Keeping track of both Codeforces round and online team contest was a doozy, so this is only the best draft of the editorial I have. Missing problems will gradually be added, and existing explanations may improve over time. Hope you enjoyed the problems, and let me know if anything can be explained better!

**UPD**: okay, all of the problems are out, and most of the bugs are fixed (I hope). By the way, we had a nice discussion with Errichto on his stream about Div. 2 problems, which some of you may find more approachable. Be sure to check it out, as well as other stuff Errichto creates!

Tutorial is loading...

Tutorial is loading...

Tutorial is loading...

Tutorial is loading...

Tutorial is loading...

Tutorial is loading...

Tutorial is loading...

*(Kudos to Golovanov399 for his neat grid drawing tool)*

Tutorial is loading...

https://codeforces.com/contest/1459/submission/101773389 Why my solution fails in test case 21 . problem div2c

i don't know that as i was just doing div2 B Now can you help me with div2 B. in very simple language as i didn't get editorial posted here.

See my solution I recommend this video. It helped me.

thanks for your reply , and specially for attached video. it helped a lot me too.

No clue what you're trying to do with the mod 3s. Try checking other's solutions for a simpler answer.

5 1 1 5 9 3 3 1

Try this test with your code and you'll get 4 as result, which is not correct. The reason it doesn't work is because you didn't count a[3] — a[2] while calculating the gcd of all (a[i] — a[i — 1])

thanks i realize the mistake.

Can someone explain why the property used in Problem C is also valid for multiple arguments?

We can show these properties by definition of $$$\gcd$$$.

Then we can extend our claim to multiple arguments by mathematical induction.

Therefore $$$\gcd(a_1 + b_j, a_2 + b_j, \cdots, a_{n-1} + b_j, a_n + b_j)$$$ is equal to $$$\gcd(a_1 + b_j, a_2-a_1, \cdots, a_{n-1}-a_{n-2}, a_n-a_{n-1})$$$.

Update: We do not need to choose adjacent differences. $$$\gcd(a_1, a_2, \cdots, a_{n-2}, a_{n-1}, a_n)$$$ is also equal to $$$\gcd(a_1, a_2-a_1, \cdots, a_{n-1}-a_1, a_n-a_1)$$$, which can be proved similarly. The important step is to express all but one integers as $$$a_i - a_j$$$ to negate $$$+b_j$$$.It would be very helpful if you share your implementation..

If you're looking for good implementations, the best place to go is usually the top of the leaderboard (https://codeforces.com/contest/1458/standings). For instance, this is tourist's solution: https://codeforces.com/contest/1458/submission/101720547

Thank you !, I just implemented it

The only part I was not able to understand is, gcd(a,b,c) = gcd(a,gcd(b,c)) => gcd(a,gcd(b,c-b)) = gcd(a,b,c-b). How are we taking gcd(b,c-b) just as b,c-b. That way we could just write gcd(a,b) = a,b as well.

gcd(a,b,c) means gcd of a,b and c. So

`gcd(a,b,c)`

will be equal to`gcd(a,gcd(b,c))`

.Examplegcd(8,20,40)=4 gcd(20,40)=20 and gcd(8,20)=4 So`gcd(8,gcd(20,40)) = gcd(8,20) = 4 = gcd(8,20,40)`

This was great, Thanks a lot!

why someone's submissions been skipped and receive no message

it's the case of plagiarism bruh

I didn't know the proof for problem B's solution but i solved it.

Endagorion Please add implementations to the problems as well it will be really helpful.

In problem B, when n is odd, k = (n+1)/2.

For n=3, k=2, ans=2*2*3=12.

For n=5, k=3, ans=2*3*4=24.

For n = k + (k-1), if we start with a horizontal step, we can reach :

k+1 positions horizontally

k positions vertically

The total is k(k+1).

For example, if n = 3 + 2, if we start with a horizontal step, we can reach the positions with a cross :

x . x . x . x

x . x . x . x

x . x . x . x

Solution sequence for Div2B is also available on OEIS. https://oeis.org/A241496. But I didn't understand the general formula. Can someone explain?

.

https://youtu.be/1OWGTQpfk5Y — recording of post-contest discussion with Endagorion with analysis of all div2 problems (so up to div1-D)

can you please provide the code for the above explaination of question D?

