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	\(n\brace 0\)	\(n\brace 1\)	\(n\brace 2\)	\(n\brace 3\)	\(n\brace 4\)	\(n\brace 5\)	\(n\brace 6\)	\(n\brace 7\)	\(n\brace 8\)	\(n\brace9\)
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 \({n\brace m}\) 表示将 \(n\) 件物品分成 \(m\) 个子集（非空）的方案数。

有

\[{n\brace m} = {n-1\brace m-1} + m{n-1\brace m} \]

第一类斯特林数

n	\(n\brack 0\)	\(n\brack 1\)	\(n\brack 2\)	\(n\brack 3\)	\(n\brack 4\)	\(n\brack 5\)	\(n\brack 6\)	\(n\brack 7\)	\(n\brack 8\)	\(n\brace9\)
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 \(n\brack m\) 表示将 \(n\) 件物品分成 \(m\) 个轮换（非空，圆排列）的方案数。

有

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

上升幂，下降幂，普通幂

\[\begin{aligned} &x^{\underline{n}} = \frac{x!}{(x-n)!}\\ &x^{\overline{n}} = \frac{(x+n-1)!}{(x-1)!}\\ &x^n = x^n\\ &x^{\overline{n}} = (x+n-1)^{\underline{n}}\\ &x^{\underline{n}} = (x-n+1)^{\overline{n}}\\ \end{aligned} \]

斯特林数的性质

\[\begin{aligned} \sum_{i=0}^{n}{n\brace i}x^{\underline{i}} &= \sum_{i=0}^{n}(i{n-1\brace i}+{n-1\brace i-1})x^{\underline{i}}\\ &= \sum_{i=0}^{n}i{n-1\brace i}x^{\underline{i}}+\sum_{i=0}^{n}{n-1\brace i-1}x^{\underline{i}}\\ &= \sum_{i=0}^{n}i{n-1\brace i}x^{\underline{i}}+\sum_{i=1}^{n}{n-1\brace i-1}x^{\underline{i}}\\ &= \sum_{i=0}^{n}i{n-1\brace i}x^{\underline{i}}+\sum_{i=1}^{n}{n-1\brace i-1}x^{\underline{(i-1)+1}}\\ &= \sum_{i=0}^{n}i{n-1\brace i}x^{\underline{i}}+\sum_{i=0}^{n-1}{n-1\brace i}x^{\underline{i+1}}\\ &= \sum_{i=0}^{n-1}i{n-1\brace i}x^{\underline{i}}+\sum_{i=0}^{n-1}{n-1\brace i}x^{\underline{i}}(x-i)\\ &= \sum_{i=0}^{n-1}i{n-1\brace i}x^{\underline{i}}+{n-1\brace i}x^{\underline{i}}(x-i)\\ &= x\sum_{i=0}^{n-1}{n-1\brace i}x^{\underline{i}}\\ &= x \times x^{n-1} = x^{n}\\ \end{aligned} \]

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

\[\begin{aligned} x^{\underline{n}} &= (-1)^n(-x)^{\overline{n}}\\ &= (-1)^n\sum_{i=0}^{n}{n\brack i}(-x)^i\\ &= \sum_{i=0}^{n}{n\brack i}(-1)^{n-i}x^i\\ \end{aligned} \]

\[\begin{aligned} x^n &= (-1)^n(-x)^n\\ &= (-1)^n\sum_{i=0}^{n}{n\brace i}(-x)^{\underline{i}}\\ &= (-1)^n\sum_{i=0}^{n}{n\brace i}(-1)^ix^{\overline{i}}\\ &= \sum_{i=0}^{n}{n\brace i}(-1)^{n-i}x^{\overline{i}}\\ \end{aligned} \]

整理一下

\[\begin{aligned} &x^n = \sum_{i=0}^{n}{n\brace i}x^{\underline{i}}\\ &x^n = \sum_{i=0}^{n}{n\brace i}(-1)^{n-i}x^{\overline{i}}\\ &x^{\overline{n}} = \sum_{i=0}^{n}{n\brack i}x^i\\ &x^{\underline{n}} = \sum_{i=0}^{n}{n\brack i}(-1)^{n-i}x^i\\ \end{aligned} \]

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~~：

\[\begin{aligned} x^n &= \sum_{i=0}^{n}{n\brace i}(-1)^{n-i}x^{\overline{i}}\\ &= \sum_{i=0}^{n}{n\brace i}(-1)^{n-i}\sum_{m=0}^i{i\brack m}x^{m}\\ &= \sum_{m=0}^{n}x^{m}\sum_{i=m}^{n}{n\brace i}{i\brack m}(-1)^{n-i} \end{aligned} \]

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

\[\begin{aligned} x^{\underline{n}} &= \sum_{i=0}^{n}{n\brack i}(-1)^{n-i}x^{i}\\ &= \sum_{i=0}^{n}{n\brack i}(-1)^{n-i}\sum_{j=0}^{i}{i\brace j}x^{\underline{j}}\\ &= \sum_{j=0}^{i}x^{\underline{j}}\sum_{i=j}^{n}{n\brack i}{i\brace j}(-1)^{n-i}\\ \end{aligned} \]

