Collecting Bugs POJ - 2096 (expected dp)

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.
A system has n sub-system and a bug m type, the system will be a bug (belonging to a subsystem type and a bug) per day , bug types are equally probable, bug also occurs with equal probability in each subsystem. All subsystems are expected to ask the number of days and all of the bug bug types appear appear.
INPUT
INPUT File the contains TWO Integer Numbers, n-and S (0 <n-, S <=. 1 000).
The Output
the Output The Expectation of The Ivan apos Working Days needed to Call The Program Disgusting, Accurate to. 4 digits After The decimal Point.
The Sample INPUT
2. 1
the Sample the Output
3.0000

Although it is summer written questions, but I can not write. . .

Ideas:
the definition dp [i] [j] for the i-th system bug, J types bug occurs, a desired number of days for all systems for all types bug to the bug occurs.
Then dp [n] [m] = 0.

Sub-state is:

1. 1, representing spent one day with probability 1.
2. dp [i] [j], yes, that is their own! Dp desirable to have to consider their probability p1 for the i / n * j / m
3. dp [i + 1] [j], a system has more bug, p2 is the probability of (1 - i / n) * j / m
4. dp [i] [j + 1], more than one type of bug, a probability p3 of i / n * (1 - j / m)
5. dp [i + 1] [j + 1], more than one system in the bug, a type of multi-bug, p4 probability of (1 - i / n) * (1 - j / m).

The result is
dp [i] [j] = 1 + p1 * dp [i] [j] + p2 * dp [i + 1] [j] + p3 * dp [i] [j + 1] + p4 * dp [ i + 1] [j + 1 ].
Note that there are answers to "their" we "own" key to shift to the left, get
dp [i] [j] = (1 + p2 * dp [i + 1] [j] + p3 * dp [ i] [j + 1] + p4 * dp [i + 1] [j + 1]) / (1 - p1)

A noteworthy point is: why the state is defined as a state (i, j) to the target state expected for several days, causing We must reverse dp?

Why not defined as dp [i] [j] is the number of days to the state (i, j) of the target, then the result is dp [n] [m].

Cool analysis, which defines the way according to assumptions, then proceeds to dp [n] [m] will appear when p1 = 1. Then the divisor becomes 0! It has become the limit. Thus reverse this situation can be effectively avoided dp

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

double dp[1005][1005];

int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        dp[n][m] = 0;
        for(int i = n;i >= 0;i--)
        {
            for(int j = m;j >= 0;j--)
            {
                if(i == n && j == m)continue;
                double ii = (double)i,jj = (double)j,nn = (double)n,mm = (double)m;
                double p1 = (double)(ii / nn) * (jj / mm);
                double p2 = (double)(ii / nn) * (1.0 - jj / mm);
                double p3 = (double)(1.0 - ii / nn) * (jj / mm);
                double p4 = (double)(1.0 - ii / nn) * (1.0 - jj / mm);
                dp[i][j] =  1.0 + p2 * dp[i][j + 1] + p3 * dp[i + 1][j] + p4 * dp[i + 1][j + 1];
                dp[i][j] /= (1.0 - p1);
            }
        }
        printf("%.4f\n",dp[0][0]);
    }
    
    return 0;
}