题意：袋子里有w只白鼠和b只黑鼠。龙和公主轮流从袋子里抓老鼠。谁先抓到白色老师谁就赢。公主每次抓一只老鼠，龙每次抓完一只老鼠之后会有一只老鼠跑出来。每次抓老鼠和跑出来的老鼠都是随机的。如果两个人都没有抓到白色老鼠则龙赢。公主先抓。问公主赢的概率。
做了这么多概率dp的题目了，本来接的差不多了，结果一做还是不会。。。。。。
下面是看了别人的思路
win[i][j] = i * 1.0 / (i + j); //i只白老鼠j只黑老鼠时公主选白老鼠
win[i][j] += lost[i][j-1] * j * 1.0 / (i + j); //i只白老鼠j只黑老鼠时公主选黑老鼠，但公主选完黑老鼠后龙还是输了
lost[i][j] = j * 1.0 / (i + j) * win[i-1][j-1] * (i * 1.0 / (i + j - 1)); //i只白老鼠j只黑老鼠时龙选黑老鼠，选完后跳出去只白老鼠
lost[i][j] += j * 1.0 / (i + j) * win[i][j-2] * ((j - 1) * 1.0 / (i + j - 1)); //i只白老鼠j只黑老鼠时龙选黑老鼠，选完后跳出去只黑老鼠
Description
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 Input
Input
1 3
Output
0.500000000
Input
5 5
Output
0.658730159
Hint
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<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
double win[1100][1100],lost[1100][1100];
int main()
{
int w,b;
while(~scanf("%d %d",&w,&b))
{
memset(win,0,sizeof(win));
memset(lost,0,sizeof(lost));
for (int i = 1; i <= w; ++i)
win[i][0] = 1.0;
for(int i=1; i<=w; i++)
for(int j=1; j<=b; j++)
{
win[i][j] = i * 1.0 / (i + j) + lost[i][j-1] * j * 1.0 / (i + j);//这个是他本次赢或者本次不赢但是下次龙输到最后一定赢
lost[i][j]=j*1.0/(i+j)*win[i-1][j-1]*i*1.0/(i+j-1);//龙本次输了肯定是选了黑色的，所以要用下次公主赢得乘上对应跑白色或者黑色的概率
lost[i][j]+=j*1.0/(i+j)*win[i][j-2]*(j-1)*1.0/(i+j-1);
}
printf("%.9lf\n",win[w][b]);
}
return 0;
}