2023-04-16 17:27阅读: 220评论: 0推荐: 0

斯特林数，上升幂，下降幂学习笔记

斯特林数，上升幂，下降幂，普通幂的定义

第二类斯特林数

n	${\binom{n}{0}}$	${\binom{n}{1}}$	${\binom{n}{2}}$	${\binom{n}{3}}$	${\binom{n}{4}}$	${\binom{n}{5}}$	${\binom{n}{6}}$	${\binom{n}{7}}$	${\binom{n}{8}}$	${\binom{n}{9}}$
0	$1$	$0$	$0$	$0$	$0$	$0$	$0$	$0$	$0$	$0$
1	$0$	$1$	$0$	$0$	$0$	$0$	$0$	$0$	$0$	$0$
2	$0$	$1$	$1$	$0$	$0$	$0$	$0$	$0$	$0$	$0$
3	$0$	$1$	$3$	$1$	$0$	$0$	$0$	$0$	$0$	$0$
4	$0$	$1$	$7$	$6$	$1$	$0$	$0$	$0$	$0$	$0$
5	$0$	$1$	$15$	$25$	$10$	$1$	$0$	$0$	$0$	$0$
6	$0$	$1$	$31$	$90$	$65$	$15$	$1$	$0$	$0$	$0$
7	$0$	$1$	$63$	$301$	$350$	$140$	$21$	$1$	$0$	$0$
8	$0$	$1$	$127$	$966$	$1701$	$1050$	$266$	$28$	$1$	$0$
9	$0$	$1$	$255$	$3025$	$7770$	$6951$	$2446$	$462$	$36$	$1$

定义 ${\binom{n}{m}}$ 表示将 $n$ 件物品分成 $m$ 个子集（非空）的方案数。

有

{\binom{n}{m}} = {\binom{n - 1}{m - 1}} + m {\binom{n - 1}{m}}

第一类斯特林数

n	$[\binom{n}{0}]$	$[\binom{n}{1}]$	$[\binom{n}{2}]$	$[\binom{n}{3}]$	$[\binom{n}{4}]$	$[\binom{n}{5}]$	$[\binom{n}{6}]$	$[\binom{n}{7}]$	$[\binom{n}{8}]$	${\binom{n}{9}}$
0	$1$	$0$	$0$	$0$	$0$	$0$	$0$	$0$	$0$	$0$
1	$0$	$1$	$0$	$0$	$0$	$0$	$0$	$0$	$0$	$0$
2	$0$	$1$	$1$	$0$	$0$	$0$	$0$	$0$	$0$	$0$
3	$0$	$2$	$3$	$1$	$0$	$0$	$0$	$0$	$0$	$0$
4	$0$	$6$	$11$	$6$	$1$	$0$	$0$	$0$	$0$	$0$
5	$0$	$24$	$50$	$35$	$10$	$1$	$0$	$0$	$0$	$0$
6	$0$	$120$	$274$	$225$	$85$	$15$	$1$	$0$	$0$	$0$
7	$0$	$720$	$1764$	$1624$	$735$	$175$	$21$	$1$	$0$	$0$
8	$0$	$5040$	$13068$	$13132$	$6769$	$1960$	$322$	$28$	$1$	$0$
9	$0$	$40320$	$109584$	$11824$	$67284$	$22449$	$4536$	$546$	$36$	$1$

定义 $[\binom{n}{m}]$ 表示将 $n$ 件物品分成 $m$ 个轮换（非空，圆排列）的方案数。

有

[\binom{n}{m}] = (n - 1) [\binom{n - 1}{m}] + [\binom{n - 1}{m - 1}]

上升幂，下降幂，普通幂

\begin{aligned} x^{\underset{―}{n}} = \frac{x!}{(x - n)!} \\ x^{\overset{―}{n}} = \frac{(x + n - 1)!}{(x - 1)!} \\ x^{n} = x^{n} \\ x^{\overset{―}{n}} = (x + n - 1)^{\underset{―}{n}} \\ x^{\underset{―}{n}} = (x - n + 1)^{\overset{―}{n}} \end{aligned}

