洛谷 [P3338] 力

FFT

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

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

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

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

我们发现这是一个卷积的形式, 可以用 FFT 处理

cpp #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(...)  评论(...编辑  收藏