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.
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.
先说一说题意把:一个软件有 s 个子系统，存在 n 种 bug。某人一天能找到一个 bug。问，在这个软件中找齐 n 种 bug，并且每个子系统中至少包含一个 bug 的时间的期望值（单位：天）。注意：bug 是无限多的，每个 bug 属于任何一种 bug 的概率都是 1/n；出现在每个系统是等可能的，为 1/s。
同样先开一个二维数组f[i][j]，代表现在有了i种bug在j个系统中要到达目标的期望天数。
那么再分四种：1.找到的bug在j个系统中且在i种内。此时递推式:f[i][j]=f[i][j]*i/n*j/s.
2.在j系统内不在i种内，此时:f[i][j]=f[i+1][j]*(n-i)/n*j/s.
3.不在在j系统内在i种内，此时:f[i][j]=f[i][j+1]*(s-j)/s*i/n.
4.不在两者中，此时:f[i][j]=f[i+1][j+1]*(s-j)/s*(n-i)/n.
根据期望定义，我们合并式子:f[i][j]=f[i+1][j]*(n-i)/n*j/s+f[i][j+1]*(s-j)/s*i/n+f[i+1][j+1]*(s-j)/s*(n-i)/n+f[i+1][j+1]*(s-j)/s*(n-i)/n。
大家再随便化简一下吧。。(把f[i][j]移到左边去)
还有这题卡精度。lf过不了的f都能过。呵呵。
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm> using namespace std; double f[][]; int main()
{
int n,s;
while(~scanf("%d%d",&n,&s))
{
memset(f,,sizeof(f));
for(int i=n;i>=;i--)
for(int j=s;j>=;j--)
{
if(n==i&&j==s)continue;
f[i][j]=(double)(f[i+][j]*(n-i)*j+f[i][j+]*i*(s-j)+f[i+][j+]*(n-i)*(s-j)+n*s)/(n*s-i*j);
}
printf("%.4f\n",f[][]);
}
return ;
}