题面

题解

考虑分治

假设我们已经求出\(A'^2\equiv B\pmod{x^n}\)，考虑如何计算出\(A^2\equiv B\pmod{x^{2n}}\)

首先肯定存在\(A^2\equiv B\pmod{x^n}\)

然后两式相减\[A'^2-A^2\equiv 0\pmod{x^n}\]

\[(A'-A)(A'+A)\equiv 0\pmod{x^n}\]

我们假设\(A'-A\equiv 0\pmod{x^n}\)，然后两边平方\[A'^2-2A'A+A^2\equiv 0\pmod{x^{2n}}\]

（关于平方之后模数变化的原因可以看我多项式求逆那篇文章，里面有写）

又因为\(A^2\equiv B\pmod{x^{2n}}\)，代入得\[A'^2-2A'A+B\equiv 0\pmod{x^{2n}}\]

\[A\equiv\frac{A'^2+B}{2A'}\pmod{x^{2n}}\]

//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 C=-1,Z=0;inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}void print(R int x){    if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;    while(z[++Z]=x%10+48,x/=10);    while(sr[++C]=z[Z],--Z);sr[++C]=' ';}const int N=(1<<18)+5,P=998244353,Gi=332748118,inv2=499122177;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[19][N],rt[2][N<<1],lim,d;void Pre(){    fp(d,1,18)fp(i,1,(1<
      
       >1]>>1)|((i&1)<<(d-1));    for(R int t=(P-1)>>1,i=1,x,y;i<=262144;i<<=1,t>>=1){        x=ksm(3,t),y=ksm(Gi,t),rt[0][i]=rt[1][i]=1;        fp(k,1,i-1)            rt[1][i+k]=mul(rt[1][i+k-1],x),            rt[0][i+k]=mul(rt[0][i+k-1],y);    }}inline void init(R int len){lim=1,d=0;while(lim
       
        <<=1,++d;}void NTT(int *A,int ty){    fp(i,0,lim-1)if(i
        
         >1);    static int A[N],B[N];init(len<<1);    fp(i,0,len-1)A[i]=a[i],B[i]=b[i];    fp(i,len,lim-1)A[i]=B[i]=0;    NTT(A,1),NTT(B,1);    fp(i,0,lim-1)A[i]=mul(A[i],mul(B[i],B[i]));    NTT(A,0);    fp(i,0,len-1)b[i]=dec(add(b[i],b[i]),A[i]);    fp(i,len,lim-1)b[i]=0;}void Sqrt(int *a,int *b,int len){    if(len==1)return b[0]=1,void();    Sqrt(a,b,len>>1);    static int A[N],B[N];    fp(i,0,len-1)A[i]=a[i];Inv(b,B,len);    init(len<<1);fp(i,len,lim-1)A[i]=B[i]=0;    NTT(A,1),NTT(B,1);    fp(i,0,lim-1)A[i]=mul(A[i],B[i]);    NTT(A,0);    fp(i,0,len-1)b[i]=mul(add(b[i],A[i]),inv2);    fp(i,len,lim-1)b[i]=0;}int A[N],B[N],n;int main(){//  freopen("testdata.in","r",stdin);    Pre();    n=read();    fp(i,0,n-1)A[i]=read();    int len=1;while(len
         
          <<=1; Sqrt(A,B,len); fp(i,0,n-1)print(B[i]); return Ot(),0;}