awoo's blog

By awoo, history, 22 months ago, translation, In English

1476A - K-divisible Sum

Idea: vovuh

Tutorial
Solution (adedalic)

1476B - Inflation

Idea: adedalic

Tutorial
Solution (adedalic)

1476C - Longest Simple Cycle

Idea: adedalic

Tutorial
Solution (adedalic)

1476D - Journey

Idea: BledDest

Tutorial
Solution 1 (BledDest)
Solution 2 (BledDest)

1476E - Pattern Matching

Idea: BledDest

Tutorial
Solution 1 (awoo)
Solution 2 (awoo)

1476F - Lanterns

Idea: BledDest

Tutorial
Solution (BledDest)

1476G - Minimum Difference

Idea: Neon

Tutorial
Solution (Neon)
 
 
 
 
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22 months ago, # |
  Vote: I like it +42 Vote: I do not like it

Problem E was beautiful! Thanks :D

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22 months ago, # |
  Vote: I like it -9 Vote: I do not like it

Finally a contest with less ad-hoc!

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22 months ago, # |
  Vote: I like it +89 Vote: I do not like it

contest is good but statement of E was awful

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22 months ago, # |
  Vote: I like it +3 Vote: I do not like it

In problem C , How the value of the following test case will be 6? please explain..

input
3
2 4 2 
-1 2 4 
-1 2 1 
output
6
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    22 months ago, # ^ |
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    Excuse the use of Paint lol, but it looks something like this (the blue edges are the ones forming the cycle):

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22 months ago, # |
  Vote: I like it +6 Vote: I do not like it

I think Problem D is simpler to use the Disjoint-set. Am I right?

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    22 months ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    It can be done simply by moving from current city to Right to find the largest pattern of RL...RL or RL...RLR and similarly on left from current city to left as LR...LR or LR...LRL and length of both the patterns + 1 will be the max cities visited from current city

    You can see my solution : https://codeforces.com/contest/1476/submission/106090614

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

This was a very nice contest after a while. Kudos to the authors, each and every problem in the contest that I solved / up-solved taught me something new. Hope for more such rounds :)

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22 months ago, # |
Rev. 2   Vote: I like it +8 Vote: I do not like it

Can someone help me figure out the error of code for submission below? My Code Here is my code for problem F. It gets wrong answer on test 5, case 1259. However, I can't get that sample.

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22 months ago, # |
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Can Anyone tell me how to solve question D recursively and doing memoization?

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22 months ago, # |
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For B, can anyone explain why there's a k-1 added to 100ll * p[i] - k * pSum?

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    22 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    It is basically done to take ceil of the value.

    Let me make it more clear — Ceil(a/b) == (a+(b-1))/b

    You can check this by taking two cases -

    1. a = k*b. Both functions return k

    2. a = k*b + x (0<x<b). Both functions return k+1;

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      22 months ago, # ^ |
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      but in editorial he didn't mention the ceil thing i understand it from the code but i dont understand the purpose of it, if u can explain please

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        22 months ago, # ^ |
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        It is mentioned in the editorial that -

        $$$x \ge \left\lceil \frac{100 \cdot p_j - k \cdot (p_0 + \dots + p_{j - 1})}{k} \right\rceil$$$

        Notice the bracket, that is for denoting ceil function.

        Now the next question is why do we even need that — because x is a non negative integer and right hand side acc. to maths will return a floating value (which may not be integer).

        Let's take an example — You know x is an integer and after solving RHS you get -

        $$$x \geq 1.2 $$$

        So definitely you know you should pick next integer (ceil value) as answer i.e. x = 2.

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22 months ago, # |
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A constant optimization for 1476G - Minimum Difference -- observing that the upper bound of value $$$c$$$ always equals to the lower bound of $$$(c + 1)$$$, we can store only the lower bounds (but not both as in the model solution),

You can see my submission 106048247 for the detail.

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    22 months ago, # ^ |
      Vote: I like it +13 Vote: I do not like it

    There is no need of segment trees for 1476F - Lanterns. We need only two operations:

    • The one is to query $$$\max\{(j + p_j), \dots, (i + p_i)\}$$$ for increasing $$$i$$$.
    • The other one is to find the minimum $$$j$$$ where $$$\mathrm{dp}[j] \geq i - p_i - 1$$$.

    The two operations can be supported by two standard (monotonic) stack and std::lower_bound.

    My submission 106219219.

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      22 months ago, # ^ |
        Vote: I like it -39 Vote: I do not like it

      forget about this stupid problem is your profile photo-real ??

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      12 months ago, # ^ |
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      Hello, can you take a look at my submission for problem F Lanterns.

      I have not used a segment tree but a sparse table and for each lantern(let us say i) that points to the left, I have kept a pointer to the last lantern(let us say j) such that the lanterns 1 to j cover some continuous segment of lanterns not covered by the lantern i.

      I am failing at some 1200+ subtest of test case 5 and will be much thankful if you can help me point out where my algorithm fails. Or if you can provide some test case where it fails?

      Any help will be much appreciated. Thanks!

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22 months ago, # |
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For E, why can't I topsort it?

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22 months ago, # |
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In E you can also find a match converting the string to integer. The number of possible strings is $$$27^k$$$, so every pattern can be indexed.

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22 months ago, # |
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Question for problem E

Input

  • 4 4 4
  • aaaa
  • aaaa
  • aaaa
  • aaaa
  • aaaa 2
  • aaaa 2
  • aaaa 2
  • aaaa 2

The solution's output is "NO", but I think 2 1 3 4 is right, do I understand the problem wrong?

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    22 months ago, # ^ |
    Rev. 2   Vote: I like it +3 Vote: I do not like it

    Your input is not correct. All patterns should be pairwise distinct.

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13 months ago, # |
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In the second problem i.e inflation, for the test case : 4 1 20100 1 202 202

why is the answer 99? Shouldn't it be zero coz all the elements satisfy the given condition? Please help me out. Thank you!!

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    2 months ago, # ^ |
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    It shouldn't be zero...

    Check the condition like that :-

    p[i]*100 <= k*sum

    Where sum = p[0]+p[1]+...+p[i-1]

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12 months ago, # |
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My code to F Lanterns fail at some 1200+ test of test case 5. Can someone take a look at my submission and guide me to where my code's failing? Or provide a test case where it fails?

Help will be much appreciated. Thanks!

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3 months ago, # |
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How A is binary search? somebody, please explain.