# HDU 4609 3-idiots

http://acm.hdu.edu.cn/showproblem.php?pid=4609

```#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;

const double PI = acos(-1.0);
struct complex
{
double r,i;
complex(double _r = 0,double _i = 0)
{
r = _r; i = _i;
}
complex operator +(const complex &b)
{
return complex(r+b.r,i+b.i);
}
complex operator -(const complex &b)
{
return complex(r-b.r,i-b.i);
}
complex operator *(const complex &b)
{
return complex(r*b.r-i*b.i,r*b.i+i*b.r);
}
};
void change(complex y[],int len)
{
int i,j,k;
for(i = 1, j = len/2;i < len-1;i++)
{
if(i < j)swap(y[i],y[j]);
k = len/2;
while( j >= k)
{
j -= k;
k /= 2;
}
if(j < k)j += k;
}
}
void fft(complex y[],int len,int on)
{
change(y,len);
for(int h = 2;h <= len;h <<= 1)
{
complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h));
for(int j = 0;j < len;j += h)
{
complex w(1,0);
for(int k = j;k < j+h/2;k++)
{
complex u = y[k];
complex t = w*y[k+h/2];
y[k] = u+t;
y[k+h/2] = u-t;
w = w*wn;
}
}
}
if(on == -1)
for(int i = 0;i < len;i++)
y[i].r /= len;
}

const int MAXN = 400040;
complex x1[MAXN];
int a[MAXN/4];
long long num[MAXN];//100000*100000会超int
long long sum[MAXN];

int main()
{
int T;
int n;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
memset(num,0,sizeof(num));
for(int i = 0;i < n;i++)
{
scanf("%d",&a[i]);
num[a[i]]++;
}
sort(a,a+n);
int len1 = a[n-1]+1;
int len = 1;
while( len < 2*len1 )len <<= 1;
for(int i = 0;i < len1;i++)
x1[i] = complex(num[i],0);
for(int i = len1;i < len;i++)
x1[i] = complex(0,0);
fft(x1,len,1);
for(int i = 0;i < len;i++)
x1[i] = x1[i]*x1[i];
fft(x1,len,-1);
for(int i = 0;i < len;i++)
num[i] = (long long)(x1[i].r+0.5);
len = 2*a[n-1];
//减掉取两个相同的组合
for(int i = 0;i < n;i++)
num[a[i]+a[i]]--;
//选择的无序，除以2
for(int i = 1;i <= len;i++)
{
num[i]/=2;
}
sum[0] = 0;
for(int i = 1;i <= len;i++)
sum[i] = sum[i-1]+num[i];
long long cnt = 0;
for(int i = 0;i < n;i++)
{
cnt += sum[len]-sum[a[i]];
//减掉一个取大，一个取小的
cnt -= (long long)(n-1-i)*i;
//减掉一个取本身，另外一个取其它
cnt -= (n-1);
//减掉大于它的取两个的组合
cnt -= (long long)(n-1-i)*(n-i-2)/2;
}
//总数
long long tot = (long long)n*(n-1)*(n-2)/6;
printf("%.7lf\n",(double)cnt/tot);
}
return 0;
}```

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