Bestcoder#5 1003

Poor RukawTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24    Accepted Submission(s): 11

Problem Description

Last time, Hanamichi lost the basketball battle between him and Rukaw. So these days he had a burning desire to wreak revenge. So he invented a new game and asked Rukaw to play with him. What’s more, the loser should confess his ignorance and on the other hand the winner can go on a trip with Haruko.

Hanamichi knows the game well (as he invented it), and he always chooses the best strategy, however Rukaw is not so willing to play this game. He just wants it to be finished as soon as possible so that he can go to play basketball. So he decides not to think about the best strategy and play casually.

The game’s rules are here. At first, there are N numbers on the table. And the game consists of N rounds. Each round has a score which is the number of the numbers still left on the table. And Each round there will be one number to be removed from the table. In each round, two players take turns to play with these numbers.To be fair, Rukaw plays first in the first round. If there’s more than 1 numbers on the table, players can choose any two numbers they like and change them to a number abs(x-y). This round ends when there’s only one number left on the table, and if this number is an odd number, Rukaw wins, otherwise Hanamichi wins. The score of this round will be add to the winner. After that, all numbers will be recovered to the state when this round starts. And the loser of this round has the right to remove one number and he also has the right to play first in the next round. Then they use the remaining numbers to start next round. After N rounds, all numbers removed and this game ends. The person who has more scores wins the whole game.

As you know, Rukaw has already decided to play casually, that is to say, in his turn, he chooses numbers randomly, each numbers left on the table has the same possibility to be chosen. When a round ends, if Rukaw is the loser, he also randomly chooses a number to remove. And Hanamichi will always choose numbers or remove numbers to maxmium his final total score. Here comes the question:
Given the N numbers on the table at the beginning, can you calculate the expectation of the final score of Hanamichi. (We don’t care about who wins the whole game at all.)

Input

This problem contains multiple tests.
In the first line there’s one number T (1 ≤ T ≤ 200) which tells the total number of test cases. For each test case, there a integer N (1 ≤ N ≤ 1000) in the first line, and there are N intergers Ai , i = 1, 2, … , N (1 ≤ Ai ≤ 100000), in the second line which are the numbers at the beginning.

Output

This problem is intended to use special judge. But so far, BestCoder doesn’t support special judge. So you should output your answer in the following way. If the expectation you get is X, output \([3\times X+0.5]\) in a line. Here, [A] means the largest integer which is no more than A.

Sample Input

222 421 2

Sample Output

93

Hint

In the first example, Hanamichi will always win two rounds and the score of two rounds will be 2 and 1. So the answer is 3. (And you should output 9.) In the second example, Rukaw wins the first round. And after that Hanamichi has the right to choose a number from 1 and 2 to remove. (Because this round started with this two numbers.) And We know that he will choose 1 to maximum his final total score. So when the second round starts, there’s only one number 2 left on the table and Hanamichi plays first. He immediately wins this round and got 1 point. Then the game ends. So the answer is 1.(And you should output 3.)

错误点:概率dp,因为cnt没初始化,RE

思路:想到了奇偶关系以及转化关系,但是没有想到如何处理那么多状态,其实后面的状态是重复的,所以dp从后向前考虑就可以得到答案。。是不是概率dp都是这个思路??

 #include <vector>

 #include <cstdio>

 #include <cstring>

 #include <iostream>

 #include <algorithm>

 using namespace std;

 const int N = ;

 double func[][];

 int main()

 {

     func[][] = ;

     for(int i = ;i<N;i++)

         for(int j = ;j<=i;j++)

     {

         if(j%==)

         {

             func[i][j] = func[i-][j-];

         }

         else

         {

             func[i][j] = i+func[i-][j]*(i-j)*1.0/i+func[i-][j-]*j*1.0/i;

         }

     }

     int T,n,cnt=;

     scanf("%d",&T);

     while(T--)

     {

         cnt = ;

         scanf("%d",&n);

         int a;

         for(int i = ;i<n;i++)

         {

             scanf("%d",&a);

             if(a%==)

                 cnt++;

         }

         //cout<<n<<' '<<cnt<<endl;

         int ans = *func[n][cnt]+0.5;

         printf("%d\n",ans);

     }

     return ;

 }

Bestcoder#5 1003的更多相关文章

  1. HDU 4859(Bestcoder #1 1003)海岸线(网络流之最小割)

    题目地址:HDU4859 做了做杭电多校,知识点会的太少了.还是将重点放在刷专题补知识点上吧,明年的多校才是重点. 这题题目求的最长周长.能够试想一下,这里的海岸线一定是在"."和 ...

