Codeforces Round #511

6 лет назад, # |

+45

why am i so bad lol

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caojielong

6 лет назад, # |

+11

another mathforces or digitforces?

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Adhami

6 лет назад, # |

Looks like I'm gonna get hacked so hard :) . Please test cases be strong.

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NaZaR_IO

6 лет назад, # |

Больше математики!!! Больше цифр!!!

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Jhin

6 лет назад, # |

← Rev. 2 →

that moment when you waste one hour trying to find what's the mistake in your algorithm and in the end you just wrote one letter by mistake

edit:I'm gonna get Tle for a stupid '}' I'm out :)

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Rahul

6 лет назад, # |

← Rev. 2 →

+29

C gets hacked... Meanwhile this!

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GullyBoy

6 лет назад, # ^ |

for what reason was div2c getting hacked ?

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Adhami

6 лет назад, # ^ |

+29

Bugs

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EbraM96

6 лет назад, # |

The hell is the solution for problem C ?

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Fayaz

6 лет назад, # ^ |

Strip the gcd off all the numbers. Now just find the largest set of numbers with gcd > 1. You can do this by looking for numbers divisible by a common prime factor.

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AkiLotus

6 лет назад, # |

read 4 first problems Mathforces confirmed.

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eisen_shinzou

6 лет назад, # |

Anybody knows test 8 problem D(of course div2)??

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mxh1999

6 лет назад, # ^ |

+13

I think that is 2 7.

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eisen_shinzou

6 лет назад, # ^ |

What's the answer?? 12?? So why my code gets WA??

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dungdq

6 лет назад, # |

+44

IMO 2019?

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AghaTizi

6 лет назад, # |

+27

Every time a chinese guy prepares a contest, I say to myself this chinese contest won't be as bad as the last one. And yet here we are!

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Deathly_Hallows

6 лет назад, # |

+36

Problem D div2 / B div1 was a very bad problem for me, what is the point of problems with no ideas but just conditions ?

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P___

6 лет назад, # ^ |

This problem included the idea, to figure out the conditions.

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shahed_ahmed

6 лет назад, # |

+24

constraints of C is too strict. :(

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pleb

6 лет назад, # ^ |

-6

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shahed_ahmed

6 лет назад, # ^ |

I was unable to think a good solution... Can you tell me what is complexity of the expected solution?

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dush1729

6 лет назад, # ^ |

← Rev. 2 →

max = 1.5 * 10 ^ 7
complexity
max * log(log(max)) to find least prime divisor
and
n * log(max) for finding prime divisor of every number
So complexity = O( max * log(log(max)) + n * log(max) )

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pajenegod

6 лет назад, # ^ |

Here is my solution. The idea is to find the cheapest cost for making the gcd a prime factor p greater, for each prime p ≤ a_max. This is done by dividing all numbers by the initial gcd, and then counting how many numbers are missing a factor p.

Factorization of the numbers can be done by precalculating an O(a_max) prime sieve to find the smallest prime factor (SPF) of every number ≤ 1.5 * 10⁷. With SPF precalculated it is possible to factorize numbers in $\text{[math]}$ .

All in all it has the time complexity $\text{[math]}$ .

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rvmzes

6 лет назад, # |

+76

I thought I registered for Codeforces contest, not Project Euler...

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uzzal_rahman

6 лет назад, # |

← Rev. 2 →

Can anyone tell me what is test case 8 for Problem D (Div-2) ?

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Shaykhutdinov-T-I

6 лет назад, # ^ |

Try 3 7 and 2 7

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faishol27

6 лет назад, # |

How to solve Div2C and Div2D?

