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Hello! Codeforces Round 653 (Div. 3) will start at Jun/28/2020 17:35 (Moscow time). You will be offered 6 or 7 problems (or 8) with expected difficulties to compose an interesting competition for participants with ratings up to 1600. However, all of you who wish to take part and have rating 1600 or higher, can register for the round unofficially.

The round will be hosted by rules of educational rounds (extended ACM-ICPC). Thus, during the round, solutions will be judged on preliminary tests, and after the round it will be a 12-hour phase of open hacks. I tried to make strong tests — just like you will be upset if many solutions fail after the contest is over.

You will be given 6 or 7 (or 8) problems and 2 hours to solve them.

Note that **the penalty** for the wrong submission in this round (and the following Div. 3 rounds) is **10 minutes**.

Remember that only the *trusted participants of the third division* will be included in the official standings table. As it is written by link, this is a compulsory measure for combating unsporting behavior. To qualify as a *trusted participants of the third division*, you must:

- take part in at least two rated rounds (and solve at least one problem in each of them),
- do not have a point of 1900 or higher in the rating.

**Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you.**

Thanks to MikeMirzayanov for the platform, help with ideas for problems and for coordination of my work. Thanks to my good friends Daria nooinenoojno Stepanova, Mikhail awoo Piklyaev, Maksim Neon Mescheryakov and Ivan BledDest Androsov for help in round preparation and testing the round. Also thanks to Artem Rox Plotkin and Dmitrii _overrated_ Umnov for the discussion of ideas and testing the round!

Good luck!

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**UPD**: Also thanks to ma_da_fa_ka for testing the round and special thanks to Dmitrii _overrated_ Umnov, Artem Rox Plotkin and, of course, Mike MikeMirzayanov Mirzayanov for discussing ideas and great help with round preparaion!

**UPD2**: Editorial is published!

vovuh orz!

This is missing from the announcement after everyone thought our man vovuh retired...

You know I'm back like I never left (I never left)Another sprint, another step (another step)Another day, another breath (another breath)Been chasing dreams, but I never slept (I never slept)This has been happening in the last couple of months :)

Edit: Ok. Only vovuh fans missed him And I am one of them.

Would love to know what vovuh feels after reading such cringey comments/memes.

I also want to know, but i think i have not enough ability to get attention of him. Cause he has so many fans like me.

div3 by vovuh are best

Deleted

what if i asked who is this vovuh and why is he referred here ..so many times

Deleted

I am not able to register for the contest. When I click on the CONTESTS options above, I am able to see only ICPC challenge 2020 registration option.

Go to this Link or wait for ending ICPC Challenge 2020 then can easily register from contest option..

I don't see any difference between vovuh problems/non-vovuh problems.

Its because you have not participated in any contests and had less than 10 submissions....How will you differentiate then???

.

I cannot believe what you said.

.

Vovuh orz!!!!!!!!!! vovuh is back baby!!!! Missed you Vovuh ❤❤

Spoiler.

I think it's notorious coincidence

Cause both of you copied those From above 3 comments.

Starting RN I'm gonna explain my family members to keep quiet during 8:00 — 10:00 Tomorrow ! Maybe they'll understand this time

Mine will probably start bugging me on purpose if I say that.

vovuh : "The man the myth the legend !"

Div-3 and vovuh perfect combination.

Finally vovuh is back :)

Try some new idea :))

Like

This comment has been deleted due to negative feedback!

I don't know about this man vovuh but going through these comments makes me feel there's a good contest ahead ! Wishing it's true ... All the best to all participants

So, was it true?

I was stuck at E and I feel Yes is the answer from me

meanwhile vovuh counting all the money he made setting div 3 contests!

MikeMirzayanov No more Div.4 rounds?

Conspiracy theory :p

I can nearly always solve div2 C but I can't solve div3 D should div3 D be harder than div2 C ?

Stupid comparison

I compare div4B and div1B. It's not stupid.

u r stupid

Mind the color gapI will be orange soon!!! and this stupid community will upvote me for my shittiest comment these greys and greens used to cry and downvote instead of solving problems!!

no u won't

:) ok

You surely will.. Apes together ... Strong.

why it's stupid div1 A is div2 C so div3 D shouldn't be harder than div2 C but I guess it is

What's your feeling today can solved not only D also E1 .

I didn't solve div 3 D since like a year this time it's easier maybe

Does Div-4 contests are discontinued?

Cringe overloaded

"It wasn't important"

pls can someone tell me what happens in hacks ?

