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 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

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

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

1 2
Sample Output

3.0000

首先我要先懺悔一波這學期概率論沒好好聽課….
哎沒想到真的用的上…
我是真的傻逼

期望可以粗略理解為 可能的結果*對應的概率
那麼本題的dp[a][b] 就可以理解為是a個系統b種bug
因此dp[a][b] 可以從4種狀態轉移
dp[a+1,b] dp[a,b] dp[a,b+1] dp[a+1,b+1]
首先說一下為什麼是從後向前轉移
因為後面是已經算好的結果，也就是可能的結果，是已經確定的
因此需要乘以對應概率…
以上四個狀態的概率是好求的就不寫了..
問題在於他在給自己賦值轉移狀態的時候用到了他自己…
這個尷尬的問題…
其實也是好解決的..雖然這裡的等號是賦值
但是也可以理解為是相等..
那麼通過移項來把等式右邊的部分給弄掉…
除以係數就求出來了

#include<iostream>
#include<algorithm>
#include<string>
#include<map>
#include<cmath>
#include<cstdio>
#include<queue>
using namespace std;
double dp[1002][1002];
int main()
{
    int n, m;
    cin >> n >> m;
//  dp[n][m] = 1;
    for (int a = n;a >= 0
 
;a--)
    {
        for (int b = m;b >= 0;b--)
        {
            if (a == n&&b == m)continue;
            //dp[a][b] = 1;
            dp[a][b] += (n*m+dp[a][b + 1] * (m - b)*a + dp[a + 1][b] * (n - a)*b + dp[a + 1][b + 1] * (n - a)*(m - b)) / (n*m-a*b);
        }
    }
    printf("%.4f", dp[0][0]);
    return 0;
}