I can finally participate in a contest without staying up late as a Chinese.

There is a hacking phase too in this round so we will be very busy around 1205(UTC) lol

Also if there something happens, they will have to increase contest time. Which is unfortunate from every angle.

Yep! We remember well what happened last time this statement was not included :) http://codeforces.com/blog/entry/59720?#comment-434084

Don't worry step_by_step has :P.

Div.3 is just a part of Div.2 so it is rated for Div.3.

Is it faked?

the "education" in educational rounds stands for that.

If I think 4 times, contest would already be over :p

"Educational rounds are rated for users with rating below 2100, but it doesn't mean other users are unofficial. I think, In those rounds all users are welcome to participate officialy."

That's true. Also it is for all people living in the Far East)

I agree with you, but there is a spelling mistake.. Think of "int mian" "inlcude" "isotream" and so on.

The differences are the hacking phase afterwards and full retest on the final set of tests with the hacks included.

I think answer is NO for n = 2 and a=1, b=1, also n=3, a=1,b=1, and min(a,b)>1.

I think its NO when neither of them is 1 or if both of them are 1 (except when n=1) The rule here seems to be that either of G and H will be connected (either of a and b must be 1) ^Not sure if it's entirely true though

I forgot when the a==1 and b==1... how foolish I was!

1

)(

5 1 1

none. in educational rounds there is no points for hacking

Тhe style of these codes is also different.

Some coders such as tourist use this strategy sometimes.

Fortza_Gabi_Tulba is a joke account.... Gabi Tulba is purple on codeforces. Take the joke

Yes, sorry, there was a mistake in the statement. Luckily, no one got problems on that test during the contest.

Just open any solution which has reached to test case 10.

You must not include cases such as )( or )((() in your answer

Just precalculate for each street index the index just lower or equal where you can place the lamp-post. Then iterate over all possible values of powers and compute for all values the mincost by directly jumping to the place where you place next lamp-post and take minimum over all these. 39119701

First check the maximum count of consecutive blocked positions, if this count (let this count's name be maxcon) is >=k or if the first position (0) is blocked then the answer is -1. Then search for the cheapest power starting from maxcon+1 to k. For power i, you will calculate its cost by putting its lamps greedily starting from j=0 and moving with j+=i, but if you reached some blocked j you should change j to the value of the last unblocked position before j (as if you were putting this lamp in the last unblocked position before j not in j itself).

This should work, because there are only log(ai) different values of gcd, so complexity will be n*log(ai)^2.

Only trailing spaces are ignored(I think)

I don't understand your solution

what I did in C was ignored all cases of brackets which gave )( this sort of thing Do you ignore "()( )( "? Then your program won't count ()( )( + )( )() as a regular bracket sequence.

At last I found it. Your program gives wrong answer to this test case. I hope it will help:)

Answer is YES http://codeforces.com/contest/990/submission/39099219 #6

forgive me, I misunderstand it.

That is my strategy!!!

Once this slowed my solution down to make it get TLE though so it's not a strictly good idea

Connecting vertex to all others means that this vertex will make isolated component in graph's complement. So if b > 1, you need b - 1 vertex connected with other vertexes in main graph. If a > 1, you need complete graph on n - a + 1 vertexes, and a - 1 isolated ones.

Yes.

Probably not because the constraint on the input is a[i] <= 1e6 so there aren't many numbers to create collisions.

Unordered map doesn't use a fixed hash fucntion so it's "improbable" to find a hack.

The code has an uninitialized variable. In test 3, there's nothing assigned to the variable 'got' in the binary search code. You can initialize it with a specific value and see what happens.

you're right since your code has a mistake 2nd submission should be ignored.

after the hacks

I don't think op[N]={} initializes an array with zeroes, but I think op[N]={0} do.

You are writing in an empty string (st) without changing its size, so it's undefined behavior. When stuff like this happens it's usually because you allocated the memory badly.

Notice that if we add a cyclic flow on a valid flow, it's still valid. So, if the graph has cycles, we can use cyclic flow to zero an edge on the cycle, effectively removing it. Repeatedly remove edges from cycles until the graph becomes a tree, and then the problem is pretty simple.(recursively solve it from leaves) Complexity is O(n).

I didn't solve F at contest but shortly after the contest I figured this out. Hope that I am correct.

Update: confirmed it works

If you scroll down, you can literally see the test together with the message "wrong answer The number of components in the complement graph should be 84, but found 1"

after hacks

Yes i worked out problem E as i found it was greedy and almost wrote the code but just needed 5 mins to do some case testing.

Brother i don't code in java but as i have seen several times in codeforces comments that arrays.sort() has a worst case complexity of O(n*n).

Sorting Primitive types in Java uses quick sort so it takes O(N^2) in worst case. It's actually common for some reason to always include these anti quick sort tests in codeforces problems.

To get around these you can do two things, declare your types as wrapper class (i.e Integer instead of int) so it becomes an object so runs in O(nlogn), or shuffle the input array.

So it turn out we can shift the lamppost to its left end in this segment to cover [x, x + 2 * l] only if [x + l] was not blocked and the problem just reduces to the same version in here.

This idea was suggested by lrvideckis :)

Fck myself. I was trying to solve this problem during contest. And couldn't create solution. So I decided to solve G instead.

Only after your comment I reread the stamenent. (((

Until the admins of the contest wake up! We need to wait for final tests.

Well it's 10:10pm here so I will simply go sleep soon and check the next morning instead of worrying about it now. :P Good night!

One bracket sequence can be:

1. Good.

2. Need some left brackets ( to make it good.

3. Need some right brackets ) to make it good.

4. Can never be good.

Then you can do bracket matching algorithm to determine which type one sequence is and how many left or right brackets that can make it good. You can see my code at http://codeforces.com/contest/990/submission/39098111

A bracket sequence will be regular if, starting count from first character, for any position of that sequence, number of opening brackets "(" is more or equal to the number of closig bracket ")" ,and finaly number of opening bracket is equal to the number of closing bracket.

Now , if you want to make a regular sequence by concatenating two sequences , there will be some options

1.choose two regular sequence

2.choose one sequence that have n more opening brackets(here n=total number of opening brackets-total number of closing bracket) .and for any position of that sequence number of opening brackets are not less than number of closing brackets.and choose another sequence that have n more closing brackets than opening brackets (here n is similar to previous one).and for any position of this sequence number of closing brackets should not be more than n+number of opening brackets. now you can concatenate these two sequences in one way , 1st+2nd will be a regular sequence .

[Don't know how much I could explain , sorry for my bad English :) ]

We have to wait for the problem maker to wake up.

Do you know that there is a chrome extension for codeforces? You can check the expected rate change quite precisely.