[codeforces438E]The Child and Binary Tree

试题描述

Our child likes computer science very much, especially he likes binary trees.

Consider the sequence of n distinct positive integers: $c_1, c_2, \cdots , c_n$. The child calls a vertex-weighted rooted binary tree good if and only if for every vertex v, the weight of v is in the set $\{c_1, c_2, \cdots , c_n\}$. Also our child thinks that the weight of a vertex-weighted tree is the sum of all vertices' weights.

Given an integer $m$, can you for all $s (1 \le s \le m)$ calculate the number of good vertex-weighted rooted binary trees with weight $s$? Please, check the samples for better understanding what trees are considered different.

We only want to know the answer modulo $998244353$ ($7 \times 17 \times 223 + 1$, a prime number).

给出 $n$ 种点的点权，定义一棵二叉树的权值等于它所有点的点权和。求对于 $[1, m]$ 中的 $s$，权值为 $s$ 的不同的二叉树有多少种。两棵二叉树不同当且仅当它们的左子树、右子树或根节点点权不同。一棵二叉树中可以出现多个点权相同的点。

输入

The first line contains two integers $n, m (1 \le n \le 10^5; 1 \le m \le 10^5)$. The second line contains n space-separated pairwise distinct integers $c_1, c_2, ..., c_n$. $(1 \le c_i \le 10^5)$.

输出

Print $m$ lines, each line containing a single integer. The $i$-th line must contain the number of good vertex-weighted rooted binary trees whose weight exactly equal to $i$. Print the answers modulo $998244353$ ($7 \times 17 \times 2^{23} + 1$, a prime number).

输入示例

3 10

9 4 3

输出示例

数据规模及约定

见“输入”

题解

首先看看暴力 dp 怎么解决这个问题。设 $f_k$ 表示权值为 $k$ 的二叉树的数目，那么有转移方程（注意 dp 边界）：

\[f_k = \sum_{i=1}^n { \sum_{j=0}^{k-c_i} f_{k-j-c_i} \cdot f_j } \\\\
f_0 = 1
\]

然后搞生成函数，令 $C(x) = \sum_{i=1}^n { x^{c_i} }$，$F(x) = \sum_{i=0}^{+ \infty} { f_i \cdot x^i }$。

然后我们发现里面的 sigma 是一个卷积，然后把式子缩一点：

\[[x^k]F(x) = \sum_{i=1}^n { 1 \cdot [x^{k-c_i}]F^2(x) } \\\\
[x^0]F(x) = 1
\]

然后前面那个 $1$，由于只在幂是 $c_i$ 的时候出现，可以想象 $C(x)$ 又在和 $F^2(x)$ 做卷积，即

\[[x^k]F(x) = \sum_{i=1}^n { [x^{c_i}]C(x) \cdot [x^{k-c_i}]F^2(x) } \\\\
[x^0]F(x) = 1
\]

然后我们发现可以化成初中学过的二元一次方程的形式：

\[F(x) = C(x)F^2(x) + 1
\]

用 $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ 求根公式解一下上面这个关于 $F(x)$ 的方程，得到

\[F(x) = \frac{1 \pm \sqrt{1-4C(x)}}{2C(x)}
\]

两个解，怎么办呢？

初中老师告诉我们：检验！

怎么检验？我们从 $[x^0]F(x) = 1$ 入手，可以发现这就是在 $x = 0$ 的时候，$F(x) = 1$。

但是由于 $C(x)$ 常数项为 $0$，且它在分母，所以显然有

\[\lim_{x \rightarrow 0} \frac{1 + \sqrt{1-4C(x)}}{2C(x)} = + \infty
\]

