[ZJOI 2014]力

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

Sample Output

-16838672.693
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 ;
 }