codeforces rules.the contest's email mentions it.

Это будет обычный раунд codeforces для двух дивизионов

You can actually miss your exam for codeforces round because it is more important. sorry for my bad english.

Exam on Sunday? Quite strange. Sunday is generally a holiday.

You can virtual participate in this contest when you are free, though it is unrated for virtual participating.

Mine too!!! :)

So copied comment got downvotes? Roger that.

Good luck and Have fun in div.1

I think that second option is much less painful, though the first one will result in better performance.

2b||!2b

Well, I just woke up at 5 am. (You could never stay up until 5 am :D)

Это будет обычный раунд codeforces для двух дивизионов

Oh Got it !

Why this ^^ comment can't be seen! It has +44 vote. It's quite strange! May be codeforces have some technical issues.

Now the contest is pending... So you can't watch the problem now. After about 3 hours the contest will start. You can participate it if you have register this contest.

Just visit this when the contest starts Codeforces Round 274 (Div. 2)

Sorry for my pure English

I forgive :D

Pure means perfect. I think you misuse it. May be you wanted to say Sorry for my bad/poor English. BTW never mind :) We should learn from our mistakes :)

And I problems were also interesting, great contest today. I had a little problem to understand C today, but that's my problem only...

I hope you are hungry for problems.

There's no requirement that the number of operations used is minimum.

This problem took me 20 minutes. I misunderstood it. I thought like you that the second number should be minimal. And then I saw this example... . It took me far more time to see that second number shouldn't be minimal.

Then you should change your avatar as well :)

+1

You have to lock your problem first on the Dashboard page. Only then can you hack other solutions.

You cannot do that if the input is so long, you have generate input when you try to hack.

there is a file upload option also, you could use that.

i think it is, cuz my solution with segment tree was hacked

My solution is O(nk), but yes, it's DP.

Well I think so, pretty sure that there is no soln in less than O(nk) and O(nk) surely exists.

I just sorted them in descending order by a[i]. Have a variable that keeps track of the current day, let's name it day. Starting case is day = b[0] (zero indexed, already sorted) Now for all elements (starting from the second):

if b[i]>day, day=b[i]

else if b[i]<day, day=a[i]

Just print day at the end. It's a simple O(NlogN) solution that I believe works (hasn't been hacked, can't know for sure before grading).

EDIT: Forgot about the sorting when mentioning complexity.

Here's my solution

3
3 2
4 1
6 5

Many people assumed that when the pairs are sorted by a, the resulting permutation of b must also be sorted in order for the answer to be b_n (otherwise a_n). This is not the case, as shown by the test case above. (The answer is 5; many people give 6.)

it's like gambling, you play, you lose but play again hoping to win what you have lost, until you end up being completely homeless.

At least, Rostislav_the_great can now be sure his D submission is correct.

Well, my solution didn't pass pretests when I wasn't checking if new added dot < L, so maybe it's only the left if it's positive

I found this (which also failed my locked submission):

6 20 14 15
0 9 10 16 17 20

Some people that made searching whether a distance exists in the array made it a function and thus only returned the first occurrence. In this case, the first occurrence of y - x is (9, 10), and with bounds they go over the ruler. However, there is another occurrence (16, 17), which fits; a new mark can go to 2, so the answer is 1 new mark. Those unlucky people (read: me and a couple I hacked) answered 2.

My solution had the problem that I was introducing only the new values of arr[i]+x or arr[i]+y. A case like

3 9 1 4
0 6 9

would fail.

Well, maybe the setters wanted that even a O(N^2) sort can do.

You can just take k times,in each time you work for the maxium-1 and the minium+1,notes the ans. And then it'll be solved:)

Solution can be 0,1 or 2(its obvious). Now, you can check if answer is 0(There exists both x and y). After, you can check if you can add a dot somewhere and make both x and y. If you cannot, just print 2 x y. Time Complexity: O(NlogN) logN is for using map/set/binarysearch. Space Complexity: O(N)