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DeeaX99

6 лет назад, # ^ |

GCD is min prime factorization for all number example : gcd (9 , 4 , 2) = 2^min(2 , 1 , 0) * 3^(min(2 , 0 , 0)) * 5 ^min(0 , 0 , 0) .. etc so to make gcd bigger you need to loop over every prime factorization and the answer for this prime frac is the number of number that not have this prime you can find it by find every prime frac for all number and add it for freq array then the ans for prime will we just in case n != freq[prime fac] n — freq[prime fac] — ( number of one ) ans = min for all frac don't forget to handle this case 1 2 3 4 => 1 remove (1) 3 3 3 9 9 => 3 remove (3 , 3,3) my solution https://ideone.com/zHLllz i wish it will not get WA at main test and sorry for my bad english

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RDDCCD

6 лет назад, # |

It's a pity that i can not hack myself.

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Omar_Morsi

6 лет назад, # |

$\text{[math]}$ Math Problems.

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EbraM96

6 лет назад, # |

+19

Look around, life has many interesting problems. But dude, you chose math :D

Good contest by the way.

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6 лет назад, # |

+18

how do you solve B?

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6 лет назад, # ^ |

+21

nvm, i see it is all casework like i expected, too bad

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6 лет назад, # ^ |

+13

can't believe i wasted an hour and a half on that and still got it wrong LOL

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aid

6 лет назад, # ^ |

+42

There is no need to check all the tricky cases by hand. You can write bruteforce and find all of them.

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Rahul

6 лет назад, # ^ |

I don't understand how you get all tricky cases by bruteforce. Can you explain briefly? (maybe with code?)

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zscoder

6 лет назад, # ^ |

Recursive backtracking should be enough for very small grids.

Alternatively, you can also find the max matching of the bipartite graph where each vertex is a square of the grid and two squares are connected iff their Manhattan Distance is 3 via Max Flow or any other matching algorithm.

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Deathly_Hallows

6 лет назад, # ^ |

+26

just a bunch of if conditions

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sggutier

6 лет назад, # ^ |

What is the better way of training for solving these problems?

Also, what philosophy does one need to follow when solving them? Doing a bruteforce solution, then trying a bunch of test cases until one finds a pattern?

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Deathly_Hallows

6 лет назад, # ^ |

+16

i am really not the good person to ask :V but anyway if i didn't find any reasonable solution to a problem i make a bruteforce solution ans observe the pattern if there is one

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zktxdy

6 лет назад, # |

← Rev. 3 →

It seems the pretests of problem div2.C are weak...

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6 лет назад, # ^ |

You are scaring me:|

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Reziba

6 лет назад, # ^ |

Very, very weak

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BaaaaaaaaaaaaaaarkingDog

6 лет назад, # |

"that of all integers" Nah..

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AghaTizi

6 лет назад, # |

+46

And just like that, you wasted 2 hours of my life!

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6 лет назад, # |

← Rev. 6 →

In div1C, does the following hold?

We can divide the kingdom into sections with each having sum X, iff sum(tree)%X = 0, and exactly sum(tree) / X subtrees T have sum(T)%x = 0. The division can be done by cutting edges to the parent from such nodes.

I couldn't find a counterexample, but unfortunately didn't have time to code either.

Edit: Proof:

Clearly we need X divides sum(tree), and that there exist at least sum(tree) / X such subtrees, for X to work. Now assume that there are K such subtrees. Cut their parents. Now we have K different parts (one of the subtrees was our root), each of them has sum divisible by X, and of course holds $\text{[math]}$ . If K > sum(tree) / X, we have a contradiction. If $\text{[math]}$ , we have a contradiction. Therefore the sum of every part is X, and we're done.

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Noam527

6 лет назад, # ^ |

I've also thought of it, but I couldn't see a way to use this.

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6 лет назад, # ^ |

← Rev. 3 →

I think if that holds, you can solve the problem pretty easily.

