component cnblogs form algo 可以轉化 -s set != reac

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

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

dp求期望
逆著遞推求解
題意：
   一個軟件有s個子系統，會產生n種bug
   某人一天發現一個bug,這個bug屬於一個子系統，屬於一個分類
   每個bug屬於某個子系統的概率是1/s,屬於某種分類的概率是1/n
   問發現n種bug,每個子系統都發現bug的天數的期望。
求解：
         dp[i][j]表示已經找到i種bug,j個系統的bug，達到目標狀態的天數的期望
         dp[n][s]=0;要求的答案是dp[0][0];
         dp[i][j]可以轉化成以下四種狀態:
              dp[i][j],發現一個bug屬於已經有的i個分類和j個系統。概率為(i/n)*(j/s);
              dp[i][j+1],發現一個bug屬於已有的分類，不屬於已有的系統.概率為 (i/n)*(1-j/s);
              dp[i+1][j],發現一個bug屬於已有的系統，不屬於已有的分類,概率為 (1-i/n)*(j/s);
              dp[i+1][j+1],發現一個bug不屬於已有的系統，不屬於已有的分類,概率為 (1-i/n)*(1-j/s);
        整理便得到轉移方程

移項搞一搞就可以了。

 1 #include<cstdio>
 2 #include<cmath>
 3 #include<algorithm>
 4 #include<iostream>
 5 #include<cstring>
 6 #define N 1007
 7 using namespace std;
 8 
 9 int n,s;
10 double f[N][N];
11 
12 int main()
13 {
14     while(~scanf("%d%d",&n,&s))
15     {
16         memset(f,0
 
,sizeof(f));
17         for (int i=n;i>=0;i--)
18             for (int j=s;j>=0;j--)
19                 if (i!=n||j!=s) f[i][j]=(i*(s-j)*f[i][j+1]+(n-i)*j*f[i+1][j]+(n-i)*(s-j)*f[i+1][j+1]+n*s)/(n*s-i*j);
20         printf("%.4f\n",f[0][0]);
21     }
22 }

POJ2096 Collecting Bugs(概率DP,求期望)