Jerry's Protest
Andrew and Jerry are playing a game with Harry as the scorekeeper. The game consists of three rounds. In each round, Andrew and Jerry draw randomly without replacement from a jar containing n balls, each labeled with a distinct positive integer. Without looking, they hand their balls to Harry, who awards the point to the player with the larger number and returns the balls to the jar. The winner of the game is the one who wins at least two of the three rounds.
Andrew wins rounds 1 and 2 while Jerry wins round 3, so Andrew wins the game. However, Jerry is unhappy with this system, claiming that he will often lose the match despite having the higher overall total. What is the probability that the sum of the three balls Jerry drew is strictly higher than the sum of the three balls Andrew drew?
Input
The first line of input contains a single integer n (2 ≤ n ≤ 2000) — the number of balls in the jar.
The second line contains n integers ai (1 ≤ ai ≤ 5000) — the number written on the ith ball. It is guaranteed that no two balls have the same number.
Output
Print a single real value — the probability that Jerry has a higher total, given that Andrew wins the first two rounds and Jerry wins the third. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if 
.
.
Examples
input
2 1 2
output
0.0000000000
input
3 1 2 10
output
0.0740740741
Note
In the first case, there are only two balls. In the first two rounds, Andrew must have drawn the 2 and Jerry must have drawn the 1, and vice versa in the final round. Thus, Andrew's sum is 5 and Jerry's sum is 4, so Jerry never has a higher total.
In the second case, each game could've had three outcomes — 10 - 2, 10 - 1, or 2 - 1. Jerry has a higher total if and only if Andrew won2 - 1 in both of the first two rounds, and Jerry drew the 10 in the last round. This has probability 
.
.
---------------------------------EDITORIAL-------------------------------
this problem is some what related to the problem not a triangle in which we have given a set of edge and we need to find maximu number of triangle we can form ,....
now coming to this problem , in each turn both player can pic two different balls, so there are bascically nc2 ways to pic the balls ,
suppuose in turn 1 player 1 lead by value x1 and in turn 2 player 1 lead by value x2 than in next turn player 2 have to gain by a value >x1+x2,
in the last turn game if playe 1 pic c1 and player pic c2 than
c2-c1> x1+x2,
if initilly compute all possible pairs difference than now we just need to find all possible triplets a,b,c such tha sum of 2 must be < than the sum of the third ,
but number of triplets will be n*n ie 10^6, now it is impossible to fixx all possible pair a , b and find c such that
a+b<c,
but main observation in the problem is that values on each ball could be maximum 5000, means there can be maximum 5000 difference , and we can easily iterate through all there to make pair a ,b and find c such that a+b<c, special care should be need for the while fixing a and b because , supopose i am fixing a as 5 than there can be many 5 so count of number of times of occurance must multiplied.......
---------------------------------code-----------------------------------------
#include<bits/stdc++.h>
using namespace std;
typedef long long int lli ;
lli has[1000000];
int arr[1000000];
int main()
{
int t;
int n;
cin>>n;
for(int i=0;i<n;i++)
{
cin>>arr[i];
}
set<int> s;
map<int,int> ma;
for(int i=0;i<n;i++)
{
for(int j=i+1;j<n;j++)
{
s.insert(abs(arr[i]-arr[j]));
has[abs(arr[i]-arr[j])]++;// will be use for creating a forward array
ma[abs(arr[i]-arr[j])]++;
}
}
for(int i=1;i<=10010;i++)
{
has[i]=has[i-1]+has[i];
}
vector<int> dis;
set<int>:: iterator it;
for(it=s.begin();it!=s.end();it++)
{
dis.push_back(*it);
}
int siz=dis.size();
unsigned long long int ans=0;
for(int i=0;i<siz;i++)
{
for(int j=0;j<siz;j++ )
{
int len1=dis[i];
int len2=dis[j];
int val=len1+len2;
ans+=(has[10010]-has[val])*ma[len1]*ma[len2];
}
}
double ff=(double) ans;
ff=ff/(double)((n*(n-1))/2);
ff=ff/(double)((n*(n-1))/2);
ff=ff/(double)((n*(n-1))/2);
printf("%.12lf",ff);
}
