luogu P3338 [ZJOI2014]力

传送门

首先化简原式$$F_j=sum_{i<j}frac{q_iq_j}{(i-j)^{2}-sum_{i>j}frac{q_iq_j}{(i-j)}2},E_j=F_j/q_j$$

把所有(q_j)提出来,则显然$$E_j=sum_{i<j}frac{q_i}{(i-j)^{2}-sum_{i>j}frac{q_i}{(i-j)}2}$$$$E_j=...-frac{q_{j-2}}{2^{2}-frac{q_{j-1}}{1}2}+0+frac{q_{j+1}}{1^{2}+frac{q_{j+2}}{2}2}...$$

然后设多项式(A,B),其中(A[i]=q_{i+1}(0le i<n))(,B[i]=frac{1}{(i-(n-1))|i-(n-1)|}(0le i<2n-1)(B[n-1]=0)),然后把(A,B)乘起来,答案要求的(E_i)也就是多项式的第(n-2+i)项的系数

#include<bits/stdc++.h>
#define LL long long
#define db double
#define il inline
#define re register
  
using namespace std;
const int N=100000+10,M=550000+10;
const db pi=acos(-1);
il int rd()
{
  int x=0,w=1;char ch=0;
  while(ch<'0'||ch>'9') {if(ch=='-') w=-1;ch=getchar();}
  while(ch>='0'&&ch<='9') {x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}
  return x*w;
}
struct comp
{
  db r,i;
  comp(){r=i=0;}
  comp(db nr,db ni){r=nr,i=ni;}
  il comp operator + (const comp &bb) const {return comp(r+bb.r,i+bb.i);}
  il comp operator - (const comp &bb) const {return comp(r-bb.r,i-bb.i);}
  il comp operator * (const comp &bb) const {return comp(r*bb.r-i*bb.i,r*bb.i+i*bb.r);}
}a[M],b[M];
int n,m,nn,l,rdr[M];
void fft(comp *a,int op)
{
  comp W,w,x,y;
  for(int i=0;i<nn;++i) if(i<rdr[i]) swap(a[i],a[rdr[i]]);
  for(int i=1;i<nn;i<<=1)
    {
      W=comp(cos(pi/i),sin(pi/i)*op);
      for(int j=0;j<nn;j+=i<<1)
        {
          w=comp(1,0);
          for(int k=0;k<i;++k,w=w*W)
            {
              x=a[j+k],y=w*a[j+k+i];
              a[j+k]=x+y,a[j+k+i]=x-y;
            }
        }
    }
    
}
  
int main()
{
  n=rd();
  for(int i=0;i<n;++i) scanf("%lf",&a[i].r);
  m=-1;
  for(int i=-n+1;i<=n-1;++i) b[++m].r=1/(db)i/fabs(i);
  b[n-1].r=0;
  m+=n-1;
  for(nn=1;nn<=m;nn<<=1) ++l;
  for(int i=0;i<nn;++i) rdr[i]=(rdr[i>>1]>>1)|((i&1)<<(l-1));
  fft(a,1),fft(b,1);
  for(int i=0;i<nn;++i) a[i]=a[i]*b[i];
  fft(a,-1);
  for(int i=n-1;i<n+n-1;i++) printf("%.3lf
",a[i].r/nn);
  return 0;
}