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 .

 

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 .

  简单dp一下解决。

 #include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int N=;
int n,a[N];
double dp[N],p[N],ans;
int main(){
scanf("%d",&n);
for(int i=;i<=n;i++)
scanf("%d",&a[i]);
sort(a+,a+n+);
for(int i=;i<=n;i++)
for(int j=;j<i;j++)
p[a[i]-a[j]]+=2.0/(n*(n-));
for(int i=;i<=;i++)
for(int j=;j<=;j++)
if(i+j<=)dp[i+j]+=p[i]*p[j];
for(int i=;i<=;i++)
for(int j=;j<i;j++)
ans+=p[i]*dp[j];
printf("%.10lf\n",ans);
return ;
}

数学(概率)CodeForces 626D:Jerry's Protest的更多相关文章

  1. Codeforces 626D Jerry's Protest(暴力枚举+概率)

    D. Jerry's Protest time limit per test:2 seconds memory limit per test:256 megabytes input:standard ...

  2. Codeforces 626D Jerry's Protest 「数学组合」「数学概率」

    题意: 一个袋子里装了n个球,每个球都有编号.甲乙二人从每次随机得从袋子里不放回的取出一个球,如果甲取出的球比乙取出的球编号大则甲胜,否则乙胜.保证球的编号xi各不相同.每轮比赛完了之后把取出的两球放 ...

  3. CodeForces 626D Jerry's Protest

    计算前两盘A赢,最后一盘B赢的情况下,B获得的球的值总和大于A获得的球总和值的概率. 存储每一对球的差值有几个,然后处理一下前缀和,暴力枚举就好了...... #include<cstdio&g ...

  4. 8VC Venture Cup 2016 - Elimination Round D. Jerry's Protest 暴力

    D. Jerry's Protest 题目连接: http://www.codeforces.com/contest/626/problem/D Description Andrew and Jerr ...

  5. Codeforces Round #370 (Div. 2) E. Memory and Casinos (数学&&概率&&线段树)

    题目链接: http://codeforces.com/contest/712/problem/E 题目大意: 一条直线上有n格,在第i格有pi的可能性向右走一格,1-pi的可能性向左走一格,有2中操 ...

  6. codeforces626D . Jerry's Protest (概率)

    Andrew and Jerry are playing a game with Harry as the scorekeeper. The game consists of three rounds ...

  7. codeforces 711E E. ZS and The Birthday Paradox(数学+概率)

    题目链接: E. ZS and The Birthday Paradox. time limit per test 2 seconds memory limit per test 256 megaby ...

  8. CodeForces 621C 数学概率期望计算

    昨天训练赛的题..比划了好久才想出来什么意思 之前想的是暴力for循环求出来然后储存数组 后来又想了想 自己萌的可以.. 思路就是求出来每个人与他的右边的人在一起能拿钱的概率(V(或)的关系)然后*2 ...

  9. 【数学】Codeforces 707C Pythagorean Triples

    题目链接: http://codeforces.com/problemset/problem/707/C 题目大意: 给你一个数,构造其余两个勾股数.任意一组答案即可,没法构造输出-1. 答案long ...

随机推荐

  1. 用Filezilla往ubuntu虚拟机上传文件

    也许不用这么复杂,但就这么干了 1.安卓ubuntu虚拟机 2.虚拟机安装ssh服务:sudo apt-get openssh-server 3.虚拟机新建目录test 4.修改test文件夹的访问权 ...

  2. Java获取项目路径

    参考博客.自己就不写了.我觉得他写得很详细 http://blog.csdn.net/hpf911/article/details/5852127

  3. pat_1009

    1009. 说反话 (20) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 8000 B 判题程序 Standard 作者 CHEN, Yue 给定一句英语,要求你编写程序,将句中 ...

  4. Java SE Java EE Java ME 的区别

    Java SE(Java Platform,Standard Edition) Java SE 以前称为 J2SE.它允许开发和部署在桌面.服务器.嵌入式环境和实时环境中使用的 Java 应用程序.J ...

  5. 邓白氏编码(duns number)申请入口的路径-苹果开发者申请必

    http://tieba.baidu.com/p/3861287522 这个网址有详细的介绍

  6. C++单元测试2

    这里再对上一篇<C++单元测试>进行技巧补充. 我们知道对动态链接库(lib和dll)的测试是比较简单的,我这里主要对需要注意的地方说明一下. 1.建议单独创建单元测试解决方案(不是创建项 ...

  7. 单页应用引擎的写法artTemplate

    使用到了ba-haschange.js <script src="../style/js/plugin/template-native-debug.js"></s ...

  8. 基础-函数3(IIFE立即执行函数)

    参考链接: http://benalman.com/news/2010/11/immediately-invoked-function-expression/#iife http://segmentf ...

  9. 好用的自适应表格插件-bootstrap table (支持固定表头)

    最近工作中找到了一款十分好用的表格插件,不但支持分页,样式,搜索,事件等等表格插件常有的功能外,最主要的就是他自带的冻结表头功能,让开发制作表格十分容易,不过网上大多都是英文文档,第一次使用会比较麻烦 ...

  10. ERROR 2003 (HY000): Can't connect to MySQL server

    http://blog.csdn.net/longxibendi/article/details/6363934 一.问题的提出 /usr/local/webserver/mysql/bin/mysq ...