In problem D (div. 2) isn't the complexity O(n * k * n * A) where A the maximum capacity of a glass? Which in turn is 100^4 (since max n = 100 max k = 100 and max A = 100 )? 100^4 = 100 000 000 dp cells, so the solution should pass yes (with the memory optimization). But your complexity is O(n * k * n * A) I don't get how you got A^2 in there. Please someone explain, thanks.

Good point, thanks. I'll fix it soon.

That fix created some confusion.

A line above that, $$$A$$$ was used to denote the 'total capacity of the subset'. In the next line, $$$A$$$ is used to denote the 'maximum capacity of the glass'.

Thanks for pointing it out, I made it more clear now.

In Problem-B's explanation,

Last Line:

Shouldn't it be

`where k=n/2 rounded up`

, instead of`where k=n/2 rounded down`

?Endagorion

Someone, do correct me if I am wrong.

Yes, thanks for pointing this out! Will fix soon.

Endagorion I think that in Problem-E's(div2) explanation

$$$p'$$$ should be equal to $$$v_0 + iv_x + jv_y + kv_z$$$ instead of $$$v_0 + xv_x + yv_y + zv_z$$$

this is part of the paragraph before the last one

Sorry if I am wrong.

You are correct! I'll be fixing all mistakes soon once I finish the editorial for div1F.

Why did the rating change again as it was? Endagorion

No idea honestly. Let's just let CF staff do their thing.

Ignoring the short contest time and difficulty gap, I think your rounds are generally of high quality.

Thanks for the round :)

Thanks a lot! There's always room to improve for the next time huh.

Auto comment: topic has been updated by Endagorion (previous revision, new revision, compare).Is there a thought process on how to obtain the key observation of 'Latin Square'? It feels very clever and beautiful, but hard to come up with at the same time

In problem D(Glass half spilled), can anybody tell me , why we are considering taken glasses from n to 1 order.

Edit : The order of $$$k$$$ doesn't matter too, since we allocate $$$n$$$ 2D dp arrays for each glasses. Sorry!

We need to calculate $$$dp(i,k,A)$$$ from $$$k=n$$$ to $$$k=1$$$ and from $$$A=sum$$$ to $$$A=0$$$ order because if we do the opposite, 'duplication' will happen in case you use the same $$$dp$$$ table for all glasses. The order of $$$i$$$ doesn't matter, I think you can do either 1 to $$$n$$$ or vice versa.

So, suppose that we are going from 1 to n ans we have taken 3 glasses. Then it may happens that we are taking the same glasses to extract water and then put it in the same glass. Am i getting correct? And can you please explain with an example?

Can someone explain this? I also dont understand the need to iterate backwards.

101822812 Saw this submission, this proceeds in forward direction. So I guess we can proceed both ways.

yeah, can anyone please explain why do we need to iterate backwards? Thanks in advance :)

Would the order of $$$A$$$ matter even if you're using $$$n$$$ 2D Arrays?

I'm able to get the solution following the correct ordering with only one 2D array, but I thought that the order wouldn't matter if I used a separate 2D array for each i (since there is nothing to duplicate / overwrite).

Can someone explain why this submission with $$$n$$$ 2D arrays gets a wrong answer? https://codeforces.com/contest/1459/submission/101994708

Never mind, figured it out by looking at the submission mentioned above.

The order indeed doesn't matter but I need to make sure that I copy over all the values from the previous array.

Can we solve the Row Gcd problem with the given below logic?

(1) sort the array 1

(2) take the array[0] and array[1]

(3) iterate through the array2 and take gcd of array[0]+array[j] and array[1]+array[j]

(4) check divisiblity with all array1[i]+b[j]

(5) if divisible then we got the gcd else dividing the current gcd with prime factor list of gcd and goto step 4 until gcd!=1 or all elements are divisible with gcd.

I had implemented this logic in the contest but I was getting errors in test case 3 on the 29th element. How can I download the test case of the above problem to debug my code? ThankYou

my solution can anyone pls tell me how to optimize my code. Thank you.

my logic is:

initially I can move in any four direction north south east west so I made four function calls

then if I move in x-direction(taken care by movex variable) then I have to move in y-direction either positive or negative

if I move in y-direction(taken care by movey variable) then I have to move in x-direction either positive or negative

then if I ran out of moves I insert the coordinates in a set

then I finally print the size of set.

