Descrption
Link.
对于每一个 (i)，求出：
[sum_{j=1}^{i-1}frac{a_{j}}{(i-j)^{2}}-sum_{j=i+1}^{n}frac{a_{j}}{(i-j)^{2}}
]
Solution
令 (f(i)=a_{i},g(i)=frac{1}{i^{2}})。
然后
[sum_{j=1}^{i-1}f(j) imes g(i-j)-sum_{j=i+1}^{n}f(j) imes g(j-i)
]
可以 FFT 了。
#include<bits/stdc++.h>
using namespace std;
const double bh_pi=acos(-1);
int n;
namespace Poly
{
typedef complex<double> comp;
typedef vector<complex<double> > poly;
#define len(x) (int((x).size()))
int lim,rev[400010];
void fft(poly &f,int op)
{
for(int i=0;i<lim;++i) if(i<rev[i]) swap(f[i],f[rev[i]]);
for(int len=2;len<=lim;len<<=1)
{
comp bas(cos(2*bh_pi/len),op*sin(2*bh_pi/len));
for(int fr=0;fr<lim;fr+=len)
{
comp now(1,0);
for(int ba=fr;ba<fr+(len>>1);++ba,now*=bas)
{
comp tmp=now*f[ba+(len>>1)];
f[ba+(len>>1)]=f[ba]-tmp;
f[ba]+=tmp;
}
}
}
if(op==-1) for(int i=0;i<lim;++i) f[i]/=lim;
}
poly mulPoly(poly f,poly g)
{
int n=len(f)+len(g)-1;
for(lim=1;lim<=n;lim<<=1);
for(int i=0;i<lim;++i) rev[i]=(rev[i>>1]>>1)|((i&1)?(lim>>1):0);
f.resize(lim),g.resize(lim);
fft(f,1),fft(g,1);
for(int i=0;i<lim;++i) f[i]*=g[i];
fft(f,-1),f.resize(n);
return f;
}
}using namespace Poly;
int main()
{
poly f,g;
scanf("%d",&n);
f.resize(n+1),g.resize(n+1);
for(int i=1;i<=n;++i)
{
double x;
scanf("%lf",&x);
f[i]=comp(x,0);
}
for(int i=1;i<=n;++i) g[i]=comp(1.0/i/i,0);
poly onetmp=mulPoly(f,g);
reverse(next(f.begin()),f.end());
poly anotmp=mulPoly(f,g);
for(int i=1;i<=n;++i) printf("%.3lf
",onetmp[i].real()-anotmp[n-i+1].real());
return 0;
}