Just as reminder, the round is scheduled at 16:00 UTC. Top 1000 contestants advance to Round 2.
Just as reminder, the round is scheduled at 16:00 UTC. Top 1000 contestants advance to Round 2.
| # | User | Rating |
|---|---|---|
| 1 | Benq | 3676 |
| 2 | Kevin114514 | 3659 |
| 3 | jiangly | 3640 |
| 4 | turmax | 3450 |
| 5 | tourist | 3408 |
| 6 | strapple | 3360 |
| 7 | Um_nik | 3354 |
| 8 | squareOf105 | 3348 |
| 9 | maroonrk | 3312 |
| 10 | idontwannadothis…e | 3191 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 154 |
| 2 | nik_exists | 151 |
| 3 | maspy | 149 |
| 4 | AmShZ | 146 |
| 5 | Um_nik | 143 |
| 6 | Errichto | 139 |
| 7 | BledDest | 136 |
| 8 | adamant | 135 |
| 9 | maroonrk | 133 |
| 10 | cry | 130 |
| Name |
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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