CodeForces 626D Jerry's Protest+思维题

Description

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 .

Sample Input

Input

2

1 2

Output

0.0000000000

Input

3

1 2 10

Output

0.0740740741

Hint

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 won 2 - 1 in both of the first two rounds, and Jerry drew the 10 in the last round. This has probability .

题意：给你n个数，，两人比赛，，每次取两个求。前两场A胜利，最后一场B胜利，，但有3场比赛B的数目总和比A的总和大的可能性，求这个可能性的概率。。

first[i]表示一场比赛差值为i的样本数，second[i]表示两场比赛差值为i的样本数

首先我们先把一场比赛所有可能的差的个数记录下来。。first[a[j]-a[i]]++;那么两场比赛的总和的差，也就可以用相同的方法表示出来了，second[i+j]+=first[i]*first[j];，，2+4=3+3，所以second是累加的

所以第三场比赛只要找出一场比赛的结果，大于前两场和的情况数。。sum+=first[i]*second[j];这里i>j

样本空间总数=3次 从n个数里任取两个。。。

注意的是first，second可能要炸int…（wa了几发）

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<queue>
#include<vector>
#include<map>
#include<stack>
#include<set>
#define pi acos(-1.0)
#define EPS 1e-6    //log(x)
#define e exp(1.0); //2.718281828
#define mod 1000000007
#define INF 0x7fffffff
#define inf 0x3f3f3f3f
#pragma comment(linker,"/STACK:102400000,102400000")
typedef long long LL;
using namespace std;
int main()
{
int n;
int a[2222]={0};
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
long long int first[10005],second[10005];
memset(first,0,sizeof(first));
memset(second,0,sizeof(second));
sort(a+1,a+n+1);
for(int i=1;i<=n;i++){  //差值为a[j]-a[i]的个数
for(int j=i+1;j<=n;j++){
first[a[j]-a[i]]++;
}
}
for(int i=1;i<=5555;i++){   //第二次差值是两次值的乘积和
for(int j=1;j<=5555;j++){   //多种可能性为i+j的值
second[i+j]+=first[i]*first[j];
}
}
long long int sum=0;
for(int i=0;i<5001;i++){  //第三次的值大于请两次和的次数
for(int j=0;j<i;j++){
sum+=first[i]*second[j];
}
}
long long int mu=n*(n-1)/2;
long long int m=mu*mu*mu; //总的可能性次数。3次。n里任取2个
double out=double(sum)/m;
printf("%.10f\n",out);
return 0;
}

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