Is it faked?

6 лет назад, # |

Done with semesters finally!! Wishing higher ratings to everyone including myself :P

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6 лет назад, # |

For me, it is first time contest starts in reasonable time zone. (19:05 KST) I was struggle on midnight-beginning contest heretofore :)

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masterbios

6 лет назад, # |

← Rev. 3 →

+10

float ship; // must be a float otherwise ship sinks

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agent623

6 лет назад, # |

+23

I couldn't help but notice that people with rating >= 2100 are not marked as out of competition.

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mredigonda

6 лет назад, # ^ |

← Rev. 2 →

-7

"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."

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Democy

6 лет назад, # |

← Rev. 2 →

+10

This time is friendly for Chinese!

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overhyped

6 лет назад, # ^ |

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

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ouuan

6 лет назад, # ^ |

← Rev. 2 →

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

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Democy

6 лет назад, # ^ |

oops...I didn't notice that spelling mistake.

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IvayloS

6 лет назад, # |

Can someone explain what are the extended ACM ICPC rules? In what do they differ from the regular ACM ICPC?

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6 лет назад, # ^ |

+13

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

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EbraM96

6 лет назад, # |

For problem D.. the only "NO" testcase is n > 1 and b > 1 .. is this true ?

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Len

6 лет назад, # ^ |

← Rev. 4 →

+36

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.

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6 лет назад, # ^ |

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

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fauzdar65

6 лет назад, # ^ |

Both of them can be connected at the same time also. (Consider the graph 1--2--3--4 with 4 vertices and 3 edges)

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6 лет назад, # ^ |

Oh yeah you're right

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KOH

6 лет назад, # |

I wonder what's the matter with problem D?

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KOH

6 лет назад, # ^ |

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

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Ahs_58

6 лет назад, # |

Give me some strong test case for problem C

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alwerty

6 лет назад, # ^ |

)(

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6 лет назад, # |

What's the 3rd test case for problem D?

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veschii_nevstrui

6 лет назад, # ^ |

5 1 1

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6 лет назад, # ^ |

The Answer for this is NO right?

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Firastic

6 лет назад, # ^ |

Chain 1-2-3-4-5 satisfy this i think

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rishi_07

6 лет назад, # ^ |

Answer is yes. (1-2, 2-3, 3-4, 4-5)

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6 лет назад, # |

Can anyone tell me how many points will I get for successful hack and how many I will lose for an unsuccessful one?

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nimom

6 лет назад, # ^ |

none. in educational rounds there is no points for hacking

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6 лет назад, # ^ |

Then I don't get what's the point of giving 12 hours for the hack, if there is no point for hacking, I mean at last they will check each code on their test code then why 12 hours for hacking. @Infinite_M can you please explain. I am actually a beginer here.

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6 лет назад, # ^ |

+10

There is a point. You can increase your rank if you hack solutions of people ranked above you ;)

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6 лет назад, # ^ |

All successful hacks will be added to system tests. existing test cases don't guarantee that program is correct.

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6 лет назад, # ^ |

[user:killer_good] Can you please clarify this point? I was in an impression that codeforces provide some basic test cases to check code during contest.But they do have a test file including all corner test cases(like other websites , codechef,etc) after the contest they run every code through these test files(including corner test cases). is it true?? Or They just prepare base test cases on which they check submissions during contest and let participants find the corner test cases during hack??I mean don't they have test files for possible corner cases?

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6 лет назад, # ^ |

for normal contests what you said is true. but for educational round system test will consist of successful hacks from users that's why we have a separate hacking phase unlike other contests.

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sansae

6 лет назад, # |

← Rev. 2 →

+74

User Fortza_Gabi_Tulba submitted F(39097227) at 00:28, and submitted E(39097570) at 00:29.

Since it is (almost) impossible for human to solve and code problem F in only one minute,

does it mean he submitted problem F while coding problem E?

Or he coded E and F, and then submitted F and E?

Or, is he a genius?

same thing for problem B(39091953) and D(39092747)...

I doubt if Fortza_Gabi_Tulba coded with friends...

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312352145215

6 лет назад, # ^ |

Тhe style of these codes is also different.

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6 лет назад, # ^ |

-7

Some coders such as tourist use this strategy sometimes.

