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Gol_D's blog

By Gol_D, history, 2 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
  • Vote: I like it
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2 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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2 years ago, # |
  Vote: I like it +4 Vote: I do not like it
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    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Can you explain ur solution

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      23 months 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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2 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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23 months ago, # |
  Vote: I like it 0 Vote: I do not like it

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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    23 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Most likely this is due to integer overflow with arrays.

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

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        23 months 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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          23 months ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Wow, now I get that! Thanks very much!

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23 months 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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22 months ago, # |
  Vote: I like it 0 Vote: I do not like it

How can I solve C. Get an Even String with dp ?

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    20 months ago, # ^ |
    Rev. 2   Vote: I like it 0 Vote: I do not like it

    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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20 months 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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7 months ago, # |
  Vote: I like it 0 Vote: I do not like it
import java.util.*;

public class A_Vasya_and_Coins {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt();
        while(t-->0)
        {
            int a = sc.nextInt();
            int b = sc.nextInt();

            int a1 = a*1;
            int a2 = b*2;

            if(a==0)
            System.out.println(1);
            else 
            {

            int ans = (a1+a2) + 1;

            System.out.println(ans);
            }


        }
    }
}

nice