Codeforces Round #435 (Div. 2)

7 years ago, # |

-27

Pls do something about the long queues. Yesterday's contest was just an example of it but in practice also these days the subs take too much time to judge.

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saisurya027

7 years ago, # ^ |

← Rev. 2 →

+14

Mike told that it was his mistake and there was no internal fault(that's what I understood from his explanation :) sorry if I am wrong). Yes many times while practicing there are situations of LONGGGG queues... Anyways hope it will be a good RATED round :)

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sergon2002

7 years ago, # |

-54

is it rated?

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Hasan

7 years ago, # ^ |

-31

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7 years ago, # |

hope this time in the questions the fullstops and multiplication signs will be different! last time they used '.' for both and it made the question A look way complicated XD lol!

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isgulkov

7 years ago, # ^ |

Do you mean you can't distinguish this: "·" from this: "."?

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7 years ago, # ^ |

even if it was that it is still confusing right? :p XD lol.

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UpdateRatingOrRIOT

7 years ago, # |

-15

Gl & Hf to everyone!

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Knight_of_Thirteen

7 years ago, # |

+53

The queue problem isn't going away. Even in yesterday's NEERC Subregional, the queue was too long and it affected the contest. When you submit a code, and you get a WA after 7-8 minutes, it doesn't feel good. In a contest, every second counts. Also, you can't concentrate properly on another problem even if you want to while waiting for a queued verdict. It's frustrating. I hope this doesn't occur in today's round. If it does, then it'd be really disheartening. CF is the best, we want it to remain the same way.

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sm_hemel

7 years ago, # |

Is it mathematical?

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redocyz

7 years ago, # ^ |

+10

he is also the author of this round. so maybe you can try it to find out?

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mohammedehab2002

7 years ago, # ^ |

+23

In fact, Most of the problems in round #396 are mine while most of the problems in this round are his .

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KOH

7 years ago, # |

Hope the server will work effiently

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saisurya027

7 years ago, # |

← Rev. 2 →

+16

My first contest in CODEFORCES is Round 396 which is conducted by you and I am very happy that again I am participating in your contest.

Thank you.

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7 years ago, # |

-16

why the last contest was unrated?

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amanbol

7 years ago, # ^ |

Technical reasons

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hugewarriors

7 years ago, # |

Score Distribution ? Not Declared Yet , 1 & Half Hour Left

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enoone

7 years ago, # ^ |

already one

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mahmoudbadawy

7 years ago, # |

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

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ouch__casquinha

7 years ago, # |

-20

if repeted problem sucks

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wrick

7 years ago, # |

Scala support is completely broken and couple of users reported it for a while. See: http://mirror.codeforces.com/blog/entry/54283 http://mirror.codeforces.com/blog/entry/54245

I also messaged the admins about it and its still not fixed :(

Hope someone from CodeForces notices this!

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MikeMirzayanov

7 years ago, # ^ |

I think it is fixed now.

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wrick

7 years ago, # ^ |

+21

Thank you MikeMirzayanov — I verified it works now!! If it is possible, can you reset my ratings to before the past couple of contests? All my Scala solutions failed during contest due to this bug (I verified they would have passed tests otherwise now).

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Amaryllis

7 years ago, # |

Am I the only one that gets stuck on infinite loading when trying to submit? Have been trying to submit C for 15 min already with no success.

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7 years ago, # ^ |

maybe the problem is with your computer

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7 years ago, # ^ |

or internet

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Amaryllis

7 years ago, # ^ |

Yeah it turned out to be the wifi, I did the rest of the contest connected to my phone's hotspot.

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getSchwifty

7 years ago, # |

How to solve B ?

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amrsaeedhosny

7 years ago, # ^ |

Count the number of nodes in the odd levels of the tree (first set) and count the number of nodes in the even levels of the tree (second set). Then multiply both of them and subtract the edges that are already exist.

