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

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

1 | tourist | 3557 |

2 | Radewoosh | 3468 |

3 | Um_nik | 3429 |

4 | Petr | 3354 |

5 | Benq | 3286 |

6 | mnbvmar | 3280 |

7 | LHiC | 3276 |

8 | wxhtxdy | 3258 |

9 | ecnerwala | 3214 |

10 | yutaka1999 | 3190 |

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

1 | Errichto | 191 |

2 | Radewoosh | 180 |

3 | tourist | 173 |

4 | Vovuh | 166 |

4 | antontrygubO_o | 166 |

6 | PikMike | 164 |

7 | rng_58 | 160 |

8 | majk | 156 |

9 | Um_nik | 155 |

9 | 300iq | 155 |

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

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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