[LOJ2409][生成函数][NTT]THUPC2017：小L的计算题

LOJ2409

设$F(x)$为$f$的OGF
则$$F(x)=\sum _{i=0}^{\infty }f[i]x^i$$
$$=\sum _{i=0}^{\infty }{x}^i\sum _{j=1}^n{a_j}^i$$
$$=\sum _{j=1}^n\sum _{i=0}^{\infty }(x\ast a_j{)}^i$$
$$=\sum _{i=1}^n\frac{1}{1-a_{i}x}$$
$$=\sum _{i=1}^{n}1+\frac{a_{i}x}{1-a_{i}x}$$
$$=n-\sum _{i=1}^n\frac{-a_{i}x}{1-a_{i}x}$$
$$=n-x\sum _{i=1}^n\frac{-a_i}{1-a_{i}x}$$
然后右边就可以愉快的求导了
$$=n-x(ln(\prod _j(1-a_{j}x){)}^{\prime }$$
则分治NTT+多项式ln即可

Code：

#include<bits/stdc++.h>
#define poly vector<ll>
#define ll long long
#define int long long
#define pb push_back
#define mod 998244353
using namespace std;
inline int read(){
	int res=0,f=1;char ch=getchar();
	while(!isdigit(ch)) {if(ch=='-') f=-f;ch=getchar();}
	while(isdigit(ch)) {res=(res<<1)+(res<<3)+(ch^48);ch=getchar();}
	return res*f;
}
inline ll add(ll x,ll y){x+=y;if(x>=mod) x-=mod;return x;}
inline ll dec(ll x,ll y){x-=y;if(x<0) x+=mod;return x;}
inline ll mul(ll a,ll b){return ((a*b)-(ll)((long double)a/mod*b)*mod+mod)%mod;}
inline void Mul(ll &x,ll y){x=mul(x,y);}
inline void inc(ll &x,ll y){x+=y;if(x>=mod) x-=mod;}
inline ll ksm(ll a,ll b){ll res=1;for(;b;b>>=1,Mul(a,a)) if(b&1) Mul(res,a);return res;}
namespace Ntt{
	const int N=1e6+5;
	int *w[22],rev[N<<2];
	inline void init(int n){for(int i=0;i<n;i++) rev[i]=(rev[i>>1]>>1)|((i&1)*(n>>1));}
	inline void init_w(){
		for(int i=1;i<=21;i++) w[i]=new int[1<<(i-1)];
		int wn=ksm(3,(mod-1)/(1<<21));
		w[21][0]=1;
		for(int i=1;i<(1<<(20));i++) w[21][i]=mul(w[21][i-1],wn);
		for(int i=20;i;i--)
			for(int j=0;j<(1<<(i-1));j++) w[i][j]=w[i+1][j<<1];
	}
	inline void ntt(poly &f,int n,int kd){
		for(int i=0;i<n;i++) if(i>rev[i]) swap(f[i],f[rev[i]]);
		for(int mid=1,l=1;mid<n;mid<<=1,l++){
			for(int i=0;i<n;i+=(mid<<1)){
				for(int j=0,a0,a1;j<mid;j++){
					a0=f[i+j],a1=mul(f[i+j+mid],w[l][j]);
					f[i+j]=add(a0,a1);f[i+j+mid]=dec(a0,a1);
				}
			}
		}
		if(kd==-1 && (reverse(f.begin()+1,f.begin()+n),1))
			for(int inv=ksm(n,mod-2),i=0;i<n;i++) Mul(f[i],inv);
	}	
	inline poly operator -(poly a,poly b){
		poly c;int lim=max(a.size(),b.size());c.resize(lim);
		for(int i=0;i<lim;i++)c[i]=dec(a[i],b[i]);return c;
	}
	inline void poly_mul(poly &a,int b){for(int i=0;i<a.size();i++) a[i]=mul(a[i],b);}
	inline poly operator *(poly a,poly b){
		int m=a.size()+b.size()-1,n=1;
		if(m<=128){
			poly c(m,0);
			for(int i=0;i<a.size();i++)
				for(int j=0;j<b.size();j++) inc(c[i+j],mul(a[i],b[j]));
			return c;	
		}
		while(n<m) n<<=1;
		init(n);
		a.resize(n);ntt(a,n,1);
		b.resize(n);ntt(b,n,1);
		for(int i=0;i<n;i++) Mul(a[i],b[i]);
		ntt(a,n,-1);a.resize(m);
		return a;
	}
}
using namespace Ntt;

ll val[N];
inline poly solve(int l,int r){
	if(l==r){poly c;c.pb(1);c.pb(-val[l]);return c;}
	int mid=l+r>>1;
	return solve(l,mid)*solve(mid+1,r);
}

inline poly Inv(poly a,int n){
	poly c,b(1,ksm(a[0],mod-2));
	for(int lim=4;lim<(n<<2);lim<<=1){
		init(lim);
		c=a;c.resize(lim>>1);
		c.resize(lim);ntt(c,lim,1);
		b.resize(lim);ntt(b,lim,1);
		for(int i=0;i<lim;i++) Mul(b[i],dec(2,mul(b[i],c[i])));
		ntt(b,lim,-1);b.resize(lim>>1);
	}
	b.resize(n);return b;
}

inline poly deriv(poly a){
	for(int i=0;i<a.size()-1;i++) a[i]=mul(a[i+1],i+1);
	a.pop_back();return a;
}

ll inv[N];

inline poly integ(poly a){
	a.push_back(0);
	for(int i=a.size()-1;i;i--) a[i]=mul(a[i-1],inv[i]);
	a[0]=0;
	return a;
}

inline poly ln(poly a,int n){
	a=integ(deriv(a)*Inv(a,n)),a.resize(n);
	return a;
}

poly a,ans;
signed main(){
	inv[1]=1;
	for(int i=2;i<N;i++) inv[i]=((mod-mod/i)*inv[mod%i])%mod;
	init_w();
	int t=read();
	while(t--){
		int n=read();
		for(int i=1;i<=n;i++) val[i]=read();
		a=solve(1,n);
		poly b;
		b.pb(0);b.pb(1);
		ans=deriv(ln(a,a.size()+1))*b;
		poly_mul(ans,-1);
		ll finalans=0;
		for(int i=1;i<=n;i++) finalans^=ans[i];
		cout<<finalans<<"\n";
	}
	return 0;
}