常系数齐次线性递推板子

板子是抄的

不过他m=1的情况有bug要特判

#include <bits/stdc++.h>
using namespace std;
#define rep(i,h,t) for (int i=h;i<=t;i++)
#define dep(i,t,h) for (int i=t;i>=h;i--)
#define me(x) memset(x,0,sizeof(x))
#define IL inline
#define rint register int
#define ll long long
using namespace std;
const int N=150150,P=998244353; // len=32000--->n=150150 
struct F{
  #define FF for (int i=0;i<L;i++)
  #define CON(a,b) DFT(a,L);FF a[i]=1ll*a[i]*b[i]%P;IFT(a,L);
  int n,m,K,L,o,A,Q;
  int rv[N],W[N],p[N],f[N],t[N],a[N],b[N],C[N];
  unsigned long long B[N];
  int qpw(int x,int y){int z=1;for(;y;y>>=1,x=1ll*x*x%P)if(y&1) z=1ll*z*x%P;return z;}
  void ini(int n){
      for(L=1;L<=n;L<<=1,o++);
      FF rv[i]=rv[i>>1]>>1|(i&1)<<(o-1);
      int V=qpw(3,P>>o);W[L>>1]=1;
      rep(i,(L>>1)+1,L-1) W[i]=1ll*W[i-1]*V%P;
      dep(i,(L>>1)-1,1) W[i]=W[i<<1];
  }
  void DFT(int*A,int L){
      int u=o-__builtin_ctz(L),t;
      FF B[i]=A[rv[i]>>u];
      for(int i=1;i<L;i<<=1)for(int j=0,s=i<<1;j<L;j+=s) rep(k,0,i-1)
          t=B[i+j+k]*W[i+k]%P,B[i+j+k]=B[j+k]+P-t,B[j+k]+=t;
      FF A[i]=B[i]%P;
  }
  void IFT(int*A,int L){
      reverse(A+1,A+L);DFT(A,L);
      int V=P-(P-1)/L;
      FF A[i]=1ll*A[i]*V%P;
  }
  int upl(int n){return 1<<(32-__builtin_clz(n));}
  void INV(int*A,int*B,int n){
      static int L;
      if(!n) return B[0]=qpw(A[0],P-2),void();
      INV(A,B,n>>1);L=upl(n<<1);
      FF C[i]=i>n?0:A[i];
      DFT(C,L);DFT(B,L);
      FF B[i]=(2-1ll*C[i]*B[i]%P+P)*B[i]%P;IFT(B,L);
      FF B[i]=i>n?0:B[i],C[i]=0;
  }
  void MUL(int*A,int*B){
      FF C[i]=A[i];DFT(C,L);
      CON(B,C);
      FF C[i]=i>K?0:B[n-i];
      CON(C,t);
      FF C[i]=i>K?0:C[i];
      reverse(C,C+K+1);
      CON(C,p);
      FF (B[i]+=P-C[i])%=P;
  }
// f[i]=p[j]*f[i-j](1<=j<=m) f[0..n-1] Q:f[Q]
  void solve(){
      cin>>Q>>m;
      ini(m<<1);n=m-1<<1;K=n-m;
      rep(i,1,m) { cin>>p[i]; p[i]=(p[i]%P+P)%P;}
      rep(i,0,m-1) { cin>>f[i]; f[i]=(f[i]%P+P)%P;}
      //if (m=1) special judge
      if (m==1)
      {
          cout<<((f[0]*qpw(p[1],Q)%P)+P)%P<<endl;
          return 0;
      }
      p[0]=P-1;
      INV(p,t,K);DFT(t,L);
      reverse(p,p+m+1);DFT(p,L);
      a[1]=b[0]=1;
      for(;Q;Q>>=1,MUL(a,a))if(Q&1) MUL(a,b);
      rep(i,0,m-1) (A+=1ll*b[i]*f[i]%P)%=P;
      cout<<A<<'
';
  }
} F; 
int main()
{
    ios::sync_with_stdio(false);
    F.solve();
    return 0;
}