POJ 2096 Collecting Bugs probability DP

collecting Bugs

Time Limit: 10000MS		Memory Limit: 64000K
Total Submissions: 3379		Accepted: 1672
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 with all bugs categories, he calls the program disgusting, publishes this spreadsheet on his home page, an D forgets completely about the program.
companies, Macrosoft and Microhard is in tight competition. Microhard wants to decrease sales of one Macrosoft program. They hire Ivan to prove, the program in question is disgusting. However, Ivan has a complicated problem. This new program have S subcomponents, and finding bugs of all types in each subcomponent would take too long before the TA Rget could be reached. So Ivan and Microhard agreed to use a simpler criteria---Ivan should find at least one bugs in each subsystem and at LEAs T one bug of each category.
Macrosoft knows about these plans and it wants to estimate the time that's required for Ivan to call their program DISG Usting. It's important because the company releases a new version of soon, so it can correct its plans and release it quicker. Nobody would is interested in Ivan's opinion about the reliability of the obsolete version.
A Bug found in the program can is of any category with equal probability. Similarly, the bug can be found in any given subsystem with equal probability. Any particular bugs cannot belong to the different categories or happen simultaneously in both different subsystems. The number of bugs in the program was almost infinite, so the probability of finding a new bug of some category in some sub System does not reduce after finding any number of bugs of the that category in the that subsystem.
Find an average time (in days of Ivan's work ) required to name the program disgusting.

Input

Input file contains the integer numbers, N and s (0 < N, s <= 1).

Output

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

Sample Input

1 2

Sample Output

3.0000

Source

Northeastern Europe 2004, Northern subregionTest Instructions: A software has s subsystem, there are n kinds of bugs. Someone can find a bug in one day. Q, the expected value of gathered up n bugs in this software and the time of at least one bug in each subsystem (in days). Note: Bug is infinite, each bug belongs to any one of the probability of a bug is 1/n; appears on each system is as possible, for 1/s. The Dp[i][j] indicates that the bugs,j of the I subsystem remains the expected value of the bugs not found. initialization: dp[0][0]=0.0request: Dp[s][n] state transition equation:in the status (I,J), find a new bug, there are 4 kinds of possible1. New subsystems, new kinds of2. New subsystems, old types of3. Old subsystems, new kinds of4. Old subsystems, old typesfor each of the probabilities:to create an axis with S,n as a point on the X, Y axis, the geometry is approximate (that is, the proportion of the total area):1. I * J/(n * s)2. I * (n-j)/(n * s)3. (s-i) * J/(n * s)4. (s-i) * (N-J)/(n * s)then we can get the state transfer equation, move the term, and push the solution.

1#include <cstdio>2#include <cstring>3#include <algorithm>4 5 using namespacestd;6 7 Const intmaxn=1010;8 Const intmaxs=1010;9 Ten DoubleDP[MAXS][MAXN]; One  A intn,s; -  -InlineDoubleRetDoubleXDoubley) the { -     returnx*y/((Double) ns); - } -  + intMain () - { +      while(~SCANF ("%d%d",&n,&s)) A     { atdp[0][0]=0.0; -          for(intI=0; i<=s;i++) -         { -              for(intj=0; j<=n;j++) -             { -dp[i][j]=0.0; in                 if(i>0&&j>0) -Dp[i][j]+=ret (I,J) *dp[i-1][j-1]; to                 if(i>0) +Dp[i][j]+=ret (I,N-J) *dp[i-1][j]; -                 if(j>0) theDp[i][j]+=ret (j,s-i) *dp[i][j-1]; *                 if(i!=0|| j!=0) $                 {Panax Notoginsengdp[i][j]+=1.0; -dp[i][j]=dp[i][j]/(1.0-ret (s-i,n-j)); the                 } +             } A         } theprintf"%.4f\n", Dp[s][n]); +     } -     return 0; $}

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POJ 2096 Collecting Bugs probability DP