【LOJ #2409】【THUPC 2017】—小L的计算题（多项式Ln+生成函数）

传送门

先推一波式子
构造生成函数$OGF$
$f(x)= sum_{i=0}^{infty}f_ix^i \ = sum_{i=0}^{infty}sum_{j=1}^{n}a_j^ix^i \ =sum_{j=1}^{n}sum_{i=0}^{infty}(a_jx)^i \ =sum_{j=1}^{n}frac{1}{1-a_jx}$

这个东西也不好直接求
继续化

$f(x)=sum_{i=1}^{n}1-frac{-a_jx}{1-a_jx}\ =n- xsum_{i=1}^{n}frac{-a_j}{1-a_jx}$

考虑对$1-a_jx$求导就是$-a_j$
所以后面一坨东西可以化成$Ln$求导得到的

$f(x)=n-x(sum_{i=1}^{n}Ln(1-a_jx))' \ =n-x[Ln(prod_{i=1}^n(1-a_jx))]'$

分治$ntt$后做$Ln$即可

$a$读入时先取模

#include<bits/stdc++.h>
using namespace std;
#define gc getchar
inline int read(){
	char ch=gc();
	int res=0,f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
#define re register
#define pb push_back
#define cs const
#define pii pair<int,int>
#define fi first
#define se second
#define ll long long
cs int mod=998244353,G=3;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline void Dec(int &a,int b){(a-=b)<0?(a+=mod):0;}
inline int mul(int a,int b){return 1ll*a*b>=mod?1ll*a*b%mod:a*b;}
inline void Mul(int &a,int b){a=mul(a,b);}
inline int ksm(int a,int b,int res=1){
	for(;b;b>>=1,a=mul(a,a))(b&1)&&(res=mul(res,a));return res;
}
inline void chemx(ll &a,ll b){a<b?a=b:0;}
inline void chemn(int &a,int b){a>b?a=b:0;}
cs int N=(1<<20)|1,C=21;
#define poly vector<int>
#define bg begin
poly w[C+1];
inline void init_w(){
	for(int i=1;i<=C;i++)w[i].resize(1<<(i-1));
	int wn=ksm(G,(mod-1)/(1<<C));
	w[C][0]=1;
	for(int i=1;i<(1<<(C-1));i++)w[C][i]=mul(w[C][i-1],wn);
	for(int i=C-1;i;i--)
	for(int j=0;j<(1<<(i-1));j++)
	w[i][j]=w[i+1][j<<1];
} 
int rev[N<<1],inv[N<<1];
inline void init_rev(int lim){
	for(int i=0;i<lim;i++)rev[i]=(rev[i>>1]>>1)|((i&1)*(lim>>1));
}
inline void ntt(poly &f,int lim,int kd){
	for(int i=0;i<lim;i++)if(i>rev[i])swap(f[i],f[rev[i]]);
	for(int a0,a1,mid=1,l=1;mid<lim;mid<<=1,l++)
	for(int i=0;i<lim;i+=(mid<<1))
	for(int j=0;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.bg()+1,f.bg()+lim);
		for(int i=0;i<lim;i++)Mul(f[i],inv[lim]);
	}
}
inline poly operator *(poly a,poly b){
	int deg=a.size()+b.size()-1,lim=1;
	if(deg<=128){
		poly c(deg,0);
		for(int i=0;i<a.size();i++)
		for(int j=0;j<b.size();j++)
		Add(c[i+j],mul(a[i],b[j]));
		return c;
	}
	while(lim<deg)lim<<=1;
	init_rev(lim);
	a.resize(lim),b.resize(lim);
	ntt(a,lim,1),ntt(b,lim,1);
	for(int i=0;i<lim;i++)Mul(a[i],b[i]);
	ntt(a,lim,-1),a.resize(deg);
	return a;
}
inline int F(cs poly &f,int x){
	int res=0;
	for(int i=0,t=1;i<f.size();i++,Mul(t,x))Add(res,mul(f[i],t));
	return res;
}
inline poly deriv(poly f){
	for(int i=0;i<f.size()-1;i++)f[i]=mul(f[i+1],i+1);
	f.pop_back();return f;
}
inline poly integ(poly f){
	f.pb(0);
	for(int i=f.size()-1;i;i--)f[i]=mul(f[i-1],inv[i]);
	f[0]=0;return f;
}
inline void init_inv(){
	inv[0]=inv[1]=1;
	for(int i=2;i<N*2;i++)inv[i]=mul(mod-mod/i,inv[mod%i]);
}
inline poly Inv(poly a,int deg){
	poly c,b(1,ksm(a[0],mod-2));
	for(int lim=4;lim<(deg<<2);lim<<=1){
		c=a,c.resize(lim>>1);
		init_rev(lim);
		b.resize(lim),ntt(b,lim,1);
		c.resize(lim),ntt(c,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(deg);return b;
}
inline poly Ln(poly f,int deg){
	f=integ(deriv(f)*Inv(f,deg)),f.resize(deg);
	return f;
}
poly f[N<<2];
#define lc (u<<1)
#define rc ((u<<1)|1)
#define mid ((l+r)>>1)
inline void build(int u,int l,int r,int *v){
	if(l==r){f[u].clear();f[u].pb(1),f[u].pb(mod-v[l]);return;}
	build(lc,l,mid,v),build(rc,mid+1,r,v);
	f[u]=f[lc]*f[rc];
}
#undef lc
#undef rc
#undef mid
int n,a[N];
int main(){
	#ifdef Stargazer
	freopen("1.in","r",stdin);
	#endif
	int T=read();
	init_inv();
	init_w();
	while(T--){
		n=read();
		for(int i=1;i<=n;i++)a[i]=read()%mod;
		build(1,1,n,a);
		poly res=Ln(f[1],n+1);
		res=deriv(res);
		res.pb(0);
		for(int i=res.size()-1;i;i--)res[i]=mod-res[i-1];
		res[0]=n;
		int ans=0;
		for(int i=1;i<res.size();i++)ans^=res[i];
		cout<<ans<<'
';
	}
}