Checkout this blog post

wait, doesn't the hacker also get 100 points or something?

In div3 and educational no but in div2,div1,global yes

not in div 3

I know here is not a good place to ask this question but what happened for div4 contests ??

are they removed from codeforces contests or we can see them back soon ??

as I know you are a candidate (maybe I am wrong)

I think you don't need Div. 4 contests :|

Conspiracy theory

It is high time to introduce filters to comments section on codeforces -_-.

Harder than usual div 3 :)

In my opinion no. A-E1 obvious, E2, F — impossible.

Why do you have Wrong Submissions on obvious questions?

Jokes aside, obviously, if you know a concept or saw some observation, it is easy for you. Don't discourage others.

lol you are grey... there is no problem that obvious for you

And I didn't claim any problem is either?

why u guys friending my alt acc for div3/div4 rounds?

The gap between E1 and E2??

Help me with D after the contest...

D problem was giving TLE when i used unordered_map and when i changed it to map it got accepted. You just have to calculate the minimum number which we have to add to the numbers of the array after which they are divisible by k.

This can be calculated by taking mod.

for example the numbers are 5,1,3,4 and value of k = 3

so for the first element we need to add 1 to get the divisible number

this number can be calculated by taking mod of a[i]-k & k.

-> (k + (a[i]-k)%k)%k;

now similarly for other numbers also we'll calculate in the same way so -> 1,2,0,2 is the required array.

Now Notice that at a time we can select only one element from the above calculated array. It means that when we see 2 or more elements of same type we can't give the same x to them.

It means we have to calculate the next greater number which must be added so that the given number is divisible by k.

the next number can be calculated by multiplying k with the number of times this number has previously occurred, which means we have to maintain the hash table for it.

Wow, just tried this and confirm the TLE with unordered_map. Does anyone know why this happens?

unordered_map worst case is linear, so you can assume that your algorithm time complexity is O(n * n) in some cases

don't use unordered_map when it's not needed and you can solve the problem with map :)

same happened with me in problem D

My approach:

for all the array elements calculate the value required to make it a number divisible by k and store it in map and keep incrementing its value if it repeats.Then, find the max "value" in the key-value pair and ans= key+(value-1)*k+1. If two keys have the max value, take the max key.

eg: second example, 5 repeats 3 times. so ans= 5+(3-1)*6+1=18.

Can you explain why this works ?

To make

`a`

such that`a%k==0`

you have to increase`a`

by`k-(a%k)`

. Now if there is another element with the value of`a`

, minimum increment to make`a%k==0`

is`k-(a%k)`

. But according to statement we can not choose multiple indices to increment by same x. So the next minimum value is`k-(a%k)+k`

. If there is another element with the value of`a`

then we have to increment it by`k-(a%k)+k+k`

. So generally,`required minimum increment for a value 'a' = k-(a%k) + (frequency of a - 1)*k`

Maximum of these value for each distinct element is the optimal answer. (As values of other element are less than the answer we can increment them accordingly on the way to the answer )

I think your this approach might not work.

Consider the case 2 6 3 9 for 3 the minimum required increment = 3 for 9 the minimum required increment = 3

And max of (3, 3) will give 3.

But the answer here would be 9

Yes. I'm sorry i forgot to mention a really important thing. when calculating the frequency of each element we must take the mod of them. Because as in your case the increment needed for both 3 and 9 is 3. As we are focusing mainly on the x (increment) 3 and 9 is same for us. (

`3%6 = 9%6 = 3`

).Then the answer would be,

`6 - 3 + (2 - 1) * 6 = 9`

https://mirror.codeforces.com/contest/1374/submission/85372487

Isn't this linear ... why it's giving TLE?

Use map instead and then try your luck

Can you please tell me the logic behind this?

My code gave TLE when I used unordered_map but got accepted when I used map.

Because worst case time complexity for unordered map is in order of n.

I am confused about where to use map and where to use unordered_map

Always use map, never use unordered_map.

Or if you wish to use unordered_map, you can try this custom hash

I think these would easily be div2 E/F.

raj1307

Now I see why vovuh rounds are so amazing. Thoroughly enjoyed the round!

What Exactly you find amazing in them??

Easy solution to E2: Copy a solution to this problem https://mirror.codeforces.com/contest/799/problem/E which is the exact same except you don't have to reconstruct the indices. Then add boring code to reconstruct the indices

nice, 2500 problem for div3 contestants

I tried that but I was failing at test 12. Well if this worked for you then I definitely made some really stupid mistake.

https://ideone.com/oNqhoj

My code is obviously wrong (fails sample test case 2 lol).. But I spent more then an hour can't find where my logic is wrong :/.. I'm almost convinced my 24 IS the correct answer haha..

