多项式求逆 模板

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 300000
#define md 998244353
#define ll long long
ll f[N], g[N], f0[N], g0[N];
int rev[N];
ll ksm(ll x, ll y) {
	if(!y) return 1;
	ll l = ksm(x, y / 2);
	if(y % 2) return l * l % md * x % md;
	return l * l % md;
}
void ntt(ll *a, int n, int r) {
	for(int i = 0; i < n; i++) {
		rev[i] = rev[i / 2] / 2 + (i & 1) * n / 2;
		if(i < rev[i]) swap(a[i], a[rev[i]]);
	}
	for(int i = 2; i <= n; i *= 2) {
		ll t = ksm(3, (md - 1) / i);
		if(r == -1) t = ksm(t, md - 2);
		for(int j = 0; j < n; j += i) {
			ll s = 1;
			for(int k = 0; k < i / 2; k++) {
				ll A = a[j + k], B = a[j + i / 2 + k] * s % md;
				a[j + k] = (A + B) % md;
				a[j + i / 2 + k] = (A - B + md) % md;
				s = s * t % md;
			}
		}
	}
	if(r == -1) {
		ll v = ksm(n, md - 2);
		for(int i = 0; i < n; i++) a[i] = (a[i] + md) * v % md;
	}
}
int main() {
	int n, i, j, k;
	scanf("%d", &n);
	for(i = 0; i < n; i++) scanf("%lld", &f[i]);
	
	int ns = 1;
	while(ns <= n) ns *= 2;
	
	g[0] = ksm(f[0], md - 2);
	for(i = 2; i <= ns; i *= 2) {
		for(j = 0; j < i; j++) f0[j] = f[j], g0[j] = g[j], g0[j + i] = f0[j + i] = 0;
		
		ntt(f0, 2 * i, 1);
		
		ntt(g0, 2 * i, 1);
		
		for(j = 0; j < 2 * i; j++) g0[j] = g0[j] * g0[j] % md * f0[j] % md;
		
		ntt(g0, 2 * i, -1);
		
		for(j = 0; j < i; j++) g[j] = (g[j] * 2 - g0[j] + md) % md;
	}
	for(i = 0; i < n; i++) printf("%lld ", g[i]);
	return 0;
}