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BadSector

7 years ago, # ^ |

I can`t understand why the following code gives wrong answer

int one = 0;
    queue<int> q;
    q.push(1);
    v[1] = true;
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        if(v[u])
        {
            one++;
        }
        vector<int>::iterator iter;
        for(iter = e[u].begin(); iter != e[u].end(); iter++)
        {
            v[*iter] = !v[u];
            q.push(*iter);
        }
    }

    int count = (one * (n-one)) - (n-1);

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ForgotToUpsolve

7 years ago, # ^ |

It might be because of overflows. Change int to long long may be.

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KOH

7 years ago, # ^ |

← Rev. 2 →

here's my solution.Maybe it's make sense.first,we can mark every nodes with zero or one following the the rules of Bipartite Graph，and we can figure out how many ones and zeros, and the result is the count of zeros times ones' ,at last,we should minus the loops.

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7 years ago, # |

← Rev. 2 →

can someone help me figure out why my solution gives tle? I used a random algorithm, the expected running time should be small and it runs fast in my computer. I cant pass pretest 6.

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7 years ago, # ^ |

Same here. Try test: 2 0

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7 years ago, # ^ |

Hi, gives no in 0.003 seconds.

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micangela

7 years ago, # ^ |

Maybe RAND_MAX is 32767?

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7 years ago, # ^ |

Yes, that was it! Thanks a lot. Its good I learned this while not actually participating. In my computer MAX_RAND is way bigger so it worked fine.

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7 years ago, # ^ |

So in general it would be wise to first take the residues mod 32767? and then use them? In case RAND_MAX is bigger, so you dont get overflow

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KieranHorgan

7 years ago, # ^ |

To get around it you could go (rand()<<15)|rand()

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7 years ago, # ^ |

Hello, thanks for the comment, can you explain the logic behind this when RAND_MAX is 2^31-1?

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KieranHorgan

7 years ago, # ^ |

Well, the whole point of using this is that Codeforces is run on Windows servers, which means that RAND_MAX is in fact 2^15-1, so this of course combines two random numbers to get one that is ~2^30. If you were using a *nix system then there would be no need for the bit shift, however if you did, then it would overflow. It's probably safer then to have an if statement along the lines of

if(RAND_MAX < 1<<16)
    x = (rand()<<15)|rand();
else
    x = rand();

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7 years ago, # ^ |

Oh, I understand. That safe solution makes a lot of sense, Ill do that from now on, thanks for the help. fare well and prosper.

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Vasiljko

7 years ago, # |

← Rev. 2 →

+10

How to solve C? Good problem in my opinion

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madn

7 years ago, # ^ |

I did this : if n <= 4 then you can just brute force, also in 2 0 case answer is NO, else i just find xor of all numbers from 1 to n — 2, the you just have to find two numbers, such that their xor is equal to (xor from 1 to n — 2) ^ x. Also, if (xor from 1 to n — 2) ^ x is 0, I delete n — 2 and add 0 to answer, so I find two numbers, such that their xor is equal to (xor from 1 to n — 3) ^ x

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sinus_070

7 years ago, # ^ |

← Rev. 4 →

NO only if the case is 2 0.

if n>=2 then let the first number be 2^18. Then keep filling numbers starting from 1, incrementing by 1 or 2[by 2 if the xor so far turns out to be 2^18], till there's one number left to be filled. The last number = xor(xor so far, x).

UPD: failed systests :(

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newCEA

7 years ago, # ^ |

When n ≥ 3, define a_k = k for k = 0, ..., n - 3. Calculate p = x^0^1^...^(n - 3). Then define a_n - 2 = 2¹⁸ and a_n - 1 = 2¹⁸ + p or similar. This will guarantee that there are no two identical numbers in the sequence a_i and all elements are smaller than 2¹⁹ - 1 < 10⁶

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yassin_

7 years ago, # ^ |

You failed systest because p can be zero right?

