【模板】FFT

FFT 的本质是在 (O(nlogn)) 的时间内进行点值表达和系数表达之间的转换，并在 (O(n)) 的时间内计算结果，故总时间复杂度为 (O(logn))。
FFT注意事项

• 由于 tot 采用的是倍增的方式，因此实际内存开销可能是 (2(n+m))。
• 多项式的项数和次数不同。

update at 2019.10.11
代码如下

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

const LL mod = 998244353, g = 3, ig = 332748118;

inline LL fpow(LL a, LL b) {
	LL ret = 1 % mod;
	for (; b; b >>= 1, a = a * a % mod) {
		if (b & 1) {
			ret = ret * a % mod;
		}
	}
	return ret;
}

void ntt(vector<LL> &v, vector<int> &rev, int opt) {
	int tot = v.size();
	for (int i = 0; i < tot; i++) if (i < rev[i]) swap(v[i], v[rev[i]]);
	for (int mid = 1; mid < tot; mid <<= 1) {
		LL wn = fpow(opt == 1 ? g : ig, (mod - 1) / (mid << 1));
		for (int j = 0; j < tot; j += mid << 1) {
			LL w = 1;
			for (int k = 0; k < mid; k++) {
				LL x = v[j + k], y = v[j + mid + k] * w % mod;
				v[j + k] = (x + y) % mod, v[j + mid + k] = (x - y + mod) % mod;
				w = w * wn % mod;
			}
		}
	}
	if (opt == -1) {
		LL itot = fpow(tot, mod - 2);
		for (int i = 0; i < tot; i++) v[i] = v[i] * itot % mod;
	}
}
vector<LL> convolution(vector<LL> &a, int cnta, vector<LL> &b, int cntb, const function<LL(LL, LL)> &calc) {
	int bit = 0, tot = 1;
	while (tot <= 2 * max(cnta, cntb)) bit++, tot <<= 1;
	vector<int> rev(tot);
	for (int i = 0; i < tot; i++) rev[i] = rev[i >> 1] >> 1 | (i & 1) << (bit - 1);
	vector<LL> foo(tot), bar(tot);
	for (int i = 0; i < cnta; i++) foo[i] = a[i];
	for (int i = 0; i < cntb; i++) bar[i] = b[i];
	ntt(foo, rev, 1), ntt(bar, rev, 1);
	for (int i = 0; i < tot; i++) foo[i] = calc(foo[i], bar[i]);
	ntt(foo, rev, -1);
	return foo;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	int n, m;
	cin >> n >> m;
	vector<LL> a(n + 1), b(m + 1);
	for (int i = 0; i <= n; i++) {
		cin >> a[i];
	}
	for (int i = 0; i <= m; i++) {
		cin >> b[i];
	}
	vector<LL> c = convolution(a, n + 1, b, m + 1, [&](LL a, LL b) {
		return a * b % mod;
	});
	for (int i = 0; i <= n + m; i++) {
		cout << c[i] << " ";
	}
	return 0;
}