[Luogu P4726] 【模板】多项式指数函数

洛谷传送门

题目描述

给出 $n-1$ 次多项式 $A(x)$，求一个 $\mathrm{m}\mathrm{o}\mathrm{d}{x}^n$ 下的多项式 $B(x)$，满足 $B(x)\equiv e^{A(x)}$.

输入输出格式

输入格式：

第一行一个整数 $n$.

下一行有 $n$ 个整数，依次表示多项式的系数 $a_0,a_1,\cdots ,a_{n-1}$.

保证 $a_0=0$.

输出格式：

输出 $n$ 个整数，表示答案多项式中的系数 $a_0,a_1,\cdots ,a_{n-1}$.

输入输出样例

输入样例#1：

6
0 927384623 817976920 427326948 149643566 610586717

输出样例#1：

1 927384623 878326372 3882 273455637 998233543

说明

对于 $100\%$ 的数据，$n\le 10^5$.

解题分析

为啥取$ln$很简单， $exp$这么难啊QAQ， 还得先去学牛顿迭代和泰勒展开。

推荐一篇讲泰勒展开的文章：戳这里

牛顿迭代是指这个：

假设我们已经知道了一个函数$G(x)$， 要求满足要求的$F(x)$， 满足$G(F(x))\equiv 0(mod{x}^n)$。

这是有套路的。设$F_0(x)$满足$G(F_0(x))\equiv 0(mod{x}^{\lceil \frac{n}{2}\rceil })$， 那么怎么求得$F(x)$？ 我们可以在$F_0(x)$处泰勒展开， 得到：
$$G(F(x))=G(F_0(x))+\frac{G^{\prime }(F_0(x))}{1!}(F(x)-F_0(x))+\frac{G^{\prime \prime }(F_0(x))}{2!}(F(x)-F_0(x){)}^2+...+\frac{G^{\infty }(F_0(x))}{\infty }(F(x)-F_0(x){)}^{\infty }$$
注意到$F(x)-F_0(x)$的最低项都是$x^{\lceil \frac{n}{2}\rceil }$， 所以后面的几项在模意义下都是0。即：
$$G(F(x))=G(F_0(x))+G^{\prime }(F_0(x))(F(x)-F_0(x))$$
又因为$G(F(x))\equiv 0(mod{x}^n)，$移项得到：
$$F(x)=\frac{-G(F_0(x))}{G^{\prime }(F_0(x))}+F_0(x)$$
递归下去， 当$n=1$时暴力计算即可。

那么这里就有：
$ln(B(x))-A(x)\equiv 0(mod{x}^n)$。

设$G(B(x))=ln(B(x))-A(x)$，那么根据上面的套路就有：
$$\begin{gathered} B(x)=\frac{A(x)-ln(B_0(x))}{\frac{1}{B_0(x)}}+B_0(x) \\ =B_0(x)(1+A(x)-ln(B_0(x))) \end{gathered}$$
于是只需要写个求$ln$， 积分， 求导， 求逆即可。