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newCEA

7 years ago, # ^ |

← Rev. 4 →

Actually, no. I forgot the case n = 2 and x = 0. I took care of the case p = 0 by redefining a₀ = 2¹⁹ - 1, a_n - 2 = 3 × 2¹⁷, a_n - 1 = 2¹⁷ - 1. I considered that corner case, and yet forgot abound n = 2 and x = 0. Forgot to mention this in the original comment.

Most recent submission : 30524864

Submission during contest : 30508844

Only difference is the case n = 2.

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Darui99

7 years ago, # ^ |

n — 2 random numbers and then generate last two.

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Kaszper

7 years ago, # ^ |

stupid (n == 2 && x == 0) if :(

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7 years ago, # |

← Rev. 2 →

Nice problems, how to solve C?

I had an idea that if a is odd number then aXOR(a + 1)XOR(a + 2)XOR(a + 3) = a - 1, so we generate first 4 * k numbers and find remaining ones with some bruteforce.

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Morphy

7 years ago, # ^ |

I used the same but when a is even a^(a+1)^(a+2)^(a+3)=0

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7 years ago, # ^ |

My problem was NO case :(

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Amaryllis

7 years ago, # ^ |

OK, I couldn't debug my solution during contest ( cause of stupid Wi-Fi problems) but I did now and it got AC, I will try to explain it:

We have 2 cases that we need to deal with:

1) N is odd

2) N is even

Let P be a power of 2 > 1e5. And in every case we will use an auxiliary array "numbers" which has n numbers such that not one of them is x.

Case 1 solution: let the first n-1 numbers be ans[i] = P + numbers[i] and their xor be _xor. our last number will be ans[n]=xor ^ _xor

Case 2 solution: let the first n-2 numbers be ans[i] = P + numbers[i] and their xor be _xor. If _xor = x, we will just change one of ans[i] to be a diffrent and number and ans[n-1] = _xor ans[n] = x;

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HaabyHings

7 years ago, # ^ |

here is a countercase for your claim. 3, 4, 5, 6 by your claim, answer is 2 but correct answer is 4.

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madn

7 years ago, # |

Can answer be "NO" in C except 2 0 case?

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C137

7 years ago, # ^ |

-11

6 0

10 0

n%4==2 0

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up-and-down

7 years ago, # ^ |

+38

0 1 2 262147 4 262148

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7 years ago, # ^ |

Oh with this case I expect lots of people will be failing. I myself got hacked.

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yassin_

7 years ago, # ^ |

← Rev. 2 →

6 0

YES
1 2 3 4 8 12

doesn't it work?

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madn

7 years ago, # ^ |

My program gives 1 2 3 4 8 12 to 6 0 and i think its correct

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keshav

7 years ago, # ^ |

1 2 4 8 16 31

satisfies 6 0 case

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Darui99

7 years ago, # ^ |

YES 29637 1088 18455 13546 1 2937

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achaitanya.sai

7 years ago, # ^ |

For 6, 0 my code prints:
781959 11593 584443 901753 630736 488604
Which when xor'ed gives 0. Am I missing something?

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C137

7 years ago, # ^ |

Now i don't have to wait System testing, C is WA :(

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LoDThe

7 years ago, # |

How to solve D?

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7 years ago, # ^ |

Binary search. Even tho I expect that my code runs in an infinite loop for some cases.

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HBK_Wasi

7 years ago, # ^ |

-11

Use srand(time(0)) and rand() to generate 15 different integers in range 1 to n. Then Check By Changing the position of a string of n '0'es . Then Print The Answer.

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7 years ago, # ^ |

What's probability of finding correct answer in your solution?

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Darui99

7 years ago, # |

And Don't lie that you didn't try to send random solve for C.

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eagle93

7 years ago, # |

Will the random solution pass for problem C? (If not, how to hack it?)

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qscqesze8

7 years ago, # ^ |

my solution is random solution , but i can not prove its time complexity.

