i] == 'b') or (s[i] == 'b' and t[i] == 'c'). for any other combination you can't arrive to t.

2. now let's take eg: s[i] == 'a' and t[i] == 'b', now I found nearest 'b' that is mispalced in s, let it be at index j.

key point to note is there shouldn't be any 'c' between index i and j, because if there is 'c' in between you can't swap 'a' and 'b'. If there is no 'c' then you can simply write s[i] = 'b' and s[j] = 'a'

I checked if there is any c in between by prefix sum method and I maintained a que for a,b,c seperately which has misplaced indexes of a,b,c respectively and took out first indexes of a and b ques, if i could swap them both.

Could someone say me where my 160349731 failed for problem C. I collected all those places where there is a mistake in string. They should be even in number because there if a letter could be corrected by swapping, then the other letter in pair of it should also be in wrong position. Now going from smallest index to largest index (in the collected indexes). Suppose if 2nd index letter is 'b' and 1st is 'a' without any c between them (I calculated it by prefix sum difference between these 2 positions is 0) then I'll swap them and remove those two indexes from my que. Similarly for c and b pair. Note: if 2nd element in que is 'a' then you can't swap element in 1st position to correct place.

difference between x and y could be 2*(10^5), and number of operations could be 2*(10^5). So, yeah prefix sum is expected solution.

There you go, You are back to Expert again. Good Luck in today's Contest

Why is this contest suddenly unrated for me all of sudden? (Although I spoiled this contest very badly)

I like your confidence.

Oh god! I'm getting nervous for A and it's setting up or spoiling my mood for the rest of contest based on whether pretests of it are passed or not.

Free Advice: 30mins left in contest, even if you start solving now, if you could do D you'll be on top 200 :)

WOW!, Contest Announcement itself is neat, clean and precise. Looking forward for my first MIDNIGHT contest

nope, it's not reflected in todays contest!

swaping 1 and 2 is sufficient ans = 1&2 = 0

3 6 5 6 3

a = 2 1 2 a' = 2 1 2 (reversed)

second and third element form LIS edit: Boy but you got it in the Contest? Hopefully it's only due to luck

A should have been little easier :( I wasted more than 20 mins thinking there should be easier solution for A by having right one in my mind. Lesson learnt: Solve what comes to your mind first for A atleast

I didn't use DP. Refer it if it passes system test :|

Hopefully, I pass system test for all questions for which my pretests are passed. Last two times it didn't :(

Thank you! Can you also please suggest other efficient ways to do it with same logic(like checking if all any one of columns or rows are filled) if possible. Or is there any other logic for this

Can someone see my solution for C I find nothing time consuming here. Can any minor changes make it acceptable.

what if both zeros are equally far from both sides :(

you could just avoid div 4 and participate in div 2 and 3. Why leaving codeforces all together?