Codeforces Round #609 — Editorial

5 years ago, # ^ |

← Rev. 3 →

+21

Let me give it a shot!

Well, if you color it this way, you can see that each cell has adjacent cells only of a different color. Also, note that if we place a tile on the diagram, it will always cover 1 white ane 1 black square.

Now, let's say that the white squares are less than the black ones. We can see that if try an example!

So, if for example, the white squares are less than the black ones, it makes sense that if you try to connect them to blacks, we'll run out of white ones before blacks. So, intuitively the answer is $$$min(numOfBacks, numOfWhites)$$$!

→ Reply

prerak872

5 years ago, # ^ |

← Rev. 4 →

+17

If we use domino structure as given in image and keep it's both endpoints on the columns containing an odd number of lengths. now both endpoints require even number of the block left to be filled and middle columns are unaffected(on basis of parity).but the length of this structure should be even. this way we can joint at most min(columns with an odd block at odd indexes, columns with an odd block at even indexes) columns. ~~~~~ int tot = 0,odd = 0,even = 0; for(i=1;i<=n;i++) { tot+=a[i]/2; if(a[i]%2) { if(i%2) odd++; else even++; } } ans = tot + min(odd,even) ~~~~~

→ Reply

prerak872

5 years ago, # ^ |

IMAGE LINK — https://drive.google.com/open?id=1gSMW0RAXRMz8LtSVUREzNiI0lYAYfAHd

→ Reply

sparshKat

5 years ago, # |

+28

So, i am a newbie to giving contests. This was my 5th contest and i wasn't able to solve any one of them this time too. Is that normal and will i be able to become good at this anytime soon if i practice. I have been practicing a2oj ladders lately. Everyone's feedback and support is appreciated.

→ Reply

lemongrab

5 years ago, # ^ |

My only advice in this respect would be to try and solve most of the questions that you couldn't solve in this contest. If you can solve questions out of the contest, then atleast you will have something to do during the contest.

You may make simple targets like solving 1 or 2 in order of difficulty which you couldn't solve during the contest. :)

→ Reply

Scopula

5 years ago, # ^ |

I'm new too and getting crushed in my first 4 contests. I solve mostly problem A in div 2, maybe B and I never have time to attempt C or harder problems.

If you look at the rating graphs of other users, even the top ones, they often have a dip (falling rating) in the beginning so I think it is normal.

→ Reply

shivang_garg99

5 years ago, # ^ |

You should participate in maximum contest and try to do questions after contest you couldn't solve and then analyse it why u couldn't solve them in contest.

As you can see my graph initially i solved 1-2 questions and i gradually improved these things requires patience.

→ Reply

5 years ago, # |

For div2D, why doesn't this solution work?

We start from the first column and we move to the right. There are several cases:

If our current column has even squares we add it as it is (ie. $$$ans += arr[i] / 2$$$)
If it has odd, we do the following: We try to find the next column with odd squares and if in between them there are even columns, we can use all of them and get an optimal answer (ie. $$$ans += (sum-of-all-squares-we-found) / 2$$$). And then we move to the next column! (you can see that if you try an example)
Otherwise, if there are odd (or it is the last column that has odd squares), there is no way to use the 1 square left (or am I wrong?), so we do the same as the 1st case (ie. we add ans += $$$arr[i] / 2$$$) and we move forward.

I could not find any test cases that break this greedy. Can you help me? Thanks!

Here is my code: https://mirror.codeforces.com/contest/1269/submission/67367653

→ Reply

JoseSoto

5 years ago, # ^ |

← Rev. 2 →

+17

I was thinking something like that :(

Check this case:

7 6 5 4 4 3 2 1

The answer is 16.

→ Reply

5 years ago, # ^ |

Thanks for the test case! Now I can see why it doesn't work! :)

→ Reply

abj02

5 years ago, # ^ |

You should use long long int instead of int and then try again because in your code, you have used int and the answer in some cases may exceed the range of int. Hope, it helps.

→ Reply

5 years ago, # ^ |

← Rev. 3 →

Here is my code with long long: https://mirror.codeforces.com/contest/1269/submission/67371998

It still fails, because the greedy I described is wrong.

→ Reply

0cHild8oFF0the8dEvil

5 years ago, # |

Problem A. I spent half an hour on you. Good round.Thank you 300iq!!!

→ Reply

NotGivingUp

5 years ago, # |

Can someone hack my Div2C / Div1A or is it actually O(n) code?

→ Reply

Swistakk

5 years ago, # |

+15

Don't you think that samples in Div2A are a bit too suggestive xD?

→ Reply

5 years ago, # ^ |

Oh... You know, I had other answers in the "tests" test set. But then I copied that to "pretests" and forgot about it...

→ Reply

5 years ago, # ^ |

And right now I can't stop my laugh because I didn't even suspect that I have these answers XD

→ Reply

bufftowel

5 years ago, # |

← Rev. 2 →

Can we solve Div2D using dp?

