P4238 【模板】多项式求逆

\(\color{#0066ff}{ 题目描述 }\)

给定一个多项式\(F(x)\) ，请求出一个多项式 \(G(x)\)， 满足 \(F(x) * G(x) \equiv 1 ( \mathrm{mod\:} x^n )\),系数对 \(998244353\) 取模。

\(\color{#0066ff}{输入格式}\)

首先输入一个整数 \(n\) 表示输入多项式的次数。 接着输入 \(n\) 个整数，第 \(i\) 个整数 \(a_i\) 代表 \(F(x)\) 次数为 \(i-1\) 的项的系数。

\(\color{#0066ff}{输出格式}\)

输出 \(n\) 个数字，第 \(i\) 个整数 \(b_i\) 代表 \(G(x)\) 次数为 \(i-1\) 的项的系数。

\(\color{#0066ff}{输入样例}\)

5
1 6 3 4 9

\(\color{#0066ff}{输出样例}\)

1 998244347 33 998244169 1020

\(\color{#0066ff}{数据范围与提示}\)

\(1\leq n \leq 10^5,0\le ai\le 10^9\)

\(\color{#0066ff}{ 题解 }\)

牛顿迭代推一波

给出\(A(x)\),求\(B(x)\),使\(A(x) \equiv B(x) ( \mathrm{mod\:} x^n )\)

设\(F(x)=B(x)\)

设\(G(F(x))= \frac{1}{A(x)*F(x)}-1\equiv 0( \mathrm{mod\:} x^n )\)

则\(G'(F(x))=-\frac{1}{A(x)*F^2(x)}\)

根据牛顿迭代，有\(F(x)\equiv F_0(x)-\frac{G(F_0(x))}{G'(F_0(x))} ( \mathrm{mod\:} x^n )\)

于是 \(F(x)\equiv F_0(x)-\frac{\frac{1}{A(x)F_0(x)}-1}{-\frac{1}{A(x)F_0^2(x)}} ( \mathrm{mod\:} x^n )\equiv 2*F_0(x)-A(x)*F_0^2(x) ( \mathrm{mod\:} x^n )\)

每次长度二分，递归求\(F_0\)即可

#include<bits/stdc++.h>
#define LL long long
LL in() {
	char ch; LL x = 0, f = 1;
	while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
	for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
	return x * f;
}
using std::vector;
const int mod = 998244353;
const int maxn = 4e5 + 10;
int r[maxn], len;
LL ksm(LL x, LL y) {
	LL re = 1LL;
	while(y) {
		if(y & 1) re = re * x % mod;
		x = x * x % mod;
		y >>= 1;
	}
	return re;
}
void FNTT(vector<int> &A, int flag) {
	A.resize(len);
	for(int i = 0; i < len; i++) if(i < r[i]) std::swap(A[i], A[r[i]]);
	for(int l = 1; l < len; l <<= 1) {
		int w0 = ksm(3, (mod - 1) / (l << 1));
		for(int i = 0; i < len; i += (l << 1)) {
			int w = 1, a0 = i, a1 = i + l;
			for(int k = 0; k < l; k++, a0++, a1++, w = 1LL * w * w0 % mod) {
				int tmp = 1LL * A[a1] * w % mod;
				A[a1] = ((A[a0] - tmp) % mod + mod) % mod;
				A[a0] = (A[a0] + tmp) % mod;
			}
		}
	}
	if(flag == -1) {
		std::reverse(A.begin() + 1, A.end());
		int inv = ksm(len, mod - 2);
		for(int i = 0; i < len; i++) A[i] = 1LL * A[i] * inv % mod;
	}
}
vector<int> operator * (const vector<int> &A, const vector<int> &B) {
	int tot = A.size() + B.size() - 1;
	vector<int> C = A, D = B;
	for(len = 1; len <= tot; len <<= 1);
	for(int i = 0; i < len; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) * (len >> 1));
	FNTT(C, 1), FNTT(D, 1);
	vector<int> ans;
	ans.resize(len);
	for(int i = 0; i < len; i++) ans[i] = 1LL * C[i] * D[i] % mod;
	FNTT(ans, -1);
	ans.resize(tot);
	return ans;
}
vector<int> operator - (const vector<int> &A, const vector<int> &B) {
	vector<int> ans;
	for(int i = 0; i < (int)std::min(A.size(), B.size()); i++) ans.push_back(A[i] - B[i]);
	if(A.size() < B.size()) for(int i = A.size(); i < (int)B.size(); i++) ans.push_back(-B[i]);
	if(A.size() > B.size()) for(int i = B.size(); i < (int)A.size(); i++) ans.push_back(A[i]);
	return ans;
}
vector<int> inv(const vector<int> &A) {
	if(A.size() == 1) {
		vector<int> ans;
		ans.push_back(ksm(A[0], mod - 2));
		return ans;
	}
	int n = A.size(), _ = (n + 1) >> 1;
	vector<int> ans, B = A;
	ans.push_back(2);
	B.resize(_);
	B = inv(B);
	ans = B * (ans - A * B);
	ans.resize(n);
	return ans;
}
int main() {
	int n = in();
	vector<int> a;
	for(int i = 0; i < n; i++) a.push_back(in());
	a = inv(a);
	for(int i = 0; i < n; i++) printf("%d%c", a[i], i == n - 1? '\n' : ' ');
	return 0;
}