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VLamarca

6 лет назад, # ^ |

+35

tourist does not do this in codeforces. It only makes sense to do this in judges where the penalty is calculated by the time of the last submission, which is not the case of codeforces where the penalty is calculated as the sum of times of all accepted submissions.

So here it is better to submit a problem as soon as you coded it, and not doing that is suspicious.

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AlexArdelean

6 лет назад, # ^ |

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

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LuckStillWins

6 лет назад, # |

Bloody build stupid graph for a=1,b=1 and n>=1,damm i should have check more carefully for problem D :(.

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IvayloS

6 лет назад, # |

Test number 31 on problem F is not correct. It has n = 1, while this is not possible. According to the statement m >= 1 and you can not have non-zero m for n = 1

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6 лет назад, # ^ |

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

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6 лет назад, # |

test case 10 in C?

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6 лет назад, # ^ |

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

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6 лет назад, # ^ |

It is almost impossible to understand from someone's solution what I am failing at as my logic can be different than to the referred code.

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6 лет назад, # ^ |

← Rev. 2 →

When you open a solution you can see test cases it has reached:

in this case

Perhaps that will help.

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6 лет назад, # ^ |

If you open a solution, you can view all of the test cases that solution has reached. Just scroll down. That's what Emil_PhysMath meant.

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6 лет назад, # ^ |

thanks, I did not know that!

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6 лет назад, # ^ |

Do note that it only works out of contest :)

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ameyanator

6 лет назад, # ^ |

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

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6 лет назад, # ^ |

yes i have done that.

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Oreo_85

6 лет назад, # |

First contest during travelling ^_^

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6 лет назад, # |

I was suffered from the word in problem A "commentary delegation burle demolish blahblah" I should study english more :(

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rishi_07

6 лет назад, # |

+12

How to solve E?

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amnesiac_dusk

6 лет назад, # ^ |

← Rev. 2 →

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

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veschii_nevstrui

6 лет назад, # ^ |

Why O(nlogn)? I can prove only O(nsqrt(n)).

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amnesiac_dusk

6 лет назад, # ^ |

← Rev. 4 →

+30

Proving that for power i it needs O(n/i) should be enough. Let us say that we are computing for power i and length n. Now, say there is a lamppost at position l and the next lamppost be placed at r now even next lamppost must be placed at least l+i, clearly except in the case that it is not possible to illuminate the street. So there should be atmost about ~(2*n)/i iterations.So, the complexity should be O(k+nlogn). Please correct me if something is wrong.

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mohamedeltair

6 лет назад, # ^ |

← Rev. 4 →

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).

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Rogue33

6 лет назад, # ^ |

For the 1st test case,

6 2 3

1 3

1 2 3

Can't we use the lamp with power 1 to light the whole street?

If we put it at position 0 , it will cover [0,1] . then at 2 covering [2,3] , then at 4 covering [4,5] and at 5 covering [5,6]

That way the total cost will be 4, which is cheaper.

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mohamedeltair

6 лет назад, # ^ |

← Rev. 2 →

You need also to cover the segments [1, 2] and [3, 4]. There was an announcement regarding this during the contest which was:

"You have to illuminate the whole segment, not only its integer coordinates. For example, if n = 3, illuminating [0, 1] and [2, 3] is not enough."

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Rogue33

6 лет назад, # ^ |

Oh! Didn't notice. Thanks

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hyunynim

6 лет назад, # |

can I get the rating in this round?

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DongwonShin

6 лет назад, # |

← Rev. 2 →

Does anyone know solution for G? My solution keeps getting TLE on test case 95, and I think mine has O(n log^2 n) time complexity and I can't think of any better solution. I used centroid decomposition

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Len

6 лет назад, # ^ |

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

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DongwonShin

6 лет назад, # ^ |

Thank you. I was thinking too complicated solution, you are very smart

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pk842

6 лет назад, # ^ |

can you please explain the idea a little bit.

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Kognition

6 лет назад, # ^ |

← Rev. 2 →

Intuitively I felt that there should only be around log(ai) different values of gcd, but I can't think of a good formal argument for why that is true. Do you have any good proof for it?

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tfg

6 лет назад, # ^ |

← Rev. 2 →

Hacked huh. There are actually O(divisors) gcd values for paths (think about a leaf for every divisor and the rest of the tree has the same value). What you were thinking about is another thing, there are at most O(log) CHANGES of gcd in a path.

So your solution is actually O(N * maxDivisors(2e5)^2) along with bad constant from using map (mostly chache misses I guess).

