题目链接
\[\Huge{E_i=\sum_{j=1}^{i-1}\frac{q_j}{(i-j)^2}-\sum_{j=i+1}^{n}\frac{q_j}{(i-j)^2}}\]
设\(A[i]=q[i]\),\(B[i]=\frac{1}{i^2}\),\(A\times B\)就能得到第一个\(\sum\)，把\(A\)反过来再\(\times B\)就能得到第二个\(\sum\)，相减即可。
用\(FFT\)算。
#include
#include
#include
#define re register
using namespace std;
const int MAXN = 300010;
const double PI = M_PI;
struct complex{
double x, y;
complex(double xx = 0, double yy = 0){ x = xx; y = yy; }
}a[MAXN], b[MAXN];
inline complex operator + (complex a, complex b){
return complex(a.x + b.x, a.y + b.y);
}
inline complex operator - (complex a, complex b){
return complex(a.x - b.x, a.y - b.y);
}
inline complex operator * (complex a, complex b){
return complex(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x);
}
double q[MAXN], e[MAXN];
int r[MAXN], n, m;
void FFT(complex *f, int mode){
for(re int i = 0; i < m; ++i) if(i < r[i]) swap(f[i], f[r[i]]);
for(re int p = 2; p <= m; p <<= 1){
re int len = p >> 1;
re complex tmp(cos(PI / len), mode * sin(PI / len));
for(re int l = 0; l < m; l += p){
re complex w(1, 0);
for(re int k = l; k < l + len; ++k){
re complex t = w * f[len + k];
f[len + k] = f[k] - t;
f[k] = f[k] + t;
w = w * tmp;
}
}
}
}
int main(){
scanf("%d", &n);
for(re int i = 1; i <= n; ++i) scanf("%lf", &q[i]);
for(re int i = 1; i <= n; ++i) a[i].x = q[i], b[i].x = 1.0 / (1.0 * i * i);
for(m = 1; m <= (n << 1); m <<= 1);
for(re int i = 1; i < m; ++i) r[i] = r[i >> 1] >> 1 | ((i & 1) * (m >> 1));
FFT(a, 1); FFT(b, 1);
for(re int i = 0; i < m; ++i) a[i] = a[i] * b[i];
FFT(a, -1);
for(re int i = 1; i <= n; ++i) e[i] = a[i].x;
for(re int i = 0; i < m; ++i) a[i].x = b[i].x = a[i].y = b[i].y = 0;
for(re int i = 1; i <= n; ++i) a[i].x = q[n - i + 1], b[i].x = 1.0 / (1.0 * i * i);
FFT(a, 1); FFT(b, 1);
for(re int i = 0; i < m; ++i) a[i] = a[i] * b[i];
FFT(a, -1);
for(int i = 1; i <= n; ++i) printf("%.3lf\n", (e[i] - a[n - i + 1].x) / m);
return 0;
}