给N个木棍问任选三个可组合出三角形的概率

以前写的，整理整理存个档

//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include<stdio.h>
#include<algorithm>
#include<queue>
#include<string.h>
#include<iostream>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<iomanip>
using namespace std;
#define ll long long
#define pb push_back
#define FOR(a) for(int i=1;i<=a;i++)
const double PI = acos(-1.0);struct complex
{double r, i;       //real and imagecomplex(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 = 4e5 + 6;
complex x1[maxn];
int arr[maxn >> 2];
ll num[maxn]; ll sum[maxn];int main() {int T, n;scanf("%d", &T);while (T--) {memset(num, 0, sizeof num);scanf("%d", &n);for (int i = 0; i<n; i++) {scanf("%d", &arr[i]);num[arr[i]]++;}sort(arr, arr + n);int len1 = arr[n - 1] + 1;                  //数域范围int len = 1;while (len<2 * len1)len <<= 1;             //扩充成最小2的幂次，即多项式项数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);                   //应用2n阶的fft计算出A(x)的点值表达for (int i = 0; i<len; i++)x1[i] = x1[i] * x1[i]; //逐点相乘,O(n)fft(x1, len, -1);                    //对2n个点值应用fft，计算其逆dftfor (int i = 0; i<len; i++) {num[i] = (long long)(x1[i].r + 0.5);}len = 2 * arr[n - 1];                       //从2的幂次缩回原来那么大
/************************************************************************************************************///减掉相同组合for (int i = 0; i<n; i++)num[arr[i] + arr[i]]--;//选择无序，地位等价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];ll cnt = 0;for (int i = 0; i<n; i++) {cnt += sum[len] - sum[arr[i]];cnt -= (long long)(n - 1 - i)*i;cnt -= n - 1;cnt -= (long long)(n - 1 - i)*(n - i - 2) / 2;}ll tot = 1ll * n*(n - 1)*(n - 2) / 6;printf("%.7lf\n", (double)cnt / tot);}
}