Codeforces 438E.The Child and Binary Tree

首先把 \(c\) 看做一个集合,然后定义 \(C(x) = \sum\limits_{i \in c} x^i\)
再设 \(f_m\) 为总权值和为 \(m\) 时的答案。

根据某些 DP 套路,可以得到 \[ f_m = 1 + \sum\limits_{i \in c}\sum\limits_{j = 0}^{m - i} f_j f_{m-i-j} \]

\(F(x)\)\(f\) 的生成函数,则根据上式有 \[ \begin{align*} F(x) &= CF^2(x) + 1 \\ CF^2(x) - F(x) + 1 &= 0 \\ F(x) &= \frac{1 \pm \sqrt{1 - 4C}}{2C} \end{align*} \]

由于 \(0 \not\in c,f_0 = 1\) 所以 \(C(0) = 0,F(0) = 1\)
据此可知应取负号,即 \[ F(x) = \frac{1 - \sqrt{1 - 4C}}{2C} \]

然后就发现似乎就可做了,但是 \(C\) 的常数项为 \(0\),有点麻烦。
考虑用类似分母有理化的方法对分子有理化试试。
易得 \[ F(x) = \frac2{1+\sqrt{1-4C}} \]

然后就完全可做了。
贴个多项式开根和多项求逆就做完了。

代码:

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#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <utility>
#include <algorithm>
#define add(a,b) (a + b >= mod ? a + b - mod : a + b)
#define dec(a,b) (a < b ? a - b + mod : a - b)
using namespace std;
const int N = 1 << 18;
const int mod = 998244353;
const int G = 3;
inline int fpow(int a,int b)
{
int ret = 1;
for(;b;b >>= 1)
(b & 1) && (ret = (long long)ret * a % mod),a = (long long)a * a % mod;
return ret;
}
struct poly
{
int a[N + 5];
inline const int &operator[](int x) const
{
return a[x];
}
inline int &operator[](int x)
{
return a[x];
}
inline void clear(int x = 0)
{
memset(a + x,0,(N - x + 1) << 2);
}
} f;
int m,len,k,n,lg2[N + 5];
int rev[N + 5],fac[N + 5],ifac[N + 5],inv[N + 5];
int rt[N + 5],irt[N + 5];
inline void init(int len)
{
for(n = 1;n < len;n <<= 1);
for(register int i = 2;i <= n;++i)
lg2[i] = lg2[i >> 1] + 1;
int w = fpow(G,(mod - 1) / n);
rt[n >> 1] = 1;
for(register int i = (n >> 1) + 1;i <= n;++i)
rt[i] = (long long)rt[i - 1] * w % mod;
for(register int i = (n >> 1) - 1;i;--i)
rt[i] = rt[i << 1];
fac[0] = 1;
for(register int i = 1;i <= n;++i)
fac[i] = (long long)fac[i - 1] * i % mod;
ifac[n] = fpow(fac[n],mod - 2);
for(register int i = n;i;--i)
ifac[i - 1] = (long long)ifac[i] * i % mod;
for(register int i = 1;i <= n;++i)
inv[i] = (long long)ifac[i] * fac[i - 1] % mod;
}
inline void ntt(poly &a,int type,int n)
{
type == -1 && (reverse(a.a + 1,a.a + n),1);
int lg = lg2[n] - 1;
for(register int i = 0;i < n;++i)
rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg),
i < rev[i] && (swap(a[i],a[rev[i]]),1);
for(register int w = 2,m = 1;w <= n;w <<= 1,m <<= 1)
for(register int i = 0;i < n;i += w)
for(register int j = 0;j < m;++j)
{
int t = (long long)rt[m | j] * a[i | j | m] % mod;
a[i | j | m] = dec(a[i | j],t),a[i | j] = add(a[i | j],t);
}
if(type == -1)
for(register int i = 0;i < n;++i)
a[i] = (long long)a[i] * inv[n] % mod;
}
inline void mul(poly &a,const poly &b,int n)
{
static poly x,y;
int lim = 1;
x.clear(),y.clear();
for(;lim < (n << 1);lim <<= 1);
x = a,y = b;
x.clear(n),y.clear(n);
ntt(x,1,lim),ntt(y,1,lim);
for(register int i = 0;i < lim;++i)
x[i] = (long long)x[i] * y[i] % mod;
ntt(x,-1,lim);
x.clear(n),a = x;
}
inline poly inverse(const poly &f,int n)
{
static int s[30];
static poly g,h,q;
int lim = 1,top = 0;
g.clear();
for(;n > 1;s[++top] = n,n = (n + 1) >> 1);
g[0] = fpow(f[0],mod - 2);
for(;top;--top)
{
n = s[top];
for(;lim < (n << 1);lim <<= 1);
q = g,h = f,h.clear(n);
ntt(g,1,lim),ntt(h,1,lim);
for(register int i = 0;i < lim;++i)
g[i] = (long long)g[i] * g[i] % mod * h[i] % mod;
ntt(g,-1,lim);
for(register int i = 0;i < n;++i)
g[i] = dec(add(q[i],q[i]),g[i]);
g.clear(n);
}
return g;
}
inline void derivative(poly &f,int n)
{
for(register int i = 1;i < n;++i)
f[i - 1] = (long long)f[i] * i % mod;
f[n - 1] = 0;
}
inline void integral(poly &f,int n)
{
for(register int i = n - 1;~i;--i)
f[i + 1] = (long long)f[i] * inv[i + 1] % mod;
f[0] = 0;
}
inline poly ln(const poly &f,int n)
{
static poly g;
g = f,derivative(g,n),mul(g,inverse(f,n),n),integral(g,n);
return g;
}
inline poly exp(const poly &f,int n)
{
static int s[30];
static poly g,h;
int lim = 1,top = 0;
g.clear();
for(;n > 1;s[++top] = n,n = (n + 1) >> 1);
g[0] = 1;
for(;top;--top)
{
n = s[top];
for(;lim < (n << 1);lim <<= 1);
h = g,g = ln(g,n);
for(register int i = 0;i < n;++i)
g[i] = dec(f[i],g[i]);
g[0] = add(g[0],1);
ntt(g,1,lim),ntt(h,1,lim);
for(register int i = 0;i < lim;++i)
g[i] = (long long)g[i] * h[i] % mod;
ntt(g,-1,lim);
g.clear(n);
}
return g;
}
inline poly power(const poly &f,int k,int n)
{
static poly g;
g = ln(f,n);
for(register int i = 0;i < n;++i)
g[i] = (long long)g[i] * k % mod;
g = exp(g,n);
return g;
}
namespace Mod_sqrt
{
typedef pair<int,int> cp;
int w;
inline cp operator*(const cp &a,const cp &b)
{
return cp(((long long)a.first * b.first % mod + (long long)a.second * b.second % mod * w % mod) % mod,((long long)a.first * b.second % mod + (long long)a.second * b.first % mod) % mod);
}
inline cp pow(cp a,int b)
{
cp ret(1,0);
for(;b;b >>= 1)
(b & 1) && (ret = ret * a,1),a = a * a;
return ret;
}
inline int mod_sqrt(int x)
{
int y = rand() % mod;
for(;fpow(w = ((long long)y * y % mod - x + mod) % mod,mod - 1 >> 1) <= 1;y = rand() % mod);
cp ret = pow(cp(y,1),mod + 1 >> 1);
return min(ret.first,mod - ret.first);
}
}
using Mod_sqrt::mod_sqrt;
inline poly sqrt(const poly &f,int n)
{
static int s[30];
static poly g,h;
int top = 0;
g.clear();
for(;n > 1;s[++top] = n,n = (n + 1) >> 1);
g[0] = mod_sqrt(f[0]);
for(;top;--top)
{
n = s[top];
for(register int i = 0;i < n;++i)
h[i] = add(g[i],g[i]);
h = inverse(h,n),mul(g,g,n);
for(register int i = 0;i < n;++i)
g[i] = add(g[i],f[i]);
mul(g,h,n);
}
return g;
}
int main()
{
scanf("%d%d",&m,&len),init(++len << 1);
int x;
for(;m;--m)
scanf("%d",&x),x < len && (f[x] = 1);
for(register int i = 0;i < len;++i)
f[i] = mod - 4LL * f[i] % mod;
f[0] = add(f[0],1),f = sqrt(f,len),f[0] = add(f[0],1),f = inverse(f,len);
for(register int i = 0;i < len;++i)
f[i] = 2LL * f[i] % mod;
for(register int i = 1;i < len;++i)
printf("%d\n",f[i]);
}