所以可以排除这个解了，但为了严谨，我们当然还要验证一下另一个解，但是另一个解的检验比较棘手，因为我们会得到一个 $0$ 除以 $0$ 的形式，这时候就需要用洛必达法则了（$\leftarrow$ 戳它进入百度百科）

\[\lim_{x \rightarrow 0} \frac{1 - \sqrt{1-4C(x)}}{2C(x)} \\\\
= \lim_{x \rightarrow 0} \frac{1 - \sqrt{1-4x}}{2x} \\\\
= \lim_{x \rightarrow 0} \frac{\frac{\mathrm{d}(1 - \sqrt{1-4x})}{\mathrm{d}x}}{\frac{\mathrm{d}(2x)}{\mathrm{d}x}} \\\\
= 1
\]

（以上直接跳过求导过程，读者不妨仔细手算一下）正确了！

那么接下来搞一个多项式求逆、开方$^*$就好啦。

注意，上面的式子不能直接算，因为 $C(x)$ 常数项为 $0$，不存在逆元！不过没关系，我们可以分子有理化一下：

\[\frac{1 - \sqrt{1-4C(x)}}{2C(x)} \\
= \frac{1 - (1 - 4C(x))}{2C(x) (1 + \sqrt{1 - 4C(x)})} \\
= \frac{2}{1 + \sqrt{1 - 4C(x)}}
\]

这样就可以直接求逆元啦！

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <algorithm>

using namespace std;

#define rep(i, s, t) for(int i = (s); i <= (t); i++)

#define dwn(i, s, t) for(int i = (s); i >= (t); i--)

const int BufferSize = 1 << 16;

char buffer[BufferSize], *Head, *Tail;

inline char Getchar() {

	if(Head == Tail) {

		int l = fread(buffer, 1, BufferSize, stdin);

		Tail = (Head = buffer) + l;

	}

	return *Head++;

}

int read() {

	int x = 0, f = 1; char c = Getchar();

	while(!isdigit(c)){ if(c == '-') f = -1; c = Getchar(); }

	while(isdigit(c)){ x = x * 10 + c - '0'; c = Getchar(); }

	return x * f;

}

#define maxn 524288

#define MOD 998244353

#define Groot 3

#define LL long long

int Pow(int a, int b) {

	int ans = 1, t = a;

	while(b) {

		if(b & 1) ans = (LL)ans * t % MOD;

		t = (LL)t * t % MOD; b >>= 1;

	}

	return ans;

}

int brev[maxn];

void FFT(int *a, int len, int tp) {

	int n = 1 << len;

	rep(i, 0, n - 1) if(i < brev[i]) swap(a[i], a[brev[i]]);

	rep(i, 1, len) {

		int wn = Pow(Groot, MOD - 1 >> i);

		if(tp < 0) wn = Pow(wn, MOD - 2);

		for(int j = 0; j < n; j += 1 << i) {

			int w = 1;

			rep(k, 0, (1 << i >> 1) - 1) {

				int la = a[j+k], ra = (LL)w * a[j+k+(1<<i>>1)] % MOD;

				a[j+k] = (la + ra) % MOD;

				a[j+k+(1<<i>>1)] = (la - ra + MOD) % MOD;

				w = (LL)w * wn % MOD;

			}

		}

	}

	if(tp < 0) rep(i, 0, n - 1) a[i] = (LL)a[i] * Pow(n, MOD - 2) % MOD;

	return ;

}

void Mul(int *A, int an, int *B, int bn) {

	int n = an + bn, N = 1, len = 0;

	while(N <= n) N <<= 1, len++;

	rep(i, 0, N - 1) brev[i] = (brev[i>>1] >> 1) | ((i & 1) << len - 1);

	FFT(A, len, 1); FFT(B, len, 1);

	rep(i, 0, N - 1) A[i] = (LL)A[i] * B[i] % MOD;

	FFT(A, len, -1);

	return ;

}

int tmp[maxn];

void inverse(int *f, int *g, int n) { // module x^n

	if(n == 1) return (void)(f[0] = Pow(g[0], MOD - 2));

	inverse(f, g, n + 1 >> 1);

	int N = 1, len = 0;

	while(N <= (n << 2)) N <<= 1, len++;

	rep(i, 0, N - 1) brev[i] = (brev[i>>1] >> 1) | ((i & 1) << len - 1);

	rep(i, n + 1 >> 1, N - 1) f[i] = 0; rep(i, 0, n - 1) tmp[i] = g[i]; rep(i, n, N - 1) tmp[i] = 0;

