poj 2096(dp數學期望)

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

題意是找出n個bugs，m個subcomponents的數學期望（就是尋找的平均次數）

首先介紹一下

對隨機變數A、B,有 數學期望E(aA+bB)=aE(A)+bE(b);
有了這個公式你就可以進行將連續的期望問題，轉化為獨立的狀態了，機率就相當於a，b；A，B為變數，那麼來分析一下這個問題，對於下一次尋找，只可能有4種狀態的轉移，找到bugs，找到subcomponents，都找到和都找不到。因為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]表示已經找到i個bugs，找到j個subcomponents距離n個bugs和m個subcomponents還需要的平均尋找次數（數學期望），加1，是加上這次尋找），顯然e[n][m]=0;
這樣設計的巧妙之處在於狀態方程，與下個狀態（i+1和j+1）有關，所以我們將它倒過來做，當然就變成這樣定義了。
下面是My Code
#include<stdio.h> 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;}