Find sum(T) for every subtree T in the tree. Store these in vector subtree-sums
subtree-sums <- gcd(subtree-sums, sum(tree))
find all divisors of sum(tree), and index them
Create subtree-counts, where subtree-counts[j] is the amount of appearances of divisor j of sum(tree) in subtree-sums
do SOS-dynamic programming over this set, to set subtree-counts[i] <- amount of elements in subtree-sums such that divisor i divides them
make array subtree-ans, where subtree-ans[i] = 1 iff subtree-counts[i] = sum(tree) / (divisor i)
loop over all divisors i of sum(tree) from smallest to largest, and do subtree-ans[j] += subtree-ans[i] for all j, such that divisor-j / divisor-i = p for some prime p. Remember to take mod
Output subtree-ans[i], where divisor-i = sum(tree)

We use a few facts here:

sum(tree) has at most like 1e5 divisors
If X is a possible number, and Y is a possible number, and X % Y = 0, then we can first divide into X, then into Y. This follows from the way we select the nodes whose parents we cut.
The way to cut the parents is unique for any value X

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6 лет назад, # ^ |

← Rev. 3 →

It's even more easier. Let number x be "good" if you can divide T with subtree size sum(T) / x. dp[i] = 0 if i is not good, $\text{[math]}$ otherwise. Time complexity is $\text{[math]}$ using harmonic sums.

Checking if $\text{[math]}$ is good can be done with similar way in $\text{[math]}$ , after you know gcd(sum(subtree), sum(T))

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6 лет назад, # ^ |

Doesn't doing the harmonic sum naively with something like unordered maps give you complexity O(sum(tree) log(sum(tree))? How do you get it to n log n?

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6 лет назад, # ^ |

+10

The number in form of sum(T) / i for $\text{[math]}$ are only relevant, so you don't need unordered map, just an array of size N.

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6 лет назад, # ^ |

Ohh okay, I get it now. Nice solution!

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https://stackoverflow.com/questions/39779880/c-int-vs-long-long-in-64-bit-machine

6 лет назад, # ^ |

Yes, your statement holds.

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n00bie

6 лет назад, # ^ |

The idea is probably correct, I have used it before on another problem. I couldn't really do anything with it though.

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saihemanth9019

6 лет назад, # |

← Rev. 2 →

Div2C Converting all ll divisions to int divisions drastically changed the running time... How does it even affect though?

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Reziba

6 лет назад, # ^ |

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saihemanth9019

6 лет назад, # ^ |

Fair enough... Never considered memory level optimizations.

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6 лет назад, # |

What happens if I resubmit a solution to a problem which I have already submitted a solution which passed the pretests?

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saihemanth9019

6 лет назад, # ^ |

← Rev. 2 →

The first solution which passes the system tests is considered and any WA before that will be an unsuccessful submission as usual.

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cbosch_carlgauss

6 лет назад, # ^ |

That is not accured, the system ignore all submmits but the last one, you shoould read Skipped in the previos pretests passed.

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6 лет назад, # |

What will be the approach for problem Div2 C (Enlarge GCD). I had TLE in 8th pretest.

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contest_dhuye_pani_khai

6 лет назад, # ^ |

Did something like sieve. Only marginally avoided TLE. Wonder what will happen after systest.

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6 лет назад, # ^ |

I used sieve and then calculated the minimum power of each prime in each number so as to evaluate the minimum number of terms to delete to get an enlarged GCD. How did you solve @from_bd_with_depression??

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contest_dhuye_pani_khai

6 лет назад, # ^ |

Instead of calculating power, I just checked how many numbers are divisible by for all numbers > gcd

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Vijaykrishna

6 лет назад, # ^ |

Yes I also had the issue that if I use all primes upto MAX=1.5x10^7, then I got TLE on test case 8, whereas if I use numbers upto sqrt(MAX), then the pretests passed (but the solution is wrong, you have to take all primes upto MAX). Why do they keep such tight time limits?

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RayPatterson

6 лет назад, # ^ |

Use the O(NloglogN) sieve. I used it and ran C in merely 0.5 seconds. NlogN sieve will obviously be tight.

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6 лет назад, # ^ |

I used sieve of eratosthenes, its complexity is nloglogn.

