Make it unrated

2 years ago, # |

Finally a contest on the weekend that I can do!

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2 years ago, # |

← Rev. 2 →

Hope For a Big +ve delta to everybody in the contest

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Ghulam_Junaid

2 years ago, # ^ |

-7

That's not actually possible because rating is not calculated like this, let's suppose everybody solved all problems then its not that everybody will have +ve delta, delta is based on the people which have higher rating than you but performed less than you, then you will have a +ve delta...like sort the registered participants by rating and then see on which number you are coming if your rank in the contest is more than that then you will have +ve delta (assuming everybody who has registered participates).

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Satrpx

2 years ago, # |

← Rev. 2 →

+16

A round with no tester comments! That's pretty astonishing. Edit: Now there are some.

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coderdhanraj

2 years ago, # ^ |

← Rev. 2 →

Many users(including me) would be sad as the last round had too many FSTs and Hacks due to bad testing in problem B and C! :(

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2 years ago, # ^ |

hope good pretests this time, lost too much points cus of fst

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_Gawd_

2 years ago, # ^ |

What is FST?

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coderdhanraj

2 years ago, # ^ |

Failed System Testing

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triple__a

2 years ago, # |

+61

As a tester, this contest has a couple of nice problems, and the overall quality is much better than average!

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TimDee

2 years ago, # ^ |

So can you guarantee that I will go to BOI remaining purple?

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Satrpx

2 years ago, # ^ |

← Rev. 3 →

-17

[Deleted]

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dwight_schrute15

2 years ago, # ^ |

do you use some extension? How there is 3 extra columns in the end ?

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xypnx

2 years ago, # ^ |

yea, i cant find that extension, can u pls share the name

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Aphrim

2 years ago, # ^ |

It is 'Carrot'

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majiamajia66

2 years ago, # |

As a tester, I must say that the problems are really great and I highly recommend everyone to come.

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ffao

2 years ago, # ^ |

+106

Press X to Doubt

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Abdullah_Schneider

2 years ago, # |

This gonna be my first contest ever, I'm so excited

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ArashJafariyan

2 years ago, # |

← Rev. 2 →

Hopefully i dont get destroyed by the problems like the last contest :*).

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Aphrim

2 years ago, # |

Hoping for +141 delta

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Jhin

2 years ago, # |

← Rev. 2 →

-9

meme

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Faizal

2 years ago, # |

My first competition

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leonlian

2 years ago, # |

Good Luck Everyone!!! Hope everyone has a fun time :)

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rushatonight

2 years ago, # |

+39

You could find someone to teach you if you don't know how to write a problem description, and about what information shuold be emphasized.

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poremax738

2 years ago, # |

-32

Solution for problem D has been leaked on a discord group by a user named Instrospection. Solution is in Python

ATTENTION, please look into it coordinators and MikeMirzayanov

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2 years ago, # ^ |

but how did you find that out

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Bo0tstrap

2 years ago, # ^ |

ahahhahahahha he was searching for it

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ArashJafariyan

2 years ago, # ^ |

+21

i don't think you should comment this under the blog (where everyone can see it) in the middle of the contest :/

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2 years ago, # ^ |

Strongly agree with you.

Still, he couldn't have known if it was leaked on a "discord group" if he wasn't in that group.

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AnotherTwinkle

2 years ago, # |

The problem statements were shit.

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Shihab2

2 years ago, # ^ |

-10

D is a good problem, but the problem statements are very bad.

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Faizal

2 years ago, # ^ |

I could not understand problem B at all

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0htoAi

2 years ago, # |

I spent too much time understanding the question, which is unfair to foreigners like me. Maybe my English is too poor.

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2 years ago, # |

← Rev. 6 →

+17

I appreciate illustrations to the problem statements to make them more understandable. Without them it might've been crazy

A — 24mins, still don't know what it was about, just submitted the obvious formulae from the samples after giving up on thinking.
B — 16mins,
C — 31mins, I feel like my implementation is crap but the task itself is quite decent.
D — 24mins (just like A). Surprisingly easy. I enjoyed this kind of problem but wonder if it is too straightforward. Guess could've solved it even faster.