Print for some values and observe the pattern. In your code, even if you apply dp, it will give tle.

ok, thank you I will try and get back to you.

sorry, I can't come up with any pattern can you help me? thanks

The first 15 numbers are: 4,4,12,9,24,16,40,25,60,36,84,49,112,64, 144

1-indexing if n is even ans is p*p where p = n/2+1 Now take odd numbers 4,12,24,40,60,84,112,144 Now you can see that the difference between two numbers is getting increased by 4. You can see my solution for implementation.

You can also watch errichto(youtube channel) soln for easier soln

In problem B, it is given a dp tag .Though there is a nice O(1) solution but can anyone please describe about how to get a dp solution ?

You can check my submission for the DP Implementation.. I figure it out by observing the pattern through recursive implementation.

thanks bro :D . but can you describe that ??

seems like it is an iterative solution. Where is your recursive part and you did it in math,I didn't find any dp there.Please can you explain ?

For odd numbers, my dp state is calculated as below :

So, I agree that it is a formula only since the previous state of an odd number is an even number only which can be calculated directly through a formula, but it can still be viewed as recursion, although weakly. I could not find a DP flavour more than this amount to this problem

Dear Endagorion,

The heading of

Problem 1459C(Row GCD) is not translated to English. I could make out that it's the solution of Row GCD based off the description that follows. Kindly fix it!Thank You!

How is the graph structured in Flip and Reverse F? The editorial makes it look like something on a straight line, which cant be true

That's exactly what it looks like though: vertices are balance values in the real line.

Can Div2D/Div1B be solved using a top-down approach? Recursion with memoization?

Is almost the same as in the tutorial, think it like knapsack: If we fix a total capacity

`A_s`

, a total number of glasses that we are going to fill`K_s`

and a current glass`i`

.We have to choices:Include the glass among the ones we are going to fill: This is possible if we haven't include all the glasses already and we are not exceeding the capacity. In this case we should decrease`k_s`

by 1, because we are including one more glass, and also decrease`A_s`

by the capacity of the current glass. Then just call the recursive function adding the amount of water of the current glass (just like the dp in the tutorial).Not including it: In this case we ignore the current glass, thus we just call the recursion with a different glass.In order to find the solution for a

`k`

just iterate over all possible capacities (I did it from 0 to the total capacity of all glasses) and the answer will be the maximum value of`min(current_capacity, dp[start_index][current_k][current_capacity] / 2 + total_water_of_all_glasses / 2)`

. Here is my submission 102646662In problem D, declaring the arrays globally makes the same solution pass which otherwise was failing due to runtime error when I had declared them locally. Could someone elaborate on this point or am I messing up something else?

Can anyone explain B's editorial with an example? how the editorial is working?

I honestly don't understand the editorial either. but i've observed the pattern through brute force. i also visualized it using R for better understanding about the pattern.

here's the pdf file

hope it helps!

Finally finished my alternative solution for div1D.

Can someone please help me understand the proof why we can convert any two Eulerian pathS with the operation of reversing a cycle ?

In problem b if r = 51 and b = 61, then blue will win. But when r = 51 and b = 16 then they will end up equal. Am I missing something?

does anyone know the proof of this property of gcd?

`GCD(x,y)=GCD(x−y,y)`

or any resource for exlanation?http://people.cs.ksu.edu/~schmidt/301s14/Exercises/euclid_alg.html

Can anyone explain me the DP solution of div2-B...Please

105154843

You can draw the final possible answers for each step in the grid, and then you can summarize the rules.

In problem D why is the total available capacity necessary as a state ?

Maybe elaborate on why you think it shouldn't be.

Okay one reason I found is that since I am aiming to gather max water from a set I have to also know the upper bound of what I can carry. That is why Total sum is needed. Tell me if I am right.

For 1459D - Glass Half Spilled

When I used int for integer variables, I got a runtime error. 107625846

However, when I replaced all int's with int16_t, it gave AC. 107680750

This is the first time I've encountered something like this. Is there a way to avoid runtime error without using int16_t?

You use too much memory with

`int`

.what will be recursive solution to div2b

Problem C can be solved using following identities

Instead $$$L$$$ write it as $$$R^{-1}$$$ and instead of $$$U$$$ write it as $$$D^{-1}$$$

Using these identities, we can reduce the whole expression to an expression where there is atmost one $$$I$$$ or $$$C$$$ and at most one transpose and rest $$$R$$$, $$$D$$$ and add operations which freely commute with each other.