http://codeforces.com/problemset/problem/148/D

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.

Sample test(s)
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

/*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。
王妃每次抓一仅仅老鼠,龙每次抓完一仅仅老鼠之后会有一仅仅老鼠跑出来。
每次抓老鼠和跑出来的老鼠都是随机的。
如果两个人都没有抓到白色老鼠则龙赢。 王妃先抓。
问王妃赢的概率。 分析:如果dp[i][j]表示轮到王妃抓老鼠时面对剩余i仅仅白鼠和j仅仅黑鼠的胜率
则dp[i][j]能够转化到下面四种情况:
1.王妃胜利,转化概率为i/(i+j)
2.dp[i-1][j-2]---王妃抓黑鼠,龙抓黑鼠,逃跑白鼠,转化概率是j/(i+j) * (j-1)/(i+j-1) * i/(i+j-2)
3.dp[i-1][j-1]---王妃抓到黑鼠,龙抓到白鼠,输! ,转化概率为j/(i+j) * i/(i+j-1)//这不能到达,到达就输了
4.dp[i][j-3]--王妃抓到黑鼠,龙抓到黑鼠,逃跑黑鼠,转化率为j/(i+j) * (j-1)/(i+j-1) * (j-2)/(i+j-2)
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <iomanip>
#define INF 999999999
typedef long long LL;
using namespace std; const int MAX=1000+10;
int w,b;
double dp[MAX][MAX]; int main(){
while(cin>>w>>b){
for(int i=1;i<=w;++i)dp[i][0]=1;//有白鼠无黑鼠胜率为1
for(int i=0;i<=b;++i)dp[0][i]=0;//无白鼠胜率为0
for(int i=1;i<=w;++i){
for(int j=1;j<=b;++j){
dp[i][j]=i*1.0/(i+j);
//dp[i][j]+=j*1.0/(i+j) * i*1.0/(i+j-1) * dp[i-1][j-1];
if(j>=2)dp[i][j]+=j*1.0/(i+j) * (j-1)*1.0/(i+j-1) * i*1.0/(i+j-2) * dp[i-1][j-2];
if(j>=3)dp[i][j]+=j*1.0/(i+j) * (j-1)*1.0/(i+j-1) * (j-2)*1.0/(i+j-2) * dp[i][j-3];
}
}
printf("%.9f\n",dp[w][b]);
}
return 0;
}

codeforces 148D之概率DP的更多相关文章

  1. CodeForces 602E【概率DP】【树状数组优化】

    题意:有n个人进行m次比赛,每次比赛有一个排名,最后的排名是把所有排名都加起来然后找到比自己的分数绝对小的人数加一就是最终排名. 给了其中一个人的所有比赛的名次.求这个人最终排名的期望. 思路: 渣渣 ...

  2. codeforces 696C PLEASE 概率dp+公式递推+费马小定理

    题意:有3个杯子,排放一行,刚开始钥匙在中间的杯子,每次操作,将左右两边任意一个杯子进行交换,问n次操作后钥匙在中间杯子的概率 分析:考虑动态规划做法,dp[i]代表i次操作后的,钥匙在中间的概率,由 ...

  3. Codeforces 229E Gifts 概率dp (看题解)

    Gifts 感觉题解写的就是坨不知道什么东西.. 看得这个题解. #include<bits/stdc++.h> #define LL long long #define LD long ...

  4. 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 ...

  5. 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 ...

  6. codeforces 148D 概率DP

    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. 王妃每次抓一仅仅老鼠,龙每次抓完一仅仅老鼠之后会有一仅仅老鼠跑出来. 每次抓老鼠和跑出来的老鼠都是随 ...

  7. Codeforces #548 (Div2) - D.Steps to One(概率dp+数论)

    Problem   Codeforces #548 (Div2) - D.Steps to One Time Limit: 2000 mSec Problem Description Input Th ...

  8. Codeforces Round #301 (Div. 2) D. Bad Luck Island 概率DP

    D. Bad Luck Island Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/540/pr ...

  9. codeforces 768 D. Jon and Orbs(概率dp)

    题目链接:http://codeforces.com/contest/768/problem/D 题意:一共有k种球,要得到k种不同的球至少一个,q个提问每次提问给出一个数pi,问概率大小大于等于pi ...

随机推荐

  1. [Java] 字符流Reader,读取字符数据

    package test.stream; import java.io.BufferedReader; import java.io.FileNotFoundException; import jav ...

  2. [Linux]常用命令与目录全拼

    命令缩写: ls:list(列出目录内容)cd:Change Directory(改变目录)su:switch user 切换用户rpm:redhat package manager 红帽子打包管理器 ...

  3. crontab Linux定时器工具

    要使用crontab定时器工具,必须要启动cron服务: service cron start crontab的语法,以备日后救急.先上张超给力的图: crontab各参数说明: -e : 执行文字编 ...

  4. Vue基础开发笔记

    以下实例代码地址:https://github.com/NewBLife/VueDev 1,Vue组件导入 新建组件:Header.vue <template> <div> & ...

  5. java基础 (三)之ConcurrentHashMap(转)

    一.背景: 线程不安全的HashMap     因为多线程环境下,使用Hashmap进行put操作会引起死循环,导致CPU利用率接近100%,所以在并发情况下不能使用HashMap.   效率低下的H ...

  6. css3---2D效果 ---3D效果

    CSS3边框: CSS3圆角:border-radius(**px 或 **%) 属性——创建边框线的圆角 CSS3盒子阴影:box-shadow属性——创建阴影 box-shadow:30px 0p ...

  7. 【转载】 5G+边缘计算,着眼可见的未来 【边缘计算】

    原文地址: https://www.cnblogs.com/upyun/p/10641489.html ------------------------------------------------ ...

  8. gentoo wireshark 安装

    安装软件 emerge --ask net-analyzer/wireshark 把用户加入 wireshark 组. gpasswd -a $USER wireshark 如果不像重新登录就可以使用 ...

  9. Java编程的逻辑 (1) - 数据和变量

    ​本系列文章经补充和完善,已修订整理成书<Java编程的逻辑>,由机械工业出版社华章分社出版,于2018年1月上市热销,读者好评如潮!各大网店和书店有售,欢迎购买,京东自营链接:http: ...

  10. [dts]TI-am437x dts

    imx6 可以参考http://blog.csdn.net/shengzhadon/article/details/49908439 参照文件: Documentation/devicetree/bi ...