Codeforces Round #760 (Div. 3)

3 years ago, # |

← Rev. 3 →

-13

Looking forward to this round!

Wish everyone positive delta

→ Reply

ARNAB.VCC

3 years ago, # ^ |

-6

Though I have solved 1st problem in 22 min in 1st attempt yet why I'm not ratted?

→ Reply

3 years ago, # ^ |

-7

The rating changes will be published after 12hrs long hacking phase, where you can hack others' solutions.

→ Reply

tirtha_indian

3 years ago, # ^ |

what is hacking?

→ Reply

Brajesh77

3 years ago, # ^ |

After the end of this contest round. The hacking phase is of 10-11 hours in which the participants and view the submission of other participants and and can give hack(i.e some test case in which code will fail but code has passed system test). Giving a correct hack will add to your rating and giving wrong hack will deduce to your rating.

→ Reply

3 years ago, # ^ |

Actually in Div3 there are no points for hacking, no add and no reduce.

→ Reply

TheBrownWolf

3 years ago, # ^ |

But if someone's code is hacked he will obviously lose marks

→ Reply

binary_vijay01

3 years ago, # ^ |

For reference Check my submissions, I got hacked on F :)

→ Reply

Techphile

3 years ago, # ^ |

when will the rating change??

→ Reply

3 years ago, # ^ |

idk, hope that it changes soon

→ Reply

tarun__m

3 years ago, # ^ |

im waiting so eagerly

→ Reply

Brick_Braker

3 years ago, # |

← Rev. 2 →

+15

→ Reply

3 years ago, # |

+15

Aah ,i miss those days of rated div3's for me.

→ Reply

nebula

3 years ago, # ^ |

+13

Though, not too long ago.

→ Reply

3 years ago, # ^ |

Thats why i ain't used to it ,hence i miss them

→ Reply

Brick_Braker

3 years ago, # |

← Rev. 2 →

BledDest and what if the people who participated in All-Russian Olympiad for schoolchildren in Saratov and the Saratov region take part in this Div 3?

→ Reply

dushenkov

3 years ago, # ^ |

Just solve more and it wont affect you.

→ Reply

Your_Bunny_Wrote

3 years ago, # |

1199r. I hope i will be green after this round

→ Reply

pkpawan

3 years ago, # ^ |

← Rev. 2 →

-34

Sorry to say but your rank in last round look suspisious .

UPD :- There is nothing wrong in your rank and performance , I misunderstood your rank which you acheived in Technocup round 3 as your rank in CF round 759 . I recently came across blog stating 22-40 rank in previous contest are all cheaters and lot of cheating happend in last contest and then I saw you rank and it look suspisious and I didn't saw the contest name . I apologize for that .

→ Reply

3 years ago, # ^ |

← Rev. 2 →

what's wrong? what if a person just performed well for their rating and reached 1199?

Upd.: ok,now it seems ok;)

→ Reply

Your_Bunny_Wrote

3 years ago, # ^ |

My rating was 1199. I remember. Techno cube gave me 1199 r

→ Reply

Small bug in codeforces ...

3 years ago, # ^ |

← Rev. 2 →

maybe you could turn green today itself after cheaters are removed.

Upd: sadly you lost 1 point

→ Reply

gaserashraf

3 years ago, # |

+12

First unrated round for me :)

→ Reply

Mohamed_Alaa

3 years ago, # |

Looking forward to be unrated on this contest

→ Reply

Silly__coder__

3 years ago, # |

+17

I wish i could gain 200+ in this round.

→ Reply

Randall

3 years ago, # ^ |

+19

Wow, I am very impressed by your persistence on cp(by looking at your graph). Keep it up!

→ Reply

Silly__coder__

3 years ago, # ^ |

+30

Thanks bro.

→ Reply

Mahim

3 years ago, # |

← Rev. 2 →

Last 2 rounds had very bad organization. But I think this will be good.

→ Reply

Sobzi

3 years ago, # ^ |

BledDest and awoo have written about a hundred educational rounds and they are pretty good at their work. So I think (and hope) we can expect a great round :))

→ Reply

b4nanacream

3 years ago, # |

The round will contain 8 problems, which are mostly suited for participants with rating below 1600 (or we hope so).

we hope that our tests are strong enough, so there won't be too many solutions hacked during this phase

I hope you'll enjoy solving the problems we have prepared. hopeforces?

