codeforces 148 D. Bag of mice 概率dp

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.

題意：

原來袋子裏有w只白鼠和b只黑鼠 
龍和王妃輪流從袋子裏抓老鼠。誰先抓到白色老師誰就贏。 
王妃每次抓一隻老鼠，龍每次抓完一隻老鼠之後會有一隻老鼠跑出來。 
每次抓老鼠和跑出來的老鼠都是隨機的。 
如果兩個人都沒有抓到白色老鼠則龍贏。王妃先抓。 
問王妃贏的概率。 

思路：

設dp[i][j]表示有i個白鼠，j個黑鼠時王妃贏的概率，則dp[1][0]爲必勝點。

dp[i][0]=1，3種情況，推一遍就出來了

1.兩人都拿黑的並且逃跑的也是黑的：dp[i][j-3]*(1.0*j/(i+j))*(1.0*(j-1)/(i+j-1))*(1.0*(j-2)/(i+j-2))

2.兩人都拿黑的逃跑的是白的：dp[i-1][j-2]*(1.0*j/(i+j))*(1.0*(j-1)/(i+j-1))*(1.0*i/(i+j-2))

3.贏了：1.0*i/(i+j)

注意i j爲整形，要轉化爲double才能運算。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
using namespace std;
const int N=1005;
const double eps=1e-5;
int w,b;
double dp[N][N];
int main()
{
    scanf("%d%d",&w,&b);
    memset(dp,0,sizeof(dp));
    dp[1][0]=1;
    for(int i=0;i<=w;i++){
        for(int j=0;j<=b;j++){
            if(i==1&&j==0)continue;
            if(j>=3)
            dp[i][j]+=dp[i][j-3]*(1.0*j/(i+j))*(1.0*(j-1)/(i+j-1))*(1.0*(j-2)/(i+j-2));
            if(i>=1&&j>=2)
            dp[i][j]+=dp[i-1][j-2]*(1.0*j/(i+j))*(1.0*(j-1)/(i+j-1))*(1.0*i/(i+j-2));
            if(i>=1)
            dp[i][j]+=1.0*i/(i+j);
        }
    }
    printf("%.10f\n",dp[w][b]);

    return 0;
}