https://code.google.com/codejam/contest/10214486/dashboard
As Cost[t] ¡Ü Cost[t+1]+1, we cannot stop at any city.
So, just for each start time, bfs to find the minimal time.
As p/e*e'/t=P/Q, we can calculate all possible e/e' in the extra
gears. Then, for each p in Np and t in Nt, check if Pt/pQ is possible.
Calculate all factors of n with only one prime. 9075=3*25*121, etc.
Then dfs with memorization to solve it!
It's a really, really difficult problem.
DP, dp[i][j][k][l] where i means the current character, j means
the number of a, k means the number of b, and l means if we have
started to deal with c/d.
Check D.java for more details.