题意

给出\(n\)个数\(q_i\),给出\(Fj\)的定义如下：

令\(E_i=F_i/q_i\)，求\(E_i\).

题解

一开始没发现求\(E_i\)... 其实题目还更容易想了...

这个东西就是转换成求两个一样的东西就行了。

就是求$$sum_i=\sum \limits_{j<i} \frac{q_j}{(i-j)^2}$$.

这个就是可以转换成求一个卷积形式就行了。

注意多项式乘法格式是这样的：

令\(A\)与\(B\)的卷积为\(C\),则$$C_i=\sum \limits {j \le i}A_j*B{i-j}$$.

发现\(i-j\)那个形式似乎就可以满足本题的形式。

所以令\(B_i=\frac{1}{i^2}\)就行了，然后\(A_i=q_i\).

对于这个求两边卷积就行了23333

注意有的细节要处理一下，就是要清空一些数组，

注意一下下标（思维要清楚），而且也要令\(A_0=B_0=0\)。

而且之前求\(B_i\)的时候，\(i^2\)会爆long long ！

代码

/**************************************************************
    Problem: 3527
    User: zjp_shadow
    Language: C++
    Result: Accepted
    Time:3688 ms
    Memory:32012 kb
****************************************************************/
#include <bits/stdc++.h>
#define For(i, l, r) for(register int i = (l), _end_ = (int)(r); i <= _end_; ++i)
#define Fordown(i, r, l) for(register int i = (r), _end_ = (int)(l); i >= _end_; --i)
#define Set(a, v) memset(a, v, sizeof(a))
using namespace std;
 
bool chkmin(int &a, int b) {return b < a ? a = b, 1 : 0;}
bool chkmax(int &a, int b) {return b > a ? a = b, 1 : 0;}
 
inline int read() {
    int x = 0, fh = 1; char ch = getchar();
    for (; !isdigit(ch); ch = getchar() ) if (ch == '-') fh = -1;
    for (; isdigit(ch); ch = getchar() ) x = (x<<1) + (x<<3) + (ch ^ '0');
    return x * fh;
}
 
void File() {
#ifdef zjp_shadow
    freopen ("P3527.in", "r", stdin);
    freopen ("P3527.out", "w", stdout);
#endif
}
 
struct Complex {
    double re, im;
};
 
inline Complex operator + (const Complex &lhs, const Complex &rhs) {
    return (Complex) {lhs.re + rhs.re, lhs.im + rhs.im};
}
 
inline Complex operator - (const Complex &lhs, const Complex &rhs) {
    return (Complex) {lhs.re - rhs.re, lhs.im - rhs.im};
}
 
inline Complex operator * (const Complex &lhs, const Complex &rhs) {
    return (Complex) {lhs.re * rhs.re - lhs.im * rhs.im, lhs.re * rhs.im + rhs.re * lhs.im};
}
 
const int N = 1 << 19;
int n_, n;
double f[N], g[N];
const double Pi = acos(-1.0);
 
int r[N];
 
void FFT(Complex P[], int opt) {
    For (i, 0, n - 1) if (i < r[i]) swap(P[i], P[r[i]]);
    for (int i = 2; i <= n; i <<= 1) {
        Complex Wi = (Complex) {cos(2 * Pi / i), opt * sin(2 * Pi / i)};
        int p = i / 2;
        for (int j = 0; j < n; j += i) {
            Complex x = (Complex) {1.0, 0.0};
            For (k, 0, p - 1) {
                Complex u = P[j + k], v = x * P[j + k + p];
                P[j + k] = u + v;
                P[j + k + p] = u - v;
                x = x * Wi;
            }
        }
    }
}
 
int m;
void Mult(Complex a[], Complex b[]) {
    int cnt = 0;
    for (n = 1; n <= m; n <<= 1) ++ cnt;
    For (i, 1, n - 1)
        r[i] = (r[i >> 1] >> 1) | ((i & 1) << (cnt - 1) );
    FFT(a, 1); FFT(b, 1);
    For (i, 0, n - 1) a[i] = a[i] * b[i];
    FFT(a, -1);
    For (i, 0, n - 1) a[i].re = a[i].re / n;
}
 
double ans[N];
Complex a[N], b[N];
 
int main () {
    //int n1 = read(), n2 = read(),
    File();
    n_ = read();
    m = n_ + n_;
    For (i, 1, n_) {
        scanf("%lf", &f[i]);
        g[i] = (double)1.0 / ((long long)i * (long long)i);
    }
    For (i, 0, n_) a[i].re = f[i], a[i].im = 0;
    For (i, 0, n_) b[i].re = g[i], b[i].im = 0; 
    Mult(a, b);
    For (i, 1, n_)
        ans[i] += a[i].re;
 
    reverse(f + 1, f + 1 + n_);
    For (i, 0, n - 1) a[i].re = f[i], a[i].im = 0;
    For (i, 0, n - 1) b[i].re = g[i], b[i].im = 0;
 
    Mult(a, b);
 
    For (i, 1, n_)
        ans[n_ - i + 1] -= a[i].re;
 
    For (i, 1, n_)
        printf ("%.4lf\n", ans[i]);
    return 0;
}