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Len

6 лет назад, # ^ |

It works, just forgot to clear maps after using ;w;

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tfg

6 лет назад, # ^ |

← Rev. 5 →

+16

http://mirror.codeforces.com/contest/990/submission/39135974 I submited the exact same solution that got hacked and it passed in the same time as before the hack hmmm.... Looks like that hack didn't get into systests lol

Edit: we can see the hack now http://mirror.codeforces.com/contest/990/hacks/458169/test. I tried to test it locally and the second code crashed in my computer (probably couldn't handle the memory lol). Maybe someone could add this test to the test set? This is a intuitive solution that has already appeared in codeforces but with downward paths only iirc, I wouldn't be surprised if a similar solution passed systests just because it didn't got hacked.

Edit2: I tested it in a custom invocation and the second submission passes in a bit under 4 seconds (ignoring input) and without the clear it's MLE. Can someone prove a lower bound than what I said for this solution?

Edit3: the hacker that did this did most hacks for G http://mirror.codeforces.com/contest/990/hacks?chosenProblemIndex=G and the test didn't get added. Hopefully he didn't miss anyone :P

Edit4: this case was added!

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-osdajigu-

6 лет назад, # ^ |

Hey, can you explain your approach?

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tfg

6 лет назад, # ^ |

For each i, compute h(i) = number of paths that have value multiple of i. For each i, dp[i] = number of paths that have value exactly i = h(i) — dp[i * 2] — dp[i * 3] — dp[i * 4] — dp[i * 5] — ...

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ujzwt4it

6 лет назад, # |

← Rev. 6 →

-34

I was getting wrong answer in D coz I printed
YES
0 1 0 0 0
1 0 1 0 0
0 1 0 1 0
0 0 1 0 1
0 0 0 1 0
instead of
YES
01000
10100
01010
00101
00010

basically the same result with spaces. Shouldn't that be allowed to pass?

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dush1729

6 лет назад, # ^ |

Only trailing spaces are ignored(I think)

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6 лет назад, # |

← Rev. 2 →

what I did in C was ignored all cases of brackets which gave )( this sort of thing, and kept a count of regular expression brackets, then if suppose (())), then this one has 1 closing bracket more, hence it will require any bracket sequence which has 1 opening bracket (such as (, (() or many like this)

I iterated for every sequence if it was a regular one, I added the total number of regular ones in the list given. If it was something like (())), which means it requires one open one, hence I added number of sequences with one open bracket, and when I came to the bracket with one open required, I ignored, hence I only calculated for the regular and closed ones, as open would be covered while doing for closed ones.

Can anyone suggest me what am missing?

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6 лет назад, # ^ |

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.

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6 лет назад, # ^ |

Nah, I ignored the ones which has )( both

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6 лет назад, # ^ |

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

input

right answer

your programs output

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BeardAspirant

6 лет назад, # ^ |

thank you. I was also stuck on the same test case.

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6 лет назад, # ^ |

You're welcome:) I hope it will help you to correct your code.

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xish

6 лет назад, # |

I saw one of the test case of Problem D is 1 1 1, which means a graph with just one node.

The correct answer is "NO", can anyone explain why there is no answer when n = 1?

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6 лет назад, # ^ |

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

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xish

6 лет назад, # ^ |

forgive me, I misunderstand it.

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6 лет назад, # ^ |

It's okay XD

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ouuan

6 лет назад, # |

Remember to use long long everywhere, even though you think that it can't be over 2^31. Especially when you can't find your mistake anywhere.

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6 лет назад, # ^ |

That is my strategy!!!

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WolfBlue

6 лет назад, # ^ |

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

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6 лет назад, # ^ |

Yeah you are right. But once on a USACO contest I got WA. Then I changed only one variable(I thought only it could cause a WA) to long long and again got WA. After the contest I read the editorial and the solution was suspiciously similar to mine. I changed all variables and got AC:(

After that changing int to long long is the first(or maybe the second) thing I do after getting WA.

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01124699

6 лет назад, # |

What is the solution for D?

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DoomzGay

6 лет назад, # ^ |

← Rev. 2 →

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.

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mohamedeltair

6 лет назад, # ^ |

For the second case I think you meant complete graph on n-a+1 and a-1 isolated ones.

