luoguP4389 完全揹包計數
題意就是完全揹包計數，也就是求出\(\prod\frac{1}{1-x^{a_i}}\)
這個東西肯定不能直接做，我們考慮ln
\[\ln(\frac{1}{1-x^v}) =-\ln(1-x^v)\\=-\int\frac{(1-x^v)'}{1-x^v}\\=-\int vx^{v-1}\sum_{i=0} x^{iv}
\\=-\int \sum_{i=1} vx^{iv-1}\\=-\sum_{i=1} \frac{1}{iv} vx^{iv}\\=-\sum_{i=1} \frac{1}{i} x^{iv}
\]
最後歸歸類加起來最後exp就行了，複雜度是一個log
#include<bits/stdc++.h>
using namespace std;
int read(){
int x=0,pos=1;char ch=getchar();for(;!isdigit(ch);ch=getchar()) if(ch=='-') pos=0;
for(;isdigit(ch);ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';return pos?x:-x;
}
#define FOR(i,a,b) for(int i=a;i<=b;++i)
const int N = 400201;
#define ROF(i,a,b) for(int i=a;i>=b;--i)
int n;
const int mod = 998244353;
const int LSZ=21,SZ=1<<LSZ;
int ksm(int a,int b){
int res=1;
while(b){
if(b&1) res=1ll*res*a%mod;
a=1ll*a*a%mod,b>>=1;
}return res;
}
int tr[SZ],omg[SZ][2];
void init(int n){
int lim=0;while((1<<lim)<n) lim++;
FOR(i,0,n-1) tr[i]=tr[i>>1]>>1|((i&1)<<(lim-1));
omg[0][0]=omg[0][1]=1;
omg[1][0]=ksm(3,(mod-1)/n);
omg[1][1]=ksm(omg[1][0],mod-2);
FOR(i,2,n) omg[i][0]=1ll*omg[i-1][0]*omg[1][0]%mod,omg[i][1]=1ll*omg[i-1][1]*omg[1][1]%mod;
}
void dft(int *f,int n,int opt){
//assert((int)f.size()<=n);
for(int i=0;i<n;++i) if(i<tr[i]) swap(f[i],f[tr[i]]);
for(int l=2;l<=n;l<<=1){ int m=l/2;
for(int *g=f;g!=f+n;g+=l){
for(int i=0;i<m;i++){
int t=1ll*omg[n/l*i][opt]*(g[i+m])%mod;
g[i+m]=(g[i]-t+mod)%mod;g[i]=(g[i]+t)%mod;
}
}
}
if(opt) for(int i=0,iv=ksm(n,mod-2);i<n;i++) f[i]=1ll*f[i]*iv%mod;
}
void dao(int *a,int n){
FOR(i,1,n-1) a[i-1]=1ll*a[i]*i%mod;a[n-1]=0;
}
void jifen(int *a,int n){
ROF(i,n,1)a[i]=1ll*a[i-1]*ksm(i,mod-2)%mod;a[0]=0;
}
int MA[N],MB[N];
void mul(const int *a,const int *b,int *ans,int n){
FOR(i,0,n-1) MA[i]=a[i],MB[i]=b[i];
int len=1;while(len<n) len*=2;init(len);
dft(MA,len,0),dft(MB,len,0);
FOR(i,0,len-1) ans[i]=1ll*MA[i]*MB[i]%mod;
dft(ans,len,1);
FOR(i,0,len-1) MA[i]=MB[i]=0;
}
int FV[N];
inline void getinv(const int *a,int *b,int n){
if(n==1) return b[0]=ksm(a[0],mod-2),void();
getinv(a,b,(n+1)/2);
int len=1;while(len<n*2) len*=2;init(len);
FOR(i,0,n-1) FV[i]=a[i];FOR(i,n,len-1) FV[i]=0;
dft(FV,len,0);dft(b,len,0);
FOR(i,0,len-1) b[i]=(2ll-1ll*FV[i]*b[i]%mod+mod)%mod*b[i]%mod;
dft(b,len,1);FOR(i,n,len) b[i]=0;
}
int DLN[N];
void out(int *ans){
FOR(i,0,n) printf("%d ",ans[i]);putchar(10);
}
inline void getln(const int *a,int *ans,int n){
getinv(a,ans,n);
FOR(i,0,n-1) DLN[i]=a[i];dao(DLN,n);
mul(ans,DLN,ans,2*n);
jifen(ans,n);ans[n]=0;
}
int ET[N];
inline void getexp(const int *a,int *ans,int n){
if(n==1) return ans[0]=1,void();
getexp(a,ans,(n+1)/2);
FOR(i,0,n-1) ET[i]=0;getln(ans,ET,n);
ET[0]=(1ll+a[0]-ET[0]+mod)%mod;
FOR(i,1,n-1) ET[i]=(0ll+a[i]-ET[i]+mod)%mod;
mul(ans,ET,ans,n*2);FOR(i,n,n*2-1) ans[i]=0;
}
int ST[N];
inline void getsqrt(const int *a,int *ans,int n){
if(n==1) return ans[0]=1,void();
getsqrt(a,ans,(n+1)/2);
FOR(i,0,n-1) ST[i]=0;getinv(ans,ST,n);
mul(ST,a,ST,n*2);int i2=ksm(2,mod-2);
FOR(i,0,n-1) ans[i]=(1ll*i2*((ans[i]+ST[i])%mod))%mod;
}
int a[N],ans[N],v[N];
int f[N],g[N],inv[N],m;
void init1(){
f[0]=1;FOR(i,1,N-1) f[i]=1ll*f[i-1]*i%mod;
g[N-1]=ksm(f[N-1],mod-2);ROF(i,N-2,0) g[i]=1ll*g[i+1]*(i+1)%mod;
FOR(i,1,N-1) inv[i]=1ll*g[i]*f[i-1]%mod;
}
int main(){
n=read();m=read();m++;
init1();
for(int _=0;_<n;_++){
int x=read();v[x]++;
}
FOR(x,1,m){
if(!v[x]) continue;
for(int i=1;i*x<m;i++){
a[i*x]=(a[i*x]+1ll*inv[i]*v[x]%mod)%mod;
}
}
getexp(a,ans,m);
FOR(i,1,m-1){
printf("%d\n",ans[i]);
}
return 0;
}