	FFT(f, len, 1); FFT(tmp, len, 1);

	rep(i, 0, N - 1) f[i] = (2ll - (LL)tmp[i] * f[i] % MOD + MOD) * f[i] % MOD;

	FFT(f, len, -1);

	rep(i, n, N - 1) f[i] = 0;

	return ;

}

int inv[maxn], _inv[maxn];

void p_sqrt(int *f, int *g, int n) { // g[0] = 1

	if(n == 1) return (void)(f[0] = 1);

	p_sqrt(f, g, n + 1 >> 1);

	rep(i, 0, (n + 1 >> 1) - 1) _inv[i] = (f[i] << 1) % MOD; rep(i, n + 1 >> 1, n - 1) _inv[i] = 0;

	inverse(inv, _inv, n);

	int N = 1, len = 0;

	while(N <= n + 1) N <<= 1, len++;

	rep(i, 0, N - 1) brev[i] = (brev[i>>1] >> 1) | ((i & 1) << len - 1);

	rep(i, n + 1 >> 1, N - 1) f[i] = 0; rep(i, 0, n - 1) tmp[i] = g[i]; rep(i, n, N - 1) tmp[i] = 0;

	FFT(f, len, 1); FFT(tmp, len, 1);

	rep(i, 0, N - 1) f[i] = (tmp[i] + (LL)f[i] * f[i]) % MOD;

	FFT(f, len, -1);

	rep(i, n, N - 1) f[i] = 0;

	Mul(f, n - 1, inv, n - 1);

	N = 1; while(N <= (n - 1 << 1)) N <<= 1;

	rep(i, n, N - 1) f[i] = 0;

	return ;

}

int num[50], cntn;

void putint(int x) {

	if(!x) return (void)puts("0");

	cntn = 0;

	while(x) num[++cntn] = x % 10, x /= 10;

	dwn(i, cntn, 1) putchar(num[i] + '0'); putchar('\n');

	return ;

}

int n, val[maxn], C[maxn], c[maxn];

int main() {

	int n = read(), m = read(), mxv = 0;

	rep(i, 1, n) val[i] = read(), mxv = max(mxv, val[i]);

	rep(i, 1, n) if(val[i] <= m) C[val[i]]++;

	rep(i, 1, m) C[i] = MOD - 4ll * C[i] % MOD; C[0] = 1;

	p_sqrt(c, C, m + 1); (c[0] += 1) %= MOD;

	inverse(C, c, m + 1);

	rep(i, 0, m) (C[i] <<= 1) %= MOD;

	rep(i, 1, m) putint(C[i]);

	return 0;

}

$^*$多项式开方，可以用前文（请翻到最后一题）的方法，但要用到多项式求 ln 和 exp，很麻烦，常数也大。这里我们可以利用一下开根的次数为 $2$ 这个特殊性质优化一下。（以下多项式都省略后面的 $(x)$）

还是考虑倍增，令 $f_0^2 \equiv g (\mathrm{mod}\ x^{\lceil \frac{n}{2} \rceil})$，用 $f_0, g$ 表示出 $f^2 \equiv g (\mathrm{mod}\ x^n)$ 的 $f$。

显然有

\[f - f_0 \equiv 0 (\mathrm{mod}\ x^{\lceil \frac{n}{2} \rceil})
\]

两边平方一下就好啦

\[(f - f_0)^2 \equiv 0 (\mathrm{mod}\ x^n) \\\\
f^2 - 2f_0f + f_0^2 \equiv 0 (\mathrm{mod}\ x^n) \\\\
g - 2f_0f + f_0^2 \equiv 0 (\mathrm{mod}\ x^n)
\]

于是得到

\[f \equiv \frac{g + f_0^2}{2f_0} (\mathrm{mod}\ x^n)
\]