hat gre 表示 format nio icu collect home one

Collecting Bugs

Time Limit: 10000MS		Memory Limit: 64000K
Total Submissions: 5760		Accepted: 2853
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 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


題目很難看懂，我是看了挺久題解才看懂的，
我還是用一下大牛的解釋吧，

題意：有n種bug和s種子系統，bug的數量是無限的，一個程序員每天都可以發現一個bug，現在求的是發現n中bug並存在
於s個子系統中的平均天數（期望）。題意有點難懂，讀了好就遍了。
假設dp[i][j]表示已經發現了i種bug並存在於j個子系統中（註意是已經發現了），那麽dp[n][s] 就是 0了。
從dp[i][j]已經一天會出現4中情況
1、dp[i+1][j+1],出現一個新的bug並存在一個新的子系統中，概率是(n-i)*(s-j)/(n*s)
2、dp[i][j+1],在一個子系統中出現一個bug，不過這個bug種類是之前出現過的，概率是(s-j)*i/(n*s)
3、dp[i+1][j],出現一個新的bug，但是在子系統的已經出現過了，概率是(n-i)*j/(n*s)
4、dp[i][j],在已經發現過的bug中發現一個bug，概率是i*j/(n*s)
然後根據E(aA+bB+cC+dD+...)=aEA+bEB+....;//a,b,c,d...表示概率,A,B,C...表示狀態
所以dp[i][j]=p1*dp[i+1][j+1]+p2*dp[i+1][j]+p3*dp[i][j+1]+p4*dp[i][j]+1;//dp[i][j]表示的就是到達狀態i,j的期望 
=>dp[i][j]=(p1*dp[i+1][j+1]+p2*dp[i+1][j]+p3*dp[i][j+1]+1)/(1-p4);
其實看久點也還是可以看懂的，
 

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 
 5 using namespace std;
 6 double dp[1010][1010];
 7 int main(){
 8     int n,m;
 9     while(scanf("%d%d",&n,&m)!=EOF){
10         memset(dp,0,sizeof(dp));
11         for(int i=n;i>=0;i--){
12             for
 
(int j=m;j>=0;j--){
13                 if(i==n&&j==m){
14                     continue;
15                 }else{
16                     dp[i][j] = (i*(m-j)*dp[i][j+1]+(n-i)*j*dp[i+1][j]+(n-i)*(m-j)*dp[i+1][j+1]+n*m)/(n*m-i*j);
17 
18                 }
19             }
20         }
21         printf("%.4f\n",dp[0][0]);//註意保留四位小數
22     }
23     return 0;
24 }

Collecting Bugs