So to me C>A>D>B . But from the number of submissions the round seems well balanced in general except E which has a very few submissions but it is fine for most people I guess.

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2 years ago, # ^ |

how do you do D? Why is it easy?

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2 years ago, # ^ |

I guess as always for me I either think in the right direction or not. And here I was lucky.

First think about how to find all triplets. Notice that two cards uniquely identify the third one. So you can calculate all the triplets for n*n*k. After that it becomes a basic combinatorics

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balalaika

2 years ago, # ^ |

1) By two cards you can determine third to be in set
2) So if two cards determine the third, then there cannot be two sets with strictly two common cards
3) It means that any meta-set is two sets with one common card
4) This gives O(n^2 * k) solution (store all meta-sets by common card and some combinatorics)

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TheBhediyaOfDalalStreet

2 years ago, # ^ |

-8

lol idk why so many ppl have such weird opinions on the problem set. Sure, the statements were unclear (especially for B, where it was blatantly incomplete). Regardless, I think the ideas required for the first 3 problems were what you would expect in a Div2

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2 years ago, # ^ |

ok then maybe ur better than me then. I find it's super challenging to think (even if the problem is easy) after spending 90% of brain stamina to understand the statements.

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TheBhediyaOfDalalStreet

2 years ago, # ^ |

-9

I wasn't replying to you. But I can tell you for sure that the English used for A was perfect, not sure about B. It's just that A is a difficult to understand problem in itself, maybe they could have reworded it better to make it sound less mathematically rigorous?

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2 years ago, # ^ |

← Rev. 2 →

k. Ur an english genius then.

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DisconnectedGraph

2 years ago, # |

← Rev. 3 →

-17

The gap between D and E is too big.
And WHY TIMELIMIT OF D is FOUR SECOND???
Even the most brute force $$$\mathcal{O}(n^3k)$$$ can pass??? Why???
(Submission 174397296 and My hack)
Just why 4 sec??? Is there a standard solution that can`t run in 2 sec???

a very bad round for me. not worthy of the time.

wow upvotes are dropping very fast:) expected it to happen:)

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2 years ago, # ^ |

Wuf. Even my python solution passed for half a second

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KikyouKushida

2 years ago, # ^ |

If $$$O(n^3k)$$$ can pass then it is the fault of the weak pretests not the loose time limit.

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2 years ago, # ^ |

I'm actually very surprised that brute force would pass, even with the generous 4 second time limit, since it's like, $$$10^{10}$$$ operations. My guess is that the pretests are actually not too strong, and we might see TLE hacks and/or FST on $$$O(n^3k)$$$ approaches.

I think the standard approach would take $$$O(n^2 k \log n)$$$ time (find sum of all pairs, store them in a map; then for each card, count how many instances of its additive inverse is present as a pair sum and add the count choose 2 to the answer). Mine took 623ms, so it should comfortably fit 4 seconds with slower languages or approaches with poorer constant factors (though I don't think mine was anywhere near optimized to begin with). Maybe they wanted to disallow unordered map hacks?

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Satrpx

2 years ago, # ^ |

n^3*k passed main tests...

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wangxiwen

2 years ago, # |

+10

I think the problems is really good and I enjoy it, but the descriptions of them is so unclear that I waste lots of time. Hope the next time the words will be much clearer.

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2 years ago, # |

did u have to do graphs to detect cycle in C?

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2 years ago, # ^ |

-13

well, it's just one cycle so you can check indegree and outdegree but I was too lazy so I used a DSU xd

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2 years ago, # ^ |

can u show me ur submission by pasting it here or somewhere else with a link? i wanna see but submission are not viewable rn

