NTT

存一发模板233

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
using namespace std;
typedef long long ll;
const int N = 4000100,mod = 998244353;
ll read() {
	ll x=0,f=1;char c=getchar();
	while(c<'0'||c>'9') {
		if(c=='-') f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9') {
		x=x*10+c-'0';
		c=getchar();
	}
	return x*f;
}
int qm(ll x,int y) {
	ll ret = 1;
	for(;y;y >>= 1,x = x * x % mod)
		if(y & 1) ret = ret * x % mod;

	return ret;
}
ll A[N],B[N];
int rev[N];
void NTT(ll *a,int n,int xs) {
	for(int i = 0;i <= n;++i)
		if(rev[i] > i) swap(a[i],a[rev[i]]);
	for(int m = 2;m <= n;m <<= 1) {
		ll w1 = qm(3,(mod - 1) / m);
		if(xs == -1) w1 = qm(w1,mod - 2);

		for(int i = 0;i < n;i += m) {
			ll w = 1;
			for(int k = 0;k < (m >> 1);++k) {
				ll u = a[i + k],t = w * a[i + k + (m >> 1)] % mod;
				a[i + k] = (u + t) % mod;a[i + k + (m >> 1)] = (u - t) % mod;
				w = w * w1 % mod;
			}
		}
	}
}
int main() {
	int n = read(),m = read();
	for(int i = 0;i <= n;++i) A[i] = read();
	for(int i = 0;i <= m;++i) B[i] = read();

	int tot = 1;
	while(tot <= n + m) tot <<= 1;
	for(int i = 0;i <= tot;++i) 
		rev[i] = (rev[i >> 1] >> 1) | (i & 1 ? (tot >> 1) : 0);

	NTT(A,tot,1);NTT(B,tot,1);
	for(int i = 0;i <= tot;++i) A[i] = A[i] * B[i] % mod;
	NTT(A,tot,-1);
	int tmp = qm(tot,mod - 2);
	for(int i = 0;i <= n + m;++i)
		printf("%lld ",(A[i] + mod) % mod * tmp % mod);

	return 0;
}