快速傅里叶（FFT）+推式子
#include<bits/stdc++.h>
using namespace std;
const int N=4e6; //2^22=4194304
const double PI=acos(-1.0);
typedef complex<double> Complex;
double A[N], B[N], C[N], L[N], R[N]; // FFT加速卷积，输入：a[0..na], b[0..nb]，输出：c[0..na+nb]
void Mul(double *a, double *b, double *c, int na, int nb) {int len = 1;while (len < na + nb - 1) len <<= 1; // 补齐到 2 的幂static Complex A[1 << 22], B[1 << 22]; // 注意：临时数组大小需根据实际最大len调整for (int i = 0; i < na; i++) A[i] = Complex(a[i], 0);fill(A + na, A + len, Complex(0, 0));for (int i = 0; i < nb; i++) B[i] = Complex(b[i], 0);fill(B + nb, B + len, Complex(0, 0));vector<int> rev(len); // 位逆序置换数组for (int i = 0; i < len; i++)rev[i] = (rev[i >> 1] >> 1) | ((i & 1) ? (len >> 1) : 0);auto fft = [&](Complex *a, int type) { // FFT 变换（type=1 为正变换，-1 为逆变换）for (int i = 0; i < len; i++)if (i < rev[i]) swap(a[i], a[rev[i]]);for (int mid = 1; mid < len; mid <<= 1) {Complex wn(cos(PI / mid), type * sin(PI / mid)); // 单位根for (int i = 0; i < len; i += (mid << 1)) {Complex w(1, 0);for (int j = 0; j < mid; j++) {Complex x = a[i + j], y = w * a[i + j + mid];a[i + j] = x + y;a[i + j + mid] = x - y;w *= wn;}}}if (type == -1) {for (int i = 0; i < len; i++) a[i] /= len;}};fft(A, 1); fft(B, 1);for (int i = 0; i < len; i++) A[i] *= B[i];fft(A, -1);for (int i = 0; i < na + nb - 1; i++) c[i] = A[i].real(); // 取实部
}int main() {int n;cin >> n; for(int i=1; i <= n; i++) {cin >> A[i];C[n-i]=A[i];B[i]=(double)(1.0/i/i);}Mul(A, B, L, n+1, n+1);Mul(C, B, R, n+1, n+1);for(int i=1;i<=n;++i)printf("%.3lf\n",L[i]-R[n-i]);return 0;
}