# 洛谷 [P3338] 力

## FFT

$E_i = F_i / q_i = \sum_{i<j} \frac {q_j} {(i - j)^2} - \sum _{ i > j} \frac{q _ j} {(i - j)^2}$

$p _ i = q_{n - i}$ $g(i) = \frac 1 {i^2}$

$E_i = \sum_{j=1}^{i-1} q _ i g(j - i) - \sum _ {j = i + 1} ^ n q_j g(i - j)$

$E_i = \sum_{j=1}^{i-1} q_i g(i-1) - \sum_{j=1}^{n - i} p_i g(i - j)$

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int MAXN = 400005;
const double PI = acos(-1);
struct Complex {
double x, y;
Complex(double xx = 0.0, double yy = 0.0) {
x = xx; y = yy;
}
Complex operator + (const Complex &u) const {
return Complex(x + u.x, y + u.y);
}
Complex operator - (const Complex &u) const {
return Complex(x - u.x, y - u.y);
}
Complex operator * (const Complex &u) const{
return Complex(x * u.x - y * u.y, x * u.y + y * u.x);
}
}a[MAXN], b[MAXN], c[MAXN], buf[MAXN];
int n, lim = 1, limcnt, rev[MAXN];
void fft(Complex a[], int opt) {
for(int i = 0; i <= lim; i++) {
if(i < rev[i]) swap(a[i], a[rev[i]]);
}
for(int mid = 1; mid < lim; mid <<= 1) {
Complex wn = Complex(cos(PI / mid), opt * sin(PI / mid));
for(int R = mid << 1, j = 0; j < lim; j += R) {
Complex w = Complex(1.0, 0.0);
for(int k = 0; k < mid; k++) {
Complex x = a[j + k], y = w * a[j + mid + k];
a[j + k] = x + y;
a[j + mid + k] = x - y;
w = w * wn;
}
}
}
if(opt == -1) {
for(int i = 0; i <= lim; i++) a[i].x = a[i].x / lim;
}
}
int main() {
cin >> n;
for(int i = 1; i <= n; i++) {
scanf("%lf", &a[i].x);
b[n - i].x = a[i].x;
c[i].x = (1.0 / (double)i) / (double)i;
}
while(lim <= (n << 1)) lim <<= 1, limcnt++;
for(int i = 0; i <= lim; i++) {
rev[i] = (rev[i>>1]>>1) | ((i&1)<<(limcnt - 1));
}
fft(a, 1);fft(b, 1); fft(c, 1);
for(int i = 0; i <= lim; i++) a[i] = a[i] * c[i];
for(int i = 0; i <= lim; i++) b[i] = b[i] * c[i];
fft(a, -1); fft(b, -1);
for(int i = 1; i <= n; i++) {
printf("%.4f\n", a[i].x - b[n - i].x);
}
return 0;
}`
posted @ 2018-05-24 08:43  Mr_Wolfram  阅读(115)  评论(1编辑  收藏