Educational Codeforces Round 171 [Rated for Div. 2]

9 days ago, # ^ |

-10

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g_x_j

9 days ago, # ^ |

+38

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equation_tracker

9 days ago, # ^ |

-12

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Asuna_Yuuki

9 days ago, # ^ |

-12

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Marubozu_

9 days ago, # ^ |

-12

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yashsinha129

9 days ago, # ^ |

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crashsaladd

9 days ago, # ^ |

-8

Haha, I broke your silly chain.

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prathamalu1

9 days ago, # ^ |

NO you didn't

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deepdblm

9 days ago, # |

I got a question, the penalty of time i.e. 10 minutes is the format of div. 3 and it is in this contest which is div. 2 so will the hacks have -ve points on unsuccessful attempts?

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BiggestOtaku

9 days ago, # ^ |

No, hacks are done after contest ends so there is no penalty

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deepdblm

9 days ago, # ^ |

but it is written 12 hours hacking phase

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dark111

9 days ago, # ^ |

12 hours after the contest ends

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bihnehminduh

9 days ago, # |

glhf

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9 days ago, # |

+19

If I get minus again in this contest,I will leave CP.

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Vitaly503

9 days ago, # ^ |

+111

Did you know that 95% of people leave codeforces before their biggest delta ever?

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equation_tracker

9 days ago, # ^ |

That's why I am still CPing after getting $$$-35$$$ and $$$-8$$$ in a row. (Got $$$+4$$$ is the last contest :))

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9 days ago, # ^ |

I am still going to give this contest although got -102 :( in last contest

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bjjvsrjjt

9 days ago, # ^ |

-140 and -54 in a row :(, for me

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papa-ka-para

8 days ago, # ^ |

means, +250 is waiting for you, only if you practice hard.

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picoton

9 days ago, # ^ |

only -43? lol

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CaveNightingale

9 days ago, # ^ |

lol. I solved $$$1$$$ problem and got $$$-253$$$ in a contest CF1930

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equation_tracker

9 days ago, # ^ |

← Rev. 2 →

Thanks. You guys motivated me.

Edit: Motivation overloaded :). Lost $$$37$$$ in the last round (Solved only one question which is worse than my first ever contest in CF)

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stakudog

9 days ago, # ^ |

100% actually, even tourist

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shxyan

9 days ago, # ^ |

CP is not for rating just telling you, be better you will be higher rated. Its a hard skill, keep practicing.

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9 days ago, # ^ |

Don't leave! I'll touch your acc and help you.

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pratyush_24

9 days ago, # ^ |

help me also!!

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9 days ago, # ^ |

Are you planning to leave CP too? :(

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pratyush_24

9 days ago, # ^ |

+10

nah mom didn't raise a quitter :)

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9 days ago, # ^ |

Yes, I need your help! :((

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9 days ago, # ^ |

Can confirm, wltt touching my account helped me

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Osama_Fahim

9 days ago, # ^ |

Please guide me as well what should i do to improve i tried different approaches but It has become a cycel positive delta in few contests and then a biggest -130 or something, and this loop continues for a long time, Please guide me your what can i do looking at my current situation ?

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pirate__hunter

9 days ago, # ^ |

-59 in the last 3 contests 😎

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9 days ago, # ^ |

-297 in the last 4 contests xD

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9 days ago, # ^ |

take a rest/leave when it's no longer feels good. Comeback when you feel it or just leave it be.

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9 days ago, # ^ |

Will I leave CP cause I am sure that I will get minus... I solved 0 problem...

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Aritro_Sarkar

9 days ago, # ^ |

Take a break

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frissky9

9 days ago, # ^ |

Leave CF. Go to LC and atc spend 3 months solving mediums and A-D (on atc) there, and when you come to CF stop solving 800-1000, push yourself, gl

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dynato24

9 days ago, # ^ |

meow meow meow meow (what i was made for tune)

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8 days ago, # ^ |

i didn't solve a single question in this Edu round, I'm easily having -150 in this contest.

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kagelov

9 days ago, # |

I haven't lately been able to see any of the other users' submissions. Everything I see is "N/A" for source code, even for problems I have solved. Any solution to this very annoying issue? Thanks!

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wille_und_vorstellung

2 days ago, # ^ |

same for me, could you solve the problem??

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brave-kid

9 days ago, # |

Today is my debut contest with Rainboy strategy , gonna start from $$$D$$$

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9 days ago, # ^ |

get ready for -250 XD jk bro all the best hope u do best

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kovidsaivemuri

9 days ago, # ^ |

hmm.. brave-kid ohhh ok!

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9 days ago, # ^ |

It's extended ICPC rules, it's not beneficial to do this.

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Aritro_Sarkar

9 days ago, # ^ |

beneficial if you can make sure that you will solve it and if it is a points giving contest.

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brave-kid

9 days ago, # ^ |

I am not caring about rating for now, for now my goal is to get full grasp of $$$D$$$ as soon as possible so that I can become yellow fast

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8 days ago, # ^ |

nice I am also gonna use it now in today's codechef

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9 days ago, # ^ |

← Rev. 2 →

at least u did it, nice try xD I'm also doing it.

