hdu2021多校7

链接

1011

题意

给一序列 $a_1,a_2\cdots a_n$,求子序列的和的乘积对 $998244353$ 取模后的结果。

保证 $\sum a_i\le 10^5$。

题解

只需求出有多少个子序列满足子序列和等于 $x$ 的,设为 $b_x$。答案即为 $\prod_{i=1}^{10^5}i^{b_i}$。

而 $b_m$ 显然等于 $[x^m]\prod_{i=1}^n(1+x^{a_i})$。

分治FFT即可。

但是 $b_m$ 是在模 $\varphi(998244353)=998244352$ 意义下的,于是需要使用拆系数FFT。

拆系数FFT详见链接

程序

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#include <bits/stdc++.h>
using namespace std;
#define FO(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout)
#define fo(i,j,k) for(int i=(j),end_i=(k);i<=end_i;i++)
#define ff(i,j,k) for(int i=(j),end_i=(k);i< end_i;i++)
#define fd(i,j,k) for(int i=(j),end_i=(k);i>=end_i;i--)
#define DEBUG(x) cerr<<#x<<"="<<x<<endl
#define all(x) (x).begin(),(x).end()
#define cle(x) memset(x,0,sizeof(x))
#define lowbit(x) ((x)&-(x))
#define ll long long
#define ull unsigned ll
#define db double
#define lb long db
#define pb push_back
#define mp make_pair
#define fi first
#define se second
inline int read()
{
int x=0; char ch=getchar(); bool f=0;
for(;ch<'0'||ch>'9';ch=getchar()) if(ch=='-') f=1;
for(;ch>='0'&&ch<='9';ch=getchar()) x=(x<<3)+(x<<1)+(ch^48);
return f?-x:x;
}
#define CASET fo(___,1,read())

const ll mod=998244353;
inline ll Add(ll x,ll y){x+=y; return (x<mod)?x:x-mod;}
inline ll Dec(ll x,ll y){x-=y; return (x<0)?x+mod:x;}
inline ll Mul(ll x,ll y){return x*y%mod;}
inline ll Pow(ll x,ll y)
{
y%=(mod-1);ll ans=1;for(;y;y>>=1,x=x*x%mod)if(y&1) ans=ans*x%mod;
return ans;
}
const ll mo=998244352;

db pi=acos(-1.);
struct P{
db x,y;
P(db _x=0,db _y=0) {x=_x,y=_y;}
friend inline P operator +(const P&A,const P&B)
{
return (P){A.x+B.x,A.y+B.y};
}
friend inline P operator -(const P&A,const P&B)
{
return (P){A.x-B.x,A.y-B.y};
}
friend inline P operator *(const P&A,const P&B)
{
return (P){A.x*B.x-A.y*B.y,A.x*B.y+A.y*B.x};
}
friend inline P operator /(const P&A,const db &x)
{
return (P){A.x/x,A.y/x};
}
};
P conj(P A) {return (P){A.x,-A.y};}

const int M=1<<20;
P W[M]; int R[M];
inline void PolyInit()
{
for(int i=1;i<M;i<<=1)
fo(j,0,i-1)
W[i+j]=(P){cos(pi*j/i),sin(pi*j/i)};
}
typedef vector<P> Poly;
inline void ntt(P *a,int n,int t)
{
fo(i,0,n-1)
{
R[i]=(R[i>>1]>>1)|((i&1)*(n>>1));
if(i<R[i]) swap(a[i],a[R[i]]);
}
P w;
for(int i=1;i<n;i<<=1)
for(int j=0;j<n;j+=(i<<1))
for(int k=0;k<i;k++)
w=W[i+k]*a[i+j+k],
a[i+j+k]=a[j+k]-w,
a[j+k]=a[j+k]+w;
if(t==1) return;
reverse(a+1,a+n);
fo(i,0,n-1) a[i]=a[i]/n;
}
inline void ntt(Poly &A,int n,int t){ntt(&A[0],n,t);}
inline Poly operator +(Poly A,Poly B)
{
A.resize(max(A.size(),B.size()));
fo(i,0,B.size()-1) A[i]=A[i]+B[i];
return A;
}
#define Real(A) ((ll)floor(A.x+0.5))
#define Imag(A) ((ll)floor(A.y+0.5))
inline Poly operator *(Poly A,Poly B)
{
static Poly C,D;
int n=A.size(),m=B.size(),k=n+m-1,len=1;
for(;len<k;len<<=1);
C.resize(len); D.resize(len); A.resize(len); B.resize(len);
ff(i,0,len) C[i]=(P){Real(A[i])&32767,Real(A[i])>>15},D[i]=(P){Real(B[i])&32767,Real(B[i])>>15};
ntt(C,len,1); ntt(D,len,1);
int j;
ff(i,0,len)
{
P d4,d0,d1,d2,d3;
j=(len-i)&(len-1);
d4=conj(C[j]);
d0=(d4+C[i])*P(0.5,0);
d1=(d4-C[i])*P(0,0.5);
d4=conj(D[j]);
d2=(d4+D[i])*P(0.5,0);
d3=(d4-D[i])*P(0,0.5);
A[i]=d0*d2+d1*d3*P(0,1);
B[i]=d0*d3+d1*d2;
}
ntt(A,len,-1); ntt(B,len,-1);
ff(i,0,len) C[i]=(Real(A[i]) + (Imag(A[i]) % mo << 30) + (Real(B[i]) % mo << 15))%mo;
C.resize(k);
return C;
}

const int N=100010;

Poly A;
int n,sum;
int a[N];

Poly solve(int l,int r)
{
if(l==r)
{
Poly B;
B.resize(a[l]+1);
fo(i,0,a[l]) B[i]=0;
B[0]=1; B[a[l]]=1;
return B;
}
int mid=(l+r)>>1;
return solve(l,mid)*solve(mid+1,r);
}

int main()
{
PolyInit();
CASET
{
n=read(); sum=0;
fo(i,1,n) sum+=(a[i]=read());
bool flag=0;
fo(i,1,n) if(a[i]==0) {flag=1; break;}
if(flag){puts("0"); continue;}
A=solve(1,n);
ll ans=1;
fo(i,1,sum) ans=ans*Pow(i,Real(A[i]))%mod;
printf("%lld\n",ans);
}
return 0;
}