bzoj3527 [Zjoi2014]力

题目链接：bzoj3527

题目大意：

给出n个数qi，定义Fj:

Fj=∑i<jqiqj(i−j)2−∑i>jqiqj(i−j)2

令Ei=Fiqi。求Ei。

题解：

FFT

代入可得Ei=∑j<iqiqj(i−j)2−∑j>iqiqj(i−j)2qi=∑j<iqj(i−j)2−∑j>iqj(i−j)2

设f(i)=1i2，1≤i≤n。

所以Ei=∑j<iqj×f(i−j)−∑j>iqj×f(i−j)

卷积啊卷积的形式啊。

把f(i)在[−n,n]范围内的值都求出来就直接E=q∗f了

可以发现i<0时f(i)=−1i2；i=0时f(i)=0；i>0时f(i)=1i2。

把下标都+n弄成正的，然后就套FFT。

答案就是[n,n+n]那部分的。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
#define maxn 800010

const double pi=acos(-1);
struct Complex
{
double x,y;
Complex(){x=y=0;}
Complex(double x,double y):x(x),y(y){}
friend inline Complex operator + (Complex x,Complex y){return Complex(x.x+y.x,x.y+y.y);}
friend inline Complex operator - (Complex x,Complex y){return Complex(x.x-y.x,x.y-y.y);}
friend inline Complex operator * (Complex x,Complex y){return Complex(x.x*y.x-x.y*y.y,x.x*y.y+x.y*y.x);}
}q[maxn],f[maxn];
void DFT(Complex *c,int ln,int t)
{
for(int i=0,j=0;i<ln;i++)
{
if(i>j) swap(c[i],c[j]);
for(int k=ln>>1;(j^=k)<k;k>>=1);
}
for (int i=1;(1<<i)<=ln;i++)
{
int m=(1<<i);
Complex wn(cos(2*pi/m),t*sin(2*pi/m));
for (int k=0;k<ln;k+=m)
{
Complex w(1,0);
for (int j=0;j<(m>>1);j++,w=w*wn)
{
Complex a0=c[k+j];
Complex a1=w*c[k+j+(m>>1)];
c[k+j]=a0+a1;
c[k+j+(m>>1)]=a0-a1;
}
}
}
}
double ans[maxn];
void FFT(int ln)
{
int fn=1;while (fn<=ln) fn<<=1;
DFT(q,fn,1);DFT(f,fn,1);
for (int i=0;i<=fn;i++) q[i]=q[i]*f[i];
DFT(q,fn,-1);
for (int i=0;i<=ln;i++) ans[i]=q[i].x/fn;
}
int main()
{
//freopen("a.in","r",stdin);
//freopen("a.out","w",stdout);
int n,m,i;
scanf("%d",&n);n--;
for (i=0;i<=n;i++) scanf("%lf",&q[i].x);
m=2*n;f
.x=0;
for (i=1;i<=n;i++) f[i+n].x=(double)1/i/i,f[-i+n].x=-(double)1/i/i;
FFT(n+m);
for (i=n;i<=n+n;i++) printf("%.3lf\n",ans[i]);
return 0;
}