Разбор задач Технокубок 2018 — Отборочный Раунд 3 (и открытые рейтинговые раунды Codeforces Round 445 Div.1, Div.2)

7 лет назад, # |

I am not able to understand div-2 D clearly. Can anyone please explain?

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ironsoul

7 лет назад, # ^ |

Your resultant string can not contain similar letters. For example, if you have "aba" where 'a' appears two times, it is already impossible to construct a string where "aba" will be the most frequent string. Now, using this fact, let's build a graph on letters of our strings. If we have a string "curd", that means we need to create 3 edges {'c' -> 'u', 'u' -> 'r', 'r' -> 'd'}. If such directed graph does not contain loops, just restore strings and output them in lexicographical order

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geniucos

7 лет назад, # |

← Rev. 3 →

+11

After reading this three times, it still doesn't make any sense "Let us note that projection of set of points to line move center of mass of initial set to center of mass of initial set to center of mass of projections multiset" I think it should be that "Let us not that the projection of the set of points to the line moves the mass center of the initial set to the mass center of the projected points". This is in the solution of div1D

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arasmus

7 лет назад, # ^ |

were you able to understand it ? if so please share

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jo_on

7 лет назад, # |

The definition of D(N) in 886E is a bit confusing. It may be better if it is written in the following way.

Define D(n) as the number of good permutations of length n that have p_n = n.

Anyway, thanks for the nice editorial. It is too bad that the contest is unrated(undated?), but I liked the problems.

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RobeZH

7 лет назад, # |

← Rev. 7 →

When I got my code on Problem D accepted after the contest, it was more than 100 lines. I was thinking that I finally got accepted and was satisfied. But then when I saw another one's code: http://codeforces.net/contest/890/submission/32272491 It has just 40 lines!!! I was shocked. However, reading that piece of code made me think about the logic of my code and found that there was a lot of redundancy. Then I improved my code and it became less than 60 lines. I feel really lucky about the fact that we can read others' code on Codeforces.

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tantam75

7 лет назад, # |

In problem D div 2, how should we print the anwser if we got the graph?

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ironsoul

7 лет назад, # ^ |

If there are cycles, answer is NO. Otherwise, restore all strings and output them in lexocgraphical order. 32280888

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tantam75

7 лет назад, # ^ |

Can you explain how to restore all string? Reading code is hard for me.

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ironsoul

7 лет назад, # ^ |

Well, your graph should be like a chain. If it is, just start to restore a string from a top of this chain one by one.

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orlon.

7 лет назад, # |

A small addition to the 886C editorial, you also need to add 1 to the summation because the first room is NOT incorporated in the summation. This is because every term of the summation is a result of adding an "extra" room when a particular 'time' reappears in the array, ans since the first room is not added as "extra" room, it is therefore not a term in the summation. So the answer should be: ans = ans + 1

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nautiyogi

7 лет назад, # |

"If there is cycle in such graph then good string doesn't exist." in Div2 Problem D.Didn't quite get it .... Can anyone explain ?

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ameyanator

7 лет назад, # ^ |

If it were so then the occurrence of a character would be more than once.

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Siriuslight

7 лет назад, # |

← Rev. 2 →

In Div I B, why 2nd test case can't have kekpreceqcheburek as answer. If I am not wrong this string contains kek, preceq and cheburek one time which is the most frequent count.

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Single_Ready_To_Mingle

7 лет назад, # ^ |

According to the question , A string (not necessarily from this set) is called good if all elements of the set are the most frequent substrings of this string.

Here "kek" , "preceq" and "cheburek" are not the most frequent substrings of "kekpreceqcheburek". The most frequent substring is "k" with frequency 3. Hence it is not good

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Siriuslight

7 лет назад, # ^ |

OK, it means that any substring will have part in maximum frequency. I missed that part. If you have solved can you explain me your solution.

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7 лет назад, # |

You have a typo in analysis for 886 E. The last factorial is right only in the second line, in the first and the third you should fix it to (h-1)!

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7 лет назад, # ^ |

How do you derive that equation? Can you show us?

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7 лет назад, # ^ |

+10

First addendum is the number of permutations where number n — 1 is among the first n — k — 1 numbers, all of this permutations are good. There are (n — k — 1) ways to place n — 1 and (n — 2)! ways to place every other number from 1 to n — 2.

The second addendum is the number of permutations, where n — 1 comes at the h-th place, h >= n — k. Such permutation is good if its prefix of size h is good. So when we decide which numbers appear among the first h — 1, the number of such permutations is D(h) * (n — h — 1)!, because the n — h — 1 numbers between n — 1 and n can come in any order. There are C_n-2_h-1 ways to pick h — 1 numbers from the n — 2. So the result of multiplying is D(h) * (n — h — 1)! * (n — 2)! / (h — 1)! / (n — h — 1)! = D(h) * (n — 2)! / (h — 1)!

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7 лет назад, # ^ |

I will also translate my another comment.

