Probability dp poj2096

1  / * *
 2      dp find the desired title.
3      meaning of the questions: s a software sub-systems, will generate n kinds of bug.
4      someone one day find a bug, the bug belongs to some kind of bug, occurs in a subsystem.
5      seek n find all kinds of bug, and each subsystem find bug, so expect to be a number of days.
6      Note that: the number of bug is infinite, so find a bug, appeared in the probability of a subsystem is 1 / S,
 7      probability of belonging to a certain type is 1 / n.
8      Solution:
 . 9      DP [i] [j] denotes the i have found bug, and is present in the j-th system, to achieve the desired target state of the number of days.
10      Obviously, dp [n] [s] = 0, because the target has been reached. The dp [0] [0] is the answer to our requirements.
. 11      DP [i] [j] can be converted into the following four state:
 12 is          DP [i] [j] belongs to a bug found that the i and j th bug system has been found in
 13 is          DP [i +. 1] [j] found a bug belongs to a new kind of bug, but belong to the j seed systems have found
 14         dp [i] [j + 1 ] found that a bug has been found belonging to the i-bug, but belongs to a new subsystem
 15          DP [i +. 1] [+ J. 1] found a new bug belongs to a new bug and a subsystem
 16      probability of these four were:
 . 17      P1 * I = J / (S * n-)
 18 is      P2 = (Ni) * J / (S * n-)
 . 19      P3 = I * (SJ) / (n-* S)
 20 is      P4 = (Ni) * (SJ) / (n-* S)
 21 is      another: desired may be decomposed into a plurality of sub desired weighted sum of weights of sub desired probability of occurrence, i.e., E (aA + bB + ... ) = aE (A) -bE + (B) + ...
 22 is      therefore:
 23 is      DP [I, J] * DP = P1 [I, J] + P2 * DP [I +. 1, J] + P3 * DP [ I, J +. 1] + P4 * DP [I +. 1, J +. 1] +. 1;
 24      order was:
 25      DP [I, J] = (. 1 + P2 * DP [I +. 1, J] + P3 * DP [I, J +. 1] + P4 * DP [I +. 1, J +. 1]) / (. 1-P1)
 26 is             = ( 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 )
27 **/
28 #include <cstdio>
29 #include <iostream>
30  
31 using namespace std;
32  
33 double dp[1005][1005];
34  
35 int main()
36 {
37     int n, s, ns;
38  
39     cin >> n >> s;
40     ns = n*s;
41     dp[n][s] = 0.0;
42     for (int i = n; i >= 0; i--)
43         for (int j = s; j >= 0; j--)
44         {
45             if ( i == n && j == s ) continue;
46             dp[i][j] = ( ns + (n-i)*j*dp[i+1][j] + i*(s-j)*dp[i][j+1] + (n-i)*(s-j)*dp[i+1][j+1] )/( ns - i*j );
47         }
48     printf("%.4lf\n", dp[0][0]);
49  
50     return 0;
51 }