斯特林数的性质

\begin{aligned} \sum_{i = 0}^{n} {\binom{n}{i}} x^{\underset{―}{i}} & = \sum_{i = 0}^{n} (i {\binom{n - 1}{i}} + {\binom{n - 1}{i - 1}}) x^{\underset{―}{i}} \\ = \sum_{i = 0}^{n} i {\binom{n - 1}{i}} x^{\underset{―}{i}} + \sum_{i = 0}^{n} {\binom{n - 1}{i - 1}} x^{\underset{―}{i}} \\ = \sum_{i = 0}^{n} i {\binom{n - 1}{i}} x^{\underset{―}{i}} + \sum_{i = 1}^{n} {\binom{n - 1}{i - 1}} x^{\underset{―}{i}} \\ = \sum_{i = 0}^{n} i {\binom{n - 1}{i}} x^{\underset{―}{i}} + \sum_{i = 1}^{n} {\binom{n - 1}{i - 1}} x^{\underset{―}{(i - 1) + 1}} \\ = \sum_{i = 0}^{n} i {\binom{n - 1}{i}} x^{\underset{―}{i}} + \sum_{i = 0}^{n - 1} {\binom{n - 1}{i}} x^{\underset{―}{i + 1}} \\ = \sum_{i = 0}^{n - 1} i {\binom{n - 1}{i}} x^{\underset{―}{i}} + \sum_{i = 0}^{n - 1} {\binom{n - 1}{i}} x^{\underset{―}{i}} (x - i) \\ = \sum_{i = 0}^{n - 1} i {\binom{n - 1}{i}} x^{\underset{―}{i}} + {\binom{n - 1}{i}} x^{\underset{―}{i}} (x - i) \\ = x \sum_{i = 0}^{n - 1} {\binom{n - 1}{i}} x^{\underset{―}{i}} \\ = x \times x^{n - 1} = x^{n} \end{aligned}

\begin{aligned} \sum_{i = 0}^{n} [\binom{n}{i}] x^{i} & = \sum_{i = 0}^{n} ((n - 1) [\binom{n - 1}{i}] + [\binom{n - 1}{i - 1}]) x^{i} \\ = \sum_{i = 0}^{n} (n - 1) [\binom{n - 1}{i}] x^{i} + \sum_{i = 0}^{n} [\binom{n - 1}{i - 1}] x^{i} \\ = \sum_{i = 0}^{n} (n - 1) [\binom{n - 1}{i}] x^{i} + \sum_{i = 1}^{n} [\binom{n - 1}{i - 1}] x^{(i - 1) + 1} \\ = \sum_{i = 0}^{n} (n - 1) [\binom{n - 1}{i}] x^{i} + \sum_{i = 0}^{n - 1} [\binom{n - 1}{i}] x^{i + 1} \\ = (n - 1 + x) \sum_{i = 0}^{n} [\binom{n - 1}{i}] x^{i} \\ = (x + n - 1) x^{\overset{―}{n - 1}} \\ = x^{\overset{―}{n}} \end{aligned}

\begin{aligned} x^{\underset{―}{n}} & = (- 1)^{n} (- x)^{\overset{―}{n}} \\ = (- 1)^{n} \sum_{i = 0}^{n} [\binom{n}{i}] (- x)^{i} \\ = \sum_{i = 0}^{n} [\binom{n}{i}] (- 1)^{n - i} x^{i} \end{aligned}

\begin{aligned} x^{n} & = (- 1)^{n} (- x)^{n} \\ = (- 1)^{n} \sum_{i = 0}^{n} {\binom{n}{i}} (- x)^{\underset{―}{i}} \\ = (- 1)^{n} \sum_{i = 0}^{n} {\binom{n}{i}} (- 1)^{i} x^{\overset{―}{i}} \\ = \sum_{i = 0}^{n} {\binom{n}{i}} (- 1)^{n - i} x^{\overset{―}{i}} \end{aligned}

