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 $A‘_i$ 。
1 //It is made by Awson on 2018.1.28 2 #include <set> 3 #include <map> 4 #include <cmath> 5 #include <ctime> 6 #include <queue> 7 #include <stack> 8 #include <cstdio> 9 #include <string> 10 #include <vector> 11 #include <complex> 12 #include <cstdlib> 13 #include <cstring> 14 #include <iostream> 15 #include <algorithm> 16 #define LL long long 17 #define dob complex<double> 18 #define Abs(a) ((a) < 0 ? (-(a)) : (a)) 19 #define Max(a, b) ((a) > (b) ? (a) : (b)) 20 #define Min(a, b) ((a) < (b) ? (a) : (b)) 21 #define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b)) 22 #define writeln(x) (write(x), putchar(‘\n‘)) 23 #define lowbit(x) ((x)&(-(x))) 24 using namespace std; 25 const int N = 100000*4; 26 const double pi = acos(-1.0); 27 28 int n, m, s, L, R[N+5]; 29 double q[N+5], sum, ans[N+5]; 30 dob a[N+5], b[N+5]; 31 32 void FFT(dob *A, int o) { 33 for (int i = 0; i < n; i++) if (i > R[i]) swap(A[i], A[R[i]]); 34 for (int i = 1; i < n; i <<= 1) { 35 dob wn(cos(pi/i), sin(pi*o/i)), x, y; 36 for (int j = 0; j < n; j += (i<<1)) { 37 dob w(1, 0); 38 for (int k = 0; k < i; k++, w *= wn) { 39 x = A[j+k], y = w*A[i+j+k]; 40 A[j+k] = x+y, A[i+j+k] = x-y; 41 } 42 } 43 } 44 } 45 void work() { 46 scanf("%d", &n); n--; s = n; 47 for (int i = 0; i <= n; i++) scanf("%lf", &q[i]), a[i] = q[i]; 48 for (int i = 1; i <= n; i++) b[i] = 1./i/i; 49 m = n<<1; for (n = 1; n <= m; n <<= 1) L++; 50 for (int i = 0; i < n; i++) R[i] = (R[i>>1]>>1)|((i&1)<<(L-1)); 51 FFT(a, 1), FFT(b, 1); 52 for (int i = 0; i <= n; i++) a[i] *= b[i]; 53 FFT(a, -1); 54 for (int i = 0; i <= s; i++) ans[i] = a[i].real()/n; 55 for (int i = 0; i <= n; i++) a[i] = 0; 56 for (int i = 0; i <= s; i++) a[i] = q[s-i]; 57 FFT(a, 1); 58 for (int i = 0; i <= n; i++) a[i] *= b[i]; 59 FFT(a, -1); 60 for (int i = 0; i <= s; i++) printf("%lf\n", ans[i]-a[s-i].real()/n); 61 } 62 int main() { 63 work(); 64 return 0; 65 }
