error

3 years ago, # ^ |

+53

I think that problemsetter should tell competitors about statement-update quickly.

→ Reply

errorgorn

3 years ago, # ^ |

← Rev. 7 →

+22

I edited the statements for problems A,D,E only (and a removed problem). Problem B was added very late into testing so I did not look at it. I wrote problem E with only using $$$T$$$, I am not sure why it is changed as well. (I believe it was added quite late into testing as well so please do not blame the rest of the testers.)

I tried to do my best as a tester but I did not want to interfere too much with the content of the problems. I personally felt that I was pushing it when I removed some backstories in the statements (you will notice the english statement for A is much shorter than the russian statement).

I don't really have anything to defend myself but I did not know the statements were changed from what I edited it to until today.

I can understand your frustration but there is only so much a tester can do...

If there are any comments that are actually about translation and making the statements easy to read, I would be happy to read them. The problems you listed have nothing to do with translation. Do you really think they are the translators fault? They also appear in the russian version of the statements you know.

→ Reply

3 years ago, # ^ |

← Rev. 2 →

+23

That's not phyzzmat's or errorgorn's mistake, because they made translation before some edits in english texts.

So, sorry for such terrible statements :(

→ Reply

unknown0711

3 years ago, # ^ |

← Rev. 5 →

Someone please give hint for Problem D1

→ Reply

Kevin114514

3 years ago, # ^ |

use dp

also , $$$\sum_{i=1}^n(\lfloor\frac{n}{i}\rfloor)$$$ is about $$$n \ln n$$$

→ Reply

atreides_

3 years ago, # ^ |

← Rev. 4 →

-11

error

I don't know, when will the editorial release so, I am commenting here.

In problem B I have just used

  sort(all(a), [&](auto &a, auto &b)
         {
             if (a.first != b.first)
                 return a.first > b.first;
             else
                 a.second < b.second;
         });

to sort the array and I got TLE.

Even if I remove else a.second < b.second;, still I am getting TLE.

I used this sort(all(a), greater<pair<ll,ll>>()); after the contest, the solution got accepted.

Solution1

Solution2

If possible, kindly resolve this issue, as according to me this should be a problem.

→ Reply

errorgorn

3 years ago, # ^ |

Eh, I am only a translator... I guess i can help fix your code

The issue is with the code below,

sort(all(a), [&](auto &a, auto &b)
     {
         if (a.first != b.first)
             return a.first > b.first;
         else
             a.second < b.second;
     });

you have missed out the return statement in the else I think changing it to the edited code below should work

sort(all(a), [&](auto &a, auto &b)
     {
         if (a.first != b.first)
             return a.first > b.first;
         else
             return a.second < b.second;
     });

→ Reply

FHVirus

3 years ago, # ^ |

You missed a return in the compare function in solution 1.

→ Reply

Sprdalo

3 years ago, # |

← Rev. 2 →

+27

This is my first time testing the round and I hope my feedback helped to balance it! I encourage you to read all of the problems — as they are very interesting.

→ Reply

3 years ago, # |

-10

A great time for Chinese people. Hope everyone can gain rating.

→ Reply

yeetholmes619

3 years ago, # ^ |

-10

Why downvote this? The comment is pretty wholesome.

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Closed_

3 years ago, # |

Note the unusual timing

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kchwoony210

3 years ago, # |

-14

D1>D2? That's weird. I hope everyone can solve them.

→ Reply

3 years ago, # ^ |

← Rev. 2 →

I think that easy version points $$$>$$$ hard version points is usual, for example Codeforces Raif Round 1 (Div. 1 + Div. 2) and Codeforces Round 750 (Div. 2).

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kchwoony210

3 years ago, # ^ |

-10

but why is easy version points are greater than hard version points?

→ Reply

3 years ago, # ^ |

I think that if you solved hard version, you almost can solve easy version.

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Logical_Operator

3 years ago, # |

← Rev. 2 →

I always see, when the contest is held in unusual time, there are relatively less people who participate in the contest. And that is why we get less rating changes as compared to a usual time contest (because rating change also depends on no. of participants).

→ Reply

reportplag

3 years ago, # ^ |

-35

Ok so you participate in contests just to get some ratings?

