Hello Codeforces!

On May/28/2020 17:35 (Moscow time) Educational Codeforces Round 88 (Rated for Div. 2) will start.

Series of Educational Rounds continue being held as Harbour.Space University initiative! You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.

This round will be **rated for the participants with rating lower than 2100**. It will be held on extended ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.

You will be given **6 or 7 problems** and **2 hours** to solve them.

The problems were invented and prepared by Roman Roms Glazov, Adilbek adedalic Dalabaev, Vladimir vovuh Petrov, Ivan BledDest Androsov, Maksim Neon Mescheryakov and me. Also huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.

Good luck to all the participants!

Our friends at Harbour.Space also have a message for you:

*Hey Codeforces!*

*As you may remember, we have a free webinar series starring our all-star faculty members who share valuable content and insiders’ knowledge that you don’t get to learn about in traditional classrooms.*

*Join us tomorrow, Thursday, May 28 at 12h (BCN) / 17h (BKK) to watch Sergey Gordeychik, CIO of the Inception Institute of Artificial Intelligence, share his expertise and insights in his “Digital Lockdown: AI against COVID-19” session. Sergey will discuss how AI is being used both positively and negatively during the COVID-19 global pandemic.* *Tune in for some practical examples of how companies are using AI to innovate and disrupt during a time of crisis, exploring topics like Medical Imaging for CT analysis, diagnosis and mass surveillance.*

*By participating in this webinar you will get a certificate of participation, a special digital gift from Sergey, and stand a chance to win a FREE 3-week module at Harbour.Space University depending on the availability and prerequisites of the course.*

*See you tomorrow, and good luck on your round!*

Congratulations to the winners:

Rank | Competitor | Problems Solved | Penalty |
---|---|---|---|

1 | 244mhq | 6 | 174 |

2 | bmerry | 6 | 219 |

3 | dlalswp25 | 6 | 233 |

4 | hepth | 6 | 238 |

5 | Volkov_Ivan | 6 | 251 |

Congratulations to the best hackers:

Rank | Competitor | Hack Count |
---|---|---|

1 | Hideki_Ryuga_L | 37 |

2 | KonaeAkira | 19:-1 |

3 | ujjwalsingh30 | 18:-1 |

4 | veteran_ | 14 |

5 | hashlib | 13 |

And finally people who were the first to solve each problem:

Problem | Competitor | Penalty |
---|---|---|

A | andryusha_na_knopke | 0:01 |

B | thech0sen1 | 0:03 |

C | IAKWF | 0:11 |

D | Kerim.K | 0:06 |

E | HeHere | 0:06 |

F | user202729_ | 0:44 |

**UPD:** Editorial is out

First comment! Thank you Codeforces and Harbour.Space University for these rounds in this tough time!

Don't know the reason of such hatred shown by the community here. I just thanked the organizing panel for their efforts, that's it, as a good gesture. No wonder why the community hated it so much. Maybe because I am a specialist and not a red, not really sure about it.

Don't worry about downvotes too much.

Your positivity will reach the people no matter what.

This is one big problem with cf. On one side, there is Mike who is bringing out new div4 contests for not so high rated participants and on other side are these guys who boast a lot about their ratings.

If everyone showed thanked the organizer in the comments, then we'd just have a thread of 19,000 people saying thank you. No one wants that, if you want to thank them give them an upvote.

Almost everyone is sad about this lock down (including me), and if these contests give hope to people or put a smile on their face or excite them, there's no harm in saying a "thank you" as a good gesture. And I think that it's nearly impossible for such a large number of people to post it in an announcement. It's just a few, which doesn't flood these announcements. In my case, I just got excited seeing the announcement and that there isn't any comment yet, so I thought why not write a good thing for the organizers, there's nothing bad in it. It gives hope (or just puts a smile) to some, if not all.

I don't want to hurt anyone with my comments, and if I did, I am sorry about it.

Hey, don't worry both of the groups supporting/disagreeing with your comment are right in their point of view. The first group that supported you was for your pleasant thanksgiving and the other group that disagreed on you was because they want that more important messages/helpful messages are to be shown so that it would give way to more important messages in a bunch of good messages. That thanksgiving gesture was very pleasant hope it will spread positivity in this environment.

I guess you've been downvoted for your 'First comment'. That's totally useless.

Where's the unusual time ??

It's unusual that it's not unusual.

Whoops, you saw nothing

Nice ...

Your rating curve is inspiration to many

I like to watch other people's rating graph when cf predictor says mine is going to be -50

Well it's still unusual for some people!

But this is the standard time of CodeForces

Googleforces

Let's goooo!

Rude

Hope this educational round will be much better than previous one^_^

Looking forward to more geometry problems! Especially, ones with subtasks <3

Me too.

Lets hope we get to see problems involving both math and graph theory. It was long back when we had problems involving math as well as graph theory.

Let's hope your dream never comes true

Thanks for having back to back contests! Giving something to cheer for.

Looking for a data structures problem ..

I rarely see a Div2C or Div2D a graph theory problem I think in a while i have seen two only one livonia and kingdom and one from ehab and bla bla .... I am looking for good Problems rather than standard ones with unusual time limit and memory limit.

