Codeforces 148D Bag of mice (概率dp)

 

D. Bag of mice time limit per test：2 seconds memory limit per test：256 megabytes

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 containsw white andb 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 integersw andb (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：公主和龍抓老鼠，老鼠有w只白的，b只黑的，先抓到白老鼠的贏，龍每次抓完一只老鼠後會多跑出去一只老鼠，一旦沒有老鼠或者沒人抓到白老鼠，則算公主輸，公主和龍抓的老鼠是黑或白是等可能的，龍抓完跑掉的老鼠是黑或白也是等可能的，現在公主先抓，要求公主贏的概率

 

 

題目分析：比較好想的一道概率dp，dp[i][j]表示還剩i只白老鼠和j只黑老鼠時公主贏的概率，這裡只需要推公主贏的情況即可，顯然有三種情況公主會贏或者還有贏的可能

注意每一輪是按順序執行的，即公主先抓，龍再抓，老鼠再跑，注意每次還剩多少

第一：公主抓到白老鼠 dp[i][j] += i / (i + j)

第二：公主和龍都抓到黑老鼠，跑出去的是黑老鼠 dp[i][j] = ( j / (i + j) ) * ( (j - 1) / (i + j - 1) * (j - 2) / (i + j - 2) ) * dp[i][j - 3] (減3是因為這種情況下需要至少三只黑老鼠)

第三：公主和龍都抓到黑老鼠，跑出去的是白老鼠 dp[i][j] = ( j / (i + j) ) * ( (j - 1) / (i + j - 1) * i / (i + j - 2) ) * dp[i - 1][j - 2]

注意初始化，沒有白老鼠，dp[0][i] = 0，沒有黑老鼠dp[i][0] = 1，一只老鼠都沒有dp[0][0] = 0

最後答案是dp[w][b]

 

 

#include 
#include 
int const MAX = 1005;
double dp[MAX][MAX];

int main()
{
    int w, b;
    memset(dp, 0, sizeof(dp));
    scanf(%d %d, &w, &b);
    for(int i = 0; i <= b; i ++)
        dp[0][i] = 0;
    for(int i = 1; i <= w; i++)
        dp[i][0] = 1.0;
    for(int i = 1; i <= w; i++)
    {
        for(int j = 1; j <= b; j++)
        {
            dp[i][j] += (double) i / (i + j);
            if(j >= 2)
                dp[i][j] += (double) j / (i + j) * (double) (j - 1) / (i + j - 1) * (double) i / (i + j - 2) * dp[i - 1][j - 2];
            if(j >= 3)
                dp[i][j] += (double) j / (i + j) * (double) (j - 1) / (i + j - 1) * (double) (j - 2) / (i + j - 2) * dp[i][j - 3];
        }
    }
    printf(%.9f
, dp[w][b]);
}