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9 days ago, # |

As a contestant, I wish to test the problems live

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9 days ago, # |

Doing contest, problem B feels weird AF so imma skip it. Please correct me if it's skill issue.

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Chess.Master

9 days ago, # ^ |

It's too standard problem, Just Binary Search the possible values of K, Solve more problems and you will find that it's a very common standard problem.

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nguyenquocthao00

9 days ago, # ^ |

Binary search is possible but not necessary. This can be solved in O(n)

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Chess.Master

9 days ago, # ^ |

Yes but it a standard binary search problem, The O(n) solution will be hard to implement and we don't need this at all,

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9 days ago, # ^ |

Á đù hội anh em KHTN.
Figured that the O(N) solution would be just "stitching" a[i-1] and a[i+1] together for every i and check, right?

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nguyenquocthao00

8 days ago, # ^ |

KHTN

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dynato24

9 days ago, # ^ |

elaborate please

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GUNNER19_2.0

9 days ago, # ^ |

Can you please tell how to solve it in O(n)

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9 days ago, # ^ |

O(N) solution for (B): 288645656 I refuse to elaborate, I hate (B).

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Ranlen

9 days ago, # ^ |

← Rev. 3 →

I solved it with prefixes in and I think its pretty clear O(n) 288565752

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9 days ago, # ^ |

N is only 2000 so i was tired with it and decided to go "Fk it" and yolo with O(N^2).

> inb4 "O(N) exist", I know but I'm too tired with B and I just want to get over with it.

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9 days ago, # |

$$$A$$$ is hard ._.

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9 days ago, # ^ |

"black circles" flashbacks... 2002C - Black Circles

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8 days ago, # ^ |

I was able to solve black circle pretty fast that time this one was hard for div2 A

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8 days ago, # ^ |

I learned my lesson after failing black circles.

Do guessing sometimes, trade off between a penalty vs a quick solve + spare time for other problems.

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8 days ago, # ^ |

u know what in the black circle problem I was able to visualize how it would be moving so I was able to come up with the distance formula solution but with this one I couldn't visualize what was happening so it took me time

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9 days ago, # ^ |

← Rev. 6 →

$$$B$$$ is also hard ._.
Sucked without seeing small $$$n$$$.

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9 days ago, # ^ |

Not quite, B is just tricky

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9 days ago, # ^ |

Not quite, A is just output.

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9 days ago, # ^ |

A is guess-forces

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9 days ago, # ^ |

no, the constraint has k<=1414, hinting a factor of root of 2 occurred someherewhich gives an hint towards the formation of square with side length = min(x,y)

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9 days ago, # ^ |

well, at least i guessed and got it right

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9 days ago, # ^ |

literally

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8 days ago, # ^ |

Idk why i'm getting TLE in A. the constraints were pretty lenient for O(N * N) solution 288664487

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2288

8 days ago, # ^ |

Actually, the constraints are on X,Y and K in each test case, not on the sum of them over all test cases. So time complexity of O(XY) (or something else that is second order) requires a worst case of 1000*1000*5000 calculations, exceeding the time limit

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8 days ago, # ^ |

Yeah, i realised that after fucking up my contest :(

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Violet___Evergarden

9 days ago, # |

What the hell is this Problem B?

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kmth.7z

9 days ago, # |

+11

took 2 minutes for A and 2 hours for B

hardest B i've ever seen fr

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the_last_smilodon

9 days ago, # ^ |

Actually I found A harder than B.

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9 days ago, # ^ |

I feel like you either observe the solution instantly for A or you don't at all.

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9 days ago, # ^ |

i literally kept testing everything I could think of on A and somehow it passed

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8 days ago, # ^ |

Well, you just had to fix any two points of the two lines at the corners of the (x * y) rectangle, then vary the other two on x and y axes. I got TLE tho 288664487

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FatihCihan

8 days ago, # ^ |

5e3 * 1e3 * 1e3 = 5e9 , obviously TLE , you should always aim for <1e9 (unless the time limit is huge).

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8 days ago, # ^ |

Oh damn, I forgot that test cases complexity too counts.

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8 days ago, # ^ |

you did some overthinking, if y>k and x>k you can just make an "L" on the grid, and if k>x or k>y you make an X on the grid xd

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8 days ago, # ^ |

← Rev. 2 →

I had that idea but idk why i messed up. This costed me the whole round, because i gave up on A.

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_Unseen_

9 days ago, # ^ |

Ahh, so it's not just me? I literally have NO idea how people managed to come up with B's solution in such less time :(

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anishray042

9 days ago, # ^ |

I guess since n was low enough for n^2, for odd n we can just brute force as we need to insert one extra number anywhere in the given array. Obviously, only at max 2*n possible values as we want new number to be closest to each a[i].

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_FFT_

9 days ago, # ^ |

B is a standard problem, and it took me nothing to figure out the solution.

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-dub-otrezkov-

9 days ago, # |

how to solve E?

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Chess.Master

9 days ago, # ^ |

How to solve D? Is it a Segment Tree?