整理一下

\begin{aligned} x^{n} = \sum_{i = 0}^{n} {\binom{n}{i}} x^{\underset{―}{i}} \\ x^{n} = \sum_{i = 0}^{n} {\binom{n}{i}} (- 1)^{n - i} x^{\overset{―}{i}} \\ x^{\overset{―}{n}} = \sum_{i = 0}^{n} [\binom{n}{i}] x^{i} \\ x^{\underset{―}{n}} = \sum_{i = 0}^{n} [\binom{n}{i}] (- 1)^{n - i} x^{i} \end{aligned}

表示普通幂时用第二类斯特林数，表示上升幂、下降幂时用第一类斯特林数，用大数表示小数时用 $(- 1)^{n - i}$ 。

考虑公式的嵌套，~~让斯特林数更毒瘤~~：

\begin{aligned} x^{n} & = \sum_{i = 0}^{n} {\binom{n}{i}} (- 1)^{n - i} x^{\overset{―}{i}} \\ = \sum_{i = 0}^{n} {\binom{n}{i}} (- 1)^{n - i} \sum_{m = 0}^{i} [\binom{i}{m}] x^{m} \\ = \sum_{m = 0}^{n} x^{m} \sum_{i = m}^{n} {\binom{n}{i}} [\binom{i}{m}] (- 1)^{n - i} \end{aligned}

\sum_{i = m}^{n} {\binom{n}{i}} [\binom{i}{m}] (- 1)^{n - i} = [n = m]

\begin{aligned} x^{\underset{―}{n}} & = \sum_{i = 0}^{n} [\binom{n}{i}] (- 1)^{n - i} x^{i} \\ = \sum_{i = 0}^{n} [\binom{n}{i}] (- 1)^{n - i} \sum_{j = 0}^{i} {\binom{i}{j}} x^{\underset{―}{j}} \\ = \sum_{j = 0}^{i} x^{\underset{―}{j}} \sum_{i = j}^{n} [\binom{n}{i}] {\binom{i}{j}} (- 1)^{n - i} \end{aligned}

\sum_{i = m}^{n} [\binom{n}{i}] {\binom{i}{m}} (- 1)^{n - i} = [m = n]

既然有二项式反演，也有斯特林反演！

\begin{aligned} g (n) = \sum_{i = 0}^{n} {\binom{n}{i}} f (i) & ⟺ f (n) = \sum_{i = 0}^{n} [\binom{n}{i}] (- 1)^{n - i} g (i) \\ g (n) = \sum_{i = 0}^{n} [\binom{n}{i}] f (i) & ⟺ f (n) = \sum_{i = 0}^{n} {\binom{n}{i}} (- 1)^{n - i} g (i) \end{aligned}

~~证明略~~。

在来个柿子：

x^{n} = \sum_{i = 0}^{x} {\binom{n}{i}} (\binom{x}{i}) i!

考虑组合意义来解释， $x^{n}$ 表示将 $n$ 个小球放进 $x$ 个盒子的方案数， $\sum_{i = 0}^{x}$ 表示有几个盒子非空， $(\binom{x}{i})$ 表示选哪几个盒子非空， $i!$ 表示非空盒子的顺序， ${\binom{n}{i}}$ 表示 $n$ 个小球划分成 $i$ 个集合的方案数。

二项式反演可得：

x^{n} = \sum_{i = 0}^{x} {\binom{n}{i}} (\binom{x}{i}) i! ⟺ {\binom{n}{m}} m! = \sum_{i = 0}^{m} (\binom{m}{i}) (- 1)^{m - i} i^{n}

\begin{aligned} {\binom{n}{m}} & = \sum_{i = 0}^{m} \frac{(\binom{m}{i}) (- 1)^{m - i} i^{n}}{m!} \\ = \sum_{i = 0}^{m} \frac{m!}{i! (m - i)! m!} (- 1)^{m - i} i^{n} \\ = \sum_{i = 0}^{m} \frac{i^{n} (- 1)^{m - i}}{i! (m - i)!} \\ = \sum_{i = 0}^{m} \frac{i^{n}}{i!} \times \frac{(- 1)^{m - i}}{(m - i)!} \end{aligned}

