Codeforces Round #863 (Div. 3)

19 months ago, # |

As a tester...! wait a minute, something ain't right

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RollBack

19 months ago, # |

Hope to become Expert again

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Salyone

19 months ago, # |

At last a Div. 3, a good opportunity for improve rating.

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Math_is_Good

19 months ago, # ^ |

Lol, it didn't age well.

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Taanvirr

19 months ago, # |

← Rev. 2 →

+12

Need more frequent Div.3 contests for beginners.

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_.baba_yaga._

19 months ago, # |

lessgo..

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19 months ago, # |

+38

As a tester, I feel that the problems are very interesting and everyone should participate in this round.

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d_sqaure

19 months ago, # ^ |

Just asking to confirm, there wil be no interective problem,right ?.As, in a previous contest there was interective problem and didn't mentioned in blog.

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19 months ago, # ^ |

You can trust this blog.

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The_Dragon_Warrior

19 months ago, # |

Hope to become pupil today!

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fangqing

19 months ago, # ^ |

Hello, can we make frined? Or, we can make progress each other.

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sheverlu

19 months ago, # |

hoping for Expert

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AVdovin

19 months ago, # |

+11

Is it Div2.25?

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Cooper1214

19 months ago, # |

+18

came back here in the middle of the contest, just to downvote. what kind of div 3 is that?

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twosat

19 months ago, # |

-25

I don't think this round is too hard for a Div. 3 Round after AKing it, I think it normal and DE is even too easy for their place.

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JEE_MAIN_LOOSER

19 months ago, # ^ |

I am not so sure About D .

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cockatooo

19 months ago, # ^ |

+43

I like how you submitted C and G in the same minute, with completely different templates lmao. Imagine cheating then bragging online about it ICANT.

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twosat

19 months ago, # ^ |

See if I'll be skipped then.

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Muhammad-Daniyal

19 months ago, # ^ |

So do you really think that all the people here are little babies and they don't know how a code can be changed after being copied. Such a fool you are!

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twosat

19 months ago, # ^ |

I'm not foolish enough to copy one code without editing the template.

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Lwchicky0

19 months ago, # ^ |

← Rev. 2 →

it seem unrated.

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Harshjain4001

19 months ago, # |

worst Div. 3 contest ever.

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Loic_

19 months ago, # |

-20

Nice problems except for F. F is simply a trash problem.

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WannaSleep43

19 months ago, # ^ |

LoL, nice problem...

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SleepingThread

19 months ago, # ^ |

← Rev. 2 →

For F: You need to get all the simple cycles in the graph,

check if there is exactly one cycle where its nodes degree is 4

And for all other cycles, the degree of its nodes should be 2 except for only one node (wich is part of the main cycle and its degree is 4).

Easy :)

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19 months ago, # |

Is there any way to solve "Living Sequence" without digit dp ?

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EleCursity

19 months ago, # ^ |

+26

consider the number k in base 9.

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MrOtter

19 months ago, # ^ |

+21

https://oeis.org/A052406

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jalanprakhar

19 months ago, # ^ |

Can you tell me your digit dp approach?

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19 months ago, # ^ |

← Rev. 2 →

First you need to binary search for the number, let's say x.

Then, the check(x) function is pretty straightforward if you know the standard digit dp approach. Just skip "4" at all the order from $$$0$$$ to $$$log_{10}(n)$$$ in x, and do calculations for other numbers as usual.

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jalanprakhar

19 months ago, # ^ |

I am really sorry if the question sounds dumb , but in the questions that I've done based on digit dp, we are always given an upper bound , which we dont have here, how can I compute the kth number

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TheBhediyaOfDalalStreet

19 months ago, # ^ |

You need to count number of numbers less than or equal to k, that have a 4 in them (or don't have a 4 in them).

This is you recursive function. You can use it along with binary search to solve the problem

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19 months ago, # ^ |

Yeah sorry, check the edited answer now.

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jalanprakhar

19 months ago, # ^ |

Thank you so much man, helped a lot!! :)

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adityasah09

19 months ago, # ^ |

My digit DP solution was giving TLE E Solution

I did binary search for the answer with low=1 and high = 5*k, and in checker function i used Digit DP which tells me how many numbers are there which has digit as '4'. According to me in worst case no of operations : 100 (for binary search)* 50(checker fn Digit DP)*10000 (t) = 10^7. Am I missing out something or is there any way this approach can be optimized. Thanks

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Just change endl to '\n'.

