I found a topics that nCr % P can be calculated in o(P) pre processing and lg(n) query.
I have no idea how it is done. Can anybody help me regarding this topics. Any idea, or any link. Thanks in advance.
I found a topics that nCr % P can be calculated in o(P) pre processing and lg(n) query.
I have no idea how it is done. Can anybody help me regarding this topics. Any idea, or any link. Thanks in advance.
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CodeChef January Challenge problem "MTRICK"?
If my post somehow relates to that problem then I am sorry. It was not my intention. actually I am still stuck on this problem: http://mirror.codeforces.com/blog/entry/9697 Getting TLE. then I after a long time google search I found a topic that states something like this " nCr % P can be calculated in o(P) pre processing and lg(n) query." — Only states doesn't say how. I have a same post on TC forum "http://apps.topcoder.com/forums/?module=Thread&threadID=807841&start=0&mc=4#1828320". but the replies wasn't enough for me to get the idea. And that post on TC was quite before the contest you are talking about. Thank you.
I wrote a detailed blog on finding nCr mod a^p. It also contains details on Euclids and Chinese Remainder Theorems. Do give a good look and give a thumbs up if you like it. www.divyanshkumarsblog.wordpress.com
"No blog entry" it shows. Could you please update it?
I dont know its not working. Maybe a bug. Just copy paste for the time being.
What is nCr?
Also can be written C(n,r) which represent number of ways to choose r things from n things
sorry bad explanation . ya ,right , nCr means C(n,r).
For prime P it is easy to calculate — just apply Lucas theorem. (O(p·log(p)) preprocessing and log(n) by query)
Otherwise, express P as p1a1·...·pkak, calculate C(n, k) modulo piai and get the answer by chinese remainder theorem.
Calculating C(n, k) mod pa, as far as I understood from this paper, are not so easy.
Are you sure that modulo is not prime?
Yes I am sure the modulo is not prime. http://www.lightoj.com/volume_showproblem.php?problem=1318 This is the actual problem where i was thinking to implement that idea. Coz, I have tried all possible C(n,r)%P that i knew, but got TLE. In lucas theorem and all others theorems I have seen , all starts with assuming P prime. :(
OK, maybe it will help:
C(n, k) mod pa can be computed in O(p·log(n)) time:
Calculate the largest b, such that C(n, k) = 0(mod pb)
How to calculate n!! mod pa where n!! is the largest divisor of n! coprime with pa?
n!! = (1·2·...·p - 1·1)·(1·2·...·p - 1·2)·...(1·2·...·n mod p)
We can see that n!! mod pa = (p - 1)![n / p]·(1·2·...·n mod p)·([n / p]!! mod pa)
Answer is C(n, k) = pb · n!! · inverse((n - k)!!) · inverse(k!!)
Finally, you can precalculate all factiorial up to (p - 1)! modulo pa and answer the queries in O(log(n)2) time.
Thanks all ... specially PavelSavchenkov. Solved and learned.
if we take p=2, won't n!!mod p^a always give 1??(irrespective of even a)
How calculate the largest b, such that C(n, k) = 0(mod p^b) ?
You can count, how often the prime p appears in C(n, k). C(n, k) = n! / k! / (n-k)!. So you just need to count, how often p appears in n!, in k! and in (n-k)!.
The expression from --Pavel--'s comment about n!! is incorrect
n!! = (1·2·...·p - 1·1)·(1·2·...·p - 1·2)·...(1·2·...·nmodp)
It doesn't make sense here to rewrite all the numbers mod p if you want to get the answer mod pa.
Any idea for (n!)modpa, where p is a prime number??
Found it: https://es.wikipedia.org/wiki/Teorema_de_Wilson
I could not find anything that proves
n!! = (1·2·...·p - 1·1)·(1·2·...·p - 1·2)·...(1·2·...·nmodp)
I am not getting why we are taking mod p when we need to find n! mod p^a.
How to find inverse modulo (pa). Please, someone explain it. Edit : Got it. a - 1 ≡ (a mod m) - 1 (mod m). So we can calculate a mod m and then use extended euclid's to find the inverse.
how do you calculate when p is not square free ?
I_Love_Shristi what is m here? Is it p^a ?
can you take an example to explain?
What is the logic behind taking n!!?
Great approach based on Wilson's theorem but how can i find inverse(n)!!
The generalized Lucas Theorem isn't all that hard after all. After three days of pondering and reading I finally implemented it. Take a read here.
I submitted this problem in the codechef problem SANDWICH from May challenge and got AC.
"For prime P it is easy to calculate — just apply Lucas theorem (O(p·log(p)) preprocessing ". How ?? precalculating this will take O(p^2)memory and time.
This blog is related to problem in running contest.
This blog created 3 years ago....
But people are discussing now......
백준 1등 쿠사가님 평소에 블로그 잘 보고 있습니다. pavel.savchenkov의 코멘트에 대해서 혹시 설명해주실 수 있으신가요..? 지금 거의 이틀째 nCr mod p^q에 대해서만 보고 있는데 이해가 될만한 문서를 찾지를 못하겠네요.
도와주세요! ㅠㅠ 어떻게 하면 저 과정으로 정답이 성립되는 걸까요? n!! = (1·2·...·p - 1·1)·(1·2·...·p - 1·2)·...(1·2·...·n mod p) << 왜 이런식을 가지는지랑 we can see that으로 넘어가는 부분이 이해가 잘 안가네요. 제발 도와주세요.
쿠사가님 블로그덕분에 이항계수 5번문제까지는 풀었는데, 6번문제를 풀려면 이 이론을 알아야하는데 쉽지가 않네요. 감사합니다.
I didn’t tried to understand pavel.savchenkov‘s reply, and I don’t know why you are asking me here.
Also, you don’t need that knowledge to solve that problem.
I'm sorry for being rude. Because of my bad English and Because I learned binomial coefficient on your tistory, I was glad to see you on this post. That's it. Sorry. Thank you for your comment.
Really? I think this is bad when the question directly solves the problem or it's too obvious that someone is asking about the problem, but this not happen here ... o wait maybe it happen now, you posted the problem!
I wrote a detailed blog on finding nCr mod a^p. It also contains details on Euclids and Chinese Remainder Theorems. Do give a good look and give a thumbs up if you like it.
Divyansh Kumar's Blog