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RayPatterson

6 лет назад, # ^ |

← Rev. 2 →

Same as me then. How on earth did you TLE that out. I ran that sieve on custom test and it resulted in a mere 138ms EDIT: Nvm I just saw your complexity. Obviously you will TLE with that O(N * max(A)) algorithm, even with 5 seconds

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6 лет назад, # ^ |

Okay I saw the editorial. I get the difference now. Thank you so much for the suggestions DrumpfTheGodEmperor :)

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alexey_kuldoshin

6 лет назад, # |

+106

Sorry, but it was one the worst contest I've ever seen (I saw rounds of GreenGrape)

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nikanqilaizhenhaoxiao

6 лет назад, # |

+31

I used custom test for my E before submitting it. It ran more than 10s when n = 21 on custom test. Then I spent about 30 min on how to squeeze my code in to TL, but failed. After that, I decided to submit one similar to my first code and it got PP in 499ms. Anyone knows what had just happened to me?

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P___

6 лет назад, # ^ |

-8

From what you said, looks like you lost 300 pts, 30 min * 10 pts per minute and you got 9'th instead of 3'rd. That's what happened to you.

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https://ideone.com/6yWmn9

6 лет назад, # |

I was using this as generator to hack Div2 C:

It kept giving me an invalid input error with the following:

"Validator 'validator.exe' returns exit code 3 [FAIL Expected EOLN (stdin, line 2)]"

Can someone tell me how to fix it?

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BaaaaaaaaaaaaaaarkingDog

6 лет назад, # ^ |

In my memory, you should give newline as endl(instead of '\n') in hack data generator code.

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misael

6 лет назад, # ^ |

I use printf("\n") without problem. The problem is that it should have a "return 0;"

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6 лет назад, # ^ |

In this case cf should give a better error :/ Is there any way to test the generator now?

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6 лет назад, # ^ |

Oh wow, I didn't know there was any difference between the two. Is there anyway I can test if replacing "\n" with endl would fix the generator now?

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TianyiChen

6 лет назад, # ^ |

You wrote if(i!=n); so it generates an extra space at the end of the second line.

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6 лет назад, # ^ |

I don't get it? That was written specifically to avoid printing an extra space at the end of the last (nth) element.

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Len

6 лет назад, # ^ |

+13

if(1 > 2);
  cout << "1 is greater than 2";

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6 лет назад, # ^ |

Oh wow, just saw it, that was pretty stupid lol. Thank you guys for pointing it out!

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C137

6 лет назад, # |

+46

DIV. 1 B is the kind of problem that when you submit it you just pray you didn't forget any if -_-

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otaliptus

6 лет назад, # |

← Rev. 2 →

Pretests of Div2 C was weak I think. I see some codes(including mine) which only calculates primes to sqrt(1.5*10^7), and they all passed. But the case above makes them broken:

3
100069 100069 1000103

These codes gives answer as "-1", while it is "1".

Hope to see a round with better constraints, and better pretests, of course.

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prabal_1997

6 лет назад, # |

← Rev. 3 →

what is wrong in this submission(DIV 2 B).43208324

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soul_voyage

6 лет назад, # ^ |

Test 3 4 0 5 You are actually calculating max distance from origin rather than max of sum of x+y because since two sides are equal the equation of third side is of the form x/a+y/a=1 we have to find a.

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Priyam2k

6 лет назад, # ^ |

All you had to do was print the max possible sum of coordinates.

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6 лет назад, # |

How to solve Div2 D?

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ouuan

6 лет назад, # |

← Rev. 2 →

+36

Really strange thing happened...

When I submitted A, it said I had submitted the same code before:

Then I tried to modify my code, and add some annotations, but it's no use. I tried B, still the same.

I don't know why but I came up with the idea of using VPN, and I found myself not registered yet, but I remembered that I had registered. However, I registered again, and I was able to submit my codes.

Later, I turned off the VPN, and I could not submit my codes again. I turned it on and I could submit again.

The most incredible thing is, I found that I had registered twice..

I just wanna know what happened?

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Adhami

6 лет назад, # ^ |

Codeforces got hacked after System Tests. Mike should find the bug of his code.

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delete1

6 лет назад, # |

← Rev. 2 →

I am confusing...

I tried to hack the problem, but there no success.