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2 years ago, # ^ |

-18

Here you go

#include<iostream>
#include<algorithm>
#include<set>
#include<functional>
using namespace std;

int main()
{
	int q;cin>>q;
	while(q--)
	{
		string db(256,0);
		string dbr(256,0);
		int dsu[256];
		for(int i=0;i<256;i++)dsu[i]=i;
		int n;cin>>n;
		string s;cin>>s;
		
		function<int(int)>find=
		[&](int k){if(k==dsu[k])return k;return dsu[k]=find(dsu[k]);};
		
		function<void(int,int)>merge=[&](int a,int b)
		{a=find(a);b=find(b);if(a!=b)dsu[b]=a;};
		
		
		string unq;
		set<char>cs;
		for(char&c:s)if(!cs.count(c)){cs.insert(c);unq+=c;}
		
		for(char&c:unq)
		{
			for(char t='a';t<='z';t++)
			{
				if(c==t)continue;
				if((find(c)!=find(t)||
				(size(unq)==26&&unq[25]==c))
				&&dbr[t]==0&&db[t]!=c)
				{
					db[c]=t;
					dbr[t]=c;
					merge(c,t);
					break;
				}
			}
		}
		
		for(char&c:s)
		{
			cout<<db[c];
		}
		cout<<"\n";
	}
}

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https://codeforces.net/contest/1735/submission/174413223

2 years ago, # ^ |

thanks

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Termii

2 years ago, # ^ |

No graph. I used a Map to track which character corresponds to which character (obviously empty in the beginning).

Filling that map is the crux of the problem. I used a List-array of size 26 to store all connections.

Example

String: bafdcesigh

Lists:

init
[[a], [b], [c], [d], [e], [f], [g], [h], [i], [j], ..] 

connect b to a
[[a, b], [], [c], [d], [e], [f], [g], [h], [i], [j], ..]

connect a to c
[[], [], [c, a, b], [d], [e], [f], [g], [h], [i], [j], ..]

connect f to b
[[], [], [c, a, b, f], [d], [e], [], [g], [h], [i], [j], ..]

connect d to e
[[], [], [c, a, b, f], [], [e, d], [], [g], [h], [i], [j], ..]

connect c to d
[[], [], [], [], [e, d, c, a, b, f], [], [g], [h], [i], [j], ..]

connect e to g
[[], [], [], [], [], [], [g, e, d, c, a, b, f], [h], [i], [j], ..]

connect s to f
[[], [], [], [], [], [], [g, e, d, c, a, b, f, s], [h], [i], [j], ..]

connect i to h
[[], [], [], [], [], [], [g, e, d, c, a, b, f, s], [h, i], [], [j], ..]

connect g to i
[[], [], [], [], [], [], [], [h, i, g, e, d, c, a, b, f, s], [], [j], ..]

connect j to h
[[], [], [], [], [], [], [], [], [], [j, h, i, g, e, d, c, a, b, f, s], ..]

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2 years ago, # ^ |

thanks, thats a really smart method

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2 years ago, # ^ |

I used DSU with path compression. I was about to implement union by rank, when I recalled that it's only size 26, which is definitely not worth that much effort.

flakes24 it should be enough to just store a parent array or even a map to indicate which character precedes another. Then when you want to make a new mapping, you can simply check all ancestors to make sure this new successor doesn't form a cycle, unless you already have a chain of all 26 letters, so the only remaining relation is to complete the cycle. I didn't actually do this (I used DSU instead), but I think this should still pass the time limit, provided that the rest of your algorithm is efficient enough.

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2 years ago, # ^ |

I thought about doing the second method (cycle method also striked but im not that comfortable with graphs yet), but it felt too implementation heavy to check 26 ancestors. But i had about 110 minutes after doing B and should have given it a try atleast. Thanks for the help tho.

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2 years ago, # ^ |

btw, do u know any good resource to learn dsus in this context?

i dont know why youre getting downvoted tho

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2 years ago, # ^ |

-18

I just went to cp-algorithms during the competition for this

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2 years ago, # ^ |

← Rev. 2 →

can you link that particular page / piece of code please

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2 years ago, # ^ |

-18

Here you go

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2 years ago, # |

+13

5/10 for problem quality.

1/10 for problem description.

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2 years ago, # |

Solved D 30 seconds before the end of the contest b/c I had thought my sol was incorrect.

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Termii

2 years ago, # |

← Rev. 2 →

I did not like A at all. I liked B a lot. The hardest about C and D were understanding the problem statements. I am grateful the testcases were good, else I would not have understood anything.

