CodeForces 148D Bag of mice —— 概率DP

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

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.

题意是有w只白老鼠，b只黑老鼠，公主和龙轮流抽老鼠，先抽到白老鼠的获胜，公主先抽，龙每次抽取时都会造成一只老鼠逃出袋子，问你公主获胜的概率。

思路是公主在i只白老鼠，j只黑老鼠的状态下获胜的概率可由三种状态转移而得：

一：直接抽中白老鼠。

二：公主取到黑鼠，龙取到黑鼠，公主取到黑鼠，转移到i只白鼠，j - 3 只黑鼠获胜的状态。

三：公主取到黑鼠，龙取到黑鼠，公主取到白鼠，转移到i - 1只白鼠，j - 2只黑鼠获胜的状态。

/*
ID: xinming2
PROG: stall4
LANG: C++
*/
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>
#include <cassert>
using namespace std;
///#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 1000 + 50;
const int sigma_size = 26;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-8
const int mod = (int)1e9 + 7;
typedef long long LL;
const LL MOD = 1000000007LL;
const double PI = acos(-1.0);

typedef pair<int , int> pi;
#define Bug(s) cout << "s = " << s << endl;
///#pragma comment(linker, "/STACK:102400000,102400000")
double dp[MAXN][MAXN];///dp[i][j]表示在有i只白老鼠和j只黑老鼠的情况下公主获胜的概率
int main()
{
for(int i = 0 ; i <= 1000; i++)
{
dp[i][0] = 1.0;
dp[0][i] = 0.0;
}
for(int i = 1 ; i <= 1000 ; i++)
{
for(int j = 1 ; j <= 1000 ; j++)
{
///公主取到白鼠
dp[i][j] += (double)i / (i + j);
///公主取到黑鼠，龙取到黑鼠，公主取到黑鼠
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];
///公主取到黑鼠，龙取到黑鼠，公主取到白鼠
if(i >= 1 && j >= 2)dp[i][j] += (double)j / (i + j) * (double)(j - 1) / (i + j - 1) * (double)i / (i + j - 2) * dp[i - 1][j - 2];
//            cout << dp[i][j] << endl;
}
}
int w , b;
scanf("%d%d" , &w , &b);
printf("%.9lf\n" , dp[w][b]);
return 0;
}