洛谷P5205 【模板】多项式开根（FFT）

题面

传送门

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<bits/stdc++.h>
#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;
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<<d)-1)r[d][i]=(r[d][i>>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<len)lim<<=1,++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(rt[ty][mid+k],A[j+k+mid])),
                A[j+k]=add(A[j+k],t);
    if(!ty)for(R int i=0,inv=ksm(lim,P-2);i<lim;++i)A[i]=mul(A[i],inv);
}
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];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<n)len<<=1;
    Sqrt(A,B,len);
    fp(i,0,n-1)print(B[i]);
    return Ot(),0;
}

转载于:https://www.cnblogs.com/bztMinamoto/p/10567954.html