→ Reply

5 years ago, # ^ |

I don't think so.

→ Reply

bufftowel

5 years ago, # ^ |

Ok, Thanks!

→ Reply

Jonathan1234

5 years ago, # ^ |

I AC Div2D using a greedy approach, please tell me if it is wrong. Looping over pillars from left to right. For each pillar, try to place as many blocks vertically from top to down. If it have one space empty in the bottom, then we occupy this space greedily by placing a block horizontally. When processing pillars, if (len[i]-horizontal_height)%2==1, then increase horizontal blocks height by one, else decrease by one(it's pretty obvious if you draws it). If the horizontal blocks height is larger than maximum allowed height, then calculate how many space is wasted.

→ Reply

hoke_t

4 years ago, # ^ |

genius

→ Reply

toph3110

3 years ago, # ^ |

What's len[i] here ??

→ Reply

Jonathan1234

3 years ago, # ^ |

$$$a_i$$$

→ Reply

5 years ago, # |

Is Div1B Div2D editorial not completed?

→ Reply

5 years ago, # ^ |

No, its completed.

→ Reply

5 years ago, # ^ |

+24

Then, something wrong with it:

If the Young diagram has two equal rows (or columns) you can delete one domino, and the diagram will still have an equal number of white and black cells.

First, it may have not equal number of white and black cells from beginning. Second, I think it should be stressed out that after deletion it is still Young diagram.

so you can find the answer by algorithm described below.

Where to look at? Should I look into other task editorial? I don't understand.

→ Reply

5 years ago, # ^ |

The algorithm is: find two equal rows or columns and delete domino on the.

For example, if you have

7 6 6 5 4 3

You can delete one domino, to get another young diagram.

7 5 5 5 4 3

→ Reply

5 years ago, # ^ |

+22

I claim that editorial is incomplete. First quoted by me sentence is simply wrong if you don't say that number of black and white cells were equal.

Next, if you say "assume that number of black and white cells are equal" then later where you say "So we have a contradiction!" Contradiction with what? With equal number of black and white? Yes. In short: assume we have equal -> no, we don't! This is not connected to anything else.

You could say following: you can place one domino and unfilled cells would still have equal number of black and white cells. So, lets continue fill cells by domino like this until you can't. This means that there is no two equal rows or columns. Thus, we're left with 'basic' diagram. But in basic diagram number of black and white cells are not equal. Therefore we should fill everything using this algorithm.

In short, we showed that if we have equal number of black and white cells, then we can fill all cells by domino.

Now, about not equal:

Just because if you have more white cells (case with more black case is symmetrical), and there are no equal rows and columns...

You talk about case when no equal rows and columns. But what about case when there are equal rows and columns? I think you want just do the same algorithm until you have 'basic' diagram. But what if there is filling that is better because it does something different?

There is simple proof that you can't do better than minimum, but you didn't include it. As someone already noted in comments: no matter how you place domino, you always "use" single black and single white cell. So, you can't place more than $$$min(black, white)$$$. Then, it's easy to see that the same algorithm doesn't reduce this minimum, but in the end you have 'basic' diagram that can be processed as you said.

If you don't want complete editorial, then alright. I just don't like when missing parts of editorial is placed in comments.

→ Reply

BanazadehAria

5 years ago, # ^ |

The editorial does not need to have every detail in depth its just the general idea so you can get the details yourself.

→ Reply

5 years ago, # ^ |

+30

At least I would require editorial to be coherent. I mean, it should have narative order.

But, let me just disagree with you. And I'll also tell why.

Task solutions should be proven. Otherwise, we can't claim that there is no better output. In other words, task should be correct. Because if solution or checker is wrong, then how can you judge?

Now. Imagine there is no proof in editorial. Then, all we have to do is: prove it by yourselves, or simply belive that author has correct proof.

I was looking into editorial, because I was need a proof, and I had problems with mine. And, instead of connected proof I saw puzzle of sentences, and had to figure out what is related to what.

Yes, I was able to figure out, but only after reply with example what exactly means "remove domino".

And, in the end: I think, editorial should not have "simply wrong" statements, and also there should be no places where you should guess right meaning of what author wanted to say.

So, in short: I don't belive that problem is correct unless I have a proof. I don't require proofs of simple facts. But when fact is not simple, I would like to have detailed explanation. Simplicity of fact is based on difficulty to find proof, not by difficulty of fact or difficulty of proof. Fact can be simple, but proof can be hard. Fact and proof can be both simple, but if it's very hard to find a proof — it should be explained.

→ Reply

_PortgasD.Ace_

5 years ago, # ^ |

← Rev. 2 →

I find a test that I use author's algorithm is wrong: 16 2 2 1 2 1 2 2 2 2 1 1 1 1 1 2 1 If using author's algorithm, we will have 12 domino, it's mean we have to fill all cells, but how we fill the 15th and 16th col

→ Reply

5 years ago, # ^ |

This test is invalid. Sequence should be non increasing.