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9 days ago, # ^ |

It's just some prefix sum stuffs

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-dub-otrezkov-

9 days ago, # ^ |

prefsums + binary search. you can check my sol https://codeforces.me/contest/2026/submission/288582700

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123gjweq2

9 days ago, # ^ |

damn that's a lot of prefix sums

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9 days ago, # ^ |

i gave up after trying too much prefix

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9 days ago, # ^ |

I used $$$4$$$ different prefixes in total ._.

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Alenygam

9 days ago, # ^ |

+10

Yea I just upsolved it and this problem was a pain. It took me quite a bit to implement it and iron out all the bugs...

Also not very "educational". It doesn't really teach you to think differently, just "implement harder".

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neal27

9 days ago, # ^ |

← Rev. 2 →

(assuming 1-based indexing everywhere) consider the b array as divided into blocks as this: (n size block) (n — 1 size block) (n — 2 size block)..... now consider one block which has starting element of pair as i. what is the sum of this block? s(i, i) + s(i, i + 1) + s(i, i + 2) + ... s(i, n) where s(l, r) is sum of a from l to r. you will see that in this expression, a[i] is added n + 1 — i times. so calculate suffix array on a[i] by this rule (a[i] * (n + 1 — i))

now if i want sum of k blocks, i need sum of block 1, sum of block 2, ... sum of block k i.e. suff[1] + suff[2] + .... suff[k] so create a prefix array on the suffix array created before.

now consider this problem: we want to find sum 1 to x, in array b:

first find number of blocks that are completely covered say k — 1 (use binary search) add pref[k — 1] then we are at block k, now if this block has l as left point of the pairs. we want to find (l, l) + (l, l + 1) +.... (l, l1) (l1 depends on where x lies in this block. now for this we can add suff[l] — suff[l1 + 1] directly. but note : this followed rule (a[i] * (n + 1 — i), but now a[i] has not been added (n + 1 — i) times. you will see that now a[i] has been added (n + 1 — i) — (n — l1) times. so you need to subtract s(l, l1) * (n — l1). again have a prefix array on original a for this say this answer we calculated for sum of b from 1 to x, is given by f(x), then given query is answered by f(r) — f(l-1).

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9 days ago, # ^ |

I saw lots of max flow solutions.

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-dub-otrezkov-

9 days ago, # ^ |

me too, but why them work?

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jackson.souza

9 days ago, # ^ |

The solution is a direct application of Project Selection. Here is blog about this: https://codeforces.me/blog/entry/101354

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Rising-coder

9 days ago, # ^ |

← Rev. 2 →

+11

Why is div2E max flow :(
I hope there are simpler solutions

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shouldorshouldnot

9 days ago, # ^ |

is flow hard?

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qwertylkjhg

9 days ago, # ^ |

answer is n — max pair matching in the graph, where left side vertices are 1,2...n and right side 0,1,2...59 ,edges i <-> j where a[i]'s j-th bit is set.

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9 days ago, # |

← Rev. 2 →

B?!!!! Why is n < 2000? Is it really n^2 for odd n? How to solve B?

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MonoLith

9 days ago, # ^ |

Yepp just brute force for every index For every index I push a[I] +1 just after a[I] and then check the minimum among all maximum You can see my code

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9 days ago, # ^ |

Thanks, can you give a case where the picking the either last or first doesn't work for odd n? I couldn't come up with such a case

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MonoLith

9 days ago, # ^ |

← Rev. 3 →

For n= 7

1 2 3 1000 1001 1002 1003 It's optimal to add answer is 1 If I add 4 after 2nd index (0 based) And then pair (1 2) (3 4) (1000 1001) (1002 1003) But If you add in begining or end it will be 997

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9 days ago, # ^ |

Thanks got it!

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anishray042

9 days ago, # ^ |

← Rev. 2 →

1 2 20 25 26

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9 days ago, # ^ |

yeah, for even N you just sort and pair up, and for odd you test all possible Ai that you could pair up with the extra cell

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nivedkml

9 days ago, # ^ |

You can do in O(n + log2(max(ai)) using binary search. You can check this comment.

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ab_malek

9 days ago, # ^ |

I did it with o(n^2). You can actually remove each element and cut the array at that position to make the array two even array and solve it.

I think this intuition is less complex to solve the problem. By also accepting o(n^2) authors try to make the problem simpler.

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Alenygam

9 days ago, # ^ |

I did it in O(N). You can check out my solution: 288539510

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7 days ago, # ^ |

Nice, using suffix and prefix max

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lhlephuocdao

7 days ago, # ^ |

Could you explain your solution?

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Subarna1

9 days ago, # |

What is the solution for B?

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nivedkml

9 days ago, # ^ |

You can binary search for each k from low = 0 and high. = (arr[n-1] — arr[0])

Code

#include <bits/stdc++.h>
using namespace std;

#define pii pair<int, int>
typedef long long ll;
typedef long double ld;

const int mod = 1e9 + 7;

bool check (ll k, vector<ll>& a) {
  int n = a.size();
  int cnt = 0;
  for(int i = 1; i < n; i++) {
    if (a[i] - a[i-1] <= k) {
      cnt++;
      i++;
    }
  }
  
  return cnt >= n/2;
}

void solve () {
  int n;
  cin>>n;
  
  vector<ll> a(n);
  for (int i = 0; i < n; i++)
    cin>>a[i];
    
  ll low = 1, high = a[n-1] - a[0];
  
  while (low < high) {
    ll mid = low + (high - low)/2;
    
    if (check(mid, a))
      high = mid;
    else
      low = mid + 1;
  }
  
  cout<<low<<"\n";
}

int main() 
{
    ios_base::sync_with_stdio(false);
	  cin.tie(NULL);
	  
	  int t;
    cin>>t;
    
    while (t--)
      solve();
}

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sachin.13

9 days ago, # ^ |

why return cnt >= n/2;

how you got this idea?