→ Reply

anindiangeek

3 years ago, # |

I hoe... I become LGM this round. teh heh

→ Reply

fonmagnus

3 years ago, # ^ |

+40

You... what..?

→ Reply

NM_Mehedy

3 years ago, # ^ |

+18

You can copy 'p' from here and paste it.

→ Reply

163onmyneck

3 years ago, # |

+19

i hate comments under this post so much.

like i hope to reach pupil/expert or i hope everyone positive delta

there are so many comments like these, they are just rubbish

read this: https://codeforces.me/blog/entry/68909

→ Reply

divyansh_sahu

3 years ago, # |

+16

→ Reply

fonmagnus

3 years ago, # ^ |

+13

Also me when it's accepted

→ Reply

ac2hu

3 years ago, # ^ |

FSsT

→ Reply

taogehtsilalos

3 years ago, # |

← Rev. 2 →

EDIT : It has been fixed, sorry

→ Reply

Sacchidananda

3 years ago, # |

Educational Div3 Round

→ Reply

Birthday1703

3 years ago, # |

← Rev. 2 →

How Many problems should I do in div 3 as a newbie to become pupil?

→ Reply

3 years ago, # ^ |

Increase by how much ?

→ Reply

rishisgsits_22

3 years ago, # ^ |

Depends on rank not no. Of problems

→ Reply

3 years ago, # ^ |

Also on number of participants

→ Reply

andreyDagger

3 years ago, # ^ |

I think three or four will be enough

→ Reply

20bcs030

3 years ago, # ^ |

solve just like my alt hi123bye he is again pupil now

→ Reply

abhilash_8642

3 years ago, # ^ |

130 day streak ... How did you get that much patience?

→ Reply

20bcs030

3 years ago, # ^ |

inspired by the legend of all times prashant_th18

→ Reply

bjy

3 years ago, # |

Last few contests have had issues, there are some new-to-me names in list of setters, and the site is bugging out as I try and see who's registered (in addition to the usual last-contest-colorchange bug)...

Div3 all-problems-equal-points ups the risk for bricking and I haven't had a good night's sleep since advent of code started...

Screw it, I'm in. I don't want to start caring about keeping my gamified color-number artificially high from contests I don't participate in anyway...

May this comment age like rancid eggnog.

→ Reply

binary_vijay01

3 years ago, # |

OkaY Bois, Hope this one will be f*cking awesome Enjoy solving

→ Reply

pankov_batya

3 years ago, # |

let's make me the most downvoted user on cf !

→ Reply

Brick_Braker

3 years ago, # |

1 minute left. Good luck everyone :)

→ Reply

vrintle

3 years ago, # |

RecoveryForces

→ Reply

Joshc

3 years ago, # |

+15

Announcement blog says there are 8 problems, but there are only 7?

→ Reply

BledDest

3 years ago, # ^ |

+22

We decided that the contest was difficult enough with 7 problems, so one problem was excluded.

→ Reply

saksham441

3 years ago, # |

-39

worst E

→ Reply

Siuuuuu_7

3 years ago, # ^ |

+32

Why?

→ Reply

justTrolling

3 years ago, # |

My smoll brain takes forever to understand Problem E's statement ;-;

→ Reply

3 years ago, # ^ |

just read the sample explanation. I didn't understand the statement either.

→ Reply

marat0210

3 years ago, # |

could someone explain problem D to me? my greedy solution gives WA 2. Is there another way to solve a problem?

→ Reply

3 years ago, # ^ |

← Rev. 2 →

Your solution was difficult to understand. But basically, you want to match the last $$$2k$$$ elements, $$$a_n$$$ with $$$a_{n-k}\ $$$, $$$a_{n-1}$$$ with $$$a_{n-k-1}\ $$$ and so on..

