题面
题解
首先你得会多项式开根->这里
其次你得会解形如
\[x^2\equiv a\pmod{p}\]
的方程
这里有两种方法，一个是\(bsgs\)（这里），还有一种是\(Cipolla\)（施工）（不过这个只能用来解二次剩余就是了）
代码里留着的是\(bsgs\)，注释掉的是\(Cipolla\)
如果用\(Cipolla\)的话注意这里需要求的是较小的那个解
//minamoto
#include<bits/stdc++.h>
#include<tr1/unordered_map>
#define R register
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(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;
using namespace std::tr1;
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,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))(y&1)?res=mul(res,x):0;
return res;
}
unordered_map<int,int>mp;
int bsgs(int x){
int m=sqrt(P)+1;mp.clear();
for(R int i=0,res=x;i<m;++i,res=mul(res,3))mp[res]=i;
for(R int i=1,tmp=ksm(3,m),res=tmp;i<=m;++i,res=mul(res,tmp))
if(mp.count(res))return i*m-mp[res];
}
int exgcd(int a,int b,int &x,int &y){
if(!b)return x=1,y=0,a;
int d=exgcd(b,a%b,y,x);
return y-=a/b*x,d;
}
int Get(int p){
int c=bsgs(p),x,y,d=exgcd(2,P-1,x,y);
if(c%d)return -1;
int t=abs((P-1)/d);x=(1ll*x*c/d%t+t)%t;
return ksm(3,x);
}
//struct cp{
// int x,y;
// inline cp(R int xx,R int yy):x(xx),y(yy){}
//};
//inline cp mul(R cp a,R cp b,R int w){return cp(add(mul(a.x,b.x),mul(w,mul(a.y,b.y))),add(mul(a.x,b.y),mul(a.y,b.x)));}
//int ksm(R cp x,R int y,R int w){
// cp res(1,0);
// for(;y;y>>=1,x=mul(x,x,w))if(y&1)res=mul(res,x,w);
// return res.x;
//}
//int Get(int p){
// srand(time(0));
// if(ksm(p,(P-1)>>1)==P-1)return -1;
// while(true){
// int a=mul(rand(),rand()),w=dec(mul(a,a),p);
// if(ksm(w,(P-1)>>1)==P-1){
// int x=ksm(cp(a,1),(P+1)>>1,w);
// return min(x,P-x);
// }
// }
//}
int r[21][N],w[2][N],lg[N],inv[21];
void Pre(){
fp(d,1,18){
fp(i,0,(1<<d)-1)r[d][i]=(r[d][i>>1]>>1)|((i&1)<<(d-1));
lg[1<<d]=d,inv[d]=ksm(1<<d,P-2);
}
for(R int t=(P-1)>>1,i=1,x,y;i<262144;i<<=1,t>>=1){
x=ksm(3,t),y=ksm(332748118,t),w[0][i]=w[1][i]=1;
fp(k,1,i-1)
w[1][i+k]=mul(w[1][i+k-1],x),
w[0][i+k]=mul(w[0][i+k-1],y);
}
}
int lim,d;
void NTT(int *A,int ty){
fp(i,0,lim-1)if(i<r[d][i])swap(A[i],A[r[d][i]]);
for(R int mid=1;mid<lim;mid<<=1)
for(R int j=0,t;j<lim;j+=(mid<<1))
fp(k,0,mid-1)
A[j+k+mid]=dec(A[j+k],t=mul(w[ty][mid+k],A[j+k+mid])),
A[j+k]=add(A[j+k],t);
if(!ty)fp(i,0,lim-1)A[i]=mul(A[i],inv[d]);
}
void Inv(int *a,int *b,int len){
if(len==1)return b[0]=ksm(a[0],P-2),void();
Inv(a,b,len>>1);
static int A[N],B[N];lim=(len<<1),d=lg[lim];
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]=Get(a[0]),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);
lim=(len<<1),d=lg[lim];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);
n=read(),Pre();
fp(i,0,n-1)A[i]=read();
int len=1;while(len<n)len<<=1;
Sqrt(A,B,len);
fp(i,0,n-1)print(B[i]);
return Ot(),0;
}