[Poj2096]Collecting Bugs（入门期望dp）

Collecting Bugs

Time Limit: 10000MS		Memory Limit: 64000K
Total Submissions: 6237		Accepted: 3065
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

Sample Output

3.0000

题意：

有n个系统，s种bug。问在n个系统种都找到bug，s种bug都找到的期望天数是多少？

期望值：

简单说下什么是期望值。就是的得到的结果 * 概率的总和。

举个简单的例子，掷骰子的期望为： 1 * 1/6 + 2 * 1/6 + 3 * 1/6 + 4 * 1/6 + 5 * 1/6 + 6 * 1/6 = 3.5

分析：

一般求期望Dp都是倒着推的。

先定义dp[i][j]表示已经在n个系统中找到i个系统的bug，在s种bug中找到j种bug的期望值。

我们先看顺着推可以：

由dp[i][j] 推得 dp[i + 1][j] ，概率为p1 = （n - i）/ n * j / s；(表示在接下来一天新找到一个系统bug，没找到新的种类bug）

由dp[i][j] 推得 dp[i][j +1] ，概率为p2 = i / n* （s - j）/ s；(表示在接下来一天没找到新的一个系统bug，找到新的种类bug）

由dp[i][j] 推得 dp[i + 1][j +1] ，概率为p3 = （n - i）/ n *（s - j）/ s；(表示在接下来一天找到新的一个系统bug，找到新的种类bug）

由dp[i][j] 推得 dp[i + 1][j] ，概率为p4 = i / n * j / s；(表示在接下来一天新没找到新的一个系统bug，没找到新的种类bug）

四个情况的概率相加是等于 1 。

可以得到dp方程 dp[i][j] = （dp[i + 1][j] + 1） * p1 + （dp[i][j + 1] + 1）* p2 + （dp[i + 1][j + 1] + 1）* p3 + （dp[i][j] + 1）* p4。

移项化简得到：dp[i][j]= ( n*s + (n-i)*j*dp[i+1,j] + i*(s-j)*dp[i,j+1] + (n-i)*(s-j)*dp[i+1,j+1] )/( n*s - i*j ) 。

因为dp[n][s]时所有都已经找到期望为 0.最后倒着推出dp[0][0]即可

（因为此题特别坑，输出不能用%.4lf 要用%.4f Wa了三次，看评论才知道Ac了）

感谢：http://www.cnblogs.com/flipped/p/5230359.html为我提供了思路，本来期望dp很弱的我只有慢慢学了。

附上AC代码：

# include <iostream>

# include <cstdio>

# include <cstring>

using namespace std;

const int N = ;

double dp[N][N];

int n,s;

int main(){

    scanf("%d %d",&n,&s);

    for(int i = n;i >= ;i--){

        for(int j = s;j >= ;j--){

            if(i != n || j != s){

                dp[i][j] += dp[i + ][j] * (n - i) * j + dp[i][j + ] * i * (s - j);

                dp[i][j] += dp[i + ][j + ] * (n - i) * (s - j) + n * s;

                dp[i][j] = dp[i][j] / ((n * s - i * j));

            }

        }

    }

    printf("%.4f",dp[][]);

}