Any help is appreciated. (Would be awesome if you don't spoil the problem but hint me where I went wrong)

Try this test case

Edit: answer should be 70. Your code gives 108.

Ty! I finally was able to get rid of WA but then I was struck by TLE :(

https://mirror.codeforces.com/contest/1374/submission/85394497

Could you identify my bottleneck ? I think the solution is nlog(n) and should pass ..?

Your code has worst case complexity of O(n^2), having worst case when all elements are same.

See this part in your code

For each iteration of inner loop, it will search multiset the number of times value of

`arr[i]`

has appeared before in the array. if`n`

same elements have appeared before, then inner loop runs`n`

times, giving complexity of O(n) for inner loop. The outer loop runs`n`

times, giving total complexity of O(n^2).Hints for F?

Just simulate. If it does not end in correct ordering, then do one triple swap including two same elements from a[]. The simulate again. If still not correct ordering it is impossible.

Thanks for the solution! I thought of this solution, but I am still figuring how to prove that it will be impossible, in case both simulations fail.

Seems that it does not work good :/

Here is the submission of #1 85347474 He does count invariants, and if odd parity, does one swap in two same elements.

Then sorts with triple swaps, but including the indices in comparison, which does not change the order of elements, but makes the parity even.

The idea behind proving whether it is possible/impossible is based on the following idea:

Define f(P) for some permutation P to be the number of pairs of indices (i, j) such that i and j are the 'wrong way round' — that is, i < j but in the permutation p, j is to the left of i.

For example if n = 4 and P = {1, 2, 3, 4} then f(P) = 0. f({2, 1, 3, 4}) = 1 and f({4, 3, 2, 1}) = 6 (everything is the wrong way round).

We claim that each cyclic shift is invariant mod 2 for f(P) — that is, f(P) will change by -2, 0 or 2 with each shift. This is because if [a, b, c] --> [c, a, b] then the pairs (a, c) and (b, c) have reversed (now a, b are to the right of c when before they were to the left). So since there are two changes, then f(P) goes up by 2, stays constant or down by 2.

This also explains why we need to do two simulations — if two elements in the original array are the same, then we label them (x, x') in one simulation and (x', x) in the other simulation. Then for the starting configuration, f(P) will be odd for one of them and even for the other, and at least one of them should work.

From this we can conclude that it's impossible if both below conditions are met:

Got it! Thanks a lot for the wonderful explanation.

Really enjoyed this round. hail vovuh

What's the limit for compilation time? I submitted my code for E2 nearly the end of contest and got this verdict. "Can't compile file: Compilation process timed out." I guessed the reason is that my code is too long, but I don't think I can make it significantly shorter. Do you know how to make it compile?

85381670

I don't think your code is too long xD https://mirror.codeforces.com/contest/1373/submission/85026680

We don't specify the exact time limit for compilation, but your solution can't be compiled on my laptop in 30 seconds or even in 1 minute. Probably, it is a bug of GCC or how your solution is written. Try to fix it and submit again.

I manage to make it compile now. Thanks for reply! The reason is I wrote this in my segment tree code.

erase "= {}" part fix the issue. I write this since I think it can prevent UB sometimes, probably it causes me UB this time...

Can someone tell me where my code is wrong for E1? https://mirror.codeforces.com/contest/1374/submission/85384974

what if they only read from both books. what would be your answer, you will print INT_MAX which is wrong

so add this to your code

Can you give an example test case? I'm not sure I understand what you are saying because the loop would take care of the case where they read from both books (when i = k)

check out this case, you code will produce negative value. because $$$k - i$$$ is negative and you index the vector with negative indices.

Spoilerwhat if both is empty?

sorry, i misread it

Please a fast editorial , i need to upsolve

What is the test 4 of problem E like ?

I tried to use priority_queue a(only a like), b(only b like), c(both a and b like)

Then with ca(number of chosen a), cb(number of chose b)

If (ca < k), I will take min if exist (a.top() and c.top())

If (cb < k), I will take min if exist (b.top() and c.top())

If (c.top()) is selected I will increase both (ca) and (cb) else I only increase 1 (a.top() -> ca and b.top() -> cb)

Even I did the same priority_queue thing. But got stuck at test case 5.