The difficulty felt nice, except for A. I think A was too strange, took me 30min to write 1 correct line of code. Solved A, B, C and D

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Alex.Svanidze

2 years ago, # |

-12

Upvotes on this blog is gonna drop faster than my rating

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OccDreamer

2 years ago, # |

The problems is really good but the descriptions of these problems are not clear.

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Kennybench

2 years ago, # |

+10

got a stroke reading/trying to understand problem C

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Furukawa

2 years ago, # |

+11

E is a bit boring, because there are too many details to pay attention to.

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adityagamer

2 years ago, # |

+34

In problem B, " You want that for each pair of pieces, their sizes differ strictly less than twice. "

Twice of what?

In problem C, " each letter in s was replaced with the one that follows in clockwise order ",

Follows what in clockwise order? should have been "replaced with one that follows it in circle in clockwise order"

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yashdabhade283

2 years ago, # ^ |

Exactly ! Not specified clearly

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2 years ago, # ^ |

Nice comment!

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2 years ago, # ^ |

Twice of what?

I think it is pretty clear the size of two items differ strictly less than twice means the bigger of the two is less the two times bigger than the smaller.

Idk

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2 years ago, # ^ |

It definitely was not clear. My initial interpretation was that if you compare every possible pair, you would find strictly less than two instances in which the the result was that the elements were different. Of course, this would imply that all elements must be equal unless there are $$$\leq 2$$$ elements, so I was very confused at how simple this was, until I saw the test cases.

The visual explanation helped clear up my mistake, but it doesn't change the fact that the initial wording was very poor. Thankfully, a clarification was issued shortly at the start, so they at least responded properly to inquiries.

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2 years ago, # |

← Rev. 2 →

I think C is easy for people who knows graphs (dfs) and can write very hard code

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haochenkang

2 years ago, # ^ |

+47

I think all questions are easy for people who know how to solve it

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2 years ago, # ^ |

You don't need graphs and dfs for C

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2 years ago, # ^ |

Yes, but i don't say that graphs is only solution

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2 years ago, # ^ |

Also come on. Stop mentioning dfs as some advanced technique that few people know

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2 years ago, # ^ |

Yes, a while loop will do.

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haochenkang

2 years ago, # |

All the problems were insanely good, and I especially liked C. One of the most balanced + well prepared round I have ever participated in. Congrats to authors!

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Gary2005

2 years ago, # |

+41

Problem F is a good problem!

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yukai

2 years ago, # |

← Rev. 2 →

I misread the shit problem statements many times. My entry experience was extremely bad.

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Codula

2 years ago, # |

+25

Is there anybody else who struggled in understanding problem statements today?

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2 years ago, # ^ |

I struggled a lot

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Ivan_len

2 years ago, # |

+35

Very confusing description... Had to guess the meaning based on given test cases for the first few problems :-)

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KAN

2 years ago, # ^ |

Can you specify what exactly was not clear?

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Ivan_len

2 years ago, # ^ |

Perhaps "very confusing" is a bit over, but I think I do need to carefully look over every word in the description to understand them, while in previous contests just skimming over is enough. For example, "For this reason, all 26 lowercase English letters were arranged in a circle in some order, afterwards, each letter in s was replaced with the one that follows in clockwise order, in that way the string t was obtained." It seems unclear to me what each character in s is replaced with, and I had to go over the TCs to make sure what I am thinking is correct.

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adityagamer

2 years ago, # |

How to solve D?

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Phoenix_8

2 years ago, # ^ |

← Rev. 2 →

Ok, this was the first time I solved D, and to do so I broke this problem into different parts, so the first significant observation was no 2 sets can have 2 same cards, as it will force the third card to be same and hence 2 sets to be same. So let's say card 1 was in 6 sets then I can combine any 2 sets and that will be a meta-set, so the total meta-set would be 6C2 (choose any 2 sets among all possible), so, if I am able to find in how many sets each card is in then I can sum all their results (cnt[i]C2). To find a set, for each pair of cards I found out which was the required card so that they form a set (this card will always be unique) and checked if that card existed in our n cards, to do so in log(n) I stored each card in the map. so total time complexity was O(n*n*k*log(n)).