I hope to see Tree , Data structure , Math Problem But i don't want to see geometry .... i am very weak of that

everything is fair until it's not one click away on google. specially when it's prob A/B/C/D. because that makes contest unfair for 80% rated participants.

How did you calculate the percentage of the unfairness? Why not 0.4% or 5% or 15% or 146%? Please, provide some proof or stop speculating.

Anyway, if you deserve your rating you'll gain it in no time. Otherwise, if you need to google A/B/C/D in ER you probably don't have the rating you can be proud of.

Btw how did you generate such numbers .4%,5%,15%,146%. Any formulas for that or random. He is saying correct those who wasted time in solving that suffered. Googlers got Ac instantly. At least it should be not present on google

No one :

Literally no one :

People who get negative delta after the contest :

Educational codeforces rounds should be unrated.

I enjoy every contest even if I get negative delta. I don't know why people complain a lot with the rating! Everyone should enjoy contest even if you get negative or possitive delta! I think that's the authors purpose and what really matters!

Yes, even if there would be no rating system i would have given same or more number of contests. Its the number of questions i do make me feel good than positive or negative delta .

Why score distribution is not given in any educational round announcements?

Because all problem has equal score in educational rounds.

RUSH RUSH！！

[deleted]

More than 3 years and you're still a newbie. It's your fault, not due to the statement

I never wrote that I have lower ratings due to the problem statements. I just gave an opinion what I felt and that's it.

I think the best solution is just practice because reading is also an important skill

Who else is trying to solve their first div2 c problem

Not the first. But it's been a while since I solved my last C >.>

Problem C:This Problem Can be Solved Mathematically using inequalities.Ideahot cup = cold cup = k, we have temp of barrel = (h+c)/(2).hot cup!=cold cupthen let the no. ofhot cups=n, therefore no. of cold cups = n-1 (As we started pouring from hot cup). So, after n hot cup and n-1 cold cup the temperature of barrel = (n*h + (n-1)*c) / (n+n-1).Solutionf(n) = (n*h + (n-1)*c) / (2*n-1) , where n = no. of hot cup. This is a decreasing function and also bounded below by (h+c)/2 (As we increase n, denominator increases more rapidly as compared to numerator and also as n->inf , f(n)->(h+c)/2 or we can check using derivatives also) .t <= (h+c)/2it is optimal to only pour one hot and one cold cup. So answer in this case is2.(as our function cannot reach below or at (h+c)/2) .t>(h+c)/2, we can find smallest n for which function takes value <= t using thet >= (n*h + (n-1)*c) / (2*n-1)which givesn >= (t-c) / (2*t-h-c) .difference from tfor thisn(remember hot cups) and n-1 will be therefore cold cups (why?) and forn-1(hot cups because it will give the lowest value of function >= t ) and answer will be the2 * ( one which will give minimum difference ) — 1 .Link to my solution : 81821029

Woohooo Vovuh is back!

Hope there are no geometry problems!

sry, I LOVE MATH!

I love math...! :) Happy coding...

I guess should have enjoyed solving C question then?

It was just an inequality question.

why has no one yet asked whether the round is rated or not!!

It is already written in

bold, rated for participants with rating lower than 2100me .... literally after every "is it rated "comment in every contest

I feel like it's my first contest even if it's my 85th contest. Can you please tell me if the contest is rated or not.

XD

pretty clever username dude.

I give up Codeforces and leave this site because of problem C. I am 100% sure that my code is correct but I got 7 WA. I give up. Bye.

If you got WA then maybe, just maybe, it is not 100% correct..

You could instead try E which is easier than C

i place my trust on Ross over Chandler

just make sure he is not on break lol

If only i knew it...

I passed pretests on second attempt. Its a pen and paper problem. Looked like binary search first, but its O(1) solution with some messy algebra, theory of increasing/decreasing functions, case work, simplifying fractions.

Actually, C can be solved using binary search, I used it during the contest.

You can see my solution for the same -> https://codeforces.com/contest/1359/submission/81760272

Yes.. Very nice observation that it can be solved in O(1) using some mathematics. I just solved it after looking at your comment and turns out, we don't need any fancy mathematics. Just simple algebra works.

After that just check at values (x-1), x, and (x+1)

https://codeforces.com/contest/1359/submission/81874188

thank's , but letters not understand

please explain more

check this video for detail explanation C problem https://youtu.be/Ts6to-_N-Y0

can you tell why it's only (n*H + (n-1)*C) and not (n*C + (n-1)*H)

I figured out that expected precision = 1e-15, I got it accepted when i incurred that change

I tried to read your code and saw this: 81775456. No pity you leave, since you would be banned for intentional code disguise anyway.

what with your weird way of writing code. it's like cancer, you write code using define.

I've never laughed so much looking at code (except maybe ArnoldC). Now I have.

1937 : There will be flying cars in 2020

Meanwhile in 2020 : brainless pupils and Newbies exist.

Please change your profile picture, it irritates me every time I see it.

For fuck's sake, change the dp otherwise don't comment!

Codeforces should introduce

reportfeature.Speedforces!

Last time I was lucky to notice the pattern in C task which resulted in formula answer, this time C revenged! :D

Video Tutorial for today's C

A great contest with beautiful problems and short statements for which Educational rounds are generally known.

Can C be solved with the help of binary search ? If yes, how ??