可以发现这是个卷积， $N T T$ 即可计算同一行的第二类斯特林数。

例题

P6620 [省选联考 2020 A 卷] 组合数问题

遇到不可以直接计算的多项式、组合数计数题时可以考虑普通幂转下降幂或上升幂。

\begin{aligned} (\sum_{k = 0}^{n} f (k) x^{k} (\binom{n}{k})) mod p & = \sum_{i = 0}^{m} \sum_{k = 0}^{n} b_{i} k^{\underset{―}{i}} x^{k} (\binom{n}{k}) \\ = \sum_{i = 0}^{m} \sum_{k = 0}^{n} b_{i} n^{\underset{―}{i}} x^{k} (\binom{n - i}{k - i}) \\ = \sum_{i = 0}^{m} b_{i} n^{\underset{―}{i}} \sum_{k = 0}^{n} (\binom{n - i}{k - i}) x^{k} \\ = \sum_{i = 0}^{m} b_{i} n^{\underset{―}{i}} \sum_{k = 0}^{n - i} (\binom{n - i}{k - i}) x^{k} 1^{n - i - k} \\ = \sum_{i = 0}^{m} b_{i} n^{\underset{―}{i}} x^{i} \sum_{k = 0}^{n - i} (\binom{n - i}{k}) x^{k} 1^{n - i - k} \\ = \sum_{i = 0}^{m} b_{i} n^{\underset{―}{i}} x^{i} (x + 1)^{n - i} \end{aligned}

其中

\begin{array}{r} f (x) = \sum_{i = 0}^{m} b_{i} x^{\underset{―}{i}} \end{array}

有

\begin{aligned} f (x) & = \sum_{i = 0}^{m} a_{i} x^{i} \\ = \sum_{i = 0}^{m} a_{i} \sum_{j = 0}^{i} {\binom{i}{j}} x^{\underset{―}{j}} \\ = \sum_{j = 0}^{m} x^{\underset{―}{j}} \sum_{i = j}^{m} {\binom{i}{j}} a_{i} \end{aligned}

第二类斯特林数直接暴力 $m^{2}$ 计算即可，总时间复杂度为 $O (m^{2})$ 。

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 3001;
ll stirling[N][N], MOD;
void get_stirling(ll tt){
  stirling[0][0]=1;
  for(ll i = 1; i <= tt; i++)
    for(ll j = 1; j <= i; j++)
      stirling[i][j] = 1ll*(1ll*stirling[i-1][j]*j%MOD+1ll*stirling[i-1][j-1]) % MOD;
}
ll my_pow(ll x, ll y){
  ll res = 1;
  for(;y;y>>=1,x=x*x%MOD)
    if(y&1)res=res*x%MOD;
  return res;
}
ll n, X, m;
ll a[N], b[N];
int main(){
  cin>>n>>X>>MOD>>m;
  get_stirling(m);
  for(ll i = 0; i <= m; i++)
    cin>>a[i];
  for(ll i = 0; i <= m; i++)
    for(ll j = i; j <= m; j++)
      b[i] = (b[i]+stirling[j][i]*a[j]%MOD)%MOD;
  ll ans = 0;
  for(ll i = 0, now = 1; i <= m; i++){
    ll res = b[i] * now % MOD * my_pow(X, i) % MOD * my_pow(X+1, n-i) % MOD;
    ans = (ans + res) % MOD;
    now = now * (n-i) % MOD;
  }
  cout<<(ans+MOD)%MOD<<endl;
  return 0;
}

P5395 第二类斯特林数·行

x^{n} = \sum_{i = 0}^{x} {\binom{n}{i}} (\binom{x}{i}) i!

二项式反演可得：

x^{n} = \sum_{i = 0}^{x} {\binom{n}{i}} (\binom{x}{i}) i! ⟺ {\binom{n}{m}} m! = \sum_{i = 0}^{m} (\binom{m}{i}) (- 1)^{m - i} i^{n}

\begin{aligned} {\binom{n}{m}} & = \sum_{i = 0}^{m} \frac{(\binom{m}{i}) (- 1)^{m - i} i^{n}}{m!} \\ = \sum_{i = 0}^{m} \frac{m!}{i! (m - i)! m!} (- 1)^{m - i} i^{n} \\ = \sum_{i = 0}^{m} \frac{i^{n} (- 1)^{m - i}}{i! (m - i)!} \\ = \sum_{i = 0}^{m} \frac{i^{n}}{i!} \times \frac{(- 1)^{m - i}}{(m - i)!} \end{aligned}

