第一 / 二类 Stirling 数的一些平庸求法

2020-01-28

综述

看到洛谷上有求 Stirling 数的神奇板子, 于是就想写 = =

~~虽然不知道会有什么用处, 其实是不想写题又不敢颓废吧~~

大概是一些式子推推推然后拉板子算算算…

第一类 Stirling 数

本文采用 $\begin{bmatrix} n \ m \end{bmatrix}$ 表示第一类 Stirling 数.

Part 1 行

题目链接: https://www.luogu.com.cn/problem/P5408

我们知道, $x$ 的 $n$ 次上升幂就是第 $n$ 行第一类 Stirling 数的生成函数, 即

$x^{\overline n} = \prod_{i=0}^{n-1} (x + i) = \sum_{i=0}^n \begin{bmatrix} n \\ i \end{bmatrix} x^i$

论据如下, 归纳法易证

$\begin{bmatrix} n \\ m \end{bmatrix} = \begin{bmatrix} n-1 \\ m-1 \end{bmatrix} + (n-1) \cdot \begin{bmatrix} n-1 \\ m \end{bmatrix}$

此时通过分治 FFT 可以在 $O(n \log^2 n)$ 的时间复杂度内计算.

但是用倍增可以做到 $O(n \log n)$.

考虑倍增, 有

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

写成这个形式鬼知道怎么推, 于是

$x^{\overline{2n}} = \prod_{i=0}^{n-1} (x+i) \prod_{i=0}^{n-1} (x+i+n)$

设 $f(x) = \prod\limits_{i=0}^{n-1} (x+i) = \sum\limits_{i=0}^n a_i x^i$, 如果能根据 $f(x)$ 算 $f(x+n)$ 就可以了.

$f(x+n) = \sum_{i=0}^n a_i (x+n)^i = \sum_{i=0}^n a_i \sum_{j=0}^{i} \binom{i}{j} x^i n^{i-j}$

交换 $i$, $j$ 枚举顺序, 得

$f(x+n) = \sum_{i=0}^n x^i \sum_{j=i}^n \binom{j}{i} n^{j-i} a_i$

展开, 有

$f(x+n) = \sum_{i=0}^n \frac{x^i}{i!} \sum_{j=i}^n \frac{n^{j-i}}{(j-i)!} j!\ a_j$

最后一个和式的卷积形式并不显然, 还要将其中一项翻转… 不过还是很套路的

代码实现

UPD on 2020.5.22 重写了这部分的代码

> 点击显示 / 隐藏

// Luogu P5408
// DeP
#include <cstdio>
#include <algorithm>
using namespace std;

const int LOG = 19, MAXN = 1 << LOG | 1, P = 167772161, G = 3;

int W[LOG][MAXN];
int inv[MAXN], ifac[MAXN], fac[MAXN];

int fpow(int base, int b) {
  int ret = 1;
  while (b > 0) {
    if (b & 1) ret = 1LL * ret * base % P;
    base = 1LL * base * base % P, b >>= 1;
  }
  return ret;
}

namespace Poly {
  int r[MAXN];
  inline void init(const int& Lim, const int& L) {
    for (int i = 1; i < Lim; ++i) r[i] = (r[i>>1] >> 1) | ((i & 1) << (L-1));
  }

  void NTT(int* f, const int& Lim, const int& type) {
    for (int i = 1; i < Lim; ++i) if (i < r[i]) swap(f[i], f[r[i]]);
    for (int k = 0, Mid = 1; Mid < Lim; ++k, Mid <<= 1) {
      const int* w = W[k];
      for (int i = 0; i < Lim; i += Mid << 1)
        for (int j = 0; j < Mid; ++j) {
          int f0 = f[i+j], f1 = 1LL * w[j] * f[i+j+Mid] % P;
          f[i+j] = (f0 + f1) % P, f[i+j+Mid] = (f0 - f1 + P) % P;
        }
    }
    if (type < 0) {
      int iv = fpow(Lim, P - 2);
      for (int i = 0; i < Lim; ++i) f[i] = 1LL * f[i] * iv % P;
      reverse(f + 1, f + Lim);
    }
  }

