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 | tourist | 3690 |
2 | jiangly | 3647 |
3 | Benq | 3581 |
4 | orzdevinwang | 3570 |
5 | Geothermal | 3569 |
5 | cnnfls_csy | 3569 |
7 | Radewoosh | 3509 |
8 | ecnerwala | 3486 |
9 | jqdai0815 | 3474 |
10 | gyh20 | 3447 |
# | User | Contrib. |
---|---|---|
1 | maomao90 | 174 |
2 | awoo | 165 |
3 | adamant | 161 |
4 | TheScrasse | 160 |
5 | nor | 158 |
6 | maroonrk | 156 |
7 | -is-this-fft- | 152 |
8 | orz | 146 |
9 | SecondThread | 145 |
9 | pajenegod | 145 |
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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