→ Reply

_PortgasD.Ace_

5 years ago, # ^ |

Oh !!! Sorry, my mistake

→ Reply

Swistakk

5 years ago, # |

"Just because if you have more white cells (case with more black case is symmetrical), you can take the first column with more white cells than black cells and delete the last cell of this column" — I think it is not true if you have equal columns since resulting diagram can no longer be Young one. But if you preced that with argument of dealing with equal columns then you will be fine I think.

→ Reply

5 years ago, # ^ |

+16

Oh, yes, of course.

→ Reply

5 years ago, # |

← Rev. 2 →

+16

Excuse me, there might be a small problem in the editorial of 1269E:

[After that, let's say that $$$x≤k$$$ is zero, and $$$x>k$$$ is one.]

[Then you need to calculate the smallest number of swaps to make segment $$$1,1,…,1$$$ of length $$$k$$$ appear in the permutation.]

Isn't it segment $$$0,0,...,0$$$ of length $$$k$$$? I'm confused.

UPD: Fixed.

→ Reply

5 years ago, # ^ |

+13

Yes, indeed, sorry.

→ Reply

Godspeeed

5 years ago, # |

Can anyone explain me div2 D

→ Reply

shreya479

5 years ago, # |

can someone exlain me Div2B?

→ Reply

pritishn

5 years ago, # ^ |

← Rev. 4 →

Basically we are first finding all the possible values of x for which elements in the array a can be converted totally into array b, We are doing this by comparing each element in array a with b1

For example

let m=5, a={4,2,3,2,1} and b={2,3,3,4,5}

so first comparing

a1 with b1 we get x=3,

a2 with b1 we get x=0,

a3 with b1 we get x=4,

a4 with b1 we get x=0,

a5 with b1 we get x=1,

then we can add each of these Xs to array a and check if we are getting array b or not, for all valid Xs , we store the minimum X

I did it using 2 loops(one for getting X and then temporarily increasing array a with x and comparing it with b) with O(n^2) and it passed the testcases

→ Reply

5 years ago, # ^ |

Thanks, how is it guaranteed that the value of X will be from one of these only and not anything else since we didn't consider b2, b3, b4... in this?

→ Reply

pritishn

5 years ago, # ^ |

Think about it this way.. Suppose you know the correct X, then adding the X to the array A will result in ONE of the element of A to become equal to b1.

I mean for a correct X, a1 may become b1 or a2 may become b1 and so on.. I mean every element of A has chance to become b1.. So we can just compare all elements of A with b1, store all the values for which elements in A become b1.. For a correct X all elements in A will map to corresponding elements in B.

→ Reply

retry__

5 years ago, # |

in div2 E what is "si"?

→ Reply

Wild4Stone

5 years ago, # ^ |

← Rev. 3 →

I think $$$s_i$$$ here represents the array formed after replacing x<=k by one and x>k by zero.

Then for $$$s_i$$$ = 0 means that it should not be present in the subsegment so minimum number of swaps required to remove it would be min($$$p_i$$$, k-$$$p_i$$$).

→ Reply

SevenDeadlySins

5 years ago, # |

What was test case 7 for Div2C?

→ Reply

Maxymilian

5 years ago, # ^ |

← Rev. 2 →

I believe it was larger version of:

9 3

123113133

Answer should be:

123123123

Correct me if i'm wrong

→ Reply

SevenDeadlySins

5 years ago, # ^ |

It kind of was...

actually there was a mistake in the part of code which compares numbers...

Thanks for helping!!!

→ Reply

towrist

5 years ago, # ^ |

← Rev. 2 →

-8

What a coincidence! I made the same error and Maxymilian's test-case illuminated me. I won't forget you when I become a grandmaster. :)

→ Reply

Tusher_CSE

5 years ago, # |

Can someone explain div2 B to me , i didn't understand the editorial

→ Reply

5 years ago, # ^ |

Please see my reply below

→ Reply

NoobCoder1998

5 years ago, # |

Pls explain Div 2 B (with code if possible) I'm stuck:(

→ Reply

5 years ago, # ^ |

Hi, In this question, listA is at some distance from listB and author guarantee that there is a solution i,e if you will add some non-negative number X then listA will become listB. Also, this means every number of listA will map out to some number of B after addition of this number X. Now, let's take first element of A (i,e A[0]) and look for it's all possible distances from listB by iterating over elements of listB. One of this distance is definitely guaranteed to be X. So, just iterate over all possible distance and for each distance, calculate new listA (let's call it listA') and compare listA' and listB (by sorting). Finally, output minimum distance for which listA' is equal to listB. Submission Link

Hope this helps

→ Reply

tanmay157

5 years ago, # ^ |

unable to understand this part

ll curr_diff = (B[i] - A[0] + m) % m;

why are we adding a +m before dividing by mod m?