  void Mul(int* f, const int& n, int* g, const int& m, int* h) {
    static int A[MAXN], B[MAXN];
    int Lim = 1, L = 0;
    while (Lim < n + m - 1) Lim <<= 1, ++L;
    for (int i = 0; i < Lim; ++i)
      A[i] = (i < n)? f[i]: 0, B[i] = (i < m)? g[i]: 0;
    init(Lim, L), NTT(A, Lim, 1), NTT(B, Lim, 1);
    for (int i = 0; i < Lim; ++i) h[i] = 1LL * A[i] * B[i] % P;
    NTT(h, Lim, -1);
  }
}

void PolyPre(int N) {
  inv[1] = 1;
  for (int i = 2; i <= N; ++i)
    inv[i] = 1LL * inv[P % i] * (P - P / i) % P;
  ifac[0] = fac[0] = 1;
  for (int i = 1; i <= N; ++i) {
    fac[i] = 1LL * fac[i - 1] * i % P;
    ifac[i] = 1LL * ifac[i - 1] * inv[i] % P;
  }
  for (int w, k = 0, Mid = 1; k < LOG; ++k, Mid <<= 1) {
    W[k][0] = 1, w = fpow(G, (P - 1) / (Mid << 1));
    for (int i = 1; i < Mid; ++i)
      W[k][i] = 1LL * w * W[k][i - 1] % P;
  }
}

namespace Stirling {
  void offset(int* f, const int& Mid, const int& k, int* g) {
    static int A[MAXN], B[MAXN];
    for (int pwk = 1, i = 0; i < Mid; ++i) {
      A[Mid - i - 1] = 1LL * f[i] * fac[i] % P;
      B[i] = 1LL * pwk * ifac[i] % P, pwk = 1LL * pwk * k % P;
    }
    Poly::Mul(A, Mid, B, Mid, A);
    for (int i = 0; i < Mid; ++i)
      g[i] = 1LL * A[Mid - i - 1] * ifac[i] % P;
  }

  void solve(int* f, const int& n) {
    static int f0[MAXN];
    if (!n)
      return void( f[0] = 1 );
    int Mid = n / 2;
    // 得到 f(x + Mid)
    solve(f, Mid), offset(f, Mid + 1, Mid, f0);
    Poly::Mul(f, Mid + 1, f0, Mid + 1, f0);
    if (n & 1) {
      // n 为奇数时, 有乘 (x + (n-1)) 的情况, 此时直接乘就好了.
      for (int i = 0; i <= n; ++i)
        f[i] = (((i > 0)? f0[i - 1]: 0) + 1LL * (n - 1) * f0[i] % P) % P;
    } else
      for (int i = 0; i <= n; ++i) f[i] = f0[i];
  }
}

int n;
int f[MAXN];

int main() {
#ifndef ONLINE_JUDGE
  freopen("input.in", "r", stdin);
#endif
  scanf("%d", &n);
  PolyPre(n), Stirling::solve(f, n);
  for (int i = 0; i <= n; ++i)
    printf("%d%c", f[i], " \n"[i == n]);
  return 0;
}

Part 2 列

题目链接: https://www.luogu.com.cn/problem/P5409

首先有第一类 Stirling 数的一个性质, 即

$x^{\underline n} = \prod_{i=0}^{n-1} (x-i) = \sum_{i=0}^n (-1)^{n-i} \begin{bmatrix} n \\ i \end{bmatrix} x^i$

同样也是归纳法易证.

对于 $(1+x)^t$, 由牛顿二项式定理, 可知

$(1+x)^t = \sum_{i=0}^\infty \binom{t}{i} x^i$

其中 $\binom{t}{i} = \frac{t (t-1) \cdots (t-i+1)}{i!}$, 不妨写作

$(1+x)^t = \sum_{i=0}^\infty \frac{t^{\underline i}}{i!} x^i$

愉快套公式, 得

$(1+x)^t = \sum_{i=0}^\infty \frac{1}{i!} x^i \sum_{j=0}^i (-1)^{i-j} \begin{bmatrix} i \\ j \end{bmatrix} t^j$

交换枚举顺序, 得

$(1+x)^t = \sum_{i=0}^\infty t^i \sum_{j=i}^\infty \frac{1}{j!} x^j (-1)^{j-i} \begin{bmatrix}j \\ i \end{bmatrix}$

同时, 有

$(1+x)^t = \exp (t \cdot \ln (1+x)) = \sum_{i=0}^\infty \frac{1}{i!} \ln^i (1+x)\ t^i$

对应项系数相等, 得

$\frac{1}{i!} \ln^i (1+x) = \sum_{j=i}^\infty \frac{1}{j!} (-1)^{j-i} \begin{bmatrix}j \\ i \end{bmatrix} x^j$

可以看到, 其中 $i$ 为给定的值, 利用多项式幂函数算就好了.

