You need to have qualified from the Round 1.

How do you compute it in O(26n) time? A solution with O(26^2 * n) time complexity passes anyway, but how to get rid of another alphabet size multiplier?

I made the same observation in the first 5 minutes and kept trying to find out how to compute g till the very end. If G(i) were undecreasing, it'd work, but it is not. For example string abc has g: 2 1 0

Yes, angel wins if cheapest flow of size D+1 is of cost <=A, on a complete graph where edge's cost is 1 iff it doesn't exist in initial graph and 0 otherwise. Demon wins if D>=n-1.

@authors Yeah, it would be great thing to put constraints like A, D ≥ 2 in the statement next time.

I also work hard in handling the case A=1...

By the way, A = 1 is solvable.

Consider the foxes ordered in increasing order (by weight). Notice that swapping any two adjacent elements A(i) and A(i + 1) (where A(i) < A(i + 1)) can increase the answer by at most 1. So, keep doing the following:

numberOfWins = ; // Number of Wins when foxes are in increasing order (by weight)
while(numberOfWins < k) {
    int i = ; // smallest index such A(i) < A(i + 1)
     swap(A(i), A(i + 1));
}
return numberOfWins == k ? A : vector<int>();

I wrote an optimised exhaustive search. At each step the minimum and maximum result was calculated with remaining foxes, sorting foxes in decreasing and increasing order, respectively. If k was between the minimum and maximum, then continue search in this direction, otherwise go back. When minimum is equal to the maximum we found the answer. I had little confidence that would pass the system tests, but it did.

If anybody has some time to spend here's my code: https://paste.ubuntu.com/25147848/ (I know this is shit code)

Link : https://community.topcoder.com/stat?c=problem_solution&rm=330384&rd=16951&pm=14651&cr=23303116

because I don't know how to hack on normal arena and only possibility to learn is during the contest

Nope. You can hack in practice rooms. Maybe you can submit crap and try to hack yourself to learn, but I'm sure there's plenty to hack even if not.

Last time I remember, the web arena had a lot of bugs. Like randomly vanishing codes.

&gt; is HTML for > (greater than), &lt; is < (less than) to avoid parsing as HTML tags. Trying to parse one thing in many different ways is absolute cancer, you'll get stuff like having to write 4 backslashes in a row in order to render something invisible.

My plugin has this issue as well and I found it before the contest, but I thought it was just because my plugin sucked, and didn't pay much attention. This may be the reason why administrators didn't find it either.

If I type '<', it will fail because '<' is special in HTML. That's why I used &gt.

Sorry for wasting your time.

The applet shows a poor stack-trace sometimes for Python programs. It points to some line in their backend's Python wrapper. I understand that the error message can be used to trace back and figure out where my code screwed up, but maybe a cleaner solution could be possible to point the origin of the error in my code.

BTW, if you use arena applet, you can contact admins by using the chat box at the bottom. By clicking the ">>" button on the left side you can set your receiver as admins.

The way I thought is similar and shows why this works.

We're trying to maximize maxflow after adding A edges. The edge (0, n - 1) is easy. If the path 0, i, n - 1 exists for any i, then we can send flow through that path — if, after adding edges, one edge from that path was used in the flow and the other wasn't, then we can redirect the flow; if both were used in different paths from 0 to n - 1, we can switch parts of those paths at vertex i.

Now, let's find some paths to complete the maxflow. As long as w.l.o.g. the degree of 0 is less than of n - 1, we can find an edge (i, n - 1) that isn't in one of those paths, meaning (0, i) didn't exist and we can add it, increasing the maxflow by 1.

As soon as this isn't possible, the maxflow can be increased by 1 only at cost  ≥ 2 (since the degrees of 0 and n - 1 are equal to the current maxflow). Cost 2 is also sufficient since we can add edges (0, i), (i, n - 1) if they didn't exist or (0, j), (i, n - 1) if there was a path 0, i..j, n - 1.