Firstly, we denote the multiplicative inverse of x mod p as inv(x,p).
- use dp method to calculation x! mod p for x=1 ~ n (1<=n<p, p is some prime)
- calculate inv(n!,p) utilize Extended Euclidean algorithm.
- use dp again to calculate inv(x!,p) for x=n-1 ~ 1 with the fact inv(x!,p) * x = inv((x-1)!, p)
- now, if we want to now inv(x,p) for some x in [1,n], we only need to calculate (x-1)! * inv(x!,p)
Or just do the following:
Great! I've seen that approach once before and since then I was trying to find it again. It's some kind of magic, but I'm going to understand how that works...
Maybe this will help :)
It's quite easy to explain how it works. Let's take simple equation and do some transformations with it:
UPD: too slow
but doesnt a muplicative inverse of x mod p only exist when gcd(x,p) == 1,how do you find the muplicative inverse of a range if some of them dont exist
We can only use this approach if we have prime modulo p and want to calculate inverse in range [1, p).
i have seen this approach in rng_58's solution.
I know that this is like 6 years old, but this post shows up a lot on Google, and I wanted to mention that you can omit the last
% p
.Challange: p = 113513 look at the value of inv[3] :)
That's just because the multiplication between p/i and inv[p%i] overflows, I think.
If so, another good warning for future searches is: Don't forget to declare inv as a long long array (or cast to a long long before multiplying)!
2. calculate inv(x!,p)
Do you mean inv(n!,p)?
we only need to calculate (x-1)! * inv(x,p)
Do you mean (x-1)! * inv(x!,p)?
Yes, you are right. :p Now I fix it.
Nice method! The also exists O(NloglogN) dp solution, which is slower, but seems to be very easy to implement. Let f[x] be the smallest prime, which divides x. We can calculate f[x] for all x in range 1...n using sieve of Eratosthenes. Then we can calculate inv[x] using formula inv[x]=(inv[x/f[x]]*inv[f[x]])%mod.
If write good sieve of Eratosthenes, it works O(n).
Yes, you are right.
Can you share your most optimized code about sieve Eratosthenes?
Is the following function correct to calculate nCr? ( n Choose r )
(inv[x] is Modular Multiplicative Inverse of x!(x factorial) and fac[x] is x!. i64 is macro of long long. modL is large prime number.)
Yes it is.
(((333333336*500000004)%1000000007)*120)%1000000007 is 20.
I tried to calculate 5C2 and found it 20 from this function.
The correct output is 10.
this and this sites are used for calculation.
Well you don't need inverse of 3 and 2 you need the inverse of 3! = 6 and 2! = 2 (because nCr is equal to n! / ( r! * (n — r)! ) so 5C2 is ( 5! * inverse[6] * inverse[2] ) % 1000000007 = ( 120 * 166666668 * 500000004 ) % 1000000007 = 10.
sorry, i wrote this under wrong blog announcement