值得注意的是, $[x^0] \ln (1+x)$ 有不等于 $1$ 的可能, 所以要用鲁棒性更强的板子 = =

时间复杂度是带常数的 $O(n \log n)$, 大概是跑不过 $O(n \log^2 n)$ 的带常数吧

代码实现

> 点击显示 / 隐藏

// Luogu P5409
// DeP
#include <cstdio>
#include <algorithm>
using namespace std;

const int MAXN = 131075 << 1;
const int P = 167772161, G = 3, iG = 55924054;

int fac[MAXN], ifac[MAXN], inv[MAXN];

int fpow(int base, int b, int m = P) {
    int ret = 1;
    while (b > 0) {
        if (b & 1) ret = 1LL * ret * base % m;
        base = 1LL * base * base % m, b >>= 1;
    }
    return ret;
}

namespace Poly {
    int r[MAXN];
    inline void init(const int& Lim, const int& L) {
        for (int i = 1; i < Lim; ++i) r[i] = (r[i>>1] >> 1) | ((i & 1) << (L-1));
    }

    void NTT(int* f, int Lim, int type) {
        for (int i = 1; i < Lim; ++i) if (i < r[i]) swap(f[i], f[r[i]]);
        for (int Mid = 1; Mid < Lim; Mid <<= 1) {
            int unit = fpow(type > 0? G: iG, (P-1) / (Mid << 1));
            for (int i = 0; i < Lim; i += Mid << 1) {
                int w = 1;
                for (int j = 0; j < Mid; ++j, w = 1LL * w * unit % P) {
                    int f0 = f[i+j], f1 = 1LL * w * f[i+j+Mid] % P;
                    f[i+j] = (f0 + f1) % P, f[i+j+Mid] = (f0 - f1 + P) % P;
                }
            }
        }
        if (type < 0) {
            int inv = fpow(Lim, P-2);
            for (int i = 0; i < Lim; ++i) f[i] = 1LL * inv * f[i] % P;
        }
    }

    void Inv(int* f, int* g, int n) {
        static int A[MAXN], B[MAXN];
        g[0] = fpow(f[0], P-2);
        for (int L = 0, Lim = 1, Mid = 2; Mid < 2*n; Mid <<= 1) {
            while (Lim < 2*Mid) Lim <<= 1, ++L;
            for (int i = 0; i < Mid; ++i) A[i] = f[i], B[i] = g[i];
            for (int i = Mid; i < Lim; ++i) A[i] = B[i] = 0;
            init(Lim, L), NTT(A, Lim, 1), NTT(B, Lim, 1);
            for (int i = 0; i < Lim; ++i)
                g[i] = ((B[i] + B[i]) % P - 1LL * A[i] * B[i] % P * B[i] % P + P) % P;
            NTT(g, Lim, -1);
            for (int i = min(Mid, n); i < Lim; ++i) g[i] = 0;
        }
    }

    void Der(int* f, int* g, int n) {
        for (int i = 1; i < n; ++i) g[i-1] = 1LL * i * f[i] % P;
        g[n-1] = 0;
    }
    void Int(int* f, int* g, int n) {
        g[0] = 0;
        for (int i = n-1; i; --i) g[i] = 1LL * inv[i] * f[i-1] % P;
    }
    
    void Ln(int* f, int* g, int n) {
        static int A[MAXN], B[MAXN];
        Der(f, A, n), Inv(f, B, n);
        int Lim = 1, L = 0;
        while (Lim < 2*n) Lim <<= 1, ++L;
        for (int i = n; i < Lim; ++i) A[i] = B[i] = 0;
        init(Lim, L), NTT(A, Lim, 1), NTT(B, Lim, 1);
        for (int i = 0; i < Lim; ++i) A[i] = 1LL * A[i] * B[i] % P;
        NTT(A, Lim, -1), Int(A, g, n);
    }