This mentality of just gaining ratings and not enjoying CP as a sport has led to so many cheating cases these days!

→ Reply

Logical_Operator

3 years ago, # ^ |

+64

haha...go and check my profile...still stuck on specialist after giving around 70 contests. If I were a cheater I'd have been orange or red till date after giving around 70 contests. Moreover I have 192 days problem solving streak(and counting) till date. If I were a cheater, why would I have been practicing CP this much. These things are enough for anyone to understand that I don't cheat.

Reward is something people wants when they works for it. People first need expectation to gain something in life provided that they are working fair. Yes I give contest for rating, but that is not greed. Because your school teacher has not taught you the meaning of greed. Greed is something that we want without working, and expectation is something that we want when we work hard for it. That is called 'achieving', not 'cheating'. I hope god made you smart enough to understand this.

→ Reply

agarwala2512

3 years ago, # ^ |

op(2)

→ Reply

3 years ago, # ^ |

-64

Looks like you have cheater mentality of just giving contest for Greed(ratings). Stop it or you will be banned..

→ Reply

TechnobladeNeverDies

3 years ago, # ^ |

+43

who are you, the thought police? no one's getting banned for their mentality

→ Reply

Logical_Operator

3 years ago, # ^ |

← Rev. 2 →

+11

→ Reply

wargang

3 years ago, # ^ |

respect++

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LetterC67

3 years ago, # |

← Rev. 3 →

+31

First time being a tester, so excited (^.^). Although the start time is unusual again, hope you guys could spend time participating the well-prepared round with interesting problems >~<

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WalterWhite52

3 years ago, # ^ |

your graph is really motivating :)

→ Reply

LetterC67

3 years ago, # ^ |

Thanks! I have focused on CP since last year in order to have an opportunity to participate in the national OI in my country. If I could proceed to the national round, my next target is reaching yellow color on CF ;)

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WalterWhite52

3 years ago, # ^ |

I hope you do!, all the best

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INeedCarb

3 years ago, # ^ |

solo ?

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ChickenInKitchen

3 years ago, # ^ |

look at my graph and be demotivated again lol

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turzo_sawroop

3 years ago, # |

← Rev. 3 →

Hope that my streak of underperformance ends here

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vishaaaal

3 years ago, # |

I hope to solve four problems in this round. Looking forward to it :).

→ Reply

Md_yz

3 years ago, # ^ |

maybe 5?

→ Reply

vishaaaal

3 years ago, # ^ |

But that's after I solve 4 :).

→ Reply

FairyWinx

3 years ago, # |

+39

As a tester I'm a tester

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mac_n_cheese_pog

3 years ago, # ^ |

zamn

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mon0pole

3 years ago, # |

+62

Meme

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raj_saurabh

3 years ago, # ^ |

+13

Hopeforces

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evermore

3 years ago, # |

fingers crossed for some non negative delta

→ Reply

3 years ago, # |

Are Ratings going to update before the round starts ?? Any Idea ??

→ Reply

WAonPretest

3 years ago, # |

Can we participate in a contest without rating update from the previous contest?

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yoyohalka

3 years ago, # |

Thxx for the unusual timing. Its better for us (russians)

→ Reply

quagmire

3 years ago, # |

hey I have confusion as previous contest rating is not updated yet so this contest rating will be calculated with change in rating or rating before previous contest ? please someone clearify

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MohamedAboOkail

3 years ago, # |

The contest was delayed by 10 minutes!

→ Reply

Nyamdavaa.U

3 years ago, # ^ |

+20

Maybe in the meantime, the rating will be changed

→ Reply

peek4b00

3 years ago, # ^ |

done

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Sacchidananda

3 years ago, # |

Why delay??

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MikeMirzayanov

3 years ago, # |

+237

Sorry, I postponed the round. I would like to have time to update the rating based on the results of yesterday's Div3. I'm in a hurry!

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-AmrKhaled-

3 years ago, # |

Good decision to increase 10 minutes , i didn't drink coffee

→ Reply

xcx0902

3 years ago, # |

-7

Why it's start time is 19:15(in China)?

It's usually 22:45(in China).

