CodeForces 148D Bag of mice 概率DP

龙和公主从一个袋子里抓老鼠，谁先抓到白老鼠谁赢，如果最后谁都没有抓到白老鼠龙赢，龙每次抓老鼠都会又一只老鼠从袋中逃跑，问最后公主胜利的概率。

假设dp[j]表示袋中有i只白老鼠，j只黑老鼠时公主胜利的概率，那么对于这个局面，会有4种不同的可能：
公主抓到白老鼠；

公主抓到黑老鼠，龙抓到白老鼠；（输，不可向后推进）

公主抓到黑老鼠，龙抓到黑老鼠，跑掉黑老鼠；

公主抓到黑老鼠，龙抓到黑老鼠，跑掉白老鼠；

于是，我们只需要计算1，3，4事件的概率和即可，因为这三个事件都可以继续推进状态的转换。

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 [i]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.

#include <iostream>
#include <cstdio>
using namespace std;

const int MAX_N = 1000 + 10;

double dp[MAX_N][MAX_N];
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;
}

 

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