→ Reply

sergeev.PRO

3 years ago, # |

How to solve E? Looks very interesting, dissapointed about not solving that

→ Reply

3 years ago, # ^ |

+12

sum(a) = sum(b) / (n * (n — 1) / 2) b[i] — b[i — 1] = sum(a) — n * a[i]

→ Reply

aopo

3 years ago, # ^ |

I think it should be sum(a)=sum(b)/((n*(n+1))/2).

→ Reply

LoboLobo

3 years ago, # ^ |

← Rev. 3 →

+13

I love it, the we have two main observation, first is:

Let S(b) sum of all bi and S(a) sum of all ai

the total time of every singer is ai + 2*ai + ... + n*ai = ai*n*(n+1)/2

S(b) = sum of total time of every singer -> S(b) = a1*n*(n+1)/2 + ... an*n*(n+1)/2 -> S(b) = S(a)*n*(n+1)/2

Using 0 indexed, the second observation is:

bi = 1*ai + 2*a(i-1)%n + ... n*a(i+1)%n
b(i+1) = 2*ai + 3*a(i-1)%n + ... 1*a(i+1)%n
b(i+1) — bi = ai + a(i-1)%n + ... — (n-1)*a(i+1)%n
b(i+1) — bi = ai + a(i-1)%n + ... + a(i+1)%n — n*a(i+1)%n
b(i+1) — bi = S(a) — n*a(i+1)%n
a(i+1)%n = (b(i+1) — bi — S(a))/n

To check if the answer is YES or NO, you have to be sure that all divisions leave 0 remainder and all ai > 0.

My code: https://pastebin.com/iaxwvTSq

→ Reply

LofiGirl

3 years ago, # |

Thank you for this amazing contest! I really like the problems, all of them were perfect for their place, especially problem E!

→ Reply

Amrit_4343

3 years ago, # |

Is problem D a mixture of greedy and sorting?? btw i was not able to solve it... can Anyone tell me the correct approach

→ Reply

3 years ago, # ^ |

yes. sorting in decrease order then greedy use a[i + k] / a[i] for each i from 1 to k

→ Reply

3 years ago, # ^ |

← Rev. 2 →

We must/want remove the k*2 biggest numbers.

So sort descending, and pair a[i],a[i+k] for cost of 0 if these both numbers differ, or cost=1 if same. Then add the other (n-k*2) smallest numbers.

→ Reply

3 years ago, # |

For me B and C where more complecated than D,E,F. No idea how to solve C, even cannot think of a randomized solution.

→ Reply

faraz16iqbal

3 years ago, # ^ |

Take the GCDs of all the elements at odd and even indices, let's say G1 and G2. If any element at an odd index is divisible by G2 and any element at even index by G1, the answer does not exist. Else answer is either G1 or G2.

→ Reply

3 years ago, # ^ |

get gcd of all odd-index elements, then check if any even-index element is divisble by it. Then do reversely.

→ Reply

3 years ago, # ^ |

Ah, ok. I somehow misinterpreted this one. I searched for a d that divides the array into two sets of numbers with same size. :/

→ Reply

NeverGiveDiv3sInLife

3 years ago, # |

Just wondering why did the authors gave n <= 100 for problem D , was it just to confuse , or the correct solution is O(n*2) indeed ?

→ Reply

3 years ago, # ^ |

It's O(nlogn). I am also confused.

→ Reply

gleb32

3 years ago, # |

How to solve F?

→ Reply

3 years ago, # ^ |

← Rev. 2 →

I generated all possible binary numbers as strings up to length 62. Thats fairly quick since we basically can only append a '1' at begin and reverse.

→ Reply

3 years ago, # ^ |

at first operation, if you add 0, then all the 0s at the end of the original number is removed. Then you can only add 1 to both sides of the original number to get the target number. Add 0 is no-op.

→ Reply

swayingdream

3 years ago, # ^ |

← Rev. 2 →

for F it's kind of understandable if you can break it down. how i did it was:

If you add a zero, essentially you are saying, I can flip the string and trim all zeroes at the front and back. So essentially I can always reverse it whenever I want.

If you add a one, you can think of it as I can add a one on the left or right of the string without much thought because I can flip whenever I want

So the only caveat left is, did I have to trim the zeroes to do this reversing operation in the first place?