19 months ago, # ^ |

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adityasah09

19 months ago, # ^ |

Thank you so much for the reply. But why that is happening I am curious to know

ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);

Isn't it this piece of code takes care of that. Sorry if I am bothering you Thanks

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fuwad

19 months ago, # ^ |

std::endl prints a new line and also flushes the output stream, where '\n' just prints a new line, that is the difference as far as I know.( which is why endl takes more time comparably)

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Abito

19 months ago, # ^ |

Define $$$f(x)$$$ as the number of integers that don't have 4 in them. Now for each power $$$i \le 15$$$ we can find $$$f(10^i)$$$ which is $$$9^i$$$. Now for any other integer $$$x$$$ we can find $$$f(x)$$$ by going through each digit $$$i$$$ of value $$$x_i$$$ and adding to the answer $$$f(10^i)*x_i$$$ if $$$x_i \le 3$$$ or else adding $$$f(10^i)*(x_i -1)$$$. Now we can do binary search to find the answer.

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Wolowitz

19 months ago, # ^ |

We can have 9 digits. So just check if the number can fit if my ith digit was k. Then subtract the weight of it. Do this for every possible digit

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SilverWing05

19 months ago, # ^ |

https://math.stackexchange.com/questions/3958169/find-nth-number-which-does-not-contain-the-digit-c

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19 months ago, # ^ |

Use digit dp too but got hacked. The time is too tight for this approach. :(

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19 months ago, # ^ |

what was your complexity? I used digit dp for E. Time complexity was $$$O(logMX*20)$$$ per test case Where $$$MX=1e18$$$

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19 months ago, # ^ |

I set Mx to 5e12

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19 months ago, # ^ |

the $$$MX$$$ refers to the upper bound of the binary search... how did you pass samples with setting it to $$$5e12$$$?

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19 months ago, # ^ |

When k = 1e12 the answer is about 3e12, so I set it to 5e12

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19 months ago, # ^ |

oh mb, I calculated wrongly lol.

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19 months ago, # ^ |

I find a litte mistake which leads to a big constant. It's my fault:(

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https://codeforces.me/contest/1811/submission/200755771

19 months ago, # ^ |

oh :(

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Hmys

19 months ago, # ^ |

You can look at my submission. I just thought of the elements of the sequence as base 9 numbers, and then converted back. Very short code. Though I could see it being confusing (for non IMs haha) if you didn't think of that way of viewing the problem.

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tanmaytaneja2

19 months ago, # |

khatam

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Wolowitz

19 months ago, # |

← Rev. 2 →

-22

[.]

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serenhade00

19 months ago, # |

How to solve C?

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M_A_Noman

19 months ago, # ^ |

← Rev. 2 →

Just check the minimum of the consecutive elements. The first and last element will be as it is. For example

3 4 4 5 3 will remain as it is then the sequence will be min(3,4),min(4,4),min(4,5) and the last element 5 will be as it is

so the ans will be 3,3,4,4,5

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asr_003

19 months ago, # ^ |

← Rev. 2 →

what is the logic behind this?

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codefforces

19 months ago, # ^ |

← Rev. 2 →

$$$min(3,4)$$$ is the largest value of $$$a_2$$$ that you can pick (otherwise $$$b_1$$$ will become more than 3, a contradiction). It turns out this suffices as well (after copying the edge values).

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bad_imagination

19 months ago, # |

G1 was much easier than C/D/F

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Pensive

19 months ago, # |

← Rev. 3 →

+24

This round was pretty interesting for me. Found C much harder than D. Although I did copy E after googling from here, had an insane last minute adrenaline rush trying to guess/brute-force necessary conditions for flower graph in F!

Thanks for the interesting problems — will go and prove F now :blobsweat: Update: hacked lmao

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rohanmitra2420

19 months ago, # ^ |

same, had to google for E as well

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19 months ago, # ^ |

-24

kill yourselves

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19 months ago, # ^ |

i stand by what i said

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0gnjen1

19 months ago, # ^ |

Woah edgy boy

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19 months ago, # ^ |

vužgal bum te ognjene

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19 months ago, # |

← Rev. 2 →

+13

chromate00 Orz for the great performance!

