CodeForces - 148D Bag of mice 概率DP

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 bblack 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.

Example

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.

题目大意：说公主和龙在玩一个游戏。这个游戏是说在一个盒子中中有w只白色老鼠，b只黑色老鼠。

两人轮流从里面随机取出一只老鼠。谁第一个取出白色老鼠谁就赢得了最终的胜利。

如果说最后盒子中的老鼠数目为零并且没有任何人拿到过白色老鼠，那么龙获胜。

问最后公主取得胜利的概率是多少。。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <map>
#include <set>
#include <sstream>
#include <queue>
#include <stack>
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a));
#define For(a,b) for(int i = a;i<b;i++)
#define ll long long
#define MAX_N 100010

using namespace std;

double dp[1005][1005];
int main()
{
int n,m;
while(~scanf("%d %d",&n,&m))
{
mem(dp,0);
// 没有黑鼠的时候胜率为1
for(int i = 1; i<=n; i++)
{
dp[i][0] = 1.0;
}
// 没有白鼠的时候胜率为0
for(int i = 1; i<=m; i++)
{
dp[0][i] = 0;
}
for(int i = 1; i<=n; i++)
{
for(int j = 1; j<=m; j++)
{
dp[i][j] += (double)i / (i+j); // 第一次就胜利
// 黑 黑 跑出黑鼠
if(j>=3)
dp[i][j] += (double)j/(j+i)*(double)(j-1)/(j+i-1)*(double)(j-2)/(j+i-2)*dp[i][j-3];
// 黑 黑 跑出白鼠
if(j>=2)
dp[i][j] += (double)j/(j+i)*(double)(j-1)/(j+i-1)*(double)i/(i+j-2)*dp[i-1][j-2];
}
}
printf("%.9f\n",dp
[m]);
}
return 0;
}