POJ 2096:collecting Bugs probability DP seeking expectation

Collecting Bugs Topic Connection:

http://poj.org/problem?id=2096

Test instructions

Ivan likes to collect bugs, he will find a bug every day, found this bug has a property and belongs to a subsystem, bug total n Properties, subsystem total s (0<n, s≤1000), find Ivan set up n bug and each subsystem has the expectation of bug.

Exercises

The first one asks for the desired question, so that dp[i][j] indicates that the system already has all the J bugs of I system and expects to get the number of days of all bugs, so dp[n][s]=0, and dp[0][0] is the answer.

DP[I][J] to the next day there will be four cases dp[i][j] (found an existing bug and the bug belongs to the subsystem has a bug), Dp[i+1][j] (found a new bug but the bug belongs to the subsystem has a bug), dp[i][j+1] (an existing bug was found, but the system of bugs exists before this bug), Dp[i+1][j+1] (a new bug has been found and the subsystem to which the bug belongs has not yet been bug).

The probabilities of obtaining these four cases are:

p1=i*j/(n*s);

P2= (n-i) *j/(n*s);

p3=i* (S-J)/(n*s);

p4= (n-i) * (s-h)/(n*s);

By the expected linear nature can know dp[i][j]=p1*dp[i][j]+p2*dp[i+1][j]+p3*dp[i][j+1]+p4*dp[i+1][j+1]; because it is the next day, so Dp[i][j] also add 1.

　　　　　　　　　　　　　　

Code

#include<stdio.h>Constint N=1002;Double DP[N][n],p1, p2, p3, p4Ns;IntMain(){int n, S;While(~scanf("%d%d", &n, &s)){NS=n*s;dp[n][s]=0.0;For(int I=n; I>=0;--I.)For(Int J=s; j>=0;--j)If(I!=n|| J!=s){P1=i*j/(NS);p 2= (n-I.) *j/(NS);p 3=i* (S-j)/(NS);p 4= (n-I.) * (s-j)/(NS); Dp[I][j]=(1.0+p2*dp[I+1][j]+p3*dp[i][j< Span class= "Sh_number" >+1]+p4*dp[i+1][j+1])/(1.0-p1} printf ( "%.4f \n ",dp[< Span class= "Sh_number" >0][0]); }}          

　　

POJ 2096:collecting Bugs probability DP seeking expectation