[To] combination number modulus Lucas theorem

For C (n, m) mod p. Here the N,m,p (P is the prime number) are very large cases. The formula for C (n, m) = C (n-1,m) + C (n-1, m-1) can no longer be recursive.

The Lusac theorem is used here.

For non-negative integers m and N and a prime p, the following congruence relation holds:

where

and

is the base p expansions of m and n respectively.

For a separate C (NI, MI) mod p, the known C (n, m) mod p = n!/(m! ( N-M)!) MoD p. Obviously, the division takes the mold, and here we use the m!. (n-m)! The inverse of the element.

According to Fermat theorem:

Known (A, p) = 1, then ap-1≡1 (mod p), so a*ap-2≡1 (mod p).

i.e. (m! ( N-M)!) The inverse is (m! ( N-M)!) P-2;

Code:

#include <iostream>
using namespace Std;
typedef long Long LL;

ll Exp_mod (ll A, ll B, ll P) {    ll res = 1;    while (b! = 0) {        if (b&1) res = (res * a)% P;        A = (a*a)% p;        b >>= 1;    }    return res;} ll Comb (ll A, ll B, ll p) {    if (a < b)   return 0;    if (a = = b)  return 1;    if (b > a)   b = a A;    LL ans = 1, CA = 1, cb = 1;    for (LL i = 0; i < b; ++i) {        CA = (CA * (a-i))%p;        CB = (CB * (b-i))%p;    }    Ans = (Ca*exp_mod (CB, P-2, p))% P;    return ans;} ll Lucas (int n, int m, int p) {     ll ans = 1;     while (N&&m&&ans) {        ans = (Ans*comb (n%p, m%p, p))% P;        n/= p;        M/= p;     }     return ans;} int main () {    Read ();    int n, m, p;    while (~SCANF ("%d%d%d", &n, &m, &p)) {        printf ("%lld\n", Lucas (N, M, p));    }    return 0;}

[To] combination number modulus Lucas theorem