codeforces 148D Bag of mice(概率dp)

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最新推荐文章于 2021-08-05 21:10:16 发布

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在一个游戏中，公主和龙通过轮流从袋中抽取白鼠和黑鼠来决定胜负，公主先抽。若一方首先抽到白鼠，则该方获胜。若所有鼠被抽完仍未有人抽到白鼠，则龙获胜。求公主获胜的概率。

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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.

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：袋子里有w只白鼠，b只黑鼠。公主和龙谁先抓到第一只白鼠谁就获胜，都抓不到白鼠，算龙获胜。两人轮流抓，每次只能抓一只，公主先抓。龙在抓老鼠的时候，老鼠会因为害怕而每次从袋子中爬出来一只。问公主获胜的机会有多大？

题解：我们可以用dp[i][j]表示在有i只白鼠，j只黑鼠的时候公主获胜的概率。于是我们能够得到以下几点：

1.当 b=0时， 0<i<=w，dp[i][0]=1；当w=0时，0<=i<=b，dp[0][i]=0 ；

2.公主直接抓到一只白鼠时，王妃赢了，概率为 i/(i+j) ；

3.公主抓到一只黑鼠，龙抓到一只白鼠，公主输了，概率为 j/(i+j)*(i)/(i+j-1) ；由于dp[i][j]表示公主获胜的概率，这种情况就不用考虑进入状态转移，我们接着往下看；

4.公主抓到一只黑鼠，龙抓到一只黑鼠，从袋子中跑了一只黑鼠，概率为 j/(i+j)*(j-1)/(i+j-1)*(j-2)/(i+j-2) ，转态就转移到了dp[i][j-3]，当然出现这种情况的限制在于 j>=3 ；

5.公主抓到一只黑鼠，龙抓到一只黑鼠，从袋子中跑出来一只白鼠，概率为 j/(i+j)*(j-1)/(i+j-1)*i/(i+j-2) ，转态就转移到了 dp[i-1][j-2]，出现这种情况必须 j>=2 ；

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 1010
double dp[maxn][maxn];
int main()
{
	int w,b;
	int i,j;
	while(scanf("%d%d",&w,&b)!=EOF)
	{
		memset(dp,0,sizeof(dp));
		for(i=1;i<=w;++i)
			dp[i][0]=1;
		for(i=1;i<=w;++i)
		{
			for(j=1;j<=b;++j)
			{
				dp[i][j]+=(double)i/(i+j);
				if(j>=3)//公主抓到黑鼠，龙也抓到黑鼠，从袋子中跑了一只黑鼠 
					dp[i][j]+=(double)j/(i+j)*(j-1)/(i+j-1)*(j-2)/(i+j-2)*dp[i][j-3];
				if(j>=2)//公主抓到黑鼠，龙也抓到黑鼠，从袋子中跑了一只白鼠 
					dp[i][j]+=(double)j/(i+j)*(j-1)/(i+j-1)*i/(i+j-2)*dp[i-1][j-2];
			}
		}
		printf("%.9lf\n",dp[w][b]);
	}
	return 0;
}