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. 认识javascript作用域

    JavaScript的作用域链 这是一个非常重要的知识点了,了解了JavaScript的作用域链的话,能帮助我们理解很多‘异常’问题. 下面我们来看一个小例子,前面我说过的声明提前的例子. var n ...

  2. hibernate使用sql语句查询实体时,要写上addEntity

    abDAO.getSession().createSQLQuery(hql).addEntity(对象.class).list(); 参考http://blog.csdn.net/vacblog/ar ...

  3. 4 C#和Java 的比较

    2007年11月1日    1.访问控制方面:C#有public.internal.protected.private,比java多了个internal,其实它跟java的包访问差不多,interna ...

  4. html5 + css3 + zepto.js实现的微信广告宣传页

    最新学习html5 + css3, 参考微信的一个推广页写出一个实例巩固自己知识,自己已经将原实例打包到自己博客文件当中,但是不知道如何提供下载,如有需要的朋友可以联系我qq309666726

  5. java基础(死循环退出选项)

    java程序中为了程序正常运行,需要给无限循环加入一个退出选项,保证程序的可执行性. import java.util.Scanner; public class { public static vo ...

  6. POJ 3181 Dollar Dayz(高精度 动态规划)

    题目链接:http://poj.org/problem?id=3181 题目大意:用1,2...K元的硬币,凑成N元的方案数. Sample Input 5 3 Sample Output 5 分析: ...

  7. 流形(Manifold)初步【转】

    转载自:http://blog.csdn.net/wangxiaojun911/article/details/17076465 欧几里得几何学(Euclidean Geometry) 两千三百年前, ...

  8. windows8一直更新不了的问题————解决方案

    以下是微软官方工程师的详细解答: 尊敬的佐先生: 您好! 感谢您联系微软技术支持!我是微软技术支持工程师,我姓张.我将协助您解决有关问题.您的问题编号是SRX 1274238225 对于您当前的更新问 ...

  9. bootstrap中的动态加载出来的图片轮播中的li标签中的class="active"的动态添加移除

    //该方法是在slide改变时立即触发该事件, $('#myCarousel').on('slide.bs.carousel', function () { $("#myCarousel o ...

  10. VC杂记

    获得Combobox的状态:向ComboBox发送CB_GETDROPPEDSTATE消息. 格式化字串:char buff[10] ; sprintf(buff,"1+1=%d" ...