hdu 4609 3-idiots —— FFT

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef double db;
typedef long long ll;
int const xn=(1<<20),xm=1e5+5;
db const Pi=acos(-1.0);
int n,rev[xn],lim,num[xm];
struct com{db x,y;}a[xn];
com operator + (com a,com b){return (com){a.x+b.x,a.y+b.y};}
com operator - (com a,com b){return (com){a.x-b.x,a.y-b.y};}
com operator * (com a,com b){return (com){a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x};}
int rd()
{
  int ret=0,f=1; char ch=getchar();
  while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=0; ch=getchar();}
  while(ch>=‘0‘&&ch<=‘9‘)ret=ret*10+ch-‘0‘,ch=getchar();
  return f?ret:-ret;
}
void fft(com *a,int tp)
{
  for(int i=0;i<lim;i++)
    if(i<rev[i])swap(a[i],a[rev[i]]);
  for(int mid=1;mid<lim;mid<<=1)
    {
      com wn=(com){cos(Pi/mid),tp*sin(Pi/mid)};
      for(int j=0,len=(mid<<1);j<lim;j+=len)
    {
      com w=(com){1,0};
      for(int k=0;k<mid;k++,w=w*wn)
        {
          com x=a[j+k],y=w*a[j+mid+k];
          a[j+k]=x+y; a[j+mid+k]=x-y;
        }
    }
    }
  if(tp==1)return; 
  for(int i=0;i<lim;i++)a[i].x=a[i].x/lim;
}
int main()
{
  int T=rd();
  while(T--)
    {
      n=rd(); int mx=0;
      memset(num,0,sizeof num);
      for(int i=1,x;i<=n;i++)x=rd(),num[x]++,mx=max(mx,x);
      lim=1; int l=0;
      while(lim<=mx+mx)lim<<=1,l++;
      for(int i=0;i<lim;i++)
    rev[i]=((rev[i>>1]>>1)|((i&1)<<(l-1)));
      for(int i=0;i<lim;i++)a[i].x=0,a[i].y=0;
      for(int i=0;i<=mx;i++)a[i].x=num[i];
      fft(a,1);
      for(int i=0;i<lim;i++)a[i]=a[i]*a[i];
      fft(a,-1);
      for(int i=2;i<lim;i+=2)a[i].x=(ll)(a[i].x+0.5)-num[i/2];
      ll sum=(ll)n*(n-1)*(n-2),ans=sum; ll pre=0;
      for(int i=0;i<=mx;i++)//mx
    {
      pre+=3*(ll)(a[i].x+0.5);
      if(num[i])ans-=num[i]*pre;//num[i]*...!
    }
      printf("%.7f\n",1.0*ans/sum);
    }
  return 0;
}