For even number of cups, the resulting temperature is always the same.

For odd number of cups, the resulting temperature decreases as we add new cups, but never becomes less than or equal to $$$\frac{h + c}{2}$$$. So if $$$t > \frac{h + c}{2}$$$, we can use binary search to find the first moment it becomes less than $$$t$$$, and check this moment and the moment $$$2$$$ cups before.

Okay so i was doing everything right but failed to realize that i need to use binary search only for t>(h+c)/2. Changed that and accepted. Thanks !

I suppose I have a constant time solution.

You are not alone.

of course man..Its basic math and error analysis

just saw your solution, can you explain why exactly:

boo is the value so that the error is absolutely 0.. i is the floor of boo

Please correct me if I'm wrong, but consider $$$h=10$$$ and $$$c=1000000$$$

Wouldn't the average temperature for odd cups be strictly increasing in this case?

In the statement, it says: "1≤c<h≤10^6". The temperature of the hot cup must be higher/larger than the temperature of the cold cup.

Oh, thank you!

Considering the case 999977 17 499998, where hot temperature is 999977 and cold temperature is 17, target being 499998. Using 499979 pours, and, while using 499981 pours, we get the same difference with the target.

So shouldn't the answer be 499979. While in the solution, the answer is 499981. Can anyone let me know what I am missing here?

Try reading other comments. Comment

If we consider the case where we have $$$X + 1$$$ hot and $$$X$$$ cold, the temperature is a decreasing function, so binary search could work here. However you can also directly calculate the value of $$$X$$$ for which the intersection occurs and then check $$$\lfloor X \rfloor$$$ and $$$\lceil X \rceil $$$ for whichever gives a closer temperature.

Yes I did the same way by directly calculating the intersection value.

Can $$$ \lfloor X \rfloor $$$ and $$$ \lceil X \rceil $$$ and then temperatures at these points be calculated with good precision ?

You do not need to calculate them as double, you can calculate as integers directly: either [t / (h + c + 1 — 2 * t)] or 1 + [t / (h + c + 1 — 2 * t)] where / is integer division. even if X is an integer, it would not hurt to calculate temperature difference at X + 1

Yeah, did the same when I finally got an AC

How did you get this formula? As far as I know, it should be [(h-t)/(h+c-2t)]

I used double division followed by floor and passed fine. However you can calculate it exactly using integer division.

How did you derive the formula?

First wlog let $$$c = 0$$$. Let $$$x$$$ be the number of cold cups. Then for the odd case we want $$$h(x+1)/(2x+1) = t$$$. This intersection point only exists if $$$t/h >1/2$$$. Otherwise we take an even number of hot and cold cups. Now just check if the floor or ceil of $$$x$$$ gives a closer answer.

I did the same. I think I am having precision issues. Help please. https://codeforces.com/contest/1359/submission/81752804

I did the same but still got WA in the third case can somebody tell why? My submission for C- 81790933

same problem subcase 66 (getting 2 instead of 1).

You have most probably missed out on cases where the the difference between

handtis the smallest possible difference and therefore 1 should be the output.Tc:

1

7 1 6

Your code fails for this and other similar test cases the answer should be 1 and not 3.

here expected 1 found 3.

Seems I forgot to exclude equal case here 1 and 3 both have difference 1 but minimum was to be considered. Thanks

Yes. Notice that the more cups you take (assuming it's an odd number), the closer you will be to the average of hot and cold temperatures. Use this fact to binary-search for the smallest number of cups such that the temperature is not more than the target. To deal with any precision issues, check 2-3 numbers above and below what you found and select the best one.

Yes. I solved it with Binary Search.

However, my first few submissions for C got WA because of Double precision issues! If this hadn't happened, I would have reached master. :(

Check out my submission here: 81789311

LOL The rating predictor was wrong, you reached master :v

What is the testcase #4 of problem C ?

I just changed double to long double and it worked.

Do you know what is the solution in Python for this issue? Submission failed on Test case 4

I am failing on the test case 999977 17 499998

I tried using the 'Decimal' module in Python for more precision. But, I am still getting an error.

The Fraction class, fractions.Fraction worked for me (https://codeforces.com/contest/1359/submission/81820143) but only after the contest.

Testcase #4 of problem C is some testcase which causes Double precision issues to cause WA>

Yes, Fu**ing precision error in python got me that answer wrong and that question sucked soul out of me.

Yes, it sucked the soul out of my rating :(

I would have finished around rank 100 if I had solved C immediately without any errors...but instead I got 347 :(

Can relate. I wasn't able to overcome the precision error until after the time was over. :(

Yes, same problem. Discovered after the contest that the easiest way round this is to use the Python Fraction class (fractions.Fraction), see https://codeforces.com/contest/1359/submission/81820143. I wasn't aware of this class before, so I actually learnt something from this "Educational" round!

Not sure what one would do in other languages.

I was getting TLE on this test, so I think there is some stress-tests, for example:

30000

999999 2 500001

999997 2 500000

... and more

How to solve D?

for each element in array .. assume the current element as maximum for segment under consideration.

then find max_left_sum[i] and max_right_sum[i] (crucial part).

ans will be max(all(max_left_sum[i] + max_right_sum[i]))

can be done in O(N). once try for yourself, anything more i explained will be a spoiler.

