概率与期望 POJ 2096 Collecting Bugs
来源:互联网 发布:js 扫描条形码 编辑:程序博客网 时间:2024/06/14 05:13
题目:
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.
Sample Input
1 2
Sample Output
3.0000
相当于求max(X,Y)的期望,其中X和Y都是超几何分布。
不过,用数学方法求解是非常难的(貌似是可以求的,不过组合的计算实在是复杂)
这个题目要用递推式来求解,
代码:
#include<iostream>#include<iomanip>#include<string.h>using namespace std;int n, s;double list[1001][1001];double f(int i, int j){if (list[i][j] >= 0)return list[i][j];if (i > n || j > s)return 0;double d = n*s;d += i*(s - j)*f(i, j + 1);d += (n - i)*j*f(i + 1, j);d += (n - i)*(s - j)*f(i + 1, j + 1);list[i][j] = d / (n*s - i*j);return list[i][j];}int main(){cin >> n >> s;memset(list, -1, sizeof(list));list[n][s] = 0;cout << fixed << setprecision(4) << f(0, 0);return 0;}
2 0
- 概率与期望 POJ 2096 Collecting Bugs
- Poj 2096 Collecting Bugs (概率与期望)
- 【POJ 2096】Collecting Bugs(概率与期望+dp)
- poj 2096 Collecting Bugs (概率与期望DP)
- Collecting Bugs+POJ 2096+概率期望dp
- POJ 2096 Collecting Bugs (概率期望)
- POJ 2096 Collecting Bugs 概率DP(期望)
- POJ 2096Collecting Bugs(概率期望dp)
- POJ 2096 Collecting Bugs(概率DP求期望)
- [ACM] poj 2096 Collecting Bugs (概率DP,期望)
- poj 2096 Collecting Bugs 概率dp求期望
- poj 2096 Collecting Bugs 概率dp(期望)
- POJ 2096 - Collecting Bugs(概率DP 求期望)
- poj 2096 Collecting Bugs 概率DP求期望(简单)
- POJ 2096 Collecting Bugs 概率dp 求期望 入门
- Poj 2096 Collecting Bugs(概率(期望)+dp)
- poj 2096 Collecting Bugs 【概率DP】【逆向递推求期望】
- POJ 2096 Collecting Bugs(概率DP求期望)
- 在Dubbo中开发REST风格的远程调用(RESTful Remoting)
- 解决VS Code编译调试中文输出乱码
- reverse list (recursive)
- 模拟实现部分string函数
- eclipse console 看不到全部的输出解决办法
- 概率与期望 POJ 2096 Collecting Bugs
- System.getenv()
- 我搞ACM的心得
- Java学习3---多态
- java常量池概念
- a[i++]
- cover letter 和response letter的写法
- 深入浅出Dubbo(一)
- 洛谷 P1541 [NOIP2010 T2] 乌龟棋