Please see code link, this solution should be time limit exceeded, but codeforces system said everything ok and I got wrong hack attempt with this test 1 10^9 10^9. I tested it in my local computer, the code isn't working in IDE.

How can I understand?

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AkiLotus

6 лет назад, # ^ |

There were many cases source code in C++ works properly with complexity of O(n), when n ≤ 10⁹.

Try with 10⁵ points instead.

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nns2009

6 лет назад, # |

Is it just me or do you also think this one for Div.1 was extra hard?

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majk

6 лет назад, # ^ |

+17

You can usually anticipate super hard contest with Chinese authors. I especially like problem E, but my bit mojo is weak.

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nns2009

6 лет назад, # ^ |

← Rev. 2 →

I especially like problem E...

I have only read A, B and C and solved nothing during the contest anyway. I think I came up with a solution for B during the contest and finished writing it a few minutes after the contest finished but I can only check after the system testing.

Anyway, (if I were to finish and submit 'B' in time) I would probably only lose rating for the single problem solved few minutes before the finish.

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striver_79

6 лет назад, # |

For C, just divide the array elements by initial GCD, and then https://www.geeksforgeeks.org/largest-subsequence-gcd-greater-1/

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aryaman

6 лет назад, # ^ |

This method is the most intuitive IMO, thanks for this short solution :)

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Bojack

6 лет назад, # |

How to solve Div2-C?

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ag_orz

6 лет назад, # ^ |

For C just find gcd and then use sieve to find numbers of multiples in given array for every number above this gcd. Now the maximum number of multiples among these numbers will remove minimum integers from array. However due to strict time limit we cant use this algo. So to optimize it we see that if we find number of multiples for a number then we dont need to check this for its multiples as number of their multiples will be less than or equal to this answer. this is my solution

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Alpha_Q

6 лет назад, # |

+41

Solved exactly div 1 E a few days back but didn't read it the whole contest. :)

Solution if anyone's interested: https://people.csail.mit.edu/rrw/presentations/subset-conv.pdf

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HBK_Wasi

6 лет назад, # ^ |

Thanks, Bae

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NMouad21

6 лет назад, # |

Div2C constraints aren't fair for Java users x)

Same solution in C++ passes in around ~600ms & around ~990ms in JAVA.

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6 лет назад, # ^ |

+67

Java isn't fair for Java users

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ainta

6 лет назад, # |

+98

I got TL on test 15 in A :( isn't it too harsh to set n = 300000, MAX = 15000000 for A?

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sun123zxy

6 лет назад, # |

So many math problems... This contest is not friendly to me at all

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SirirNicheBirirDokan

6 лет назад, # ^ |

+13

This is not friendlyToMeContestForces.

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Lawali

6 лет назад, # |

+20

I'm not good in Mathematics, so I used bipartite matching in div1B. It was terrible for me. :(

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9646516

6 лет назад, # |

+26

Div2C/Div1A should have had 1 more second time limit

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Heisenberg

6 лет назад, # |

← Rev. 3 →

+20

After C — that memory in right corner

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Masalmah

6 лет назад, # |

+10

really you can define array of size 1.5 * 1e7 when i am define array with size 1e6 the codeblocks became slow, That's why I did not think about sieve and get TLE in case 10!!!

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metamorphosis00

6 лет назад, # |

I'm the only one who solved the problem div2 B with Binary Search by answer? I dont understand solutions

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Praxis

6 лет назад, # ^ |

You have to search the point further to the point (0,0) whit this point (x,y) we need to know which is the begining of the hipotenuse of the triangle that passes fot that point and this is (0,x+y) and the end is (x+y,0)

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racsosabe

6 лет назад, # ^ |

It's very simple, you need to cover all n points with an isosceles triangle that uses X and Y axis as sides, so you will use a line y = mx + b such that the points (0, x) and (x, 0) belong to it for any x.

We see that x = b and m = - 1, and also notice that the sides in the axis are the smallest ones from the triangle, so we only need to find b and it will be our answer.

Now, if we set an "order" to the points by using the b they would generate, we notice that the maximum b generated will cover all of the points, so that's what we do.

