Hello, Codeforces! Today I've faced the following problem: is it possible to calculate N! modulo M (M is prime) in polynomial (relative to length of M) time? I thought it would be pretty popular problem, but I couldn't google anything useful. Does anybody know such algorithm or proofs that it does not exist?
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Время на сервере: 26.04.2024 09:57:15 (j1).
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You can try this algo, if M is not too large.
Thanks, but I'm just curious if the polynomial algorithm exists.
Actually, you can calculate $$$n! \bmod p$$$ in $$$O(\sqrt{p} \log p)$$$.
See this paper Linear recurrences with polynomial coefficients and application to integer factorization and Cartier-Manin operator for more details.
wrong
а)
polynomial (relative to length of M) time.
if $$$N>=M$$$ then $$$N!$$$ $$$mod$$$ $$$M$$$ $$$=$$$ $$$0$$$.
Else calculate it in $$$O(N)$$$ :P
Am I missing something?
Maybe your's is not in Length of m complexity.
Then my questions is what is length of m? Can you please define it?
Is this no of digits in M?
Maybe saying $$$O(logM) == O(logM/log10)$$$ would have been better.
I think so. He may want the solution to be in O(log m).blyat can you confirm?
Yes, I want solution to be in O(log^k m) for some k.
This is the problem you are describing, maybe with few extra constraints: https://www.spoj.com/problems/BORING/en/
maybe with few extra constraints
Extra constraints sometimes make life easier.
Notice the constraint Abs(N-P) < 10^4.
Lemma — $$$(P-1)! \% P = 1$$$.
If the only solns of $$$x^2\%P=1$$$ are $$$-1,1$$$ for prime $$$P$$$ then this lemma holds. $$$(P-D)! \% P = (-1)^{(D-1)}*(P-1)! / (D-1)! \% P$$$
If I'm not mistaken this shall be the problem (even though the constrains are solvable with approach suggested by zimpha ==> so not polynomial in length ^_^ )