→ Reply

5 years ago, # ^ |

To avoid negative difference. Adding modulus M doesn't affect the significance of the current difference. For example, number "-1" and "2" both are same number with respect to modulus 3. Make sense?

→ Reply

_gsdt

5 years ago, # |

+10

More explain on Div.2 problem E, please.

→ Reply

5 years ago, # ^ |

May be this will help: https://mirror.codeforces.com/blog/entry/72358?#comment-566415 I would like to see proof though, about independency of gathering and inversions.

→ Reply

_gsdt

5 years ago, # ^ |

← Rev. 3 →

Could you please explain to me the formula in Endagorion 67338561

ans += x * L - L * (L - 1) / 2 - ls;
ans += rs - x * R - R * (R + 1) / 2;

→ Reply

5 years ago, # ^ |

← Rev. 2 →

$$$lh, rh$$$ — sets of positions of left and right parts of sequence.
$$$ls, rs$$$ — sums of all elements in corresponding sets.
$$$L, R$$$ — count of elements in corresponding sets. Also it is length of arithmetic sequences.
In first line $$$ls$$$ is first query in BIT from my explanation in linked comments section. Other stuff from first line is first sum of arithmetic progression: $$$ [x - L + 1, x] $$$
In second line $$$rs$$$ is second query in BIT from my explanation. Other stuff in second line is sum of second arithmetic progression: $$$ [x + 1, x + R] $$$

→ Reply

_gsdt

5 years ago, # ^ |

← Rev. 3 →

Thank you, but why do we have L * (L - 1) / 2 and R * (R + 1) / 2 here? In masalskyi's code (67434904), he has another way:

left = left_am * mid - left - left_am*(left_am+1)/2;
right = right - right_am * mid - right_am*(right_am+1)/2;

→ Reply

5 years ago, # ^ |

He don't include mid for both of ranges.

$$$[x-L, x-1]\;[x+1, x+R]$$$

→ Reply

tooDumbToThink

5 years ago, # |

Can anyone prove that editorial solution of div2C gives correct answer?

→ Reply

towrist

5 years ago, # ^ |

Yes, I can. Consider the set of all beautiful integers. For example, 123123123 can be written as (123)*(10**(0)+10**(3)+10**(6)) So any beautiful integer in general can be written as (a1a2a3..ak)*(sum of powers of 10). Now, the key thing to realise is if the number a1a2a3...ak is lesser than the first k digits of the input number x, then the beautiful integer of the same length so formed is definitely less than x. This can be easily shown.

Since we have to find y>=x, we start our search from the first k integers of x.

Note again that if a1a2a3...ak is greater than the first k digits of x, then the beautiful integer so formed is definitely greater than x. This can also be shown similarly and easily.

So we start our search from a1a2a3...ak, the first k integers of x. If the beautiful solution so formed is greater than or equal to x, we can print y. (We are sure that this is indeed the smallest, since any smaller and y definitely becomes less than x).

If y is smaller, we add 1 to the number a1a2...ak and print the beautiful integer so formed. This is definitely greater than x and no number smaller than this is greater than x. Hence proved.

If you have any doubt, feel free to post it.

→ Reply

cuebits

5 years ago, # |

Could anyone here help me figure out what's wrong with my solution for Div 2 (C)? I have tried to follow the editorial here.
Link.

→ Reply

sandeep_kumar_e

5 years ago, # ^ |

for (int i = 1; i <= n; i++) if (num1[i] > num2[i]) flag++;

this part of the code is causing the error

run this testcase: 6 3 427129 your solution — 428428 but the correct ans 427427.

→ Reply

cuebits

5 years ago, # ^ |

Yes, I figured it out. I was comparing the numbers stored in the arrays incorrectly.

for (int i = 1; i <= n; i++) 
    if (num1[i] > num2[i]) 
        flag++;

This is not the correct way to check if num1 > num2. Thanks!

→ Reply

AECDNhau

5 years ago, # |

← Rev. 3 →

-31

i think the word 'ciclycally' is a russian word. as a non-russian speaker, i find the use of the word confusing. you should have used a better word, such as 'cyclically.'

edit: the typo was fixed. i don't think my comment warrants such many downvotes, i was pointing out the typo in a somewhat sarcastic way.

→ Reply

5 years ago, # ^ |

+51

that's not a Russian word, that's just a typo.

→ Reply

-deleted-

5 years ago, # |

For Div2 C, we can take the first k digits as a number and repeat it for length n, and then add one to that number and repeat it again for length n . This is optimal and i know it works.

But i am unable to proof that why it'll always work, can anyone help me with a proof?

→ Reply

gupta_samarth

5 years ago, # ^ |

← Rev. 4 →

Make new number with first $$$k$$$ digits same as given number.

Firstly, check if new number ( with first $$$k$$$ repeated till $$$n$$$ ) is greater or equal to original number. If yes, then obviously print it.