    void Exp(int* f, int* g, int n) {
        static int A[MAXN], B[MAXN], lng[MAXN];
        g[0] = 1;
        for (int L = 0, Lim = 1, Mid = 2; Mid < 2*n; Mid <<= 1) {
            Ln(g, lng, Mid);
            while (Lim < 2*Mid) Lim <<= 1, ++L;
            for (int i = 0; i < Mid; ++i) A[i] = (f[i] - lng[i] + P) % P, B[i] = g[i];
            A[0] = (A[0] + 1) % P;
            for (int i = Mid; i < Lim; ++i) A[i] = B[i] = 0;
            init(Lim, L), NTT(A, Lim, 1), NTT(B, Lim, 1);
            for (int i = 0; i < Lim; ++i) g[i] = 1LL * A[i] * B[i] % P;
            NTT(g, Lim, -1);
            for (int i = min(Mid, n); i < Lim; ++i) g[i] = 0;
        }
    }

    void Pow(int* f, int* g, int n, int K) {
        static int lng[MAXN];
        int pos = 0;
        while (!f[pos]) ++pos;
        int Mid = n - pos;
        for (int i = 0; i < Mid; ++i) g[i] = f[i+pos];
        int base = g[0], inv = fpow(base, P-2);
        for (int i = 0; i < Mid; ++i) g[i] = 1LL * g[i] * inv % P;
        Ln(g, lng, Mid);
        for (int i = 0; i < Mid; ++i) lng[i] = 1LL * lng[i] * K % P;
        Exp(lng, g, Mid);
        base = fpow(base, K);
        for (int i = 0; i < Mid; ++i) g[i] = 1LL * g[i] * base % P;
        pos = min(pos * K, n);
        for (int i = n-1; i >= pos; --i) g[i] = g[i-pos];
        for (int i = 0; i < pos; ++i) g[i] = 0;
    }
}

int n, K;
int f[MAXN];

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.in", "r", stdin);
#endif
    // input
    scanf("%d%d", &n, &K);
    // init
    int N = n;
    ifac[0] = fac[0] = inv[1] = 1;
    for (int i = 2; i <= N; ++i) inv[i] = 1LL * (P - P / i) * inv[P % i] % P;
    for (int i = 1; i <= N; ++i) fac[i] = 1LL * i * fac[i-1] % P;
    ifac[N] = fpow(fac[N], P-2);
    for (int i = N; i; --i) ifac[i-1] = 1LL * i * ifac[i] % P;
    // solve
    f[0] = f[1] = 1;
    Poly::Ln(f, f, n+1); Poly::Pow(f, f, n+1, K);
    for (int i = 0; i <= n; ++i) f[i] = 1LL * f[i] * ifac[K] % P;
    for (int i = 0; i <= n; ++i)
        f[i] = 1LL * f[i] * fac[i] % P * (((i-K) & 1)? P-1: 1) % P;
    // output
    for (int i = 0; i <= n; ++i) printf("%d%c", f[i], " \n"[i==n]);
    return 0;
}

第二类斯特林数

本文采用 $\begin{Bmatrix} n \ m \end{Bmatrix}$ 表示第二类 Stirling 数.

Part 1 行

题目链接: https://www.luogu.com.cn/problem/P5395

~~感觉这个是最简单的了~~

考虑第二类 Stirling 数的组合意义, 也就是把 $n$ 个有区别的小球放入 $m$ 个无区别的盒子里的方案数.

其实这个东西是可以用容斥算的, 可以得到

$\begin{Bmatrix} n \\ m \end{Bmatrix} = \frac{1}{m!} \sum_{k=0}^n (-1)^{k} \binom{m}{k} (m-k)^n$

右边和式的意义为, 在 $m$ 个有区别的盒子里挑 $k$ 个空盒子, 然后把 $n$ 个小球丢到盒子里. 因为容斥时假设盒子有区别, 最后还要除以 $m!$.

把这个式子展开就可以看到卷积的形式了.

$\begin{Bmatrix} n \\ m \end{Bmatrix} = \sum_{k=0}^n \frac{(-1)^k}{k!} \frac{(m-k)^n}{(m-k)!}$

通过 NTT 就可以在 $O(n \log n)$ 的时间复杂度内计算.