So there are 2 cases:

A) String is [1?????0].

B) String is [1?????1].

For case A: I can add a 1 at the back, and only then can I add the 1s however I want. Or, I can trim the zeroes, and then add the 1s however I want.

For case B: I can add 1s at the front or back, however I want.

So your final answer will look like this:

Case A:

Possible if string itself is [1?????0]
Possible if string is of the form ...1[1?????01]1...
Possible if string is of the form ...1[10?????1]1...
Possible if string is of the form ...1[1????1(trimmed)]1...
Possible if string is of the form ...1[1????1(trimmed)(flipped)]1...

Case B:

Possible if string is of the form ...1[1?????1]1...
Possible if string is of the form ...1[1?????1(flipped)]1...

and thats the final answer, maybe a little complicated now that i type it out

edit: removed redundant statement for clarity

→ Reply

dryice

3 years ago, # |

← Rev. 2 →

Submitted AC solution of F just after the contest :)

E was a nice problem.

→ Reply

Maango16

3 years ago, # |

+19

An amazing round ruined in the end by leaked E and leaked D. Mass submission at the end was the indicator.
Please take appropriate actions BledDest MikeMirzayanov

→ Reply

pyrogramer

3 years ago, # ^ |

-7

very educational channel

→ Reply

snorkel

3 years ago, # ^ |

Very anti-educational channel

→ Reply

elliaK

3 years ago, # |

**Any One what is wrong in my Code C **

Code

→ Reply

3 years ago, # ^ |

I think it is because you are taking maximum of the two gcds. Instead you should try both.

→ Reply

hyoseok

3 years ago, # ^ |

← Rev. 5 →

sorry for my bad case

→ Reply

tabakade

3 years ago, # |

Could someone tell me how did you guys solve B and C? It took me forever to understand the question and could not come up with the solution. Even a hint would be enough, this is my second contest and I am trying to improve.

→ Reply

3 years ago, # ^ |

First you have a string s equal to the first bigram. for B, if one bigram's second character and the next bigram's first character is the same, you can always treat them as adjacent, add the next bigram's last character to s. Otherwise, add the next bigram to s. After you have iterated all the bigrams, if length of s is smaller than N, add one 'a'.

→ Reply

rsFalse

3 years ago, # ^ |

After you read a bigram into a string, you can delete all spaces with a consecutive letter if that space has the same letters on both sides. After that you delete the last space left or add any letter otherwise. Regex code: 139328475

→ Reply

shivam565

3 years ago, # |

one doubt about F problem, can we add multiple '0' or '1' and then perform a reverse operation?

→ Reply

3 years ago, # ^ |

no. only one for each operation.

→ Reply

plotemy

3 years ago, # |

-9

I could not solve any problem. It was so disappointing. Although I can solve division-2's problem in archive but I couldn't solve any problem here in real time contest.

→ Reply

pandeyshivam10

3 years ago, # ^ |

DO practice my friends !!!!!

→ Reply

SourabhCh

3 years ago, # |

what is wrong with my code? please help!

my approach:- I divided the array into two arrays a(all Even Indexed Elements) and b(all odd indexed). then either the ans is min(a) or min(b) or 0. if min(a) divide entire a and doesn't divide a single element in b then min(a) is ans, and vice versa for b. if none of the above two conditions are satisfied then ans is 0. https://codeforces.me/contest/1618/submission/139292894

→ Reply

3 years ago, # ^ |

not min(a), it's gcd(a).

→ Reply

3 years ago, # ^ |

35 6 14 9. No element divides any other, but $$$d=3,7..$$$ are answers.

→ Reply

anujdhillxn

3 years ago, # |

I solved F recursively with some early termination conditions. Hacks would be appreciated

→ Reply

NeverGiveDiv3sInLife

3 years ago, # |

I noticed this relation in problem E that

b[i] — b[i+1] = n*a[i + 1] — sum(aj)[j = 1..n]

b[i — 1] — 2*b[i] + b[i + 1] = n*(ai — ai+1)

and then i calculated a1 — a2 , a1 — a3 ,...... a1 — an

but i couldn't get which value of a1 give the solution

→ Reply

3 years ago, # |

+40

my first time to solve all problems in a contest, although it's a div. 3. Thanks to the authors!

