斯特林数学习笔记

第一类斯特林数

生成函数

式子

(f(x)=prodlimits_{i=0}^{n-1}(x+i))

证明

使用数学归纳法。

递推式

式子

(left[egin{matrix}n\kend{matrix} ight]=(n-1)left[egin{matrix}n-1\kend{matrix} ight]+left[egin{matrix}n-1\k-1end{matrix} ight])

证明

假设最后一个元素自己成为一个圆排列，或者最后一个元素插到前面的 (n-1) 的元素的前面(后面类似)，形成圆排列，也就是前面的每一个长度为 (n) 的圆排列提供了 (n) 个不同的方案来插入。

性质

1

(n!=sumlimits_{i=0}^nleft[egin{matrix}n\iend{matrix} ight])

2

(left[egin{matrix}n\n-2end{matrix} ight]=2 imes C_n^3+3 imes C_N^4)

求行

倍增求解，多项式卷积，可以看洛谷的题解。

例题

题目1

P5408 【模板】第一类斯特林数·行

注意

卡常。

另外一定记得哪些函数需要取反，一个是

(F(x)=LastFuntion_x imes x!)

(LastFunction) 表示上一次倍增的函数

一个是

(H(x)=sumlimits_{i=0}^nF(i) imes G(n-i))

其中

(G(x)=frac{n^x}{x!})

代码

#include <cstdio>

using namespace std;

inline void swap (int &a, int &b) {
	int c = a;
	a = b,
	b = c;
}
const int N = (262144 + 5) << 1, mod = 167772161, inver = 55924054;
inline int add (int u, int v) { return u + v >= mod ? u + v - mod : u + v; }
inline int mns (int u, int v) { return u - v < 0 ? u - v + mod : u - v; }
inline int mul (int u, int v) { return 1LL * u * v % mod; }
inline int qpow (int u, int v) {
	int tot = 1, base = u % mod;
	while (v) {
		if (v & 1) tot = mul (tot, base);
		base = mul (base, base);
		v >>= 1;
	}
	return tot;
}

int rev[N], sz, cnt2;
inline void NTT (int *a, int n, int bit, int flag) {
	for (int i = 0; i < n; ++i) {
		rev[i]  = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
		if (i > rev[i]) swap (a[i], a[rev[i]]);
	}
	for (int l = 2; l <= n; l <<= 1) {
		int wi = qpow (flag ? inver : 3, (mod - 1) / l),
		m = l / 2;
		for (int *k = a; k != a + n; k += l) {
			int w = 1;
			for (int i = 0; i < m; ++i) {
				int tmp = mul (k[i + m], w);
				k[i + m] = mns (k[i], tmp);
				k[i] = add (k[i], tmp);
				w = mul (w, wi);
			}
		}
	}
	int tmp = qpow (n, mod - 2);
	for (int i = 0; i < n && flag; ++i) {
		a[i] = mul (a[i], tmp);
	}
}

inline void MUL (int *a, int *b, int *c, int la, int lb) {
	static int mula[N << 1], mulb[N << 1], n, bit;
	n = 1,
	bit = 0;
	while (n < la + lb - 1) n <<= 1, ++bit;
	for (int i = 0; i < la; ++i) mula[i] = a[i]; for (int i = la; i < n; ++i) mula[i] = 0;
	for (int i = 0; i < lb; ++i) mulb[i] = b[i]; for (int i = lb; i < n; ++i) mulb[i] = 0;
	NTT (mula, n, bit, false), NTT (mulb, n, bit, false);
	for (int i = 0; i < n; ++i) mula[i] = mul (mula[i], mulb[i]);
	NTT (mula, n, bit, true);
	for (int i = 0; i < la + lb - 1; ++i) c[i] = mula[i];
}

int fac[N], invfac[N];
inline void init () {
	fac[0] = 1, fac[1] = 1;
	for (int i = 2; i <= (sz << 1); ++i) fac[i] = mul (fac[i - 1], i);
	invfac[sz << 1] = qpow (fac[sz << 1], mod - 2);
	for (int i = (sz << 1) - 1; i >= 0; --i) invfac[i] = mul (invfac[i + 1], i + 1);
}

inline void solve (int *a, int len) {
	if (len == 1) { a[0] = 0, a[1] = 1; return ; }
	static int f[N << 1], g[N << 1], h[N << 1], xd, l, r;
	solve (a, len >> 1);
	xd = len >> 1;
	for (int i = 0; i <= xd; ++i) f[xd - i] = mul (a[i], fac[i]);
	int tmp = 1;
	for (int i = 0; i <= xd; ++i) g[i] = mul (tmp, invfac[i]), tmp = mul (tmp, xd);
	MUL (f, g, h, xd + 1, xd + 1);
	l = 0, r = xd; while (l < r) swap (h[l], h[r]), ++l, --r;
	for (int i = 0; i <= xd; ++i) h[i] = mul (h[i], invfac[i]);
	MUL (a, h, a, xd + 1, xd + 1);
	if (len % 2) {
		g[0] = len - 1, g[1] = 1;
		MUL (a, g, a, len, 2);
	}
}

int main () {
	scanf ("%d", &sz);
	init ();
	static int ans[N << 1];
	solve (ans, sz);
	for (int i = 0; i <= sz; ++i) printf ("%d ", ans[i]);
	return 0;
}