CodeFoces 148D Bag of mice 题解：概率DP

概率 DP 比起期望 DP，可是容易多了～ CodeFoces 148D Bag of mice 题目链接

Problem

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 exceed $10^{- 9}$

Examples

Input #1

1 3

Output #1

0.500000000

Input #2

5 5

Output #2

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.

Translation

题目大意是：有一个公主和一条龙，龙想看美女跳舞但是公主想去睡觉，他们要“达成一项友好协议”，所以他们要玩个游戏来决定……（OI题目的套路）有一个有 w 只白鼠和 b 只黑鼠的袋子，他们要随机拿出老鼠，公主先拿，先拿出白鼠的获胜。每拿出一只老鼠，袋子里老鼠就会惊慌失措，就会有随机一只老鼠窜出来。如果他们都没拿到白鼠，就龙赢。问你最后公主赢的概率。

Solution

可以想到一个 DP：$F_{i,j}$ 表示袋子里还剩下 i 只白鼠和 j 只黑鼠，现在轮到公主拿，公主赢的概率。枚举当前状态，我们要考虑两种情况：

1. 公主拿的时候，一拿就拿到白鼠了，概率：i / (i+j)；
2. 公主没拿到白鼠，龙也没拿到白鼠。略微复杂，考虑跑出的老鼠的颜色。

Code

下面是我超级详细注释（废话连篇）的代码

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn=1005;
int n,m;
double f[maxn][maxn];
int main(){
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n;i++) f[i][0]=1.0; // 如果等到公主拿，只剩白鼠，那公主肯定赢
    for (int i=0;i<=n;i++){ // F[i][j]: 等到公主拿，袋子里还剩下i只白鼠j只黑鼠，公主赢的概率
        for (int j=1;j<=m;j++){ // 枚举黑鼠  drg(Dragon):当时龙赢的几率
            f[i][j]=(double)i/(i+j); // 考虑公主直接赢的情况
            double drg=(double)j*(double)(j-1)/(i+j)/(i+j-1); // 公主没有赢，龙继续拿也没有赢的情况
            if (j>=3) f[i][j]+=drg*(double)(j-2)/(i+j-2)*f[i][j-3]; // 跑出来的两只都是黑鼠，公主继续拿，赢了的概率
            if (i>=1&&j>=2) f[i][j]+=drg*(double)i/(i+j-2)*f[i-1][j-2]; // 跑出来一只黑一只白，...
        }
    }
    printf("%.10f\n",f[n][m]);
    return 0;
}