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7 years ago, # ^ |

← Rev. 6 →

Yes， random solution works.
I randomly pick up (n — 1) different numbers from 0 to 524287 , calculate there xor sum S, then check if S xor x is used. if not, then these (n — 1) numbers and S xor x together is a solution (S xor x < 524288, obviously); if so, then make another try. Since there are plenty of numbers to choose, the probability to success is about 90% percent, maybe higher (I test n = 100000 several times, and they all succeed at the first attempt). So, if you tried many many times, (say, 100 times) and failed, then you can assume this is impossible and print "NO".
I passed system test, used 46ms/2100Kb, it works.

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Sahashoo

7 years ago, # |

← Rev. 2 →

Can anyone say this code bug?(Div 2/c)

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int maxn=1e5+7;
int a[maxn];

int32_t main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int n,x;cin>>n>>x;
	int sum=0;
	for(int i=0;i<n;i++)a[i]=i,sum^=i;
	int nwx=x^sum;
	a[0]=nwx;
	if(nwx>0&&nwx<n)a[nwx]+=(1<<15),a[(nwx==1?2:nwx-1)]+=(1<<15);
	for(int i=0;i<n;i++)cout<<a[i]<<" "
}

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iriszero

7 years ago, # ^ |

It does not print "YES" neither "NO" You should print "NO" iff N=2, X=0

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Sahashoo

7 years ago, # ^ |

tank you.

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Fred_Flintstone

7 years ago, # |

I wanna cry, it was 5 seconds left before I press submit and it tells me contest is over((( Pretty sure the code i was about to submit was correct

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songzy12

7 years ago, # ^ |

Same here hhh. I just adds "stdout.flush()" 5 seconds before it ends.

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DanAlex

7 years ago, # |

← Rev. 2 →

For problem E, I have O(N + M + (M — N) log (M — N) + Q log (M — N)). Is that right?

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radoslav11

7 years ago, # |

+24

Here is a $\text{[math]}$ solution to F:

First of there is a basic observations which is important for the solution.

lcp(s_l, s_l + 1, ..., s_r) = min(lcp(s_l, s_l + 1), lcp(s_l + 1, s_l + 2), ..., lcp(s_r - 1, s_r))

Now lets use a_i as lcp(i, i + 1) ( a_i = lcp(i, i + 1) ).

The problem becomes:

We have an array a_i. We have queries for finding the maximum value of min(a_i, a_i + 1, ..., a_j - 1) * (j - i + 1) for some i and j in a given range. We also have a query for updating a value at a given position.

We also have some additional constraints. There are at most $\text{[math]}$ different values a_i — it's easy to prove.

So the main idea is to store $\text{[math]}$ different segment trees (for every value of a_i). In the segment tree for a value X we will consider only elements with values ≥ X. For simplicity let's assume that in the segment tree all positions with value ≥ X are set to 1 and all other to 0. Then if we want to find the maximum length of an interval [i;j] inside our query interval such that min(a_i, a_i + 1, ..., a_j - 1) > = X we can simply find the length of the longest interval in the segment tree for X with all elements equal to 1.

Then the query can be done by considering every segment tree, doing query for each of them and then getting max X * T_X.query(l, r).

The update can be done by simply preforming at most $\text{[math]}$ updates in segment trees.

The only problem with this solution is the memory. It will consume $\text{[math]}$ memory. Fortunately we can use a persistent segment tree instead of a regular one and this way the memory will be $\text{[math]}$ .

PS: I was to lazy to write it during the contest, but is should pass.

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ivan100sic

7 years ago, # ^ |

+15

That is so brutal! I hope there is a nicer solution!

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lnzva

7 years ago, # ^ |

← Rev. 2 →

-8

Wow. ._. I thought of this solution as well, but didn't code it cause I was sure it would get TLE >_<

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dush1729

7 years ago, # |

← Rev. 3 →

ignore*

found my mistake in div2c

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yassin_

7 years ago, # ^ |

← Rev. 2 →

> 10^5 is ok, just <= 10^6

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LoDThe

7 years ago, # ^ |

Solution must print value <= 1000000

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Cro-Marmot

7 years ago, # |

-8

How 2 solve D ? I do not even understand why there is mention of fflush

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songzy12

7 years ago, # ^ |

fflush is to let the system know that you have print something so it can response to it. (Or under normal condition all the outputs are sent together at the end to save time.)

