推公式发现（这不是水题吗，这要推吗）
[E_i=Sigma^{i-1}_{j=1} frac{q_j}{(i-j)^2} - Sigma^{n}_{j=i+1} frac{q_j}{(i-j)^2}
]
[设A[i] = q[i], B[i] = frac{1}{i^2}，FFT将A，B相乘可以得到E_i的前半部分
]
[后半部分就把数组反过来FFT就可以了
]
也可以合起来FFT（懒得想）
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(2e6 + 10);
const double Pi(acos(-1));
struct Complex{
double real, image;
IL Complex(){ real = image = 0; }
IL Complex(RG double a, RG double b){ real = a; image = b; }
IL Complex operator +(RG Complex B){ return Complex(real + B.real, image + B.image); }
IL Complex operator -(RG Complex B){ return Complex(real - B.real, image - B.image); }
IL Complex operator *(RG Complex B){ return Complex(real * B.real - image * B.image, real * B.image + image * B.real); }
} A[_], B[_];
int n, N, M, l, r[_];
double q[_], E[_];
IL void FFT(RG Complex *P, RG int opt){
for(RG int i = 0; i < N; ++i) if(i < r[i]) swap(P[i], P[r[i]]);
for(RG int i = 1; i < N; i <<= 1){
RG Complex W(cos(Pi / i), opt * sin(Pi / i));
for(RG int p = i << 1, j = 0; j < N; j += p){
RG Complex w(1, 0);
for(RG int k = 0; k < i; ++k, w = w * W){
RG Complex X = P[k + j], Y = w * P[k + j + i];
P[k + j] = X + Y; P[k + j + i] = X - Y;
}
}
}
}
int main(RG int argc, RG char *argv[]){
scanf("%d", &n);
for(RG int i = 1; i <= n; ++i) scanf("%lf", &q[i]);
for(RG int i = 1; i <= n; ++i) A[i].real = q[i], B[i].real = 1.0 / (1.0 * i * i);
for(N = 1, M = 2 * n; N <= M; N <<= 1) ++l;
for(RG int i = 0; i < N; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
FFT(A, 1); FFT(B, 1);
for(RG int i = 0; i < N; ++i) A[i] = A[i] * B[i];
FFT(A, -1);
for(RG int i = 1; i <= n; ++i) E[i] = A[i].real;
for(RG int i = 0; i < N; ++i) A[i].real = A[i].image = B[i].real = B[i].image = 0;
for(RG int i = n; i; --i) A[n - i + 1].real = q[i], B[i].real = 1.0 / (1.0 * i * i);
FFT(A, 1); FFT(B, 1);
for(RG int i = 0; i < N; ++i) A[i] = A[i] * B[i];
FFT(A, -1);
for(RG int i = 1; i <= n; ++i) E[i] -= A[n - i + 1].real;
for(RG int i = 1; i <= n; ++i) printf("%.3lf
", E[i] / N);
return 0;
}