代码实现

> 点击显示 / 隐藏

// Luogu P5395
// DeP
#include <cstdio>
#include <algorithm>
using namespace std;

const int MAXN = 1e6+5;
const int P = 167772161, G = 3, iG = 55924054;

int fpow(int base, int b, int m = P) {
    int ret = 1;
    while (b > 0) {
        if (b & 1) ret = 1LL * ret * base % m;
        base = 1LL * base * base % m, b >>= 1;
    }
    return ret;
}

namespace Poly {
    int r[MAXN];
    inline void init(const int& Lim, const int& L) {
        for (int i = 1; i < Lim; ++i) r[i] = (r[i>>1] >> 1) | ((i & 1) << (L-1));
    }

    void NTT(int* f, int Lim, int type) {
        for (int i = 1; i < Lim; ++i) if (i < r[i]) swap(f[i], f[r[i]]);
        for (int Mid = 1; Mid < Lim; Mid <<= 1) {
            int unit = fpow(type > 0? G: iG, (P-1) / (Mid << 1));
            for (int i = 0; i < Lim; i += Mid << 1) {
                int w = 1;
                for (int j = 0; j < Mid; ++j, w = 1LL * w * unit % P) {
                    int f0 = f[i+j], f1 = 1LL * w * f[i+j+Mid] % P;
                    f[i+j] = (f0 + f1) % P, f[i+j+Mid] = (f0 - f1 + P) % P;
                }
            }
        }
        if (type < 0) {
            int inv = fpow(Lim, P-2);
            for (int i = 0; i < Lim; ++i) f[i] = 1LL * f[i] * inv % P;
        }
    }
}

int n;
int f[MAXN], g[MAXN];
int fac[MAXN], ifac[MAXN];

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.in", "r", stdin);
#endif
    // input
    scanf("%d", &n);
    // init
    fac[0] = ifac[0] = 1;
    for (int i = 1; i <= n; ++i) fac[i] = 1LL * fac[i-1] * i % P;
    ifac[n] = fpow(fac[n], P-2);
    for (int i = n; i; --i) ifac[i-1] = 1LL * ifac[i] * i % P;
    // solve
    for (int i = 0; i <= n; ++i) {
        f[i] = (i & 1)? P - ifac[i]: ifac[i];
        g[i] = 1LL * fpow(i, n) * ifac[i] % P;
    }
    int Lim = 1, L = 0;
    while (Lim < 2*n) Lim <<= 1, ++L;
    Poly::init(Lim, L), Poly::NTT(f, Lim, 1), Poly::NTT(g, Lim, 1);
    for (int i = 0; i < Lim; ++i) f[i] = 1LL * f[i] * g[i] % P;
    Poly::NTT(f, Lim, -1);
    // output
    for (int i = 0; i <= n; ++i) printf("%d%c", f[i], " \n"[i==n]);
    return 0;
}

Part 2 列

题目链接: https://www.luogu.com.cn/problem/P5396

设第二类 Stirling 数一列的生成函数为 $S_m$, 即

$S_m (x) = \sum_{i=0}^\infty \begin{Bmatrix} i \\ m \end{Bmatrix} x^i$

根据第二类 Stirling 数递推式, 可以得到

$S_m (x) = \sum_{i=1}^\infty \begin{Bmatrix} i-1 \\ m-1 \end{Bmatrix} x^i + m \sum_{i=1}^\infty \begin{Bmatrix} i-1 \\ m \end{Bmatrix} x^i$ $S_m (x) = x \cdot S_{m-1} (x) + m x \cdot S_m (x)$

移项并整理, 得

$S_m (x) = \frac{x}{1-mx} S_{m-1} (x) = \frac{x^m}{\prod_{i=1}^m (1-ix)}$

分式的分母可以使用分治 FFT 得到, 然后求逆并乘一个单项式 (也就是把系数右移) 即可.

所以我们就得到了 $O(n \log^2 n)$ 的做法, 然而存在 $O(n\log n)$ 且小常数的倍增优秀做法, 只是我学不动了…

佐以 O2 食用更佳.

