題干：

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

題目大意：

一個軟件有s個子系統(tǒng)，會產(chǎn)生n種bug。某人一天一定會發(fā)現(xiàn)一個bug，這個bug屬于某種bug，發(fā)生在某個子系統(tǒng)中。求找到所有的n種bug，且每個子系統(tǒng)都找到bug，這樣所要的天數(shù)的期望。

解題報告：

需要注意的是：bug的數(shù)量是無窮大的，所以發(fā)現(xiàn)一個bug，出現(xiàn)在某個子系統(tǒng)的概率是1/s，屬于某種類型的概率是1/n。

又有：期望可以分解成多個子期望的加權(quán)和，權(quán)為子期望發(fā)生的概率，即 E(aA+bB+...) = aE(A) + bE(B) +...

以下題解來自kuangbin：

dp[i][j]表示已經(jīng)找到i種bug,j個系統(tǒng)的bug，達(dá)到目標(biāo)狀態(tài)的天數(shù)的期望dp[n][s]=0;要求的答案是dp[0][0];dp[i][j]可以轉(zhuǎn)化成以下四種狀態(tài):dp[i][j],發(fā)現(xiàn)一個bug屬于已經(jīng)有的i個分類和j個系統(tǒng)。概率為(i/n)*(j/s);dp[i][j+1],發(fā)現(xiàn)一個bug屬于已有的分類，不屬于已有的系統(tǒng).概率為 (i/n)*(1-j/s);dp[i+1][j],發(fā)現(xiàn)一個bug屬于已有的系統(tǒng)，不屬于已有的分類,概率為 (1-i/n)*(j/s);dp[i+1][j+1],發(fā)現(xiàn)一個bug不屬于已有的系統(tǒng)，不屬于已有的分類,概率為 (1-i/n)*(1-j/s);整理便得到轉(zhuǎn)移方程

注意不能用注釋的那個代碼，因?yàn)闆]有整理，所以等號左側(cè)和右側(cè)都有dp[i][j]，這樣是不行的。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define F first #define S second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5; int n,s; double dp[1555][1555]; int main() {cin>>n>>s;dp[n][s]=0;for(int i = n; i>=0; i--) {for(int j = s; j>=0; j--) {if(i == n && j == s) continue; // dp[i][j] = 1 + dp[i+1][j] * (n-i)*j/n/s + dp[i][j] * i*j/n/s + dp[i][j+1] *i*(s-j)/n/s + dp[i+1][j+1]*(n-i)*(s-j)/n/s;dp[i][j]=((n-i)*j*dp[i+1][j]+(n-i)*(s-j)*dp[i+1][j+1]+i*(s-j)*dp[i][j+1]+n*s)/(n*s-i*j);}}printf("%.4f\n",dp[0][0]);return 0 ; }

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