D. Bag of mice
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard 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.

Input

The only line of input data contains two integers w and b (0 ≤ w, b ≤ 1000).

Output

Output the probability of the princess winning. The answer is considered to be correct if its absolute or relative error does not exceed10 - 9.

Examples
input
1 3
output
0.500000000
input
5 5
output
0.658730159
Note

Let'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.

大佬题解:http://blog.csdn.net/swust_Three/article/details/68941926

#include<cstdio>
using namespace std;
double dp[][];
int main()
{
int w,b;
scanf("%d%d",&w,&b);
for(int i=;i<=w;i++) dp[i][]=1.0;
for(int i=;i<=w;i++)
for(int j=;j<=b;j++)
{
dp[i][j]=1.0*i/(i+j);
if(j>=) dp[i][j]+=1.0*j/(i+j)*(j-)/(i+j-)*(j-)/(i+j-)*dp[i][j-];
if(j>=) dp[i][j]+=1.0*j/(i+j)*(j-)/(i+j-)*i/(i+j-)*dp[i-][j-];
}
printf("%.9lf",dp[w][b]);
}

Codeforces 148 D Bag of mice的更多相关文章

  1. CodeForces:148D-D.Bag of mice

    Bag of mice time limit per test 2 seconds memory limit per test 256 megabytes Program Description Th ...

  2. 【codeforces 148D】 Bag of mice

    http://codeforces.com/problemset/problem/148/D (题目链接) 题意 包中有w个白鼠,b个黑鼠.公主和龙轮流画老鼠,公主先画,谁先画到白鼠谁就赢.龙每画完一 ...

  3. 【Codeforces 105D】 Bag of mice

    [题目链接] http://codeforces.com/contest/148/problem/D [算法] 概率DP f[w][b]表示还剩w只白老鼠,b只黑老鼠,公主胜利的概率,那么 : 1. ...

  4. 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 ...

  5. 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 ...

  6. Bag of mice(CodeForces 148D )

    D. Bag of mice time limit per test 2 seconds memory limit per test 256 megabytes input standard inpu ...

  7. 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 ...

  8. CF 148D. Bag of mice (可能性DP)

    D. Bag of mice time limit per test 2 seconds memory limit per test 256 megabytes input standard inpu ...

  9. 【CF148D】 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 ...

随机推荐

  1. Thunder——互评beta版本

    基于NABCD和spec评论作品 Hello World!:http://www.cnblogs.com/vector121/p/7922989.html 欢迎来怼:http://www.cnblog ...

  2. SQL连接的方法

    1.创建连接字符串 string con = "Data Source=127.0.0.1;Initial Catalog=dingdan;Integrated Security=True& ...

  3. 类的static成员变量和成员函数能被继承吗

    1.   父类的static变量和函数在派生类中依然可用,但是受访问性控制(比如,父类的private域中的就不可访问),而且对static变量来说,派生类和父类中的static变量是共用空间的,这点 ...

  4. 1st 结对编程:简易四则运算

    结对编程:简易四则运算 功能:进行简易的四则运算,并根据给出的结果判断正误. 实现:使用java的图形化界面实现. 导入包库 package six; import javax.swing.*; im ...

  5. PHP 多文件打包下载 zip

    <?php $zipname = './photo.zip'; //服务器根目录下有文件夹public,其中包含三个文件img1.jpg, img2.jpg, img3.jpg,将这三个文件打包 ...

  6. 虚拟机下安装CentOS6.5系统教程

    虚拟机下安装CentOS6.5系统教程 时间:2014-12-09 01:40来源:linuxdown.net 作者:linuxdown.net 举报 点击:15315次 其实通过VM安装虚拟机还是蛮 ...

  7. sublime插件时间

    import datetime import sublime_plugin class AddCurrentTimeCommand(sublime_plugin.TextCommand): def r ...

  8. PHP中与类有关的几个魔术常量

    与类有关的魔术常量: 以前学过的魔术常量: __FILE__ __DIR__ __LINE__ 现在: __CLASS__: 代表当前其所在的类的类名: __METHOD__:代表其当前所在的方法名:

  9. DNS原理及解析过程

    本文主要参考自:http://369369.blog.51cto.com/319630/812889 并做了小幅修改 什么是DNS? 因特网上的主机和人类一样,也可以使用多种方式进行识别.主机的一种识 ...

  10. python while 学习

    while True: reply = input('please input:') if reply == 'stop': break else: print (reply.upper())