还是搞不明白生成函数是什么东西……

首先设对于体积为\(v\)的物品，它的生成函数为\(f(x)=\sum_{i\geq 0} x^{vi}\)，那么答案的生成函数就是所有的物品的生成函数的乘积，复杂度为\(O(nm\log n)\)

于是考虑把所有生成函数取\(\ln\)相加再\(\exp\)回去，设\(g(x)=\ln(f(x))\)，因为有\(f(x)=\frac{1}{1-x^v}\)

所以\[g'(x)=\frac{f'(x)}{f(x)}\]

\[g'(x)=(1-x^v)f'(x)\]

\[g'(x)=(1-x^v)\sum_{i\geq 0}vix^{vi-1}\]

\[g'(x)=\sum_{i\geq0}v(i-(i-1))x^{vi-1}\]

\[g'(x)=\sum_{i\geq0}vx^{vi-1}\]

然后积分\[g(x)=\sum_{i \geq 1} \frac{1}{i}x^{vi}\]

然后用多项式\(\exp\)求出\(f\)就可以了

//minamoto#include
    
     #define R register#define fp(i,a,b) for(R int i=a,I=b+1;i
     
      I;--i)#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)using namespace std;char buf[1<<21],*p1=buf,*p2=buf;inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}int read(){    R int res,f=1;R char ch;    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');    return res*f;}char sr[1<<21],z[20];int K=-1,Z=0;inline void Ot(){fwrite(sr,1,K+1,stdout),K=-1;}void print(R int x){    if(K>1<<20)Ot();if(x<0)sr[++K]='-',x=-x;    while(z[++Z]=x%10+48,x/=10);    while(sr[++K]=z[Z],--Z);sr[++K]='\n';}const int N=5e5+5,P=998244353,Gi=332748118;inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}int ksm(R int x,R int y){    R int res=1;    for(;y;y>>=1,x=mul(x,x))if(y&1)res=mul(res,x);    return res;}int r[N],O[N],A[N],B[N],C[N],D[N],F[N],G[N],H[N];int n,m,x,cnt[N];void NTT(int *A,int ty,int len){    int lim=1,l=0;while(lim
      
       <<=1,++l;    fp(i,0,lim-1)r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));    fp(i,0,lim-1)if(i
       
        >1);fp(i,0,len-1)A[i]=a[i],B[i]=b[i];    NTT(A,1,len<<1),NTT(B,1,len<<1);    fp(i,0,(len<<1)-1)A[i]=mul(A[i],mul(B[i],B[i]));    NTT(A,-1,len<<1);fp(i,0,len-1)b[i]=dec(mul(b[i],2),A[i]);    fp(i,0,(len<<1)-1)A[i]=B[i]=0;}void Direv(int *a,int *b,int len){    fp(i,1,len-1)b[i-1]=mul(a[i],i);b[len-1]=0;}void Inter(int *a,int *b,int len){    fp(i,1,len-1)b[i]=mul(a[i-1],ksm(i,P-2));b[0]=0;}void Ln(int *a,int *b,int len){    Direv(a,C,len),Inv(a,D,len);    NTT(C,1,len<<1),NTT(D,1,len<<1);    fp(i,0,(len<<1)-1)C[i]=mul(C[i],D[i]);    NTT(C,-1,len<<1),Inter(C,b,len);    fp(i,0,(len<<1)-1)C[i]=D[i]=0;}void Exp(int *a,int *b,int len){    if(len==1)return (void)(b[0]=1);    Exp(a,b,len>>1),Ln(b,F,len);    F[0]=dec(a[0]+1,F[0]);fp(i,1,len-1)F[i]=dec(a[i],F[i]);    NTT(F,1,len<<1),NTT(b,1,len<<1);    fp(i,0,(len<<1)-1)b[i]=mul(b[i],F[i]);    NTT(b,-1,len<<1);fp(i,len,(len<<1)-1)b[i]=F[i]=0;}int main(){//  freopen("testdata.in","r",stdin);    n=read(),m=read();    fp(i,1,n)x=read(),++cnt[x];    fp(i,1,m)if(cnt[i]){        for(R int j=i;j<=m;j+=i)        G[j]=add(G[j],mul(cnt[i],mul(i,ksm(j,P-2))));    }int len=1;while(len<=m)len<<=1;    Exp(G,H,len);fp(i,1,m)print(H[i]);    return Ot(),0;}