Spoileri will show how to compute the max_left_sum array you can figure out the construction of max_right_sum array from this.

there are two for loops used. although you can do it single loop.

in the first for loop we are calculating for each number in the given array the maximum we can get until we reached some x >= current element.

In the second loop we are using the computation done some x in the left portion x == current element.(this step is to ensure that we achieve the O(N) complexity).

Once go through the code with an example you will get it.

this is another question which can be solved using similar technique — Largest Rectangle from Hackerrank

I overkilled D using 3 Segment Trees, that ultimately did exactly what you said...

Are you using Kandanes algorithm both side ? Can you please tell your solution since what you left is the hardest part . Your can put it as a spoiler .

I have updated the comment with spoiler section(with explanation and code) and a bonus similar problem.

Is the complexicity of for loop is n*m where m is no of different value that array element can take m=60 in this case..

Note that there is no need to check the negative values, since if the biggest value of the subarray is negative, it is better to use a single element subarray which results in sum of 0.

Yupp, that largest rectangle problem uses stack to get the next and previous greatest elements. I find stacks tricky hence used segment tree + binary search to find the next, prev greatest element

for a decreasing array, the complexity is O(n^2) in the first loop itself. This code will get TLE. right?? Or, please explain why the complexity should be O(n).

there is the condition for input -30<=ai<=30.

this is useful to maintain O(N) if the diversity increases this algorithm reached O(N^2) as you said.

Can we use stacks to solve this problem??

You can also use DP with dp[i][j] := "maximum sum of subarray a[?:i] with max value j" which has pretty trivial transitions, and then maximize dp[i][j]-j, looks like this: 81795283

How to solve F?

My idea:

Forget about starting at any time for a moment. Let all cars start moving at $$$t = 0$$$. For any $$$t$$$, Consider the line segments of each car whose endpoints are starting position of that car and position of that car at time $$$t$$$. Find the least $$$t$$$ for which there is an intersection of any two line segments. Then the answer is $$$t$$$, because if line segments corresponding to cars $$$A$$$ and $$$B$$$ intersect, then I can start one of them late and make them crash exactly at $$$t$$$. So binary search on $$$t$$$ and use line sweep to check if there is any intersection.

I solved C using ternary search . can some one please proof how the shape will be similar to parabola (I just assumed) .

You can use just Binary Search in the odd number of movements.

Can you plz tell me how? . I am able to find out that we need to search only in odd position but didnt find the how to solve it as brute gives tle and you said binary search . But I didnt get it how??

Here for every cup of cold water you have already poured a cup of hot water already. So whenever the even number of cups poured, the temperature will be constant which is the

`avg(h, c)`

.After you pour the 1st cup of hot water. The temperature will be h. After you pour the 2nd cup of hot water. The temperature will be slightly low since you have poured one cup of cold water before.

Therefore, the temperature will be decreasing monotonically towards the

`avg(h, c)`

when ever you pour a cup of hot water. So you can use this property to binary search the appropriate value.Link to my submission : 81771518 . Hope you find this useful

No the shape would not be parabolic.

I meant convex or concave.

Assume you put x hot cups and x — 1 cold cups. Then middle temperature is (xh + (x-1)c) / (2x-1) If you put x + 1 hot cups and x cold cups then middle temperature is ((x+1)h + xc) / (2x + 1).

Subtract second from first and you get (h-c)/(4x^2 — 1) which is always positive. So temperature is strictly descending function from number of hot cups.

I can be proved that $$$ \frac {x} {2 * x - 1} $$$ (x = 1, 2, ...) is decreasiong.

Proof:

$$$ 2x^2 + x > 2x^2 + x - 1 $$$

$$$ x *(2x + 1) > (2x - 1) * (x + 1) $$$

$$$ \frac{x}{2x-1} > \frac{x + 1}{2(x + 1)-1} $$$

stock svg images

I just plotted the graph, and you can see it is a decreasing function for odd number of cups.

and you can see it is constant for even number of cups.

The ordering of problems was very bad E was even easy from C

I think it was easy to guess but not easy to prove .

A somewhat loose proof for the solution to E is that if you consider a list of numbers [2,2x,2y,2z,...] such that every term divides the smallest number (for example, 2), no matter how you re-arrange the numbers, eventually, you will have to take it modulo the minimum number in the array, which will cause the result to become 0. Then, it is zero all the way to the end.

How to prove that if there is more than one number not the multiple of $$$k$$$ then the final modulo is different?

Suppose that there is a number $$$a_i$$$ that is not divisible by $$$a_1$$$.

Let $$$x$$$ be equal to $$$a_i$$$, and let's take two following orders: $$$[a_1, a_2, a_3, \dots, a_k]$$$ and $$$[a_i, a_1, a_2, \dots, a_{i-1}, a_{i+1}, \dots, a_k]$$$. Then in the first case, $$$x \bmod a_1$$$ is non-zero (but less than all $$$a_i$$$, so it is the resulting value), and in the second case, $$$x \bmod a_i = 0$$$.

If the property isn't true, then there is some number (call it m), which isn't a multiple of k.

Now, feed in X=k to the machine. If the permutation of A is [k,m,kx,ky,kz,...] then our answer is clearly 0.

