Codeforce 148D

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

这道题是一道概率dp的题，看起来规则挺复杂，其实可以快速构建其状态转化的方程，我们可以这样理解：无论有多少只老鼠，其实都是女王和龙在无数次没能抽到白老鼠的情况下最后变成只有白老鼠的情况，而每次变化的是当前能让女王赢的概率。那么状态的变化就可以得出，记得要一个概率一个概率的乘起来，不然容易出错。#include<iostream>
#include<cstdio>
#include<cmath>
#include<memory.h>
using namespace std;
int w,b;
double dp[1010][1010];
int main()
{
while(scanf("%d%d",&w,&b)!=EOF){
memset(dp,0.0,sizeof(dp));
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);//当前状态让女王赢
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];
}//自然出一只黑老鼠
if(j>=2&&i>=1){
dp[i][j]+=((double)j/(i+j))*((double)(i)/(i+j-1))*((double)(j-1)/(i+j-2))*dp[i-1][j-2];
}//自然出一只白老鼠

//转移成数量少的情况
}

}
printf("%.9lf\n",dp[w][b]);
}
return 0;
}