Check this out: https://mirror.codeforces.com/contest/1374/submission/85356629

I did the same with priority queue:)

I failed at tc 4 too with priority_queue too because i forgot to convert it into the ascending one.

E2 would have been

~~better~~easier if it didn't have to traceback indexes.If you really think so, then try this

Hi! Thank you so much for sharing the problem! It feels good to know my approach for E2 was correct!

Though had to add a few extra steps because of coordinate compression.

Here's my submission for 799-E -- 85410030 (solved using two BITs , same as E2)

Although my thought still remains, tracing back through BIT would have sucked :p

Thanks!

What a huge implemention on E1.

Three pointers and so many if-else, I am about to not to solve it. Luckily, before the end of the contest I finally solved it by huge implemention :), hope not to be FST.

I think instead of using 3 pointers, you can just iterate over how many books are you taking which Alice and Bob both like. Then it is obvious that you have to take k — (both likes) books from 1 0 and 0 1. Just precalculate the prefix sums.

I have done the same but it gives WA on test case 5. can you please check it 85387758

upd : missed case (0,0)

Instead of

`else b.push_back(t)`

, you have to use`else if(B==1) b.push_back(t)`

Because you are also pushing 0 0 values into b, where no one likes the book. Instead, you should push 0 1 values into b, where Bob likes the book.

I slightly changed your code according to the above logic. Check 85389515.

Yup, there are many ways to solve E1 with easy implementation. You can even just take the minimum number of commonly liked books initially, then keep replacing the two longest individually liked selections with the shortest one commonly liked one remaining if it makes the answer better.

You could have implemented it directly using priority queue, one for each of the possible cases.

Then greedily go through the input :)

Hi! I think the approach for E1 was fairly straighforward. (maybe)

I stored time for

(0,1),(1,0)and(1,1)in different arrays and sorted them by non decreasing order of time.Also we don't need to store (0,0)

(why?)Say we can take

xamount of (1,1) then we know we have to takek-xfrom (0,1) and (1,0).Let's iterate over all possible case of x and the minimum possible time is answer.

Here's my submission : 85342055

Thanks!

Thanks. Best solution for me at least.

Tried doing something different but on the same lines

Took Three separate vectors for (0,1) , (1,0) and (1,1) , and sorted them

Then took a variable which would store the answer , Now

Iterate over three vectors at the same time , keeping variable i for 1,1 and j for 0,1 and 1,0

if time amount of (1,1) at i is greater than time amount of ((0,1) + (1,0)) at j then add (0,1) + (1,0) at j and j++ else add (1,1) and i++

But getting WA on test 8

Here's my code 85373064

My implemention is:

1. Can (1,1) replace (0,1)+(1,0)?

(1) (1,1) <=(0,1)+(1,0)

(2) One of them run out of their "own like" set which means the book only he likes, but he still needs to read book, so just put (1,1) dont need to check anything.

2. Can (1,1) replace (0,1) or (1,0)

In this situation, one of them has got k books to read but the other is not. Just check (1,1)<=(0,1) or(1,1)<=(1,0), then just choose (1,1)

3. Otherwise we just choose the book in their "own like" set

PS: we also have to check whether the set (1,1) has run out, if so we just go to 3, and get the number of books we need, because we have checked the exist of answer before, which is put all the referenced books in the shared set and check whether both of them have k books to read.

Thanks for the answer , but I dont think you saw my submission

There was only one typo in a loop (wrote i instead of i1) and Nothing wrong with the logic , got AC

Iam not completely sure if I understand your 2nd point though

Thanks for the effort of commenting though

I have also done similar thing. Here's my video Editorial for E1 Video Link

You overkilled it.

https://mirror.codeforces.com/contest/1374/submission/85352661

How to Solve F ?

Why am I getting TLE on test case 8? 85389867

may be your while(1) loop run for long time(>2000ms)

Even now I'm getting TLE. I've tried using very large random test cases. It works for them, but I can't figure out what's causing TLE here

try with map instead of unordered_map

Try using map instead of unordered_map. unordered_map takes O(n) time to search in the worst case but map takes O(log n)

Yeah, works now! :(

D: Could someone please tell why this is giving TLE?

http://mirror.codeforces.com/contest/1374/submission/85383846

inline bool c(ll a, ll b) { return (a >= b); }

int main() { ll t; scanf("%lld", &t); while(t--) { ll n, k; scanf("%lld %lld", &n, &k);

} }

Try using map instead of unordered_map. Because unordered map has a worst case complexity of O(n), your code has a worst case complexity of O(n^2).

Wohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh, sorry to spam, it blew my mind, Thanks a lot sir, could you please tell when to use map or unordered_map?

PS: I got accepted

http://mirror.codeforces.com/contest/1374/submission/85388466

In my experience, always use map instead of unordered map, since it gives you a complexity of O(lgn) every time.

Also you would like to read this: https://mirror.codeforces.com/blog/entry/62393

It uses a custom hash, to avoid getting TLE when using unordered map :)

thank you sir :)

Didn't realise this stupidity of mine. Worked around it by using

`reserve()`

upto`1e4`

. I guess it will be hacked.I am so stupid I thought unordered_map is obviously faster than map always :|

I see C easier than A ;) :v

Can you explain me C I feel so stupid :(

Here check out my video solution. It is time stamped so just skip to C.

https://mirror.codeforces.com/contest/1374/submission/85372487

I think this solution is linear. Anyone explain the reason for TLE?

use map instead of unordered_map

I was thinking that unordered_map is faster than map? Can anyone make me clear?

Check this out https://mirror.codeforces.com/blog/entry/79433?#comment-652083

unordered map can take anywhere between o(1) to o(n). while map takes o(logn) always.

Codeusing map instead of unordered_map, got it Accepted, But could anyone pls explain why did it happen?

Don't use unordered_map or unordered_set they blow up for certain inputs , if you want a detailed explanation read this blog

What's wrong with this code. (Problem D) It gives MLE at test case 5. Thanks in advance.

because you create many map keys. For Example fre = {{1, 4}, {100000, 3}}, k = 1000000 You are created keys 2, 3, 4 .... 1000000. if(fre[num] > 0) created this keys

Thanks got it.

You must refer to the next element. By the way, such a solution will not work even in time. use binary search to find the element closest to the current element x. After it has reached the required value. So the asymptotic will become nlog (n), not kn

How to solve D if the first operation can be performed any number of times on each $$$i$$$.

You mean [k = 3, a = {2, 2, 2, 2}]?

I mean let's say $$$k$$$ = 4 and $$$a$$$ = $$$[1,4,8]$$$. In original problem answer would be 4 because we can perform operation $$$a_i = a_i + x$$$ on a particular $$$i$$$ at most once. How to solve it if we can perform it any number of times. Like in this case answer would be 3 as we can perform $$$a_1 = a_1 + 1$$$ and $$$a_1 = a_1 + 2$$$.

If x is answer to the current question, then an O(x).k solution can be easily found for the modified question. We maintain a frequency array of size k, to store the remainders and another frequency array of size k to store currently available remainders . While processing x, we can check if it can make a remainder which is needed, and we can modify accordingly.

I initially didn't read that line and was getting hopeless. I mean there won't be any fast algorithm, I have a felling that there won't be any polynomial time algorithm

My A, B, C, D, E1 solution`s

A: 85388328

B: 85388390

C: 85388427

D: 85388457

E1:85388485

If you have questions regarding the logic of the solution or the principle of the code — write here

Why am I getting TLE in this submission for Question D?My Submission

Because worst case time complexity of unordered map in order of n . Solve it by using map

I didnt fully understand the reason, but its in unordered_map. Use the common map and it will work. I will now try to understand why you cannot use unordered_map.

P.S. I understood the reason. In some cases, unordered_map works for a long time

In E1, is it always necessary to take A and B in pair?

Yes. Since we write the values in set, this will be the most profitable solution.

Tutorial of problem C:First, solve this, then assume your output is $$$ans$$$. Now, back to C, and print $$$(n-ans)/2$$$ for each testcase. I wish I've made a clear solution and stay tuned for my YouTube channel :'DCan someone help me in finding bug in my E2(Hard Version) solution?

Here is my solution 85387601

It is failing in 12th test case. Thanks in advance.

Our solution seems to have the same idea. Even my solution is failing on test 12, with the same answer. If someone could help it would be great, thanks

Can anyone please tell me what's wrong with my solution in E2??

giving RE on test case 5.

https://mirror.codeforces.com/submissions/Lord_Saga

Next time use spoiler or link to share code.

I like this comment

`//FIRST THINK THEN CODE.`

So big comment,hard to read for every-one. Edit it using spoiler & block option or give the code link..

Unordered map can run in O(n) per query in the worst case if there are enough collisions. It is trivial to generate cases that cause such behavior for unordered map without custom hashes and would make the solution run in O(t * n ^ 2).