If it is not greater, then increasing the number formed by the first $$$k$$$ digits by one is the smallest increase which enables you to have a number greater than the original, since now your number has first $$$x$$$ ( depending on number of $$$9$$$s at end of the first $$$k$$$ digits ) digits same as original and next $$$k-x$$$ digits greater than original. So already, the prefix of first $$$k$$$ digits is greater than given number. Then repeat it to make answer till $$$n$$$ digits.

→ Reply

samio

5 years ago, # |

you have 300iq s

→ Reply

Muhammad98

5 years ago, # |

In problem B div2 I just generated all possible x's and took the most frequent one. Is that right? 67350595

→ Reply

AkaiShuuichi9866

5 years ago, # |

I really liked 1269D. To me, the solution seemed very non-intuitive though. (Like I couldn't understand how that train of thought could originate in the first place).

One question I had about it was, wouldn't this chessboard coloring logic be applicable even if the heights columns were not in a non-increasing order? I'd like to understand why this logic would fail in that case.

Any help would be appreciated.

→ Reply

PuRpLe_FoReVeR

5 years ago, # ^ |

← Rev. 3 →

1 0 0 1

1 2 1 1 2 1

it's possible to get non-connected diagram.

→ Reply

AkaiShuuichi9866

5 years ago, # ^ |

Ah I see! It was quite silly of me to miss that. Thanks for pointing it out. Btw, I still find the idea of coloring the diagram in white and black quite elegant. Would you happen to know if this is a well known idea (like are there other problems which use similar techniques) or did people come up with the solution on the fly during this round?

→ Reply

PuRpLe_FoReVeR

5 years ago, # ^ |

I don't know another problems like this, but coloring board with chess order is common idea. Another idea with 1x2, 2x1 tiles is bipartite matching of graph with vertices as board cells and edges between cells that have common side. Its equivalent problems. Interesting thing about domino tilings is Aztec diamond. But have nothing common with div1B problem )

→ Reply

Swistakk

5 years ago, # ^ |

← Rev. 2 →

Coloring chessboards and packing there some weird shapes is one of the most popular genres of mathematical olympic problems and packing dominoes into a 8x8 chessboard with two opposite corners removed is the most classical problem in it (which I learned in first half of my life)

→ Reply

ghoshsai5000

5 years ago, # |

← Rev. 2 →

I always look forward to contests with a single writer behind them and 300iq did not disappoint with this. The questions were really good !

How to solve the bonus question of $$$B$$$ ?

→ Reply

5 years ago, # |

I have written code for div2C as per logic: 1) First, simply repeat 1st K characters and compare 2) If new_string is >= old string, print it 3) Otherwise, look for the index from first K character (starting from last one) for which if you iterate over K-sequence of (index, index+k..), then old_str > new_str. If so, increase the character value at that found_idx and at other position in its K-sequence order. After this, make all character K-sequences in between found_idx and K to '0'. I couldn't find the flaw in above logic and thus couldn't see why am I getting WA. Test case on which it's failing is pretty HUGE itself. I looked for other test cases from accepted solutions but still got the correct answer. Can someone please help me out of this trauma? My submission Link

→ Reply

evenharder

5 years ago, # ^ |

Consider this input.

5 3
12113

The answer is 5 12212, while your code produces 5 13013.

→ Reply

5 years ago, # ^ |

Thanks a ton for replying with a test case..:)

→ Reply

5 years ago, # ^ |

Sorry for trouble! I have solved it.

→ Reply

mohamedeltair

5 years ago, # |

← Rev. 3 →

+44

Clarifying the editorial's $$$Div1B$$$:

Suppose you mapped the diagram to a chessboard, where the white cells count is $$$w$$$ and $$$b$$$ is the black cells count. We can see that the upper bound of dominoes we can put is $$$min(w,\, b)$$$. But what is the proof that we can always achieve it?

Let $$$a_i$$$ be number of available cells in the $$$i^{th}$$$ column. Starting with the original diagram, try always to place a vertical domino in a column i where $$$a_i>a_{i+1}+1$$$ (decreasing $$$a_i$$$ by $$$2$$$), or a horizontal domino on top of 2 columns $$$i$$$ and $$$i+1$$$ where $$$a_i=a_{i+1}$$$ and $$$a_{i+1}>a_{i+2}$$$ (decreasing both $$$a_i$$$ and $$$a_{i+1}$$$ by $$$1$$$).

Both of the previous $$$2$$$ operations will not change the property of $$$a_i\geq a_{i+1}$$$, and will not change $$$w-b$$$. If you can't do any of the previous 2 operations any more, then you have reached a configuration with values of $$$a_i$$$ being $$$t$$$, $$$t-1$$$, ..., $$$2$$$, $$$1$$$. In this configuration $$$b\neq w$$$, this means that $$$b\neq w$$$ from the beginning (So if $$$b=w$$$ from the beginning, you could have filled the entire diagram).

