題意：

有s個(gè)系統(tǒng)，n種bug，小明每天找出一個(gè)bug，可能是任意一個(gè)系統(tǒng)的，可能是任意一種bug，即是某一系統(tǒng)的bug概率是1/s，是某一種bug概率是1/n。 求他找到s個(gè)系統(tǒng)的bug，n種bug，需要的天數(shù)的期望。~~題意很難懂··真的很難懂···~~

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題目：

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.

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Sample Input

1 2

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Sample Output

3.0000?

分析：

基礎(chǔ)的概率dp。
令f[i][j]為當(dāng)前已經(jīng)找出i種bug，j個(gè)系統(tǒng)的，要到達(dá)目標(biāo)狀態(tài)（n,s)的期望天數(shù)。顯然f[n][s]=0；，因?yàn)橐呀?jīng)達(dá)到目標(biāo)了。而dp[0][0]就是我們要求的答案。
考慮一下狀態(tài)的轉(zhuǎn)移根據(jù)題意，每個(gè)dp[i][j] 之后可以達(dá)到的狀態(tài)有四種。
1.轉(zhuǎn)移到[i+1][j]的概率為(n-i)/n*?j/s
2.轉(zhuǎn)移到[i][j+1]的概率為i/n*(s-j)/s
3.轉(zhuǎn)移到[i+1][j+1]的概率為（n-i)/n*(s-j)/s
4.轉(zhuǎn)移到[i][j]的概率為i/n*j/s
f[i][j]=(n-i)/n*j/s*f[i+1][j]+i/n*(s-j)/s*f[i][j+1]+f[i+1][j+1]*(n-i)/n*(s-j)/s+f[i][j]*i/n*j/s
移一下項(xiàng)
(1-i/n*j/s)*f[i][j]=(n-i)/n*j/s*f[i+1][j]+i/n*(s-j)/s*f[i][j+1]+f[i+1][j+1]*(n-i)/n*(s-j)/s
然后把左邊的除過去得
f[i][j]=((n-i)/n*j/s*f[i+1][j]+i/n*(s-j)/s*f[i][j+1]+f[i+1][j+1]*(n-i)/n*(s-j)/s)/(1-i/n*j/s)
然后整理一下得
f[i][j]=(f[i+1][j]*(n-i)*j+f[i][j+1]*(s-j)*i+f[i+1][j+1]*(n-i)*(s-j)+n*s)/(n*s-i*j);

AC代碼

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#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int M=1000+100; double f[M][M]; int n,s; int main() {while(~scanf("%d%d",&n,&s)){memset(f,0,sizeof(f));for(int i=n; i>=0; i--)for(int j=s; j>=0; j--){if(i==n&&j==s)continue;f[i][j]=(f[i+1][j]*(n-i)*j+f[i][j+1]*(s-j)*i+f[i+1][j+1]*(n-i)*(s-j)+n*s)/(n*s-i*j);}printf("%.4f\n",f[0][0]);}return 0; }

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