However, if our permutation of A is [m,k,kx,ky,kz,...] then clearly since m is not divisible by k (by definition), then the value will be (k%m), and the value after the second step will also be non-zero. The value of the result won't change after the second step, because kx,ky,kz > k. Therefore, the final result in this state is non-zero.

So our array that included m was therefore unstable.

We can prove using mathematical induction. As BledDest showed that only those sequences may work which contain multiples of smallest number .

Notation : $$$x$$$ mod $$$(a_1,a_2,a_3)$$$ = $$$((x$$$ mod $$$a_1)$$$ mod $$$a_2)$$$ mod $$$a_3$$$

Base case : $$$size=2$$$ , consider $$$[a_i,v.a_i]$$$ where $$$v>1$$$ . Take any number $$$x$$$. Then $$$x = p.(v.a_i) + r$$$ , where $$$r<v.a_i$$$. Then $$$(x$$$ mod $$$v.a_i)$$$ mod $$$a_i$$$ = $$$r$$$ mod $$$a_i$$$ ,$$$(x$$$ mod $$$a_i )$$$ mod $$$(v.a_i)$$$ = $$$(r$$$ mod $$$a_i)$$$ mod $$$(v.a_i)$$$ = r mod $$$a_i$$$ .

Assumption : Suppose above is true for size $$$k$$$ i.e $$$[a_1,a_2 ... a_k]$$$. Note that all all elements are multiple of $$$a_1$$$.

Let us prove for size $$$k+1$$$ i.e $$$[a_1,a_2 ... a_{k+1}]$$$. consider $$$x$$$ mod $$$(a_i,...a_{k+1} ..a_j)$$$ (type 1) and $$$i$$$ not equal to $$$k+1$$$.It will make no difference since $$$a_{k+1}$$$ is not at first position.consider $$$x$$$ mod $$$(a_{k+1} ....a_j)$$$ (type 2) i.e $$$a_{k+1}$$$ is at first position. Then by assumption it will be same for all permutation where $$$a_{k+1}$$$ is at first position. Now we only need to prove that answer for any particular permutation of type 1 and type 2 are same. Let us choose $$$x$$$ mod $$$(a_{k+1},a_1,a_2,a_3.....a_k)$$$ and $$$x$$$ mod $$$(a_1,a_{k+1},a_2,a_3....a_k)$$$. Since $$$x$$$ mod $$$(a_{k+1},a_1)$$$ = $$$x$$$ mod $$$(a_1,a_{k+1})$$$ (from base case proof) hence both are equal . Thus answer is same for all permutation of length $$$k+1$$$.

I will be thankful if some one points out mistake in above proof.

suppose the gcd you'll take is g, then x=g*q+r, where r is x%g. now if you take x%(n*g) then x%(n*g)=g*(q%n)+r which is some g*q'+r. after repeated divisions when you finally do %g the answer will be q.

couldn't agree more

The round was very similar to AtCoder Beginner Contest, except for F (which I don't dare to even read lol)

Why D gets WA on testcase 7.

my submission : 81782286

use kadane from left and then again right taking sum-max...i was stuck there as well

Why?

Whaaat??? What the hell?? I just did the left and got struck. But, well. Why we need to do again from the right??

pointless...its wrong..my submission is hacked

Because the largest number taken at each step (subtracted from the sum) maybe different when taking from the right.

Example:

When taken only from left, the answer is 3 but when taken from right, the answer is 4 (3 2 2).

However, even this solution doesn't work for cases like,

Couldn't solve it, but i changed my code for this testcase

n = 6

[9, 1, -9, 2, 2, 2]

I didn't come up with a counterexample like this >_<

Screw up +120 like that. No!!!

it is ok, if you use kadanes from left and right still you will end up in this case :P

7

30 -20 5 1 3 -20 30

ans would be 4 but by using two kadanes we get 0 :(

C was really interesting...

Solotion of C!!! Please!

I used two ternary-search (one for hot == cold + 1 and one for hot == cold) and compared two final results from each ternary search, but I think there are easier solutions

You don't need to do any search for case hot == cold since the answer for that case must be

`(h + c) / 2`

.

For all the odd pouring, the temperature at ith pour would be ((i+1)*h/2+(i-1)*c/2) and this has to be equal to t. Now on finding i, the answer would be 2*i-1.

C can be solved in O(1) time.

This is my submission: 81757911

If h == t, clearly the answer is 1. If h + c >= 2 * t, the answer is 2 as the most you can lower the average to the target is via the first cold cup.

Now we consider the case where the average of h and c is less than t.

We make an observation that now there will be one more hot cup than cold cup, so let the number of cold cups poured in be k. Inclusive of the latest hot cup, the average temperature would now be ((h+c)k + h) / (2k + 1). We would like to find the value of k such that the fraction is closest to t.

We first solve the equation where ((h+c)k + h) / (2k + 1) == t. We can then obtain h-t = k(2t-h-c) and k = (h-t)/(2t-h-c). We can see that either floor(k) or ceil(k) will give us the closest value to the target, hence we check the difference between ((h+c)k + h) / (2k + 1) and t for floor(k) and ceil(k), thus solving the problem.

Hi, I am unable to understand why when h+c>=2t, answer will be 2. Can you please explain?