→ Reply

Venn

3 years ago, # |

+14

Congratulations to 4qqqq to get 107 ratings. XD

→ Reply

Phileein

3 years ago, # |

Good luck to everyone

→ Reply

QuietBeautifulThoughts

3 years ago, # |

-6

Is it div.3?

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Sacchidananda

3 years ago, # ^ |

No its Div93849329 you're too noob

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hbarp

3 years ago, # ^ |

+29

Is it some IGM alt ?

→ Reply

3 years ago, # ^ |

Maybe LGM.

→ Reply

soumyadeep_pal_21

3 years ago, # ^ |

← Rev. 2 →

list, coding style seems very close

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phir

3 years ago, # ^ |

-6

ow that's rude

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dXqwq

3 years ago, # ^ |

+120

It is div.trash

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froggyzhang

3 years ago, # ^ |

+58

It is div.114514

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Lets_crack_it

3 years ago, # |

Another mathforces round :(

→ Reply

Next_semester_kopay_dimu

3 years ago, # |

+13

Retired_MiFaFaOvO participating .. after a long time

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rgnerdplayer

3 years ago, # |

The amount of ad hoc contained in problem C frightened me.

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ChickenInKitchen

3 years ago, # |

+13

pretest 2 of problem c has cost me my mouse and sanity

→ Reply

thalapathi

3 years ago, # ^ |

How do you solve it? Pretest 2 killed me

→ Reply

3 years ago, # ^ |

In my case I missed the input like:

1

1 1

1 1 1073741823

it took my last 40 minutes during the contest...

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pranavvinchurkar

3 years ago, # |

How to solve D2? How to optimise from O(Max(ai)*log(max(ai)))?

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hxu10

3 years ago, # |

← Rev. 2 →

The problems themselves are good, but the contest is unbalanced. There are so many math problems. Problems A and B are too easy, and problems D E are too hard. If you have some problems with problem C, like me, wasting too much time for mistaking "OR" by "XOR", you will be finished.

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NimbleNeuron

3 years ago, # ^ |

I saw it after seeing your comment I thought it's xor of subsegment.. I'm doomed

→ Reply

_Enginor_

3 years ago, # ^ |

Same thing happened to me also :(.

→ Reply

shreyas.mandade

3 years ago, # |

Hint for C I wasted my 30 minutes in thinking this logic. Could have googled it early and got a decent rank!

→ Reply

Yahia_Emara

3 years ago, # ^ |

Same

→ Reply

Yahia_Emara

3 years ago, # ^ |

Two Friends Of Mine Tried Solving It Using Lazy Propagation xD, only one got ACC

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thalapathi

3 years ago, # ^ |

I was able to figure this out — but what was the logic for finding the frequency/number of each bit present in the array? e.g. in array [1, 2] — 0th bit freq is 1, and 1st bit freq is 1. How do I get that info from the subsegments?

→ Reply

3 years ago, # ^ |

← Rev. 3 →

if an element is covered by more than 1 subsegments, then its largest possible value is the bitwise AND of all values corresponding to the subsegments that are covering it.

For example, if we have n=3 and subsegments 1 2 7, 2 3 14, then the largest possible value of a_2 is 7 & 14 = 6. Moreover, a possible value must only have the bits occur in the binary representation of the largest possible value. But we can safely just use the largest possible value here.

I used a segment tree (range update + point query) to process this information, and I wonder whether there exists a simpler approach.

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sherlockedcp

3 years ago, # ^ |

What i did was, took OR of all the OR's given (irrespective of l,r) and just multiplied it with pow(2,n-1). This passed pretests as well as sys tests.

→ Reply

3 years ago, # ^ |

It's like we don't need to reconstruct a valid sequence, am I right?

