Today, Saturday, September 1, 2012 at 20:00 (Moscow-time).
Good luck && have fun!
# | 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 |
Name |
---|
прикольные задачи, если не считать, что я ушел в минус по взломам, то раунд удался
47 to all.
How to solve 1000 div2?
Dynamic dp[h+1][first][second][third][fourth][bad_cubes] = (dp[h+1][first][second][third][fourth][bad_cubes] + dp[h][first1][second1][third1][fourth1][bad_cubes-q]) % mod
The question is still open :)
Actually it's quite easy. Our subproblem is: How many ways can we build the tower if its height is H and the last four pieces have the color a, b, c, d and we are still allowed to have k adjacent cubes. Answer: dp[H][k][a,b,c,d] += dp[H-1][Q][w,x,y,z] where [w,x,y,z...C(all permutations)] // C number of colors and Q is the number of adjacent cubes. Base case dp[0][k][abcd] = 1
Look at my source code in the practice room :)