Gol_D's blog

By Gol_D, history, 3 years ago, In English

1660A - Vasya and Coins

Idea: MikeMirzayanov

Tutorial
Solution

1660B - Vlad and Candies

Idea: Vladosiya

Tutorial
Solution

1660C - Get an Even String

Idea: MikeMirzayanov

Tutorial
Solution

1660D - Maximum Product Strikes Back

Idea: Aris

Tutorial
Solution

1660E - Matrix and Shifts

Idea: myav, MikeMirzayanov

Tutorial
Solution

1660F1 - Promising String (easy version)

Idea: MikeMirzayanov

Tutorial
Solution

1660F2 - Promising String (hard version)

Idea: MikeMirzayanov

Tutorial
Solution
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3 years ago, # |
  Vote: I like it +6 Vote: I do not like it

If you are/were getting a WA/RE verdict on problems from this contest, you can get the smallest possible counter example for your submission on cfstress.com. To do that, click on the relevant problem's link below, add your submission ID, and edit the table to increase/decrease the constraints.

I've also added a new feature to view progress of your judgement in near real-time. (For example, the current state of your ticket, how many inputs were evaluated, etc).

If you are not able to find a counter example even after changing the parameters, reply to this thread, mentioning the contest_id, problem_index and submission_id.

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3 years ago, # |
  Vote: I like it +4 Vote: I do not like it
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    3 years ago, # ^ |
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    Can you explain ur solution

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      3 years ago, # ^ |
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      in problem I made simple observation that whenever number of negative is greater

      let say r = number of negative in subarray — number of pos in subarray

      if r >0 and r%3 == 0 then subarray will be promissing

      I used divide and conquer to calculate those subarray

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3 years ago, # |
Rev. 2   Vote: I like it +6 Vote: I do not like it

There is a solution in the f2 order_set or fenw_tree.This is awesome

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3 years ago, # |
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Can anyone please point out for me the difference between these two submissions:

152864441(I used vector and got WA)

152864364 (I used array and got AC)

Thanks in advance!!!

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    3 years ago, # ^ |
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    Most likely this is due to integer overflow with arrays.

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      3 years ago, # ^ |
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      Can you elaborate more for me ? I'm still confused. Thank you very much!

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        3 years ago, # ^ |
        Rev. 3   Vote: I like it 0 Vote: I do not like it

        Look at expression in line 21:

        if((n == 1 && vec[0] > 1) || (vec[n — 1] — vec[n — 2] > 1))

        The condition behind the OR doesn't check for vector size, so for n == 1, then you are evaluating this: (vec[0] — vec[-1] > 1)

        Imagine this input:

        1

        1

        1

        Then:

        (n == 1 && vec[0] > 1) || (vec[0] — vec[-1] > 1)

        (1 == 1 && 1 > 1) || (1 — vec[-1] > 1)

        (true && false) || (1 — vec[-1] > 1)

        false || (1 — ?? > 1)

        So basically you are accessing vec[-1]. Depending on the value in that "illegal" memory position, it may be true or false. Same happens for arr[-1], but you just go lucky there with the memory value there at the time of the submit.

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          3 years ago, # ^ |
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          Wow, now I get that! Thanks very much!

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3 years ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

Does the solution of F1 take into account the adjacent situation?

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2 years ago, # |
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How can I solve C. Get an Even String with dp ?

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    2 years ago, # ^ |
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    To solve 1660C with DP we should start by finding the recursive solution: for each i in the strings there are 2 options: I either remove the current element, or I remove elements until I reach the next occurrence of this character in the string (if there is any). We recur on both options and take the minimum result.

    The naive approach to finding out how many elements are until current element and next occurrence of character is a simple linear scan. We can speed it up by a LOT by first scanning the string and then storing the positions of the characters in the string in a vector<set<int>> pos(26), where index stores the character type, and the elements in the set store the indexes where this element occurs.

    After that simple (but most likely necessary) optimization, we can notice one thing: if I happen to be in the same index again, the best solution is going to be the same! So I simply store it in a memo vector.

    This is my submission for additional clarity: 163585297

    It's gonna be pretty simple translating it to a simple iterative bottom-up approach! This exercise is left for you.

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2 years ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

I think there is a mistake in tutorial for problem C. Instead of used it should be prev.

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8 months ago, # |
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Can someone give me the DP solution of number C? I think it will help me get intuition faster to this types of problems.

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7 months ago, # |
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Can anyone give counter test case for my solution for D here it is:257423322 getting wrong ans on test 2

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    7 weeks ago, # ^ |
    Rev. 3   Vote: I like it 0 Vote: I do not like it

    I tried to analyse your code : pair<ll,pair<ll,ll>> give_mx(vll &v,ll i,ll j){ ll prod=1; for(int k=i;k<=j;++k)prod*=v[k]; suppose i = 2 and j = n-1; so you will make j-i iterations which is n-2 and if n=1e5 so the you will have product of pow(2,1e5) which is impossible because as we know that max value of long integer cant be more than (in this range 2^62) so in the process you will get bad behaviour of overflow , just if I'm wrong just tell me maybe i couldn't clearly understood your code