CF438E The Child and Binary Tree

首先想了直接上生成函数，经过一些推导后，一直找不到复杂度对的算法，于是尝试 \(DP\)

设 \(f_i\) 表示权值为 \(i\) 的合法二叉树的个数，\(g_i\) 表示 \(i\) 这个值能不能选

枚举根节点权值和左右儿子的权值，得出一个 \(dp\) 式子

\[f_k = \sum_{i = 0} ^ k g_i \sum_{j = 0} ^ {k - i} f_j f_{k - i - j} \]

发现是一个卷积式子

\[F = G * F^2 + 1 \]

后面加一个 1 是因为 \(f_0\) 应该是 1，后面的卷积根节点必须有值

然后用求根公式求出 \(F\)

\[F = \frac {1 \pm \sqrt{1 - 4G}} {2G} \]

考虑当 \(\lim\limits_{x \to 0}\) 时，\(\frac {1 + \sqrt{1 - 4G}} {2G} \to \infty\) 0次项显然不符和，所以抛弃

\[ans = \frac {1 - \sqrt{1 - 4G}} {2G} \]

然后我们就想多项式开根求逆，但是 \(2G\) 没有 0 次项，也就是没有逆元，所以要把式子再化一下

上下同乘 $ (1 + \sqrt{1 - 4G} ) $ 式子变为

\[\frac 2 {1 + \sqrt{1 - 4G}} \]

