[洛谷P4726]【模板】多项式指数函数

题目大意：给出$n-1$次多项式$A(x)$，求一个 $mod{x^n}$下的多项式$B(x)$，满足$B(x) equiv e^{A(x)}$。

题解：(by Weng_weijie)

泰勒展开：
$$
f(x)=f(x_0)+dfrac{f'(x_0)(x-x_0)}{1!}+dfrac{f''(x_0)(x-x_0)^2}{2!}+dots
$$

牛顿迭代：

$$
解关于 F(x) 的方程使得 G(F(x))equiv 0pmod{x^n} \
假设 G(F_0(x)) equiv 0 pmod{x^{iglceildfrac{n}{2}ig ceil}} \
对 G(F(x)) 在 F_0(x) 处泰勒展开得 \
G(F(x)) equiv G(F_0(x)) + dfrac{G'(F_0(x))(F(x)-F_0(x))}{1!}+dots pmod{x^n} \
又F(x)-F_0(x)equiv 0pmod{x^{iglceildfrac{n}{2}ig ceil}} \
(F(x)-F_0(x))^2equiv 0pmod{x^n} \
egin{align*}
herefore G(F(x))&equiv G(F_0(x)) + G'(F_0(x))(F(x)-F_0(x))\
&equiv 0pmod{x^n} \
end{align*}\
得到 F(x)=F_0(x)-dfrac{G(F_0(x))}{G'(F_0(x))}
$$

多项式指数函数：

$$
设F(x)=e^{A(x)}, ln F(x)=A(x), G(F(x))=ln F(x)-A(x) \
于是就是解 G(F(x))=0，代入牛顿迭代公式得：\
F(x)=F_0(x)(1-ln F_0(x)+A(x))
$$
卡点：无

C++ Code：

#include <cstdio>
#include <algorithm>
#define maxn 1 << 18 | 3
const int mod = 998244353, G = 3;
inline int pw(int base, int p) {
	int ans = 1;
	for (; p; p >>= 1, base = 1ll * base * base % mod) if (p & 1) ans = 1ll * ans * base % mod;
	return ans;
}
inline int Inv(int x) {return pw(x, mod - 2);}
namespace Poly {
	int lim, ilim, s, rev[maxn];
	int Wn[maxn + 1], inv[maxn], __invnum;
	#define i __invnum
	inline int getinv(int n) {
		while (i < n) {i++; inv[i] = 1ll * inv[mod % i] * (mod - mod / i) % mod;}
		return inv[n];
	}
	inline void INIT() {inv[i = 1] = 1;}
	#undef i
	inline void init(int n) {
		lim = 1, s = -1; while (lim < n) lim <<= 1, s++; ilim = getinv(lim);
		for (int i = 0; i < lim; i++) rev[i] = rev[i >> 1] >> 1 | (i & 1) << s;
		int t = pw(G, (mod - 1) / lim);
		Wn[0] = 1; for (int i = 1; i <= lim; i++) Wn[i] = 1ll * Wn[i - 1] * t % mod;
	}
	
	inline void up(int &a, int b) {if ((a += b) >= mod) a -= mod;}
	inline void NTT(int *A, int op) {
		for (int i = 0; i < lim; i++) if (i < rev[i]) std::swap(A[i], A[rev[i]]);
		for (int mid = 1; mid < lim; mid <<= 1) {
			int t = lim / mid >> 1;
			for (int i = 0; i < lim; i += mid << 1) {
				for (int j = 0; j < mid; j++) {
					int W = op ? Wn[t * j] : Wn[lim - t * j];
					int X = A[i + j], Y = 1ll * A[i + j + mid] * W % mod;
					up(A[i + j], Y), up(A[i + j + mid] = X, mod - Y);
				}
			}
		}
		if (!op) for (int i = 0; i < lim; i++) A[i] = 1ll * A[i] * ilim % mod;
	}
	inline void DER(int *A, int *B, int n) {
		B[n - 1] = 0; for (int i = 1; i < n; i++) B[i - 1] = 1ll * A[i] * i % mod;
	}
	inline void INT(int *A, int *B, int n) {
		B[0] = 0; for (int i = 1; i < n; i++) B[i] = 1ll * A[i - 1] * inv[i] % mod;
	}
	int C[maxn];
	void INV(int *A, int *B, int n) {
		if (n == 1) {B[0] = Inv(A[0]); return ;}
		INV(A, B, n + 1 >> 1);
		init(n << 1);
		for (int i = 0; i < n; i++) C[i] = A[i];
		for (int i = n; i < lim; i++) C[i] = 0;
		NTT(B, 1), NTT(C, 1);
		for (int i = 0; i < lim; i++) B[i] = (2 + mod - 1ll * B[i] * C[i] % mod) * B[i] % mod;
		NTT(B, 0);
		for (int i = n; i < lim; i++) B[i] = 0;
	}
	int D[maxn];
	inline void LN(int *A, int *B, int n) {
		DER(A, D, n), INV(A, B, n);
		init(n << 1);
		NTT(B, 1), NTT(D, 1);
		for (int i = 0; i < lim; i++) D[i] = 1ll * D[i] * B[i] % mod;
		NTT(D, 0);
		INT(D, B, n);
		for (int i = n; i < lim; i++) B[i] = 0;
	} 
	int E[maxn], F[maxn];
	void EXP(int *A, int *B, int n) {
		if (n == 1) {B[0] = 1; return ;}
		EXP(A, B, n + 1 >> 1);
		for (int i = 0; i < n << 1; i++) E[i] = F[i] = 0;
		LN(B, E, n);
		for (int i = 0; i < n; i++) F[i] = A[i];
		NTT(B, 1), NTT(E, 1), NTT(F, 1);
		for (int i = 0; i < lim; i++) B[i] = (1ll + mod - E[i] + F[i]) * B[i] % mod;
		NTT(B, 0);
		for (int i = n; i < lim; i++) B[i] = 0;
	}
}
int a[maxn], b[maxn], n;
int main() {
	scanf("%d", &n);
	Poly::INIT();
	for (int i = 0; i < n; i++) scanf("%d", a + i);
	Poly::EXP(a, b, n);
	for (int i = 0; i < n; i++) printf("%d ", b[i]); puts("");
	return 0;
}