Collecting Bugs
Time Limit: 10000MS | Memory Limit: 64000K | |
Total Submissions: 6361 | Accepted: 3113 | |
Case Time Limit: 2000MS | Special Judge |
Description
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
Output
Sample Input
1 2
Sample Output
3.0000
Source
题意
每次发现一个bug,求找到所有种类的bug,且每个子系统都要有找到一个bug,这样所要的次数的期望。
分析
dp[i][j]表示已经找到i种bug,并存在于j个子系统中,要达到目标状态的天数的期望。
显然,dp[n][s]=0。并且dp[0][0]就是我们要求的答案。
dp[i][j]状态可以转化成以下四种:
dp[i][j]: 发现一个bug属于已经找到的i种bug和j个子系统中,概率(i/n)*(j/s)
dp[i+1][j]: 发现一个bug属于新的一种bug,但属于已经找到的j种子系统,概率(1-i/n)*(j/s)
dp[i][j+1]: 发现一个bug属于已经找到的i种bug,但属于新的子系统,概率(i/n)*(1-j/s)
dp[i+1][j+1]: 发现一个bug属于新的一种bug和新的一个子系统,概率(1-i/n)*(1-j/s)
性质:期望可以分解成多个子期望的加权和,权为子期望发生的概率,即 E(aA+bB+...) = aE(A) + bE(B) +...
所以:将上面的概率变为
p1 = i*j / (n*s)
p2 = (n-i)*j / (n*s)
p3 = i*(s-j) / (n*s)
p4 = (n-i)*(s-j) / (n*s)
dp[i,j] = p1*dp[i,j] + p2*dp[i+1,j] + p3*dp[i,j+1] + p4*dp[i+1,j+1] + 1;
整理得:
dp[i,j] = ( 1 + p2*dp[i+1,j] + p3*dp[i,j+1] + p4*dp[i+1,j+1] )/( 1-p1 )
= ( n*s + (n-i)*j*dp[i+1,j] + i*(s-j)*dp[i,j+1] + (n-i)*(s-j)*dp[i+1,j+1] )/( n*s - i*j )
code
1 #include<cstdio> 2 3 const int N = 1010; 4 5 double dp[N][N];//dp[i][j]表示已经找到i种bug,j个系统的bug,达到目标状态的天数的期望 6 7 int main() { 8 9 int n,s;// n种类, s子系统 10 while (~scanf("%d%d",&n,&s)) { 11 dp[n][s] = 0; 12 for (int i=n; i>=0; --i) 13 for (int j=s; j>=0; --j) { 14 if (i==n && j==s) continue; 15 dp[i][j] = (i*(s-j)*dp[i][j+1]+(n-i)*j*dp[i+1][j]+(n-i)*(s-j)*dp[i+1][j+1]+n*s)/(n*s-i*j); 16 } 17 printf("%.4lf ",dp[0][0]); 18 } 19 return 0; 20 }
参考:http://www.cnblogs.com/jackge/archive/2013/05/21/3091757.html