README.md

APAC 2017 Round D

https://code.google.com/codejam/contest/5264486/dashboard

Problem A: Vote

Catalan Number.
Consider of a way with n upstrokes and m downstrokes. It's a route from (0, 0) -> (n+m, n-m), totals numbers are (n+m)Cm.
However, the first vote must be A, and afterwards the way cannot reach the horizontal line. That is, (1, 1) -> (n+m, n-m) where the whole route is above the horizontal line.
All possible routes number of (1, 1) -> (n+m, n-m) is (n+m-1)C(n-1).
All invalid routes number is (n+m-1)C(m-1) (Reflection, check here.)
Finally, the result is ((n+m-1)C(n-1)-(n+m-1)C(m-1))/((n+m)Cn) = (n-m)/(n+m).

Problem B: Sitting

Assume row<=col.
If row<=2 then sit as ## ## ## ## ##.... Else, sit as:

## ## ## ## ## ...
# ## ## ## ##  ...
 ## ## ## ## # ...
## ## ## ## ## ...
...

Problem C: Codejamon Cipher

Sort each word, save to HashMap.
Then for each string, DP to find the number of ways.

Problem D: Stretch Rope

DP & Segment Tree.
dp[i][j] means the minimal money needed, to get length j in first i ropes.
As the i-th rope may has [A[i], B[i]] length, dp[i][j]=min(dp[i-1][j], p[i]+dp[i-1][k]) where j-B[i]<=k<=j-A[i].
So, we can build a segment tree for dp[i-1], to query the minimal of dp[i-1][j-A[i]] ~ dp[i-1][j-B[i]] in O(log l) time.
Total time complexity: O(n)*(O(l*log(l))+O(l*log(l)))=O(n*l*log(l)).
Check D.java for more details.