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nivedkml

9 days ago, # ^ |

I will try to explain it here, let me know if you get it or not.

cnt is the count of pairs, number of pairs should be n/2, as 2 elements make up a pair. Also imp thing is we only have to check diff between conseq elements as least diff of any j with current i will be when j = i + 1 as array is increasing. eg: if n = 6. there should be 3 pairs. if n = 7, there should be 3 pairs and the remaining 1 we can match by adding our available "atmost one addition case".

n is even : Lets say a = {1, 5, 8, 10, 13, 14} and k = 3 here 1 and 14 would not get a pair (as (5-1) = 4 > k & 14 doesn't have a pair to match with), whereas (5, 8) and (10, 13) can make up pair. So cnt = 2. As we can only add atmost 1 number, we can't make pair for both 1 and 14. So k = 3 is not feasible. if cnt was 3 (= n/2) then that means all the numbers have pairs. So its feasible.

n is odd : Lets say a = {1, 5, 8, 10, 13} and k = 3 here 1 would not get a pair (as 5-1 = 4 > k), whereas (5, 8) and (10, 13) can make up pair. So cnt = 2 (= n/2). As we can only add atmost 1 number, we can make a pair for 1 by adding 0. So k = 3 is feasible.

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sachin.13

9 days ago, # ^ |

ok got ,good explanation man :)

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9 days ago, # ^ |

observe if we have n==even , then we cant take the extra element as that can't be compensated by any other element while making pairs so just find the max difference b/w consecutive elements, if n==odd , we have the choice to leave one element and find an extra new one for it, we can do this for each element and store the max diff to the right of it and to the max diff b/w consecutive elements to the left of it and take the max of both the overall answer for this case would be minimum of all such cases

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9 days ago, # ^ |

Solution can be seen here : https://pastebin.com/WyE4dwSL

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Subarna1

9 days ago, # ^ |

My idea is also similarly. But I increment the iterator for even case with i++ instead of i+=2, and I spend whole 2 hours thinking some edge cases for odd case.

Now I get the mistake.

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louisjohns32

9 days ago, # |

c's gonna make me cry when i see the case im failing on

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Shoeib

9 days ago, # |

please don't write problems like B again

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9 days ago, # ^ |

Actually, I think it's a very good and tricky problem. No one on my friends list got it on the first try.

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Shoeib

9 days ago, # ^ |

after the first WA2, i waste about 50mins. at WA5 bcos i were minimizing the ans with 1e15 :(

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m_nature

9 days ago, # ^ |

What? It's just a standard binary search problem.

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9 days ago, # |

after the contest, i felt like it is somehow harder than the last div1+2

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123gjweq2

9 days ago, # ^ |

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9 days ago, # |

Many low-rated people solved D in the last 15 minutes, I don't know if it's my skill issue but I don't think it is an easy problem. So please skip those who cheated.

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FrostBytee

9 days ago, # ^ |

Same with B ig

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tickbird

9 days ago, # ^ |

← Rev. 2 →

no wonder, D is ChatGPT solvable. in fact, first AC was done by hardstuck gray account.

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shviv1

9 days ago, # |

I attempted only the first two problems and found both of them weird, or maybe I am making some blunders; I don't know.

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9 days ago, # |

couldn't solve any thing I'm mad but at least the problems are not guessing problem

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zfnu

9 days ago, # |

+33

I mean I like prefix sums, but prefix sums of prefix sums of prefix sums is too much man.

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alpha1215

9 days ago, # |

AC D at 1:59 :)

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chlalstjr

9 days ago, # ^ |

+25

AC D at 2:01 :(

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christmas964

9 days ago, # |

Thank you very much for contest.

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MonoLith

9 days ago, # |

-7

I spent close to 38 minutes for problem A and got one wrong on B just in hurry of choices Struggle is real kn geometry problems

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9 days ago, # |

today D feels like a buffed version of 2009F - Firefly's Queries.

Which I regret haven't upsolve the problem yet...

But idk maybe I'd stuck on this anyway.

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9 days ago, # ^ |

They are not similar at all.

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YocyCraft

9 days ago, # |

← Rev. 3 →

+28

E: Build a bipartite graph with n vertexs on the left and 60 vertexs on the right. For each 1<=i<=n and 0<=j<60, we add an edge if a[i] has j-th bit set on. For any subset of left nodes, the score is (number of chosen left vertexs )-(number of right vertexs with edge connected to chosen left vertexs ) = (number of chosen left vertexs )+(number of right vertexs without connection to chosen left vertexs )-60, then the answer is (Size of maximum independent set)-60. By the Konig theorem: In any bipartite graph, the size of maximum matching is same as minimum vertex corver, and (size of minimum vertex corver)+(size of maximum independent set)=n+60, we can calculate answer by any maximum matching algorithm.

