Description
给出n个数qi，给出Fj的定义如下：
$$F_j = \sum_{i<j}\frac{q_i q_j}{(i-j)^2 }-\sum_{i>j}\frac{q_i q_j}{(i-j)^2 }$$
令Ei=Fi/qi，求Ei.
Input
第一行一个整数n。
接下来n行每行输入一个数，第i行表示qi。
n≤100000，0<qi<1000000000
Output
n行，第i行输出Ei。与标准答案误差不超过1e-2即可。
Sample Input
5
4006373.885184
15375036.435759
1717456.469144
8514941.004912
1410681.345880
4006373.885184
15375036.435759
1717456.469144
8514941.004912
1410681.345880
Sample Output
-16838672.693
3439.793
7509018.566
4595686.886
10903040.872
3439.793
7509018.566
4595686.886
10903040.872
题解
约掉 $q_i$ $$E_j = \sum_{i<j}\frac{q_j}{(i-j)^2 }-\sum_{i>j}\frac{q_j}{(i-j)^2 }$$
我们拿出 $A_i=\sum\limits_{i<j}\frac{q_j}{(i-j)^2 }$ 讨论。
构造第一个多项式系数依次为 $q_i,i\in[0,n)$ ，第二个多项式系数 $\begin{cases}0 &i=0\\ \frac{1}{i^2} &i\in[1,n)\end{cases}$
卷积之后第 $i$ 项就是所求的 $A_i$ 。之后的类似，对于 $A'_i=\sum\limits_{i>j}\frac{q_j}{(i-j)^2 }$ 只要把第一个多项式翻转，卷积后第 $n-1-i$ 项就是所求的 $A'_i$ 。
//It is made by Awson on 2018.1.28
#include <set>
#include <map>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define dob complex<double>
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('\n'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int N = *;
const double pi = acos(-1.0); int n, m, s, L, R[N+];
double q[N+], sum, ans[N+];
dob a[N+], b[N+]; void FFT(dob *A, int o) {
for (int i = ; i < n; i++) if (i > R[i]) swap(A[i], A[R[i]]);
for (int i = ; i < n; i <<= ) {
dob wn(cos(pi/i), sin(pi*o/i)), x, y;
for (int j = ; j < n; j += (i<<)) {
dob w(, );
for (int k = ; k < i; k++, w *= wn) {
x = A[j+k], y = w*A[i+j+k];
A[j+k] = x+y, A[i+j+k] = x-y;
}
}
}
}
void work() {
scanf("%d", &n); n--; s = n;
for (int i = ; i <= n; i++) scanf("%lf", &q[i]), a[i] = q[i];
for (int i = ; i <= n; i++) b[i] = ./i/i;
m = n<<; for (n = ; n <= m; n <<= ) L++;
for (int i = ; i < n; i++) R[i] = (R[i>>]>>)|((i&)<<(L-));
FFT(a, ), FFT(b, );
for (int i = ; i <= n; i++) a[i] *= b[i];
FFT(a, -);
for (int i = ; i <= s; i++) ans[i] = a[i].real()/n;
for (int i = ; i <= n; i++) a[i] = ;
for (int i = ; i <= s; i++) a[i] = q[s-i];
FFT(a, );
for (int i = ; i <= n; i++) a[i] *= b[i];
FFT(a, -);
for (int i = ; i <= s; i++) printf("%lf\n", ans[i]-a[s-i].real()/n);
}
int main() {
work();
return ;
}