首先我们知道\(\displaystyle E_j=\sum_{i<j}\frac{q_i}{(i-j)^2}-\sum_{i>j}\frac{q_i}{(i-j)^2}\),
设\(\displaystyle g[i]=\frac{1}{i^2}\),因为\(g\)是偶函数，所以\(\displaystyle E_j=\sum_{i=0}^{j-1} q_i g[j-i]-\sum_{i=j+1}^n q_ig[j-i]\)。
前面这东西很明显就是卷积，处理后面就要发挥人类智慧了，将\(q_i\) 翻转成新数组\(q_i'\)，原来的式子就成了
\(\displaystyle E_j=\sum_{i=0}^{j-1} q_i g[j-i]-\sum_{i=0}^{j-1} q_i'g[j-i]\),
后面这东西也变成卷积啦，用FFT处理一下就ok啦。
时间复杂度O(n log n).
#include<iostream>
#include<cstdio>
#include<cmath>
#define DB double
using namespace std;
int n,lim=1,tmp;
const DB PI=acos(-1);
const int N=400010;
struct lmaginary
{
DB x,y;
lmaginary(double X=0,double Y=0){x=X,y=Y;}
friend lmaginary operator +(const lmaginary &a,const lmaginary &b)
{return (lmaginary){a.x+b.x,a.y+b.y};}
friend lmaginary operator -(const lmaginary &a,const lmaginary &b)
{return (lmaginary){a.x-b.x,a.y-b.y};}
friend lmaginary operator *(const lmaginary &a,const lmaginary &b)
{return (lmaginary){a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x};}
}f1[N],f2[N],g[N];
DB q[N];
int r[N];
void FFT(lmaginary *A,int lim,int opt)
{
for(int i=0;i<lim;++i)
if(i<r[i])swap(A[i],A[r[i]]);
for(int mid=1;mid<lim;mid<<=1)//长度的一半
{
lmaginary wn(cos(PI/mid),opt*sin(PI/mid));
for(int len=mid<<1,j=0;j<lim;j+=len)
{
lmaginary w((DB)1,(DB)0);
for(int k=j;k<mid+j;++k,w=w*wn)
{
lmaginary a=A[k],b=w*A[k+mid];
A[k]=a+b;
A[k+mid]=a-b;
}
}
}
}
void YYCH()
{
for(int i=1;i<=n;++i)
{
g[i].x=(1.0/i/i);
f1[i].x=f2[n-i+1].x=q[i];
}
while(lim<2*n)lim<<=1,++tmp;
for(int i=0;i<lim;++i)
r[i]=(r[i>>1]>>1)|((i&1)<<(tmp-1));
}
int main()
{
cin>>n;
for(int i=1;i<=n;++i)scanf("%lf",&q[i]);
YYCH();
FFT(f1,lim,1);FFT(f2,lim,1);FFT(g,lim,1);
for(int i=0;i<lim;++i)
f1[i]=f1[i]*g[i],f2[i]=f2[i]*g[i];
FFT(f1,lim,-1);FFT(f2,lim,-1);
for(int i=1;i<=n;++i)
printf("%f\n",(f1[i].x-f2[n-i+1].x)/lim);
return 0;
}