HDU_4609_3-idiots

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>

#define LL long long
#define mt(a,b) memset(a,b,sizeof(a))

using namespace std;

const int M = 1e5+100;

struct Complex {///复数结构体
double x,y;///实部和虚部 x+yi
Complex(double _x=0,double _y=0) {
x=_x;
y=_y;
}
friend Complex operator -(const Complex &a,const Complex &b) {
return Complex(a.x-b.x,a.y-b.y);
}
friend Complex operator +(const Complex &a,const Complex &b) {
return Complex(a.x+b.x,a.y+b.y);
}
friend Complex operator *(const Complex &a,const Complex &b) {
return Complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);
}
};
class FFT {///快速傅里叶变换
Complex u,t;
void change(Complex y[],int len) { ///len必须是2的幂
for(int i=1,j=len>>1,k; i<len-1; i++) {
if(i<j) swap(y[i],y[j]);
k=len>>1;
while(j>=k) {
j-=k;
k>>=1;
}
if(j<k) j+=k;
}
}
public:
void fft(Complex y[],int len,int on) { ///on==1时是DFT，on==-1时是IDFT
change(y,len);
double tmp=-on*2*acos(-1.0);
for(int h=2; h<=len; h<<=1) {
Complex wn(cos(tmp/h),sin(tmp/h));
for(int j=0; j<len; j+=h) {
Complex w(1,0);
int h2=h>>1;
for(int k=j; k<j+h2; k++) {
u=y[k];
t=w*y[k+h2];
y[k]=u+t;
y[k+h2]=u-t;
w=w*wn;
}
}
}
if(on==-1) {
for(int i=0; i<len; i++) {
y[i].x/=len;
}
}
}
} g;

Complex x[M*4];
int a[M*2];
LL flag[M*4];
LL sum[M*4];

int main() {
int _;
int n;
scanf("%d",&_);
while(_--) {
mt(flag,0);
scanf("%d",&n);
for(int i=0; i<n; i++) {
scanf("%d",&a[i]);
flag[a[i]]++;
}

sort(a,a+n);

int len = 1;
while(len<(a[n-1]+1)*2) len <<= 1;

for(int i=0; i<len; i++) {
x[i] = Complex(flag[i],0);
}

g.fft(x,len,1);

for(int i=0; i<len; i++) {
x[i] = x[i] * x[i];
}

g.fft(x,len,-1);

for(int i=0; i<len; i++) {
flag[i] = (LL)(x[i].x+0.5);
}

len = a[n-1]*2;

for(int i=0; i<n; i++) {
flag[a[i]+a[i]] -- ;
}

///注意这边是1~len
for(int i=1; i<=len; i++) {
flag[i] /= 2;
}

sum[0] = 0;
for(int i=0; i<len; i++) {
sum[i+1] = sum[i] + flag[i+1];
}

LL cnt = 0;
for(int i = 0; i < n; i++) {
cnt += sum[len]-sum[a[i]];
cnt -= (LL)(n-1-i)*i;
cnt -= (n-1);
cnt -= (LL)(n-1-i)*(n-i-2)/2;
}

//            printf("%d\n",cnt);

LL all = (LL)n*(n-1)*(n-2)/6;
printf("%.7f\n",cnt*1.0/all);

}

return 0;
}
/**
2
4
1 3 3 4
4
2 3 3 4

*/