→ Reply

milfhunter2007

3 years ago, # |

Please, can somebody explain why is my code for problem D wrong or give any counterexample? 139306202

→ Reply

https://dunjudge.me/analysis/problems/1166/

3 years ago, # ^ |

1 2 4 4, I think.

→ Reply

milfhunter2007

3 years ago, # ^ |

Thanks. Such a stupid mistake, it's a shame.

→ Reply

ThePhoenix

3 years ago, # |

Guys I need help with question C. I keep getting wrong answer on test 3, and I don't understand why my code fails. if you can, please help me understand why my code fails and how to properly solve it? my submission: https://codeforces.me/contest/1618/submission/139297986 thank you in advance.

→ Reply

milfhunter2007

3 years ago, # ^ |

← Rev. 3 →

Think about this case:

1 4 3 6 5 The answer is 2 but your code says that there is no answer. Divisor can be not presented in array, you should check gcd of all odd and then for even positions as a d.

→ Reply

somerandomaltacc3034

3 years ago, # |

+24

An extension of problem A if anyone wants to try :

→ Reply

Nasrul97

3 years ago, # |

Hello everyone, I'm new to competitive programming, and this is my first official round. I have a problem with Prob A (yes, A). I think my answer is correct but marked as a wrong answer. You can check this submission 139305580 for more details.

I want to know, if possible, where my mistake is so I won't make the same mistake in the upcoming rounds.

Thanks!

→ Reply

Techphile

3 years ago, # ^ |

Check ur solution for value of a=[ 1 2 4 ]

→ Reply

3 years ago, # ^ |

it is wrong .I also made the same error at the first glance. actually if you are looking for sums you know that 2 smallest values will be your first and 2nd value but you don't know about the third value because if 1+2=3<4 ,so we know that sum of all the values is going to be the maximum ( last value in the given input ) so if you are taking given input in array soln will look like this. a[0] ,a[1], a[6]-a[0]-a[1]

→ Reply

3 years ago, # |

Hi ,Can anyone hack my problem F ? i wrote it in a hurry and want to know if someone can hack it.Thanks

→ Reply

harshit135

3 years ago, # |

+13

How to solve G ?

→ Reply

altynai

3 years ago, # ^ |

If you concat array A and B (let's call it array C), after sorting the array C, the problem becomes: for each K, you can connect C[i ... j] if each i <= x < j, and C[x + 1] — C[x] <= K, so after choosing some elements in each range, what's the maximum sum?

So we need to maintain:

What are the leftmost and rightmost index of the range that C[i] belongs to? (hint: DSU)
How many numbers I can choose in this range? (hint: since array C is sorted, we should always start picking the largest one first in greedy)
How do I merge 2 ranges?

Turns out our best sum only changes when the merge happened, so I took the offline approach by sorting the queries first and used DSU to maintain the information.

Hope it's clear, you can check my submission: 139325375.

→ Reply

thanhchauns2

3 years ago, # ^ |

← Rev. 2 →

A simple way to solve: use two sets, both store vectors of four elements: the leftmost index of the segment, the rightmost index of the segment, how many numbers we take from it, and the distance between it and the next segment on the array. The first set will sort by the distance, the second set will sort by their first elements.

So how to use them? At first, merge the two arrays a and b into 1 array, let's call it TTL, use a map to save if we are currently have something or not. Then for each element, we will make a segment based on them: the leftmost and rightmost is their index, the number of elements we will take is either 1 or 0 depending on the map I described above, the distance will be the difference between it and the next number in TTL.

What will we do now? Solve the problem offline, for each query merge some of the segments which have the smallest "distance" into some other segments, so there will be no more than N merges. Since we are using sets, the complexity is O(NlogN).

To calculate the sum efficiently, use a prefix array.

