[ZJOI2014]力题解

最新推荐文章于 2019-12-17 10:13:25 发布

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最新推荐文章于 2019-12-17 10:13:25 发布

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分类专栏： FFT 文章标签：数学卷积 FFT

本文链接：https://blog.youkuaiyun.com/qq_39677783/article/details/89285538

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本文介绍了一种利用快速傅立叶变换(FFT)解决特定数学问题的方法，该问题涉及计算序列的特殊求和形式。通过将序列转换和求和问题转化为FFT的卷积操作，提供了一个高效且新颖的解决方案。

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题意：给出 $n$ 个数 $q_i$ ，定义 $F_j=\sum\limits_{i<j}\frac{q_iq_j}{(i-j)^2}-\sum\limits_{i>j}\frac{q_iq_j}{(i-j)^2}$ ，令 $E_i=\frac{F_i}{q_i}$ ，求 $E_i$

$E_i=\sum\limits_{j=1}^{i-1}\frac{q_j}{(i-j)^2}-\sum\limits_{j=i+1}^n\frac{q_j}{(i-j)^2}$
设 $A_i=\sum\limits_{j=1}^{i-1}\frac{q_j}{(i-j)^2},B_i=\sum\limits_{j=i+1}^n\frac{q_j}{(i-j)^2}$ ，分开求
设 $f_i=\frac{1}{i^2}$ ，并规定 $f_0=0$ ，那么 $A_i=\sum\limits_{j+k=i}q_jf_k$ ，FFT求卷积即可
如果设 $p_i=q_{n-i}$ 即把 $q$ 翻转，那么
$B_i=\sum\limits_{j=1}^{n-i}\frac{q_{j+i}}{j^2}=\sum\limits_{j=1}^{n-i}\frac{p_{n-i-j}}{j^2}=\sum\limits_{j+k=n-i}p_jf_k$ ，FFT求卷积即可

#include <cstdio>
#include <cmath>
#include <algorithm>

const int maxn = 1e5 + 207;
const double pi = acos(-1.0);

struct Complex {
    double a, b;
    Complex(double x, double y) : a(x), b(y) {}
    Complex() : Complex(0.0, 0.0) {}
};
inline Complex operator+(const Complex &lhs, const Complex &rhs) {
    return Complex(lhs.a + rhs.a, lhs.b + rhs.b);
}
inline Complex operator-(const Complex &lhs, const Complex &rhs) {
    return Complex(lhs.a - rhs.a, lhs.b - rhs.b);
}
inline Complex operator*(const Complex &lhs, const Complex &rhs) {
    return Complex(lhs.a * rhs.a - lhs.b * rhs.b, lhs.a * rhs.b + lhs.b * rhs.a);
}

int r[maxn << 2], lim, l, n;
Complex a[maxn << 2], b[maxn << 2], c[maxn << 2];

inline void fft(Complex *A, int tp) {
    for (int i = 0; i < lim; ++i)
        if (i < r[i]) std::swap(A[i], A[r[i]]);
    for (int mid = 1; mid < lim; mid <<= 1) {
        Complex wn = Complex(cos(pi / mid), tp * sin(pi / mid));
        for (int j = 0; j < lim; j += mid << 1) {
            Complex w = Complex(1, 0);
            for (int k = 0; k < mid; ++k, w = w * wn) {
                Complex x = A[j + k], y = w * A[j + k + mid];
                A[j + k] = x + y;
                A[j + k + mid] = x - y;
            }
        }
    }
    if (tp == -1) {
        for (int i = 0; i < lim; ++i)
            A[i] = Complex(A[i].a / lim, 0);
    }
}

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%lf", &a[i].a);
        b[n - i].a = a[i].a;
    }
    for (int i = 1; i <= n; ++i)
        c[i].a = 1.0 / i / i;
    for (lim = 1; lim <= n << 1; lim <<= 1, ++l);
    for (int i = 0; i < lim; ++i)
        r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
    fft(a, 1); fft(b, 1); fft(c, 1);
    for (int i = 0; i <= lim; ++i)
        a[i] = a[i] * c[i], b[i] = b[i] * c[i];
    fft(a, -1); fft(b, -1);
    for (int i = 1; i <= n; ++i)
        printf("%.5lf\n", a[i].a - b[n - i].a);
    return 0;
}

[ZJOI2014]力 题解

[ZJOI2014]力题解