#  User  Rating 

1  tourist  3757 
2  jiangly  3647 
3  Benq  3581 
4  orzdevinwang  3570 
5  Geothermal  3569 
5  cnnfls_csy  3569 
7  Radewoosh  3509 
8  ecnerwala  3486 
9  jqdai0815  3474 
10  gyh20  3447 
#  User  Contrib. 

1  maomao90  171 
2  awoo  164 
2  adamant  164 
4  TheScrasse  159 
5  nor  154 
5  maroonrk  154 
7  isthisfft  152 
8  Petr  147 
9  orz  146 
10  pajenegod  145 
On
george_stelian →
"Adolescent Grigore Moisil" (AGM) International Programming Contest 2022, 2 years ago
0
No :( 
On
george_stelian →
"Adolescent Grigore Moisil" (AGM) International Programming Contest 2022, 2 years ago
0
Several teams registered until now. 
On
george_stelian →
"Adolescent Grigore Moisil" (AGM) International Programming Contest 2022, 2 years ago
0
the editorials were live; the problems were presented after the contest by the proposers 
On
george_stelian →
"Adolescent Grigore Moisil" (AGM) International Programming Contest 2022, 2 years ago
+12
You can also message me or George if any question about our contest. 
On
george_stelian →
"Adolescent Grigore Moisil" (AGM) International Programming Contest, 3 years ago
0
yes. thank you 
On
george_stelian →
"Adolescent Grigore Moisil" (AGM) International Programming Contest, 3 years ago
0
We made a live editorial on google meets :( 
On
george_stelian →
"Adolescent Grigore Moisil" (AGM) International Programming Contest, 3 years ago
+8
We have just extended the registration period until 19 February, 5 PM GMT Time. Reason: 20 teams signed up in the last 24 hours 
On
george_stelian →
"Adolescent Grigore Moisil" (AGM) International Programming Contest, 3 years ago
0
Hi! We will upload the contest on the GYM Section 
On
george_stelian →
"Adolescent Grigore Moisil" (AGM) International Programming Contest, 3 years ago
0
cool! hope you will enjoy the competition 
On
george_stelian →
"Adolescent Grigore Moisil" (AGM) International Programming Contest, 3 years ago
+1
Hi! We will upload the contest on the GYM Section 
On
george_stelian →
"Adolescent Grigore Moisil" (AGM) International Programming Contest, 3 years ago
+1
Yes, you can 
+3
Europe, Romania: bicsi , theodor.moroianu , livlivi 
0

10
Unfortunately no, but we will post the statements on our website 
18
Update: 2 more countries involved in competition 
10
9 AM UTC time 
+3
Almost 70 teams registered. New registration deadline until 21:59 UTC today 
+3
No problem, buddy 
+3
Yes, just like I said, it will be ICPC (number of problems, then prnalty and submission time) 
12
registrations have been extended: until 7th February 23:59 
+3
If you have other questions, don't hesitate 
+6
Thank you, buddy! 
+3
You will have 5 hours to solve the problems 
+3
Auto comment: topic has been updated by Usu (previous revision, new revision, compare). 
+6