You can check my implementation if you want:

Implementation

        ll a, b, q;
	cin >> a >> b >> q;
	vec(ll) ttl;
	map<ll, ll> C, first, last;
	ll cnt = 0;
	
	FOR(i, 0, a - 1)
	{
		ll x;
		cin >> x;
		cnt += x;
		ttl.pb(x);
		C[x]++;
	}
	FOR(i, 0, b - 1)
	{
		ll x;
		cin >> x;
		ttl.pb(x);
	}
	
	SORT(ttl);
	FOR(i, 0, ttl.size() - 1) last[ttl[i]] = i;
	FORD(i, ttl.size() - 1, 0) first[ttl[i]] = i;
	vector<ll> PREF = ttl;
	FOR(i, 1, PREF.size() - 1) PREF[i] += PREF[i - 1];
	set<vector<ll> > SORT_BY_DIST; // DIST -> TAKEN_ELEMENTS -> LAST_ELE -> FIRST_ELE
	set<vector<ll> > SORT_BY_FIRST_ELEMENT; // FIRST_ELE -> LAST_ELE -> TAKEN_ELEMENTS -> DIST 
	
	FOR(i, 0, ttl.size() - 2)
	{
		if (ttl[i] == ttl[i + 1]) continue;	
		ll x = ttl[i];
		SORT_BY_DIST.insert({ttl[i + 1] - ttl[i], C[x], last[x], first[x]});
		SORT_BY_FIRST_ELEMENT.insert({first[x], last[x], C[x], ttl[i + 1] - ttl[i]});
	}
	SORT_BY_DIST.insert({(ll)1e18, C[ttl.back()], last[ttl.back()], first[ttl.back()]});
	SORT_BY_FIRST_ELEMENT.insert({first[ttl.back()], last[ttl.back()], C[ttl.back()], (ll)1e18});
	
	vec(ll) ans(q);
	vector<p<ll, ll> > QUERIES(q);
	FOR(i, 0, q - 1)
	{
		cin >> QUERIES[i].f;
		QUERIES[i].s = i;
	}
	SORT(QUERIES);
	
	function<ll(ll, ll)> SUB = [&](ll LAST, ll TAKEN)
	{
		ll l = LAST - TAKEN;
		ll tmp = 0;
		if (l == -1)
		{
			tmp += PREF[0];
			l = 0;
		}
		tmp += PREF[LAST] - PREF[l];
		return tmp;
	};
	
	for (auto x : QUERIES)
	{
		vector<ll> K = *SORT_BY_DIST.begin();
		while (K[0] <= x.f)
		{
			vector<ll> Q = K;
			revrs(Q);
			vector<ll> T = *SORT_BY_FIRST_ELEMENT.upper_bound(Q);
			vector<ll> P = T;
			revrs(P);
			SORT_BY_DIST.erase(K);
			SORT_BY_DIST.erase(P);
			SORT_BY_FIRST_ELEMENT.erase(Q);
			SORT_BY_FIRST_ELEMENT.erase(T);
			cnt -= SUB(Q[1], Q[2]);
			cnt -= SUB(T[1], T[2]);
			cnt += SUB(T[1], Q[2] + T[2]);
			T[2] = Q[2] + T[2];
			T[0] = Q[0];
			SORT_BY_FIRST_ELEMENT.insert(T);
			revrs(T);
			SORT_BY_DIST.insert(T);
			K = *SORT_BY_DIST.begin();
		}
		ans[x.s] = cnt;
	}
	
	printVector(ans);	
	cout << '\n';

Hope that my defines are not difficult to understand.

→ Reply

sasta_madara

3 years ago, # |

made it to double digit for the first time,got my f hacked ffs *cry_emoji *cry_emoji

→ Reply

turzo_sawroop

3 years ago, # |

how to solve F?

→ Reply

3 years ago, # ^ |

There are only few numbers you can generate from a given number. Generate all and check if one equals the searched value.

→ Reply

MateiKing80

3 years ago, # |

Hello, at my source's verdict writes hacked. Can you please tell me what's wrong with it? It's code is 139283121. Thank you!

→ Reply

Senior

3 years ago, # |

← Rev. 2 →

Bye 2021

→ Reply

xudian

3 years ago, # |

nice round, first time solve all problems in div3

→ Reply

_seg_error_

3 years ago, # |

If I am successfully able to hack someone's code in Educational or DIV3 rounds , then what will happen ? my penalty will decrease or not ?