CF blog that discusses this.

"Definitely , There is a big scandal behind all this"Maybe in your mind.

unordered maps are not always faster than maps, as ordered maps can take anywhere between o(1) to o(n) for indexing while maps always take o(logn).

So, How come don' I see any attempt at this contest in your account?

what a bad day for me, my rating continously decreasing from past 4-5 contest and my current rating is just 1615 and this contest is unrated for me in which my rank would be in top 10. hope this would never happened to anyone

is vovuh from BTS?

I'm going to assume that my solution for problem E2 using treap is not the intended one, because that would be very weird for a div3 contest. But I usually try to kill a mosquito with a bazooka during contests, so I'm sure that there is a simpler and nicer way to solve the problem :P

Can you please explain your treap solution?

Fix the amount of books of type 1-1 that you will take. You can calculate how many books of type 1-0 and 0-1 you will need to force that both people have at least k books. Obviously, you will take the ones that have lower cost for each group of books. This works for E1.

Now for E2, you should keep the same strategy, but maybe you have to use a few more books to complete the M books you need to take. But both people already have at least K books, so you can take any other book you want (even if it is a 0-0 book). Obviously, you will take the smallest amount of books you need (let's say X). Here is where treap comes up to help us: you can use it to calculate the sum of the first X smallest values in the remaining books that you didn't take before.

but why didn't u implement it??????

You can see my submission during contest here :)

If you are looking for a clean implementation, don't even waste time with my code. I was on a hurry because the contest was reaching its end, so I had to code everything quickly :P

But how do you save the indices of solution set of books? I could calculate the minimum answer using set, but couldn't figure how to find the indices of final answer.

You can keep track of the number of 1-1 books in the optimal solution, sort the unchosen books, and take the smallest m-k books, print them along with the chosen books

Can someone explain how to solve E2? I saw some people using treap to solve the problem.

I think the main part of this problem is how to find k-smallest elements sum. Usually you can do this by binary search tree, so treap can do the job. The easiest way to implement for me is to use segment tree that each node store pair of (number of element, sub of element) in subtree. To get sum of k-smallest elements. Check if left node size greater than k or not. If yes, try searching on left subtree, otherwise subtract sum of left subtree and search on right subtree. Fenwick tree can also do the job if you keep two fenwick trees that one store sum and one store number of elements and use binary lifting or binary search to find sum of k-smallest elements.

I saw many people solve using priority_queue or multiset, but I don't know how it works though.

Edit: I think I get it now. It’s similar to finding median by priority_queue or multiset. Notice that k is always increasing. So you can simply do the same thing. When you encounter query, just transfer elements between 2 multiset respected to value of k. The total number of operations will still be $$$O(n)$$$

Thanks for this contest! Through this contest, I realized that my implementation skill is very very weaker than others In F, I observed the main idea very quickly, but that is all.. I failed to write accurate code...

can you tell your idea for solving F?

For(int e=1;e<=n-2;e++), we find the Farthest minimum value that located in [e~n]. And located in e After that, if the array is sorted, finished. but if not, last two number is not sorted. so in this case, we find two same value in [1~(n-2)] and maximum in last two number is going to that ( we can go either of two same value ) then we only go fowrad one move -> then maximum value can go to last of array

thanks, I was thinking about the same idea

Your code for E2 is a mess!

my code?

No, TaeHo0o00o0N's.

i can't solve E2

he didn't solve E2

I also face the same issue, for questions rated above 1500-1600. Can anyone provide some tips?

practice question of difficulty :1600 or above.

look at the code of more experienced people. They usually know how to make it in a simple way.

https://mirror.codeforces.com/contest/1374/submission/85392451 its showing TLE for using map also. pls someone find the fault.

K can be up to $$$10^{9}$$$

ok yaa. thanks

Hey, You are running a for loop from 1 to k (1 <= k <= 10^9) This is the reason for TLE. Instead of going through all the numbers from 1 to k just check for those which are present in the map.

ya got it. Thanks.

Can anyone tell me why I am getting this runtime error in TC 5 for E1 ?

Link for submission — https://mirror.codeforces.com/contest/1374/submission/85375093

I am pretty sure this has to do something with my comparator function for sort but I cannot figure it out.

This might be what you need.

Can someone help me with D? I am not able to understand what's wrong with my code. Main logic below

`lli == long long int`

You should skip elements with

`diff = 0`

.`diff=0`

when`(k - x%k)`

. I don't think it will ever be 0Consider the case $$$x \bmod k = 0$$$.

