3527: [Zjoi2014]力
先除去qiq_iqi，令g(i)=1i2g(i)=\frac{1}{i^2}g(i)=i21式子变成Ei=∑j=1i−1qi∗g(i−j)−∑j=i+1nqi∗g(j−i)E_i=\sum_{j=1}^{i-1} q_i*g(i-j)-\sum_{j=i+1}^nq_i*g(j-i)Ei=∑j=1i−1qi∗g(i−j)−∑j=i+1nqi∗g(j−i)
然后将后一半翻转Ei=∑j=1i−1qi∗g(i−j)−∑j=1n−iqn−i∗g(j−i)E_i=\sum_{j=1}^{i-1} q_i*g(i-j)-\sum_{j=1}^{n-i}q_{n-i}*g(j-i)Ei=∑j=1i−1qi∗g(i−j)−∑j=1n−iqn−i∗g(j−i)
就是两个卷积相减辣，直接FFT
#include
#include
#include
using namespace std;
const int MAXN=200005;
const double PI=acos(-1.0);
int n,Len,lg2,rev[MAXN<<1];
struct CP{
double x,y;
CP(double X=0,double Y=0){x=X,y=Y;}
CP operator +(const CP b)const{return CP(x+b.x,y+b.y);}
CP operator -(const CP b)const{return CP(x-b.x,y-b.y);}
CP operator *(const CP b)const{return CP(x*b.x-y*b.y,x*b.y+y*b.x);}
};
CP G[MAXN<<1],B1[MAXN<<1],B2[MAXN<<1],Ans1[MAXN<<1],Ans2[MAXN<<1];
void FFT(CP *A,int opt){
for(int i=0;i>1]>>1)|((i&1)<<(lg2-1));}
int main(){
scanf("%d",&n);n--;
for(int i=0;i<=n;i++) scanf("%lf",&B1[i].x),B2[n-i].x=B1[i].x;
for(int i=1;i<=n;i++) G[i].x=1.0/i/i;
for(Len=1,lg2=0;Len<=n<<1;Len<<=1,lg2++);Getrev();
FFT(G,1),FFT(B1,1),FFT(B2,1);
for(int i=0;i