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simp_pro

2 years ago, # |

-96

Make it unrated

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2 years ago, # ^ |

Why? That's all legal. Statements hard to understand but you can understand

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Boring_Day

2 years ago, # ^ |

This round must be unrated because we can't understand the problems. So without understanding we can't solve.

some people are not good in english so you better use simple words/sentences.

"their sizes differ strictly less than twice." this is shit.

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2 years ago, # ^ |

1.Read carefully 2.Watch at examples and notes 3.Use translator

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2 years ago, # ^ |

I think, your problem isn't "understanding problem"

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Devo

2 years ago, # |

← Rev. 2 →

How much for a good problem statement, please? ,

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HLqwq

2 years ago, # |

why engish in this round so hard to understand qnq

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2 years ago, # |

← Rev. 2 →

-6

Is there anyone like me? When I'm confused about a problem statement, my brain is fried and can't think anymore for the rest of the contest.

I was super confused about A and D problem statements.

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beethoven97

2 years ago, # ^ |

I spend about 20 minutes in understanding problem A and decided to leave the contest.

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tandonjiii

2 years ago, # ^ |

I spend around 25 min in finding a approach with wrong submission , It is more logical than B

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Lol2004

2 years ago, # |

I quickly found a solution for E, but unfortunately, because i was stuck on C for a long time because of implementation issue, i had no time to implement E. Solved A,B,C,D

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wuhudsm

2 years ago, # |

Solution for D:

Firstly,let's find all "sets":

Enumerate each pair $$$(card[i],card[j])$$$,we can calculate a unique $$$card[k]$$$ according to $$$card[i],card[j]$$$.

Next, we only need to check whether $$$card[k]$$$ is in the given cards.

Time complexity:$$$O(n^2k)$$$.

Next, we notice that the intersection of two "sets" has at most one card.

$$$Proof$$$:assume the intersection of two "sets" has 2 cards $$$card[i],card[j]$$$,we can calculate a unique $$$card[k]$$$ according to $$$card[i],card[j]$$$,which leads to contradiction.

For each $$$card[i]$$$,suppose it exists in $$$k[i]$$$ sets,if $$$k[i] \geq 2$$$,we can select two from the $$$k[i]$$$ "sets" to form a "meta-set".

It can also be proved that all selected "meat-sets" are distinct.

So the answer is $$$\Sigma {C^{k[i]}_{2}}$$$.

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physics0523

2 years ago, # |

+37

D and E are amazing!
D ... All CPer who have played SET must think "From $$$2$$$ cards I can identify the last card" at least once, I guess.
E ... We must use matching to solve this? Nope.

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tfg

2 years ago, # ^ |

rainboy and others tried doing that and most got TLE.

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darya_navard

2 years ago, # |

good contest!!!

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DF_Pite

2 years ago, # ^ |

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Kath1031

2 years ago, # ^ |

dag wall how eye knee

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uchiha22

2 years ago, # |

Problem description left the chat.

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ArashJafariyan

2 years ago, # |

-10

i spent more time understanding problem statements than actually solving them.

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yash_15

2 years ago, # |

CF rounds helping us practice reading comprehension as well. nice.

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2 years ago, # ^ |

yes lol

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codefforces

2 years ago, # |

"their sizes differ strictly less than twice." I bet it'll be fun to solve the "once" version of it. Twice fun in fact.

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2 years ago, # ^ |

It was a weird way to write the statement ;/

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2 years ago, # |

← Rev. 2 →

I solved A by test case №3 i just divided 1033/3 and found genius formula.

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Termii

2 years ago, # |

-12

Problem D

English version: "A feature for three cards is called good if it is the same for these cards or pairwise distinct."

machine translation of Russian version: "For a trio of cards, a characteristic is good if it is the same for all three cards, or all three are different in pairs"

Sad that I prefer the machine translation...

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IamMuhammadHasib

2 years ago, # |

Can anyone please explain problem C in a clearer way?