Could you briefly explain the intuition (or proof) behind this? Thanks

→ Reply

faraz16iqbal

3 years ago, # ^ |

Yeah, we don't need to construct a valid sequence.Try this link here for a formal proof

→ Reply

sherlockedcp

3 years ago, # ^ |

If X is the bitwise OR of a segment, then for each set bit of X, there exists an element with that bit set. Then consider all subsequences excluding that element (2^(n-1) of them): they either XOR to 1 or 0 times 2^(bit). If they XOR to 1, then subsequence with our set bit XORS to 0; if they XOR to 0, then subsequence with our set bit XORs to 1. Either way, we are guaranteed a contribution of exactly 2^(bit) times 2^(n-1) from that bit, i.e. its contribution to x. So we just need to know if a given bit is set anywhere in the whole sequence, and if so we add that contribution. Otherwise, the bit is skipped.

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_Enginor_

3 years ago, # ^ |

← Rev. 2 →

i am such a stupid that i started figuring out a valid sequence and I figured out how to find a valid sequence.But when i used combinatorics to find the number of subsequences i realized that we just need to know whether there is a particular bit on in any of the element in the array which can easily be figured out by just looking OR of all elements.

2^(num of on jth bit-1)*2^(n — (num of on jth bit)) you can see that num of on jth bit cancels out but you can see num of on jth bit — 1>=0 num of on jth bit >=1

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nyet

3 years ago, # ^ |

Why does this work? I got that OR of all the ORs given will be the OR of the original array. But why does it become sum of XOR of all subsets after multiplication with pow(2,n-1) ?

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ubermensch123

3 years ago, # ^ |

In binary, look at digit r (aka the bit corresponding to the place 2^r) of all numbers in the original array. Say a of them are 1s and b of them are 0s. From the set of a numbers, choose i to be in some arbitrary subsequence, and from the set of b numbers, choose j. Note that making such a choice corresponds to forming exactly one subsequence (for now, dealing with digit r only). Clearly, the bitwise XOR will yield a 1 for digit r iff i is odd. If a>0, this corresponds to aC1 + aC3 + aC5 + ... = 2^(a-1) ways of choosing a set of i numbers from the original a, where i is odd, and simply 2^b ways of choosing any j numbers from the b with 0s at digit r. Hence, when a>0, the total contribution from all subsequences to coziness by digit r is 2^(a+b-1) * 2^r = 2^(n-1) * 2^r, as a+b=n. When a=0, however, the contribution to coziness is 0. The coziness can thus be written as 2^(n-1) * sum(2^r, where a>=1 at position r). Observe that this sum is the bitwise OR of the n numbers, and we're done.

→ Reply

nyet

3 years ago, # ^ |

← Rev. 4 →

Thanks a lot. This seems fine. Although I found this another explanation by going through the comments-

If at a given place, if at least one of the numbers has its bit = 1 :

1) lets remove any number from the array whose bit is 1 at that place.

2) now, rest of the numbers(n-1) will make pow(2,n-1) sets, now all the sets with xor of bits = 1 will contribute to the solution (addition of XORs)

3) but if xor of the bits in a set is 0, the it will become 1 when the excluded number with set bit is added to all the sets, and then it will contribute to the solution.

4) So if at least one of the numbers has its bit 1 at a place, it ensures that this place will contribute place_value*pow(2,n-1) to the solution (sum of XORs)

5) Now to ensure if at least one number has its bit 1 at place, we can find OR of all the numbers and check if the it is 1 at that place.

6) OR of all the given ORs will be the OR of the array.

→ Reply

User_Carrot

3 years ago, # ^ |

I wasted a lot of time in solving this, however, I still didn't solve it. Thanks for your hint. My rating must be reduced a lot. :(

→ Reply

thalapathi

3 years ago, # ^ |

+15

Rating is temporary. Skill is permanent.

→ Reply

User_Carrot

3 years ago, # ^ |

Thanks.

→ Reply

spookywooky

3 years ago, # ^ |

+23

So its googleforces :/

→ Reply

User_Carrot

3 years ago, # ^ |

Agree.

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third_try

3 years ago, # ^ |

i just cannot become friendly with XOR, doesnt matter how much i try.

→ Reply

MijPeter

3 years ago, # ^ |

Yea me too. In theory XOR has the same abstract complexity as AND and OR. But I feel like our (at least mine) minds don't understand XOR the same, natural way they understand AND and OR. Also XOR has so many weird characteristics that aren't easy to come up with on spot.

→ Reply

tonytony

3 years ago, # |

+10

How to solve D1?

