多项式求逆

首先定义多项式的度数 \(degA\) 为多项式 \(A(x)\) 的最高次数

那么多项式 \(A(x)\) 的逆即为存在多项式 \(B(x)\) 使得条件满足：

\[A(x)B(x) \equiv 1 \pmod{x^n} \]

求解过程

假设存在多项式 \(A(x)\) ，以及其逆 \(B(x)\) 满足条件 ，那么必定有\(A(x)B(x) \equiv 1 \pmod{x^{\left\lceil\frac{n}{2}\right\rceil}} \ ... \ (1)\) ，因为 \(x^n\) 是 \(x^{\left\lceil\frac{n}{2}\right\rceil}\) 的倍数

并且存在 \(A(x)B'(x) \equiv 1 \pmod{x^{\left\lceil\frac{n}{2}\right\rceil}} \ ... \ (2)\)

\((1) - (2)\) ，得

\[\begin{aligned} B(x) - B'(x) &\equiv 0 \pmod{x^{\left\lceil\frac{n}{2}\right\rceil}} \ ... \ (3) \\ B(x)^2 - 2B(x)B'(x) + B'(x)^2 &\equiv 0 \pmod{x^n} \end{aligned} \]

上一步解释一下，是两边平方

至于为什么模数也需要平方，是因为 \((3)\) 式满足左边的多项式在其模意义下为 \(0\) ，且式子恒成立，故其除了第 \(x^{\left\lceil\frac{n}{2}\right\rceil}\) 项其余系数皆为 \(0\) ，那么现在考虑 \(0 \le x \le n - 1\) ，又因为 \(a_i = \sum\limits_{j = 1}^i a_j b_{i - j}\) ，所以必定存在 \(i\) 或者 \(i - j\) 小于 \(x^{\left\lceil\frac{n}{2}\right\rceil}\) ，故 \(a_i = 0\) ，所以平方无妨

同乘 \(A(x)\) ，移项，得

\[B(x) \equiv 2B'(x) - A(x)B'(x)^2 \pmod{x^n} \]

在递归过程中最后会递归到 \(n = 1\) ，那么此时 \(A(x)B''(x) \equiv c \pmod{x}\) ，取 \(c^{- 1}\) 就好了

那么再用 \(FFT\) 优化就可以做到 \(O (n \log n)\) 求解

代码

#include <iostream>
#include <cstdio>
#include <cstring>

#define MOD 998244353
#define g 3

using namespace std;

typedef long long LL;

const int MAXN = 4e05 + 10;

LL power (LL x, int p) {
	LL cnt = 1;
	while (p) {
		if (p & 1)
			cnt = cnt * x % MOD;
		x = x * x % MOD;
		p >>= 1;
	}
	return cnt;
}
const LL invg = power (g, MOD - 2);

int N;
LL f[MAXN], invf[MAXN]= {0};

LL A[MAXN]= {0}, B[MAXN]= {0};
int oppo[MAXN]= {0};
int limit;
void NTT (LL* a, int inv) {
	for (int i = 0; i < limit; i ++)
		if (i < oppo[i])
			swap (a[i], a[oppo[i]]);
	for (int mid = 1; mid < limit; mid <<= 1) {
		LL omega = power (inv == 1 ? g : invg, (MOD - 1) / (mid << 1));
		for (int n = mid << 1, j = 0; j < limit; j += n) {
			LL x = 1;
			for (int k = 0; k < mid; k ++, x = x * omega % MOD) {
				LL a1 = a[j + k], xa2 = x * a[j + k + mid] % MOD;
				a[j + k] = (a1 + xa2) % MOD;
				a[j + k + mid] = (a1 - xa2 + MOD) % MOD;
			}
		}
	}
}
void Inverse (int deg, LL* ps) {
	if (deg == 1) {
		ps[0] = power (f[0], MOD - 2);
		return ;
	}
	Inverse ((deg + 1) >> 1, ps);
	int n, lim;
	for (n = 1, lim = 0; n < (deg << 1); n <<= 1, lim ++);
	limit = n;
	for (int i = 0; i < limit; i ++)
		oppo[i] = (oppo[i >> 1] >> 1) | ((i & 1) << (lim - 1));
	for (int i = 0; i < limit; i ++)
		A[i] = B[i] = 0;
	for (int i = 0; i < deg; i ++)
		A[i] = f[i];
	for (int i = 0; i < deg << 1; i ++)
		B[i] = ps[i];
	NTT (A, 1), NTT (B, 1);
	for (int i = 0; i < limit; i ++)
		B[i] = B[i] * ((2ll - A[i] * B[i] % MOD + MOD) % MOD) % MOD;
	NTT (B, - 1);
	LL invn = power (n, MOD - 2);
	for (int i = 0; i < deg; i ++)
		ps[i] = B[i] * invn % MOD;
}

int getnum () {
	int num = 0;
	char ch = getchar ();

	while (! isdigit (ch))
		ch = getchar ();
	while (isdigit (ch))
		num = (num << 3) + (num << 1) + ch - '0', ch = getchar ();

	return num;
}

int main () {
	N = getnum ();
	for (int i = 0; i < N; i ++)
		f[i] = getnum ();
	Inverse (N, invf);
	for (int i = 0; i < N; i ++) {
		if (i > 0)
			putchar (' ');
		printf ("%lld", invf[i]);
	}
	puts ("");

	return 0;
}

/*
5
1 6 3 4 9
*/

/*
10
2 3 3 3 1233 211 23 3 3 322
*/