这回就可以求逆了

#include <bits/stdc++.h>
using namespace std;
#define rg register
#define gc getchar
#define rep(i, a, b) for(int i = a; i <= b; ++i)
#define per(i, a, b) for(int i = a; i >= b; --i)
#define I inline
const int N = 4e5 + 5, mod = 998244353;
I int read(){
	rg char ch = gc();
	rg int f = 0;
	rg long long x = 0;
	while(!isdigit(ch)) f |= (ch == '-'), ch = gc();
	while(isdigit(ch)) x = ((x << 1) + (x << 3) + (ch ^ 48)) % mod, ch = gc();
	return f ? mod - x : x;
}
I int ksm(int a, long long b){
	int ans = 1;
	while(b){ if(b & 1) ans = 1ll * a * ans % mod; b >>= 1; a = 1ll * a * a % mod; }
	return ans;
}
int G = 3, Gn = ksm(G, mod - 2);
int f[N], g[N], n, k, g4[N], gsq[N], invg[N];
I void fwt_or(int *f, int lim, int flag){
	for(int l = 2; l <= lim; l <<= 1)
		for(int m = l >> 1, j = 0; j < lim; j += l)
			for(int i = j; i < j + m; ++i)
				(f[j + m] += flag * f[j]) %= mod;
}
I void fwt_and(int *f, int lim, int flag){
	for(int l = 2; l <= lim; l <<= 1)
		for(int m = l >> 1, j = 0; j < lim; j += l)
			for(int i = j; i < j + m; ++i)
				(f[j] += flag * f[j + m]) %= mod;
}
const int inv2 = ksm(2, mod - 2);
I void fwt_xor(int *f, int lim, int flag){
	for(int l = 2; l <= lim; l <<= 1)
		for(int m = l >> 1, j = 0; j < lim; j += l)
			for(int i = j; i < j + m; ++i){
				int x = f[i], y = f[i + m];
				f[i] = (x + y) % mod; f[i + m] = (x + mod - y) % mod;
				if(flag == -1){
					f[i] = 1ll * f[i] * inv2 % mod; f[i + m] = 1ll * f[i] * inv2 % mod;
				}
			}
}
I int get_phi(int x){
	int len = sqrt(x);
	int res = 1;
	rep(i, 2, len){
		if(!(x % i)){
			x /= i;
			res = 1ll * res * (i - 1) % mod;
			while(!(x % i)) x /= i, res = 1ll * res * i % mod;	
		}
	}
	if(x != 1) res = 1ll * res * (x - 1) % mod;
	return res;
}
I int find_root(int x){
	int phi = get_phi(x), p = phi;
	int len = sqrt(phi);
	static int s[N], cnt;
	cnt = 0;
	rep(i, 2, len){
		if(!(p % i)){
			p /= i;
			s[++cnt] = i;
			while(!(p % i)) p /= i;
		}
	}
	if(p != 1) s[++cnt] = p;
	rep(i, 1, cnt) cout << s[i] << " "; cout << endl;
	cout << phi << endl;
	rep(i, 2, mod - 1){
		int flag = 0;
		rep(j, 1, cnt) if(ksm(i, phi / s[j]) == 1){ flag = 1; break; }
		if(!flag) return i;
	}
}
int fac[N], ifac[N];
I void get_fac(int n){
	fac[0] = ifac[0] = 1;
	rep(i, 1, n){
		fac[i] = 1ll * fac[i - 1] * i % mod;
		ifac[i] = 1ll * ifac[i - 1] * fac[i] % mod;
	}
	int inv = ksm(ifac[n], mod - 2);
	per(i, n, 1){
		ifac[i] = 1ll * ifac[i - 1] * inv % mod;
		inv = 1ll * fac[i] * inv % mod;
	}
}
struct FFT{
	int A[N], B[N], c[N], b2[N], bb[N], ib2[N], sa[N], rev[N];
	I void NTT(int *a, int lim, int len, int flag){
		rep(i, 1, lim - 1) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (len - 1));
		rep(i, 1, lim - 1) if(rev[i] > i) swap(a[i], a[rev[i]]);
		for(int l = 2; l <= lim; l <<= 1){
			const int m = l >> 1, Gi = ksm(flag == 1 ? G : Gn, (mod + 1) / l);
			for(int j = 0; j < lim; j += l){
				int g = 1;
				for(int i = j; i < j + m; ++i, g = 1ll * g * Gi % mod){
					int x = a[i], y = 1ll * g * a[i + m] % mod;
					a[i] = (x + y) % mod;
					a[i + m] = (x + mod - y) % mod;
				}
			}
		}
	}
	I void mul(int *a, int *b, int na, int nb, int *c){
		int lim = 1, len = 0;
		while(lim <= na + nb) lim <<= 1, ++len;
		memcpy(A, a, (na + 1) * sizeof(int)); memcpy(B, b, (nb + 1) * sizeof(int));
		fill(A + na + 1, A + lim, 0); fill(B + nb + 1, B + lim, 0);
		NTT(A, lim, len, 1); NTT(B, lim, len, 1);
		rep(i, 0, lim - 1) A[i] = 1ll * A[i] * B[i] % mod;
		NTT(A, lim, len, -1);
		const int inv = ksm(lim, mod - 2);
		rep(i, 0, na + nb) c[i] = 1ll * A[i] * inv % mod;
		fill(c + na + nb + 1, c + lim, 0);
	}
	I void ni_ab(int *a, int *b, int n){
		int lim = 1;
		while(lim <= n) lim <<= 1;
		b[0] = ksm(a[0], mod - 2);
		for(int xmod = 1, nlen = 2; xmod < lim; xmod <<= 1, ++nlen){
			int nlim = xmod << 2;
			memcpy(B, b, xmod * sizeof(int)); memcpy(A, a, (xmod << 1) * sizeof(int));
			fill(B + xmod, B + nlim, 0); fill(A + (xmod << 1), A + nlim, 0);
			NTT(A, nlim, nlen, 1); NTT(B, nlim, nlen, 1);
			rep(i, 0, nlim - 1) A[i] = ((B[i] << 1) % mod + mod - 1ll * A[i] * B[i] % mod * B[i] % mod) % mod;
			NTT(A, nlim, nlen, -1);
			const int inv = ksm(nlim, mod - 2);
			rep(i, 0, (xmod << 1) - 1) b[i] = 1ll * A[i] * inv % mod;
		}
		fill(b + n + 1, b + lim, 0);
	}
	I void ln(int *a, int *b, int n){//bb b2
		int lim = 1, len = 0;
		while(lim <= n) lim <<= 1, ++len;
		rep(i, 0, n - 1) bb[i] = 1ll * (i + 1) * a[i + 1] % mod;
		ni_ab(a, b2, n);
		mul(bb, b2, n - 1, n, b);
		per(i, n, 1) b[i] = 1ll * b[i - 1] * ksm(i, mod - 2) % mod; b[0] = 0;
	}
	I void sqrt(int *a, int *b, int n){//bb ib2 b2
		int lim = 1, len = 0;
		while(lim <= n) lim <<= 1, ++len;
		fill(b, b + lim, 0); fill(b2, b2 + lim, 0);
		b[0] = 1;
		for(int xmod = 1; xmod < lim; xmod <<= 1){
			rep(i, 0, xmod - 1) b2[i] = (b[i] << 1) % mod; ni_ab(b2, ib2, (xmod << 1) - 1);
			mul(b, b, xmod - 1, xmod - 1, bb);
			rep(i, 0, (xmod << 1) - 1) bb[i] = (bb[i] + a[i]) % mod;
			mul(bb, ib2, (xmod << 1) - 1, (xmod << 1) - 1, b);
		}
	}
	I void exp(int *a, int *b, int n){//ib2 bb b2 c
		int lim = 1; while(lim <= n) lim <<= 1;
		fill(b, b + lim, 0);
		b[0] = 1;
		for(int xmod = 1; xmod < lim; xmod <<= 1){
			ln(b, ib2, (xmod << 1) - 1); //ib2 = ln(b);
			rep(i, 0, (xmod << 1) - 1) c[i] = (a[i] + mod - ib2[i]) % mod;
			c[0] = (c[0] + 1) % mod;
			mul(b, c, (xmod << 1) - 1, (xmod << 1) - 1, b);
		}
		fill(b + n + 1, b + lim, 0);
	}
	I void pow(int *a, int *b, int n, int _k = k){
		int lim = 1; while(lim <= n) lim <<= 1;
		ln(a, sa, n);
		rep(i, 0, n) sa[i] = 1ll * sa[i] * _k % mod;
		exp(sa, b, n);
	}
}T;
int m;
signed main(){
	n = read(); m = read();
	rep(i, 1, n) g[read()] = 1;
	rep(i, 1, m) g4[i] = 1ll * g[i] * (mod - 4) % mod;
	//rep(i, 1, m) g4[i] = (mod - 4ll * g[i] % mod) % mod;
	g4[0] = 1;
	//rep(i, 0, m) cout << g4[i] << " "; cout << endl;
	T.sqrt(g4, invg, m);
	invg[0] = (invg[0] + 1) % mod;
	T.ni_ab(invg, f, m);
	rep(i, 1, m) printf("%d\n", 1ll * f[i] * 2 % mod);
	return 0;
}