P12265 『STA - R9』Vacuum Permittivity

Please note: the background of the problem **may not** be related to the problem. One day, Xiao Zhou was studying a very fast sieve. He was very unhappy with the notation of the Zhouge sieve’s complexity $O\left(\frac{n^{3/4}}{\log n}\right)$. Why does an algorithm that does not need to use a bitset have a division in its complexity? Xiao Zhou suddenly came up with a good idea. Since the order of the logarithm function is lower than that of power functions, let $\epsilon$ denote an infinitesimal, and then we can make $\log n = O(n^\epsilon)$! Thus, Xiao Zhou excitedly thought, we can directly let $\log n$ correspond to some specific infinitesimal $\epsilon _0$, and then the complexity of the Zhouge sieve can be written as $O(n^{3/4 - \epsilon_0})$! So pretty! At this moment, a classmate named whk came over. “Why do you people in informatics still need the vacuum permittivity?”

Given positive integers $n, m, s$, with $\gcd(n, s) = 1$. Compute $$\sum_{k = 1}^{n-1} \sin^{-2m} \left(\frac{ks}{n}\pi\right)$$ modulo $998244353$. It can be proven that the answer must be a rational number. For how to take a rational number modulo, please refer to [P2613 Rational Number Modulo](/problem/P2613).

A single line with three positive integers $n, m, s$.

A single line with one integer, representing the answer modulo $998244353$.

| $\textbf{Subtask}$ | $n\le$ | $m\le$ | $s\le$ | Score | | :-----------: | :-----------: | :-----------: | :-----------: | :-----------: | | $1$ | $3$ | $3$ | $1$ | $5$ | | $2$ | $10^5$ | $1$ | $1$ | $10$ | | $3$ | $10^5$ | $50$ | $10^{18}$ | $20$ | | $4$ | $50$ | $10^5$ | $10^{18}$ | $20$ | | $5$ | $10^5$ | $10^{18}$ | $10^{18}$ | $45$ | For all testdata, it is guaranteed that $1\le n \le 10^5$ and $1\le m,s \le 10^{18}$. Hint: contestants should pay attention to the impact of constant factors in the complexity on the program’s running efficiency. Translated by ChatGPT 5