Just as reminder, the round is scheduled at 16:00 UTC. Top 1000 contestants advance to Round 2.

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

1 | tourist | 3698 |

2 | Um_nik | 3463 |

3 | Petr | 3341 |

4 | wxhtxdy | 3329 |

5 | LHiC | 3300 |

6 | ecnerwala | 3285 |

7 | sunset | 3278 |

8 | V--o_o--V | 3275 |

9 | Benq | 3262 |

10 | mnbvmar | 3248 |

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

1 | Radewoosh | 187 |

2 | Errichto | 181 |

3 | rng_58 | 161 |

4 | PikMike | 160 |

5 | Petr | 157 |

5 | Vovuh | 157 |

7 | 300iq | 151 |

8 | Ashishgup | 149 |

8 | Um_nik | 149 |

10 | majk | 148 |

Just as reminder, the round is scheduled at 16:00 UTC. Top 1000 contestants advance to Round 2.

↑

↓

Codeforces (c) Copyright 2010-2019 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Apr/23/2019 20:06:48 (g2).

Desktop version, switch to mobile version.

Supported by

User lists

Name |
---|

I've solved problem B large but got Wrong Answer because of silly problem my code

when I divide my long double-type variable by 2 million times it becomes too small and tend to zero after that when I multiply it by some numbers it still zero.

actully what I wanted to compute is [ C(n,k)+C(n,k+1)+C(n,k+2) .... + C(n,n) ] / 2^n for some numbers n,k

fails large input file but my algorithm is correct.

What I did was compute c(n,k)=C(n,k)/2^n in a table of doubles. The combination of multiplication and division by large doubles can be tricky, so it's good practice not to use it — most of the time, it can be replaced by simple multiplication and summation.

I played with Google Charts and made the following map: Google CodeJam 1st round statistics. It's just an experiment to learn Google Charts for this occasion. It shows

`1000 * Round2 / (Round1A + Round1B)`

taken from Google CodeJam Statisics.Single person statistics can be kinda tricky, if you notice the blue countries... good job anyway