If you didn't fill the entire diagram, how can you still achieve $$$min(w,\, b)$$$? Assuming without loss of generality that the column with $$$a_i=1$$$ (the $$$t^{th}$$$ column) has a white cell, we can see that $$$w-b=\lceil \dfrac{t}{2} \rceil$$$. So if we kept putting vertical dominoes, we will end up with leaving exactly $$$\lceil \dfrac{t}{2} \rceil$$$ cells (the number of columns with odd $$$a_i$$$).

→ Reply

Akshay184

5 years ago, # |

300iq can you please elaborate on how to solve problem E of Div2

→ Reply

Akshay184

5 years ago, # ^ |

I was unable to understand the editorial

→ Reply

masalskyi

5 years ago, # |

Can anybody please explain me how to get second value for each k in div2 E how to do it with segment tree

→ Reply

gupta_samarth

5 years ago, # ^ |

← Rev. 2 →

I guess you can initially have all values set to $$$0$$$. As you increase $$$k$$$, set $$$a[k]$$$ ( this array is the one on which segment tree is build ) to $$$1$$$. Then at every k, you have $$$1$$$s on some positions. I don't know how to do it, but now the problem is, given an array of $$$0$$$ and $$$1$$$, find minimum operations ( swaps ) to get all ones together. As editorial mentions, this value should be sum of indices of $$$1$$$s to the right of the median $$$1$$$ minus sum of indices to left of median $$$1$$$ ( think of it as right side $$$1$$$s have plus sign, and left side have minus sign ). Also, on each update, atmost $$$2$$$ signs will change, so you can handle sign change on update, and then keep track of the sum of the indices ( with sign ).

→ Reply

5 years ago, # ^ |

← Rev. 2 →

I knew that it's possible to use segment tree, but I was unable to came up with it during contest.

Then, after a while, I realized key observation: all times you add $$$k$$$ ! I know, it sounds ridiculous. But listen. You add $$$k$$$ <- this is last in sequence. In other words, it's biggest number. So, you don't have any number higher than $$$k$$$. According to inversions, the impact of including $$$k$$$ is number of all inversions with it. To calculate this impact, all you have to do is find all numbers less than $$$k$$$ at positions with higher indexes than $$$k$$$. But they all already less than $$$k$$$.

This means, that if you hold in segment tree all numbers from 1 to $$$k$$$, your request is just: find all X that has index higher index of $$$k$$$.

With segment tree, it's just sum of ones from (position of $$$k$$$)+1 to $$$n$$$.

→ Reply

5 years ago, # ^ |

Oh, sorry, I thought about inversions. it was harder to me.

About swaps with segment tree. Lets note $$$p_i$$$ initial position of 1 and $$$t_i$$$ position we want. Then total swaps:

$$$\sum{\left|p_i-t_i\right|}$$$

Why? Because it's useless to swap 1 with 1, so we just 'pack' them. Now, split into two sums:

$$$\sum{\left|p_i-t_i\right|} = \sum\limits_{p_i<t_i}\left(t_i-p_i\right) + \sum\limits_{p_i\geq t_i}\left(p_i-t_i\right) = \sum\limits_{p_i<t_i}t_i - \sum\limits_{p_i<t_i}p_i + \sum\limits_{p_i\geq t_i}p_i-\sum\limits_{p_i\geq t_i}t_i$$$

First and last sum is sum of arithmetic progressions. You can find them by formula. Other two are just sum using segment tree or fenwick tree.

→ Reply

masalskyi

5 years ago, # ^ |

Ohh can you explain it more please. What is $$$t_i$$$ and what does mean 'pack'? Please )

→ Reply

5 years ago, # ^ |

← Rev. 2 →

$$$t_i$$$ is target position. Look at following example:

01234567890123456 // helper for indexing
00100010001010010 // initial
00000000111110000 // target

Then $$$p = [2,6,10,12,15], t = [8,9,10,11,12]$$$. So, sum is

$$$ |2-8|+|6-9|+|10-10|+|12-11|+|15-12| = \\ (8-2)+(9-8)+(10-10)+(12-11)+(15-12) = \\ (8+9)-(2+8)+(10+12+15)-(10+11+12) $$$

I pick word 'pack' because you can imagine this by 0 — space, and 1 for example pens, and you just move leftmost towards center and rightmost towards center and they meet together.

→ Reply

_Bishop_

5 years ago, # ^ |

Hey, in case of even number of 1s you "pack" them towards which of the two medians?

→ Reply

5 years ago, # ^ |

I pack towards best of two, calculate cost of both. It should work if indepentence is correct (I don't know proof).

→ Reply

_Bishop_

5 years ago, # ^ |

→ Reply

5 years ago, # ^ |

you can prove that both medians has same sum, using same argument that minimum should be median.