+3
yeah, this is the approach with lazy propagation. Or, could it be solved without lazy? 
+6
Can anybody tell me if this solution is correct for problem E? My solution use segment tree without lazy propagation. For each node in the segment tree I know v[nod].left = number or '(' without a pair (basically open) , v[nod].right = same but for ')' and v[nod].depth = how many colors do I need. How to find v[nod].depth if we know every information for its children? v[nod].depth = max ( v[ 2*nod ].depth, v[ 2*nod+1 ].depth) maximum depth of its children + min ( v[2*nod].left, v[2*nod+1].right) this if for: '(' CONFIGURATION + CONFIGURATION '))' so in this case I increase the depth with 1 because I make new pairs. I had a bug in implementation so I couldn't submit it, but does anybody have the same approach? 
+3
Firstly, I taught that F can be solved using dp[i][j] — number of permutations using numbers from 1 to i having the permutation's depth j, but I figured out that is not enough. Can anybody explain the dynamic with 3 states? 
+8
Yes, it helped me. Is this algorithm similar to that one: "find the number of intersections of segments, all segments parallel either Ox either Oy"? I think they are 90% the same, am I wrong? 
0
I don't have a link. I just discussed this with a friend who saw it in a printed book. Can you detail more the solution? 
+3
Thank you! This helped me 
+3
Thank you for staying awake and creating this nice solution! This really helped me. :D 
+11
Second half*, the post has been edited. The problem was given at AGM and I want to see different approaches. 
+3
Auto comment: topic has been updated by Usu (previous revision, new revision, compare). 
+3
You can solve offline the queries. 1) sort the queries in nondecreasing order regarding x; 2) build an array V with 2 parameters: the value of A[i] and i, the position, then sort V regarding first value; 3) after these operations, keep 2 pointers: you will start i from the first value from V and start with j from the first element in Q (sorted); update the values in the binary indexed tree unless V[i].first_parameter will be >= the x from actual j query, then calculate with BIT the answer like this: sum of first r values in BIT — sum of first l1 values in BIT, then print this value. 
+3
Thanks! 
+11
If "gcd(a,n)=1", you mean gcd(a,m)=1? I can not see the n 
+8
Thank you, these helped me, especially 'stars and bars' 
+2
It is not really the best ocasion to post this 
6
You didn't mention anything about the number of problems. Can you edit that please? Edited, thanks 
+1
So basically combined contest with prizes. Nice 
0
I liked problem D, but no comment for B and C 
0
Thank you all! 
0
I talked briefly with a friend about this, I wish I had a location for this problem 
0
Thank you! 
0
anyway, this round trained solvers' speed so it compensated 
0
2k accepted solutions for a Dproblem during the contest, first time when I see something like this 
On
FalseMirror →
Codeforces Round #483 [Thanks, Botan Investments and Victor Shaburov!] Editoral, 6 years ago
0
Thanks, it helped me a lot 
On
FalseMirror →
Codeforces Round #483 [Thanks, Botan Investments and Victor Shaburov!] Editoral, 6 years ago
0
Can anybody explain me please at div1 B, why f( l , r ) = f( l , r1 ) ^ f( l+1 , r) ? 
0
you can just write first line from int main(): ios::sync_with_stdio(false); and this will give you a boost 
0
I have a question: when we'll have a div1 and a div2 contest in the same time, those under 2100 can participate in both rounds or is forbidden for them to join div2? 
0
Thank you! 
0
I know, but I have problems with implementation :) 
+8
No, because hack period for this educational is only 12h length so the rating will be updated until the cf 464 will start 
6
16 contests announced. I've never seen more on the future list, keep it like this codef 
+4
Even you hacked me. This is first time in my life when I got 3 hacks on A 
+15
A round full of hacks. Even one of my friends hacked me, nice contest 
0
Yeah, that was the only mistake, I didn't notice that, but I didn't expect to get wrong answer instead of MLE or runtime error. Thanks! 
+5
That moment when you got TLE on test 105 problem C... I should change my profile photo... 
0
Thank you, too! 
+5
Thank you! 
+5
I also read a good article where NIM is well explained, but doesn't contain staircase nim and much information about mex() http://www.infoarena.ro/numerelespraguegrundy you can read it translating the whole page 
51
Highest ranked person on CF will not be able to participate 
+21
Is this the first Hello contest ever made? 
0
Sir, thank you for this excellent article. Without you I would not have been able to understand the solution of this problem 
+2
You can use Chrome and then translate automaticaly all the page 
0
4500 registrations and 8.5 hours left. I think we'll have at least 8.000 participants. I hope that the site will work 
+10
This blog enetry was post 45 mins ago and you already commented 8 times... this is funny for me 
+37
What does the Good Bye contest say: "Long time no C" 
+41
3 consecutive contests? I think this is a perfect end of year. 
0
So, in conclusion, no matter how many dimensions we have, always we can apply BIT (if the statement ask for it) 
+10
Thanks! 
On
vintage_Vlad_Makeev →
Codeforces Round 454 (and Technocup 2018 — Elimination Round 4), 6 years ago
+72
Tomorrow we'll have 2 El Clasicos: > Real Madrid — Barça * > Codeforces — AtCoder * 
+65
I really think that CF had a difficult time, but at least last 3 contest had a decent queue and nice problems. I think CF is going in a right direction, mentioning the number of contests. 
+1
IOI style means: partial scores, as many submisions as you want and it doesn't count the time when you submit or your number of submisions, everything that matters is your sum of your final scores of every problem. In these kind of contests you must firstly read all the problem and start with the easiest problems for you. If you don't know or you can not find a solution that gives you full score, you should make the brute force implementation and then make some optimizations. If you are a beginer, it's safe to do for all the brute force implementation, because you will have surely some points in your bucket with these, after that try to find better solutions. 
19
If you think (like me) that answer on D > 10^18 is a bit unfair for c++ users, upvote this in the idea that CF responsalbes will see this and will see how many we are. It was a nice contest, short statements and beautiful problems, but this thing on D deteriorate the value of the contest. 
0
How did you calculate that? In my opinion if there are cases where the answer exeed 10^18, it is not fair, because some languages support big numbers, but c++ doesn't... 
0
nice problems, can't wait the editorial. short statements <3 ++respect 
+20
I hope that the statements will be as short as those from last div2 and as interesting 
+36
5 div2 contests and 1 div1 in 9 days? already best december gift 
0
Thank you! 
0
Can someone explain me please why C(y+cnti1 ; cnti) is the corect formula? I can.t undertand from where we have this relation 
+3
Rank 612 — rating 1640 => 42 Rank 1771 — rating 1695 => 53 Why? 
+2
For those who can't understand why they got hack on B, here is a problem using Fermat's Little Theorem > see the editorial from this problem https://www.hackerrank.com/contests/infinitum18/challenges/tower3coloring; why a^b mod c is a^(b mod(c1)) mod c 
0
thanks for advices and explanations 
0
Can anybody explain me why my first div2 B submission got tle on test 9 having the complexity O(n) and using cin — cout, but after that i used scanf — printf and my second submission got accept. Why? 
0
Where can we find the official solutions? Can anyone tell me please? 
Name 
