Bag of mice CodeForces - 148D （概率dp）

CodeForces - 148D 

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 bblack 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：

Alice和Bob正在玩一个游戏：

一个袋子里一开始装着w个白球和b个黑球。Alice和Bob轮流随机抽出一个球（Alice先手）。如果抽出的球是白色的，则抽出这个球的人获胜。每当一个球被Bob取出后，会有另一个球滚出来（不算任何人抽的）。但Alice取出时很小心，不会让球滚出来。每个人抽球、和自动滚出来的球都是等概率的。那么Alice获胜率是多少呢？

如果最后袋子里没有球了，并且没有人拿到白球，那么Bob获胜。

题解：

dp【i】【j】表示存在白球i个黑球j个Alice获胜的概率，可以根据上一个状态的胜率转换到此状态的胜率，该题较为简单，不多做解释，直接看AC代码。

//#include<bits/stdc++.h>
//#include <unordered_map>
//#include<unordered_set>
//#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<set>
#include<climits>
#include<queue>
#include<cmath>
#include<stack>
#include<map>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define MT(a,b) memset(a,b,sizeof(a))
const int INF  =  0x3f3f3f3f;
const int ONF  = -0x3f3f3f3f;
const int O    =  1e5;
const int MOD  =  1e4+7;
const int maxn =  1e3+5;
const int N    =  1e9+5;
const double PI  =  3.141592653589;
const double E   =  2.718281828459;

double dp1[maxn][maxn];//先手胜
double dp2[maxn][maxn];//后手胜

int main()
{
    int n,m;scanf("%d%d",&n,&m);
    MT(dp1,0); MT(dp2,0);
    for(int i = 1; i <= n; i++) dp1[i][0]=1.0;
    for(int i=1;i<=n;i++) for(int j=1;j<=m;j++)
    {
        dp1[i][j] = (double)i/(i+j) + (double)j/(i+j) * dp2[i][j-1]; 
        // 轮到Alice时，取出白球的概率+取出黑球且Alice获胜的概率
        
        dp2[i][j] = (double)j/(i+j) * i/(i+j-1) * dp1[i-1][j-1]; 
        // 轮到Bob时，取出黑球，滚出白球且Alice获胜的概率
        
        if(j>=2) dp2[i][j] += (double)j/(i+j) * (j-1)/(i+j-1) * dp1[i][j-2];
        // 取出黑球，滚出黑球且Alice获胜的概率概率
    }
    printf("%.9f\n",dp1[n][m]);
    return 0;
}