Ceizenpok’s Formula

题目描述

Dr. Ceizenp’ok from planet i1c5l became famous across the whole Universe thanks to his recent discovery – the Ceizenpok’s formula. This formula has only three arguments: $n, k$ and $m$, and its value is a number of $k$-combinations of a set of $n$ modulo $m$.
While the whole Universe is trying to guess what the formula is useful for, we need to automate its calculation.

题意概述

求${n \choose k} \bmod m$。
数据范围：$1 \le n \le 10^{18}, \; 0 \le k \le n, \; 2 \le m \le 10^6$。

算法分析

令
$$
x={n \choose k}, \; m=p_1^{a_1}p_2^{a_2} \cdots p_q^{a_q}
$$
可以列出方程组
$$
\left\{
\begin{array}{c}
x \equiv r_1 \pmod{p_1^{a_1}} \\
x \equiv r_2 \pmod{p_2^{a_2}} \\
\cdots \\
x \equiv r_q \pmod{p_q^{a_q}}
\end{array}
\right.
$$
由于模数两两互质，所以该方程组在模$m$意义下有唯一解。
考虑如何求$r_i$。实际上，我们要求的就是${n \choose k} \bmod p_i^{a_i}$。我们知道
$$
{n \choose k}={n! \over k!(n-k)!}
$$
那么只要求出$n! \bmod p_i^{a_i}, \; k! \bmod p_i^{a_i}, \; (n-k)! \bmod p_i^{a_i}$的值，就可以用逆元求出${n \choose k} \bmod p_i^{a_i}$。
对于如何求$n! \bmod p_i^{a_i}$，令
$$
f(n)=n! \bmod p_i^{a_i}
$$
由$x \equiv x+p_i^{a_i} \pmod{p_i^{a_i}}$，可得
$$
f(n)=\left(f\left(\left\lfloor{n \over p_i}\right\rfloor\right)\cdot p_i^{\lfloor n/p_i \rfloor}\cdot\left(\prod_{i \in [1, p_i^{a_i}], \; p_i\not\mid i}i\right)^{\lfloor n/p_i^{a_i} \rfloor}\cdot\prod_{i \in [1, n \bmod p_i^{a_i}], \; p_i\not\mid i}i\right)\bmod p_i^{a_i}
$$
但是$k!, \; (n-k)!$在模$p_i^{a_i}$意义下可能不存在逆元，因此需要将$n!, \; k!, \; (n-k)!$中的$p_i$因子提取出来，求出逆元后再乘回去。
这样就得到了所有$r_i$。用中国剩余定理求解方程组即可。

代码

/*
 * Your lucky number has been disconnected.
 */

#include <algorithm>
#include <cstdio>
#include <cstring>

#define int long long

int power(int a, int b, int m) {
  int ret = 1;
  for (; b; b >>= 1)
    b & 1 && ((ret *= a) %= m), (a *= a) %= m;
  return ret;
}

void ex_gcd(int a, int b, int &x, int &y) {
  if (!b) {
    x = 1, y = 0;
    return;
  }
  ex_gcd(b, a % b, y, x), y -= a / b * x;
}

int get_inv(int a, int b) {
  int x, y;
  ex_gcd(a, b, x, y);
  return (x + b) % b;
}

int get_fac(int n, int p, int k) {
  int m = power(p, k, 1e9), ret = 1;
  for (; n; n /= p) {
    if (n / m) {
      int rec = 1;
      for (int i = 2; i < m; ++i)
        if (i % p)
          (rec *= i) %= m;
      (ret *= power(rec, n / m, m)) %= m;
    }
    for (int i = n % m; i > 1; --i)
      if (i % p)
        (ret *= i) %= m;
  }
  return ret;
}

int calc(int N, int K, int M, int p, int k) {
  int a = get_fac(N, p, k), b = get_fac(K, p, k), c = get_fac(N - K, p, k),
      cnt = 0;
  for (int i = N; i; i /= p)
    cnt += i / p;
  for (int i = K; i; i /= p)
    cnt -= i / p;
  for (int i = N - K; i; i /= p)
    cnt -= i / p;
  int m = power(p, k, 1e9),
      ret = a * get_inv(b, m) % m * get_inv(c, m) % m * power(p, cnt, m) % m;
  return ret * (M / m) % M * get_inv(M / m, m) % M;
}

signed main() {
  int N, K, M, ans = 0;
  scanf("%lld%lld%lld", &N, &K, &M);
  for (int i = 2, t = M; t > 1; ++i)
    if (!(t % i)) {
      int k = 0;
      for (; !(t % i); ++k, t /= i)
        ;
      (ans += calc(N, K, M, i, k)) %= M;
    }
  printf("%lld\n", ans);
  return 0;
}