【CF148D】 Bag of mice (概率DP)
D. Bag of micetime limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
The dragon and the princess are arguing about what to do on the New Year's Eve. The dragon suggests flying to the mountains to watch fairies dancing in the moonlight, while the princess thinks they should just go to bed early. They are desperate to come to an amicable agreement, so they decide to leave this up to chance.
They take turns drawing a mouse from a bag which initially contains w white and b black mice. The person who is the first to draw a white mouse wins. After each mouse drawn by the dragon the rest of mice in the bag panic, and one of them jumps out of the bag itself (the princess draws her mice carefully and doesn't scare other mice). Princess draws first. What is the probability of the princess winning?
If there are no more mice in the bag and nobody has drawn a white mouse, the dragon wins. Mice which jump out of the bag themselves are not considered to be drawn (do not define the winner). Once a mouse has left the bag, it never returns to it. Every mouse is drawn from the bag with the same probability as every other one, and every mouse jumps out of the bag with the same probability as every other one.
InputThe only line of input data contains two integers w and b (0 ≤ w, b ≤ 1000).
OutputOutput the probability of the princess winning. The answer is considered to be correct if its absolute or relative error does not exceed 10 - 9.
Examplesinput1 3output0.500000000input5 5output0.658730159NoteLet's go through the first sample. The probability of the princess drawing a white mouse on her first turn and winning right away is 1/4. The probability of the dragon drawing a black mouse and not winning on his first turn is 3/4 * 2/3 = 1/2. After this there are two mice left in the bag — one black and one white; one of them jumps out, and the other is drawn by the princess on her second turn. If the princess' mouse is white, she wins (probability is 1/2 * 1/2 = 1/4), otherwise nobody gets the white mouse, so according to the rule the dragon wins.
【分析】
跟前面红黑那题差不多。
f[p][i][j]表示现在是p选,已经没了i个白,j个黑,p胜的概率。
然后随便转化一下就好?
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define Maxn 1010 double f[][Maxn][Maxn]; int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=n;i>=;i--)
for(int j=m;j>=;j--)
{
if(i==n) f[][i][j]=f[][i][j]=;
else if(j==m) f[][i][j]=f[][i][j]=;
else
{
if(j+==m) f[][i][j]=1.0*(n-i)/(n+m-i-j)+1.0*(m-j)/(n+m-i-j)*(-f[][i+][j+]);
else
{
f[][i][j]=1.0*(n-i)/(n+m-i-j)+1.0*(m-j)/(n+m-i-j)*((-f[][i+][j+])*(n-i)/(n+m-i-j-)+(-f[][i][j+])*(m-j-)/(n+m-i-j-));
}
f[][i][j]=1.0*(n-i)/(n+m-i-j)+1.0*(m-j)/(n+m-i-j)*(-f[][i][j+]);
}
}
printf("%.9lf\n",f[][][]);
return ;
}
2017-04-21 19:19:06
【CF148D】 Bag of mice (概率DP)的更多相关文章
- Codeforces 148D 一袋老鼠 Bag of mice | 概率DP 水题
除非特别忙,我接下来会尽可能翻译我做的每道CF题的题面! Codeforces 148D 一袋老鼠 Bag of mice | 概率DP 水题 题面 胡小兔和司公子都认为对方是垃圾. 为了决出谁才是垃 ...
- Codeforces Round #105 (Div. 2) D. Bag of mice 概率dp
题目链接: http://codeforces.com/problemset/problem/148/D D. Bag of mice time limit per test2 secondsmemo ...
- CF 148D Bag of mice 概率dp 难度:0
D. Bag of mice time limit per test 2 seconds memory limit per test 256 megabytes input standard inpu ...
- Bag of mice(概率DP)
Bag of mice CodeForces - 148D The dragon and the princess are arguing about what to do on the New Y ...
- codeforce 148D. Bag of mice[概率dp]
D. Bag of mice time limit per test 2 seconds memory limit per test 256 megabytes input standard inpu ...
- codeforces 148D Bag of mice(概率dp)
题意:给你w个白色小鼠和b个黑色小鼠,把他们放到袋子里,princess先取,dragon后取,princess取的时候从剩下的当当中任意取一个,dragon取得时候也是从剩下的时候任取一个,但是取完 ...