When the average is greater than t, the smallest average is obtained by pouring 1 hot and 1 cold cup. The average is the same if there is an equal number of hot and cold cups, and it will increase when there is one more hot cup. Thus, 2 is the minimum number of cups for the average to be closest to t.

One of the best contests ever. Got to know for the first time(or maybe noticed for the first time) about the modulus property that had to be applied in problem E. Thanks again !!

Task C took soul out of me but never showed AcceptedFrom when Irfan Khan started coding?

after dying?

Actor like Irrfan Sir never dies, he is immortal in our hearts.

Oh I love this man too.

I was just joking <3

Legend never dies.

Your dp explains everything, Irfan Sir please comeback xD

https://codeforces.com/blog/entry/78116?#comment-632536

Please tell me what is wrong with this solouton for C 81802932 I know I've been silly there but please help me

Even if $$$\frac{h+c}{2}\neq t$$$, $$$2$$$ may still be an answer. Since $$$\frac{h+c}{2}$$$ may be closer to $$$t$$$ than the answer you use binary search to get.

that was the case I missed I forget comparing the answer of bs with the answer that I got (h+c)/2 i.e 2

Can D be solved with the DP?

Yes, I solved with DP.

Idk about dp, but i'd like to say it can be solved in nlogn independent of ai's values. Sort a list of indices based where indices with higher ai values come first, and hold a segment tree of the prefix sum values. Now just iterate through your list of indices and add them as boundaries for querying in a set as you go, and for each index find the largest prefix sum greater than and index and the least prefix sum less than an index such that the range is within all the boundaries in your set, then update ret with subtracting those prefix sums minus the current indices value.

It can be solved in O(n) as well independent of ai's value. You can maintain a stack to do so, https://codeforces.com/contest/1359/submission/81806616 .

Wow, very cool!

Thanks :)

can you please explain your solution ?

Basically for each index 'i' of array (1-based indexing), I found out what is the rightmost j for which a[j] > a[i] and j < i. Now Assuming this element has to be removed, I maintained a mxl array to maintain max. subarray sum that is ending at [i — 1] for a given 'i'. Note this includes a[i — 1], (no matter if its positive/negative). Now as per the implementation using stack, for left ('L') part, for a given 'i', s.top has 'i — 1' index, so I keep on removing indices from stack for which a[s.top()] <= a[i], but also kept on maintaining the max sum ending at 'i — 1', which includes a[i — 1]. If you observe mxl[s.top()] does not necessary means max sum goes up to indices 2nd topmost element of stack. So to find the mxl[i], you have to make the continuous sum starting from 'i — 1' to the current s.top(), and update the mlx[i] if valid.

Similar goes for right part.

And then for each element just assume you are removing it from its range, find what is the max left and right sum within the range.

Also if you are still having difficulty in understanding this, I would suggest first try solving easy part of this question — Imbalanced Array. The editorial of it have a similar idea.

I solved it in O(n) . I first used Kadane algorithm and then i consider all subsegment which do not contain negative number .submission Edit : now hacked .

Yeah, dp[i][j] = max value you can get with a segment ending at i, with maximum equal to j. The transitions are pretty straightforward (just need to offset negative values).

Submission: here

Can you explain the transitions? I don't quite get it

Initially, you can start a new segment at some index, but it will be the maximum, so

`dp[i][a[i]] = 0`

to start. But also, you could have extended some segment ending at i-1, so there are 2 cases to consider.`j`

, the old maximum was smaller than`a[i]`

. Then you get`dp[i-1][j] + j`

, because`a[i]`

becomes the new max (you already "paid" for`j`

in`dp[i-1][j]`

, so add it back in to compensate).`j`

is greater than`a[i]`

. The maximum doesn't change, so you get`dp[i-1][j] + a[i]`

I hope that helps.

Thanks! This is a really cool solution

$$$dp[index][maxval]$$$ is the maximum sum ending at $$$index$$$ containing $$$maxval$$$ as the max value. Answer is $$$max (dp[index][maxval]-maxval)$$$

isn't C binary search why alot of people wa on 2 testcase

Nope, it comes down to a simple tiny equation.

For me personally, it was using binary search even if t<(b+c)/2. As soon as i rectified that, I got it accepted.

DELETED

help me with problem C

I would love to help you, but I am not one of those 3167 people. :(

https://codeforces.com/blog/entry/78116?#comment-632536

In q C, my code gave wrong answer in C++17(64) but got accepted in C++14. Happened with anyone?

I guess you were working with double data type.

Is there any problem with double in c++17(64)? Thanks

I don't know what is actually wrong but i faced problem while comparing two double few times. So i try to compare it keeping it integer.

Say i need to compare x/a and y/b. I compare x*b and y*a (Multiply both a*b). In this case there is no chance of precession loss.

what is wrong with double data type in c++17?

See This Comment

me too! this was really frustrating as i kept on changing my code and still getting the same wrong answer

Same here.

Same is happening with me too. But I figured this out after reading your comment. Thanks for mentioning. Pupils, here I come.

I feel if E was before C it would have more solves

Can someone why my submission to problem D, received runtime error on test 2? It was working fine on my local machine.