That won't be because the code will not go to that line then. I have checked it in the line above.

Ops, you're right.

Seems that the mistake is in:

You can do this instead:

The reason is that if

`mp[diff] > mx`

and`diff < mxval`

you are not updating the`mxval`

but you must do it.Can you explain why I should update

`mxval`

when`mp[diff] > mx and diff < mxval`

. Cause`mxval`

should also be 'max' right? so should be update when`mp[diff] >= mx and diff >= mxval`

You need to take the maximum

`mp[diff]`

no matter what. Then, over all the`diff`

which have the maximum value`mp[diff]`

you should take also the maximum.Consider this case:

Correct output: 5.

That is because the correct results are

`mx = 2`

and`mxval = 1`

, but in your code it gives`mx = 2`

and`mxval = 2`

which is incorrect. If you still don't get it just do the simulation by hand.Here is my screencast of this contest, do check out it :)

Link: https://youtu.be/afn_V7YkX3U

On my channel I do educational content, so if you want to improve, make sure to subscribe to the channel!

Can Anybody explain me C please

Keep a counter, initialised to zero. Traverse through the string now, whenever you encounter ( counter++, otherwise counter--

Now if the counter becomes negative at any moment l, it means that you have a extra closing bracket, in this situation move it to the end of the string. This approach will ensure that every opening bracket is matched to some closing bracket. And also set the counter to 0.

Suppose you have variable $$$bal$$$ which stands for balance. We go through string from left to right, if we have $$$"("$$$ now then do $$$bal+=1$$$ and $$$bal-=1$$$ if we have $$$")"$$$. The answer is the global minimum value of $$$bal$$$. For example $$$s=)))((((())$$$, then balance will be $$$-1, -2, -3, -2, -1, 0, 1, 2, 1, 0$$$. The most minimal value of $$$bal$$$ was -3, thus the answer is 3. So you can fix the sequence by 3 moves. You can take 3 rightmost opening brackets and move them to the front of the string. So string $$$)))((((())$$$ will become $$$((()))(())$$$.

Hey so will the deformation always occurs in the form of )))(((, can ) or ( appear in between balanced sequences. Cause I'm unable to concretely prove it. Thanks!

not always, you can get string like $$$)())())())(((($$$ answer is 4, and you take last 4 opening brackets and move them to the beginning, you will get $$$(((()())())())$$$

How to solve problem D ?

Keep a map of values that need to be added to each element to make it divisible.Now iterate over the map and if some key has multiple count,increase it by multiples of K.Now, the maximum key will be the answer. 85341803

Here check my video editorial

Kindly explain the statement of problem E1.

Video Editorial for E1 Video Link

using min_heap (priority_queue) for D seemed more intuitive to me.

Submission

can anyone plz try to hack my first E1 submission i think it is wrong!!

upd-: never mind that solution is correct

well tried hacking it but upd it is correct

My Video solution for the contest

A,B,C The video is time stamped for your convenience (check the description).

D

E

As usual, video solutions to all problems are available at the end of my screencast of the round.

Thank you for the content. Simple explanations.

Can anyone give a proof for the solution to F?

What i used -> sort the first n — 2 part and then if the remaining 2 elements are not sorted then sort them using some element which has a duplicate. Im unable to prove why it works.

Proof at 2:17:50 here

The TLDR is that the parity of inversions doesn't change

Was just able to prove it. Amazing explanation thought. Keep up the good work!

have the people got the rating for this round anyone please help me

Ratings won't be updated until after the open hacking period is over.

In E2, I made 4 arrays, 1st is "common books", second is "books interesting Alice only", third is "books interesting Bob only" and fourth is "books both are not interested in". Then I sorted them and took corresponding books from second and third array and added them to form pairs of books that interest either one of them. After both have read 'k' books of their interest, I am checking if a total of 'm' books have been read.

Let 'num' be the books read till now.

My code is failing on test 12, and with furthur debugging, I found that this case is where num<=m. So now I am taking all the remaining books(common,interest for alice only,interest for bob only and books none of them are interested in) and sorting them by time. after this I am taking the first (m-num) books to get the final ans.

Link to the code

Please can someone help me in debugging. Thanks

That approach isn't necessarily correct. You might want to remove a book interesting to both people and replace it with two books, one interesting to each person.

at each step, I am checking if the common book has a lower time or the combined time of 2 books each interesting to one person. This approach worked for me in E1.

