题目描述

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.

输入

Input file contains two integer numbers, n and s (0<n,s<=1000).

输出

Output the expectation of the Ivan's working days needed to call the program disgusting, accurate to 4 digits after the decimal point.

样例输入

1 2

样例输出

3.0000


题目大意

共有n种bug和s个系统,每天随机发现1个系统中的1种bug,问:发现所有种类的bug,且每个系统都发现bug的期望天数。

题解

期望dp

f[i][j] = f[i+1][j+1]*(n-i)/n*(s-j)/s + f[i][j+1]*i/n*(s-j)/s + f[i+1][j]*(n-i)/n*j/s + f[i][j]*i/n*j/s + 1

移项,合并同类项,化简

#include <cstdio>
double f[1002][1002];
int main()
{
int n , s , i , j;
scanf("%d%d" , &n , &s);
for(i = n ; i >= 0 ; i -- )
for(j = s ; j >= 0 ; j -- )
if(i != n || j != s)
f[i][j] = (f[i + 1][j + 1] * (n - i) * (s - j) + f[i][j + 1] * i * (s - j) + f[i + 1][j] * (n - i) * j + (n * s)) / (n * s - i * j);
printf("%.4lf\n" , f[0][0]);
return 0;
}

【poj2096】Collecting Bugs 期望dp的更多相关文章

  1. poj2096 Collecting Bugs[期望dp]

    Collecting Bugs Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 5394   Accepted: 2670 ...

  2. POJ2096 Collecting Bugs(概率DP,求期望)

    Collecting Bugs Ivan is fond of collecting. Unlike other people who collect post stamps, coins or ot ...

  3. POJ 2096 Collecting Bugs 期望dp

    题目链接: http://poj.org/problem?id=2096 Collecting Bugs Time Limit: 10000MSMemory Limit: 64000K 问题描述 Iv ...

  4. [POJ2096] Collecting Bugs (概率dp)

    题目链接:http://poj.org/problem?id=2096 题目大意:有n种bug,有s个子系统.每天能够发现一个bug,属于一个种类并且属于一个子系统.问你每一种bug和每一个子系统都发 ...

  5. [Poj2096]Collecting Bugs(入门期望dp)

    Collecting Bugs Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 6237   Accepted: 3065 ...

  6. 【POJ2096】Collecting Bugs 期望

    [POJ2096]Collecting Bugs Description Ivan is fond of collecting. Unlike other people who collect pos ...

  7. poj2096 Collecting Bugs(概率dp)

    Collecting Bugs Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 1792   Accepted: 832 C ...

  8. POJ 2096 Collecting Bugs (概率DP,求期望)

    Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material stu ...

  9. Poj 2096 Collecting Bugs (概率DP求期望)

    C - Collecting Bugs Time Limit:10000MS     Memory Limit:64000KB     64bit IO Format:%I64d & %I64 ...

随机推荐

  1. LeetCode: 61. Rotate List(Medium)

    1. 原题链接 https://leetcode.com/problems/rotate-list/description/ 2. 题目要求 给出一个链表的第一个结点head和正整数k,然后将从右侧开 ...

  2. 封装一个CSVHelper

    public class CSVHelper { /// <summary> /// CSV转换成DataTable(OleDb数据库访问方式) /// </summary> ...

  3. spring-boot、spring-data-jpa、hibernate整合

    一.Spring Boot是由Pivotal团队提供的全新框架,其设计目的是用来简化新Spring应用的初始搭建以及开发过程.该框架使用了特定的方式来进行配置,从而使开发人员不再需要定义样板化的配置. ...

  4. Spring的定时任务(任务调度)<task:scheduled-tasks>

    Spring内部有一个task是Spring自带的一个设定时间自动任务调度,提供了两种方式进行配置,一种是注解的方式,而另外一种就是XML配置方式了.注解方式比较简洁,XML配置方式相对而言有些繁琐, ...

  5. 两个有序数组合并成一个有序数组(要求时间复杂度为O(n))

    面试题: 怎样把两个有序数组合并成有序数组呢 逻辑步骤: 1.假设两个数组为A和B 2.A和B都是从小到大的顺序进行排列 ** 1.我们可以直接比较两个数组的首元素,哪个小就把这个小元素放入可变数组. ...

  6. unity3d 角色头顶信息3D&2D遮挡解决方案(二)

    在阅读本文之前请先阅读上一篇文章:http://www.cnblogs.com/shenggege/p/4179012.html 本来一篇文章就可以说完了,但是上次只是实现了已知的一些功能 后来在实际 ...

  7. git服务器搭建及eclipse使用git

    一.搭建git服务器 1.yum install git 2.新建用户linux用户git,管理git服务 useradd git passwd git 3.初始化git仓库 git init --b ...

  8. JS实现对数组的去重

    JS实现对数组的去重 $scope.validateContect = function(text) { var arr = text; // 若传入的数据为string类型,用逗号分隔 if((ty ...

  9. C 计算时间差

    #include <stdio.h>int main(){ //新建四个变量 la 代表小时 kc代表时间 int l,k,a,c; //输入 两个时间 scanf("%d %d ...

  10. JavaScript 常用正则示例

    1. trim功能(清除字符串两端空格) String.prototype.trim = function() {  return this.replace(/(^\s+)|(\s+$)/g, '') ...