[BZOJ3527][ZJOI2014]力：FFT

分析

整理得下式：

[E_i=sum_{j<i}{frac{q_i}{(i-j)^2}}-sum_{j>i}{frac{q_i}{(i-j)^2}} ]

假设(n=5)，考虑这两个数组：

(a:q_1 quad q_2 quad q_3 quad q_4 quad q_5)

(b:-frac{1}{16} quad -frac{1}{9} quad -frac{1}{4} quad -frac{1}{1} quad 0 quad frac{1}{1} quad frac{1}{4} quad frac{1}{9} quad frac{1}{16})

容易发现(E)数组是把(a,b)数组看做多项式各项系数作卷积后一些项的系数。

FFT即可。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <algorithm>
#include <complex>
#define rin(i,a,b) for(int i=(a);i<=(b);i++)
#define rec(i,a,b) for(int i=(a);i>=(b);i--)
#define trav(i,a) for(int i=head[(a)];i;i=e[i].nxt)
using std::cin;
using std::cout;
using std::endl;
typedef long long LL;
typedef std::complex<double> Complex;

const int MAXN=100005;
const double pi=3.14159265358979;
int n,m,len;
int rev[MAXN<<3];
Complex a[MAXN<<3],b[MAXN<<3];

inline void fft(Complex *c,int dft){
	rin(i,0,n-1) if(i<rev[i])
		std::swap(c[i],c[rev[i]]);
	for(int mid=1;mid<n;mid<<=1){
		int r=(mid<<1);
		Complex u=(Complex){cos(pi/mid),dft*sin(pi/mid)};
		for(int l=0;l<n;l+=r){
			Complex v=1;
			for(int i=0;i<mid;i++,v*=u){
				Complex x=c[l+i],y=v*c[l+mid+i];
				c[l+i]=x+y;
				c[l+mid+i]=x-y;
			}
		}
	}
	if(dft==-1) rin(i,0,n-1)
		c[i]/=n;
}

int main(){
	scanf("%d",&n);
	n--;
	rin(i,0,n){
		double x;
		scanf("%lf",&x);
		a[i]=x;
	}
	m=(n<<1);
	rin(i,0,m){
		if(i<n) b[i]=-1.0/(n-i)/(n-i);
		else if(i==n) b[i]=0;
		else b[i]=1.0/(i-n)/(i-n);
	}
	int nn=n;
	for(m+=n,n=1;n<=m;n<<=1) len++;
	rin(i,1,n-1) rev[i]=((rev[i>>1]>>1)|((i&1)<<(len-1)));
	fft(a,1);
	fft(b,1);
	rin(i,0,n-1) a[i]*=b[i];
	fft(a,-1);
	n=nn;
	rin(i,n,n+n) printf("%.10lf
",a[i].real());
	return 0;
}