Imagine this testcase for E2:

4 2 1

3 1 0

4 0 1

5 0 0

6 1 1

Your approach probably will print 9 (1, 4) as 6 < 3 + 4 but the answer is 7 (1,2).

thanks for the testcase

Why map can ac

unordered_map time out problem D:

Please consider using spoiler when you want to include code.

Hello, I am stuck at this solution as to why it is giving MLE on TC 5. I am just storing the remainders in a map. Please can anyone help?

I can see that it should give TLE but why MLE?? Submission[submission:85385727]

I didn't look carefully, but if you write mp[j] and j is not a key in the map, then j will be added to the map as a key with value 0. If you don't want to add a key you can write if(mp.count(j)) instead of if(mp[j] > 0).

.

For

E1, why am I getting WA onTest case 5? 85405609My approach was that I iterated on the number of books which they should read, such that both of them liked that book. For the rest of the books, I chose greedily.

UPD:Got it fixed :)I think it was integer overflow

I have been hacked on problem A. Here is the submission. https://mirror.codeforces.com/contest/1374/submission/85298617

This is the first time I have been hacked. It says I got TLE. But my program is O(1). So I am certain that it is right. Can someone please tell me 1) If my program is wrong. 2) If the hack was judged successful incorrectly.

Please help me. This was my best performance. I will drop 3000 ranks due to this incorrect (in my opinion) hack.

Edit: My solution is definitely right. I had to switch from input() to sys.stdin.readline() for it to not be on the brim of TLE.

Note that your solution is fairly slow, the test shows

`Time: 982 ms`

So I assume the hack also used 50000 testcases, but with bigger input numbers. Which is enough to put it over 1000ms.

Your solution is logically correct but not fast enough. Python is a slower language and your solution only just passed the pretests taking 982 out of 1000 ms.

There are things you can do to speed it up without switching from python:

-150ms by removing the unused imports at the top.

-200ms by buffering the output instead of printing every line (eg 85407621 takes 592ms).

By the way, do you know any way to disable flush on a new line inside print/stdout in python? Couldn't find any From what I see the line on a terminal is always flushed when there is "\n".

To clarify. In Python, printing does not flush stdout. It is the call to input that flushes stdout. The implementation of

`input`

is that it first flushes and then calls sys.stdin.readline. So to not flush, simply use something like`input = sys.stdin.readline`

. (tested on CPython3, PyPy2 and PyPy3) .Here is a code if you want to play around with what flushes stdout.

Detect flushingBtw, just so you know, the built in IO especially in PyPy is pretty slow. During the competition I used sys.stdin.readline and it ran in

`654 ms`

85296367. Most of the time is spent on IO. Doing IO smarter makes it go down to`140 ms`

85416929.A while back I made a drop in FastIO for PyPy2 and PyPy3 that fixes these issues of slow built in IO (found https://github.com/cheran-senthil/PyRival). With it, my code runs in

`155 ms`

85417276. anoubhav's code runs in`202 ms`

85417590.Thanks I tried to determine the flush by observing how characters appear on a terminal and they always appear after "\n", but I guess I misunderstood the concept of flushing.

After giving 15 contests I can say that Yes, Ashish Gupta's round has fastest editorial . Who else agree with me?

The editorials of Div 3 rounds are given after the hacking phase.

15 contest and yet you don't know where are you stand.

Can you tell ne where I said about the Div 3 editorial, and it's the fact that Ashish Gupta's round have fastest editorial in comparison to other Div 2 contest.

You literally commented under a Div 3 announcement.

I think hacking phase is over now and there aren't any editorial till now

Is there any other way to do E1 instead of priority_queues and a lot of if-elses?

Look at this

https://mirror.codeforces.com/contest/1374/submission/85393512

Video Editorial for E1 Video Link

In question D, I used unordered_map, why it is giving TLE and Now I just changed unordered_map to map and my solution is accepted. Can anybody clarify this?

This is my code for reference.

Sometimes, the complexity in unordered map goes to O(n). However, you can define a good custom hash so that you do not blow up because of unordered map. Check the following link for reference:Link by author neal

Check out the editorials guys:

Problem E1

Problem D

Problem C

Problem A and B

tooo many if else in E1....not required... not a good solution

where is the editorial??

In Div-3 and educational round it always publish late.. I don't know the reason. But it's annoying.

Because there are open hacking phases in both.

What wrong happens if they publish editorial before the hacking phase done ???

There will be an unfair advantage to the people trying to hack. This is what someone said on this question.