# 题目

{% fold 点击显/隐题目 %}

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.

The only line of input data contains two integers w and b (0?≤?w,?b?≤?1000).

the probability of the princess winning. The answer is considered to be correct if its absolute or relative error does not exceed 10?-?9.
Example

1 3

# 题解

记忆化搜索即可

注意初始值以及除以0的情况

# 代码

{% fold 点击显/隐代码 %}```cpp Bag of mice https://github.com/OhYee/sourcecode/tree/master/ACM 代码备份

#include

const int maxn = 1005;

const double eps = 1e-12;

double dp[maxn][maxn];

void init(int w, int b) {

for (int i = 0; i <= w; i++)

for (int j = 0; j <= b; j++)

if (i == 0)

dp[i][j] = 0;

else if (j == 0)

dp[i][j] = 1;

else

dp[i][j] = -1;

}

double getAns(int w, int b, int deep) {

//for (int i = 0; i < deep; i++)

// printf("\t");

//printf("dp[%d][%d]:\n", w, b);

```
if (w < 0 || b < 0)
return 0;
if (dp[w][b] < 0) {
// get white directly
dp[w][b] = (double)w / (w + b);
if (w + b >= 3) {
// white
dp[w][b] +=
((double)b / (w + b)) * ((double)(b - 1) / (w + b - 1)) *
((double)w / (w + b - 2)) * getAns(w - 1, b - 2, deep + 1);
// black
dp[w][b] +=
((double)b / (w + b)) * ((double)(b - 1) / (w + b - 1)) *
((double)(b - 2) / (w + b - 2)) * getAns(w, b - 3, deep + 1);
}
}
//for (int i = 0; i < deep; i++)
// printf("\t");
//printf("%.9f\n", dp[w][b]);
return dp[w][b];
```

}

int main() {

int w, b;

while (scanf("%d%d", &w, &b) != EOF) {

init(w, b);

printf("%.10f\n", getAns(w, b, 1));

}

return 0;

}

{% endfold %}