[luogu P3338] [ZJOI2014]力

题目大意

$给出n个数qi,给出F_j定义如下:$
$F_i={\sum }_{i<j}\frac{q_i{q}_j}{(i-j{)}^2}-{\sum }_{i>j}\frac{q_i{q}_j}{(i-j{)}^2}$
$令E_i=\frac{F_i}{q_i}求E_i$

题解

先把式子化简一下
$E_i={\sum }_{i<j}\frac{q_j}{(i-j{)}^2}-{\sum }_{i>j}\frac{q_j}{(i-j{)}^2}$
$令F_i=q_i{g}_i=\frac{1}{i^2}$
$E_i={\sum }_{i<j}{f}_i{g}_{j-i}-{\sum }_{i>j}{f}_i{g}_{i-j}$
$=\sum _{i=0}^{j-1}{f}_i{g}_{j-i}-\sum _{i=j+1}^n{f}_i{g}_{i-j}$
左边显然是卷积的形式，现在来看右边
令$f_i^{\prime }=f_{n-i+1}$
$\sum _{i=j+1}^n{f}_i{g}_{i-j}=\sum _{i=0}^{n-j}{f}_i^{\prime }{g}_{(n-j)-i}$
$设t=n-j$
$\sum _{i=0}^t{f}_i^{\prime }{g}_{t-i}$
显然也是卷积的形式然后直接上法法踢 $fft$就行了
代码：

#include
#define N 450005
using namespace std;
const double pi = acos(-1.0);
struct complexx{
	double x, y;
	complexx(double xx = 0, double yy = 0) {x = xx, y = yy;}
}a[N], f[N], g[N], ff[N], gg[N];
double coss[N], sinn[N], q[N];
int rev[N];
complexx operator + (complexx a, complexx b) {return complexx(a.x + b.x, a.y + b.y);}
complexx operator - (complexx a, complexx b) {return complexx(a.x - b.x, a.y - b.y);}
complexx operator * (complexx a, complexx b) {return complexx(a.x * b.x - a.y * b.y, a.y * b.x + a.x * b.y);}
void fft(int len, complexx *a, int o){
	for(int i = 0; i <= len; i ++) if(i < rev[i]) swap(a[i], a[rev[i]]);
	for(int j = 1; j < len; j <<= 1){
		complexx wn = complexx(coss[j], o * sinn[j]);
		for(int k = 0; k < len; k += (j << 1)){
			complexx w0 = complexx(1, 0);
			for(int i = 0; i < j; i ++, w0 = w0 * wn){
				complexx X = a[i + k], Y = w0 * a[i + j + k];
				a[i + k] = X + Y;
				a[i + k + j] = X - Y;
			}
		}
	}
}
void cheng(int n, int m, complexx *a, complexx *b){
	int len = 1, l = 0;
	for(;len <= n + m; len <<= 1, l ++);
	for(int i = 0; i <= len; i ++) rev[i] = (rev[i >> 1] >> 1) | ((i&1) << (l - 1));
	for(int i = 1; i <= len; i <<= 1) coss[i] = cos(pi / i), sinn[i] = sin(pi / i);
	fft(len, a, 1);
	fft(len, b, 1);
	for(int i = 0; i <= len; i ++)
		a[i] = a[i] * b[i];
	fft(len, a, -1);
	for(int i = 0; i <= n + m; i ++) a[i].x = a[i].x / len;
}
int n, m;
int main(){
	scanf("%d", &n);
	for(int i = 1; i <= n; i ++) scanf("%lf", &q[i]), f[i].x = q[i], ff[n - i + 1].x = q[i];	
	for(int i = 1; i <= n; i ++) gg[i].x = g[i].x = 1.0 / i / i;
	cheng(n, n, f, g);
	cheng(n, n, ff, gg);
	for(int i = 1; i <= n; i ++) printf("%.3f\n", f[i].x - ff[n - i + 1].x);
	return 0;
}