POJ 2096 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 file contains two integer numbers, n and s (0 < n, s <= 1 000).

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

1 2

3.0000
思路：令 f[i][j] 表示已经找到了 i 种 bug，且 j 个子系统至少包含一个 bug，距离完成目标需要的时间的期望。
目标状态是 f[0][0]。再过一天找到一个 bug 可能是如下的情况：

 1. 这个 bug 的种类是 已经找到的 并且 出现在 已经找到 bug 的子系统中：f[i][j](i*j)/(n*m)

 2. 这个 bug 的种类是 已经找到的 并且 出现在 没有找到 bug 的子系统中：f[i][j+1](i*(m-j)/(n*m))

 3. 这个 bug 的种类是 没有找到的 并且 出现在 已经找到 bug 的子系统中：f[i+1][j]((n-i)*j/(n*m))

 4. 这个 bug 的种类是 没有找到的 并且 出现在 没有找到 bug 的子系统中：f[i+1][j+1]((n-i)*(m-j)/(n*m))

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
double a[1005][1005];
int n,m;
//i/n,j/s;
//E(X) = X1*p(X1） + X2*p(X2） + …… + Xn*p(Xn) = X1*f1(X1） + X2*f2(X2） + …… + Xn*fn(Xn)
//a[i][j]=p1*a[i][j]+p2*a[i+1][j]+p3*a[i][j+1]+p4*a[i+1][j+1];
//a[i][j]=(p2*a[i+1][j]+p3*a[i][j+1]+p4*a[i+1][j+1])(1-p1);
//期望的好好在看看，全忘了，不搜百度百科都不知道是啥了，差点以为超几何了。QAQ
int main()
{
    scanf("%d%d",&n,&m);
    a[n][m]=0.0;
    for(int i=n;i>=0;i--)
    {
        for(int j=m;j>=0;j--)
        {
            if(i==n && j==m) continue;
            a[i][j]=(i*(m-j)*a[i][j+1]+(n-i)*j*a[i+1][j]+(n-i)*(m-j)*a[i+1][j+1]+n*m)/(n*m-i*j);
        }
    }
    printf("%.4f\n",a[0][0]);
    return 0;
}