Can someone share the results, please?

# | User | Rating |
---|---|---|

1 | tourist | 3686 |

2 | LHiC | 3330 |

3 | wxhtxdy | 3329 |

4 | mnbvmar | 3296 |

5 | Benq | 3294 |

6 | sunset | 3279 |

7 | V--o_o--V | 3275 |

8 | ecnerwala | 3220 |

9 | yutaka1999 | 3190 |

10 | xyz111 | 3147 |

# | User | Contrib. |
---|---|---|

1 | Radewoosh | 190 |

1 | Errichto | 190 |

3 | rng_58 | 163 |

4 | PikMike | 158 |

5 | Vovuh | 157 |

6 | 300iq | 154 |

7 | Petr | 153 |

8 | Um_nik | 146 |

9 | majk | 145 |

10 | Swistakk | 143 |

Can someone share the results, please?

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Призëр?

can anyone open the link given in the pic?!

Results

There are more participants in the CMS ranking but less in the official site. Which of them will be considered while dividing medals?

In CMS rating is all participant (official and non-official), while official site count only official participants. Medal will be given to official ones

Do they follow the general rule? I mean dividing like 1/12 gold, 1/6 silver, and 1/4 bronze medals.

I think no

What is the official solution for E? I solved it onsite with dp and lazy segtree, but the number of solves confused me (in our team's opinion there were easier problems that were solved less), so I wonder if there is an easier solution.

Convert the problem to the following dp:

DP[position][answer] — maximum position j, such that the last jump was the interval [j; position] and the number of jumps was answer.

To get we will have the obvious transition from DP[pos] to DP[pos+1] (the values can stay the same) and we also will have additional transitions. We will just binary search on partial sums to find the closest position to the right, such that we can jump to it and increase the answer in the state. This way we have

O(N) transitions for each position.We can easily make the state have

O(N) by not keeping the position. Now the only last observation is that we only need to consider the largest/last 2 values in the dp and do binary search for them only. This way we get solution.hint1`pref[i] - pref[j] >= pref[j] - pref[k]`

=>`pref[i] >= 2 * pref[j] - pref[k]`

`2 * pref[j] - pref[k]`

is constant for every`j`

hint2`dp[i] >= dp[i - 1]`

hint3we should do things like in convex hull ;

`2 * pref[i] - pref[j]`

is like intersection ,`dp[i]`

is like`k`

P.S sorry for my poor English