可以发现这是个卷积， $N T T$ 即可计算同一行的第二类斯特林数。

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll MOD = 167772161;
const ll N = 1e6 + 11;
ll rev[N], w[N], gen = 3;
ll my_pow(ll x, ll y){
  ll res = 1;
  for(;y;y>>=1,x=x*x%MOD)
    if(y&1)res=res*x%MOD;
  return res;
}
void NTT(ll *a, ll Len, bool type){
  for(int i = 0; i < Len; i++){
    rev[i] = (rev[i>>1]>>1) + (i&1?Len>>1:0);
    if(rev[i]>i)swap(a[rev[i]], a[i]);
  }
  for(ll d = 1; d < Len; d <<= 1){
    ll W = my_pow(gen, (MOD-1)/(d*2));
    if(type) W = my_pow(W, MOD-2);
    w[0] = 1;
    for(ll i = 1; i < d; i++)
      w[i] = w[i-1] * W % MOD;
    for(ll fir = 0; fir < Len; fir += (d*2)){
      ll sec = fir + d;
      for(ll i = 0 ; i < d; i++){
        ll a0 = a[fir+i], a1 = a[sec+i]*w[i]%MOD;
        a[fir+i] = (a0+a1)%MOD;
        a[sec+i] = (a0-a1+MOD)%MOD;
      }
    }
  }
  if(type){
    ll invlen=my_pow(Len, MOD-2);
    for(int i=0; i<Len; i++)
      a[i] = a[i]*invlen%MOD;
  }
}
ll f[N], g[N], fac[N], ifac[N], n;
ll fu(int k){return k&1?-1:1;}
int main(){
  cin>>n;
  ifac[0]=fac[0]=1;
  for(ll i = 1; i <= n; i++) fac[i] = fac[i-1] * i % MOD;
  ifac[n] = my_pow(fac[n], MOD-2);
  for(ll i = n-1; i; i--) ifac[i] = ifac[i+1] * (i+1) % MOD;
  for(ll i = 0; i <= n; i++){
    f[i] = fu(i)*ifac[i]%MOD, f[i] = (f[i]+MOD)%MOD;
    g[i] = my_pow(i, n) * ifac[i] % MOD;
    // cout << f[i] << " " << g[i] << endl;
  }
  ll len = 1;
  while(len<=(n*2))len<<=1;
  NTT(f, len, 0), NTT(g, len, 0);
  for(int i = 0; i < len; i++)
    f[i] = f[i] * g[i] % MOD, f[i] = (f[i] + MOD) % MOD;
  NTT(f, len, 1);
  for(int i = 0; i <= n; i++)
    cout << f[i] << " ";
  cout << endl;
  return 0;
}

P5408 第一类斯特林数·行

有

x^{\overset{―}{n}} = \sum_{i = 0}^{n} [\binom{n}{i}] x^{i}

考虑倍增：

x^{\overset{―}{2 n}} = x^{\overset{―}{n}} (x + n)^{\overset{―}{n}}

假设我们求出了 $x^{\overset{―}{n}}$ 的多项式表达 $f 1$ 。

现在要求 $(x + n)^{\overset{―}{n}}$ 的多项式表达 $f 2$ 。

有:

f 1 (x) = f 2 (x - n)

设 $a_{i}$ 为 $f 1$ 的系数，有：

\begin{aligned} f 2 & = \sum_{i = 0}^{n} a_{i} (x + n)^{i} \\ = \sum_{i = 0}^{n} a_{i} \sum_{j = 0}^{i} x^{j} n^{i - j} (\binom{i}{j}) \\ = \sum_{j = 0}^{n} x^{j} \sum_{i = j}^{n} n^{i - j} a^{i} \frac{i!}{j! (i - j)!} \\ = \sum_{j = 0}^{n} \frac{x^{j}}{j!} \sum_{i = j}^{n} a^{i} i! \frac{n^{i - j}}{(i - j)!} \end{aligned}

