「多项式指数函数」

「多项式指数函数」

前置知识

导数

微积分

多项式牛顿迭代

基本问题

给定一个 (n) 次多项式 (A(x)),求 (B(x)) 满足

[B(x)equiv e^{A(x)}mod x^n ]

两遍同时用 (ln) 取对数

[ln^{B(x)}equiv A(x)mod x^n ]

[ln^{B(x)} - A(x)equiv 0mod x^n ]

套用牛顿迭代,设

[G(B_0(x))=ln^{B(x)} - A(x) ]

[B(x)=B_0(x) - frac{G(B_0(x))}{G'(B_0(x))} ]

由于 (A(x)) 是常数,所以直接消掉不影响求导

[G'(B_0(x))=frac{1}{B(x)} ]

[B(x)=B_0(x) - frac{ln^{B_0(x)} - A(x)}{frac{1}{B_0(x)}} ]

[B(x)=B_0(x) - B_0(x)ln^{B_0(x)} + B_0(x)A(x) ]

[B(x)=B_0(x)(1 - ln^{B_0(x)} + A(x)) ]

最后递归求解即可

代码
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

typedef long long ll;
typedef unsigned long long ull;

using namespace std;

const int maxn = 3e5 + 50, INF = 0x3f3f3f3f, mod = 998244353, inv3 = 332748118;

inline int read () {
	register int x = 0, w = 1;
	register char ch = getchar ();
	for (; ch < '0' || ch > '9'; ch = getchar ()) if (ch == '-') w = -1;
	for (; ch >= '0' && ch <= '9'; ch = getchar ()) x = x * 10 + ch - '0';
	return x * w;
}

inline void write (register int x) {
	if (x / 10) write (x / 10);
	putchar (x % 10 + '0');
}

int n;
int f[maxn];
int f0[maxn], rf0[maxn], nf0[maxn], lnf0[maxn];
int res[maxn], tmp[maxn], rev[maxn];

inline int qpow (register int a, register int b, register int ans = 1) {
	for (; b; b >>= 1, a = 1ll * a * a % mod)
		if (b & 1) ans = 1ll * ans * a % mod;
	return ans;
}

inline void NTT (register int len, register int * a, register int opt) {
	for (register int i = 1; i < len; i ++) if (i < rev[i]) swap (a[i], a[rev[i]]);
	for (register int d = 1; d < len; d <<= 1) {
		register int w1 = qpow (opt, (mod - 1) / (d << 1));
		for (register int i = 0; i < len; i += d << 1) {
			register int w = 1;
			for (register int j = 0; j < d; j ++, w = 1ll * w * w1 % mod) {
				register int x = a[i + j], y = 1ll * w * a[i + j + d] % mod;
				a[i + j] = (x + y) % mod, a[i + j + d] = (x - y + mod) % mod;
			}
		}
	}
}

inline void Poly_Inv (register int d, register int * a, register int * b) {
	if (d == 1) return b[0] = qpow (a[0], mod - 2), void ();
	Poly_Inv ((d + 1) >> 1, a, b);
	register int len = 1, bit = 0;
	while (len < (d << 1)) len <<= 1, bit ++;
	for (register int i = 0; i < len; i ++) res[i] = 0, rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1);
	for (register int i = 0; i < d; i ++) res[i] = a[i];
	NTT (len, res, 3), NTT (len, b, 3);
	for (register int i = 0; i < len; i ++) b[i] = ((2ll * b[i] % mod - 1ll * res[i] * b[i] % mod * b[i] % mod) % mod + mod) % mod;
	NTT (len, b, inv3);
	register int inv = qpow (len, mod - 2);
	for (register int i = 0; i < d; i ++) b[i] = 1ll * b[i] * inv % mod; for (register int i = d; i < len; i ++) b[i] = 0;
}

inline void Poly_Ln (register int d, register int * f0, register int * lnf0) {
	register int len = 1, bit = 0;
	while (len < (d << 1)) len <<= 1, bit ++;
	for (register int i = 0; i < len; i ++) rf0[i] = nf0[i] = lnf0[i] = 0, rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1);
	for (register int i = 0; i < d; i ++) rf0[i] = 1ll * f0[i + 1] * (i + 1) % mod;
	Poly_Inv (d, f0, nf0);
	NTT (len, rf0, 3), NTT (len, nf0, 3);
	for (register int i = 0; i < len; i ++) lnf0[i] = 1ll * rf0[i] * nf0[i] % mod;
	NTT (len, lnf0, inv3);
	register int inv = qpow (len, mod - 2);
	for (register int i = 0; i < d; i ++) lnf0[i] = 1ll * lnf0[i] * inv % mod; for (register int i = d; i < len; i ++) lnf0[i] = 0;
	for (register int i = d - 1; i > 0; i --) lnf0[i] = 1ll * lnf0[i - 1] * qpow (i, mod - 2) % mod; lnf0[0] = 0;
}

inline void Poly_Exp (register int d, register int * f, register int * f0) {
	if (d == 1) return f0[0] = 1, void ();
	Poly_Exp ((d + 1) >> 1, f, f0);
	Poly_Ln (d, f0, lnf0);
	register int len = 1, bit = 0;
	while (len < (d << 1)) len <<= 1, bit ++;
	for (register int i = 0; i < len; i ++) res[i] = 0, rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1);
	for (register int i = 0; i < d; i ++) res[i] = (f[i] - lnf0[i] + mod) % mod; res[0] ++;
	NTT (len, f0, 3), NTT (len, res, 3);
	for (register int i = 0; i < len; i ++) f0[i] = 1ll * f0[i] * res[i] % mod;
	NTT (len, f0, inv3);
	register int inv = qpow (len, mod - 2);
	for (register int i = 0; i < d; i ++) f0[i] = 1ll * f0[i] * inv % mod; for (register int i = d; i < len; i ++) f0[i] = 0;
}

int main () {
	n = read() - 1;
	for (register int i = 0; i <= n; i ++) f[i] = read(); Poly_Exp (n + 1, f, f0);
	for (register int i = 0; i <= n; i ++) printf ("%d ", f0[i]); putchar ('
');
	return 0;
}
原文地址:https://www.cnblogs.com/Rubyonly233/p/14212569.html