CF 148D Bag of mice【概率DP】

D. Bag of mice

time limit per test

2 seconds

memory limit per test

256 megabytes

Promblem description

 

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 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.

很有趣的概率DP，但是我一开始没有向概率的方向想，妄图找出规律，还是我太年轻了。

为什么要用DP呢，我想是这个题的前后状态是有关联的，后面的状态可以转移成前面的状态，这就是精髓所在吧。

#include<bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define ll long long
#define inf 1000000000
#define maxn 1005
#define maxm 100005
#define eps 1e-10
#define for0(i,n) for(int i=1;i<=(n);++i)
#define for1(i,n) for(int i=1;i<=(n);++i)
#define for2(i,x,y) for(int i=(x);i<=(y);++i)
#define for3(i,x,y) for(int i=(x);i>=(y);--i)
#define mod 1000000007
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') {x=10*x+ch-'0';ch=getchar();}
    return x*f;
}
double dp[maxn][maxn];//用来表示到王妃抓时有i白j黑王妃获胜的概率，剩下i个白,j个黑
int main()
{
   int n,m;
   n=read();m=read();
   for(int i=1;i<=n;++i) dp[i][0]=1;
   for(int i=0;i<=m;++i) dp[0][i]=0;
   //情况分析：
   //dp[i][j]有3种渠道 1、王妃抽中白球 2、王妃抽中黑球，龙也抽黑球，跳出白的3、跳出黑的成
   for(int i=1;i<=n;++i)
     for(int j=1;j<=m;++j)
       {
           dp[i][j]+=(double)(i)/(i+j);
           if(j>=2) dp[i][j]+=(double)(j)/(double)(i+j)*(double)(j-1)/(double)(i+j-1)*(double)(i)/(double)(i+j-2)*dp[i-1][j-2];
           if(j>=3) dp[i][j]+=(double)(j)/(double)(i+j)*(double)(j-1)/(double)(i+j-1)*(double)(j-2)/(double)(i+j-2)*dp[i][j-3];
       }
    printf("%.9lf
",dp[n][m]);
}