Comments
On AquaMoon → CodeTON Round 2, 4 years ago
-8

So will have a re-run?
On AquaMoon → CodeTON Round 2, 4 years ago
0

Cool. So it seems Java BigInt can solve it too.
On AquaMoon → CodeTON Round 2, 4 years ago
0

Cool. So mine will be ok if I replace int128 to some BigInt-like struct. Thank you.
On AquaMoon → CodeTON Round 2, 4 years ago
+3

Thank you! Learned a lot from u.
On AquaMoon → CodeTON Round 2, 4 years ago
0

how's that O(2^(n/2)) calculated? can u prove it?
On AquaMoon → CodeTON Round 2, 4 years ago
0

Because it seems to be small enough, not the case of 1 million bit, it is like we use bitset for 64 times faster. If a type can hold it, we can simulate it using O(n*n).
On AquaMoon → CodeTON Round 2, 4 years ago
0

can java BigInteger or Python pass it?
On AquaMoon → CodeTON Round 2, 4 years ago
-8

Problem E can __int128 hold all variables?
0

Is the word trick kinda glorifying? I don't think so.

And, if actually it can be proved at most a certain amount of updates lead to the minimized answer, then it's not a shenanigan. This is also why I am asking here. I'm not able to construct a case failing the solution.

Besides, early exit is sometimes useful in OI competitions.
0

Could anyone have a look at this code?

https://mirror.codeforces.com/contest/1706/submission/164815320

The solution got TLE before I added an early exit trick, then it got AC, I have no idea why. It seems that the early exit trick is wrong. So is the data too weak or the early exit is actually correct?
+8

Yeeeeep! No wonder i wrote it so smoothly.
+1

unordered_map sucks! In problem D, unordered_map sol > 2000ms array sol = 143ms, 15x faster Why am I always FST?
0

Maybe not that easy, I solved problem C and D after contest and it takes me less than 1 hour. Not even longer than the time I spend on the harder problem B that I imagine. But maybe someone else thinks otherwise too.
0

Or we can print the full answers in a compressed format, it is easy to know that the list of answers can be represented by (Ai, Ci) where i <= N+M which means the next Ci answers are all Ai. In the sample case it is (3, 2), (4, 6), (5, 4).

Don't limit your creativity bro
0

Yep, but it is POSSIBLE to make it a harder problem by querying part of the answers. For example, make the sum of all N + M not larger than 100,0000 and query no more than 100,0000 of different k.
-19

Problem B actually has a O(N+M) solution, though it is more complicated. I have spent one hour working out this O(N+M) solution because I misunderstood the hell limits of N and M!!!!!!!!!!!!!! Just didn't notice that sum of all N*M does not exceed 1e5 (it really sucks :( Then I have not enough time to submit problem C and I was almost there...
0

for D2F(D1C), it is quite easy, easier than D2D/D2E, why it is placed in the last problem, and get so much score? my solution is, firstly generate a random permutation as initialization, then brute force search an valid permutation based on this initialization permutation. the solution is very simple and source code is very short. if i know this i will definitly finish D2F earlier than D2D/D2E.