Luogu5205 【模板】多项式开根（NTT+多项式求逆）

　　https://www.cnblogs.com/HocRiser/p/8207295.html 安利！

　　写NTT把i<<=1写成了i<<=2，又调了一年。发现我的日常就是数组开小调调调，变量名写错调调调，反向判if调调调，退役吧。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define P 998244353
#define N 550000
char getc(){char c=getchar();while ((c<‘A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
	int x=0,f=1;char c=getchar();
	while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
	while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
int n,a[N],r[N],b[N],A[N],B[N],f[N],g[N],t;
int ksm(int a,int k)
{
	int s=1;
	for (;k;k>>=1,a=1ll*a*a%P) if (k&1) s=1ll*s*a%P;
	return s;
}
int inv(int a){return ksm(a,P-2);}
void DFT(int n,int *a,int g)
{
	for (int i=0;i<n;i++) r[i]=(r[i>>1]>>1)|(i&1)*(n>>1);
	for (int i=0;i<n;i++) if (i<r[i]) swap(a[i],a[r[i]]);
	for (int i=2;i<=n;i<<=1)
	{
		int wn=ksm(g,(P-1)/i);
		for (int j=0;j<n;j+=i)
		{
			int w=1;
			for (int k=j;k<j+(i>>1);k++,w=1ll*w*wn%P)
			{
				int x=a[k],y=1ll*w*a[k+(i>>1)]%P;
				a[k]=(x+y)%P,a[k+(i>>1)]=(x-y+P)%P;
			}
		}
	}
}
void IDFT(int *a,int n)
{
	DFT(n,a,inv(3));
	int u=inv(n);
	for (int i=0;i<n;i++) a[i]=1ll*a[i]*u%P;
}
void mul(int *a,int *b,int n)
{
	DFT(n,a,3),DFT(n,b,3);
	for (int i=0;i<n;i++) a[i]=1ll*a[i]*b[i]%P;
	IDFT(a,n);
}
void Inv(int *a,int *b,int n)
{
	if (n==1) {for (int i=0;i<t;i++) b[i]=0;b[0]=inv(a[0]);return;}
	Inv(a,b,n>>1);
	for (int i=0;i<n;i++) A[i]=a[i];
	for (int i=n;i<(n<<1);i++) A[i]=0;
	n<<=1;
	DFT(n,A,3),DFT(n,b,3);
	for (int i=0;i<n;i++) b[i]=1ll*b[i]*(P+2-1ll*A[i]*b[i]%P)%P;
	IDFT(b,n);
	n>>=1;
	for (int i=n;i<(n<<1);i++) b[i]=0;
}
void Sqrt(int *a,int *b,int n)
{
	if (n==1) {for (int i=0;i<t;i++) b[i]=0;b[0]=1;return;}
	Sqrt(a,b,n>>1);Inv(b,B,n);
	for (int i=0;i<n;i++) A[i]=a[i];
	for (int i=n;i<(n<<1);i++) A[i]=0;
	n<<=1;
	DFT(n,A,3),DFT(n,b,3),DFT(n,B,3);
	int inv2=inv(2);
	for (int i=0;i<n;i++) b[i]=(b[i]+1ll*A[i]*B[i]%P)*inv2%P;
	IDFT(b,n);
	n>>=1;
	for (int i=n;i<(n<<1);i++) b[i]=0;
}
int main()
{
#ifndef ONLINE_JUDGE
	freopen("sqrt.in","r",stdin);
	freopen("sqrt.out","w",stdout);
	const char LL[]="%I64d\n";
#else
	const char LL[]="%lld\n";
#endif
	n=read();
	for (int i=0;i<n;i++) a[i]=read();
	t=1;while (t<=n) t<<=1;
	Sqrt(a,b,t);
	for (int i=0;i<n;i++) printf("%d ",b[i]);
	return 0;
}

原文地址：https://www.cnblogs.com/Gloid/p/10386469.html