Codeforces 148D：Bag of mice

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 exceed 10 - 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.

题意是一个袋子里面有w个白色老鼠，b个黑色老鼠。公主和龙每一轮从袋子中随机选出一个老鼠，不放回。谁先选到白色老鼠谁获胜，然后龙每一次从袋子中选老鼠的时候，会随机从袋子中跑掉一个老鼠。如果两个人都没有人选到白色老鼠，龙获胜。公主先选。

看总结的这个笔记的时候发现的一个特点就是：到目前为止，自己好像是每一类都接触一些了，但是每一类做的题目实在是太少了，导致真要是再做一道类似的，也没有太多把握。感觉自己真心还有好多题要做。。。

这种概率问题感觉还是比较简单的，我自己反正是不太习惯从小向大递推，直接递归搜，把问题不断缩小，并且记录下来各种情况的值，就可以解出来了。

代码：

#pragma warning(disable:4996)#include <iostream>#include <algorithm>#include <cstring>#include <vector>#include <string>#include <cstdio>#include <cmath>#include <queue>#include <stack>#include <deque>#include <ctime>;#include <set>#include <map>using namespace std;typedef long long ll;#define INF 0x3fffffff#define REP(i, n) for (int i=0;i<n;++i)#define REP1(i, n) for (int i=1;i<=n;++i)const ll mod = 1e9 + 7;const int maxn = 1005;int w, b;double dp[maxn][maxn];double solve(int x, int y){if (x == 0 && y == 0)return 0;if (x < 0 || y < 0)return 0;if (dp[x][y])return dp[x][y];double xx = x, yy = y;double win = xx / (xx + yy);double lose = yy / (xx + yy);if (x == 1 && y == 0){dp[x][y] = 1;}else if (x == 0 && y == 1){dp[x][y] = 0;}else if (x == 1 && y == 1){dp[x][y] = 0.5;}else if (x == 2 && y == 0){dp[x][y] = 1;}else if (x == 0 && y == 2){dp[x][y] = 0;}else{dp[x][y] = win + lose*((yy - 1) / (xx + yy - 1))*((xx / (xx + yy - 2))*solve(xx - 1, yy - 2)+ ((yy - 2) / (xx + yy - 2))*solve(xx, yy - 3));}return dp[x][y];}int main(){//freopen("i.txt", "r", stdin);//freopen("o.txt", "w", stdout);scanf("%d%d", &w, &b);memset(dp, 0, sizeof(dp));printf("%.10f", solve(w, b));return 0;}