bzoj3527【ZJOI2014】力

题目大意：

给出n个数qi，定义 Fj为

令 Ei=Fi/qi，求Ei。

令a[i]=q[i]，b[i]=1/(i^2)，则E[i]=∑(1≤j≤i-1)a[j]*b[i-j]-∑(i+1≤j≤n)a[j]*b[j-i]。

左边就是卷积形式，直接FFT；右边和bzoj2194等价，翻转一个数组后FFT。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define F(i,j,n) for(int i=j;i<=n;i++)
#define D(i,j,n) for(int i=j;i>=n;i--)
#define ll long long
#define maxn 500005
using namespace std;
int nn,n,m,len,rev[maxn];
double q[maxn],ans[maxn];
const double pi=acos(-1.0);
struct cp
{
double x,y;
cp(double xx=0,double yy=0){x=xx;y=yy;}
inline cp operator +(cp b){return (cp){x+b.x,y+b.y};}
inline cp operator -(cp b){return (cp){x-b.x,y-b.y};}
inline cp operator *(cp b){return (cp){x*b.x-y*b.y,x*b.y+y*b.x};}
}a[maxn],b[maxn],c[maxn];
inline void fft(cp *x,int n,int flag)
{
F(i,0,n-1) if (rev[i]>i) swap(x[i],x[rev[i]]);
for(int m=2;m<=n;m<<=1)
{
cp wn=cp(cos(2.0*pi/m*flag),sin(2.0*pi/m*flag));
for(int i=0;i<n;i+=m)
{
cp w=cp(1.0);
int mid=m>>1;
F(j,0,mid-1)
{
cp u=x[i+j],v=x[i+j+mid]*w;
x[i+j]=u+v;x[i+j+mid]=u-v;
w=w*wn;
}
}
}
if (flag==-1) F(i,0,n-1) x[i].x/=n;
}
int main()
{
scanf("%d",&n);nn=n;
F(i,0,n-1) scanf("%lf",&q[i]);
n=n*2-1;m=1;
while (m<n) m<<=1,len++;
n=m;
F(i,0,n-1)
{
int x=i,y=0;
F(j,1,len) (y<<=1)|=(x&1),x>>=1;
rev[i]=y;
}
F(i,0,nn-1) a[i]=cp(q[i],0);
F(i,nn,n-1) a[i]=cp();
b[0]=cp();
F(i,1,nn-1) b[i]=cp(1.0/((double)i*(double)i));
F(i,nn,n-1) b[i]=cp();
fft(a,n,1);fft(b,n,1);
F(i,0,n-1) c[i]=a[i]*b[i];
fft(c,n,-1);
F(i,0,nn-1) ans[i]=c[i].x;
F(i,0,nn-1) a[i]=cp(q[nn-1-i],0);
F(i,nn,n-1) a[i]=cp();
b[0]=cp();
F(i,1,nn-1) b[i]=cp(1.0/((double)i*(double)i));
F(i,nn,n-1) b[i]=cp();
fft(a,n,1);fft(b,n,1);
F(i,0,n-1) c[i]=a[i]*b[i];
fft(c,n,-1);
F(i,0,nn-1) ans[i]-=c[nn-1-i].x;
F(i,0,nn-1) printf("%.5lf\n",ans[i]);
return 0;
}