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9 days ago, # ^ |

Can you explain to me what does "set bit" mean ?

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YocyCraft

9 days ago, # ^ |

If the j-th bit of a[i] is 1. Or if ((a[i]>>j)&1)==1

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Umi

9 days ago, # ^ |

+22

Or just think of it as the classical Project selection problem.

The bits are the machines.

The numbers are the projects.

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Yugandhar_Master

9 days ago, # ^ |

+10

Awesome idea

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Igor_Parfenov

9 days ago, # ^ |

+10

That is... beautiful...

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Al_Imran16731

9 days ago, # |

If i've submitted twice of one problem which one will be counted? If first one is hacked then second one counted?

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pika_._

9 days ago, # ^ |

the last submission that passes pretests will be counted

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Al_Imran16731

9 days ago, # ^ |

Thnak you!! But in current standing showing first one counted!!will this updated?

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ShardsOfNarsil

9 days ago, # |

There is nothing educational about B.

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Shoeib

9 days ago, # |

B reminds me with this problem

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Jerrasterix

9 days ago, # ^ |

It is quite similar to "socks 2" of ABC although a bit different

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9 days ago, # |

im really confused why my binary search submission failed on B, and then my brute force one passed. obviously n is 2000, so i kinda knew it was brute force, but i didnt come up with any counter test case for my binary search one, can someone give a counter example?

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9 days ago, # ^ |

← Rev. 2 →

try this

1 2 4 6 7

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9 days ago, # ^ |

i get 1 on both of my solutions

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darishkhan

8 days ago, # ^ |

Binary search would pass. I did it with binary search.

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9 days ago, # |

In problem E, what does "set bit" mean ?

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Noobish_Monk

9 days ago, # ^ |

bits that are 1 in the bitwise OR

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9 days ago, # ^ |

Thank you, I thought that it the last bit which equal to 1, so I used Dp and got wrong ans

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Yugandhar_Master

9 days ago, # ^ |

+19

But how you become CM without knowing what is set bit. I'm not insulting you just curious to know

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https://codeforces.me/contest/2026/submission/288585044

9 days ago, # ^ |

This is the first time I have ever knew about " set bit " word. I call it is "bit 1" :D

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mahendraakshansh

9 days ago, # |

can someone help why this is wrong?

Greedy solution. Logic is that I store all the i where s[i] == 1 in a greater int set. ans = (n*(n+1))/2

I iterate over them{ count = 0; I have 2 variables l and r, I iterate over l from n->0. if I find s[l] == 0, then count++; if I find s[l] == 1 then I check on count, if it is 0 then I move more left.

When my 0s are exhausted, I move to r and only check the 1s from 1->n, if there is a 1 corresponding the I make count++, if there isnt then I break,

if there is a count then I reduce the answer by the indes of the one in the starting loop}

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chaihard

9 days ago, # ^ |

← Rev. 2 →

The algorithm I used:

Greedily scan from right to left
If s[i] = 1, pair it with rightmost '0' (pos<i)
Else pair it with the leftmost '1'

In 2,3 just store the i, in these cases only, you will be saving money.

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mahendraakshansh

9 days ago, # ^ |

Yep I have done this only. Implementation is a bit different but yep

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xsyncwave

9 days ago, # ^ |

If s[i] = 0, the value $$$i + 1$$$ would be added to the cost no-matter what.

Go from right to left. The idea is to greedily exhaust the rightmost $$$1's$$$ with the first encountered $$$0's$$$ from right to left. If we exhaust in pairs, we will be able to remove as many $$$1's$$$ as possible.

Use a std::deque to store the cost generated by $$$1's$$$. Larger values are stored in the front.

if s[i] == 0,
if !dq.empty() do pop.front(); if it is empty, we can pair it with the already exhausted $$$1's$$$.
do cost += i + 1

if s[i] == 1 do dq.push_back(i + 1). Finally when the dq contains $$$1's$$$, we can greedily exhaust the largest(front) and smallest(back) values.

https://codeforces.me/contest/2026/submission/288593783

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alexshao

9 days ago, # ^ |

Consider 1100011111

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marcobonato

9 days ago, # |

Does anyone know what's wrong with my solution for B at 288576824? It seems to me like it should work, but sadly it fails on test 2.

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9 days ago, # ^ |

Check my braindead solution (that divide to 2 cases of N like you): 288562462
(Why braindead? It just bruteforce every possible choice of "removal" in case N is odd and find the "best" choice (variable "minn")).
(Case N is even, my solution is the same as yours)

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louisfghbvc

9 days ago, # |

PD I know prefix sum. But just too heavy implemention... run out of time. :(

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9 days ago, # ^ |

yea, i gave up on d too

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papa-ka-para

9 days ago, # ^ |

same.

