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^{leftlceilfrac{n}{2} ight ceil}} ... (1)) ，因为 (x^n) 是 (x^{leftlceilfrac{n}{2} ight ceil}) 的倍数

并且存在 (A(x)B'(x) equiv 1 pmod{x^{leftlceilfrac{n}{2} ight ceil}} ... (2))

((1) - (2)) ，得

[egin{aligned} B(x) - B'(x) &equiv 0 pmod{x^{leftlceilfrac{n}{2} ight ceil}} ... (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^{leftlceilfrac{n}{2} ight ceil}) 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 = sumlimits_{j = 1}^i a_j b_{i - j}) ，所以必定存在 (i) 或者 (i - j) 小于 (x^{leftlceilfrac{n}{2} ight ceil}) ，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
*/