2015
09-17

3-idiots

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.

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≤105).
The following line contains N integers a_i (1≤a_i≤105), which denotes the length of each branch, respectively.

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≤105).
The following line contains N integers a_i (1≤a_i≤105), which denotes the length of each branch, respectively.

2
4
1 3 3 4
4
2 3 3 4

0.5000000
1.0000000

num[i]表示长度为i的边有几条，求一次num与num的卷积之后，num[i]表示两条边和为i的有多少对。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
//FFT copy from kuangbin
const double pi = acos (-1.0);
// Complex  z = a + b * i
struct Complex {
double a, b;
Complex(double _a=0.0,double _b=0.0):a(_a),b(_b){}
Complex operator + (const Complex &c) const {
return Complex(a + c.a , b + c.b);
}
Complex operator - (const Complex &c) const {
return Complex(a - c.a , b - c.b);
}
Complex operator * (const Complex &c) const {
return Complex(a * c.a - b * c.b , a * c.b + b * c.a);
}
};
//len = 2 ^ k
void change (Complex y[] , int len) {
for (int i = 1 , j = len / 2 ; i < len -1 ; i ++) {
if (i < j) swap(y[i] , y[j]);
int k = len / 2;
while (j >= k) {
j -= k;
k /= 2;
}
if(j < k) j += k;
}
}
// FFT
// len = 2 ^ k
// on = 1  DFT    on = -1 IDFT
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].a /= len;
}
}
}
const int N = 100005;
typedef long long LL;
int n , a[N];
LL sum[N << 2] , num[N << 2];
Complex x1[N << 2];
int main () {
#ifndef ONLINE_JUDGE
freopen("input.txt" , "r" , stdin);
#endif
int t;
scanf ("%d", &t);
while (t --) {
memset (num , 0 , sizeof(num));
scanf ("%d", &n);
for (int i = 0 ; i < n ; i ++) {
scanf ("%d", &a[i]);
num[a[i]] ++;
}
sort (a , a + n);
int len = a[n - 1] + 1;
int l = 1;
while (l < len * 2) l <<= 1;
for (int i = 0 ; i < len ; i ++) {
x1[i] = Complex (num[i] , 0);
}
for (int i = len ; i < l ; i ++) {
x1[i] = Complex (0 , 0);
}
FFT(x1 , l , 1);
for (int i = 0 ; i < l ; i ++) {
x1[i] = x1[i] * x1[i];
}
FFT(x1 , l , -1);
for (int i = 0 ; i < l ; i ++) {
num[i] = (LL)(x1[i].a + 0.5);
}
l = 2 * a[n - 1];
for (int i = 0 ; i < n ; i ++) {
num[a[i] << 1] --;
}
for (int i = 1 ; i <= l ; i ++) {
num[i] /= 2;
}
sum[0] = 0;
for (int i = 1 ; i <= l ; i ++) {
sum[i] = sum[i - 1] + num[i];
}
double ans = 0;
for (int i = 0 ; i < n ; i ++) {
ans += sum[l] - sum[a[i]];
ans -= n - 1;
ans -= (double)i * (n - 1 - i);
ans -= (double)(n - i - 1) * (n - i - 2) / 2;
}
printf ("%.7f\n", ans * 6.0 / n / (n - 1.0) / (n - 2.0));
}
return 0;
}