【模板】多项式求逆

题目

要求出满足(F(x)∗G(x)≡1 (mod x^n))
假设已知(F(x)*H(x) ≡1 (mod x^lceil frac{n}{2} ceil))
又(F(x)*G(x) ≡1 (mod x^lceil frac{n}{2} ceil))
所以(F(x)*(G(x)-H(x)) ≡ 0 (mod x^lceil frac{n}{2} ceil))
两边平方(G(x)^2+H(x)^2-2G(x)H(x) ≡ 0 (mod x^n))
两边同时乘F(x) (F(x)G(x)^2+H(x)^2F(x)-2F(x)G(x)H(x) ≡ 0 (mod x^n))
可得(G(x) ≡ 2H(x)-H(x)^2*F(x) (mod x^n))
然后用NTT做多项式乘法...
因为第一次写这个，而且NTT也不是很熟练，所以是连写带贺的QWQ过段时间再来重构

#include <bits/stdc++.h>
using namespace std;
const int Mo = 998244353;
const int MAXN = 2100000;
int n, a[MAXN], b[MAXN], d[MAXN], rev[MAXN];
inline int read() {
    char ch; bool f = false; int res = 0;
    while (((ch = getchar()) < '0' || ch > '9') && ch != '-');
    if (ch == '-') f = true; else res = ch - '0';
    while ((ch = getchar()) >= '0' && ch <= '9') res = (res << 3) + (res << 1) + ch - '0';
    return f? ~res + 1 : res;
}
inline int ksm (int x, int y) {
	int sum = 1;
	for (; y; ) {
		if (y & 1)
			sum = 1LL * sum * x % Mo;
			y >>= 1;
			x = 1LL * x * x % Mo; 
	}
	return sum;
}
void NTT (int *a, int n, int x){
	for (int i = 0; i < n; ++ i) 
		if (i < rev[i])
			swap(a[i], a[rev[i]]);
	for (int i = 1; i < n; i <<= 1) {
		int y = ksm(3, (Mo - 1) / (i << 1));
		for (int j = 0; j < n; j += (i << 1)) {
			int x1, x2, x3 = 1;
			for (int k = 0; k < i; ++ k, x3 = 1LL * x3 * y % Mo) {
				x1 = a[j + k];
				x2 = 1LL * x3 * a[j + k + i] % Mo;
				a[j + k] = (x1 + x2) % Mo;
				a[j + k + i] = (x1 - x2 + Mo) % Mo;
			}
		}
	}
	if (x == 1) 
		return;
	int z = ksm(n, Mo - 2);
	reverse(a + 1, a + n);
	for (int i = 0; i < n; ++ i)
		a[i] = 1LL * a[i] * z % Mo;
}
void work (int c, int *a, int *b) {
	if (c == 1) {
		b[0] = ksm(a[0], Mo - 2);
		return;
	}
	work((c + 1) >> 1, a, b);
	int len = 0, x = 1;
	while (x < (c << 1))
		x <<= 1, ++ len;
	for (int i = 1; i < x; ++ i) 
		rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (len - 1));
	for (int i = 0; i < c; ++ i)
		d[i] = a[i];
	for (int i = c; i < x; ++ i)
		d[i] = 0;
	NTT(d, x, 1);
	NTT(b, x, 1);
	for (int i = 0; i < x; ++ i)
		b[i] = 1LL * (2 - 1LL * d[i] * b[i] % Mo + Mo) % Mo * b[i] % Mo;
	NTT(b, x, -1);
	for (int i = c; i < x; ++ i)
		b[i] = 0;	
}
int main() {
	n = read();
	for (int i = 0; i < n; ++ i)
		a[i] = read();
	work(n, a, b);
	for (int i = 0; i < n; ++ i)
		printf("%d ", b[i]);
	return 0;
}