- Codeforces 148D Bag of mice 概率dp(水
题目链接:http://codeforces.com/problemset/problem/148/D 题意: 原来袋子里有w仅仅白鼠和b仅仅黑鼠 龙和王妃轮流从袋子里抓老鼠. 谁先抓到白色老师谁就赢 ...
- 抓老鼠 codeForce 148D - Bag of mice 概率DP
设dp[i][j]为有白老鼠i只,黑老鼠j只时轮到公主取时,公主赢的概率. 那么当i = 0 时,为0 当j = 0时,为1 公主可直接取出白老鼠一只赢的概率为i/(i+j) 公主取出了黑老鼠,龙必然 ...
- Codeforces Round #105 D. Bag of mice 概率dp
http://codeforces.com/contest/148/problem/D 题目意思是龙和公主轮流从袋子里抽老鼠.袋子里有白老师 W 仅仅.黑老师 D 仅仅.公主先抽,第一个抽出白老鼠的胜 ...
- CF148D Bag of mice (期望dp)
传送门 # 解题思路 ~~这怕是本蒟蒻第一个独立做出来的期望$dp$的题,发篇题解庆祝一下~~.首先,应该是能比较自然的想出状态设计$f[i][j][0/1]$ 表示当前还剩 $i$个白老鼠 ...
随机推荐
- JavaScript数组的概念
数组 1.数组是什么? 数组就是一组变量存放在里面就是数组. 例如:var list=['apple','goole','alibaba',520] (1.这些数据有一些相关性的. ( ...
- Python练习-函数版-锁定三次登陆失败的用户
代码如下: # 编辑者:闫龙 if __name__ == '__main__': import UserLoginFuncation LoclCount=[]; while True: UserNa ...
- 调整扩大VMDK格式VirtualBox磁盘空间
如果虚拟机的格式是VDI格式的, 那么可以通过这篇文章来调整磁盘大小: 调整Virtual Box硬盘大小 不过楼主当初在创建虚拟机的时候,是用的VMDK格式, 以求与VMWare的兼容性.这时候要扩 ...
- GreenTrend
ExpertforSQLServer(4.7.2)和ZhuanCloud(1.0.0)工具收集内容(在个人笔记本上测试) --SZC_Info.txt :: SQL专家云 v1. :: 开始收集 :: ...
- 安卓微信、QQ自带浏览器 UXSS 漏洞
安卓微信.QQ自带浏览器 UXSS 漏洞 注:PDF报告原文下载链接 Author: hei@knownsec.com Date: 2016-02-29 一.漏洞描述 在安卓平台上的微信及QQ自带浏览 ...
- Spark RDD 窄依赖研究
1.. 简介 spark从RDD依赖上来说分为窄依赖和宽依赖. 其中可以这样区分是哪种依赖:当父RDD的一个partition被子RDD的多个partitions引用到的时候则说明是宽依赖,否则为窄依 ...
- 签名DLL
签名DLL 首先需要一个密钥文件,后缀为.snk 密钥文件使用sn.exe 创建: sn.exe /k MySingInKey.snk sn.exe 工具的具体使用,可以通过 sn.exe /h 或 ...
- Postman接口&压力测试
Postman接口与压力测试实例 Postman是一款功能强大的网页调试与发送网页HTTP请求的Chrome插件.它提供功能强大的 Web API & HTTP 请求调试. 1.环境变量和全局 ...
- SqlServr性能优化性能之层次结构(十五)
1.添加根节点: hierarchyid GetRoot()方法 --创建数据库 create table Employeeh(EmployeeID int,Name varchar(500),Ma ...
- 如何解决vuex因浏览器刷新数据消失,保持数据持久化问题?
vuex的一个全局状态管理的插件,但是在浏览器刷新的时候,内存中的state会释放.通常的解决办法就是用本地存储的方式保存数据,然后再vuex初始化的时候再赋值给state,此过程有点麻烦.因此可以使 ...