期待-kuangbin A-Collecting Bugs POJ-2096
3626 ワード
二級ディレクトリを返す——kaugnbin確率dp練習問題セット
ブログの目次
原題転送ゲート(プライベート)
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
Sample Output
各种类のbugの出现の确率は同じですべて1/nで、すべてのbugなどの确率はすべてのシステムの中ですべて1/sで现れて、すべてのシステムは无限のbugがあって、毎日1つのbugを探し当てて、すべてのシステムがすべて少なくとも1つのbugがあることを保证する前提の下で、全n种类のbugの期待の日数を探します.
既知の条件がe(end)=0であることが望ましい
dp[i][j]は、現在iつのシステムにバグが存在し、jクラスバグが見つかったことを示し、dp値が現在の状態からdp[s][n]に移行する期待であれば、dp[s]{n]=0である.
dp[i][j]=dp[i][j+1]*i*(n-j)/(s*n)+dp[i+1][j]*(s-i)*j/(s*n)+dp[i+1][j+1]*(s-i)*(n-j)/(s*n)+dp[i][j]*(i*j)/(s*n)+1;
dp[i][j]を片側解方程式に移動して最終式を得る:
dp[i][j]=dp[i][j+1]*i*(n-j)/(s*n-i*j) + dp[i+1][j]*(s-i)*j/(s*n-i*j) + dp[i+1][j+1]*(s-i)*(n-j)/(s*n-i*j)+ 1.0*s*n/(s*n-i*j)
#include using namespace std; double dp[1002][1002]; int main(){ int n,s; cin>>n>>s; //dp[i][j]=dp[i][j+1]*i*(n-j)/(s*n-i*j) + dp[i+1][j]*(s-i)*j/(s*n-i*j) + dp[i+1][j+1]*(s-i)*(n-j)/(s*n-i*j)+ 1.0*s*n/(s*n-i*j) //dp[s]{n]=0; for(int i=s;i>=0;i--){ for(int j=n;j>=0;j--) if(i!=s || j!=n) dp[i][j]=dp[i][j+1]*i*(n-j)/(s*n-i*j) + dp[i+1][j]*(s-i)*j/(s*n-i*j)+ dp[i+1][j+1]*(s-i)*(n-j)/(s*n-i*j)+ 1.0*s*n/(s*n-i*j); } printf("%.6f",dp[0][0]); }
ブログの目次
原題
原題転送ゲート(プライベート)
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 2Sample Output
3.0000に言及
各种类のbugの出现の确率は同じですべて1/nで、すべてのbugなどの确率はすべてのシステムの中ですべて1/sで现れて、すべてのシステムは无限のbugがあって、毎日1つのbugを探し当てて、すべてのシステムがすべて少なくとも1つのbugがあることを保证する前提の下で、全n种类のbugの期待の日数を探します.
構想
既知の条件がe(end)=0であることが望ましい
dp[i][j]は、現在iつのシステムにバグが存在し、jクラスバグが見つかったことを示し、dp値が現在の状態からdp[s][n]に移行する期待であれば、dp[s]{n]=0である.
dp[i][j]=dp[i][j+1]*i*(n-j)/(s*n)+dp[i+1][j]*(s-i)*j/(s*n)+dp[i+1][j+1]*(s-i)*(n-j)/(s*n)+dp[i][j]*(i*j)/(s*n)+1;
dp[i][j]を片側解方程式に移動して最終式を得る:
dp[i][j]=dp[i][j+1]*i*(n-j)/(s*n-i*j) + dp[i+1][j]*(s-i)*j/(s*n-i*j) + dp[i+1][j+1]*(s-i)*(n-j)/(s*n-i*j)+ 1.0*s*n/(s*n-i*j)
ACコード
#include using namespace std; double dp[1002][1002]; int main(){ int n,s; cin>>n>>s; //dp[i][j]=dp[i][j+1]*i*(n-j)/(s*n-i*j) + dp[i+1][j]*(s-i)*j/(s*n-i*j) + dp[i+1][j+1]*(s-i)*(n-j)/(s*n-i*j)+ 1.0*s*n/(s*n-i*j) //dp[s]{n]=0; for(int i=s;i>=0;i--){ for(int j=n;j>=0;j--) if(i!=s || j!=n) dp[i][j]=dp[i][j+1]*i*(n-j)/(s*n-i*j) + dp[i+1][j]*(s-i)*j/(s*n-i*j)+ dp[i+1][j+1]*(s-i)*(n-j)/(s*n-i*j)+ 1.0*s*n/(s*n-i*j); } printf("%.6f",dp[0][0]); }