正题

题目链接:https://www.luogu.com.cn/problem/P3338

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题目大意

$$F_j=\sum _{i=1}^{j-1}\frac{q_i\ast q_j}{(i-j{)}^2}-\sum _{i=j+1}^n\frac{q_i\ast q_j}{(i-j{)}^2}$$
$$E_j=\frac{F_j}{q_j}$$

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解题思路

显然有$$F_j=\sum _{i=1}^{j-1}\frac{q_i}{(i-j{)}^2}-\sum _{i=j+1}^n\frac{q_i}{(i-j{)}^2}$$
然后定义$a_i=q_i,b_i=\frac{1}{i^2}$，那么有
$$F_j=\sum _{i=1}^{j-1}{a}_i\ast b_{j-i}-\sum _{i=j+1}^n{a}_i\ast b_{i-j}$$
前面那个式子显然可以卷积，然后后面那个式子把$a$数组翻转之后就和前面那个一样了。

然后$FFT$优化即可，时间复杂度$O(n\log n)$。

才发现有$complex$这个库，终于不用手写复数运算了（不知道比赛能不能用）

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$code$

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<complex>
#define Pi acos(-1)
using namespace std;
typedef complex<double> comp;
const int N=3e5+10;
/*struct comp{double x,y;
};
complex operator +(complex a,complex b)
{return (complex){a.x+b.x,a.y+b.y};}
complex operator -(complex a,complex b)
{return (complex){a.x-b.x,a.y-b.y};}
complex operator *(complex a,complex b)
{return (complex){a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x};}*/
int n,r[N],z;
comp a1[N],a2[N],b1[N],b2[N];
void fft(comp *f,int op){for(int i=0;i<n;i++)if(i<r[i])swap(f[i],f[r[i]]);for(int p=2;p<=n;p<<=1){int len=p>>1;comp tmp(cos(Pi/len),sin(Pi/len)*op);
//		printf("%lf ",tmp.real()); for(int k=0;k<n;k+=p){comp buf(1,0);for(int i=k;i<k+len;i++){comp tt=buf*f[len+i];f[len+i]=f[i]-tt;f[i]=f[i]+tt;buf=buf*tmp;}}}if(op==-1)for(int i=0;i<n;i++)f[i]/=n;return;
}
int main()
{scanf("%d",&n);int l;z=n;for(int i=1;i<=n;i++){double x;scanf("%lf",&x);a1[i]=a2[n-i+1]=x;b1[i]=b2[i]=(double)1.0/i/i;}for(l=1;l<=n*2;l<<=1);n=l;for(int i=0;i<n;i++)r[i]=(r[i>>1]>>1)|((i&1)?n>>1:0);fft(a1,1);fft(b1,1);for(int i=0;i<n;i++)a1[i]=a1[i]*b1[i];fft(a1,-1);fft(a2,1);fft(b2,1);for(int i=0;i<n;i++)a2[i]=a2[i]*b2[i];fft(a2,-1);for(int i=1;i<=z;i++)printf("%.3lf\n",a1[i].real()-a2[z-i+1].real());return 0;
}