[Luogu P4725] 【模板】多项式对数函数

洛谷传送门

题目描述

给出 $n-1$ 次多项式 $A(x)$ ，求一个 $\bmod{\:x^n}$ 下的多项式 $B(x)$ ，满足 $B(x) \equiv \ln A(x)$ .

在 $\text{mod } 998244353$ 下进行，且 $a_i \in [0, 998244353] \cap \mathbb{Z}$

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 $a_0 = 1$ .

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
1 927384623 878326372 3882 273455637 998233543

输出样例#1：

0 927384623 817976920 427326948 149643566 610586717

说明

对于 $100\%$ 的数据， $n \le 10^5$ .

解题分析

两边求导可得：
$B'(x)\equiv \frac{A'(x)}{A(x)}(mod\ x^n)$
求逆，求导最后积分回来就好了。

代码如下：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <algorithm>
#include <cstdlib>
#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 rev[MX], a[MX], b[MX], A[MX], B[MX], Ainv[MX], Adr[MX], Bdr[MX];
namespace Poly
{
	IN void NTT(int *dat, R int len, R bool typ)
	{
		for (R int i = 1; i < len; ++i) if (rev[i] > i) std::swap(dat[rev[i]], dat[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;
	}
	void Getinv(R int up, R int lg)
	{
		if (up == 1) return Ainv[0] = fpow(A[0], MOD - 2), void();
		Getinv(up >> 1, lg - 1); R int len = up << 1;
		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) a[i] = b[i] = 0;
		for (R int i = 0; i < up; ++i) a[i] = Ainv[i], b[i] = A[i];
		NTT(a, len, 1), NTT(b, len, 1);
		for (R int i = 0; i < len; ++i) a[i] = (2 * a[i] % MOD - 1ll * a[i] * a[i] % MOD * b[i] % MOD + MOD) % MOD;
		NTT(a, len, 0);
		for (R int i = 0; i < up; ++i) Ainv[i] = a[i];
	}
	IN void Dr() {for (R int i = 0; i < n; ++i) Adr[i] = 1ll * (i + 1) * A[i + 1] % MOD;}
	IN void Itg() {for (R int i = 1; i < n; ++i) B[i] = 1ll * Bdr[i - 1] * fpow(i, MOD - 2) % MOD;}
	IN void Getln()
	{
		Dr(); int len, lg;
		for (len = 1, lg = 0; len < n; len <<= 1, lg++);
		Getinv(len, lg + 1);
		for (len = 1, lg = 0; len <= (n << 1); len <<= 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) a[i] = b[i] = 0;
		for (R int i = 0; i < n; ++i) a[i] = Adr[i];
		for (R int i = 0; i < n; ++i) b[i] = Ainv[i];
		NTT(a, len, 1), NTT(b, len, 1);
		for (R int i = 0; i < len; ++i) a[i] = 1ll * a[i] * b[i] % MOD;
		NTT(a, len, 0);
		for (R int i = 0; i < n; ++i) Bdr[i] = a[i];
		Itg();
		for (R int i = 0; i < n; ++i) printf("%d ", B[i]);
	}
}
int main(void)
{
	in(n);
	for (R int i = 0; i < n; ++i) in(A[i]);
	Poly::Getln();
}