题目大意
给一个大小为(n)的集合(S)，(forall x in S:x)为正整数且(xleq 10^5)。给定(m)，且对于任意正整数(sleq m)，求所有不同的合法的二叉树的方案数满足：其每个结点的权值和为(s)。方案不同当且仅当树的形态不同或者存在对应结点中权值不同。答案对(998244353)取模。
具体样例看原题。数据范围：(n,mleq 10^5)。
题解
([state])是单位函数，表示当(state)为真时，值为(1)，否则为(0)。
设(f_i)表示总和为(i)的二叉树的合法方案数。首先很好列出( ext{DP})方程：
[f_n=egin{cases}egin{aligned}sum_{i=0}^n[iin S]sum_{j=0}^{n-i}f_jf_{n-i-j}end{aligned}&n>0\1&n=0end{cases}
]
可是数据范围太大了？我们用生成函数。
定义生成函数(G(x)=sumlimits_{igeq 0}[iin S]x^i)，以及生成函数(F(x)=sumlimits_{igeq 0}f_ix^i)，根据( ext{DP})式子，我们可以得到下面的式子：
[egin{aligned}
F(x)&=sum_{igeq 0}x^ileft([i==0]+sum_{i=0}^n[iin S]sum_{j=0}^{n-i}f_jf_{n-i-j}
ight)\
&=1+sum_{igeq 0}left(sum_{i=0}^n[iin S]x^isum_{j=0}^{n-i}f_jx^jcdot f_{n-i-j}x^{n-i-j}
ight)\
&=1+G(x)F^2(x)
end{aligned}
]
解得(F=dfrac{2}{1pm sqrt{1-4G}})。代入(f_0=1)和(g_0=0)，我们选择(F=dfrac{2}{1+sqrt{1-4G}})。然后多项式操作一下就结束了。
复杂度：( ext{O}(nlog n))。
代码
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 5;
const int P = 998244353, g = 3;
int inc(int a, int b) { return (a += b) >= P ? a-P : a; }
int qpow(int a, int b) {
int res = 1;
for (int i = a; b; i = 1ll*i*i%P, b >>= 1)
if (b & 1) res = 1ll*res*i%P;
return res;
}
int W[maxn << 3], inv[maxn << 2], fac[maxn << 2], ifac[maxn << 2];
void prework(int n) {
for (int i = 1; i < n; i <<= 1) {
W[i] = 1;
for (int j = 1, Wn = qpow(g, (P-1)/i>>1); j < i; j++) W[i+j] = 1ll*W[i+j-1]*Wn%P;
}
inv[1] = fac[0] = ifac[0] = 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*fac[i-1]*i%P, ifac[i] = 1ll*ifac[i-1]*inv[i]%P;
}
void ntt(int *a, int n, int opt) {
static int rev[maxn << 2] = {0};
for (int i = 1; i < n; i++)
if ((rev[i] = rev[i>>1]>>1|(i&1?n>>1:0)) > i) swap(a[i], a[rev[i]]);
for (int q = 1; q < n; q <<= 1)
for (int p = 0; p < n; p += q << 1)
for (int i = 0, t; i < q; i++)
t = 1ll*W[q+i]*a[p+q+i]%P, a[p+q+i] = inc(a[p+i], P-t), a[p+i] = inc(a[p+i], t);
if (~opt) return;
for (int i = 0; i < n; i++) a[i] = 1ll*a[i]*inv[n]%P;
reverse(a+1, a+n);
}
struct poly {
vector<int> A;
int len;
poly(int a0 = 0) : len(1) { A.push_back(a0); }
int &operator [] (int i) { return A[i]; }
void write() {
for (int i = 0; i < len; i++) printf("%d ", A[i]);
putchar('
');
}
void load(int *from, int n) {
A.resize(len = n);
memcpy(&A[0], from, sizeof(int) * len);
}
void cpyto(int *to, int n) {
memcpy(to, &A[0], sizeof(int) * min(n, len));
}
void resize(int n = 0) {
if (!n) { while (len > 1 && !A[len - 1]) len--; A.resize(len); } else A.resize(len = n, 0);
}
} F, G;
poly poly_inv(poly A) {
poly B = poly(qpow(A[0], P-2));
for (int len = 1; len < A.len; len <<= 1) {
