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https://codeforces.me/contest/1979/submission/264408353 bruh

thanks!

YES NO forces

just found this submission for D, and to me it seems very random. Can anybody explain why this part is correct (if it is) act=(act+seguentSalt[act]++)%mida

1000000 1 1000000 1 1000000 1... causes tl for solutions that try to linear search next year every time

for (auto &i : mp), you just make a copy every time

Nice idea!

Yeah, it is confusing. But you can choose subsegment of length 1 if a[i] <= k

haha, for me it's much easier, but there're also short implementations of segment tree, so idk, let it be 100% faster =D. If you are in rush, then no worries about it, but of course I recommend you to learn it after, it's very cool data structure!

neutral is such element A that for any B, A*B = B. 0 + 5 = 5, or min(INT32_MAX, 5) = 5 for example. Yes segment trees can solve all of BIT problems, BIT is just easier to implement and faster.

There're a lot of quieries you can come up with and a lot of data strucutures for theses quieries. Segment tree supports all associative operations with neutral element. Fenwick Tree (BIT) supports only subset of these operations

Great contest! 2 problems with hello world solution is insane :D

in problem C you need to make some observations before that, so no

"rooms" are made to prevent it

I think answer is Infinity if sum of min(n, d) greatest A[i] >= C since we still can do different tasks

That is not true actually, atleast when using operator[] instead of .at()

Because of poor default hashing function. If you wanna use unordered_map or other data structures which are using hashing, write your own randomized hash function so it will be much harder to create such bad test-case for it. In this particular problem complexity nlogn of map is good enough to pass.

Try this: setvbuf(stdout, nullptr, _IOFBF, BUFSIZ);

I've played a bit with your code, and if you make #define double long double , it outputs correct answer on both compiliers. So, probably, there're some differences in the floating point treatment.

If you are stuck for a log time, you can try binary search