BZOJ 3625 多项式求逆+多项式开根

Posted 2020-09-21 SiriusRen

思路：

RT

//By SiriusRen
#include <bits/stdc++.h>
using namespace std;
const int N=1<<18,mod=998244353;
int A[N],C[N],invC[N],c[N],d[N],R[N],tmp[N],xx,len,sqrA[N],F[N];
typedef long long ll;
int power(ll x,int y){
    ll res=1;
    while(y){
        if(y&1)res=res*x%mod;
        x=x*x%mod,y>>=1;
    }return res;
}
void NTT(int *a,int n,int f){
    int m=1,L=0;
    for(;m<n;m<<=1)L++;
    for(int i=0;i<n;i++)R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));
    for(int i=0;i<n;i++)if(i<R[i])swap(a[i],a[R[i]]);
    for(int l=1;l<n;l<<=1){
        int wn=power(5,((mod-1)/(l<<1)*f+mod-1)%(mod-1));
        for(int j=0;j<n;j+=(l<<1)){
            int w=1;
            for(int k=0;k<l;k++,w=1ll*w*wn%mod){
                int x=a[j+k],y=1ll*a[j+k+l]*w%mod;
                a[j+k]=(x+y)%mod,a[j+k+l]=(x-y+mod)%mod; 
            }
        }
    }
    if(f==-1){
        int ni=power(n,mod-2);
        for(int i=0;i<n;i++)a[i]=1ll*a[i]*ni%mod;
    }
}
void get_inv(int *a,int *b,int n){
    if(n==1){b[0]=power(a[0],mod-2);return;}
    get_inv(a,b,n>>1);
    memcpy(tmp,a,sizeof(int)*n),memset(tmp+n,0,sizeof(int)*n);
    NTT(tmp,n<<1,1),NTT(b,n<<1,1);
    for(int i=0;i<n<<1;i++)b[i]=(1ll*b[i]*(2-1ll*b[i]*tmp[i]%mod)%mod+mod)%mod;
    NTT(b,n<<1,-1),memset(b+n,0,sizeof(int)*n);
}
void get_root(int *a,int *b,int n){
    if(n==1){b[0]=1;return;}
    get_root(a,b,n>>1),memset(d,0,sizeof(int)*2*n);
    get_inv(b,d,n);
    memcpy(c,a,sizeof(int)*n),memset(c+n,0,sizeof(int)*n);
    NTT(c,n<<1,1),NTT(b,n<<1,1),NTT(d,n<<1,1);
    for(int i=0;i<n<<1;i++)b[i]=(1ll*c[i]*d[i]%mod+b[i])%mod*499122177%mod;
    NTT(b,n<<1,-1),memset(b+n,0,sizeof(int)*n);
}
signed main(){
    int n,m;
    scanf("%d%d",&n,&m);
    for(len=1;len<=m;len<<=1);A[0]=1;
    for(int i=1;i<=n;i++)scanf("%d",&xx),C[xx]=1,A[xx]=mod-4;
    get_root(A,sqrA,len),sqrA[0]++,get_inv(sqrA,F,len);
    for(int i=1;i<=m;i++)printf("%d\n",(F[i]<<1)%mod);
}