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Noobish_Monk

9 days ago, # |

Do I understand right that problem F is just:

Code persistent queue on 6 stacks, in stacks we store knapsacks, so when answering the query we can just take left and right knapsacks and bruteforce the sum on each edge

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informal007

9 days ago, # |

Is large number in cpp necessary to solve problem D?

Is there a great large number template in cpp? emmmmm

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9 days ago, # ^ |

+17

nah, it is just 10 millions prefix sums

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Saivishwas

9 days ago, # |

how to solve C?

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Valenz

9 days ago, # ^ |

I transverse the array from back to front and use deque to be able to pop front efficiently. The greedy part is to match higher indexed '1' with any first '0' found and pop it. Then, if there are still '1's remaining, just pick the lowest half.

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9 days ago, # ^ |

think of a number 0 can carry the most expensive 1 (right most one)
after eliminate all the 0, the array is left with all the 1, then again, a left most 1 can carry the right most one. (*)
the answer is sum of all 0 index and all 1 index left in (*)

main take away is just greedy and 2 pointers

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8 days ago, # ^ |

I was knowing it was greedy and came up with this idea but couldn't implement again. Fck my implementation skills fk it

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8 days ago, # ^ |

← Rev. 2 →

Chill mate, look at me hard stuck in pupils for 4 months already and still die trying. You peaked in specialist is a good stuff, keep going.

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8 days ago, # ^ |

ahh mate I peaked there I am trying really hard right now solving 1300-1500 range alot hoping to improve soon

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LJY--2002

9 days ago, # |

wrong A why cant long long ? https://codeforces.me/contest/2026/submission/288534007

0 0 68724335200 68724335200 0 68724335200 68724335200 0 0 0 68724335200 68724335200 0 68724335200 68724335200 0 0 0 68724335200 68724335200 0 68724335200 68724335200 0 0 0 68724335200 68724335200 0 68724335200 68724335200 0

wrong output format Expected int32, but "68724335200" found (test case 1)

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Igor_Parfenov

9 days ago, # ^ |

Looks like weird behavior because of reading before and after sync_with_stdio. Place it at the beginning.

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LJY--2002

2 days ago, # ^ |

thanks

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bacxiuda

9 days ago, # |

+11

Can anyone try to hack this submission? I think greedy cannot pass this problem.

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BayernForce

9 days ago, # |

In these contests, is wrong answer on Test 1, considered as a penalty ?

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bhargav_roy

9 days ago, # ^ |

I don't think so. Someone has commented about it here: Link

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BayernForce

9 days ago, # ^ |

Thanks a lot!

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papa-ka-para

9 days ago, # |

Either I am getting dumb day after day, or A,B,C are actually getting difficult contest after contest.

Yesterday was Div (1 + 2) contest.

Today it is simply Div-2 contest.

I felt like today's ABC were more difficult, than Yesterday's ABC.

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9 days ago, # ^ |

Nope

Today I solved ABC, but yesterday C was giving me an insomnia...

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ciprianamza

9 days ago, # ^ |

I also found yesterday's B much easier than today's A or B. Both today's and yesterday's A were a bit more complicated written than usual. Today's C was much easier (at least for me) than yesterday's C, sadly I ran out of time. :D

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9 days ago, # |

A had a really nice hint for the idea that AB, CD are diagonals of a square with side min(x,y) i.e. K<=1414 , hinting the appearance of root of 2

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LMeyling

9 days ago, # |

+19

Problem E seems familiar

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9 days ago, # |

Could anyone please explain how to do B? I used binary search, but it didn't work.

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9 days ago, # ^ |

just brute force the point which will be paired with a point outside the given set.

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9 days ago, # ^ |

you could use binary search or brute force (idk the binary search way)

but in the bruteforce way, if n is even, you just pair up all cells, and if its odd, you test removing every cell and pairing up the others to see which way is cheaper

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9 days ago, # ^ |

Thank you. Got AC.

But I'm still wondering why my BS solution didn't work.

Anyways, thanks.

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9 days ago, # ^ |

np!

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Alosqka

9 days ago, # ^ |

Cause you have added random things in your code which is greedy when you saw the sample test cases

if(n%2!=0 && !x && (((num[i]-num[i-1]+1)/2<=mid) || i==1))

i) why i==1 is getting a free choice

ii) why n%2!=0? it makes no sense. If you approach is because you are already going to use one point to get even points then you are correct but making such mess in the same line? Cause later you are doing i==1 which means you are skipping a element at position = 0. Which again makes no sense cause you can pair any of the i so why position i=0 is getting a priority?

iii) (num[i]-num[i-1]+1)/2 <= mid makes no sense because the new point you will be placing will be not at the middle it will be anywhere ITS FREE WILL. By placing it at the middle of num[i] and num[i-1] gives no contribution cause you are meant to stick the power up for the first element that violates the <=mid thing with its next right neighbor (according to greedy approach).

I hope you understand that you have just written many wrong things and messy in a single line.

Dont be sad or angry about it. Understand it calmly and keep grinding.

If you need my bs approach (a clean code) https://codeforces.me/contest/2026/submission/288520406

Thank You!

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9 days ago, # ^ |

No I got it. I should have read the question carefully to fully understand it. Couldn’t give my best performance today.

