AISing Programming Contest 2021（AtCoder Beginner Contest 202） Announcement

We will hold AISing Programming Contest 2021（AtCoder Beginner Contest 202）.

We are looking forward to your participation!

• 5 months ago, # ^ |

  Suppose you're at i'th character and you have used z number of a's. You count the total number of combinations possible from (n-i) characters and z-1 a's. Let's call this c. If this count is more than k, that means you can afford to put 'a' at the current position, else you have to put 'b'. When you put 'b', reduce k by c.

  Submission: https://atcoder.jp/contests/abc202/submissions/22830270

• 5 months ago, # ^ |

  ← Rev. 8 →  

  +6

  Here is what I believed to be a straightforward approach:

  Understand that kth lexicographically shortest string means that there will be total—ktotal—k strings that will be lexicographically greater than it. An alternate way of saying that is that you have to leave total—ktotal—k strings behind. Let us call this number remainingremaining. It is not difficult to compute that,

  total=nCr(a+b,a)=nCr(a+b,b)total=nCr(a+b,a)=nCr(a+b,b)remaining=total−kremaining=total−k

  Now iterate over every position from 0 to n-1, and keep track of the remaining number of a and b. In every position, we have 2 possibilities. Putting either a, or b. Now if we put an 'a' in the current position, how many strings are we immediately getting ahead of? Those many in which there would be a 'b' in the current position, right? That number is,

  nCr(a+b−1,b−1)nCr(a+b−1,b−1)

  Now, if it is OK to leave behind those many strings, we put an 'a' here, and accordingly update the remaining Else, we put a 'b'. How do we understand if it is OK to put an 'a'? The condition is,

  remaining>=nCr(a+b−1,b−1)remaining>=nCr(a+b−1,b−1)

  What if once remaining becomes 0? Well, that means we have no more strings to leave behind. We can then simply create the lexicographically shortest string possible from that position. That is by printing all b's first and then the a's. But it is also fine to go with the flow and check manually as I did.

  Here is my submission following this approach.

  • 5 months ago, # ^ |

    Thank you, very detailed and understandable explanation.

  • 2 hours ago, # ^ |

    Thanks you for the explanation. This one is really helpful.

• 5 months ago, # ^ |

  ← Rev. 2 →  

  +1

  You can use a count array and solve the queries offline.
  All you need is the euler tour of the tree.
  The entire problem will be solved in linear time. code

• 5 months ago, # ^ |

  ← Rev. 2 →  

  0

  could you explain your thought process for F, I got that we have to use DP somehow with triangle decompositions and the fact that triangle area*2 is integer, but what was the main cracking hint for you for this question.

• 5 months ago, # ^ |

  can you please explain your solution and share the link to a similar problem?

• 5 months ago, # ^ |

  ← Rev. 2 →  

  +6

  I did the same. Luckily, I had solved a problem with the same technique today itself. I think it can be done with Euler tour+ binary search as well but this way seemed much more easier to code.

  PS.-Just saw that the editorial has much more cleanly implemented the Euler tour method, contrary to my expectations.

• 5 months ago, # ^ |

  For the last 10 contests I have seen that editorials are available as soon as contest ends.

• 5 months ago, # ^ |

  Hermit__OvO can you please share you submission, Thanks!

  • 5 months ago, # ^ |

    sure, (A bit ugly code though)

    Spoiler

    • 5 months ago, # ^ |

      Thanks!,It indeed is long, I will have to copy it in my editor to study :P

• 5 months ago, # ^ |

  You may follow this

  • 5 months ago, # ^ |

    Bruh how toal is C(a+b,a) why not C(a+b,b). Also i didnt understand remaining>=nCr(a+b−1,b−1). How ?

    • 5 months ago, # ^ |

      Ncr(a+b,a)== Ncr(a+b,b)

    • 5 months ago, # ^ |

      ← Rev. 2 →  

      0

      Number of remaining positions including the current one is a+ba+b, if I put a 'b' in the current position, remaining position becomes a+b−1a+b−1, and we will have to place aa number of 'a' and b−1b−1 number of 'b' in the next positions. So, that makes it nCr(a+b−1,b−1)nCr(a+b−1,b−1) If I have more strings than this to leave behind, why not do it by placing an 'a'?

• 5 months ago, # ^ |

  No. The idea is similar to finding the kth permutation which is quite a standard question.

  • 5 months ago, # ^ |

    The question is standard.But are the constraints?I don't know.Anyway that was solvable for sure.No doubt about that.

• 5 months ago, # ^ |

  probably set as per Japanese Standard Time. It's around evening for them.

• 5 months ago, # ^ |

  This is my solution on F, hopefully it will be clear enough :)

  Sort the points from the top-most one to the bottom-most one (in case of same y, sort from the right-most to the left-most). Afterwards, apply a "custom convex hull" for each prefix of points in the sorted array (the last point in the prefix is the lowest point in the convex hull). Before convex hull, mind Pick’s theorem. This tells us that we are actually interested in the parity of the number of integer points on polygon’s perimeter.

  Sort the points according to Graham Scan. Afterwards, store dp[i][j][k] — the number of partial polygons starting from the lowest point and ending with the j-i segment, having k = (number of integer coordinates on the perimeter) % 2. In fact, there will be no Graham Scan, but a dp building which can be understood easier if you refer to Graham Scan.

  For building this dp, we will iterate 0 <= k < j < i < n, and update dp[j][i][nr % 2] += power(2, interior[i][j]) * dp[k][j][0] and dp[j][i][(1+nr)%2] += power(2, interior[i][j]) * dp[k][j][1], where nr is the parity of the number of integer coordinates on the j-i segment. Also, interior[i][j] is a coefficient which denotes the number of points which can be taken in the answer set without altering the convex hull containing the i-j segment. This can be precomputed at the beginning of the convex hull. Basically, any k such that i < k < j which is interior to the polygon should be counted.

  Finally, any dp[k][j] (0 < k < j < n), such that k-j-0 is a valid ending of the Graham Scan, can be considered in the final answer. The final complexity for this solution is O(N^4) with proper precomuputation.

  • 5 months ago, # ^ |

    Wow, this is great! Thanks a lot bro ^v^~