# [BZOJ3625][Codeforces Round #250]小朋友和二叉树 多项式开根+求逆

https://www.lydsy.com/JudgeOnline/problem.php?id=3625

$F[n]=\sum_{i=0}^n \sum_{j=0}^{n-i}F[i]\cdot F[j]\cdot A[n-i-j]$

$F(x)-1=F^2(x)\cdot A(x)\\ A(x)\cdot F^2(x)-F(x)+1=0\\ F(x)=\frac{1\pm\sqrt{1-4A(x)}}{2A(x)}$

\begin{align*} F(x)&=\frac{1-\sqrt{1-4A(x)}}{2A(x)}\\ &=\frac{(1-\sqrt{1-4A(x)})(1+\sqrt{1-4A(x)})}{2A(x)(1+\sqrt{1-4A(x)})}\\ &=\frac{4A(x)}{2A(x)(1+\sqrt{1-4A(x)})}\\ &=\frac{2}{1+\sqrt{1-4A(x)}} \end{align*}

(刚开始转码风，可能有些地方不太自然，也有可能有些地方仍然保留着就码风没有注意)

#include<bits/stdc++.h>

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back
#define isbreak dbg("*")

template<typename A, typename B> inline char SMAX(A &a, const B &b) {return a < b ? a = b , 1 : 0;}
template<typename A, typename B> inline char SMIN(A &a, const B &b) {return b < a ? a = b , 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;

template<typename I>
int f = 0, c;
while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
x = c & 15;
while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
f ? x = -x : 0;
}

const int N = 4e5 + 7;
const int P = 998244353;
const int G = 3;
const int Gi = 332748118;
const int Inv2 = 499122177;

int n, m, x, a[N];

inline void SADD(int &x, int y) {x += y;x >= P ? x -= P : 0;}
inline int SMOD(int x) {return x >= P ? x - P : x;}
inline int fpow(int x,int y) {
int ans = 1;
for (; y; y >>= 1, x = (ll)x * x % P) if (y & 1) ans = (ll)ans * x % P;
return ans;
}

namespace DFT {
int A[N], B[N], C[N];

inline void NTT(int *a, int n, int f) {
for (int i = 0, j = 0; i < n; ++i) {
if (i > j) std::swap(a[i], a[j]);
for (int l = n >> 1; (j ^= l) < l; l >>= 1);
}
for (int i = 1; i < n; i <<= 1) {
int w = fpow(f > 0 ? G : Gi, (P - 1) / (i << 1));
for (int j = 0; j < n; j += i << 1)
for (int k = 0, e = 1; k < i; ++k, e = (ll)e * w % P){
int x = a[j + k], y = (ll)e * a[i + j + k] % P;
a[j + k] = SMOD(x + y); a[i + j + k] = SMOD(x + P - y);
}
}
if (f < 0) for (int i = 0, p = fpow(n, P - 2); i < n; ++i) a[i] = (ll)a[i] * p % P;
}
namespace Inv {
int A[N], B[N];
inline void GetInv(int *a, int n, int *b) {
memset(B, 0, sizeof(B)); B[0] = fpow(a[0], P - 2);
for (int deg = 2; deg < (n << 1); deg <<= 1) {
int L = deg << 1;
for (int i = 0; i < deg; ++i) A[i] = a[i];
for (int i = deg; i < L; ++i) A[i] = 0;
NTT(A, L, 1); NTT(B, L, 1);
for (int i = 0; i < L; ++i) B[i] = (ll)B[i] * (2 + P - (ll)B[i] * A[i] % P) % P;
NTT(B, L, -1);
for (int i = deg; i < L; ++i) B[i] = 0;
}
for (int i = 0; i < n; ++i) b[i] = B[i];
}
} using Inv::GetInv;

inline void GetSqrt(int *a, int n, int *b) {
B[0] = 1;
for (int deg = 2; deg < (n << 1); deg <<= 1) {
int L = deg << 1;
for (int i = 0; i < deg; ++i) A[i] = a[i];
for (int i = deg; i < L; ++i) A[i] = 0;
GetInv(B, deg, C);
NTT(A, L, 1); NTT(C, L, 1);
for (int i = 0; i < L; ++i) C[i] = (ll)A[i] * C[i] % P;
NTT(C, L, -1);
for (int i = 0; i < L; ++i) B[i] = (ll)(B[i] + C[i]) * Inv2 % P;
for (int i = deg; i < L; ++i) B[i] = 0;
}
for (int i = 0; i < n; ++i) b[i] = B[i];
}
}
using DFT::GetInv;
using DFT::GetSqrt;

int main() {
#ifdef hzhkk
freopen("hkk.in", "r", stdin);
#endif