题目传送门

题目大意

给出集合\(S\)和整数\(n\)，求出有多少个多叉树使得每个节点的孩子个数都在\(S\)中，且叶子个数为\(n\)。

思路

啊，居然没有看出来可以用拉格朗日反演，果然还是自己太菜了。。。

我们设答案的生成函数为\(F\)，\(G\)为集合\(S\)的生成函数，可以得到:

应该很显然吧，这里就懒得解释了。

我们发现这个式子可以化成:

我们如果设\(C(x)=x-G(c)\)，那我们就可以得到:

我们就发现\(C\)和\(F\)互为复合逆，于是使用拉格朗日反演可以得到：

于是，我们就可以做到\(\Theta(n\log n)\)解决这个问题了。

\(\text {Code}\)

#include <bits/stdc++.h>
using namespace std;

#define Int register int
#define mod 950009857
#define ll long long
#define MAXN 400005
#define Gi 7

int quick_pow (int a,int b,int c){
	int res = 1;for (;b;b >>= 1,a = 1ll * a * a % c) if (b & 1) res = 1ll * res * a % c;
	return res;
}

int limit = 1,l,r[MAXN];

void NTT (int *a,int type){
	for (Int i = 0;i < limit;++ i) if (i < r[i]) swap (a[i],a[r[i]]);
	for (Int mid = 1;mid < limit;mid <<= 1){
		int Wn = quick_pow (Gi,(mod - 1) / (mid << 1),mod);
		if (type == -1) Wn = quick_pow (Wn,mod - 2,mod);
		for (Int R = mid << 1,j = 0;j < limit;j += R){
			for (Int k = 0,w = 1;k < mid;++ k,w = 1ll * w * Wn % mod){
				int x = a[j + k],y = 1ll * w * a[j + k + mid] % mod;
				a[j + k] = (x + y) % mod,a[j + k + mid] = (x + mod - y) % mod;
			}
		}
	} 
	if (type == 1) return ;
	int Inv = quick_pow (limit,mod - 2,mod);
	for (Int i = 0;i < limit;++ i) a[i] = 1ll * a[i] * Inv % mod;
}

int c[MAXN];

void Solve (int len,int *a,int *b)
{
	if (len == 1) return b[0] = quick_pow (a[0],mod - 2,mod),void ();
	Solve ((len + 1) >> 1,a,b);
	limit = 1,l = 0;
	while (limit < len << 1) limit <<= 1,l ++;
	for (Int i = 0;i < limit;++ i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
	for (Int i = 0;i < len;++ i) c[i] = a[i];
	for (Int i = len;i < limit;++ i) c[i] = 0;
	NTT (c,1);NTT (b,1);
	for (Int i = 0;i < limit;++ i) b[i] = 1ll * b[i] * (2 + mod - 1ll * c[i] * b[i] % mod) % mod;
	NTT (b,-1);
	for (Int i = len;i < limit;++ i) b[i] = 0;
}

void deravitive (int *a,int n){
	for (Int i = 1;i <= n;++ i) a[i - 1] = 1ll * a[i] * i % mod;
	a[n] = 0;
}

void inter (int *a,int n){
	for (Int i = n;i >= 1;-- i) a[i] = 1ll * a[i - 1] * quick_pow (i,mod - 2,mod) % mod;
	a[0] = 0;
}

int b[MAXN];

void Ln (int *a,int n){
	memset (b,0,sizeof (b));
	Solve (n,a,b);deravitive (a,n);
	while (limit < (n << 1)) limit <<= 1,l ++;
	for (Int i = 0;i < limit;++ i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
	NTT (a,1),NTT (b,1);
	for (Int i = 0;i < limit;++ i) a[i] = 1ll * a[i] * b[i] % mod;
	NTT (a,-1),inter (a,n);
	for (Int i = n + 1;i < limit;++ i) a[i] = 0;
}

int F0[MAXN];

void Exp (int *a,int *B,int n){
	if (n == 1) return B[0] = 1,void ();
	Exp (a,B,(n + 1) >> 1);
	for (Int i = 0;i < limit;++ i) F0[i] = B[i];
	Ln (F0,n);
	F0[0] = (a[0] + 1 + mod - F0[0]) % mod;
	for (Int i = 1;i < n;++ i) F0[i] = (a[i] + mod - F0[i]) % mod;
	NTT (F0,1);NTT (B,1);	
	for (Int i = 0;i < limit;++ i) B[i] = 1ll * F0[i] * B[i] % mod;
	NTT (B,-1);
	for (Int i = n;i < limit;++ i) B[i] = 0;
}

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < ‘0‘ || c > ‘9‘){if (c == ‘-‘) f = -f;c = getchar();}while (c >= ‘0‘ && c <= ‘9‘){t = (t << 3) + (t << 1) + c - ‘0‘;c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar (‘-‘);}if (x > 9) write (x / 10);putchar (x % 10 + ‘0‘);}

int s,m,G[MAXN],G_[MAXN];

signed main(){
	read (s,m);
	for (Int i = 1,x;i <= m;++ i) read (x),G[x - 1] = mod - 1;
	G[0] = 1,Solve (s + 1,G,G_),memset (G,0,sizeof (G)),Ln (G_,s);
	for (Int i = 0;i <= s;++ i) G_[i] = 1ll * G_[i] * s % mod;
	Exp (G_,G,s + 1),write (1ll * G[s - 1] * quick_pow (s,mod - 2,mod) % mod),putchar (‘\n‘);
	return 0;
}

题解 大朋友和多叉树

原文：https://www.cnblogs.com/Dark-Romance/p/13289478.html