【刷题】BZOJ 3527 [Zjoi2014]力

Description

给出n个数qi，给出Fj的定义如下：

令Ei=Fi/qi，求Ei.

Input

第一行一个整数n。

接下来n行每行输入一个数，第i行表示qi。

n≤100000，0<qi<1000000000

Output

n行，第i行输出Ei。与标准答案误差不超过1e-2即可。

Sample Input

5

4006373.885184

15375036.435759

1717456.469144

8514941.004912

1410681.345880

Sample Output

-16838672.693

3439.793

7509018.566

4595686.886

10903040.872

Solution

推式子

\(\displaystyle E_i=\sum_{j<i}\frac{q_j}{(i-j)^2}-\sum_{j>i}\frac{q_j}{(i-j)^2}\)

令 \(g_i = \begin{cases}&\frac{1}{i^2}~~(i\neq0)\\&0 ~~~(i=0)\end{cases}\)

那么

\(\displaystyle E_i=\sum_{j=0}^{i-1}q_jg_{i-j}-\sum_{j=i+1}^{n}q_jg_{i-j}\)

\(\displaystyle ~~~~~=\sum_{j=0}^{i-1}q_jg_{i-j}-\sum_{j=0}^{n-i-1}p_jg_{n-i-j}~~~~(p_i=g_{n-i})\)

两个卷积形式，FFT就好了

#include<bits/stdc++.h>
#define ui unsigned int
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
const int MAXN=1<<19;
const db Pi=acos(-1.0);
int qn,n,m,cnt,rev[MAXN];
struct Complex{
	db real,imag;
	inline Complex operator + (const Complex &A) const {
		return (Complex){real+A.real,imag+A.imag};
	};
	inline Complex operator - (const Complex &A) const {
		return (Complex){real-A.real,imag-A.imag};
	};
	inline Complex operator * (const Complex &A) const {
		return (Complex){real*A.real-imag*A.imag,imag*A.real+real*A.imag};
	};
};
Complex q[MAXN],p[MAXN],g[MAXN];
template<typename T> inline void read(T &x)
{
	T data=0,w=1;
	char ch=0;
	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')w=-1,ch=getchar();
	while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
	x=data*w;
}
template<typename T> inline void write(T x,char ch='\0')
{
	if(x<0)putchar('-'),x=-x;
	if(x>9)write(x/10);
	putchar(x%10+'0');
	if(ch!='\0')putchar(ch);
}
template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
inline void FFT(Complex *A,int tp)
{
	for(register int i=0;i<n;++i)
		if(i<rev[i])std::swap(A[i],A[rev[i]]);
	for(register int l=2;l<=n;l<<=1)
	{
		Complex wn=(Complex){cos(2*Pi/l),sin(tp*2*Pi/l)};
		for(register int i=0;i<n;i+=l)
		{
			Complex w=(Complex){1,0};
			for(register int j=0;j<(l>>1);++j)
			{
				Complex A1=A[i+j],A2=A[i+j+(l>>1)]*w;
				A[i+j]=A1+A2;A[i+j+(l>>1)]=A1-A2;
				w=w*wn;
			}
		}
	}
}
int main()
{
	read(qn);
	m=qn+qn-1;
	for(register int i=0;i<qn;++i)
	{
		scanf("%lf",&q[i].real);
		p[qn-i].real=q[i].real;
		if(!i)g[i].real=0.0;
		else g[i].real=1.0/(1ll*i*i);
	}
	for(n=1;n<m;n<<=1)cnt++;
	for(register int i=0;i<n;++i)rev[i]=(rev[i>>1]>>1)|((i&1)<<(cnt-1));
	FFT(q,1);FFT(p,1);FFT(g,1);
	for(register int i=0;i<n;++i)q[i]=q[i]*g[i],p[i]=p[i]*g[i];
	FFT(q,-1);FFT(p,-1);
	for(register int i=0;i<qn;++i)printf("%f\n",(q[i].real-p[qn-i].real)/n);
	return 0;
}