To find the final answer let's fix where n is placed in the permutation. Let it be the i-th position in 1-indexation. Then the number of such permutation is the number of ways to pick up and place the numbers that come after the i-th position multiplied by the nomber of good permutations of size i, which is D(i). The first multiplier is C_n-1_i-1 * (n — i)! = (n — 1)! / (i — 1)! So, the answer is sum_1_n[D(i) * (n — 1)! / (i — 1)!].

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7 лет назад, # ^ |

Very thorough explanation!!! I understand it now. Thank you

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7 лет назад, # ^ |

+28

Thanks, huesoss. I just want to put your explanation in easier to read format.

First addendum is the number of permutations where number (n - 1) is among the first (n - k - 1) numbers, all of this permutations are good. There are (n - k - 1) ways to place (n - 1) and (n - 2)! ways to place every other number from 1 to (n - 2).

The second addendum is the number of permutations, where (n - 1) comes at the h^th place, (h ≥ n - k). Such permutation is good if its prefix of size h is good. So when we decide which numbers appear among the first (h - 1), the number of such permutations is D(h) * (n - h - 1)!, because the (n - h - 1) numbers between (n - 1) and n can come in any order. There are $\text{[math]}$ ways to pick (h - 1) numbers from the (n - 2).

So the result of multiplying is $\text{[math]}$

To find the final answer let's fix where n is placed in the permutation. Let it be the i^th position in 1-indexation. Then the number of such permutation is the number of ways to pick up and place the numbers that come after the i^th position multiplied by the number of good permutations of size i, which is D(i). The first multiplier is $\text{[math]}$ . So, the answer is $\text{[math]}$

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7 лет назад, # |

In div 2 D, can anyone explain why we are checking that all the characters of initial strings should be presented in our answer String. I mean that if we are doing dfs for all characters, then why we need to check it separately? If we do not check it, it is giving WA on test 24.

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7 лет назад, # ^ |

← Rev. 2 →

"A string (not necessarily from this set) is called good if ALL ELEMENTS of the set are the most frequent substrings of this string." And it looks like the path found by your dfs doesn't include ALL characters (and sequence of characters) from the set. E.g. 't' whereas the string you found is "crehpuqmbao"

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7 лет назад, # ^ |

I am asking for the same reason. Why doesn't it include all characters even though we are checking for the dfs for all character.

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7 лет назад, # ^ |

You start dfs only from the characters that have a zero number of incoming edges. If there is a cycle, then you won't start dfs on it.

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7 лет назад, # ^ |

← Rev. 2 →

But for the cycle,the answer is already "NO". So,if there has been any cycle,we won't have reach at that step. I hope you are getting my point. Edit : Understood.Thanks a lot.

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7 лет назад, # ^ |

If the test case is just string "aba", then you have a cycle with no entry point, so you won't start dfs and figure out it was a cycle. You check that the answer is not empty, but if the input is "aba, xyz" then you will get "xyz".

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coding_weeb

7 лет назад, # |

← Rev. 2 →

If some string is the most frequent then all its substrings are the most frequent too.

Where in the problem is this stated guys ?

For example if the n=1 and the string is aba why can't we construct the string as aba because the statement only requires the string in the set to be the most frequent substring

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7 лет назад, # ^ |

It's inferred, not stated in the problem. Try the opposite statement: 1. Assume a string "aba" in the set is the most frequent. Inside it there is a substring that appears more than once: 2 "a" => then "a" appears more times than "aba" in the result string. => "aba" is no longer the most frequent string. Hence, the most frequent string (or all strings in the set) must have no duplicated substring in itself, i.e. every substring in it appears at most once. In short, any substring in the most frequent string(s) is only allowed to be inside any one string at most once.

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coding_weeb

7 лет назад, # ^ |

I get it thank you

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Romansp

7 лет назад, # |

В задаче E мы нашли D(n) — количество хороших перестановок у которых на последнем месте n. Но не обязательно же что у всякой хорошей перестановке на последнем месте n. Как то можно свести к каждой?

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7 лет назад, # ^ |

Переберём, где находится в перестановке число n. Пусть, это позиция i в 1-индексации. Количество таких перестановок – это количество способов выбрать и расставить числа, которые идут после i-й позиции, умноженное на количество хороших перестановок длины i, то есть D(i). Первый множитель равен C_n-1_i-1 * (n — i)! = (n — 1)! / (i — 1)!

Значит, ответ на задачу равен sum_1_n[D(i) * (n — 1)! / (i — 1)!].

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Aaronus23

7 лет назад, # |

← Rev. 2 →

I'm having trouble with div2(D): the case:

2 ab ac

Should give a NO however, this graph does not contain cycles

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7 лет назад, # ^ |

There is a condition that all letters coming right after/before the same letter must coincide.

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ista2000

7 лет назад, # ^ |

-8

Why doesn't the editorial say so then? :/

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farmersrice

7 лет назад, # ^ |

because then the occurrence of the substring will be 0

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7 лет назад, # ^ |

It is stated in the editorial, point no.2:

"If string ab or similar is the most frequent then letter a is always followed by letter b and b always follow a."

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ista2000

7 лет назад, # ^ |

This made sense, thank you. I missed that :(

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