Блог пользователя Ari

Автор Ari, история, 6 лет назад, По-английски

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Разбор задач Codeforces Round 561 (Div. 2)
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6 лет назад, # |
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Everything was so fast :
System testing — Done
Editorials — Done
Rating Update — Done
And its just one hour after the contest.
Other CP sites should learn from this.

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6 лет назад, # |
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Good mathforces :)

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    6 лет назад, # ^ |
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    Can someone please explain what does this term "mathforces" really mean? I have seen it being used a lot in various contexts, seems like some inside joke or reference.

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      6 лет назад, # ^ |
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      Well, it is when the major part of the problem is solving an underlying math problem (incl. formula transformations, applying some theorems and common facts from algebra) rather than finding out and coding typical CP algorithms (DP, graph theory etc.). Some people argue that such problems give an unfair advantage to those who are participating in math competitions professionally. But it is well known that gaining good math skills is a part of one's development as a CP professional.

      ...or maybe I was getting it wrong all this time. Then please correct me :)

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    6 лет назад, # ^ |
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    There's no so much math I think. Anyway, very interesting round!

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      6 лет назад, # ^ |
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      Literally every task except maybe F requires solving a math formula before implementing the solution. In some, the math solution is relatively easy, say in A or B, in others its really hard to do without more advanced knowledge in maths, or maybe just having a mathematical mindset to solve the problems.

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6 лет назад, # |
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mathforces back at it again :( (thanos snapped his fingers)(my rating doesn't feel so good right now)

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6 лет назад, # |
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I spent about half an hour thinking about details about Problem C, divided it into 3 cases (by the relationship between x, y, -x and 0), only to find that it can be solved in such a simple code:

for(int i = 1; i <= n; i++)
    ans += upper_bound(a + i, a + n + 1, 2 * a[i]) - (a + i) - 1;

Can you imagine how astonished I was ?

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6 лет назад, # |
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Good contest, loved the problems! Spent quite a lot of time on Problem C just to find out the error lied in one edge case of point at 0 :facepalm:

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    6 лет назад, # ^ |
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    Take x=0, y=1. Then, |x|=0, |y|=1, |x+y|=1, and |x-y|=1. Thus, old range is [0,1] while new range is [1,1] which does not cover the entire old range. It can be seen by symmetry that the same applies when x=1, y=0.

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    6 лет назад, # ^ |
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    The new range will not be [0, 1] but it will be [1, 1] which will not cover all the points. Moreover for the case of [0, X] the new range will be [X, X] so those points will not add to our result.

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6 лет назад, # |
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I actually liked this one — the problems were mainly idea-wise and my solutions could be extremely clean and elegant even compared to my usual doing. ;)

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6 лет назад, # |
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Greetings, can someone help me figure out what's wrong with my submission to problem C ? My solution is just like the tutorial's, but I couldn't pass TestCase 20. https://mirror.codeforces.com/contest/1166/submission/54322422

Many thanks in advance ^^

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6 лет назад, # |
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How do people generally frame the questions? I am very curious to know that.

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6 лет назад, # |
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Problem E is great in constructive algorithms.But something bad is if you cannot create a valid solution,you can also get AC.

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    5 лет назад, # ^ |
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    Can a solution of E be like this ? Set the a[i] for all the stores that are common in all m days to some fixed value say 10. And all other values to 1. So , each day the lcm of visited stores is 10 and of unvisited is 1.

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      9 месяцев назад, # ^ |
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      It is completely possible for there to be NO store that occurs in all days.

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6 лет назад, # |
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In problem B, what do they mean by "filling the grid by diagonals" ?

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6 лет назад, # |
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In problem C, i have a question. How can we get x <= 2y from |x+y| >= y? i got -x <= 2y. Please someone help me to understand that.

