CF_148D_BagOfMice

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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 andb (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.

第一道概率

概率总体感觉理解上比期望难度要小些。

毕竟这个东西是正向的

这个题目状态是一个二维的，因为涉及了黑猫白猫

dp[i][j]表示的是i白j黑的时候公主赢比赛的概率

显然dp[0][j]=0,dp[i][0]=1（dp[0][0]=0）

而这也正是需要初始化的条件

剩下的无非就是公主拿白赢

公主黑，龙（dragon其实也可以翻译凶恶的人……）拿白赢

公主黑，龙黑，跑黑状态转移

公主黑，龙黑，跑白状态转移

其中只要算公主有机会赢得状态转移就可以了，注意下后两种情况的黑鼠数目关系

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;const int M=1005;double p[M][M];int main(){    int w,b;    while(scanf("%d%d",&w,&b)!=EOF)    {        memset(p,0,sizeof(p));        for(int i=1;i<=w;i++)            p[i][0]=1;        for(int i=1;i<=w;i++)            for(int j=1;j<=b;j++)            {                p[i][j]+=(double)i/(i+j);//公主抓白赢                if(j>=3)                    p[i][j]+=(double)j/(i+j)*(j-1)/(i+j-1)*(j-2)/(i+j-2)*p[i][j-3];                    //公主黑，龙黑跑黑                if(j>=2)                    p[i][j]+=(double)j/(i+j)*(j-1)/(i+j-1)*i/(i+j-2)*p[i-1][j-2];                    //公主黑，龙黑跑白            }        printf("%.9lf\n",p[w][b]);    }    return 0;}

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