[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$sumlimits_{i=0}^{+infty}frac{x^{Vi}}{i}=frac{1}{1-x^{V}}$
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=500010,mod=998244353,inv2=(mod+1)/2;
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=0; i<n; i++) printf("%d ",a[i]); puts(""); }

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

void NTT(int a[],int n,bool f){
    for (int i=0; i<n; i++) if (i<rev[i]) swap(a[i],a[rev[i]]);
    for (int i=1; i<n; i<<=1){
        int wn=ksm(3,f ? (mod-1)/(i<<1) : (mod-1)-(mod-1)/(i<<1));
        for (int p=i<<1,j=0; j<n; j+=p){
            int w=1;
            for (int k=0; 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-2);
    for (int i=0; i<n; i++) a[i]=1ll*a[i]*inv%mod;
}

void mul(int a[],int b[],int l){
    int n=1,L=0;
    for (; n<(l<<1); n<<=1) L++;
    for (int i=0; i<n; i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1));
    NTT(a,n,1); NTT(b,n,1);
    for (int i=0; i<n; i++) a[i]=1ll*a[i]*b[i]%mod;
    NTT(a,n,0); NTT(b,n,0);
}

void Inv(int a[],int b[],int l){
    if (l==1){ b[0]=ksm(a[0],mod-2); return; }
    Inv(a,b,l>>1); int n=1,L=0;
    for (; n<(l<<1); n<<=1) L++;
    for (int i=0; i<n; i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1));
    for (int i=0; i<l; i++) A[i]=a[i];
    NTT(A,n,1); NTT(b,n,1);
    for (int i=0; i<n; i++) b[i]=1ll*b[i]*(2-1ll*A[i]*b[i]%mod+mod)%mod;
    NTT(b,n,0);
    for (int i=l; i<n; i++) b[i]=0;
    for (int i=0; i<n; i++) A[i]=0;
}

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

void Inte(int a[],int b[],int l){
    for (int i=1; i<l; i++) b[i]=1ll*a[i-1]*inv[i]%mod; b[0]=0;
}

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=0; i<(l<<1); i++) D[i]=E[i]=0;
}

void Exp(int a[],int b[],int l){
    if (l==1){ b[0]=1; return; }
    Exp(a,b,l>>1); Ln(b,F,l); int n=1,L=0;
    for (; n<(l<<1); n<<=1) L++;
    for (int i=0; i<n; i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1));
    for (int i=0; i<l; i++) F[i]=(-F[i]+a[i]+mod)%mod; F[0]=(F[0]+1)%mod;
    NTT(F,n,1); NTT(b,n,1);
    for (int i=0; i<n; i++) b[i]=1ll*b[i]*F[i]%mod;
    NTT(b,n,0);
    for (int i=l; i<n; i++) b[i]=0;
    for (int i=0; i<n; i++) F[i]=0;
}

int main(){
    freopen("4389.in","r",stdin);
    freopen("4389.out","w",stdout);
    scanf("%d%d",&n,&m);
    int l=1; for (; l<=m; l<<=1); inv[0]=inv[1]=1;
    rep(i,2,l) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
    rep(i,1,n) scanf("%d",&v),cnt[v]++;
    rep(i,1,m) if (cnt[i]) for (int j=1; j*i<=m; j++) X[j*i]=(X[j*i]+1ll*cnt[i]*inv[j])%mod;
    Exp(X,Y,l);
    rep(i,1,m) printf("%d
",Y[i]);
    return 0;
}