Sometimes they do online mirrors under CONTESTS tab then move them later to the gym, I think you need to contact the coordinators for that.

Another option is to click "Update Contest" two minutes before the contest, but make sure everything is working when it's private then replace the contest.xml file as anyone in coach mode can update the contest.

5 hours.

I updated the text. Thanks!

I am semi-purple.

It is intended for solo.

"you should individually solve", so solo participation!

Why Kaimaru, there are much better places

Nope

After sorting it by L[i] + D[i], maintain a max heap. If currentAltitude ≤ L[i] then we add it in the heap and increase the answer counter. Another case is that currentAltitude - maxElementInHeap ≤ L[i] and D[i] ≤ MaxElementInHeap, so it is cool to change the max element in heap with the current element. So O(NlogN).

I thought uva 1153 is a well-known problem(?

Do people personally hate the game or something?? In fact, it was estimated to be the contest's 5th~6th easiest problem..

IMO, for fun contest, I never read long/complicated-statement problem.

Indented solution was O(NlogN) managing half-line (ray) using BBST

Editorial will be uploaded soon (soon?)

I implemented a n log (n + maxf) log maxc solution and got AC.

Author's solution was O(nlgn). I solved it in O(nlg²ⁿ) with Divide and conquer, and it's running time was not very different (1.5x from intended)

We gave 4x time limit from official solution, but it is like 2.5x in CF :/

In the on-site contest, we actually gave special prize to first-solver of Y. Maybe if you have chance to visit our campus someday ;)

If you want, I can send you a yut set. It is well made game with some luck and strategy.

Actually, there is one more additional game rule not written in statement, such as player throws Yut alternatively but there is condition for throwing Yuts twice or more in a row.

It show 404 to me.

Hope, you will add english translation.

You should solve only when k = 5, all others are much simpler with the same idea.

For k = 5, brute force the middle edge, then you just need to know the shortest distance from start to every vertex with 2 edges, and the same with destination.

You need carefully check, that there is no repetetition. I stored three minimun for each distance and then calculate answer.

O(n+m)

P.S. if k > 5, there is no solution.

For n ≤ 5, exhaust every possible path. For k > m or k ≥ n, answer is -1.

So now only k ≤ 5 left.

For k = 1, just find if edge (1, n) exist.

For k = 2, exhaust every node excluding 1 and n at the intermediate node.

For k = 3, exhaust every edge as the intermediate edge.

For k = 4, for every node v excluding 1 and n. We find the two smallest length for 1 -> x -> v. And two smallest length for n -> y -> v. Then exhaust every node v, if x ≠ y then we could update our answer.

For k = 5, it is similar for k = 4 but this time we store three smallest length. It look like this, 1 -> x -> v -> u -> y -> n. So exhaust every edge (v, u). For those 3x3 possible path. update answer if x ≠ u and y ≠ v and x ≠ y.

Any hashing that uses modulo 2^64 will fail in test 48. Hashing is hard to implement correctly, and if you implement correctly then it has very poor constant factor. I recommend you to use another approach.

Easier solution is DP. Let DP[i][j] = (is substring [i, j] a palindrome?). The answer is the number of nonzero element in DP matrix. You can maintain DP table in O(Q) time for each query. We will soon describe this approach in editorial.

Alternatively, you can use Manacher's algorithm for each string you've encountered.

overflow hash is not a good idea usually as you can see here

Data was generated using following code with Q=10,000

You can check hash collision about this string (check whether s[0:4096] and s[4096:8192] is same both hash and naive string comparison)

string s = "";
for(int i=0; i<Q; ++i)
{
	int r = (__builtin_popcount(i) % 2) * 16;
	char c = 'a' + r;
	s += c;
}
cout << Q << endl << s << endl;

Sorry, now everything is here https://kaist.run/ko/contests/