代码实现

> 点击显示 / 隐藏

// Luogu P5396
// DeP
#include <cstdio>
#include <algorithm>
using namespace std;

const int MAXN = 131075 << 1, LOG = 35;
const int P = 167772161, G = 3, iG = 55924054;

int fpow(int base, int b, int m = P) {
    int ret = 1;
    while (b > 0) {
        if (b & 1) ret = 1LL * base * ret % m;
        base = 1LL * base * base % m, b >>= 1;
    }
    return ret;
}

namespace Poly {
    int r[MAXN];
    int tmp[LOG][MAXN], ptr = -1;
    inline void init(const int& Lim, const int& L) {
        for (int i = 1; i < Lim; ++i) r[i] = (r[i>>1] >> 1) | ((i & 1) << (L-1));
    }

    void NTT(int* f, int Lim, int type) {
        for (int i = 1; i < Lim; ++i) if (i < r[i]) swap(f[i], f[r[i]]);
        for (int Mid = 1; Mid < Lim; Mid <<= 1) {
            int unit = fpow(type > 0? G: iG, (P-1) / (Mid << 1));
            for (int i = 0; i < Lim; i += Mid << 1) {
                int w = 1;
                for (int j = 0; j < Mid; ++j, w = 1LL * w * unit % P) {
                    int f0 = f[i+j], f1 = 1LL * w * f[i+j+Mid] % P;
                    f[i+j] = (f0 + f1) % P, f[i+j+Mid] = (f0 - f1 + P) % P;
                }
            }
        }
        if (type < 0) {
            int inv = fpow(Lim, P-2);
            for (int i = 0; i < Lim; ++i) f[i] = 1LL * f[i] * inv % P;
        }
    }

    void Inv(int* f, int* g, int n) {
        static int A[MAXN], B[MAXN];
        g[0] = fpow(f[0], P-2);
        for (int L = 0, Lim = 1, Mid = 2; Mid < 2*n; Mid <<= 1) {
            while (Lim < 2*Mid) Lim <<= 1, ++L;
            for (int i = 0; i < Mid; ++i) A[i] = f[i], B[i] = g[i];
            for (int i = Mid; i < Lim; ++i) A[i] = B[i] = 0;
            init(Lim, L), NTT(A, Lim, 1), NTT(B, Lim, 1);
            for (int i = 0; i < Lim; ++i)
                g[i] = ((B[i] + B[i]) % P - 1LL * A[i] * B[i] % P * B[i] % P + P) % P;
            NTT(g, Lim, -1);
            for (int i = min(Mid, n); i < Lim; ++i) g[i] = 0;
        }
    }

    void solve(int* f, int L, int R) {
        if (L == R) return f[0] = 1, void( f[1] = P - L );
        int Mid = (L + R) / 2, *f0 = tmp[++ptr], *f1 = tmp[++ptr];
        solve(f0, L, Mid), solve(f1, Mid+1, R);
        int Lim = 1, l = 0;
        while (Lim <= R-L+1) Lim <<= 1, ++l;
        init(Lim, l), NTT(f0, Lim, 1), NTT(f1, Lim, 1);
        for (int i = 0; i < Lim; ++i) f[i] = 1LL * f0[i] * f1[i] % P, f0[i] = f1[i] = 0;
        NTT(f, Lim, -1), ptr -= 2;
    }
}

int n, K;
int f[MAXN], g[MAXN];

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.in", "r", stdin);
#endif
    // input
    scanf("%d%d", &n, &K);
    // solve
    Poly::solve(f, 1, K);
    Poly::Inv(f, g, n+1);
    for (int i = 0; i+K <= n; ++i) f[i+K] = g[i];
    for (int i = 0; i < K; ++i) f[i] = 0;
    // output
    for (int i = 0; i <= n; ++i) printf("%d%c", f[i], " \n"[i==n]);
    return 0;
}

参考资料

Richard A. Brualdi. 组合数学. 机械工业出版社, 2012.4.
斯特林数及斯特林反演 by y2823774827y
题解 P5408 【模板】第一类斯特林数·行 by NaCly_Fish
题解 P5409 【模板】第一类斯特林数·列 by Great_Influence
题解 P5396 【模板】第二类斯特林数·列 by Great_Influence

DepletedPrism's Blog

第一 / 二类 Stirling 数的一些平庸求法

综述

第一类 Stirling 数

Part 1 行

代码实现

Part 2 列

代码实现

第二类斯特林数

Part 1 行

代码实现

Part 2 列

代码实现

参考资料