→ Reply

Pub

3 years ago, # ^ |

+64

Sorry for my poor English.

Let we call $$$b_i=\gcd(a_1,\cdots a_i)$$$,then we have $$$b_i|b_{i-1}$$$.

And $$$b$$$ is like $$$[c,\cdots c,d,\dots d,e,\cdots]$$$.

$$$dp_i$$$ Means the max value when you choose all $$$j$$$ That $$$i|a_j$$$.

Add some $$$i$$$ after $$$[\cdots,ki,\cdots,ki]$$$,so $$$dp_i=\max(dp_{ki}+i(t_i-t_{ki}))$$$ where $$$t_i$$$ means $$$\sum\limits_{j=1}^n[i|a_j]$$$.

Then we can solve it in $$$O(a\log a+n\sqrt a)$$$.

→ Reply

Priyansh31dec

3 years ago, # |

-45

D1 was really interesting!

→ Reply

Borjan

3 years ago, # ^ |

Really curious to know, How to do it ?

→ Reply

yaminote

3 years ago, # ^ |

-9

This is certainly not the right place to talk about it but I saw you video reviewing profile of users, I am stuck in green here. Would be glad If you could suggest something. Many thanks

→ Reply

Priyansh31dec

3 years ago, # ^ |

I don't have a lot to suggest. It seems like you left CP for a long time. Anyway, if you continue to solve problems in rating [1300:1500], you should be good to go. Also, up solving at least 1 problem after every contest is a must.

→ Reply

digishek

3 years ago, # |

← Rev. 2 →

How to solve D2 ? Does it involve the fact that we only have to consider atmost 25 distinct elements?

→ Reply

Gary2005

3 years ago, # ^ |

+15

I didn't use it . I just made some small constant optimization on the code of D1 .

If so , I think there is no need to set D2 ...

→ Reply

MGJ0007

3 years ago, # |

Problem B, i figured out total time, but how can i print the vector? Divan's office at 1 and then sort given vector in decreasing order, then v[0] at 2,v[1] at 0, v[2] at 3 v[3] at -1 ....

→ Reply

https://codeforces.me/contest/1614/submission/137006738

3 years ago, # ^ |

you can sort a vector<pair<int, int>> where the first element of each pair is the value, and the second element is the original index. In such way you can preserve the information of the original index.

→ Reply

StupidCat

3 years ago, # |

Hi everyone, sorry to bother but :( in the contests I participated in during this time, the codeforces page loads very slowly :( this prevented me from submitting my post C at the last minute can anyone give me some advice.

→ Reply

Andycraft

3 years ago, # ^ |

Try submitting using Codeforces Tool, or through these mirror sites:

→ Reply

StupidCat

3 years ago, # ^ |

I tried but it's not loading MathType.

→ Reply

Andycraft

3 years ago, # ^ |

You can use the main site for viewing problem statement, while using the mirror sites for submission :P

→ Reply

achyut_1412

3 years ago, # |

← Rev. 5 →

-11

Help needed!!!! In q.2 of today's contest I have made a submission :

Above is the code ,I surely believe that there is some system error that's why it is giving me wrong answer on test case 4 ,which passed when I made some meaningless changes,in short can someone tell me why this code is not passing because I have made this submission after 13th minute ,so if this code passes then I get a very good rank as compare to now...please help

→ Reply

Gary2005

3 years ago, # ^ |

Did you forget to use long long ?

→ Reply

achyut_1412

3 years ago, # ^ |

No i have used it. https://codeforces.me/contest/1614/submission/137006738 you can check it here @vaaven please help!!

→ Reply

Gary2005

3 years ago, # ^ |

+11

"#define ll int"

→ Reply

VaibhaveS

3 years ago, # |

← Rev. 3 →

Can someone help me understand why https://codeforces.me/contest/1614/submission/136996986 got TLE?

its O(nlogn).

EDIT — Seems like many people who used Python TLE'ed this problem.

→ Reply

azberjibiou

3 years ago, # |

D is such a funny problem. The difference between D1 and D2 is that max(a_i) difference is about 4 times bigger. Time for ConstantForces

→ Reply

3 years ago, # ^ |

Author's solutions have different asymptotics for D1 and D2.

