P3338 [ZJOI2014]力 (FFT)

题意:$$F(i)= \sum_{j=0} ^{i - 1} \frac{qi * qj}{(j - i)^{2}} - \sum_{j = i + 1}^{n} \frac{qi * qj}{(j - i)^{2}}$$
求Fi/qi
题解:可以分开来算 前面一坨 后面一坨

\[f(i) = qi \]

\[g(i)=\frac{1}{i^2} \]

\[F(i) = \sum_{j = 0}^{i - 1} f(i) * g(i - j) - \sum_{j = i + 1}^{n} f(i) * g(j - i) \]

前面已经可以算了 然后后面的g(j - i) = g(i - j) 套路的把f翻转一下

\[\sum_{j = i + 1}^{n} f(n - j + 1) * g(j - i) = F(n + 1 - i) \]

\[然后就都可以卷了! \]

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);

struct Complex {
    double x, y;
    Complex(double _x = 0.0, double _y = 0.0) {
        x = _x;
        y = _y;
    }
    Complex operator + (const Complex &b) const {
        return Complex(x + b.x, y + b.y);
    }
    Complex operator - (const Complex &b) const {
        return Complex(x - b.x, y - b.y);
    }
    Complex operator * (const Complex &b) const {
        return Complex(x * b.x - y * b.y, x * b.y + y * b.x);
    }
};

void change(Complex y[], int len) {
    int i, j, k;
    for(i = 1, j = len / 2; i < len - 1; i++) {
        if(i < j) swap(y[i], y[j]);
        k = len / 2;
        while(j >= k) {
            j -= k;
            k /= 2;
        }
        if(j < k) j += k;
    }
}

void fft(Complex y[], int len, int on) {
    change(y, len);
    for(int h = 2; h <= len; h <<= 1) {
        Complex wn(cos(-on * 2 * PI / h), sin(-on * 2 * PI / h));
        for(int j = 0; j < len; j += h) {
            Complex w(1, 0);
            for(int k = j; k < j + h / 2; k++) {
                Complex u = y[k];
                Complex t = w * y[k + h / 2];
                y[k] = u + t;
                y[k + h / 2] = u - t;
                w = w * wn;
            }
        }
    }

    if(on == -1)
        for(int i = 0; i < len; i++)
            y[i].x /= len;
}

Complex x1[400005], x2[400005], x3[400005];
int main() {
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        double x;
        scanf("%lf", &x);
        x1[i] = Complex(x, 0);
        x2[n - i + 1] = Complex(x, 0);
        x3[i] = Complex(1.0 / (1.0 * i * i), 0);
    }
    int len = 1;
    while(len <= n + n) len <<= 1;
    fft(x1, len, 1);
    fft(x2, len, 1);
    fft(x3, len, 1);
    for(int i = 0; i <= len; i++) x1[i] = x1[i] * x3[i];
    for(int i = 0; i <= len; i++) x2[i] = x2[i] * x3[i];
    fft(x1, len, -1);
    fft(x2, len, -1);
    for(int i = 1; i <= n; i++) {
        printf("%.3lf\n", x1[i].x - x2[n - i + 1].x);
    }
    return 0;
}