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; }