[problem] facebook has suspended my account because someone reported my Profile

picture maybe mimnio himself as it was his picture . anyway , if someone can

create the group and lead us plz do it.

[final edit] OK, Here we go ! :), I am preparing it from now (creating group, rules ,..) and will start to send to everyone

said OK in a comment message with more details :).

[help :) ] codeforces limit no of message to 2 per hour ! :) , so to be sure that all said OK saw the last update, if you can

send any random people message notifying them about update plz do it and write in a comment the people you sent messages to.

please everyone add me to your account (whether fake or real as you like :) ) : [the account link ]

(https://www.facebook.com/profile.php?id=100010119040189)

The Group Link .. plz all of you join this group to start stating rules

and ideas to start the training.

• Regarding this blog i am sorry but after adding all people (till Friday 11 pm ) anyone will say OK after that logically i

cant add you because the training will be already started (and by end of week/month there will be a no. of problems to be

solved or kicked out from the group (So, we will post our results here and if it succeeded with us (hope so) maybe making

the group public and available for everyone to repeat the idea.

Hi All, I have already posted an idea a while ago about helping each other but didnot know how could this be done and some top

coders said to me ok it will be good only to people at the same level, so we solve together hope all be red by the end of the

trial.

Consider it as a Training Camp (online).

Let's Start at its structure or how the process will go :

1- Will make a group anyone free to join (level preferable to be the same : under blue )

2- Will post their all the material we think is the best (not all the material there ex. : competitive programming course, good

well organized training sheets ,.. )

3- choose the best from them (only one resource from them) by voting.

4- start Solving && upsolving.

5- if problem not understandable or need a specific topic like DP , ask and all of us helps by sharing and explaining to each

other.

6- i am busy for this week finishing a project on image processing, so will start if some people said yes after this week + take

it as a period to think and prepare best resources.

7- will be some rules to make the place good and motivating like anyone not solving n problems every week/month removed from

the group.

7- hope it works perfect (hope so).

8- any suggestion by red and high rated people or anyone in general are more than welcomed.

Thanks all.

540B - School Marks

The problem has two ideas :

1- ensuring that sum does not exceed some value : x.

2- median adjustment.

__ first notice that we cant solve this problem by filling our vector with value >= y ; why ? because he said that sum must not exceed value : x , so we have to take care of our cumulative sum.

So,to solve this problem we need to keep track of where our median falls (i.e. in the required area where median is greater than or equal : y or in the "-1" rejected region where median ((n+1)/2) falls where it is less than y (and his mom would punish him).

so,to clarify more here's an example, he started by giving us k elements as follows :

k = 4 , n = 9 , y = 3;

1 3 4 5 , so here counter of elements greater than or equal to y (3) = 3 ,and lower (less than 3) equals 1 , what that means ? greater_Elements_counter + 1 > lower_Elements_counter ,which ensures that our median must falls in the area where : v [ (n+1) / 2] must be >= y , that the first part.

The second part comes as how to fill the remaining (n-k) elements in a way that we try to have out two conditions fulfilled (1 && 2 listed above). the idea is simple enough to follow : if our Greater_counter + 1 > lower_counter means we have condition 2 done and have to take care of condition 1 where sum < x , so minimize ? yes , put the smallest possible value : 1.

My code (tried to comment it as possible) : [submission:http://mirror.codeforces.com/contest/540/submission/13006828]

high Quality code : [submission:http://mirror.codeforces.com/contest/540/submission/10948313]

I tried my best , so of course anything unclear or wrong , please ask me to handle it ASAP. __

Thanks :).

Continuing the idea of helping yourself by helping others ,where each A,B problem i solve i write a full clear description about it.__

I am thinking of making it more general and motivating ; there's a total of ~300 A's and 300 B's on codeforces , if each one of us (for those who care about "blue" or A && B ) chosen randomly 150 problems (75 A's and 75 B's) for this month (i.e. till 13 Oct.) and solve them and made excellent tutorials for them (explain as good as we can by pics , examples , ..) , i think it would be great to have all A's and B's explained in more excellent way (although some tutorials there are so good).

Anyway :), i am going to do this if anyone welcomes the idea , we can motivate each other and get out something really good for our community here.

Edit : Plz if you are in :) , just comment by the range you will take [ex. : i will take from contest 200 to 275 (2*75 ) ] & no problem if there's overlapping for sure.

Edit 2 : we can push it to [A,B,C] & even D to get more people in as NikitaMikhaylov clarified :).

Edit 3 : The idea here is get better as a group of people , and get other people less than you also up (whatever less than you mean (rating ,..) ;so any one have better idea of making this like making a group of hard workers and get some work out for people.

if anyone sees this time consuming , ask me (comment or inbox) and will be deleted , no problem at all with that it is just suggestions :).

Thanks :).

I will try to make an excellent tutorial (easy to get the idea) for every A,B division 2 problem i solve ; that is the best way to help others as a Div2 guy.

540A - Combination Lock The idea of the problem is simple enough to notice that to go from '2' to '8' there's only two possible solution(directions) : 1- 2-3-4-5-6-7-8 (forward) ---> abs(s1[i] — s2[i])

2- 2-1-0-9-8 (backward) ---> 10 — abs(s1[i] — s2[i]) (same as : s1[i] — '0' + '9 — s2[i])

" Draw These two possible directions on a paper, to get the idea"

Then , take the minimum between the two and add them to your result counter. of course anything unclear or wrong , please ask me to handle it ASAP. Thanks :) .