POJ 2096 概率DP 解题报告
来源:互联网 发布:局域网设置域名 编辑:程序博客网 时间:2024/05/27 16:42
Collecting Bugs
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.
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
【解题报告】
这个题目有点繁琐。。。
题意,一个软件有s个子系统,会产生n种bug,
某人一天发现一个bug,
这个bug属于一个子系统,属于一个分类,
每个bug属于某个子系统的概率是1/s,属于某种分类的概率是1/n
问发现n种bug,每个子系统都发现bug的天数的期望
dp[i][j]表示已经找到i种bug,j个系统的bug,达到目标状态的天数的期望
dp[n][s]=0;要求的答案是dp[0][0];
dp[i][j]可以转化成以下四种状态:
dp[i][j],发现一个bug属于已经有的i个分类和j个系统。概率为(i/n)*(j/s);
dp[i][j+1],发现一个bug属于已有的分类,不属于已有的系统.概率为 (i/n)*(1-j/s);
dp[i+1][j],发现一个bug属于已有的系统,不属于已有的分类,概率为 (1-i/n)*(j/s);
dp[i+1][j+1],发现一个bug不属于已有的系统,不属于已有的分类,概率为 (1-i/n)*(1-j/s);
整理便得到转移方程
代码如下:
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define N 1010int n,s;double dp[N][N];int main(){ while(scanf("%d%d",&n,&s)!=EOF) { 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]=(i*(s-j)*dp[i][j+1]+(n-i)*j*dp[i+1][j]+(n-i)*(s-j)*dp[i+1][j+1]+n*s)/(n*s-i*j); } printf("%.4f\n",dp[0][0]); } return 0;}
- POJ 2096 概率DP 解题报告
- poj 2096 概率DP 解题报告
- HDU4089概率dp解题报告
- poj 2096 期望DP 解题报告
- HDU 3853 概率DP 解题报告
- POJ2096 Collecting Bugs 解题报告【概率DP】
- Poj 1260 DP 解题报告
- POJ 1260 DP解题报告
- POJ 1160 DP 解题报告
- poj 2096 概率dp
- poj 2096 概率dp
- POJ 2096 概率dp
- poj 2096<概率dp>
- poj 2096 概率dp
- POJ 2096 概率dp
- poj 2096 概率dp
- POJ 2096 概率DP
- poj 2096 概率dp
- 学习英语的网站
- MaterialDesign学习篇(二),Toolbar、DrawerLayout的使用
- IP判断 oj139
- ZeroClipboard实现复制功能,做到浏览器兼容
- 2017/7/24 离线赛
- POJ 2096 概率DP 解题报告
- 关于cmd界面无法启动mysql
- 今天所学知识点 (复制老师的)
- 替代glup和grunt,使用npm自动化构建项目
- 内存解析--JAVA
- Oracle 11gR2 RAC实时应用集群
- 数据库SQL优化大总结之 百万级数据库优化方案
- js练习之纸牌互换
- 发现规律&&codeforce 590A