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DoomzGay

6 лет назад, # ^ |

Yeah, sure, thx :)

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Pilot_Girl

6 лет назад, # |

Can one hack this solution with making an anti-hash test? http://mirror.codeforces.com/contest/990/submission/39097287

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BFF

6 лет назад, # ^ |

Yes.

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BFF

6 лет назад, # ^ |

I'm sure halyavin can do it.

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alwerty

6 лет назад, # ^ |

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

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6 лет назад, # ^ |

-13

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

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Em00n

6 лет назад, # ^ |

← Rev. 2 →

I once got several WAs in a problem because of a test designed to break unordered_map hash. Isn't it the same for hacks? Or does testing behaves differently using a fixed seed?

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Siriuslight

6 лет назад, # |

-24

awoo For the Problem B, I submitted this in C++ 11, it gave me a wrong answer. Then I submitted the same code in C++14 this and got accepted.

Because of wrong submission I had to face a penalty. I think this is wrong and it should be removed.

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djm03178

6 лет назад, # ^ |

← Rev. 2 →

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.

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6 лет назад, # ^ |

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

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Siriuslight

6 лет назад, # ^ |

I don't think any action would be taken by MikeMirzayanov and CF admins. I hope they take decision in my favour, if they do something. Well I enjoyed the contest, though I didn't perform well. I think next round would be favorable.

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shleodai

6 лет назад, # |

Are there any editorial now?

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ShowStopper728

6 лет назад, # ^ |

after the hacks

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darth_coder

6 лет назад, # |

← Rev. 2 →

Why this 39120549 fails while this 39119719 passes for problem C?

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MikeMirzayanov

6 лет назад, # ^ |

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

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darth_coder

6 лет назад, # ^ |

← Rev. 2 →

Even that failed. 39127177 . I am unable to understand the Diagnostics. Can someone check the Diagonostics of the above solution and explain to me. Thanks.

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http://mirror.codeforces.com/contest/990/submission/39122954

6 лет назад, # ^ |

+16

it actually does. Not an undefined behaviour.

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alwerty

6 лет назад, # ^ |

← Rev. 2 →

+20

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.

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ArchitJ

6 лет назад, # |

How to solve F? i figured i would put in a super source and a super sink to reduce it to a max flow prob. But complexity for solving flow vector is too bad

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cai_lw

6 лет назад, # ^ |

← Rev. 2 →

+24

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

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kr_abhinav

6 лет назад, # ^ |

Beautiful solution, thanks! I just submitted using your idea and can confirm that this indeed works

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Alexit

6 лет назад, # ^ |

← Rev. 3 →

why if sum !=0 test incorrect? This test looks correct.

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kr_abhinav

6 лет назад, # ^ |

Look at the bottom right node, it has a netflow of 9, and not -9. As for why the sum should be zero, you are flowing from one node to the other, so the net inflow must be equal to the net outflow

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Alexit

6 лет назад, # ^ |

oh, my mistake, ty

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sguu

6 лет назад, # |

Why My solution for D is wrong??

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6 лет назад, # ^ |

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"

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4e494b

6 лет назад, # |

when will the ratings be assigned regarding the educational round.

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6 лет назад, # ^ |

after hacks

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4e494b

6 лет назад, # ^ |

was this round rated for Div 3?

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6 лет назад, # ^ |

for everyone below rating 2100 as mentioned

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Em00n

6 лет назад, # |

Awesome educational as usual, but I think two hours might have been a bit too little for seven problems. Most people didn't have the chance to work on F or G, which is a pity considering they were good problems.

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6 лет назад, # ^ |

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.

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grekiki

6 лет назад, # |

I can't figure out why my solution hits the time limit. It should be O(nlog(n)). http://mirror.codeforces.com/contest/990/submission/39097135

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6 лет назад, # ^ |

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).

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Medo.

6 лет назад, # ^ |

← Rev. 2 →

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.

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grekiki

6 лет назад, # ^ |

Thank you, I will try that tomorrow, I have never had that kind of problem before.

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Hossam

6 лет назад, # |

How to solve E if a lamppost of power l were to cover [x - l, x + l] ??

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Hossam

6 лет назад, # ^ |

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 :)

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taow

6 лет назад, # ^ |

← Rev. 2 →

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. (((

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Hossam

6 лет назад, # ^ |

trust me you're not alone :'( I had the same misconception about this problem :|

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Princ_iple

6 лет назад, # |

+20

The tragedy of every educational round.. waiting for the final test and the rating change.

