Just as reminder, the round is scheduled at 16:00 UTC. Top 1000 contestants advance to Round 2.
# | User | Rating |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3823 |
3 | Benq | 3738 |
4 | Radewoosh | 3633 |
5 | jqdai0815 | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | ksun48 | 3390 |
10 | gamegame | 3386 |
# | User | Contrib. |
---|---|---|
1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
3 | atcoder_official | 162 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 157 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
9 | nor | 153 |
Just as reminder, the round is scheduled at 16:00 UTC. Top 1000 contestants advance to Round 2.
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