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2022-2023 Southern And Volga Russian Regional — Editorial

 1 year ago
source link: http://codeforces.com/blog/entry/109642
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1765A - Access Levels

Idea: BledDest, preparation: awoo

Tutorial
Solution (awoo)

1765B - Broken Keyboard

Idea: vovuh, preparation: vovuh

Tutorial
Solution (vovuh)

1765C - Card Guessing

Idea: DStepanenko, preparation: BledDest

Tutorial
Solution (awoo)

1765D - Watch the Videos

Idea: BledDest, preparation: DmitryKlenov

Tutorial
Solution (DmitryKlenov)

1765E - Exchange

Idea: BledDest, preparation: BledDest

Tutorial
Solution (BledDest)

1765F - Chemistry Lab

Idea: awoo, preparation: awoo

Tutorial
Solution (awoo)

1765G - Guess the String

Idea: BledDest, preparation: BledDest

Tutorial
Solution (BledDest)

1765H - Hospital Queue

Idea: Neon, preparation: Neon

Tutorial
Solution (Neon)

1765I - Infinite Chess

Idea: DmitryKlenov, preparation: dmitryme

Tutorial
Solution (awoo)

1765J - Hero to Zero

Idea: BledDest, preparation: BledDest

Tutorial
Solution (BledDest)

1765K - Torus Path

Idea: adedalic, preparation: adedalic

Tutorial
Solution (adedalic)

1765L - Project Manager

Idea: BledDest, preparation: awoo

Tutorial
Solution (awoo)

1765M - Minimum LCM

Idea: BledDest, preparation: Neon

Tutorial
Solution (Neon)

1765N - Number Reduction

Idea: Neon, preparation: Neon

Tutorial
Solution (Neon)

6 days ago, # |

Unfortunately, the editorials for two problems are not ready yet. They will appear here as soon as they're written.

6 days ago, # |

In problem Torus Path wording "Note that you can't visit all vertices on the antidiagonal (vertices .... at the same time." — is quite confusing.

6 days ago, # |

can problem N can be solved by using stack data structures ?

  • 9 hours ago, # ^ |

    Yes indeed! You need to check for all 1 <= i <= n-1, if s[i] > s[i+1] then we can delete s[i] to obtain a smaller number. You will continue doing this till k is 0. One special case, when stack's first element is greater than 0 and all the remaining values are 0, when placing a value x > 0 and x < stack.top, if k >= stack size, delete all the elements from the stack and push the current one . P.S — You need vector to do this stack operation. Check my submission

4 days ago, # |

← Rev. 2 →  

0

I am not able to get the solution of N.. Neon Can someone please explain it more ?

  • 3 days ago, # ^ |

    ← Rev. 4 →  

    0

    you should give priority to having the least number in the beginning positions(u cant have 0 at 1st position) so :- for every postion check the least number you can get that is iterate over a position vector of numbers from 0-9, and see if we can get that position by checking if it is equal to or smaller than k and once this is done for a postion update k and do this untill k != 0

    code:- https://codeforces.com/contest/1765/submission/183581472

    • 3 days ago, # ^ |

      Understood !!

8 hours ago, # |

The problem K is easily solvable using DP. But the greedy method is quite tricky, at least for me.

report

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