  2. HDU 5682/BestCoder Round #83 1003 zxa and leaf 二分+树

    zxa and leaf Problem Description zxa have an unrooted tree with n nodes, including (n−1) undirected ...

  3. BestCoder 1st Anniversary($) 1003 Sequence

    题目传送门 /* 官方题解: 这个题看上去是一个贪心, 但是这个贪心显然是错的. 事实上这道题目很简单, 先判断1个是否可以, 然后判断2个是否可以. 之后找到最小的k(k>2), 使得(m-k ...

  4. 从lca到树链剖分 bestcoder round#45 1003

    bestcoder round#45 1003 题,给定两个点,要我们求这两个点的树上路径所经过的点的权值是否出现过奇数次.如果是一般人,那么就是用lca求树上路径,然后判断是否出现过奇数次(用异或) ...

  5. BestCoder Round #81 (div.2) 1003 String

    题目地址:http://bestcoder.hdu.edu.cn/contests/contest_showproblem.php?cid=691&pid=1003题意:找出一个字符串满足至少 ...

  6. BestCoder Round #29 1003 (hdu 5172) GTY's gay friends [线段树 判不同 预处理 好题]

    传送门 GTY's gay friends Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Ot ...

  7. DP BestCoder Round #50 (div.2) 1003 The mook jong

    题目传送门 /* DP:这题赤裸裸的dp,dp[i][1/0]表示第i块板放木桩和不放木桩的方案数.状态转移方程: dp[i][1] = dp[i-3][1] + dp[i-3][0] + 1; dp ...

  8. BestCoder Round #50 (div.1) 1003 The mook jong (HDU OJ 5366) 规律递推

    题目:Click here 题意:bestcoder 上面有中文题目 分析:令f[i]为最后一个木人桩摆放在i位置的方案,令s[i]为f[i]的前缀和.很容易就能想到f[i]=s[i-3]+1,s[i ...

  9. BestCoder Round #56 1002 Clarke and problem 1003 Clarke and puzzle (dp,二维bit或线段树)

    今天第二次做BC,不习惯hdu的oj,CE过2次... 1002 Clarke and problem 和Codeforces Round #319 (Div. 2) B Modulo Sum思路差不 ...

随机推荐

  1. cefsharp设置默认语言

    cefsharp是不错的浏览器内核封装版本之一,默认语言是en-US,这个一直困扰着项目,项目好多处需修改,后来经多次尝试,才发现,原来设置默认语言这么简单. CefSharp.Settings se ...

  2. Oracle约束(Constraint)详解

    概述 约束是数据库用来确保数据满足业务规则的手段,不过在真正的企业开发中,除了主键约束这类具有强需求的约束,像外键约束,检查约束更多时候仅仅出现在数据库设计阶段,真实环境却很少应用,更多是放到程序逻辑 ...

  3. Linux吃掉我的内存

    在Windows下资源管理器查看内存使用的情况,如果使用率达到80%以上,再运行大程序就能感觉到系统不流畅了,因为在内存紧缺的情况下使用交换分区,频繁地从磁盘上换入换出页会极大地影响系统的性能.而当我 ...

  4. postman使用之三:API请求和查看响应结果

    请求 postman支持很多请求类型,界面左侧可以看到请求类型:get.post.put.patch等,右侧是发送和保存按钮,下方是请求支持的认证方式.信息头.信息体.私有脚本和测试结果.下面我们介绍 ...

  5. 卡通图像变形算法(Moving Least Squares)附源码

    本文介绍一种利用移动最小二乘法来实现图像变形的方法,该方法由用户指定图像中的控制点,并通过拖拽控制点来驱动图像变形.假设p为原图像中控制点的位置,q为拖拽后控制点的位置,我们利用移动最小二乘法来为原图 ...

  6. POJ3928Ping pong[树状数组 仿逆序对]

    Ping pong Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3109   Accepted: 1148 Descrip ...

  7. 一段拼装sql的小代码

    /** * 单表查询 * * className:实体类的名字 * vals:查询的属性 * pNames:条件的名字 * pVals:条件的值 */ @Override public List< ...

  8. mysql表复制和修改部分字段

    今天在工作中,需要造大量的加数据,1000多条数据如果都是手工输入的话,那么我今天不要干别的了,就造吧! 当时手工操作重复的事情,对程序员来说,是一件很丢人的事情,所以就上网查了一下,需要用到两个知识 ...

  9. reveal

    链接 界面调试工具Reveal Reveal使用教程 iOS分析UI利器——Reveal及简单破解方法 Reveal使用步骤和 破解Revealapp的试用时间限制 end

  10. Github 使用

    创建repository 可以在Github上无限制使用public repository进行源代码管理,创建一个repository很简单,不多说了. 获取代码到本地 首先要安装Git,然后使用命令 ...