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chromate00

19 months ago, # ^ |

+11

Thank you! The contest was very fun (Thinking about G was not, but anyways, it was fun)

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down_07

19 months ago, # ^ |

+10

Will you make miku's hair blue again after you reach expert?

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chromate00

19 months ago, # ^ |

yes actually I have saved the blue version for that

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19 months ago, # ^ |

Having saved the blue version, you should start thinking about the purple one :)

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19 months ago, # |

← Rev. 4 →

-6

I dont think problem D is a good problem that if you know the conclusion fn*fn+1=f0*f0+f1*f1+...+fn*fn, it will be very easy to solve it, otherwise it's very difficult. Although it's a comman conclusion.

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djay24

19 months ago, # ^ |

Can you prove this conclusion?

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19 months ago, # ^ |

Here it is

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djay24

19 months ago, # ^ |

Thanks, did you solved D? Approach?

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19 months ago, # ^ |

← Rev. 2 →

Even after getting to know about this during contest, i couldn't. Like removing $$$(x, y)$$$ from grid creates some regions and i was getting confused in how to divide them such that we get fibonacci squares and that too distinct.

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19 months ago, # ^ |

← Rev. 2 →

Since you konw the conclusion then every time you must cut off the biggest square,otherwise you cannont cut it into only n+1 sqaures. Then we can solve it recursively

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19 months ago, # ^ |

← Rev. 2 →

Common conclusion ? Ahh you are from china, never mind

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JEE_MAIN_LOOSER

19 months ago, # ^ |

Nope. It's not common Sir up until you aren't from china.

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Hmys

19 months ago, # ^ |

I think if you draw it up, and look at some examples it is not so bad. Basically if x can be written as sum of distinct even-numbred fibonacci numbers, and y odd ones (or opposite depending on parity of n), you get the answer. And its not that difficult to see if you look at the sample images they provided, and draw some bigger examples yourself, and try to see how the single square can move around.

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6_month_challenge

19 months ago, # |

← Rev. 5 →

Why my code tle at test 24 in G1, i think it's pretty decent enough to pass, just running 2 dp's one for maximum length and second one to calculate answers. Please help submission:200780428

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AhmedBeruny

19 months ago, # |

for problem C, How can you get output for this test case 0 1 2 1 0 ? the described constrains were incomplete I think.

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M_A_Noman

19 months ago, # ^ |

I think there will be no such case as the setter says array b is created from the array a by taking the maximum element between two consecutive elements.

So array b will be in increasing order always.

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electric_boogaloo

19 months ago, # ^ |

Not necessarily true, b doesn't have to be in increasing order as the first test contains such examples as [2, 2, 1], [0, 3, 4, 4, 3] and so on. But 0 1 2 1 0 isn't a valid input for our problem either, closest we can get is: [0, 1, 2, 2, 1, 0] from 0 0 1 2 1 0 0.

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Zuleen_2061

19 months ago, # |

This is probably my worst-ever performance in a div 3 contest, never felt more let down in my coding...lol!

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https://codeforcer.blogspot.com/2023/04/problem-863a-insert-digit-full-detailed.html

19 months ago, # |

I'm curious about there will be how many FST of problem F.

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eternal_happiness

19 months ago, # |

Problem E editorial: https://oeis.org/A052406

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deepaksolution

19 months ago, # |

← Rev. 3 →

-44

Problem :- 863A Insert Digits Solution is here with explanation and code of it

Problem :- 863C Restore the array Solution is here with explanation and code of it

https://codeforcer.blogspot.com/2023/04/problem-863c-restore-array-full.html

Problem :- 863D Umka and a Long Flight Solution is here with explanation and code of it

https://codeforcer.blogspot.com/2023/04/problem-863d-umka-and-long-flight-full.html

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kanishk779

19 months ago, # ^ |

+12

Where is the full explanation in your blogs? Do not misguide other participants.

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cosenza

19 months ago, # ^ |

He's just copying YocyCraft's explanations

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Nahida__

19 months ago, # |

E maybe little tricky,but others are interestring. And feel the difficulty of D than that of E.

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Nahida__

19 months ago, # |

E maybe little tricky,but others are interestring. And I feel the difficulty of D than that of E.

