洛谷 P3338 [ZJOI2014]力

题意简述

读入(n)个数(q_i)

设(F_j = sumlimits_{i<j}frac{q_i imes q_j}{(i-j)^2 }-sumlimits_{i>j}frac{q_i imes q_j}{(i-j)^2 })
令(E_i=frac{F_i}{q_i})，求(E_i)

题解思路

先推式子

(E_j=frac{F_j}{q_j}=sumlimits_{i<j}frac{q_i}{(i-j)^2 }-sumlimits_{i>j}frac{q_i}{(i-j)^2 })

设(T_i=i^{-2})

则(E_j=sumlimits_{i=1}^{j-1}q_iT_{j-i}-sumlimits_{i=j+1}^{n}q_iT_{i-j})

再设(p_i=q_{n-i+1})

则(E_j=sumlimits_{i=1}^{j-1}q_iT_{j-i}-sumlimits_{i=1}^{j-1}p_iT_{j-i})

可以发现，这两项都是两个多项式卷积的结果，所以可以用FFT来做

注意：如果用三次变两次优化，会掉精

代码

#include <cmath>
#include <iostream>
#include <algorithm>
const int N=400005;
const double Pi=acos(-1.0);
int n,nn,m,lim=1,l=-1,r[N];
struct Com {
	long double x,y;
	Com(long double xx=0,long double yy=0) { x=xx; y=yy; }
}p[N],q[N],t[N];
Com operator +(const Com& x,const Com& y) { return Com(x.x+y.x,x.y+y.y); }
Com operator -(const Com& x,const Com& y) { return Com(x.x-y.x,x.y-y.y); }
Com operator *(const Com& x,const Com& y) { return Com(x.x*y.x-x.y*y.y,x.x*y.y+x.y*y.x); }
inline void FFT(Com *x,const int& lim,const int& type) {
	for (register int i=0;i<lim;++i) if (i<r[i]) std::swap(x[i],x[r[i]]);
	for (register int i=1;i<lim;i<<=1) {
		Com UR(cos(Pi/i),sin(Pi/i)*type);
		for (register int j=0,I=i<<1;j<lim;j+=I) {
			Com w(1,0);
			for (register int k=0;k<i;++k,w=w*UR) {
				Com t1=x[j+k],t2=w*x[j+k+i];
				x[j+k]=t1+t2; x[j+k+i]=t1-t2;
			}
		}
	}
}
int main() {
	std::ios::sync_with_stdio(0);
	std::cin.tie(0); std::cout.tie(0); std::cout<<std::fixed;
	std::cin>>n;
	for (register int i=1;i<=n;++i) std::cin>>p[i].x,t[i].x=1.0/i/i;
	for (register int i=1;i<=n;++i) q[i].x=p[n-i+1].x;
	for (nn=n<<1;lim<=nn;lim<<=1,++l);
	for (register int i=0;i<lim;++i) r[i]=(r[i>>1]>>1)|((i&1)<<l);
	FFT(p,lim,1); FFT(q,lim,1); FFT(t,lim,1);
	for (register int i=0;i<lim;++i) p[i]=p[i]*t[i],q[i]=q[i]*t[i];
	FFT(p,lim,-1); FFT(q,lim,-1);
	for (register int i=1;i<=n;++i)
		std::cout<<(p[i].x-q[n-i+1].x)/lim<<'
';
}