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\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P3338 [ZJOI2014]力

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P3338 [ZJOI2014]力

推式子：
$$
\begin {aligned}
E _ j & = \frac {F _ i} {q _ i} \\
& = \frac {\sum _ {i = 1} ^ {j - 1} \frac {q _ i q _ j} {(i - j) ^ 2} - \sum _ {i = j + 1} ^ n \frac {q _ i q _ j} {(i - j) ^ 2}} {q _ i} \\
& = \sum _ {i = 1} ^ {j - 1} \frac {q _ i} {(i - j) ^ 2} - \sum _ {i = j + 1} ^ n \frac {q _ i} {(i - j) ^ 2}
\end {aligned}
$$
令 $j - i = k$ ，则
$$
E _ j = \sum _ {k = 1} ^ {j - 1} \frac {q _ {j - k}} {k ^ 2} - \sum _ {k = -1} ^ {j - n} \frac {q _ {j - k}} {k ^ 2}
$$
发现形式像卷积，令 $f(x) = q _ x$ ，$g(x) = \frac 1 {x ^ 2}$ ，则
$$
E _ j = \sum _ {k = 1} ^ {j - 1} f(j - k) g(k) - \sum _ {k = -1} ^ {j - n} f(j - k) g(k)
$$

将其转化为卷积形式，将两项分开考察：
$$
\begin {aligned}
& \sum _ {k = 1} ^ {j - 1} f(j - k) g(k) \\
= & \sum _ {k = 0} ^ j f(j - k) g(k) - f(0)g(j) - f(j)g(0) \\
= & F * G(j) - f(0)g(j) - f(j)g(0)
\end {aligned}
$$
通过 $f(0) = g(0) = 0$ ，直接卷积即可。

第二项的 $k$ 是负的，先把它变成正的，
$$
\begin {aligned}
& \sum _ {k = -1} ^ {j - n} f(j - k) g(k) \\
= & \sum _ {k = 1} ^ {n - j} f(j + k) g(k) \\
\end {aligned}
$$

$G(x) = \displaystyle \sum _ ig(i) x ^ i$ 将会贡献到 $i$ 上，不满足卷积了。我们希望构造一个不影响答案的另一个生成函数贡献到 $-i$ 上，且需要计算的项在正的幂上。 $\displaystyle G ^ {‘} (x) = \sum _ i g(i) x ^ {-i} = \sum _ i g(i - n - 1) x ^ {n - i + 1}$ ，实际上将答案贡献到了 $n - i + 1$ 上，所以取答案时增加 $n$ 。
$$
\begin {aligned}
& \sum _ {k = 1} ^ {n - j} f(j + k) g(k) \\
= & \sum _ {k = j + 1} ^ {n + j} f(k) g ^ {‘}(n + j - k + 1) \\
= & F * G ^ {‘} (n + j + 1) - \sum _ {k = 0} ^ j f(k) g ^ {‘}(n + j - k + 1) - \sum _ {k = n + 1} ^ {n + j} f(k) g ^ {‘}(n + j - k + 1)
\end {aligned}
$$
$f$ 和 $g$ 中大于 $n$ 的值为 $0$ 即可。

查看代码 
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#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const double PI = acos(-1);
const int N = 8e5 + 10;
struct Complex
{
    double x, y;
    Complex() {}
    Complex(double _x, double _y)
    {
        x = _x;
        y = _y;
    }
    Complex operator+(Complex k)
    {
        return Complex(x + k.x, y + k.y);
    }
    Complex operator-(Complex k)
    {
        return Complex(x - k.x, y - k.y);
    }
    Complex operator*(Complex k)
    {
        return Complex(x * k.x - y * k.y, x * k.y + y * k.x);
    }
} f[N], g[N];
double q[N];
int n, m, bit, tot, rev[N];
void fft(Complex *x, int op)
{
    for (int i = 0; i < tot; i++)
        if (rev[i] < i)
            swap(x[rev[i]], x[i]);
    for (int mid = 1; mid < tot; mid <<= 1)
    {
        Complex w1 = Complex(cos(PI / mid), sin(op * PI / mid));
        for (int i = 0; i < tot; i += (mid << 1))
        {
            Complex cur = Complex(1, 0);
            for (int j = 0; j < mid; j++, cur = cur * w1)
            {
                Complex p = x[i + j], q = cur * x[i + j + mid];
                x[i + j] = p + q, x[i + j + mid] = p - q;
            }
        }
    }
    if (op == -1)
        for (int i = 0; i < tot; i++)
            x[i].x = x[i].x / tot;
}
int main()
{
    scanf("%d", &n);
    while ((1 << bit) < (n + 1 << 1))
        bit++;
    tot = 1 << bit;
    for (int i = 1; i < tot; i++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1);
    for (int i = 1; i <= n; i++)
        scanf("%lf", &q[i]);
    static double res[N];
    for (int i = 0; i < tot; i++)
        res[i] = 0;
    for (int i = 0; i < tot; i++)
    {
        if (i > 0 && i <= n)
        {
            f[i].x = q[i];
            if (i < n)
                g[i].x = 1.0 / i / i;
        }
        else
            f[i].x = g[i].x = 0;
        f[i].y = g[i].y = 0;
    }
    fft(f, 1);
    fft(g, 1);
    for (int i = 0; i < tot; i++)
        f[i] = f[i] * g[i];
    fft(f, -1);
    for (int i = 1; i <= n; i++)
        res[i] += f[i].x;
    for (int i = 0; i < tot; i++)
    {
        if (i > 0 && i <= n)
        {
            f[i].x = q[i];
            g[i].x = 1.0 / (i - n - 1) / (i - n - 1);
        }
        else
            f[i].x = g[i].x = 0;
        f[i].y = g[i].y = 0;
    }
    fft(f, 1);
    fft(g, 1);
    for (int i = 0; i < tot; i++)
        f[i] = f[i] * g[i];
    fft(f, -1);
    for (int i = 1; i <= n; i++)
        res[i] -= f[i + n + 1].x;
    for (int i = 1; i <= n; i++)
        printf("%.3lf\n", res[i]);
    return 0;
}
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