[luogu4389]付公主的背包(多项式exp)

完全背包方案计数问题的FFT优化。
首先写成生成函数的形式：对重量为V的背包，它的生成函数为$\sum\limits_{i=0}^{+\infty}\frac{x^{Vi}}{i}=\frac{1}{1-x^{V}}$
于是答案就是$\prod \frac{1}{1-x^{V_k}}$。
直接做显然会超时，考虑使用ln将乘法变为加法。
https://www.cnblogs.com/cjyyb/p/10132855.html

 #include<cmath>
 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #define mem(a) memset(a,0,sizeof(a))
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 using namespace std;

 const int N=,mod=,inv2=(mod+)/;
 int n,m,v,cnt[N],rev[N],inv[N],X[N],Y[N],A[N],D[N],E[N],F[N];

 void Print(int a[],int n=::n){ for (int i=; i<n; i++) printf("%d ",a[i]); puts(""); }

 int ksm(int a,int b){
     int res=;
     for (; b; a=1ll*a*a%mod,b>>=)
         if (b & ) res=1ll*res*a%mod;
     return res;
 }

 void NTT(int a[],int n,bool f){
     for (int i=; i<n; i++) if (i<rev[i]) swap(a[i],a[rev[i]]);
     for (int i=; i<n; i<<=){
         int wn=ksm(,f ? (mod-)/(i<<) : (mod-)-(mod-)/(i<<));
         for (int p=i<<,j=; j<n; j+=p){
             int w=;
             for (int k=; k<i; k++,w=1ll*w*wn%mod){
                 int x=a[j+k],y=1ll*w*a[i+j+k]%mod;
                 a[j+k]=(x+y)%mod; a[i+j+k]=(x-y+mod)%mod;
             }
         }
     }
     if (f) return;
     int inv=ksm(n,mod-);
     for (int i=; i<n; i++) a[i]=1ll*a[i]*inv%mod;
 }

 void mul(int a[],int b[],int l){
     int n=,L=;
     for (; n<(l<<); n<<=) L++;
     for (int i=; i<n; i++) rev[i]=(rev[i>>]>>)|((i&)<<(L-));
     NTT(a,n,); NTT(b,n,);
     for (int i=; i<n; i++) a[i]=1ll*a[i]*b[i]%mod;
     NTT(a,n,); NTT(b,n,);
 }

 void Inv(int a[],int b[],int l){
     if (l==){ b[]=ksm(a[],mod-); return; }
     Inv(a,b,l>>); int n=,L=;
     for (; n<(l<<); n<<=) L++;
     for (int i=; i<n; i++) rev[i]=(rev[i>>]>>)|((i&)<<(L-));
     for (int i=; i<l; i++) A[i]=a[i];
     NTT(A,n,); NTT(b,n,);
     for (int i=; i<n; i++) b[i]=1ll*b[i]*(-1ll*A[i]*b[i]%mod+mod)%mod;
     NTT(b,n,);
     for (int i=l; i<n; i++) b[i]=;
     for (int i=; i<n; i++) A[i]=;
 }

 void Deri(int a[],int b[],int l){
     for (int i=; i<l; i++) b[i-]=1ll*i*a[i]%mod;
 }

 void Inte(int a[],int b[],int l){
     for (int i=; i<l; i++) b[i]=1ll*a[i-]*inv[i]%mod; b[]=;
 }

 void Ln(int a[],int b[],int l){
     Deri(a,D,l); Inv(a,E,l); mul(D,E,l); Inte(D,b,l);
     for (int i=; i<(l<<); i++) D[i]=E[i]=;
 }

 void Exp(int a[],int b[],int l){
     if (l==){ b[]=; return; }
     Exp(a,b,l>>); Ln(b,F,l); int n=,L=;
     for (; n<(l<<); n<<=) L++;
     for (int i=; i<n; i++) rev[i]=(rev[i>>]>>)|((i&)<<(L-));
     for (int i=; i<l; i++) F[i]=(-F[i]+a[i]+mod)%mod; F[]=(F[]+)%mod;
     NTT(F,n,); NTT(b,n,);
     for (int i=; i<n; i++) b[i]=1ll*b[i]*F[i]%mod;
     NTT(b,n,);
     for (int i=l; i<n; i++) b[i]=;
     for (int i=; i<n; i++) F[i]=;
 }

 int main(){
     freopen("4389.in","r",stdin);
     freopen("4389.out","w",stdout);
     scanf("%d%d",&n,&m);
     int l=; for (; l<=m; l<<=); inv[]=inv[]=;
     rep(i,,l) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
     rep(i,,n) scanf("%d",&v),cnt[v]++;
     rep(i,,m) if (cnt[i]) for (int j=; j*i<=m; j++) X[j*i]=(X[j*i]+1ll*cnt[i]*inv[j])%mod;
     Exp(X,Y,l);
     rep(i,,m) printf("%d\n",Y[i]);
     return ;
 }