[ZJOI2014]力

推公式发现（这不是水题吗，这要推吗）

\[E_i=\Sigma^{i-1}_{j=1} \frac{q_j}{(i-j)^2} - \Sigma^{n}_{j=i+1} \frac{q_j}{(i-j)^2}\]
\[设A[i] = q[i], B[i] = \frac{1}{(i-j)^2}，FFT将A，B相乘可以得到E_i的前半部分\]
\[后半部分就把数组反过来FFT就可以了\]

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\n", E[i] / N);
    return 0;
}