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Baaaraa

2 years ago, # ^ |

← Rev. 2 →

in order to get the lexicographically smallest string you should go from left to right and try to map every character by the alphabetical order from a to z,consider it as graph where there is a directed edge from a character to the character it is mapped with, here you should take care of 3 cases

1 — you can't map a character to itself (loop edge)

2 — you can't make a cycle of length less than 26

3 — you can't map a character to an already mapped one

take a look at my code , i hope i made it clear.

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Prasahnt_kr

2 years ago, # ^ |

where is your code?

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Baaaraa

2 years ago, # ^ |

174420886

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IamMuhammadHasib

2 years ago, # ^ |

Many many thanks for explaining the solution approach, but I was more into problem statement, what actually the problem said?

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Baaaraa

2 years ago, # ^ |

← Rev. 2 →

simply you initially have a string s , to obtain the string t you make your own alphabetical order of the 26 letters such that each letter in string s is replaced by the letter the goes next to it in the alphabetical order you just made

for example if your s is aqerk and your 26-string order is something like au..qw..er..rz..ko , then your t-string is uwrzo

now you have string t , you should output the lexicographically smallest string s which could lead to string t by some mapping order

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dwight_schrute15

2 years ago, # |

+21

ReadingForces

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hy2004

2 years ago, # |

Very intresting!I tried my best and I think I will get a good result.I really love these problems.

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AliFarhat

2 years ago, # |

← Rev. 3 →

I really enjoyed today's contest. Although some of the problem descriptions werent the best, problem A B and C were very interesting and the pretests were very strong. keep it up!

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HusseinFarhat

2 years ago, # ^ |

Howard Hamlin of HHM, I sincerely agree to your opinion.

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kmparvez

2 years ago, # |

Problem description is like: Go A to B then go C then back to B, again back to A then go C finally print "Helo World".

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Lucky_Sanae

2 years ago, # |

Had a contests filled weekend and I had a great time

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2 years ago, # |

The questions were really good! Thanks for good questions :)

The solutions and proofs were all very beautiful(for the questions I have solved)
The problems were generally straightforward

My only possible critism is that this round was more about finding ideas....

But I really enjoyed it!

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Furukawa

2 years ago, # |

Anyway, a good contest!

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2 years ago, # ^ |

Agree with you :)

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shaswat97

2 years ago, # |

← Rev. 2 →

can someone please help me, where am I going wrong this is solution for B

            Arrays.sort(arr);
            int div = 2*arr[0]-1;

            for (int i=0; i<n; i++) {
                if (arr[i] >= 2*arr[0]) {
                    ans += (int)Math.ceil(1.0*arr[i]/div);
                }
            }
            System.out.println(ans);

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2 years ago, # ^ |

I think you need Math.ceil(1.0*arr[i]/div)-1 instead of Math.ceil(1.0*arr[i]/div) as dividing a number in n pieces will need n-1 operations

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2 years ago, # ^ |

← Rev. 3 →

Note: ans+=(arr[i]+div-1)/div-1; That is math

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midoriyashounen

2 years ago, # |

In problem D, I used unordered map to hash from vector of int to int and it took over 1 second to run which is longer than most submissions. Anyone know why?

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2 years ago, # ^ |

hash function could be slow, there could be hash collision in your hash function, or else there is somewhere else that adds/multiplies to the time complexity.

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midoriyashounen

2 years ago, # ^ |

Is there any better data structure to store the value for each card? Because editorial also suggests us to use map/hashmap.

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2 years ago, # ^ |

I think map/hashmap is good enough for the question, but generally I don't recommend you using unordered_map on codeforces, as it could be vulnerable to hacks by deliberately causing hash collisions.

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2 years ago, # ^ |

In the case of problem D, I think it is possible to hash the set without hash collisions using 3**k and long long int; If you used this as an hash value, I think map/hashmap is going to be near to best you can use.

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midoriyashounen

2 years ago, # ^ |

Yeah, I just tried your suggestion and it still took quite long so the problem might be my implementation. Anyway, thanks a lot for detailed explanation.