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    6 лет назад, # ^ |
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    No we obtained results from |x-y| not |x+y| coz |x+y|>=x+y will always be true for positive x ........................................... |x-y| <= x ...... x-y <= x ..... y-x >= x ...... 2x <= y thus result

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      5 лет назад, # ^ |
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      |x-y|<=x

      x-y<=x & x-y>=-x

      y>=0 & y<=2x

      Here we are able to get only y<=2x, But in editorial it is written that we also obtain x<=2y. How do we obtain this equation. Please clarify

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        5 лет назад, # ^ |
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        y <= 2x is for the case when x <= y: then we have |x − y| <= x <= y <= x + y

        x can also be greater then y. then we have |x − y| <= y <= x <= x + y

        and here comes x − y <= y --> x <= 2y

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6 лет назад, # |
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for Problem E, I wanted to prove(or disprove) it. Then I found

1

2,3

4,5,6.

m = 4, (124,136,456,235) I set each a_i a prime, then I realized 2>6, 6>2...which I'm stuck, (disprove?)

Thanks to editorial.. I didn't think about set PRODUCT of prime.

next time, I should take a submission first, instead of proving myself. :(

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6 лет назад, # |
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Can someone describe how the stores would be assigned integers in problem E ?
And can the time complexity be further improved?

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6 лет назад, # |
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Is kefaa2 alt of Um_Nik?

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6 лет назад, # |
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[PROBLEM E]

Construction in problem E is so GREAT! I wonder how you can create such a nice construction like that. Can you share your MOTIVATION? Besides, problem would be a deserved E-class if it required constructing array a.

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6 лет назад, # |
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Can someone please provide a simpler explanation to Problem D. The editorial seems to be a bit difficult for me.

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    5 лет назад, # ^ |
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    Problem D is basically about the series: 1 1 2 4 8 16 32 64 128 ............ U have to turn a into b using this series. Otherwise return -1

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6 лет назад, # |
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In problem D, why are we choosing r(i) = k + d(n-1-i) ? What is the intuition behind doing this, and what exactly are we trying to show?

Also, what does k denote here?

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6 лет назад, # |
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https://mirror.codeforces.com/contest/1166/submission/54423195 I am not able to get what wrong with my code, anyone has any idea pls share with me

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6 лет назад, # |
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Can anyone please elaborate editorial of problem A. I didn't get [cnta/2] part. Thanks In advance!!

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    5 лет назад, # ^ |
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    It's basically for counting floor and ceil of [total no. of students]/2 (as it is the most evenly possible distribution) and further dividing it by 2 to get total no. of pairs!

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5 лет назад, # |
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Can someone elaborate on problem D??...I am not able to understand after writing r as d0d1...

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5 лет назад, # |
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Nice problem E, СЛАВЯНЕ!

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5 лет назад, # |
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Problem C can also be done using simple queues in O(n).

https://mirror.codeforces.com/contest/1166/submission/56070478

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5 лет назад, # |
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I read the editorial completely .There are two methods described — One writing b in a particular form : 2^(n−2)(a+k)+r(which i think is not easy to guess during contest) and the other Greedy Method . Author has not told completely to use greedy method and i tried to figure it out .After unsuccessful attempt i saw one of the submissions (http://mirror.codeforces.com/contest/1166/submission/56675841) . In this submission , first value of n is found and after that for n-2 times b is divided and changed to new value. There is another variable 'p' , which tells if we need to add 1 before dividing by 2 (basically it is alternatively) .

I want to know the thought process or how to come up with the solution that provided in that submission .Note that in other submissions also similar approach it taken .

I am new to competitive programming and i want to improve , hence it is necessary for me to solve difficult problems.

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5 лет назад, # |
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Can a solution of E be like this ? Set the a[i] for all the stores that are common in all m days to some fixed value say 10. And all other values to 1. So , each day the lcm of visited stores is 10 and of unvisited is 1.

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5 лет назад, # |
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Problem E is a straightforward copypaste from the first Iranian Team Selection Test for International Mathematical Olympiad 2018. I don't know if this was intentional, but I thought it would be worth noting. https://artofproblemsolving.com/community/c639324_2018_iran_team_selection_test

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4 года назад, # |
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Can someone explain this part of the editorial for question D?

Alternatively, after getting the formula, we can iterate on i from 2 to k and greedily choose the values of ri to be as large as we can without exceeding b. This can be easily shown to work using that the coefficients are consecutive powers of two.

I'm able to figure out the part, to check if the sequence exists or not. Also in the first solution that has been suggested, Write b as 2n−2(a+k)+r where r<2n−2, what is k here?