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Cro-Marmot

7 years ago, # ^ |

Does it mean that the input and output of this question are interactive? I first encountered this type of problem.So I need to design the query sequence?

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songzy12

7 years ago, # ^ |

Yes, you may decide what the next query will be based on what you have received.

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Vasiljko

7 years ago, # |

← Rev. 2 →

+21

So fast system testing

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vijju123

7 years ago, # |

← Rev. 7 →

+11

System Testing Pending
Refresh Page after 10min
79% done

This times system testing is running at speed of thunder lmao :D

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chb19

7 years ago, # |

upvote for the very quick testing

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Taan

7 years ago, # |

← Rev. 3 →

+18

![ ](
image upload
)

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kukyo

7 years ago, # |

-21

I could succeed to hack one's code ,if the website can run faster ,and my rank will be rise 200 and more!!! why my hack in the last minute can't do it's proper work?

By the way, how to solve the f**k problem C?

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gauss148

7 years ago, # |

← Rev. 2 →

Easy solution for C. if n<3, brute force. otherwise, xor first n-3 numbers,say res.(res = 0^1^...^(n-4)) Then 0,1,...,(n-4),(1<<18),(1<<19),((1<<18)+(1<<19))^(x^res) is answer.

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7 years ago, # ^ |

I just randomly picked all but one numbers and force the last number to fit the requirement.
No math here, great!

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stevie1024

7 years ago, # ^ |

← Rev. 2 →

You are lucky, your algorithm is incorrect. What if e.g. n=4 x=0 and your random pick is [1, 2, 3]?

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7 years ago, # ^ |

← Rev. 4 →

Read the question carefully, 0 is allowed so just 0 xor 1 xor 2 xor 3 = 0

Although the last number did have a collision, I would throw everything away and restart for another try. If it always failed ( say, failed 100 times or so), It could be pretty sure that this was impossible and should print "NO". It's the same as what Miller-Rubbin primality test do.

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stevie1024

7 years ago, # ^ |

You are right about reading carefully, the example should have been n=4 x=1 with [1, 2, 3]. In your original message you didn't mention trying '100 times or so'. You'd have to be very unlucky indeed.

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loan

7 years ago, # |

-9

Easy solution for problem C : https://ideone.com/FZ6rfw

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coding_weeb

7 years ago, # |

C with x=0 really made the problem trickier.

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7 years ago, # ^ |

← Rev. 2 →

C with n=2 and x=0 made my solution wrong :|

UPD: fortunately, I have some other bug :P

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snowwm

7 years ago, # |

← Rev. 2 →

vintage_Vlad_Makeev, KAN, MikeMirzayanov, I have accidentally sent the same hack twice (the page did not reload and I sent it from a new tab). Now I have two unsuccessful attempts. Can it be fixed?

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KAN

7 years ago, # ^ |

+46

Removed one of the attempts.

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snowwm

7 years ago, # ^ |

Thanks.

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predelnik

7 years ago, # |

Looks like I'm finally going to div.1! I'm so happy, it was quite a long journey :)

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7 years ago, # |

Hello, actually my wrong solution for problem D passed main tests. Here is my solution — http://mirror.codeforces.com/contest/862/submission/30518750 and it runs into an infinite loop in the following case : "10"

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7 years ago, # ^ |

"10" is test case #121 (last), and seems your solution works on it.

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7 years ago, # ^ |

Thanks for the good news. Idk how its working but on my machine its not running. Just felt that in case the main tests were wrong, I should have informed.

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ivan100sic

7 years ago, # |

-59

Now my C solution failed for the stupidest reason, I forgot to print YES before the answer itself in one of the cases. Why is output so poorly formatted anyway? Isn't it the tradition of problem setting to print -1 when there is no solution?

