code forces 148D Bag of mice （概率DP）

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.

题意：一个袋子里有w个白老鼠，b个黑老鼠，王妃和龙依次取，王妃先取，先取到白老鼠

的为胜者，当中龙取老鼠的时候，取出一仅仅后，会有随机的一仅仅老鼠跑出来，并且取老鼠的

时候，每仅仅老鼠取到的概率是一样的，跑出来的概率也是一样的，  让你算王妃赢的概率。

思路： dp[i][j] 表示 白老鼠为i仅仅，黑老鼠为j仅仅时，王妃赢的概率,

有四种状态：

（1）  王妃取到白鼠  。  dp[ i ][ j ] + =  i  / ( i + j ) ;

（2）  王妃取到黒鼠，龙取到白鼠 。    dp[ i ][ j ] + = 0.0 ;

（3） 王妃取到黒鼠，龙取到黑鼠 ，跑出来一仅仅黑鼠  。 dp[i][j]+=j/(i+j) * (j-1)*/(i+j-1) * (j-2)*/(i+j-2) * dp[i][j-3];

（4） 王妃取到黒鼠，龙取到黑鼠 ，跑出来一仅仅白鼠  。 dp[i][j]+=j*/(i+j) * (j-1)*/(i+j-1) * i*/(i+j-2) * dp[i-1][j-2];

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=1100;

double dp[maxn][maxn];
int n,m;

int main()
{
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for(int i=1; i<=n; i++)  dp[i][0]=1.0;
        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++)
            {
                dp[i][j]+=(i*1.0)/(i+j);
                if(j>=3)   dp[i][j]+=j*1.0/(i+j) * (j-1)*1.0/(i+j-1) * (j-2)*1.0/(i+j-2) * dp[i][j-3];
                if(j>=2)   dp[i][j]+=j*1.0/(i+j) * (j-1)*1.0/(i+j-1) * i*1.0/(i+j-2) * dp[i-1][j-2];
            }
        printf("%.9lf
",dp[n][m]);
    }
    return 0;
}