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kmparvez

19 months ago, # |

In problem statement of D. Umka and a Long Flight it states,"A checkered rectangle with a height of Fn and a width of Fn+1" means width always greater then height. Then, how can y be y > Fn at test case 3 (3 1 4 , output: YES)? n=3 x & y should x<=5 and y<=3. what did I miss?

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djay24

19 months ago, # ^ |

1 <= x <= Fn, 1 <= y <= Fn+1

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kmparvez

19 months ago, # ^ |

← Rev. 3 →

I got it. Thanks for your reply.

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kanishk779

19 months ago, # |

Can someone provide an unofficial editorial for problem C?

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eternal_happiness

19 months ago, # ^ |

Obviously, ans[1] = b[1] and ans[n] equals b[n-1], we can fill the rest by taking min(b[i], b[i+1]), because then, max(a[i], a[i+1]) = b[i] will hold

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minhle27

19 months ago, # ^ |

lol this is how you prove it?

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M_A_Noman

19 months ago, # ^ |

first and the last element will be in their place as it. Other element will be min(a[i],a[i+1]) for all i=0 to n-1

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testmayank123

19 months ago, # ^ |

what about this test case 0 1 2 1 0

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19 months ago, # ^ |

It's an invalid input. Consider for 1 2 1. If 1=max(a, b), 2=max(b, c), 1=max(c, d), then b==2 or c==2 must hold. If b==2, then 1=max(a, b)>=b==2, or if c==2, then 1=max(c, d)>=c==2. Both of them will have contradiction.

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Leftist_G

19 months ago, # |

Well, E is really tricky and F is a little trash.

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Leftist_G

19 months ago, # ^ |

BTW, I think BD is nice.

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Chimpanzee

19 months ago, # |

Nice problemset. I like task D

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19 months ago, # |

E is classical and F is kind of bad.

BCD are nice problems.

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19 months ago, # ^ |

B is the worst

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19 months ago, # |

The person who came up with problem b should be beheaded!!!

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Raks_splunk

19 months ago, # ^ |

Bro relax lmao(I totally agree).

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19 months ago, # ^ |

Tbh, B ruined my contest ... I am pretty bad at grid and these pattern problems. Had to move to C, which felt quite confusing and then somehow i got a simple solution to work. In the end even after solving E, it took me 25 mins to solve B.

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19 months ago, # ^ |

Well I solved A, B, C, D but because of the wasted time on B I couldn't solve E in time.

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19 months ago, # ^ |

why

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19 months ago, # ^ |

my solution to B:

il int find(int x, int y) {
	if (x > n / 2) x = n + 1 - x;
	if (y > n / 2) y = n + 1 - y;
	if (y <= x) return y;
	return x;
}

int main() {
	for(int T= read(); T--; ) {
		int x, y, xx, yy;
		read(n, x, y, xx, yy);
		printf("%d\n", abs(find(x, y) - find(xx, yy)));
	}
	rout;
}

it's quite easy!!

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19 months ago, # ^ |

you can shorten your find function to

int find(int x, int y, int n){
    return min({x,n-x+1,y,n-y+1});
}

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19 months ago, # ^ |

Wow that's nice

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19 months ago, # |

← Rev. 5 →

+21

A: Find the first digit less than d and insert d before it. If there's no such digit, insert at the end.

B: We need to find there are how many cycles between start and end. Let's number cycles from outside to inside 0, 1, 2, ... then (x, y) is contained in cycle min(x-1, n-x, y-1, n-y). Then the answer is the difference of cycle number of start and end.

C: b[1], min(b[1], b[2]), min(b[2], b[3]), ..., min(b[n-2], b[n-1]), b[n-1] is an answer.

D: If n<=1 answer is true. Otherwise, let h=fib[n], w=fib[n+1], we must cut a h*h square from left or right. If we cut from left and (x, y) is remained in the right part, which is a fibonacci ractangle of order n-1, we can solve the problem recursively. Similar for we cut from right and (x, y) is remained in the left part. If no matter we cut the square from left or right, (x, y) can not be remained then answer is false.

E: Represent k in base-9, and add 1 to digits >=4.

F: First for a k-flower, number of nodes n=k*k and number of edges m=k*(k+1). Then there must be k nodes with degree 4, and other nodes with degree 2. If so, 4-nodes must form a cycle of size k (we can take all edges between 4-nodes and check if they're a cycle), and all other edges must form k cycles of size k, and each cycle contains one 4-node and k-1 2-nodes.

