【luoguP4721】分治 FFT

description

给定长度为\(n-1\)的数组\(g[1],g[2],..,g[n-1]\)，求\(f[0],f[1],..,f[n-1]\)，其中

\[f[i]=\sum_{j=1}^if[i-j]g[j]
\]

边界为 \(f[0]=1\)。答案模\(998244353\)。

---

analysis

• 一道分治\(NTT\)板题

• 经历过城市规划那题的洗礼之后这题变得微不足道

• 考虑\(CDQ\)分治，求出\([l,mid]\)对\([mid+r]\)的贡献

• 把\(f[l,mid]\)拉出来，与\(g[1..r-l]\)相乘，答案数组的后\(r-mid\)位就是分别对\([mid+r]\)的贡献

• 具体可以画出两个多项式在分治过程中的相乘，结合每一个\(f\)的值就可以弄清楚

• 由于这个\(NTT\)很清真所以\(l==r\)时就直接\(return\)了，当然也没有各种阶乘逆元什么的

• 下次学多项式求逆再来做一次这题 （FLAG）

---

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 400005
#define G 3
#define mod 998244353
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)

using namespace std;

ll f[MAXN],g[MAXN];
ll a[MAXN],b[MAXN],rev[MAXN];
ll n,m;

inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
inline ll work(ll x)
{
	ll y=1;
	while (y<x)y<<=1;
	return y<<1;
}
inline ll pow(ll x,ll y)
{
	ll z=1;
	while (y)
	{
		if (y&1)z=z*x%mod;
		x=x*x%mod,y>>=1;
	}
	return z;
}
inline void ntt(ll a[],ll len,ll inv)
{
	ll bit=0;
	while ((1<<bit)<len)++bit;
	fo(i,0,len-1)
	{
		rev[i]=(rev[i>>1]>>1)|((i&1)<<(bit-1));
		if (i<rev[i])swap(a[i],a[rev[i]]);
	}
	for (ll mid=1;mid<len;mid*=2)
	{
		ll tmp=pow(G,(mod-1)/(mid*2));
		if (inv==-1)tmp=pow(tmp,mod-2);
		for (ll i=0;i<len;i+=mid*2)
		{
			ll omega=1;
			for (ll j=0;j<mid;++j,omega=omega*tmp%mod)
			{
				ll x=a[i+j],y=omega*a[i+j+mid]%mod;
				a[i+j]=(x+y)%mod,a[i+j+mid]=(x-y+mod)%mod;
			}
		}
	}
}
inline void CDQ(ll l,ll r)
{
	if (l==r)return;
	ll mid=(l+r)>>1;
	CDQ(l,mid);
	ll len=work(r-l+1),invv=pow(len,mod-2);
	fo(i,0,len-1)a[i]=b[i]=0;
	fo(i,1,mid-l+1)a[i]=f[l+i-1];
	fo(i,1,r-l)b[i]=g[i];
	ntt(a,len,1),ntt(b,len,1);
	fo(i,0,len-1)a[i]=(a[i]*b[i]%mod);
	ntt(a,len,-1);
	fo(i,0,len-1)a[i]=(a[i]*invv)%mod;
	fo(i,mid+1,r)(f[i]+=a[i-l+1])%=mod;
	CDQ(mid+1,r);
}
int main()
{
	freopen("CDQNTT.in","r",stdin);
	n=read(),m=work(n);
	fo(i,1,n-1)g[i]=read();
	f[1]=1,CDQ(1,m/2);
	fo(i,1,n)printf("%lld ",f[i]);
	printf("\n");
	return 0;
}