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sohelH

2 years ago, # |

I would like to thank:
"Aleks5d, arzhantsev64, ChthollyNotaSeniorious, DanWallgun, flowerletter, Gegege337, Imakf, LeoRiether, mejiamejia, okwedook, PO3OBAR_Bblnb, SSerxhs, TomiokapEace, triple__a, wabadabakalakaboo for testing the round and giving significant feedback."

I see that there are several testers. Do they only verify the accuracy of the judge data or do they proofread the problem descriptions as well?

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2 years ago, # |

Can Any Body explain the Solution of Problem B

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hy2004

2 years ago, # ^ |

174391163 if you see my code.I think you can understand this problem.

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2 years ago, # ^ |

← Rev. 2 →

It is no good to divide further the smallest number, as it will always result in more operations caused by having to divide up other numbers further. And therefore, we can transform the question to: given n numbers, make every numbers that are not the smallest number between the range of smallest number/2+1 ~ smallest number*2-1. Dividing each number so that all numbers are between the range will take ceil(number/(smallest number*2-1))-1 operations: and so the answer is the sum of it for all numbers.

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2 years ago, # ^ |

dhyang24 why you have taken ceil(number/((smallest number*2)-1))-1 rather than ceil(number/(smallest number*2))-1 this

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2 years ago, # ^ |

The question states that the biggest number should be strictly smaller than smallest number*2; the maximum number after the division must be (smallest number*2-1)

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2 years ago, # ^ |

It is guaranteed that if you divide a number in ceil(number/(smallest number*-1))-1 parts, when evenly distribued, it is in the range of smallest number/2+1 ~ smallest number*2 -1

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2 years ago, # ^ |

And can you tell why have take -1 after calculating ceil value

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2 years ago, # ^ |

dividing a number into n parts will take (n-1) operations, as each operation increases the total number of parts by 1, and the initial number of parts is 1.

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2 years ago, # ^ |

← Rev. 2 →

Ok I got it thanks for giving time to my Queries dhyang24 And Congrats for becoming Candidate Master

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Akulyat

2 years ago, # |

← Rev. 2 →

+53

I’m sorry that the English statements are confusing.
I tried to make them plane, but it was hard for some sentences.
We discussed the places you were confused by and these are the best formulation we eventually came to.
I would really like to see your suggestions for replacing them. I still can’t choose good options. I hope that next time I’ll be more careful and precise.

I hope you enjoyed the problem ideas after you got through the statements.
Anyway, don’t forget to check the editorial and feel free to ask me if there is something unclear there too.

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geniorzity

2 years ago, # |

Problems have pretty bad wording, otherwise it is fine I think

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thang315

2 years ago, # |

C, it correct when I tested but codef say... 174424830

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2 years ago, # ^ |

I don't understand your approach entirely (it would help if you commented what the different arrays actually meant), but it seems like you're missing the check for when all 26 letters are linked, in which case you would need to link the endpoints to form a cycle.

Instead, I think your approach would eventually try to extend beyond the 26 letters and/or try to access an array element that was not initialized, which can lead to different results. As such, even if you saw the correct answer on your own machine, there may be undefined behavior, which can lead to different results in other machines.

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mukulrawat_03

2 years ago, # |

-9

I need upvote anyone can help TT.

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geranazavr555

2 years ago, # |

+20

Ratings updated preliminarily. We will remove cheaters and update the ratings again soon!

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pulsatio

2 years ago, # |

-26

speedforces

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2 years ago, # ^ |

-16

not really, at least problems up to C were definitely not the hardest problems (in the corresponding position) I've ever seen. I even thought D might have been worth a try. (though I did lack the ideas for it)

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lightseba

2 years ago, # |

+15

Problem F is very nice, thank you!

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alive_ak4c

2 years ago, # |

In question C: why can't

badcfehgjilknmporqtsvuxwzy be the answer for abcdefghijklmnopqrstuvwxyz?

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petyb

2 years ago, # ^ |

because you must always get a big loop with length 26. But in your example you have loops: a->b->a, d->c->d and so on

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alive_ak4c

2 years ago, # ^ |

Oh yes....got it. Thanks

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petyb

2 years ago, # ^ |

You're welcome)

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