→ Reply

masalskyi

5 years ago, # ^ |

you are the best) thank you) I got accept

→ Reply

moniba

5 years ago, # ^ |

This helped me to understand it very well THANK U

→ Reply

snayan06

5 years ago, # |

What was the logic you all used in div2a i am having hard time understanding tutorail.its normal to not able to solve que i can solve a b sometimes but sometimes i have problems,

→ Reply

stefdasca

5 years ago, # ^ |

If a — b = n, then we can rewrite a as (x+1) * n and b as x * n. In order for both (x+1) * n and x * n to be composite, we need n to be at least equal to 2.

Basically, you can print every possible pair of type (x+1) * n and x * n as long as x >= 2 and (x+1) * n <= $$$10^{9}$$$

→ Reply

5 years ago, # ^ |

There is tricky test n = 1. For that case 2,3,4,5,6,7 doesn't work.

→ Reply

stefdasca

5 years ago, # ^ |

good point

→ Reply

akane_psycho

5 years ago, # |

Can anyone tell me what is wrong with dp approch in div 2 D https://mirror.codeforces.com/contest/1269/submission/67394114 I have first converted a[i] to a[i]%2 that is making the sequence 100110... 1 if odd 0 if even. Then I note compute dp[i] as the min no of squares up to column i that can go unfilled.For this I use the following observations, 1. we can use the full column of even length (dp[i]=dp[i-1] ) 2.if columns are of the form 111..m times (i.e all odd), if m is even then all the squares can be filled, if m is odd 1 square is not filled. 3.sequence of the form ( 1.. even times 0..1 ) can be completely filled.(you can take example of 3 2 2 3) Am I missing something??

→ Reply

vamaddur

5 years ago, # |

Will an editorial for all problems on the actual ByteDance contest be posted?

→ Reply

faculty

5 years ago, # |

← Rev. 7 →

In Div2 D, why do we have a contradiction?
Does the proof still work in this case?
2
1 1
I think it has an equal number of white and black cell.

→ Reply

5 years ago, # ^ |

I think that it doesn't meet the request in the editorial "If all rows and columns are different".

→ Reply

5 years ago, # |

Can anyone please tell why O(n2) solution doesn't work in Div2 B? Please have a look at my submission and help.

→ Reply

ahzong

5 years ago, # ^ |

O(n^2) submission works and your submission for the problem is accepted. What's wrong?

→ Reply

5 years ago, # ^ |

It's not accepted. I'm talking about modulo problem!

→ Reply

ahzong

5 years ago, # ^ |

Oops sorry, I looked at the wrong row

→ Reply

sigma_g

5 years ago, # |

← Rev. 2 →

Weak systests in Div2B. Some test cases where multiple x values existed, submissions have passed without taking the minimum of them. Example test case:

4 4
1 1 3 3
0 0 2 2

Here both x=1 and x=3 are valid and output should be x=1. But hacked submission outputs 3.

→ Reply

5 years ago, # ^ |

What does hacked submission mean sire?

→ Reply

sigma_g

5 years ago, # ^ |

Users with 1900+ rating can hack a submission even after the contest is over. It does not result in decrease of points of submitted code though.

→ Reply

abj

5 years ago, # |

-27

#include <bits/stdc++.h>
#define ll long long
#define ibs ios_base::sync_with_stdio(false)
#define cti cin.tie(0)
#define pb push_back
#define N 400015
ll y=0;
struct table{
    int val,id;
};
bool cmp(table a,table b)
{
    return a.val>b.val;
}
bool isPrime(ll n) 
{ 
    // Corner case 
    if (n <= 1) 
        return false; 

    // Check from 2 to n-1 
    for (ll i = 2; i < n; i++) 
        if (n % i == 0) 
            return false; 

    return true; 
} 
using namespace std;//coded by abhijay mitra
#define watch(x) cout << (#x) << " is " << (x) << endl
int main()
{
    ibs;cti;
    int n;ll m;cin>>n>>m;std::vector<ll> a(m,0),b(m,0);bool Check=0;ll x;
    for (int i = 0; i < n; ++i){
        cin>>x;a[x]++;
    }
    for (int i = 0; i < n; ++i){
        cin>>x;b[x]++;
    }
    if(m==2){
        if(a[0]==b[1] and a[1] ==b[0]){
            cout<<"1\n";return 0;
        }
    }
    if(n==1){
        if(b[0]==a[0])
            cout<<0<<"\n";
        return 0;
    }
    for (ll i = 0;; ++i){
        if(a==b){cout<<i<<"\n";return 0;}
        std::rotate(b.begin(), b.begin()+1, b.end());
    }
}

it got rte in question B. I used vector rotation. Where is issue?

→ Reply

utkarshh12

5 years ago, # |

Can someone explain Div2 D, the domino problem?

→ Reply

AjaySabarish

5 years ago, # |

In div2D/div1A, I understood the soln but what is the significance of "column lengths will be in sorted order"?Even if it's not sorted this algo shud work right?

→ Reply

Samui

5 years ago, # |

For Div2 E. I can not get how to maintain the second answer. Who can explain more, thanks.

