### SkyHawk's blog

By SkyHawk, 9 years ago,

Today, Saturday, September 1, 2012 at 20:00 (Moscow-time).

Good luck && have fun!

• +20

 » 9 years ago, # | ← Rev. 9 →   -14 прикольные задачи, если не считать, что я ушел в минус по взломам, то раунд удался
 » 9 years ago, # |   0 47 to all.
•  » » 9 years ago, # ^ |   0 How to solve 1000 div2?
•  » » » 9 years ago, # ^ | ← Rev. 3 →   +2 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
•  » » » 9 years ago, # ^ |   0 The question is still open :)
•  » » » » 9 years ago, # ^ |   +3 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] = 1Look at my source code in the practice room :)