Finding a billion factorials in 60 ms with SIMD
codeforces.com
codeforces.com
For context, in competitive programming a lot of combinatorial problems (find some formula to count something) require you to output the answer modulo some prime. This is because otherwise the answer would overflow an int and make the problem too tedious to be fun and too hard for problem setters to write good problems or checkers for.
So to prove that you still know how to count the thing, you can do it a finite field. If you use integers mod prime, you still have all the usual arithmetic operations like addition subtraction multiplication. And even division is still easy since you can calculate multiplicative inverse with Fermat's Little Theorem (a^(p-2) = a^(-1) mod p). The final answer you output is not the real thing you're counting, but just evidence that you had the right formula and did all the right operations.
Anyway, just wanted to give context for why competitive programmers care about factorial mod a prime (usually as part of a binomial or multinomial expression). And I'm kind of surprised anyone outside of competitive programming cares about it.
See also:
Tangent, but curious: what made you write it like this? I've only ever seen it written as
a^(p-1) = 1 mod p
a^p = a mod p
int FastExp(int a, int e, int p)
{
if (e == 0) return 1;
if (e == 1) return a;
if (e % 2 == 0) return FastExp((a*a)%p, e/2, p);
else return a * FastExp((a*a)%p, e/2, p);
}
That said, I'm also a bit surprised to see somebody discuss modular inverses without mentioning the extended euclidean algorithm, which is a more elementary solution.
The article computes modular inverses of a_1, ..., a_n by:
- Computing (a_1 * ... * a_i) = (a_1 * ... * a_{i-1}) * a_i recursively
- Computing (a_1 * ... * a_n)^(-1) by square-and-multiply
- Computing (a_1 * ... * a_i)^(-1) = (a_1 * ... * a_{i+1})^(-1) a_{i+1} recursively.
- Computing a_i^(-1) = (a_1 * ... * a_i)^(-1) * (a_1 * ... * a_{i-1}) for each i.
The second step is a scalar operation, so its running time is immaterial as long as you aren't doing something too silly.
For my caveman brain, both Fermat's little theorem and square-and-multiply exponentiation are pretty easy to understand. Moreover, the code is going to be "defect-evident"---if I've gotten the logic wrong or forgotten integer promotions or modular reductions as in qsort's post, it'll quickly be clear by skimming the code.
- having Colin stop by your thread is strictly an opportunity for useful information to flow from a singular source to many people
- you would hear that aloud 100 times a day in any office where serious work was being done by professionals on a deadline and think nothing of it, bet your ass places in the world where serious hackers rise only on merit and have the best gear embargoed are saying stuff like that all the time. this nepotism capture bubble is an outlier in the history of serious engineering.
Defining the rudeness threshold down to the point where cperciva clears it is one part comedy and two parts tragedy with the words Hacker News in bold at the top of the page.
Far from it. Asymptotically it's a proportion 2/log(N) of the compute cost.