Collecting Bugs

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

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

1 2

3.0000


题目很难看懂，我是看了挺久题解才看懂的，
我还是用一下大牛的解释吧，

题意：有n种bug和s种子系统，bug的数量是无限的，一个程序员每天都可以发现一个bug，现在求的是发现n中bug并存在
于s个子系统中的平均天数（期望）。题意有点难懂，读了好就遍了。
假设dp[i][j]表示已经发现了i种bug并存在于j个子系统中（注意是已经发现了），那么dp[n][s] 就是 0了。
从dp[i][j]已经一天会出现4中情况
1、dp[i+1][j+1],出现一个新的bug并存在一个新的子系统中，概率是(n-i)*(s-j)/(n*s)
2、dp[i][j+1],在一个子系统中出现一个bug，不过这个bug种类是之前出现过的，概率是(s-j)*i/(n*s)
3、dp[i+1][j],出现一个新的bug，但是在子系统的已经出现过了，概率是(n-i)*j/(n*s)
4、dp[i][j],在已经发现过的bug中发现一个bug，概率是i*j/(n*s)
然后根据E(aA+bB+cC+dD+...)=aEA+bEB+....;//a,b,c,d...表示概率,A,B,C...表示状态
所以dp[i][j]=p1*dp[i+1][j+1]+p2*dp[i+1][j]+p3*dp[i][j+1]+p4*dp[i][j]+1;//dp[i][j]表示的就是到达状态i,j的期望 
=>dp[i][j]=(p1*dp[i+1][j+1]+p2*dp[i+1][j]+p3*dp[i][j+1]+1)/(1-p4);
其实看久点也还是可以看懂的，

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 
 5 using namespace std;
 6 double dp[1010][1010];
 7 int main(){
 8     int n,m;
 9     while(scanf("%d%d",&n,&m)!=EOF){
10         memset(dp,0,sizeof(dp));
11         for(int i=n;i>=0;i--){
12             for(int j=m;j>=0;j--){
13                 if(i==n&&j==m){
14                     continue;
15                 }else{
16                     dp[i][j] = (i*(m-j)*dp[i][j+1]+(n-i)*j*dp[i+1][j]+(n-i)*(m-j)*dp[i+1][j+1]+n*m)/(n*m-i*j);
17 
18                 }
19             }
20         }
21         printf("%.4f\n",dp[0][0]);//注意保留四位小数
22     }
23     return 0;
24 }