由题意可得$f[n]=\sum _{i=1}^{n}a[i]f[n-i]$，设$f[n]$的生成函数为$F(x)$，$a[n]$的生成函数为$A(n)$
$$\begin{gathered} F(x)A(x)=\sum _{n\ge 0}{x}^n\sum _{i+j=n}{a}_i{f}_{n-i} \\ 由于a_0=0，a_0{f}_0=0，右侧卷积缺了f_0这一项 \\ F(x)A(x)=F(x)-f_0 \\ F(x)=\frac{1}{1-A(x)} \\ 多项式求个逆就好了 \end{gathered}$$

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<queue>
using namespace std;
#define ll long long
#define ull unsigned long long
#define RG register
#define MAX 288888
#define MOD (313)
const double Pi=acos(-1);
const int m=sqrt(MOD);
inline int read()
{RG int x=0,t=1;RG char ch=getchar();while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();if(ch=='-')t=-1,ch=getchar();while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();return x*t;
}
int fpow(int a,int b){int s=1;while(b){if(b&1)s=1ll*s*a%MOD;a=1ll*a*a%MOD;b>>=1;}return s;}
struct Complex{double a,b;}W[MAX],A[MAX],B[MAX],C[MAX],D[MAX];
Complex operator+(Complex a,Complex b){return (Complex){a.a+b.a,a.b+b.b};}
Complex operator-(Complex a,Complex b){return (Complex){a.a-b.a,a.b-b.b};}
Complex operator*(Complex a,Complex b){return (Complex){a.a*b.a-a.b*b.b,a.a*b.b+a.b*b.a};}
int r[MAX];
void FFT(Complex *P,int N,int opt)
{for(int i=0;i<N;++i)if(i<r[i])swap(P[i],P[r[i]]);for(int i=1;i<N;i<<=1)for(int p=i<<1,j=0;j<N;j+=p)for(int k=0;k<i;++k){Complex w=(Complex){W[N/i*k].a,W[N/i*k].b*opt};Complex X=P[j+k],Y=P[i+j+k]*w;P[j+k]=X+Y;P[i+j+k]=X-Y;}if(opt==-1)for(int i=0;i<N;++i)P[i].a/=1.0*N;
}
void Multi(int *a,int *b,int len,int *ret)
{for(int i=0;i<(len<<1);++i)A[i]=B[i]=C[i]=D[i]=(Complex){0,0};for(int i=0;i<len;++i){a[i]%=MOD;b[i]%=MOD;A[i]=(Complex){(a[i]/m)*1.0,0};B[i]=(Complex){(a[i]%m)*1.0,0};C[i]=(Complex){(b[i]/m)*1.0,0};D[i]=(Complex){(b[i]%m)*1.0,0};}int N,l=0;for(N=1;N<=len;N<<=1)++l;for(int i=0;i<N;++i)r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));for(int i=1;i<N;i<<=1)for(int k=0;k<i;++k)W[N/i*k]=(Complex){cos(k*Pi/i),sin(k*Pi/i)};FFT(A,N,1);FFT(B,N,1);FFT(C,N,1);FFT(D,N,1);for(int i=0;i<N;++i){Complex tmp=A[i]*C[i];C[i]=B[i]*C[i],B[i]=B[i]*D[i],D[i]=D[i]*A[i];A[i]=tmp;C[i]=C[i]+D[i];}FFT(A,N,-1);FFT(B,N,-1);FFT(C,N,-1);for(int i=0;i<len;++i){ret[i]=0;ret[i]=(ret[i]+1ll*(ll)(A[i].a+0.5)%MOD*m%MOD*m%MOD)%MOD;ret[i]=(ret[i]+1ll*(ll)(C[i].a+0.5)%MOD*m%MOD)%MOD;ret[i]=(ret[i]+1ll*(ll)(B[i].a+0.5)%MOD)%MOD;ret[i]=(ret[i]+MOD)%MOD;}
}
int c[MAX],d[MAX];
void Inv(int *a,int *b,int len)
{if(len==1){b[0]=fpow(a[0],MOD-2);return;}Inv(a,b,len>>1);Multi(a,b,len,c);Multi(c,b,len,d);for(int i=0;i<len;++i)b[i]=(b[i]+b[i])%MOD;for(int i=0;i<len;++i)b[i]=(b[i]+MOD-d[i])%MOD;
}
int n,a[MAX],b[MAX];
int main()
{while(n=read()){for(int i=1;i<=n;++i)a[i]=read()%MOD;for(int i=1;i<=n;++i)a[i]=(MOD-a[i])%MOD;a[0]++;int N;for(N=1;N<=n;N<<=1);Inv(a,b,N);printf("%d\n",b[n]);memset(a,0,sizeof(a));memset(b,0,sizeof(b));}return 0;
}