static int x[maxn << 2], y[maxn << 2];
for (int i = 0; i < len << 2; i++) x[i] = y[i] = 0;
A.cpyto(x, len << 1), B.cpyto(y, len);
ntt(x, len << 2, 1), ntt(y, len << 2, 1);
for (int i = 0; i < len << 2; i++) x[i] = inc(y[i], inc(y[i], P-1ll*x[i]*y[i]%P*y[i]%P));
ntt(x, len << 2, -1); B.load(x, len << 1);
}
return B.resize(A.len), B;
}
poly poly_sqrt(poly A) {
poly B = poly(1); int inv2 = P+1>>1;
for (int len = 1; len < A.len; len <<= 1) {
static int x[maxn << 2], y[maxn << 2], z[maxn << 2];
for (int i = 0; i < len << 2; i++) x[i] = y[i] = z[i] = 0;
A.cpyto(x, len << 1), B.cpyto(y, len); B.resize(len << 1); poly_inv(B).cpyto(z, len << 1);
ntt(x, len << 2, 1), ntt(y, len << 2, 1), ntt(z, len << 2, 1);
for (int i = 0; i < len << 2; i++) x[i] = (x[i]+1ll*y[i]*y[i])%P*z[i]%P*inv2%P;
ntt(x, len << 2, -1);
B.load(x, len << 1);
}
return B.resize(A.len), B;
}
poly operator + (poly A, poly B) {
if (A.len < B.len) A.resize(B.len);
for (int i = 0; i < B.len; i++) A[i] = inc(A[i], B[i]);
return A.resize(), A;
}
poly operator - (poly A, poly B) {
if (A.len < B.len) A.resize(B.len);
for (int i = 0; i < B.len; i++) A[i] = inc(A[i], P-B[i]);
return A.resize(), A;
}
int getsize(int n) { int N = 1; while (N < n) N <<= 1; return N; }
poly operator * (int k, poly A) {
for (int i = 0; i < A.len; i++) A[i] = 1ll*k*A[i]%P;
return A;
}
poly operator * (poly A, poly B) {
static int x[maxn << 2], y[maxn << 2];
int len = getsize(A.len + B.len - 1);
for (int i = 0; i < len; i++) x[i] = y[i] = 0;
A.cpyto(x, A.len), B.cpyto(y, B.len);
ntt(x, len, 1), ntt(y, len, 1);
for (int i = 0; i < len; i++) x[i] = 1ll*x[i]*y[i]%P;
ntt(x, len, -1);
return A.load(x, A.len + B.len - 1), A.resize(), A;
}
poly poly_deri(poly A) {
for (int i = 0; i < A.len - 1; i++) A[i] = 1ll*A[i+1]*(i+1)%P;
return A[A.len - 1] = 0, A.resize(), A;
}
poly poly_int(poly A) {
for (int i = A.len - 1; i; i--) A[i] = 1ll*A[i-1]*inv[i]%P;
return A[0] = 0, A;
}
poly poly_ln(poly A) {
poly B = poly_deri(A) * poly_inv(A);
return B.resize(A.len), poly_int(B);
}
poly poly_exp(poly A) {
poly B = poly(1), C;
for (int len = 1; len < A.len; len <<= 1) {
B.resize(len << 1);
C = poly(1) - poly_ln(B) + A; C.resize(len << 1);
B = B * C;
}
return B.resize(A.len), B;
}
int n, m;
int main() {
scanf("%d%d", &n, &m); m++; prework(m << 2);
G.resize(m);
for (int i = 1; i <= n; i++) {
int c; scanf("%d", &c);
if (c < m) G[c] = 1;
}
G = 1-4*G; G.resize(m);
G = 1+poly_sqrt(G); G.resize(m);
F = 2*poly_inv(G); F.resize(m);
for (int i = 1; i < m; i++) printf("%d
", F[i]);
return 0;
}