hdu 4609 3-idiots(FFT计数)

 1 #include<bits/stdc++.h>
 2 #define F(i,a,b) for(int i=a;i<=b;++i)
 3 using namespace std;
 4 
 5 const double pi=acos(-1.0);
 6 //n 必须为 2 的幂。
 7 struct comp{
 8     double r,i;
 9     comp(double _r=0,double _i=0){r=_r,i=_i;}
10     comp operator+(const comp x){return comp(r+x.r,i+x.i);}
11     comp operator-(const comp x){return comp(r-x.r,i-x.i);}
12     comp operator*(const comp x){return comp(r*x.r-i*x.i,r*x.i+i*x.r);}
13 };
14 
15 void FFT(comp a[],int n,int t){
16     for(int i=1,j=0;i<n-1;i++)
17     {
18         for(int s=n;j^=s>>=1,~j&s;);
19         if(i<j)swap(a[i],a[j]);
20     }
21     for(int d=0;(1<<d)<n;d++)
22     {
23         int m=1<<d,m2=m<<1;
24         double o=pi/m*t;comp _w(cos(o),sin(o));
25         for(int i=0;i<n;i+=m2)
26         {
27             comp w(1,0);
28             for(int j=0;j<m;j++)
29             {
30                 comp &A=a[i+j+m],&B=a[i+j],t=w*A;
31                 A=B-t,B=B+t,w=w*_w;
32             }
33         }
34     }
35     if(t==-1)for(int i=0;i<n;i++)a[i].r/=n;
36 }
37 
38 const int N=3e5+7;
39 
40 comp x1[N],x2[N];
41 
42 long long sum[N];
43 int a[N],num[N];
44 
45 int main()
46 {
47     int t,n;
48     scanf("%d",&t);
49     while(t--)
50     {
51         memset(num,0,sizeof(num));
52         scanf("%d",&n);
53         F(i,0,n-1)scanf("%d",a+i);
54         sort(a,a+n);
55         F(i,0,n-1)num[a[i]]++;
56         int len=1,tmp=a[n-1]+1;
57         while(len<2*tmp)len<<=1;
58         F(i,0,len-1)
59         {
60             if(i<tmp)x1[i]=comp(num[i],0);
61             else x1[i]=comp(0,0);
62         }
63         FFT(x1,len,1);
64         F(i,0,len-1)x1[i]=x1[i]*x1[i];
65         FFT(x1,len,-1);
66         F(i,0,len-1)sum[i]=(long long)(x1[i].r+0.5);
67         len=2*a[n-1];
68         F(i,0,n-1)sum[a[i]+a[i]]--;
69         F(i,0,len)sum[i]>>=1;
70         F(i,1,len)sum[i]+=sum[i-1];
71         long long cnt=0;
72         F(i,0,n-1)
73         {
74             cnt+=sum[len]-sum[a[i]];
75             cnt-=1ll*(n-1-i)*i;
76             cnt-=(n-1);
77             cnt-=1ll*(n-1-i)*(n-i-2)/2;
78         }
79         printf("%.7f\n",(double)cnt/(1ll*n*(n-1)*(n-2)/6));
80     }
81     return 0;
82 }