51nod 1514 美妙的序列 分治NTT + 容斥

Code:

#include<bits/stdc++.h>
#define ll long long
#define mod 998244353
#define maxn 400000
#define N 100005
#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 mid=1;mid<n;mid<<=1)
    {
    	ll wn=qpow(3, (mod-1)/(mid<<1)),x,y;
    	if(flag==-1) wn=qpow(wn,mod-2);
    	for(int i=0;i<n;i+=(mid<<1))
    	{
    		ll w=1;
    		for(int j=0;j<mid;++j)
    		{
    			x=a[i+j],y=w*a[i+j+mid];
    			a[i+j]=(x+y)%mod,a[i+j+mid]=(x-y+mod)%mod;
    			w=w*wn%mod;
    		}
    	}
    }
    if(flag==-1)
    {
    	ll rev=qpow(n,mod-2);
    	for(int i=0;i<n;++i) a[i]=a[i]*rev%mod;
    }
}
ll f[maxn],g[maxn],A[maxn],B[maxn];
void solve(int l,int r)
{
	if(l==r) return;
	int mid=(l+r)>>1,len;
	solve(l,mid);
	for(len=1;len<=(r-l+1);len<<=1);
	for(int i=l;i<=mid;++i) A[i-l]=f[i];
	for(int i=1;i<=r-l;++i) B[i-1]=g[i];
	NTT(A,len,1),NTT(B,len,1);
    for(int i=0;i<len;++i) A[i]=A[i]*B[i]%mod;
    NTT(A,len,-1);
    for(int i=mid+1;i<=r;++i) f[i]=(f[i]-A[i-l-1]+mod)%mod;
    for(int i=0;i<=len;++i) A[i]=B[i]=0;
    solve(mid+1,r);
}
void Initialize()
{
	f[1]=g[0]=1;
	for(int i=1;i<=N;++i) f[i]=g[i]=g[i-1]*i%mod;
}
int main()
{
	// setIO("input");
	Initialize();
	int T,x;
	solve(1, 100003);
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&x);
		printf("%lld\n",f[x]);
	}
	return 0;
}