# HDU 4609 3-idiots （FFT）

## Description

King OMeGa catched three men who had been streaking in the street. Looking as idiots though, the three men insisted that it was a kind of performance art, and begged the king to free them. Out of hatred to the real idiots, the king wanted to check if they were lying. The three men were sent to the king’s forest, and each of them was asked to pick a branch one after another. If the three branches they bring back can form a triangle, their math ability would save them. Otherwise, they would be sent into jail.

However, the three men were exactly idiots, and what they would do is only to pick the branches randomly. Certainly, they couldn’t pick the same branch – but the one with the same length as another is available. Given the lengths of all branches in the forest, determine the probability that they would be saved.

## Input

An integer T(T≤100) will exist in the first line of input, indicating the number of test cases.

Each test case begins with the number of branches N(3≤N≤10^5).

The following line contains N integers a_i (1≤a_i≤10^5), which denotes the length of each branch, respectively.

## Output

Output the probability that their branches can form a triangle, in accuracy of 7 decimal places.

## Sample Input

2
4
1 3 3 4
4
2 3 3 4


## Sample Output

0.5000000
1.0000000


## 思路

{0,1,0,2,1} * {0,1,0,2,1} 卷积结果为 {0,0,1,0,4,2,4,4,1}

• 因为包含某个数本身与本身的组合，所以我们要先去除这样的情况。
• 第一个选取 a1 ，第二个选取 a2 与第一个选取 a2 ，第二个选取 a1 我们认为是一样的，因此 num 数组需要整体除以 2

num 数组求前缀和，原数组 a 进行排序，接下来我们需要找第三条边，满足另外两条边的和大于这条边且标号都小于这条边。

• x,y 中有一个值大于 a[i] ，另一个值小于 a[i] ，总共有 (n-i-1)*i 种情况
• x,y 的值都大于 a[i] ，总共有 (n-i-1)*(n-i-2)/2 种情况
• x,y 中有一点落在 i 这一位置，总共有 n-1 种情况

## AC 代码

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<iomanip>
#include<cmath>
#include<cstring>
#include<vector>
typedef __int64 LL;
#define pi acos(-1.0)
using namespace std;

const int maxn = 400005;
//复数结构体
struct complex
{
double r,i;
complex(double R=0,double I=0)
{
r=R;
i=I;
}
complex operator+(const complex &a)
{
return complex(r+a.r,i+a.i);
}
complex operator-(const complex &a)
{
return complex(r-a.r,i-a.i);
}
complex operator*(const complex &a)
{
return complex(r*a.r-i*a.i,r*a.i+i*a.r);
}
};
/*
*进行FFT和IFFT前的反转变换
*位置i和i的二进制反转后位置互换，(如001反转后就是100)
*len必须去2的幂
*/
void change(complex x[],int len)
{
int i,j,k;
for(i = 1, j = len/2; i <len-1; i++)
{
if (i < j) swap(x[i],x[j]);
//交换互为小标反转的元素，i<j保证交换一次
//i做正常的+1,j做反转类型的+1，始终i和j是反转的
k = len/2;
while (j >= k)
{
j -= k;
k /= 2;
}
if (j < k) j += k;
}
}
/*
*做FFT
*len必须为2^n形式，不足则补0
*on=1时是DFT，on=-1时是IDFT
*/
void fft (complex x[],int len,int on)
{
change(x,len);
for (int i=2; i<=len; i<<=1)
{
complex wn(cos(-on*2*pi/i),sin(-on*2*pi/i));  //单位复根e^(2*PI/m)用欧拉公式展开
for (int j=0; j<len; j+=i)
{
complex w(1,0);
for (int k=j; k<j+i/2; k++)
{
complex u = x[k];
complex t = w*x[k+i/2];
x[k] = u+t;
x[k+i/2] = u-t;
w = w*wn;
}
}
}
if (on == -1)
for (int i=0; i<len; i++)
x[i].r /= len;
}

complex x[maxn];
int a[maxn];
LL num[maxn],sum[maxn];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n;
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++)
x[i]=complex(num[i],0);
for(int i=len1; i<len; i++)     // 补零
x[i]=complex(0,0);
fft(x,len,1);                   // 求值
for(int i=0; i<len; i++)        // 乘法
x[i]=x[i]*x[i];
fft(x,len,-1);                  // 插值
for(int i=0; i<len; i++)
num[i]=(x[i].r+0.5);        // 四舍五入
for(int i=0; i<n; i++)          // 去除挑选相同数字的情况
num[a[i]+a[i]]--;
for(int i=1; i<len; i++)        // 选取与顺序无关
num[i]/=2;
for(int i=1; i<=len; i++)
sum[i]=sum[i-1]+num[i];
LL ans=0;
for(int i=0; i<n; i++)
{
ans+=sum[len]-sum[a[i]];
ans-=LL(n-i-1)*i;           // 一大一小
ans-=LL(n-i-1)*(n-i-2)/2;   // 两大
ans-=n-1;                   // 其中有一点在当前位
}
LL tot=1LL*n*(n-1)*(n-2)/6;     // 所有组合
printf("%.7lf\n",(double)ans/tot);
}
return 0;
}