poj 2096(dp数学期望)

Collecting Bugs

Time Limit: 10000MS	Memory Limit: 64000K
Total Submissions: 666	Accepted: 255
Case Time Limit: 2000MS	Special Judge

Description

Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material stuff, he collects software bugs. When Ivan gets a new program, he classifies all possible bugs into n categories. Each day he discovers exactly one bug in the program and adds information about it and its category into a spreadsheet. When he finds bugs in all bug categories, he calls the program disgusting, publishes this spreadsheet on his home page, and forgets completely about the program. 
Two companies, Macrosoft and Microhard are in tight competition. Microhard wants to decrease sales of one Macrosoft program. They hire Ivan to prove that the program in question is disgusting. However, Ivan has a complicated problem. This new program has s subcomponents, and finding bugs of all types in each subcomponent would take too long before the target could be reached. So Ivan and Microhard agreed to use a simpler criteria --- Ivan should find at least one bug in each subsystem and at least one bug of each category. 
Macrosoft knows about these plans and it wants to estimate the time that is required for Ivan to call its program disgusting. It's important because the company releases a new version soon, so it can correct its plans and release it quicker. Nobody would be interested in Ivan's opinion about the reliability of the obsolete version. 
A bug found in the program can be of any category with equal probability. Similarly, the bug can be found in any given subsystem with equal probability. Any particular bug cannot belong to two different categories or happen simultaneously in two different subsystems. The number of bugs in the program is almost infinite, so the probability of finding a new bug of some category in some subsystem does not reduce after finding any number of bugs of that category in that subsystem. 
Find an average time (in days of Ivan's work) required to name the program disgusting.

Input

Input file contains two integer numbers, n and s (0 < n, s <= 1 000).

Output

Output the expectation of the Ivan's working days needed to call the program disgusting, accurate to 4 digits after the decimal point.

Sample Input

1 2

Sample Output

3.0000

题意是找出n个bugs，m个subcomponents的数学期望（就是查找的平均次数）

首先介绍一下

对随机变量A、B,有 数学期望E(aA+bB)=aE(A)+bE(b);

有了这个公式你就可以进行将连续的期望问题，转化为独立的状态了，概率就相当于a，b；A，B为变量，那么来分析一下这个问题，对于下一次查找，只可能有4种状态的转移，找到bugs，找到subcomponents，都找到和都找不到。因为e[i][j]=p1*e[i+1][j]+p2*e[i][j+1]+p3*e[i+1][j+1]+p4*e[i][j]+1;（e[i][j]表示已经找到i个bugs，找到j个subcomponents距离n个bugs和m个subcomponents还需要的平均查找次数（数学期望），加1，是加上这次查找），显然e[n][m]=0;

这样设计的巧妙之处在于状态方程，与下个状态（i+1和j+1）有关，所以我们将它倒过来做，当然就变成这样定义了。

下面是我的代码

#include 
double e[1001][1001];
int main()
{
	int n,m,i,j;
	scanf("%d%d",&n,&m);
	e[n][m]=0;
	for(i=n-1;i>=0;i--)
	{
		e[i][m]=((1.0*(n-i))*e[i+1][m]+n)/(n-i);
	}
	for(i=m-1;i>=0;i--)
	{
		e[n][i]=((1.0*(m-i))*e[n][i+1]+m)/(m-i);
	}
	for(i=n-1;i>=0;i--)
		for(j=m-1;j>=0;j--)
		{  
			e[i][j]=(e[i+1][j]*(1.0*(n-i)*j)+e[i][j+1]*(1.0*(m-j)*i)+e[i+1][j+1]*(1.0*(n-i)*(m-j))+m*n)/(m*n-i*j);
		}
		printf("%.4f\n",e[0][0]);
		return 0;
}