Codeforces 148D Bag of mice 简单概率dp_简单概率 codeforces-优快云博客

本文介绍了一个数学概率问题，公主和龙通过轮流从袋子中抽取白鼠或黑鼠来决定新年夜活动的方式，公主先抽。当一方首次抽出白鼠时即获胜。本文提供了解决该问题的动态规划算法。

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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.

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.

题目大致意思:有w个白鼠、b个黑鼠，公主和龙依次从袋中拿出一只鼠，首先拿出白鼠的赢，公主先拿，若无白鼠，则算龙赢。。龙从袋中拿出一只鼠，会有另外一只鼠跳出去，而公主不会，跳出或者拿出的鼠不会再放入袋中。从袋中拿每一只鼠的概率相等，每一只鼠从袋中跳出的概率也相等，求公主赢的概率。

定义dp[i][j]为袋中有i只白鼠、j只黑鼠时公主赢的概率。题目比较简单，直接贴代码，代码中有注释。

#include <iostream>
#include <cstdio>
using namespace std;
const int maxn = 1000 + 10;
double dp[maxn][maxn];
int w, b;
int main()
{
    while(scanf("%d%d", &w, &b) != EOF){
        for(int i = 0; i <= b; i++)
            dp[0][i] = 0.0;
        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] = 0;
                //抓住白的
                dp[i][j] += ((double)(i)) / (i + j);
                //抓住黑的然后龙抓住白的
                dp[i][j] += 0.0;
                //抓住黑的龙抓住黑的，跑出来黑的
                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];
                //抓住黑的龙抓住黑的，跑出来白的
                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];
            }
        }
        printf("%.9lf\n", dp[w][b]);
    }
    return 0;
}