可以发现里面是个卷积， $N T T$ 计算即可。

注意卷积时 $L e n$ 的长度不能大也不能小（代码第 $48$ 行）。

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll MOD = 167772161;
const ll N = 7e5, gen=3;

ll my_pow(ll x, ll y){
  ll res = 1;
  for(;y;y>>=1,x=x*x%MOD)
    if(y&1)res=res*x%MOD;
  return res;
}
ll inv(ll x){
  return my_pow(x, MOD-2);
}

ll rev[N], w[N];
void ntt(ll *a, ll Len, bool type){
  for(ll i = 0 ; i < Len; i++){
    rev[i] = (rev[i>>1]>>1) + (i&1?Len>>1:0);
    if(rev[i]>i)swap(a[rev[i]], a[i]);
  }
  for(ll d=1; d<Len; d<<=1){
    ll W = my_pow(gen, (MOD-1)/(d*2));
    if(type) W = inv(W);
    w[0]=1;
    for(ll i = 1; i < d; i++)
      w[i] = w[i-1] * W % MOD;
    for(ll fir=0; fir<Len; fir+=d*2){
      ll sec=fir+d;
      for(ll i=0; i<d; i++){
        ll a0=a[fir+i], a1=a[sec+i]*w[i]%MOD;
        a[fir+i]=(a0+a1)%MOD;
        a[sec+i]=(a0-a1+MOD)%MOD;
      }
    }
  }
  if(type){
    ll invlen=inv(Len);
    for(ll i=0; i<Len; i++)
      a[i]=a[i]*invlen%MOD;
  }
}

ll aa[N], bb[N];
void mul(ll *a, ll *b, ll alen, ll blen){
  ll len=1;
  while(len<=(alen+blen))len<<=1;//不能多也不能少！（len=xlen+ylen）
  ntt(a, len, 0);
  ntt(b, len, 0);
  for(int i=0; i<len; i++)
    a[i]=a[i]*b[i]%MOD;
  ntt(a, len, 1);
}

ll ans[N], s[N], A[N], B[N], C[N];
ll fac[N], ifac[N];

void sol_striling_one_line(ll tt){
  if(tt==1){s[1]=1;return;}
  if(tt&1){
    sol_striling_one_line(tt-1);
    for(ll i=tt; i; i--)
      s[i] = (s[i-1] + s[i]*(tt-1)%MOD)%MOD;
    s[0]=s[0]*(tt-1)%MOD;
    return;
  }
  ll slen = tt/2, res=1;
  sol_striling_one_line(slen);
  for(int i=0; i<=slen; i++)
    A[i]=s[i]*fac[i]%MOD,
    B[i]=res*ifac[i]%MOD,
    res=res*slen%MOD;
  reverse(A, A+slen+1);
  mul(A, B, slen+1, slen+1);
  for(int i=0; i<=slen; i++)
    C[i]=A[slen-i]*ifac[i]%MOD;
  mul(s,C,slen+1,slen+1);
  int len=1;
  while(len<=(tt+2))len<<=1;
  for(int i=slen+1; i<len; i++)A[i]=B[i]=C[i]=0;
  for(int i=tt+1; i<len; i++)s[i]=0;
}

void init_fac(int tt){
  fac[0]=ifac[0]=1;
  for(ll i=1; i<=tt; i++)
    fac[i]=fac[i-1]*i%MOD;
  ifac[tt]=my_pow(fac[tt], MOD-2);
  for(ll i=tt-1; i; i--)
    ifac[i] = ifac[i+1]*(i+1)%MOD;
}

ll* striling_first_line(ll tt){
  init_fac(tt<<1);
  memset(A, 0, sizeof A);
  memset(B, 0, sizeof B);
  memset(s, 0, sizeof s);
  memset(C, 0, sizeof C);
  sol_striling_one_line(tt);
  return s;
}


int main(){
  ll n;
  cin>>n;
  ll* striling=striling_first_line(n);
  for(int i=0; i<=n; i++) cout<<striling[i]<<' ';
  cout<<'\n';
  return 0;
}