It is not a bad time for Chinese!

An usual comment about unusually usual unusual time lacking articles and punctuation.

If do both,maybe you will think of rating changes of Atcoder when you do Codeforces.

By the way, thanks for the message. I previously did not hear about AtСoder.

Probably using the same rating algorithm. It should still be fair.

Since it mentions usual round, I think we can hope for hacking as well

google translate English xD

For any given time there is always some country which can't participate.
This contest is mainly intended for Russians so obviously they would choose a time that fits them first and then think about the other countries.

Problems statements will be available in Russian and English languages.

cricketers doing their job. you should do your job as a contestant too ;)

salam maybe it is advantage for us if we have good performance.

I am also not able to lock the problem

I had to click on the lock three times, but then, it worked.

Me too. I click the lock 3 times.

send*

Lol that time difference. More like instant karma xD

Same thing =) but against me..

Found a bug in my solution, but my attemps (both with different bugs) were accepted. Tests were too weak, and I used it)

P.S. hack attempt at 21:53 was really strange. Only in this moment I understood that my solution is wrong :D

if you entered the contest area, you will get a rating.

I think some people got you wrong and put minus on your comment just because they saw "will it be rated?" :D

If all its letters are consonants?

Maybe if the text consists of only words which do not have any vowels in them.

Isn't it no different than the ch allenge in the first example? It is analyzed at the bottom of the statement, and ch is specifically omitted.

If we didn't ignore words without vowels, we would formally have to answer NO to the first example, too. And an empty set of words with vowels, as in bbbb, is still a set of words with vowels.

Same as my problem ,Really annoying :( !

But at least I could get its points from hacking .

This is very unfair.

Not only there :(

Well, that is sad. Though I don't think there is any way to avoid situation that some problem appeared on some online judge — there are too many of them nowadays.

Actually, this problem happens a lot in recent rounds, so if someone has a nice suggestion on how to avoid this sort of situations — they are more than welcome.

I had exactly the same situation. I was just sitting and looking at the first sample for a very long time, trying to understand, why we can't achieve subsegment with length 3 consisting of only 1.

Well, it really was hard to understand but I think that first sample is the answer.

+1

This sentence is especially misleading:

The above operation is performed for 10^100 seconds. For each integer from 1 to m find out the longest segment found at this time.

at this time actually suggests to find the longest segment at instant t = 10^100. In Russian version of the statement says "за все это время" which means "during all that time", "throughout the course of a that time period".

bin search + greedy passed pretests.

12
20:00

Depends on your program,

I hacked a few people with: 12 30:59

Intel Hack Challenge Elimination Round

You don't need to reduce P/Q at all. This problem is harder than that one because here we need to count paths with at least 20 nodes (and in that problem we needed to count paths with at least 1).

Looks like the solution here is very similar, but I still don't know exactly how to do it in O(n²).

There's no need to reduce P/Q at all, since we do all our computations modulo a prime.

There are two words already. That means you cannot do less than 2 syllables.

You can only ignore words which consist ONLY of consonants. Hence you cant ignore "aa".

Words that consist only of consonants should be ignored!

Otherwise O(N^3) solution would pass..

Each syllable consists of exactly one vowel (yes i'm hacked because of this)

I think the statement meant to say there are only one vowel allowed in each syllable, and thus it reduces to a problem that counts the amount of vowels(eg: to has one 'o').(which makes a lot sense as a question B)

1.Nope, after locking in and trying to hack, I realized that we cannot group 2 or more vowels into one group. So the only way to split "aaaa" is "a","a","a","a" . The correct answer would be 4.

1. I guess they meant "group" instead of divide. Then you can group "to" into 1 syllable.

It can be sometimes.

Before you hack, you need to lock your solution to the problem you wish to hack. You can hack by going to your room, and double clicking on someone's solution that you wish to hack. Then you see their code, and if you think their code is wrong and doesn't work for all cases, you can click on the button hack, and give a test case for which you think the solution would produce wrong answer. If you are correct, you gain +100 points, if you are wrong you lose 50 points.

B was easy, just had to count the number of vowels for every string which could be done in a few lines of code, I would have done it if I knew how to take multiple strings in one line :/

I thought it's only my problem !

It could be hacked by:
12
35:00

I think you keep a multiset/map for the segment sums, a set that keeps the destroyed elements, and a prefix sum for quick calculations. For each destroy action you binary search for the previous and next destroyed elements and update the segment sums from the prefix sum.

Set is probably better for this. But if you looking for a long approach then segment tree (infact http://www.spoj.com/problems/GSS1/ part of it)

An O(nlogn) probably won't TLE

You can use dsu.

Here's my solution , A is the first array given in the input , B is the second array given in the input . segments is the total number of segments so far . At first you create an empty array , let ind = B[i],value = A[B[i]] . starting from the last element to be destroyed (from n to 1) : there is one element inserted in this empty array at index(ind) , so you add a new segment with :

right[segment] = ind,

left[segment] = ind,

sum[segment] = value;

now every time you insert a new element , you have to check the segment at the left index of you (ind-1) and at the right index of you (ind+1), if there exists ONLY one segment (either to your right or to your left) , you have to add the value to its sum and UPDATE its boundaries(right or left) , if you have BOTH segments you have to combine them and form a new segment with :

left[new segment] = left[segment[ind-1]];

right[new segment] = right[segment[ind+1]]];

sum[new segment] = some of BOTH segments + new value;

yes contest should be unrated :/

The same here!

Yes the contest should be made unrated.

Hmm, seems right, here's the problem link.

(to clarify, the comment is right that problems are the same, not that it automatically means unrated)