codeforce 148D 概率DP

题目链接：戳这里

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
4000
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 exceed 10 - 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.

题意：笼子里有w只白老鼠，b只黑老鼠，公主和龙玩一个游戏：公主每次随机抓一只老鼠，龙每次抓一只老鼠后会跑掉一只老鼠，谁先抓到白老鼠谁赢，如果没有白老鼠了则龙赢，问公主赢的概率。

题解：用dp[i][j]表示有i只白老鼠和j只黑老鼠的情况，该情况能转移到四种情况：

1、公主抓到白老鼠，赢，即dp[i][j]=i/(i+j)。

2、公主抓到黑老鼠，龙抓到黑老鼠，跑掉白老鼠，即dp[i][j]=dp[i-1][j-2]*j/(i+j)*(j-1)/(i+j-1)*i/(i+j-2)。

3、同2，但跑掉的是黑老鼠，即dp[i][j]=dp[i][j-3]*j/(i+j)*(j-1)/(i+j-1)*(j-2)/(i+j-2)。

4、龙抓到白老鼠，输，不能转移。

边界条件为dp[i][0]=1，dp[0][i]=0。转移即可。

代码：#include<bits/stdc++.h>
using namespace std;
typedef double db;
int read()
{
char c;int sum=0,f=1;c=getchar();
while(c<'0' || c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0' && c<='9'){sum=sum*10+c-'0';c=getchar();}
return sum*f;
}
int w,b;
db dp[1005][1005];
int main()
{
w=read();b=read();
for(int i=1;i<=b;i++)
dp[0][i]=0;
for(int i=1;i<=w;i++)
dp[i][0]=1;
for(int i=1;i<=w;i++)
for(int j=1;j<=b;j++)
{
dp[i][j]=(db)(i+0.0)/(i+j);
if(j>=2) dp[i][j]+=dp[i-1][j-2]*(j+0.0)/(i+j)*(j-1+0.0)/(i+j-1)*(i+0.0)/(i+j-2);
if(j>=3) dp[i][j]+=dp[i][j-3]*(j+0.0)/(i+j)*(j-1+0.0)/(i+j-1)*(j-2+0.0)/(i+j-2);
}
printf("%.9lf\n",dp[w][b]);
return 0;
}