写完帕秋莉的超级多项式于是正好贴个模板大汇总（带优化的那种...）

//by Judge#include
    
     #define Rg register#define fp(i,a,b) for(Rg int i=(a),I=(b)+1;i
     
      I;--i)#define ll long longusing namespace std;const int mod=998244353;const int iG=332748118;const int M=3e5+3;typedef int arr[M];#ifndef Judge#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)#endifchar buf[1<<21],*p1=buf,*p2=buf;inline int inc(int x,int y){return (x+y)%mod;}inline int dec(int x,int y){return (x-y+mod)%mod;}inline int Dec(int x,int y){return x
      
       =mod?x+y-mod:x+y;}inline int mul(int x,int y){return 1ll*x*y%mod;}inline int read(){ int x=0,f=1; char c=getchar();    for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;    for(;isdigit(c);c=getchar()) x=x*10+c-'0'; return x*f;} char sr[1<<21],z[20];int CCF=-1,Z;inline void Ot(){fwrite(sr,1,CCF+1,stdout),CCF=-1;}inline void print(int x,char chr=' '){    if(CCF>1<<20)Ot();if(x<0)sr[++CCF]=45,x=-x;    while(z[++Z]=x%10+48,x/=10);    while(sr[++CCF]=z[Z],--Z);sr[++CCF]=chr;} int n,k,d,limit; arr a,b,r[21],lg,inv,G[2];inline int qpow(Rg int x,Rg int p=mod-2,Rg int s=1){    for(;p;p>>=1,x=mul(x,x)) if(p&1) s=mul(s,x); return s;}inline void prep(int n){    inv[1]=1; for(limit=1;limit<=n;limit<<=1);    fp(i,2,limit) inv[i]=mul(mod-mod/i,inv[mod%i]);    fp(d,1,19){ lg[1<
       
        >1]>>1)|((i&1)<<(d-1));    }    for(Rg int t=(mod-1)>>1,i=1,x,y;i<262144;i<<=1,t>>=1){        x=qpow(3,t),y=qpow(iG,t),G[0][i]=G[1][i]=1;        fp(k,1,i-1) G[1][i+k]=mul(G[1][i+k-1],x),G[0][i+k]=mul(G[0][i+k-1],y);    }}inline void NTT(int* a,int tp){    fp(i,0,limit-1) if(i
        
         >1),init(n); fp(i,0,n-1) C[i]=a[i],D[i]=b[i];    fp(i,n,limit-1) C[i]=D[i]=0; NTT(C,1),NTT(D,1);    fp(i,0,limit-1) C[i]=mul(C[i],mul(D[i],D[i])); NTT(C,0);    fp(i,0,n-1) b[i]=dec(inc(b[i],b[i]),C[i]); fp(i,n,limit-1) b[i]=0;}void Sqrt(int* a,int* b,int n){    static arr D,F; if(n==1) return b[0]=sqrt(a[0]),void();    Sqrt(a,b,n>>1); fp(i,0,n<<1) F[i]=0;    Inv(b,F,n),init(n); fp(i,0,n-1) D[i]=a[i];    fp(i,n,limit-1) D[i]=0; NTT(D,1),NTT(b,1),NTT(F,1);    fp(i,0,limit-1) b[i]=mul(inc(b[i],mul(D[i],F[i])),inv[2]);    NTT(b,0); fp(i,n,limit-1) b[i]=0;    memset(D,0,limit<<2),memset(F,0,limit<<2);}inline void Direv(int* a,int* b,int n){    fp(i,1,n-1) b[i-1]=mul(a[i],i); b[n-1]=b[n]=0;}inline void Inter(int* a,int* b,int n){    fp(i,1,n-1) b[i]=mul(a[i-1],inv[i]); b[0]=0;}inline void Ln(int* a,int* b,int n){    static arr A,B; Direv(a,A,n),Inv(a,B,n);    init(n),NTT(A,1),NTT(B,1);    fp(i,0,limit-1) A[i]=mul(A[i],B[i]);    NTT(A,0),Inter(A,b,limit);    memset(A,0,limit<<2),memset(B,0,limit<<2);}inline void Exp(int* a,int* b,int n){    static arr F; if(n==1) return b[0]=1,void();    Exp(a,b,n>>1),Ln(b,F,n),init(n);    F[0]=dec(a[0]+1,F[0]); fp(i,1,n-1) F[i]=dec(a[i],F[i]);    NTT(F,1),NTT(b,1); fp(i,0,limit-1) b[i]=mul(b[i],F[i]);    NTT(b,0); fp(i,n,limit-1) b[i]=0; memset(F,0,limit<<2);}inline void Pow(int* a,int* b,int n,int k){    static arr F; memset(F,0,n<<2),Ln(a,F,n);    fp(i,0,n-1) F[i]=mul(F[i],k); Exp(F,b,n);}int main(){ prep(1<<19);    return 0;}