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.

#include<cstdio>
#include<algorithm>
typedef double DB;
using namespace std;
const int N=;
double f[N][N];
bool vis[N][N];
int n,m;
double dfs(int w,int b){
if(w<=) return ;
if(b<=) return ;
if(vis[w][b]) return f[w][b];
vis[w][b]=;
double &res=f[w][b];
res=w*1.0/(w+b);
if(b>=){
double tmp=b*1.0/(w+b);
b--;
tmp*=b*1.0/(w+b);
b--;
//取完之后的发生概率:φ*(white+black)
res+=tmp*(w*1.0/(w+b)*dfs(w-,b)+b*1.0/(w+b)*dfs(w,b-));
}
return res;
}
int main(){
scanf("%d%d",&n,&m);
printf("%.9lf",dfs(n,m));
return ;
}

codeforce 148D. Bag of mice[概率dp]的更多相关文章

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  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(概率DP)

    Bag of mice  CodeForces - 148D The dragon and the princess are arguing about what to do on the New Y ...

  7. 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 ...

  8. 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 ...

  9. 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 ...

随机推荐

  1. Xcode使用介绍

    ///// 应用程序文件的组织 Product Name:项目名字 Organization Name:组织机构名称 Company Identifier:公司唯一标识符 Bundle Identif ...

  2. Linux进程同步机制

    为了能够有效的控制多个进程之间的沟通过程,保证沟通过程的有序和和谐,OS必须提供一定的同步机制保证进程之间不会自说自话而是有效的协同工作.比如在共享内存的通信方式中,两个或者多个进程都要对共享的内存进 ...

  3. axis client error Bad envelope tag: definitions

    http://blog.csdn.net/lifuxiangcaohui/article/details/8090503 ——————————————————————————————————————— ...

  4. TensorFlow基础笔记(15) 编译TensorFlow.so,提供给C++平台调用

    参考 http://blog.csdn.net/rockingdingo/article/details/75452711 https://www.cnblogs.com/hrlnw/p/700764 ...

  5. 第三百二十节,Django框架,生成二维码

    第三百二十节,Django框架,生成二维码 用Python来生成二维码,需要qrcode模块,qrcode模块依赖Image 模块,所以首先安装这两个模块 生成二维码保存图片在本地 import qr ...

  6. 转载:30多条mysql数据库优化方法,千万级数据库记录查询轻松解决

    1.对查询进行优化,应尽量避免全表扫描,首先应考虑在 where 及 order by 涉及的列上建立索引. 2.应尽量避免在 where 子句中对字段进行 null 值判断,否则将导致引擎放弃使用索 ...

  7. Unity3D - 详解Quaternion类(一)

    一.简介 Quaternion又称四元数,由x,y,z和w这四个分量组成,是由爱尔兰数学家威廉·卢云·哈密顿在1843年发现的数学概念.四元数的乘法不符合交换律.从明确地角度而言,四元数是复数的不可交 ...

  8. make的自动变量和预定义变量

    make的自动变量 $@ 规则目标的文件名.如果目标是档案文件的一个成员,"$@"就是档案文件的名称 $% 当目标是档案文件的一个成员时,"$%"是该成员的名称 ...

  9. CS文件类头注释

    1.修改unity生成CS文件的模板(模板位置:Unity\Editor\Data\Resources\ScriptTemplates 文件名:81-C# Script-NewBehaviourScr ...

  10. 详解MathType中如何插入特殊符号

    在论文写作中,经常会用到一些特殊符号,MathType公式编辑器支持插入特殊符号,并且数量繁多,可以满足用户的需求.本教程将详解MathType如何插入特殊符号. MathType中插入特殊符号的操作 ...