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gng

6 лет назад, # |

How long do i have to wait to get my rating?

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mrz

6 лет назад, # ^ |

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

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onthefloor

6 лет назад, # ^ |

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!

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M.Ibrahim_

6 лет назад, # |

when will rating change ?

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jinlifu1999

6 лет назад, # |

Well if you cannot wait to see the final result, you may submit your code in the archive and see how it works.

→ Ответить

pappu1212

6 лет назад, # |

Can anybody explain me the strategy used to start solving Problem C or any similar question question's hint or link.

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jinlifu1999

6 лет назад, # ^ |

← Rev. 2 →

One bracket sequence can be:

Good.
Need some left brackets ( to make it good.
Need some right brackets ) to make it good.
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://mirror.codeforces.com/contest/990/submission/39098111

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tanus_era

6 лет назад, # ^ |

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 :) ]

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caojielong

6 лет назад, # |

+13

why my rating hasn't change until now? How long does the change take after the contest finished?

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Skyscrapers

6 лет назад, # ^ |

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

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nong

6 лет назад, # |

+64

Why couldn't system tests start automatically?

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sagarthecoder2

6 лет назад, # |

← Rev. 2 →

It’s really annoying to wait for rating changes after completing Hacking phase in every Educational round. please @MikeMirzayanov solve this.

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hyejinJ

6 лет назад, # ^ |

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

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quality

6 лет назад, # ^ |

The issue, system tests have not happened yet so expected rating change is not final.

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6 лет назад, # |

after round 487 there will be no contest for 5-6 days and it's not good

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arpit040199

6 лет назад, # |

Can anybody explain how to use generators for hacking ?

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djm03178

6 лет назад, # ^ |

It requires a code that prints out the test case to stdout. i.e. Use printf / cout to make a test case.

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vivace_jr

6 лет назад, # |

http://mirror.codeforces.com/contest/990/submission/39138201 can anyone help me finding whats wrong with my code for F?

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21August

6 лет назад, # |

Will the ratings beupdated after today's contest?

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6 лет назад, # |

Auto comment: topic has been updated by awoo (previous revision, new revision, compare).

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6 лет назад, # |

@ MikeMirzayanov regarding submission for the question b.microworld my submission is http://mirror.codeforces.com/contest/990/submission/39094917 it is getting correct answer(in local g++ compiler) for the test case which is showing as wrong output in the system run for one test case. this is looking very odd.

can you please look into this, and also wanted to know if any one facing same.

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6 лет назад, # ^ |

← Rev. 2 →

adedalic Roms BledDest can anyone respond to this.

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yassin_

6 лет назад, # ^ |

ar.size() - 2 can underflow (unsigned), which results in a large value and you access the array at that value, i.e. undefined behavior

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6 лет назад, # ^ |

i think that case wont be arrising. please check by running , i checked it.

test case: 1 1 1 output: 1

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6 лет назад, # ^ |

@yassin_ reply please??

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yassin_

6 лет назад, # ^ |

+12

It's undefined behavior, results can vary for different runs or compiler versions

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adedalic

6 лет назад, # ^ |

← Rev. 2 →

+12

Test 15 as you can see is: 1 1 1 and your solution gives: -1199 what is definitely UB. It seems like it is caused by ar.size() - 2.

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6 лет назад, # ^ |

← Rev. 2 →

-20

@adedalic I checked it using local compiler and some other online as well which are working well for that input. can you once check it once by running the code. since i dont feel some thing undefined should be occuring for this code.

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6 лет назад, # ^ |

who are down voting this,..please explain me whats wrong in the code.(which is working fine in other compilers )

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6 лет назад, # ^ |

← Rev. 2 →

yassin_ adedalic

so you guys are saying problem arise at ~~~~**~** while(n1>=0&&ar[n2]-ar[n1]<=k&&ar[n1]!=ar[n2]) ~~~~~ line 30 of code where compiler runs differently for this line. is it??

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6 лет назад, # |

when will ratings get updated???

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http://mirror.codeforces.com/blog/entry/20762

6 лет назад, # ^ |

ratings updated. Just saw it

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4e494b

6 лет назад, # |

How is rating change determined here?? Pls help.

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312352145215

6 лет назад, # ^ |

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kbrmqs855

6 лет назад, # |

I want to become purple...

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