【模板】多项式乘法 NTT

Code:

#include<bits/stdc++.h>
#define maxn 4000002
#define mod 998244353
#define ll long long
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
ll qpow(ll base,ll k)
{
	ll tmp=1;
	while(k)
	{
		if(k&1) tmp=tmp*base%mod;
		base=base*base%mod;
		k>>=1;
	}
	return tmp;
}
void NTT(ll *a,int n,int flag)
{
	for(int i=0,k=0;i<n;++i)
	{
		if(i>k) swap(a[i],a[k]);
		for(int j=(n>>1);(k^=j)<j;j>>=1);
	}
    for(int i=1;i<n;i<<=1)
    {
    	ll wn=qpow(3, (mod-1)/(i<<1));
    	if(flag==-1) wn=qpow(wn, mod-2);
    	for(int j=0;j<n;j+=(i<<1))
    	{
    		ll w=1,x,y;
    		for(int k=0;k<i;++k)
    		{
    			x=a[j+k], y=1ll*w*a[j+k+i]%mod;
    			a[j+k]=(1ll*x+y)%mod, a[j+k+i]=(1ll*x-y+mod)%mod;
    			w*=wn, w%=mod;
    		}
    	}
    }
    if(flag==-1)
    {
    	ll rev=qpow(n, mod-2);
    	for(int i=0;i<n;++i) a[i]=(1ll*a[i]*rev)%mod;
    }
}
ll A[maxn], B[maxn];
int main()
{
	// setIO("input");
	int n,m,len;
	scanf("%d%d",&n,&m);
	for(int i=0;i<=n;++i) scanf("%lld",&A[i]);
	for(int i=0;i<=m;++i) scanf("%lld",&B[i]);
	for(len=1;len<(n+m+1);len<<=1);
	NTT(A,len,1), NTT(B,len,1);
    for(int i=0;i<len;++i) A[i]=(1ll*A[i]*B[i])%mod;
    NTT(A,len,-1);
    for(int i=0;i<(n+m+1);++i) printf("%lld ",A[i]);
	return 0;
}