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.

題目大意：

一個袋子裡有w個白老鼠，b個黑老鼠，王子和龍依次取。王子先取，先取到白老鼠的為勝者，當中龍取老鼠的時候。取出一僅僅後。會有隨機的一僅僅老鼠跑出來，並且取老鼠的時候，每僅僅老鼠取到的概率是一樣的，跑出來的概率也是一樣的。  讓你算王子贏的概率。

思路：

概率DP，用DP[i][j] 表示 白老鼠為i僅僅，黑老鼠為j僅僅時，王子贏的概率,有兩個子狀態。一個是王子立刻就贏 還有就是王子這把不贏。這把不贏的情況還有兩種子情況（保證龍輸），一種就是取到一僅僅黑老鼠，跑出一僅僅黑老鼠。另一種就是取到一僅僅黑老鼠，跑出一僅僅白老鼠。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int maxn = 1000+10;
double dp[maxn][maxn];
bool vis[maxn][maxn];
int w,b;
double dfs(int w,int b){
    if(w==0) return 0.0;
    if(b==0) return 1.0;
    if(vis[w][b]) return dp[w][b];
    vis[w][b] = 1;
    double res = w*1.0/(w+b);
    if(b>=3)
        res += (b*1.0/(w+b))*((b-1)*1.0/(w+b-1))*((b-2)*1.0/(w+b-2))*dfs(w,b-3);
    if(b>=2&&w>=1)
        res += (b*1.0/(w+b))*((b-1)*1.0/(w+b-1))*(w*1.0/(w+b-2))*dfs(w-1,b-2);

    return dp[w][b] = res;
}
int main(){
    memset(vis,0,sizeof vis);
    while(~scanf("%d%d",&w,&b)){
        printf("%.9lf\n",dfs(w,b));
    }
    return 0;
}