Modulo the combination number

Before you look at this, we suggest you take a look at n! ......

for the number of combinations we can represent them as factorial forms : C (n,k) = . Then we might as well put these three factorial in the form of a previous topic. In this case, if the e1>e2+e3 can be divisible by p ,e1=e2+e3 cannot be p divisible. C (n,k) =a1 (a2a3)-1 in case of an inability to be divisible

1 intMod_comb (intNintKintp) {2     if(n<0|| k<0|| N&LT;K)return 0;3     intA1=mod_fact (N,P,E1), A2=mod_fact (K,p,e2), A3=mod_fact (nk,p,e3);4     if(E1&GT;E2+E3)return 0;5     returnA1*mod_inverse (a2*a3%p, p)%p;6}

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In addition, we can also use Lucas to beg.

Lucas theorem: We make n=sp+q, M=tp+r (q,r≤p) have

The concrete proof is as follows:

Let's first prove a simple formula:

C (P, f)%p= p!/(f! ( P-F)! %p because P is a prime number, and F and (P-F) on the denominator are smaller than p, that is, there are no numbers in the denominator that can take p, so%p must be equal to 0. Similarly, with a little expansion, we can get C (n,f)%p=0 (n>=p). (In fact, there is no egg for the proof of Lucas theorem)

By proving the formula, we can begin to prove Lucas ' theorem.

for (1+x) sp+q≡ (1+x) SPX (1+X) Q≡ ((1+X) p) SX (1+X) Q≡ (based on the two-polynomial theorem, and then in the formula above, so we can get) (1+XP) SX (1+X) Q≡ (again according to the two-item theorem) ≡ (mod p)

In the end we can get

Let's calculate the coefficients of xtp+r on the left and right sides:

Left =c (sp+q, tp+r), right =c (S, t) XC (q, R)

Because left = right, so we finally got Lucas theorem.

With Lucas ' theorem, we have this code, and this code is really good at understanding

1 intLucas (intNintMintp) {2     Long Longans=1;3      while(n&&m&&ans) {4Ans*=comb (n%p,m%p,p)%p;//Comb () to find the number of combinations5N/=p;m/=p;6     }7     returnans;8}

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Modulo the combination number