→ Reply

BitSane

3 years ago, # ^ |

Nothing happens, it won't improve your score or rank.

→ Reply

https://codeforces.me/contest/1618/submission/139363593

3 years ago, # ^ |

→ Reply

wargang

3 years ago, # ^ |

← Rev. 3 →

→ Reply

_seg_error_

3 years ago, # ^ |

my Friend is introvert so , he couldn't ask saransh_singh

→ Reply

jeal0uspengu1n

3 years ago, # ^ |

+11

lol

→ Reply

NeverGiveDiv3sInLife

3 years ago, # ^ |

Well your penalty won't change but if some people hacked solutions of people with rank better than you , then you might see your rank improved

→ Reply

Fanarill

3 years ago, # |

← Rev. 2 →

I got frustrated from problem B. I read statements carefully, but it seems like I still don't understand what do they want from me. Why my code gave MLE? Of course while is infinite, but still how is that possible? How can I have more than 1 "unconnected" bigrams.

→ Reply

Fanarill

3 years ago, # ^ |

← Rev. 3 →

NVM, I found an error. Now I feel more stupid. I guess I should've slept more lol

→ Reply

YPK

3 years ago, # |

-27

Why time limit for G is 4.5 seconds?

The obvious solution is $$$O(nlogn)$$$ and runs in 265 ms for me.

→ Reply

phongan105

3 years ago, # |

In Reverse (F problem) how can we turn 23 to 59?

→ Reply

Vercingetorix

3 years ago, # ^ |

0 1 0

→ Reply

sumukh7039

3 years ago, # |

I participated in this contest and got 2 questions correct but still, I am unrated. This was my first contest. So will I get ratings or not?

→ Reply

jiangbowen_

3 years ago, # ^ |

-10

Yes, you'll get rating.

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

→ Reply

jiangbowen_

3 years ago, # ^ |

-10

Wait for the system test.

→ Reply

threadpool

3 years ago, # ^ |

according to post:

To qualify as a trusted participant of the third division, you must:

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

if this is your first rated contest, I don't think you will be rated.

→ Reply

RaadSharar

3 years ago, # ^ |

untrusted != unrated

→ Reply

ayush0212

3 years ago, # |

I participated first time and solved 2 questions. Why am I unrated?

→ Reply

jiangbowen_

3 years ago, # ^ |

-10

Wait for the system test.

→ Reply

xuhao95

3 years ago, # |

← Rev. 3 →

First time to participate and solve 6 problems. Wish I can be better in the future.

→ Reply

Techphile

3 years ago, # |

how much time rating change will take after system testing??

→ Reply

Winghau

3 years ago, # ^ |

I still get the rating now,I feel so confused

→ Reply

Kath1031

3 years ago, # ^ |

i refresh page frequently o.o

→ Reply

Milex

3 years ago, # |

Hi guys I need some help with problem D.

I seem to be off by 1 in the test case and I can't seem to understand why. Much appreciated.

→ Reply

OlavBakken

3 years ago, # ^ |

Your solution seems to be matching up a_i with a_{i-1}, which is incorrect. Try this test input: 1 5 2 1 2 2 3 3

correct answer is 1, your solution gives 2.

→ Reply

hackerbhaiya

3 years ago, # |

← Rev. 3 →

I am surprised that there were very few pretests for problem D and how my 1st sol passed. The solution which I submitted for the first time should have given RTE in the pretests itself as I took it=s.rbegin() and then decremented it by 1, which is clearly outside the bounds. This solution is completely wrong, isn't it?

→ Reply

Beyrad

3 years ago, # ^ |

← Rev. 4 →

my solution is same as you

and I got FST too :(

I don't know why this logic doesn't works can someone help on this?

https://codeforces.me/contest/1618/submission/139284177

→ Reply

3 years ago, # ^ |

It is not optimal to choose neighbouring elements. Eg: 1 1 4 5.

You can refer to the editorial.

→ Reply

Beyrad

3 years ago, # ^ |

for 1 1 4 5 my algorithm answers 0

→ Reply