We compute the b by using $\text{[math]}$ , that's why all do something like this:

int ans = 0;
for(int i=0; i<n; i++){
   ans = max(ans,x[i]+y[i]);
}
printf("%d\n",ans);

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luismo

6 лет назад, # ^ |

My first thought was that too

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Aminghara

6 лет назад, # |

+19

Hey guys. This two guys cheated on problem C. Just check these two submissions. http://codeforces.com/contest/1047/submission/43207716 http://codeforces.com/contest/1047/submission/43197339 The first submission is really similar to the second

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Praxis

6 лет назад, # ^ |

Codeforces before the update of ratings search for similary solutions and if codeforces find some solutions similary all the participants except the first participant who uploaded the solution are disqualified from the contest

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6 лет назад, # |

+22

Hey guys, I really want to improve my problem solving skills. During the competition, I had no idea what to do for Div 1B/Div 2D and gave up.

I noticed a lot of accepted solutions were if-else statements. Could I kindly ask for advice on strategies to reach these solutions?

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Benq

6 лет назад, # ^ |

+19

Solve it for n, m ≤ 20 with max flow, then make a spreadsheet of all these values and recognize the pattern.

Solution

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P___

6 лет назад, # ^ |

Or just use pen and paper — thought it might be a little slower.

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6 лет назад, # ^ |

Hey, thanks for your reply! Yeah, I was trying to do that, but couldn't see any patterns... I also stopped solving it by hand at around N = 4, M = 6 -ish.

Could I ask how many examples you did by hand?

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6 лет назад, # ^ |

Awesome, thanks so much Ben! I guess I spent too much time trying to think of insights instead of analyzing the patterns haha

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zscoder

6 лет назад, # ^ |

If you don't want to analyze the pattern, you could make an educated guess that since 1 × 6 grid can be completely covered so it is likely that for larger grids you can delete 6 × m or n × 6 blocks from it and the answer will remain the same. Thus, you can keep subtracting n and m by 6 until you reach a small enough value where you can solve it via max flow.

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http://codeforces.com/contest/1047/submission/43209921

6 лет назад, # ^ |

Hey zscoder! Thanks so much for that advice, that really makes a lot of sense to me!

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SrapZark

6 лет назад, # ^ |

I got your point. But missing few things. Can you please tell me, what are source node and target node for Max Flow execution? As well as, what will be the flows on edges?

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Ahmed_Diab

6 лет назад, # |

Is there anyone solve problem C with java ?

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NMouad21

6 лет назад, # ^ |

Just barely x)

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HellAngel

6 лет назад, # |

-13

does anyone have the test case #44 of Div1 C/ Div2 A. I got WA on this test :((

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AkiLotus

6 лет назад, # |

+12

Despite losing rating again, well, I can proudly say that I've succeeded in achieving something (maybe) no Codeforces users have done before :D

(Look at the execution time of my C solution, it even overrides time limit LOL)

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6 лет назад, # |

← Rev. 3 →

-11

--Del--

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sandoval95

6 лет назад, # ^ |

-16

Yeah, you feel proud of copy-pasting GeeksforGeeks's code LOL

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6 лет назад, # ^ |

← Rev. 2 →

-8

--Del--

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Jhin

6 лет назад, # |

you know a contest is bad when 100 points=2000 position difference

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panipuri8

6 лет назад, # |

+40

The constraints for C are plain bullshit. Spoiled the contest for me.

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kristja6

6 лет назад, # ^ |

You created the sieve needlessly large (MAXN = 2e7+5). But I agree, the time limit should have definitely been 2s.

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ag_orz

6 лет назад, # ^ |

The time constraint was the only thing that made this problem interesting. With some optimization in sieve we can easily pass this problem in 1s.

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Wild_Hamster

6 лет назад, # ^ |

+19

Does copy-pasting linear sieve really make problem more interesting? Not sure about it

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tantam75

6 лет назад, # ^ |

I didn't even try to optimize my sieve, still pass

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