Thanks

Problem F is already in emaxx. https://cp-algorithms.com/geometry/intersecting_segments.html

Good thing that awoo found some tests to break that code if it is copypasted without any changes

I got AC, copying and pasting that code, just changing EPS and using long double

Devil please explain how did you assumed the limits inside for loop like

`100`

and`HI`

variable and also this part`hi - lo > 5*1e-9`

Trial_and_error XD

its ok since even mifafao was not able to solve it

In Problem C, I was worried about using floating point numbers, and then thinking about a solution using only integers ...

I used my own fraction class for the same, and it didn't even have a greater than operator, damn it sucked :(

You could have done it by holding best current value by numerator and denominator, then just comparing fractions.

Not brute forcing to find a pattern do be like that sometimes. E would have been significantly easier if it was placed as problem C.

hi guys why this solution don't work in pb b THANKS . #include<bits/stdc++.h> using namespace std; typedef long long ll;

Your solution will place a 1x2 block for this case:

The problem states that you can not break the 1x2 block, so this block can be used only when 2 dots are together, like this:

I don't think that's true. In your example, the innermost "else" triggers on each *, at which point curr has value 1, so x is added to the total and curr is reset to 0.

thanks guys for ur replyes acually , my solution work . i jut forget "(" in y*curr/2 cuz (curr/2)*y!=y*curr/2 . that why i get wrong answer.

My AC solution of problem F is wrong. Can anyone hack me? Please (>__<).

still, I can't! i m so noob ༼ ಥ_ಥ ༽

Did you think about the case when the two cars are aline?

My solution can fail on normal case. You can see other Account solution is quite different with mine.

AC problem F, just after the contest...

WHY I DIDN'T REMEMBER TO USE 1e-10 INSTEAD OF 0.0 !!!For D, i took : max of (maximum subarray sum ending at index i — max element in that subarray) for all indices i, why doesn't this work?

did you use kadane?

yes

Update: I did the same thing again from the right now it got AC.

update: Got hacked. forget this

lol that second update

Yeah man same here. Any idea why is it so?

answer is 20 (pick subarray [3, 5])

thanks!

It fails on this case

Spoiler`4`

`30 -22 6 7`

Optimal answer is 6 but your algo will give 0 as answer.

thanks!

When you missed the last exercise because you didn't round an almost zero value to zero before comparing it with zero. So sad :(

Can C be solved using ternary search?

Yes, it can be,

Here is a link to my submission: 81816118

Can someone explain the following case for problem D?

The answer is 4 but I'm getting 3 after trying it out by hand and with my code.

Edit: Figured it out. In case anyone's wondering, the max_num at each step when taken from left maybe different from the max_num when taken from the right.

Choose the subarray [8,10] 0-index. That is [3,2,2].

Segment [9, 11] gives answer 4.

Could someone tell me why https://codeforces.com/contest/1359/submission/81809462 gives TLE

Can anyone tell hack for C? TIA

Video Editorial :- Problem C

I will add editorial for problem D using segment trees soon.

Hi! So I was going through the hacks and saw something odd...

https://codeforces.com/contest/1359/submission/81798994

So this guy made a smurf, submitted his own code, and intentionally added code which would fail on certain input. And you can even see that the author's name is still the same! There aren't any points for hacks in this contest, but shouldn't there be some action against this user?

How to solve D without using dp ?

We can use sparse table and concept of NEXT GREATER ELEMENT.\

CHECKOUT :

81811299

For each element ai find the range at which it is maximum, let the range be [l,r]. The sum it would contribute if Bob chose to remove ai is maxsuffixsum(l,i-1) + maxprefixsum(i+1,r). The first part can be calculated using stack and the second part with segment tree. My submission

Even the second part can be calculated using the same stack you are using in 1st part. Refer: https://codeforces.com/contest/1359/submission/81806616 , giving you an effective O(n) solution.

See my solution, quite easy to code: 81741311

You can even replace the segment tree with sparse table and set with stack to get an O(N) solution.

For the problem C, I think there is always a way to achieve the average temperature equal to c. Let's say, x number of hot cups were poured and y number of cold cups were poured if we solve the equation assuming, the average temperature will eventually be equal to c, we have hx + cy = tx + ty from here, you get x:y = (t-c):(h-t) taking x and y in this ratio should give the answer, so final answer should x+y For example, in the second sample test case given, x = 15, y = 11, the average temperature will come to 30. So, the judgment is wrong for this question. Please advise on my understanding and correct me if i am wrong.

The problem requires that x-y is either 0 or 1.

You have to take hot and cold cup alternatingly. So either x=y or x=y+1 You are missing this condition.

can someone hack my C? 81742048

I believe it is wrong :)

I made some random tests for your code, and after 3000 tests, all answers were the same as my code. I think your code is correct! (Or we both made the same mistake)

long double is not hackable I think.

Here's a proof for $$$E$$$.

Take any index $$$i \neq 1$$$. The condition applied to $$$x = a_1a_2\dots a_k + a_i$$$ in the order $$$a_1,a_2,\dots ,a_k$$$ gives $$$a_i \pmod{a_1}$$$ on each step since this is at most $$$a_1-1$$$ and all modules are greater than that.

