http://codeforces.com/problemset/problem/148/D
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).
Output the probability of the princess winning. The answer is considered to be correct if its absolute or relative error does not exceed10 - 9.
1 3
0.500000000
5 5
0.658730159
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：原来袋子里有w仅仅白鼠和b仅仅黑鼠龙和王妃轮流从袋子里抓老鼠。谁先抓到白色老师谁就赢。王妃每次抓一仅仅老鼠，龙每次抓完一仅仅老鼠之后会有一仅仅老鼠跑出来。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。
王妃先抓。 问王妃赢的概率。 分析：如果dp[i][j]表示轮到王妃抓老鼠时面对剩余i仅仅白鼠和j仅仅黑鼠的胜率 则dp[i][j]能够转化到下面四种情况： 1.王妃胜利,转化概率为i/(i+j) 2.dp[i-1][j-2]---王妃抓黑鼠,龙抓黑鼠,逃跑白鼠,转化概率是j/(i+j) * (j-1)/(i+j-1) * i/(i+j-2) 3.dp[i-1][j-1]---王妃抓到黑鼠,龙抓到白鼠,输！
,转化概率为j/(i+j) * i/(i+j-1)//这不能到达，到达就输了 4.dp[i][j-3]--王妃抓到黑鼠,龙抓到黑鼠,逃跑黑鼠,转化率为j/(i+j) * (j-1)/(i+j-1) * (j-2)/(i+j-2) */ #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <string> #include <queue> #include <algorithm> #include <map> #include <cmath> #include <iomanip> #define INF 999999999 typedef long long LL; using namespace std; const int MAX=1000+10; int w,b; double dp[MAX][MAX]; int main(){ while(cin>>w>>b){ for(int i=1;i<=w;++i)dp[i][0]=1;//有白鼠无黑鼠胜率为1 for(int i=0;i<=b;++i)dp[0][i]=0;//无白鼠胜率为0 for(int i=1;i<=w;++i){ for(int j=1;j<=b;++j){ dp[i][j]=i*1.0/(i+j); //dp[i][j]+=j*1.0/(i+j) * i*1.0/(i+j-1) * dp[i-1][j-1]; if(j>=2)dp[i][j]+=j*1.0/(i+j) * (j-1)*1.0/(i+j-1) * i*1.0/(i+j-2) * dp[i-1][j-2]; if(j>=3)dp[i][j]+=j*1.0/(i+j) * (j-1)*1.0/(i+j-1) * (j-2)*1.0/(i+j-2) * dp[i][j-3]; } } printf("%.9f\n",dp[w][b]); } return 0; }