Editorial will be ready soon.

→ Reply

sikdangcopo

3 years ago, # ^ |

+16

I hope it isn't max(A_i)^3/2 and max(A_i)log(max(A_i)).

→ Reply

https://codeforces.me/contest/1614/submission/137006738

3 years ago, # ^ |

It is

D1: max(A_i) * log(A_i) D2: max(A_i) * log(log(A_i))

→ Reply

iaNTU

3 years ago, # |

← Rev. 3 →

+63

My D2: TLE on test 1[main tests] https://codeforces.me/contest/1614/submission/137008202

→ Reply

WalterWhite52

3 years ago, # |

Started 1 hour late
Happy that solved A,B in 15 min.
Figured out C quick
Forgot to take MOD in final answer for C

Moral: No matter what happens, even if it's a life and death situation, never forget to take MOD :(

→ Reply

achyut_1412

3 years ago, # |

Help needed!!!! In q.2 of today's contest I have made a submission :

→ Reply

Sol1

3 years ago, # ^ |

line 4

#define ll int

and your "meaningless changes" include changing it to #define ll long long int

→ Reply

Siuuuuu_7

3 years ago, # |

← Rev. 2 →

+12

Why this submission in C++ 17 gives TLE for D1

And the similar solution to it C++ 20 passes within the given time limits even though the worst case runtime for both the solutions is same.

Why changing in different version of C++ has so much big difference in runtime and if so then should the problem not have much flexible time limits as you can see it cost me an extra penalty just because I choose C++17 version initially

→ Reply

Malek_souissi

3 years ago, # |

C is segment tree range-AND for each segment with their beauty and then XOR-sum in O(n) right ? Got TLE on test case 2 sadge

→ Reply

3 years ago, # ^ |

← Rev. 3 →

Doesn't actually need to be this complicated. If X is the bitwise OR of a segment, then for each set bit of X, there exists an element with that bit set. Then consider all subsequences excluding that element (2^(n-1) of them): they either XOR to 1 or 0 times 2^(bit). If they XOR to 1, then subsequence with our set bit XORS to 0; if they XOR to 0, then subsequence with our set bit XORs to 1. Either way, we are guaranteed a contribution of exactly 2^(bit) times 2^(n-1) from that bit, i.e. its contribution to x. So we just need to know if a given bit is set anywhere in the whole sequence, and if so we add that contribution. Otherwise, the bit is skipped.

In pseudocode terms

ans = 0
For each segment:
    get(l,r,x)
    ans |= x

ans *= 2^(n-1)

(Take mods where appropriate)

→ Reply

Malek_souissi

3 years ago, # ^ |

I wonder how I miss that while still understanding we can do XOR-sum in O(n) for kinda the same argument. Thank you that was helpful.

→ Reply

3 years ago, # ^ |

I actually missed it in contest too. I did a brute force over each bit, setting only those I had to set to 0 (i.e. where we had information on a whole segment that the bit was 0). From there I applied some combinatorics.

This was all a bit of a waste of time, and my solution just scraped through in the 1s limit.

I later realised that, by symmetry, all my combinatorics effectively came out at "half of all subsequences" for any set bit.

→ Reply

Fastlaner

3 years ago, # ^ |

← Rev. 2 →

I guess for Question C, the question should have also asked for the possible values of elements of array, so that Lazy_propagation would have become necessary.

Here's my submission by which we can also print possible values of array.

→ Reply

xuanji

3 years ago, # ^ |

lazy propagation wouldn't be "necessary" strictly speaking, because you can apply all the range updates before printing the array

→ Reply

Fastlaner

3 years ago, # ^ |

Actually I used LAzy propagation in Segment tree for range updates. So I guess it's necessary.

→ Reply

xuanji

3 years ago, # ^ |

You can do offline range update (i.e. process all the range update operations, then print the array) without lazy propagation segment tree

→ Reply

Account_to_stay_unrated

3 years ago, # ^ |

I did exactly that still got WA on pretest 3. Any idea why?

→ Reply

3 years ago, # ^ |

← Rev. 2 →

Need long long, not int, I think. When you’re multiplying two ints together you’re potentially getting an overflow before you MOD them.