代码如下：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <algorithm>
#define R register
#define IN inline
#define W while
#define gc getchar()
#define MX 400500
#define MOD 998244353
#define G 3
#define Ginv 332748118
template <class T>
IN void in(T &x)
{
	x = 0; R char c = gc;
	for (; !isdigit(c); c = gc);
	for (;  isdigit(c); c = gc)
	x = (x << 1) + (x << 3) + c - 48;
}
IN int fpow(R int base, R int tim)
{
	int ret = 1;
	W (tim)
	{
		if (tim & 1) ret = 1ll * ret * base % MOD;
		base = 1ll * base * base % MOD, tim >>= 1;
	}
	return ret;
}
int n;
int A[MX], ans[MX], rev[MX], buf[6][MX];
namespace Poly
{
	IN void NTT(int *dat, R int len, R bool typ)
	{
		for (R int i = 0; i < len; ++i) if (rev[i] > i) std::swap(dat[i], dat[rev[i]]);
		R int seg, step, bd, now, cur, base, deal, buf1, buf2;
		for (seg = 1; seg < len; seg <<= 1)
		{
			base = fpow(typ ? G : Ginv, (MOD - 1) / (seg << 1)), step = seg << 1;
			for (now = 0; now < len; now += step)
			{
				deal = 1, bd = now + seg;
				for (cur = now; cur < bd; cur++, deal = 1ll * deal * base % MOD)
				{
					buf1 = dat[cur], buf2 = 1ll * dat[cur + seg] * deal % MOD;
					dat[cur] = (buf1 + buf2) % MOD, dat[cur + seg] = (buf1 - buf2 + MOD) % MOD;
				}
			}
		}
		if (typ) return; int inv = fpow(len, MOD - 2);
		for (R int i = 0; i < len; ++i) dat[i] = 1ll * dat[i] * inv % MOD;
	}
	IN void Dr(int *dat, int *dr, R int len) {for (R int i = 1; i < len; ++i) dr[i - 1] = 1ll * i * dat[i] % MOD; dr[len] = 0;}
	IN void Itg(int *dat, int *itg, R int len) {for (R int i = 1; i < len; ++i) itg[i] = 1ll * dat[i - 1] * fpow(i, MOD - 2) % MOD; itg[0] = 0;}
	void Getinv(R int up, R int lg, int *ind, int *res, int *buf1, int *buf2)
	{
		if (up == 1) return res[0] = fpow(ind[0], MOD - 2), void();
		Getinv(up >> 1, lg - 1, ind, res, buf1, buf2);
		int len = up << 1; lg++;
		for (R int i = 1; i < len; ++i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg - 1);
		for (R int i = 0; i < len; ++i) buf1[i] = buf2[i] = 0;
		for (R int i = 0; i < (up >> 1); ++i) buf1[i] = res[i];
		for (R int i = 0; i < up; ++i) buf2[i] = ind[i];
		NTT(buf1, len, 1), NTT(buf2, len, 1);
		for (R int i = 0; i < len; ++i) buf1[i] = (2 * buf1[i] % MOD - 1ll * buf2[i] * buf1[i] % MOD * buf1[i] % MOD + MOD) % MOD;
		NTT(buf1, len, 0);
		for (R int i = 0; i < up; ++i) res[i] = buf1[i];
	}
	IN void Getln(R int up, R int lg, int *ind, int *res, int *buf1, int *buf2, int *buf3, int *buf4)
	{
		for (R int i = 0; i <= up; ++i) buf1[i] = buf2[i] = 0;
		Dr(ind, buf1, up);
		Getinv(up, lg, ind, buf2, buf3, buf4);
		int len = up << 1; lg++;
		for (R int i = 1; i < len; ++i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg - 1);
		for (R int i = up; i < len; ++i) buf3[i] = buf4[i] = 0;
		for (R int i = 0; i < up; ++i) buf3[i] = buf1[i], buf4[i] = buf2[i];
		NTT(buf3, len, 1), NTT(buf4, len, 1);
		for (R int i = 0; i < len; ++i) buf3[i] = 1ll * buf3[i] * buf4[i] % MOD;
		NTT(buf3, len, 0); Itg(buf3, res, up);
	}
	IN void Getexp(R int up, R int lg)
	{
		if (up == 1) return ans[0] = 1, void();
		int half = up >> 1; Getexp(half, lg - 1);
		R int len = up << 1; lg++;
		for (R int i = 0; i < len; ++i) buf[0][i] = buf[1][i] = 0;
		for (R int i = 0; i < half; ++i) buf[0][i] = ans[i];
		Getln(up, lg - 1, buf[0], buf[1], buf[2], buf[3], buf[4], buf[5]);
		for (R int i = half; i < len; ++i) buf[2][i] = buf[3][i] = 0;
		for (R int i = 0; i < half; ++i) buf[2][i] = ans[i];
		for (R int i = 0; i < up; ++i) buf[3][i] = (A[i] - buf[1][i] + MOD) % MOD;
		buf[3][0] = (buf[3][0] + 1) % MOD;
		NTT(buf[2], len, 1), NTT(buf[3], len, 1);
		for (R int i = 0; i < len; ++i) buf[2][i] = 1ll * buf[2][i] * buf[3][i] % MOD;
		NTT(buf[2], len, 0);
		for (R int i = 0; i < up; ++i) ans[i] = buf[2][i]; 
	}
}
 int main(void)
 {
 	in(n); int len = 1, lg = 0;
 	for (; len < n; lg++, len <<= 1);
 	for (R int i = 0; i < n; ++i) in(A[i]);
 	Poly::Getexp(len, lg);
 	for (R int i = 0; i < n; ++i) printf("%d ", ans[i]);
 }