卢卡斯定理

卢卡斯定理(Lucas Law)

\(_{如无特殊说明, 则本文除法向下取整}\)

核心内容

对正整数 \(m\), \(n\), 质数 \(p\), 有

\[\binom{n}{m}~\%~p = \binom{\frac{n}{p}}{\frac{m}{p}} \binom{n~\%~p}{m~\%~p} ~\%~p \]

证明

从 \(\binom{p}{m}~(p > m)\) 开始证明

\[\binom{p}{m} = \frac{p!}{m!(p - m)!} = \frac px \frac{(p - 1)!}{(m - 1)!((p - 1) - (m - 1))!} = \frac pm \binom{p - 1}{m - 1} \]

由于是模 \(p\) 意义下的运算, 且 \(gcd(x, p) = 1\), 所以对 \(m \in (0, p)\) 有

\[\binom{p}{m}~\%~p = pInv_m \binom{p - 1}{m - 1}~\%~p = 0 \]

代入到二项式定理可得

\[(1 + x)^p~\%~p = \sum_{i = 0}^p \binom{p}{i}x^i~\%~p = (1 + x^p)~\%~p \]

将二项式指数推广到正整数 \(n\), 设 \(q_m = \frac mp\), \(q_n = \frac np, r_m = m~\%~p, r_n = n~\%~p\)

\[(1 + x)^n = (1 + x)^{q_np + r_n} = (1 + x)^{q_np}(1 + x)^{r_n} \]

使用前面的结论 \((1 + x)^p~\%~p = (1 + x^p)~\%~p\), 整理上式得

\[(1 + x)^n~\%~p = (1 + x^p)^{q_n}(1 + x)^{r_n}~\%~p \]

代入二项式定理

\[(1 + x)^n~\%~p = \sum_{i = 0}^{q_n}(\binom{q_n}{i}x^{ip}\sum_{j = 0}^{r_n}\binom{r_n}{j}x^j)~\%~p \]

整理得

\[(1 + x)^n~\%~p = \sum_{i = 0}^{q_n}\sum_{j = 0}^{r_n}(\binom{q_n}{i}\binom{r_n}{j}x^{ip+j})~\%~p \]

由于 \(j < p\), 所以任意一组 \(i\), \(j\) 对应唯一的 \(ip + j\) 值, 和 \([0, n]\) 内的每一个数一一对应, 所以将式子改写为枚举 \(m = ip + j\) 的值.

\[(1 + x)^n~\%~p = \sum_{m = 0}^{n} \binom{q_n}{q_m} \binom{r_n}{r_m}x^k~\%~p \]

所以, 模 \(p\) 意义下, \((1 + x)^n\) 的 \(m\) 次项系数 \(\binom nm\) 就是 \(\binom{q_n}{q_m} \binom{r_n}{r_m}\)

定理得证

模板Luogu3807

求 \(T\) 组 \(\binom{n + m}{n}~\%~p\)

\(1 \leq n, m, p \leq 10^5, 1 \leq T \leq 10\), \(p\) 为质数

由于 \(p\) 为质数且 \(p \leq 10^5\), 可以用 Lucas 定理 递归求解

\[\binom {n + m}n~\%~p = \binom {(n + m)~\%~p}{n~\%~p} \binom {(n + m)/p}{m / p}~\%~p \]

unsigned Lucas (unsigned x, unsigned y) {
  if(x <= Mod && y <= Mod) {
    return Binom(x, y);
  }
  return ((long long)Binom(x % Mod, y % Mod) * Lucas(x / Mod, y / Mod)) % Mod;
}

递归边界就是 \(x, y \in [0, p)\), 直接 \(O(n)\) 求出 \(\binom xy\)

需要注意的是, 当遇到取模之后 \(x < y\) 时, \(\binom xy = 0\), 因为 \((1 + x)^x\) 不存在 \(y\) 次项 (括号内的 \(x\) 是自变量, 指数 \(x\) 是函数调用的参数)

还有, 当 \(y = 0\) 的时候, \(\binom xy = 1\), 尽管不加这个特判返回值也是 \(0\), 但是不知道为什么我还是加了

unsigned Binom (unsigned x, unsigned y) {
  unsigned Up(1), Down(1);
  if (y > x) {
    return 0;
  }
  if(!y) {
    return 1;
  }
  for (register unsigned i(2); i <= x; ++i) {
    Up = ((long long)Up * i) % Mod;
  }
  for (register unsigned i(2); i <= y; ++i) {
    Down = ((long long)Down * i) % Mod;
  }
  for (register unsigned i(2); i <= x - y; ++i) {
    Down = ((long long)Down * i) % Mod;
  }
  Down = Power(Down, Mod - 2);
  return ((long long)Up * Down) % Mod;
}

函数 Binom() 中的 Power() 用来根据欧拉定理求乘法逆元, 即 \(Inv_x = x^{p - 2}\)

unsigned Power (unsigned x, unsigned y) {
  if(!y) {
    return 1;
  }
  unsigned tmp(Power(((long long)x * x) % Mod, y >> 1));
  if(y & 1) {
    return ((long long)x * tmp) % Mod;
  }
  return tmp;
}

最后放出 main(), 可以看出又多了一句无关紧要的特判, 一看就是让 julian 毒害过的可怜人

int main() {
  t = RD();
  for (register unsigned T(1); T <= t; ++T){
    Clr();
    if(!(n && m)) {
      printf("1\n");
      continue; 
    }
    printf("%u\n", Lucas(n + m, n));
  }
  return Wild_Donkey;
}