→ Reply

Account_to_stay_unrated

3 years ago, # ^ |

← Rev. 2 →

but, I have defined #define int long long in the beginning (around 5th line)

→ Reply

3 years ago, # ^ |

Sorry you're right. You shouldn't MOD your value o.

→ Reply

Sol1

3 years ago, # |

-30

worst problemset i've ever saw...

→ Reply

lexiyvv

3 years ago, # |

Very great contest,but I think the time limit for C can be 2s.

→ Reply

3 years ago, # |

+14

Finally Specialist Today !!!!

→ Reply

BugLeave

3 years ago, # |

In problem B, I failed for the error message for the pretest: "wrong answer answer is wrong: pans = 14 is not equal real_pans = 18 (test case 1)" But the example output is 14 for the test case 1. I checked many acceptable submissions, the answer is still 14 not 18. Do I miss something?

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Hiphop21

3 years ago, # ^ |

optimal answer is 14 but your output array will give 18 that's why it is showing 18..

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Chiefan

3 years ago, # |

Tester: "As a tester, I tell you that the problems are well balanced and well prepared!!"
CF User: "Ok I will participate then"

Tester IRL

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Iftekhar_Hakim_K

3 years ago, # |

+11

Passed pretests of D2 only to see failing on sample in system testing. -,-

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Ahmadsm2005

3 years ago, # ^ |

I know right. I literally did some stupid optimization and passed D2 right now. To be honest, look at the time variance of any code from any test although solution is independent of N. The time variance is just too large sometimes reaching 500ms in that problem for no reason.

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Iftekhar_Hakim_K

3 years ago, # ^ |

Just noticed system test was run for 3rd time. (Pretest passed, TLE2 on 1st ST, TLE18 on 2nd ST, AC on 3rd ST). What is going on? :|

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3 years ago, # ^ |

I guess they decided to give people a fair go (i.e. when the servers weren't overloaded)

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-_-INV4D3R-_-

3 years ago, # ^ |

My submission TLEs on different test each time I resend it... Time limits should either be tighter or looser.

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abdzag

3 years ago, # |

137041121 137027753 This is not ok. One parnthesis off

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kne3lb4m3

3 years ago, # |

Can anyone please explain why this solution of mine for problem B is getting TLE 137017820

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LastDance_NotLastOfMe

3 years ago, # ^ |

large input/output causes TLE, use this instruction before write anything in the main function:

ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL) ;

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yuBai0123

3 years ago, # |

-14

Why do you set the minimum temperature of question e to 0? Why is it not 1 or -1e9? The important boundary information is written so unobviously, fuck.

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Nabiur_Raouf

3 years ago, # |

← Rev. 3 →

I can't understand the verdict of [problem:1614B]problem B in my code? https://codeforces.me/contest/1614/submission/137041237

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Nabiur_Raouf

3 years ago, # ^ |

← Rev. 2 →

Can you check my code?

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kne3lb4m3

3 years ago, # ^ |

check your answer manually it's coming wrong..

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semicolonised

3 years ago, # ^ |

2 5 4 3 1 0

Your array outputs: (3*2)*3+(2*2)*8+(1*2)*10+(1*2)*6+(2*2)*1 = 86 which is not the correct answer.

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peti1234

3 years ago, # |

+16

What a painful lesson to learn that $$$2^{30} > 10^9+7$$$. (Problem C)

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Eclecticity

3 years ago, # |

Was this the intended solution for D1?
I used 1D DP over all the possible factors of all the numbers.
https://codeforces.me/contest/1614/submission/137040944

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NightRage

3 years ago, # |

I wrote a dp solution for probleme [problem:https://codeforces.me/contest/1614/problem/C] , while the complexity of my code should work just fine, the time constraints didn't fit [submission:https://codeforces.me/contest/1614/submission/137034593],

changing from long long to int, made the solution pass just fine [submission:https://codeforces.me/contest/1614/submission/137043656] , it's been a while since the last time I encountered such a problem.

I know It's not the intended solution, but mine should be accepted too, or what do you think?

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NightRage

3 years ago, # ^ |

NVM, both get AC now.