G: First calculate prefix-sums of occurences of each number in range [1, n]. We first run dp for max_size[i]=(max size of nice path end at i). Let max_size[0]=0, and for max_size[i], we check for each 0<=j<i, if there're at least k occurences of c[i] in range [j+1, i], we update max_size[i] with max_size[j]+k. Then we let dp[i]=(count of nice paths of size max_size[i] end at i). Let dp[0]=1 and dp[i]=0 (1<=i<=n) at first. Then for each j<i, if max_size[i]-k==max_size[j], we have dp[j] ways to choose first max_size[i]-k elements, and we have C(cnt, k-1) ways to choose other k elements (here cnt is count of occurences of c[i] in range [j+1, i-1], we can calculate it by prefix-sums).

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Chimpanzee

19 months ago, # ^ |

+29

Your solution for E is very cute

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puspendra_09

19 months ago, # ^ |

Can you explain your intuition for problem E? Thanks also just want to know if it is some type of standard problem.

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19 months ago, # ^ |

Because digit 4 is forbidden, we only have 9 available digits, so remained numbers are some modified version of base-9 numbers(012345678 --> 012356789).

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Ian_Nepomniachtchi

19 months ago, # ^ |

+16

I am just curious that suppose if there would have been two numbers forbidden then would the same approach have worked...like base-8 numbers. and if yes...then can you please tell how?

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Omar.Gawdat

19 months ago, # ^ |

in B can you explain how did you deduce that the cycle is min(x-1, n-x, y-1, n-y).

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19 months ago, # ^ |

It's the minimum distance from a cell to the boundary of square.

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NerfThis

19 months ago, # ^ |

For problem F, can you explain how you come to the conclusion that there must be k nodes with degree 4, and other nodes with degree 2 ? Are we assuming the entire graph is connected?

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19 months ago, # ^ |

That's by the defination of k-flower.

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NerfThis

19 months ago, # ^ |

I see. I misunderstood the statement of F and tried to solve a different problem.... Thanks!

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ayush_n

19 months ago, # |

How to solve Problem D

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becaido

19 months ago, # |

← Rev. 2 →

+55

Is the validator of problem A wrong?

It says $$$1\leq n\leq 2\times 10^5$$$ in the problem statement, but when I hacked with n = 2e5, it showed

" Validator 'validator.exe' returns exit code 3 [FAIL Integer parameter [name=n] equals to 200000, ...e range [1, 10000] (test case 1, stdin, line 2)] ".

Also, I think the pretests are weak too, $$$O(n^2)$$$ and $$$O(n^2\times\log(n))$$$ solution can pass.

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19 months ago, # ^ |

you'd better show your generator.

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K-423

19 months ago, # |

← Rev. 2 →

+93

What the fucking description of F!

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19 months ago, # ^ |

It's not thaat bad tbh)

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K-423

19 months ago, # ^ |

← Rev. 2 →

+44

most of my friends took 10-20 minutes to understand what k-flower actually looks like.

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Chimpanzee

19 months ago, # ^ |

-34

then most of your friends aren't very bright

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K-423

19 months ago, # ^ |

if you think so, whatever, you are right

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https://www.geeksforgeeks.org/finding-the-nth-term-in-a-sequence-formed-by-removing-digit-k-from-natural-numbers/

19 months ago, # ^ |

I thought I should determine whether there is a flower in the given graph...

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L1gma

19 months ago, # |

Problem E has striking similarity with above gfg article. Maximum users have googled it and copied the snippet of this article as it is without any changes. Please try to filter out those code solutions while checking for plagiarism. MikeMirzayanov

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codeX_pro

19 months ago, # ^ |

-18

Wolfester 200768712 Solution can be seen to copy exact helping functions from the above mentioned article. This is a serious concern for the integrity of contest. Pls ban this user from codeforces

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rohanmitra2420

19 months ago, # ^ |

+22

3rd party code posted before contest is allowed

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vaibhavsinghz

19 months ago, # ^ |

So, will these questions be taken into consideration for plagiarism?

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Valenz

19 months ago, # ^ |

Problem E? I dont think so. Div 3 is usually inteded to be more classical, so expect some similar problem ideas.

