Check for errors in your console. I guess ur browser is old.

Can u make it just visible to other people. This will not disturb you pro guys and make the weaker guys like me know the right way to practice. I think this is not that difficult. Is their any problem in that?

I wasted a lot of time bcoz of this typo in the editorial.Thank you.

Is it possible to have this feature running, it will be amazing.

I also got a runtime error on test case 92.I just changed string to char array and took the input character by character.It got AC.10776197

I am getting HTTP Status 500 — java.lang.NullPointerException in Div2 D ladder.

Please tell how do we use this platform neerc.ifmo.ru?

Best of luck to all the Indian teams. karanaggarwal and adurysk bring the best positions for our country.

It may help:Matrix Expo Tutorial

Sort the problems here of dp tag in codeforces by the number of users who solved and start doing them.

u can call it the present value in the binary search or mid.

You can go through: First:http://mirror.codeforces.com/blog/entry/16221

Second:http://mirror.codeforces.com/blog/entry/3075

You can do this by binary searching the answer which is checked valid by:

1.making a graph with cost sqrt(|x_j-x_i-L|) — R * b_j and always x_j > x_i.

2.In this graph,find shortest path from x0 to xn (take x0 as a dummy start).

i)if this value is -ve this means the mid(of binary search) is greater than the optimal value.

ii)if +ve this is opposite the above,thus take the necessary steps in else of binary search.

3.Once you get the optimal value you backtrace the waiting pts you took.

For further reading Editorial's comment

His submission for this method Submission

EDIT:Congrats for becoming the Training and Placement Representative

It is evaluated left to right so when first condition returns 1 it is left with 1<=1 which also returns 1.

Thank you sir.So is the round unrated?

I request the organisers to please make this contest unrated being full of unexpected delays and problems.

why haven't we taken the case of selecting both from the no 1s or both from both one 1s in the case of x>=1 and y>=1 thus adding (x*(x-1))/2 and (y*(y-1))/2 to x.y?

EDIT:Got it

That's great.

Can this be integrated with sublime too?

The latest version gives support for proxy.But I m not able to use it even now.There are Parsing failed again.Pl tell me if i hav to do something else except setting system proxy.

http://mirror.codeforces.com/contest/242/submission/9527768 and http://mirror.codeforces.com/contest/242/submission/2563497

The same code...i m not as lucky as the latter.

I loved it but I need to send the auth code every time i need to sign in.It neither remains signed in.

https://polygon.codeforces.com/c/contestid/english/statements.pdf Does this means if I write https://polygon.codeforces.com/c/482/english/statements.pdf i will be able to download the Round 275 Div1 problems as pdf?

I also read flashmt codes.They are the best to refer to.

Просто из любопытства.

Please make it for firefox also.

Acc to what I hav understood take every pair of i and j and see if i comes after j in all the given permutations.If yes,add edge i-j in the new graph.This makes the graph to find the longest path.Pl tell if I am wrong somewhere.

EDIT:Sorry didn't read Alexandru's reply which is the same as what I said.

As it will not be possible for sharing a spreadsheet with too many people I can't take everybody in.U can start ur own.No big deal.And those who r in stay consistent as bcoz of u someone else is not in. Interested people can still send ur ids I will consider them.

Please don't just downvote and leave.If you really don't like this post give ur suggestions for what else do u consider right instead of this.I requested for this bcoz this can save a lot of my time which everyone spend searching a nice problem of their level(including me,I also hav to spend a considerable amount of time just searching the right prob acc to no of users who solved it).

Pls explain the approach for Problem J — Product Innovation

How is getting a head second time after getting head on first throw a dependent event? Pls explain. Shouldn't both be independent events?

Thank u for the statistics.Loved watching myself here for my first hack in codeforces.

first sort the probabilities.. then suppose they are a,b,c in sorted order 1.Take a*(1-b)*(1-c)+b*(1-a)*(1-c)+c*(1-b)*(1-a) which is prob of getting one of them at a time. 2.Then leave the smallest probability a,then find b*(1-c)+c*(1-b) 3.Then leave prob b and take c. 4.Take the best value(closest to one) among the three calculated above. Using this u can calculate the answer for ur test case.

Thank you

What is meant by Non-blocked edges in Div1 C?

Is there any link to the problems too?

What is Challenge 24 and PM?

Chullu bhar pani mein doob mar yeh samajh mein aa jae to batana