→ Reply

5 years ago, # ^ |

Let $$$a_i$$$ be the number of $$$1$$$ after place $$$i$$$, $$$b_i$$$ be the number of $$$1$$$ before place $$$i$$$.

You're going to work out $$$\Sigma_{i=1}^{n}min(a_i,b_i)*[place_i=0]$$$ while each time a $$$0$$$ changes to an $$$1$$$.

It's easy to conclude that $$$a_i$$$ is non-increasing and $$$b_i$$$ is non-decreasing.

As a result, there exists an $$$i$$$, which satisfies that for all the $$$j \leq i$$$, $$$min(a_j,b_j)=b_j$$$ and for all the $$$j>i$$$, $$$min(a_j,b_j)=a_j$$$.

So just do a binary search to find $$$i$$$ (Or, according to the definition of $$$a_i$$$ and $$$b_i$$$ it is the median of all the $$$i$$$ that $$$place_i=1$$$), and the answer is $$$\Sigma_{j=1}^{i}b_i*[place_i=0]$$$+$$$\Sigma_{j=i+1}^{n}a_i*[place_i=0]$$$.

You can work it out on a segment tree.

This is my code and I hope it can help you more: 67425987

→ Reply

Samui

5 years ago, # ^ |

Thanks very much!

→ Reply

EvGen1u5

5 years ago, # ^ |

← Rev. 2 →

Can you explain how we do calculation after finding median i?

I mean.. Okay, we got that we have to add to the answer sum of Ones before 0 (if 0 is before i) and sum of ones after 0 (if 0 is after i) for all zeros

→ Reply

5 years ago, # ^ |

After finding median $$$i$$$ you realize that for all the $$$place_j=0$$$, if $$$j \leq i$$$, then the contribution of $$$j$$$ is $$$b_i$$$, else it's $$$a_i$$$.

Consider that each time a $$$0$$$ changes to $$$1$$$.

You can maintain it through the following steps: (suppose that $$$place_k$$$ changes to $$$1$$$)

1) Let $$$a_j$$$ $$$(j \in [1, k-1]) = a_j + 1$$$, $$$b_j$$$ $$$(j \in [k+1, n]) = b_j + 1$$$.

This can be realized on a segment tree.

2) Erase the label $$$k$$$ in $$${a_n}$$$ and $$${b_n}$$$.

To achieve this, you can maintain a $$$size$$$ on the segment tree node to show how many labels are still existing on the range that this node stands for.

Then when you are going to add $$$1$$$ on a whole segment tree node, you can simply add the $$$size$$$ of the node to the $$$sum$$$ of that node and leave a lazy tag there.

If you still have problems you may view my code above to better understand it.

→ Reply

sonianand966

5 years ago, # |

I see a tag of "Graph Matchings" on DIV-2 D, how the problem can be solved using such an algorithm. I couldn't see any mention of it in the editorial though.

→ Reply

5 years ago, # ^ |

All cells are node. Edges between adjacent cells. You get a bipartite graph. Find the largest matching in it.

→ Reply

_Bishop_

5 years ago, # |

Can someone please explain for Div2E how to calculate the second value(mentioned in editorial)? And why does it work(taking the median)? I understood the inversions part but unable to get the median thing!!

→ Reply

J.F.K.

5 years ago, # |

How DIV2 B can be solved in O(N)?

→ Reply

5 years ago, # |

How we can solve problem D using graph matching as given in tag ??

→ Reply

5 years ago, # ^ |

All cells are node. Edges between adjacent cells. You get a bipartite graph. Find the largest matching in it.

→ Reply

5 years ago, # ^ |

But there are too many cells so the time limit will exceed

→ Reply

5 years ago, # ^ |

Using hall's theorem you can show that the maximum matching set will have the same size as the smaller of the two bipartite sets. So you don't actually need to solve the flow problem.

→ Reply

5 years ago, # ^ |

Can you explain a little more?

→ Reply

5 years ago, # ^ |

I mean that consider the graph you get as $$$G=(X+Y, E)$$$. Creating this graph and then getting $$$X$$$ and $$$Y$$$ is easy in $$$O(N+M)$$$ and since $$$O(M) = O(N)$$$ the result is $$$O(N)$$$. Now using Hall's Theorem you can say that the maximum bipartite matching will have a size of $$$min(|X|,|Y|)$$$. You don't need the actual coverings.

As for why Hall's theorem is applicable, assume $$$X$$$ to be the smaller set with LOG. We can use induction on the size $$$A$$$ where $$$S$$$ is a subset of $$$X$$$. The base case is trivial, there must be one edge from the $$$S$$$. Now, notice that whenever we add a another node to this subset, it must add at least one new node to the neighborhood. Therefore, the proof is complete.

Why must it add at-least one node? Well, I haven't been able to prove that part yet but I'm pretty sure it's related to being the smaller set.

→ Reply