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Wolfester

19 months ago, # ^ |

← Rev. 2 →

This doesn't come under cheating. Using third-party code available before the start of contest is allowed.

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anosh21

19 months ago, # |

Catch Cheaters https://codeforces.me/blog/anosh21

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_Hridoy

19 months ago, # |

+20

F is quiet confusing.

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19 months ago, # |

-21

make it unrated

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harshbutani01

19 months ago, # |

Can someone provide a counter example for this submission?

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vgtcross

19 months ago, # ^ |

← Rev. 2 →

Input:
1
7
0 1 2 2 1 0

Correct Output (for example):
0 0 1 2 1 0 0

Your Output:
0 0 1 2 0 0 0

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harshbutani01

19 months ago, # ^ |

Thanks for the counter example!

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ASHWANTH_K

19 months ago, # |

there is some issue with testcases of A problem.
N <= 2e5
but in hacking phase, i could not generate testcases where N > 10000

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Nxxlt

19 months ago, # |

+17

Div.(2+1e-6) round.

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procoderomk

19 months ago, # |

One other way to do E is my considering 9 based numbers instead of 10 based(decimal) numbers. Convert k to a 9 based number and change all 4's to 5, 5's to 6, 6's to 7, 7's to 8 and 8's to 9

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sundaram__anand

19 months ago, # ^ |

Thanks you StackExchange !!!Hats off for working this out yourself.

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jeal0uspengu1n

19 months ago, # |

I was trying to hack A, but getting the following error!

Validator 'validator.exe' returns exit code 3 [FAIL Integer parameter [name=n] equals to 200000, ...e range [1, 10000] (test case 1, stdin, line 2)]

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ASHWANTH_K

19 months ago, # ^ |

same issue i am facing

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raghavmangla04

19 months ago, # |

-12

i guess this contest should be unrated as solution to problem E was available online and a lot of people just straight away copied the code!! Link:https://stackoverflow.com/questions/54851335/program-to-compute-n-th-number-that-doesnt-contain-given-digit

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iiand

19 months ago, # ^ |

Come on, it's Div3. There must be several easy problems in such contests, so for almost every easy problem you can find a " coincident problem "

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YENGOYAN

19 months ago, # |

I think there is a mistake in the G2 problem's tests ( 29th test ), in my opinion the answer is at least 1, but the right output for the 29th test is 0. Can anyone explain how is it possible? My submission

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nathan_luiz

19 months ago, # ^ |

The answer is probably divisible by $$$10^9 + 7$$$.

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Nokemono

19 months ago, # ^ |

← Rev. 3 →

I made the same mistake and it passed G1. Can you try hacking me with the same test case ?

My Submission

UPD: The test-case does not fit the constraints of G1 so a test-case with answer divisible by $$$10_{}^{9} + 7$$$ which satisfies $$$1 \leq k \leq n \leq 100$$$ is needed

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cockatooo

19 months ago, # |

← Rev. 4 →

An alternative solution of C, which might be easier to come up with (compared to the $$$(b_1,min(b_1,b_2)...)$$$ solution):

First note that if $$$b_i > b_{i+1}$$$, we must have $$$a_i = b_i$$$, otherwise if $$$a_{i+1} = b_i$$$ then $$$b_{i+1} \geq b_i$$$. Similarly, if $$$b_i < b_{i+1}$$$, then we must have $$$a_{i+2} = b_{i+1}$$$.

After doing one pass and dealing with these cases, there may still be some $$$a_i$$$ that are left assigned. Now we check whether for each $$$b_i$$$, at least one of $$$a_i$$$ and $$$a_{i+1}$$$ is equal to $$$b_{i}$$$. If not, we assign them. If we traverse in increasing i, we assign $$$b_i$$$ to $$$a_i$$$ first, and if $$$a_i$$$ is already filled we assign to $$$a_{i+1}$$$. In optimization terms, we make sure all constriant are active.

Finally fill the rest with 0.

This always works if the input is correct.

solution: https://codeforces.me/contest/1811/submission/200724266

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ID3

19 months ago, # |

+11

So, problems B and D ruined the contest, right?

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aqua4

19 months ago, # |

Are submissions going to be re-tested with successful hacks?

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jackylova_fan_fan_fan

19 months ago, # ^ |

Yes

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jackylova_fan_fan_fan

19 months ago, # |

So hard div3，(sad).

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