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ivan100sic

7 years ago, # ^ |

+22

Somehow I managed to accumulate more downvotes than "is it rated" comments :D

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mahmoudbadawy

7 years ago, # |

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

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suspicio

7 years ago, # |

After 17 fails and 1:49 my submission for D passed, but it failed again on 59th test. The worst feeling ever

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Erdenebayar

7 years ago, # |

nice problems, c is interesting for me and random_shuffle() solution hhe

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iCanCodeBro

7 years ago, # |

How can one proove that random brute solution for C will not time-out ?

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Deathly_Hallows

7 years ago, # |

-42

please don't make other rounds

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7 years ago, # |

← Rev. 2 →

can anyone please help me! This is my solution can someone please tell why is it a run time error one the codeforces compiler? it works in my system and had passed all the testcases! but then it gave a run time error?? thank you!

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--_--

7 years ago, # ^ |

At this statement if(fans[sizea]==0) you might be getting a runtime error as size of fans is sizea.

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7 years ago, # |

Does anyone knows what exactly compiler version and flags are used when you choose GNU C++11? I have a trouble with my solution http://mirror.codeforces.com/contest/862/submission/30520451. On my computer it answers 0 on first test using GCC 5.4.0, and on my friend's computer it returns the same answer using GCC 4.8. But test system thinks that answer is -273605408. What the problem?

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7 years ago, # ^ |

← Rev. 2 →

dude i guess they use something quite different because i had the same kind of issue! i ran your code in my computer and it gives 0. i guess codeforces has totally different compiler and flags. its just bad luck that we dont have their compiler version.

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7 years ago, # ^ |

May be we should ask admins about it? I have too few rating points to loose it that way :c

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7 years ago, # ^ |

Dude I asked KAN and he said it had nothing to do with compiler, actually I tried accessing memory that wasn't allotted so in that case it works sometimes and sometimes it doesn't! I guess something similar in your case too! Maybe but if you think there is still a problem you should approach. :)

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http://mirror.codeforces.com/blog/entry/79

7 years ago, # ^ |

Well, then i should check my code again.

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rakeen

7 years ago, # ^ |

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7 years ago, # ^ |

Thank you.

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snowwm

7 years ago, # ^ |

  vector<int> pref_l(n);
  pref_l[0] = layers[0];
  if (max_l > 0) pref_l[1] = layers[1];
  for (int i = 2; i <= max_l; ++i) {
    pref_l[i] = pref_l[i - 2] + layers[i];
  }

Here max_l can equal n and also you access pref_l[1] which might not exist if n is 1.

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7 years ago, # ^ |

Thank you.

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coding_weeb

7 years ago, # |

Can someone give me some links to good random algorithms tutorial. I see a lot of people used it to solve C.

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7 years ago, # |

i have got 237 point int this round keep doing well mahmoudbadawy thank you so much for this round and problem :)

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7 years ago, # |

+14

for problem C you can just select first n-1 elements randomly (from 0 to 2^19-1) and see if the number that you need to get the desired xor is available. try it 100 times, if none of those 100 times works then print NO. this works because the probability the nth number will be available is approximately 1/5 so the probability you get it wrong is 1/5^100. Of course this is just a heuristic, you need some sort of uniformity argument but w/e.

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7 years ago, # ^ |

thats because 2^19-1 is approximately 5*10^5, so at least four fifths of the numbers are unnocupied.

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7 years ago, # ^ |

← Rev. 2 →

Wow, I used exatly the same algorithm here, same range and also tried 100 times.

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7 years ago, # ^ |

did it work? I had to fix one little thing, I used srand and rand(). You need a small modification because rand only goes up to 2^15-1.

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7 years ago, # ^ |

Yes, I made a little function to do that. int randInt(){return (rand() * RAND_MAX + rand()) % 524288;}

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marlonbymendes

7 years ago, # ^ |

Could you please elaborate a bit more on the 1/5 probability? Specially when we don't know the number of solutions. Also why (from 0 to 2^19-1) and not [0, 10^6]? Thanks!

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