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hxu10

3 years ago, # |

← Rev. 5 →

I would like to ask if system test slower than pretest?

When I solve problem C, it says pretests passed (12 test cases) and the time cost is about 820 ms. After carefully consideration, I decided to move on.

But I finally failed the system test, and the maximum time cost for first 12 test cases is 951ms. (And it finally failed test case 15). If I get 951ms in the pretest, I will try to optimize my code.

I see in comment section that @iaNTU failed the first test case in the system test with problem D2, he must passed test case 1 in the pretest, that is why I ask this question.

UPD: Thanks god, my contest code get AC after rejudgement.

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L1ngYu233

3 years ago, # |

Problem COrigial problem at Leetcode 1863

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Scythe

3 years ago, # |

← Rev. 2 →

Was there no pretest in C for the max N? my code TLE'd on test 17. it was O(NlogN*logN) and maybe it can be optimised but I didnt try as the pretests passed. Please try to include max N in pretests. Nice problems tho

Edit: submission rejudged it is AC now

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-_-INV4D3R-_-

3 years ago, # ^ |

You should always check the max N by yourself, e.g. via the Custom Invocation. There has to be some area for hacking lol

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Scythe

3 years ago, # ^ |

my bad, the submissions are rejudged, i should not have commented before the contest was officially over

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turkhuu622

3 years ago, # |

Can there be more than 1 answer in problem C?

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jac_nikola

3 years ago, # ^ |

I don't think so. Howsoever the distribution of bits across all N elements is, in the end, it only matters if a particular bit is set in any of the N elements.

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faraz16iqbal

3 years ago, # |

How to solve D?

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Nanako

3 years ago, # |

← Rev. 2 →

+27

Tutorial video by jqdai0815 (in Chinese)

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Joshc

3 years ago, # |

Originally I wrote a solution for problem E which passed pretests, but resubmitted later because I was scared about TLE. Glad I did, because even my faster submission took 1996 ms! (https://codeforces.me/contest/1614/submission/137021260)

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_lowerBound

3 years ago, # |

+12

I am no Psychologist, but brah this Divan has some serious problems..

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LKLCK

3 years ago, # |

Lol, the number of upvotes of this post is getting fewer and fewer

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bashNewbie

3 years ago, # |

← Rev. 2 →

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Eslam_Ahmed

3 years ago, # |

-10

I have a pupil title on the site how do I cheat from newbies?

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kingkd

3 years ago, # |

← Rev. 3 →

kingkd and ayushkedia0511 are both my id because kingkd is my old it which is lost but i got it and mastakenly i gave the contest with both id, for for the voilation of rules and i will not use my old (kingkd ) id for more contests

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Indians_cheating

3 years ago, # ^ |

looking at your code, i realised that you submitted your code with kingkd untill you got accepted, and then you submitted with ayushkedia0511.

it's 100% obvious that you used two accounts in purpose but now you are still lying that it was an mistake.

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64 bit compiler is sometimes faster

3 years ago, # |

← Rev. 3 →

Please Update the Ratings . I want to see myself as a Specialist Today :_: before i go to Sleep it Late night in here in India.. Also jee main paper Doesnt decide your Future

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insurgentes

3 years ago, # |

Can someone help me figure out why these 2 similar solutions for D2 are giving different results? Is there some optimisation I'm missing?

Passing: 137049637 Not passing: 137046389

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Len

3 years ago, # ^ |

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insurgentes

3 years ago, # ^ |

Thanks! interestingly it's not just the compiler but also changing long long -> int. Seems like there were some razor thin time limits.

64 bit compiler with long long instead of int: 137063729

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nebula

3 years ago, # ^ |

There is too much unnecessary use of long long, use int instead and then you should be fine.

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naman1601

3 years ago, # |

Here are two of my exact same submissions, which differ by almost 600ms in runtime. Before this, one submission got TLEd on the first submit, and then the same code got accepted on submitting it again. 137066206 and 137066133

Are such large differences common? It seems to me that something is up with the judge.

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insurgentes

3 years ago, # ^ |

hope this gets upvoted, the D2 time limit was really too low. See also this comment

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