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

既然有二项式反演，也有斯特林反演！

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

~~证明略~~。

在来个柿子：

\[x^n = \sum_{i=0}^{x}{n\brace i}\dbinom{x}{i}i! \]

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，\(\tbinom{x}{i}\) 表示选哪几个盒子非空， \(i!\) 表示非空盒子的顺序， \({n\brace i}\) 表示 \(n\) 个小球划分成 \(i\) 个集合的方案数。

二项式反演可得：

\[x^n = \sum_{i=0}^{x}{n\brace i}\dbinom{x}{i}i! \iff {n\brace m}m! = \sum_{i=0}^{m}\dbinom{m}{i}(-1)^{m-i}i^n \]

\[\begin{aligned}{n\brace m} &= \sum_{i=0}^{m}\frac{\tbinom{m}{i}(-1)^{m-i}i^n}{m!}\\&=\sum_{i=0}^{m}\dfrac{m!}{i!(m-i)!m!}(-1)^{m-i}i^n\\&=\sum_{i=0}^{m}\dfrac{i^n(-1)^{m-i}}{i!(m-i)!}\\&=\sum_{i=0}^{m}\dfrac{i^n}{i!}\times\dfrac{(-1)^{m-i}}{(m-i)!}\end{aligned} \]

可以发现这是个卷积， \(NTT\) 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。

\[\begin{aligned} (\sum_{k=0}^{n}f(k)x^k\dbinom{n}{k}) \bmod p &=\sum_{i=0}^{m}\sum_{k=0}^{n}b_ik^{\underline{i}}x^k\dbinom{n}{k}\\ &=\sum_{i=0}^{m}\sum_{k=0}^{n}b_in^{\underline{i}}x^k\dbinom{n-i}{k-i}\\ &=\sum_{i=0}^{m}b_in^{\underline{i}}\sum_{k=0}^{n}\dbinom{n-i}{k-i}x^k\\ &=\sum_{i=0}^{m}b_in^{\underline{i}}\sum_{k=0}^{n-i}\dbinom{n-i}{k-i}x^k1^{n-i-k}\\ &=\sum_{i=0}^{m}b_in^{\underline{i}}x^{i}\sum_{k=0}^{n-i}\dbinom{n-i}{k}x^k1^{n-i-k}\\ &=\sum_{i=0}^{m}b_in^{\underline{i}}x^{i}(x+1)^{n-i}\\ \end{aligned} \]

其中

\[\begin{aligned} f(x) = \sum_{i=0}^{m}b_ix^{\underline{i}} \end{aligned} \]

有

\[\begin{aligned} f(x) &= \sum_{i=0}^{m}a_ix^i\\ &= \sum_{i=0}^{m}a_i\sum_{j=0}^{i}{i\brace j}x^{\underline{j}}\\ &= \sum_{j=0}^{m}x^{\underline{j}}\sum_{i=j}^{m}{i\brace j}a_i\\ \end{aligned} \]

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

\[x^n = \sum_{i=0}^{x}{n\brace i}\dbinom{x}{i}i! \]

二项式反演可得：

\[x^n = \sum_{i=0}^{x}{n\brace i}\dbinom{x}{i}i! \iff {n\brace m}m! = \sum_{i=0}^{m}\dbinom{m}{i}(-1)^{m-i}i^n \]

可以发现这是个卷积， \(NTT\) 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

\[x^{\overline{n}} = \sum_{i=0}^{n}{n\brack i}x^i \]

考虑倍增：

\[x^{\overline{2n}} = x^{\overline{n}}(x+n)^{\overline{n}} \]

假设我们求出了 \(x^{\overline{n}}\) 的多项式表达 \(f1\) 。

现在要求 \((x+n)^{\overline{n}}\) 的多项式表达 \(f2\) 。

有:

\[f1(x) = f2(x-n) \]

设 \(a_i\) 为 \(f1\) 的系数，有：

\[\begin{aligned} f2 &= \sum_{i=0}^{n}a_i(x+n)^i\\ &= \sum_{i=0}^{n}a_i\sum_{j=0}^{i}x^jn^{i-j}\dbinom{i}{j}\\ &= \sum_{j=0}^{n}x^j\sum_{i=j}^{n}n^{i-j}a^i\dfrac{i!}{j!(i-j)!}\\ &= \sum_{j=0}^{n}\dfrac{x^j}{j!}\sum_{i=j}^{n}a^ii!\dfrac{n^{i-j}}{(i-j)!}\\ \end{aligned} \]

可以发现里面是个卷积， \(NTT\) 计算即可。

注意卷积时 \(Len\) 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;
}

posted @ 2023-04-16 17:27 fzrcy 阅读(284) 评论(0) 收藏举报

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