Now evaluating in order $$$a_i,a_1,a_2,\dots a_{i-1}, a_{i+1}, \dots a_k$$$ evidently gives $$$0$$$ as $$$a_i | a_1\dots a_k + a_i$$$. Therefore $$$a_i \pmod{a_1}$$$ must be zero, which means $$$a_1|a_i$$$ for all indices $$$i$$$.

Proving that the smallest element dividing all others is enough is not hard. Say $$$r_1 = x \pmod{a_{p_1}}$$$, $$$r_2 = r_1 \pmod{a_{p_2}}, \dots$$$.

Notice that for any $$$r_i$$$ by definition we have $$$r_i = r_{i-1} \pmod{a_{p_i}}$$$ thus $$$a_{p_i}|r_i-r_{i-1}$$$ where we define $$$r_0 = x$$$. Since $$$a_1|a_{p_i}$$$ we must have $$$a_1|r_i-r_{i-1}$$$ for all $$$i$$$. Thus $$$a_1|(r_1-r_0)+(r_2-r_1)+\dots + (r_{k}-r_{k-1}) = r_k-r_0 = r_k-x$$$.

We also know that the result is a non negative integer at most $$$a_1$$$ because once $$$\pmod{a_1}$$$ is applied the result becomes constant as all other modules are bigger. Thus $$$r_k$$$ is always $$$x \pmod{a_1}$$$ no matter what order we do the operations in.

Therefore the problem reduces to finding for each smallest element $$$a$$$, the number of ways in which we can choose $$$a_2 < a_3 < \dots < a_k$$$ integers divisible by it in the range $$$(a,n]$$$.

This is just $$$\displaystyle \sum_{a=1}^{n} \dbinom{\left\lfloor \frac{n-a}{a} \right\rfloor}{k-1}$$$.

Today's ques C went like a Game changer for many .

what is the approach for D except seg tree! can we solve it using kadane

Apply Kadane 2 times, one from the first element and second from the last.

i am quite sure that is wrong are you sure it will work??

this solution is hacked !

I don't know if some variation of kadane can work here . But I can provide you a, without seg solution .

for each element i, we first find smallest j<=i, such that maximum in range [j,i] = ith element. and similarly, we find largest k>=i such that maximum in range [i,k] = ith element. (These both can be achieved with binary search, using any RMQ data structure, sparse table is easiest to write)

Now we need to find, left and right index in range [j,i] and [i,k] such that we get maximum sum. Again binary search and range min/max queries over prefix sum .

Take the maximum among all indices.

Solution.

In Problem C, Test Case# 4 999977 17 499998

Judgement is: wrong answer 14th numbers differ — expected: '499981', found: '499979'

But 499979 gives perfact 499998 result and it is < 499981 so why it is not correct answer?

yes you are right

Same problem here...

awoo Please explain this . I have the same problem.

The result when you pour 499979 cups is not 499998, but 499998.00000200008400352814818222.

For 499981 the result is 499997.99999799992399711189025183, which is closer to 499998.

But needing the doubles to be so accurate can be problematic. I think the problem for us is the fact we used Java, and perhaps Java calculates the doubles less accurately/differently than c++. This is a problematic property of this problem, even though I think it was somewhat interesting.``

1) Java has BigDecimal library

2) It is possible to do all calculations and comparisons in integer values

Hmm yeah you are right it can be done in integers. Thanks.

Thanks for the clarification, I guess mentioning some kind of cutoff precision in problem statement would have been helpful. Most people who use java will use double by default and will fail this.

Yeah, and the calculated differences, for me at least, are the same for both numbers: 2.0000734366476536E-6 Maybe it's a problem cuz we use Java? Maybe is c++ more accurate or something.

Some solution in c++ also faced this problem. But it can be solved not using double.

For 499979, the answer is

and for 499981, the answer is

So, 499981 is the answer.

I got 2.0000734366476536E-6 for both... Why do you think that is the case? I calculated it like this (x is the number of cups — 1 and divided by 2):

private double averageAfterTries(int x, int h, int c) {

I struggled with that Test case. When I compared 499979's value and that of 499981 using type double, these values are treated as same because of error. By compareing by fraction like this submit, the result became accepted. https://codeforces.com/contest/1359/submission/81813380

My submitted code outputs correct answer in CF but wrong answer in my machine for this test case. I tried to hack some solutions using this TC but failed :v

Is it possible to solve problem d in O(n), not using a segment tree?

https://codeforces.com/contest/1359/submission/81786125

I merged same consecutive sign elements.

Let's say chunk1 = +, chunk2 = -, chunk3 = + ~~

I judged which one is better between (chunk1 + chunk2 + chunk3) or (chunk1) or (chunk3) I couldn't find why this logic is wrong..

yes using dp we can solve it in O(n).

It can ve solved in O(n*60) with just simple implementation.

yes, see this video editorial for detail explanation

https://youtu.be/1h1D7wMbDis

Is it possible to solve problem D with Kadane Algorithm + Segment Tree to get max query ?? I tried but I'm getting WA.

E: Just tried doing it going once from forward and other backwards but now TLE.

who can tell me about the case 999977 17 499998 why 499981 is better than 499979?

499979 => 499998.0000020001 499983 => 499997.9999979999

is that cause by accuracy?

Apparently yes.

I think educational rounds are going more towards math and heavy implementation nowadays!

yes these days it seems so