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Alosqka

9 days ago, # ^ |

Its okay buddy. Sometimes we perform good and sometimes we perform bad.

Cheers that you are consistent with giving contests! Keep grinding 😇

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9 days ago, # ^ |

It's optimal to paint neighbors.

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9 days ago, # ^ |

why this case 2 and not 1 please answer


2
1 3

we can paint cell number 2 what is wrong with that ?

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9 days ago, # ^ |

You have to choose two cells at a time. In this test case the no. of cells is even. So you can't choose any cell outside this given set.

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9 days ago, # |

← Rev. 2 →

Miss chance to become master AGAIN by declaring $$$a_i$$$ of Problem E as int, not long long.

Note: shifting more than bit width is an undefined behavior, so I get WA1, not WA2 or WA3, which is confusing to me.

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dino_merlin

9 days ago, # |

+15

D is very reminiscent of Mosaic from this year's IOI.

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dg351

9 days ago, # |

Can anyone please explain why my solution fails:

Method: For even n, directly calculate the max distance between all pairs (a[0],a[1]), (a[2], a[3]).....

For odd n, calculate min(max(((a[0],a[1]), (a[2], a[3]).....), ((a[1],a[2]),(a[3],a[4]).....))).

Submission

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9 days ago, # ^ |

Consider 1 2 10 20 21

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dg351

9 days ago, # ^ |

Got it, thanks a lot.

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Aritro_Sarkar

9 days ago, # |

-41

The worst contest in codeforces history

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towrist

9 days ago, # |

Spent a lot of time thinking about E after coming up with the obvious Maximum Matching algorithm, not able to believe its actually the intended solution.

While not having a template for it cost me today's E, I am really surprised Maximum Matching was allowed to appear in E, especially when the idea that each bit should have atleast one unique entry is quite simple.

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PaciukZvichainyi

9 days ago, # ^ |

I agree that it is too much for E. But it seems to me that we have no right to complain, since this is Educ and in fact it is intended for this: you know an algorithm — 3 minutes will be enough for you, if you don't know — you are cooked for the next two hours.

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pagol

9 days ago, # |

"Euclidean or Manhattan" missing from "A"s statement.

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marcobonato

9 days ago, # ^ |

If it's not specified I think you can just assume Euclidan

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4uck_d3v

9 days ago, # |

-11

Are there any ways to solve problem E with dp? I tried to solve it with dp but get WA.

Let dp[i][j] represent the value of mask such that the number of 1-bits in mask is minimized when constructing a subsequence of length j with i as the last element of the sequence.

I think we can combine with some greedy tactics, such as save more than one mask for each dp[i][j] at a time.

Sorry for my bad English.

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Abito

8 days ago, # ^ |

-8

No it's not possible because even a mask where the bits might not be minimized might be better for bigger j and many other stuff

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9 days ago, # |

-29

The worst contest in codeforces history

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cwz2024

9 days ago, # |

-13

Using random to solve E: (This is gonna WA on test 2) (OMG, it's running on test 10!) (OMG, it's running on test 20!) (OMG! It ACCEPTED!!!) (I'm going to submit a lot of them to ensure I can pass main test) (passed 6/27) (ALL 6 AC submissions are hacked, rank 103->511 QAQ)

So is there a way to hack these submissions at a high possibility? Or it was just because I was to silly when randomizing? Can these submissions be hacked? 288611596 288611196

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cwz2024

9 days ago, # ^ |

Okay, I've come up with a hack and hacked myself:(

Spoiler

1
7
3 3 3 125 125 125 125

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9 days ago, # ^ |

thank you counterexample lmao, this one also breaks my code for some reason

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9 days ago, # |

← Rev. 3 →

for problem B this case

2
1 3

why the answer is 2 and not 1

if k=1 then we paint 1 and 2 and then we can easily paint 2 and 3 ??????????? why I'm reading the statement wrong every time ????

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marcobonato

9 days ago, # ^ |

You can only paint WHITE cells. So, after you paint 1 and 2, cell 2 will be black, so you can't use to paint 2 and 3 since you can only paint white cells.

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PaarthJuneja

9 days ago, # ^ |

i spent a few seconds thinking the same thing during the contest lol

notice that the problem statement mentions in bold letters that you can only paint white boxes. once you've done the operation on 1 and 2, 2 becomes black so you can't operate on the second box anymore.

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RDDCCD

9 days ago, # |

+19

Seems that problem F only have operation 3 in the sample. See here

code copied from jiangly's code and add one assertion on operation 3. Only have 96 tests during the contest while the verdict is RE on 97.

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awoo

9 days ago, # ^ |

← Rev. 2 →

+31

My bad :( We only discovered that during the contest. Somehow all incorrect solutions we had were incorrect even on that set of data and all correct were still correct after the generators got fixed. Luckily, not a lot of people got affected by the incomplete tests.

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jeroenodb

9 days ago, # ^ |

I now got affected due to some runtime error that only fires if there's actually operations of this type in the input.

