Right now IOI 2016 is in progress. I'm sure a lot of you guys will be watching, so this post was created to let you guys talk about the scores and the scoreboard in the comments :) The link to the scoreboard is here.

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

1 | tourist | 3882 |

2 | maroonrk | 3539 |

3 | Benq | 3513 |

4 | MiracleFaFa | 3466 |

5 | ksun48 | 3462 |

6 | ecnerwala | 3446 |

7 | slime | 3428 |

8 | Um_nik | 3426 |

9 | jiangly | 3401 |

10 | greenheadstrange | 3393 |

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

1 | awoo | 192 |

2 | -is-this-fft- | 191 |

3 | Monogon | 184 |

4 | YouKn0wWho | 182 |

4 | Um_nik | 182 |

6 | antontrygubO_o | 171 |

7 | maroonrk | 169 |

8 | kostka | 165 |

9 | SecondThread | 164 |

9 | errorgorn | 164 |

Right now IOI 2016 is in progress. I'm sure a lot of you guys will be watching, so this post was created to let you guys talk about the scores and the scoreboard in the comments :) The link to the scoreboard is here.

↑

↓

Codeforces (c) Copyright 2010-2022 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: May/24/2022 21:09:24 (h3).

Desktop version, switch to mobile version.

Supported by

User lists

Name |
---|

Bolivia is in the top Scoreboard for the first time in history! I have great expectations for this year! GO BOLIVIA!

Day 1 is over, and a lot of contestants have fully solved Task 1. What are your thoughts on the problems (and maybe your approaches)?

I think this solves 1:

`k`

smallest elements, such that their sum is less than`l`

, but if you took`k+1`

it would be larger than`u`

. (If you can't do that, the problem is either easy: a sum lands inside the interval; or impossible: the total element sum is smaller than`l`

)`k`

elements for one of the`k`

largest elements.When you change an element, the sum can never increase by more than

`w_max - w_min <= u - l`

, so there is no risk of 'jumping' over the`[l, u]`

interval.If the sum of the largest

`k`

is still smaller than`l`

, then there is no solution, because the next subset in order of sum is the`k+1`

smallest elements, which we know is too large.This solution should take time equal to sorting, which should be fine for n = 200,000.

I thought it was a fun problem :-)

your answer in good but there is a better one that is very similar instead of sorting, just take the first k elements that the sum of the first k is smaller then l and the sum of the first k+1 is higher then u. then we can use the same trick that you did for k and k+1 (if there is a solution then thre is a solution that uses k elements or k+1 elements)

You could also easily use two pointer on tge sorted array?

Yes, I think that's a nice way to accomplish the third step.

3 from China ,2 form Russia in top 5 . but how??

When are day 2 problems coming out?

When day 2 comes.

Oh, I thought today was day 2. Anyways, just a little question: how difficult are the IOI problems usually? From my point of view the first problem from day 1 would be around div1. A/B. What do you guys think?

problem 2 and 3 are much harder than Div1 A and B. a lot of those participants are red or orange and they all solve Div1 A and B easily in CF rounds of normal difficulty. but only 4 solved 2nd problem and just 1 solved the third (solved == 100 points)

Its hard and possibly impossible to compare IOI problems with CF problems due to the completely different style and format. The first problem was easy, but the other 2 were very difficult.