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9 days ago, # |

← Rev. 2 →

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9 days ago, # |

could someone provide counter example for this submission , im getting WA on test 5, and the solution seems kinda silly but it runs in O(60*n^2) in worst case scenario, and it seems correct to me. basically what i do is for every element i, lets call it valid, if for every bit that is set in in a[i], it has atleast 1 valid neighbour. first we assume all indices are valid, and then we check if there is an element that isn't and if there is such an element we delete it, and also since it may have made some other indices valid, but now doesnt anymore, we have to repeat this until no changes occur. this runs in O(60*n^2) in the worst case where we delete 1 element per iteration, and every element will be deleted, so its fast enough. but its getting WA on test 5 and i cant seem to find a simple counterexample.

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9 days ago, # ^ |

nvm, got it

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PaarthJuneja

9 days ago, # |

i went from getting a +65 in yesterday's contest to literally not being able to solve a single problem today. cf contests are SUCH a rollercoaster i swear :\

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pranavoak7698

9 days ago, # |

Sorry Demon_Master

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SilverCipher

9 days ago, # |

← Rev. 2 →

+10

Educational contest educated me that I know nothing. Nothing: No, you don't know me. Me: He knows me. He: Who, are you? Who: No, I am not "?". Not: No, your not Not. No: I know.

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invincible_rj

9 days ago, # |

C was easier than B, also A was weird. not a balanced contest tbh,

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satwika_55

9 days ago, # |

What the hell was that B question ?? After several incorrect submissions, I came to a conclusion that its a BS + DP problem.

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Hazer_BJU

9 days ago, # ^ |

For even numbers, it is only possible to paint them pairwise from left to right. For odd numbers, you can delete one element and solve it as even numbers.

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8 days ago, # ^ |

← Rev. 2 →

you can bruteforce it: for n even, just pair up all values, for n odd, fix every value and pair the others and see which way you get the minimum cost

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Hazer_BJU

9 days ago, # |

-8

I can't figure out why this code got WA on D. How mysterious.

#include <bits/stdc++.h>
#define MAXN 300005
using namespace std;
typedef long long llong;
llong n,q,L,R;
llong a[MAXN],presum[MAXN],sumsum[MAXN],press[MAXN],ppsum[MAXN];
llong find_pos(llong p) {
	llong l=1,r=n+1;
	while(r!=l+1) {
		llong mid=(l+r)/2;
		if (n*(n+1)/2-(mid-1)*mid/2>=p) l=mid;
		else r=mid;
	}
	return l;
}
llong query(llong p) {
	if (p==0) return 0;
	llong len=find_pos(p),order=n-len+1;
	llong starting=n*(n+1)/2-len*(len+1)/2+1;
	llong minus=presum[order-1];
	llong ans=ppsum[p-starting+order]-ppsum[order-1]-minus*(p-starting+1)+press[order-1];
	return ans;
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>n;
	for (llong i=1;i<=n;i++) {
		cin>>a[i];
		presum[i]=presum[i-1]+a[i];
		ppsum[i]=ppsum[i-1]+presum[i];
		sumsum[1]+=presum[i];
	}
	press[1]=sumsum[1];
	for (llong i=2;i<=n;i++) {
		sumsum[i]=sumsum[i-1]-(i-1)*(n-i+2);
		press[i]=press[i-1]+sumsum[i];
	}
	cin>>q;
	for (llong i=1;i<=q;i++) {
		cin>>L>>R;
		cout<<query(R)-query(L-1)<<'\n';
	}
	return 0;
}

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vuysztrcKJBHVIUVbkjbkhj

8 days ago, # |

← Rev. 2 →

-8

please help me to find why this code is not working for c https://codeforces.me/contest/2026/problem/C

//this is code
#include <iostream>
#include <string>
using namespace std;
 
int main() {
    int tc;
    cin >> tc;
    while (tc-- > 0) {
        int n;
        cin >> n;
        string str;
        cin >> str;
        int s = 0, e = str.length() &mdash; 1;
        long long ans = 0;
        
        while (s <= e) {
            if (e - 1 >= s && str[e - 1] == '0') {
                int idx = e - 1;
                while (idx >= s && str[idx] == '0') {
                    ans += (idx + 1);
                    idx--;
                }
                e = idx;
            } else {
                ans += (s + 1);
                s++;
                e--;
            }
        }
        cout << ans << endl;
    }
    return 0;
}

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chenhongrui

8 days ago, # |

Awesome contest!I am rank 3 and will become IM!

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8 days ago, # ^ |

orz

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Hardy_Lee

8 days ago, # |

Why unrated?

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Tannyyyyy12

8 days ago, # ^ |

Ratings will change in sometime i guess..

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Hardy_Lee

8 days ago, # ^ |

But unrated allowed maybe.

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TokaiZaopen

8 days ago, # |

+36

problem D makes me feel like I'm not participating in CP …… All of the difficulty is to write the code. Too little about algorithm and too much about coding. Well, maybe it can help us improve our coding ability ……

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ImmortaLimit

8 days ago, # ^ |

problem E is also not good. If you know Maximum Weight Closed Subgraph you can solve it quickly, but if you don't know that you are hardly solve it.

I think Div .2 D and E should have good ideas without too hard code or too hard algorithm.

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vgoofficial

8 days ago, # ^ |

+28

Isnt the point of Educational Rounds to teach these topics to a wider range of participants?

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