Poj 2096 (dp求期望 入门)】的更多相关文章

/ dp求期望的题. 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属于某种bug,发生在某个子系统中. 求找到所有的n种bug,且每个子系统都找到bug,这样所要的天数的期望. 需要注意的是:bug的数量是无穷大的,所以发现一个bug,出现在某个子系统的概率是1/s, 属于某种类型的概率是1/n. 解法: dp[i][j]表示已经找到i种bug,并存在于j个子系统中,要达到目标状态的天数的期望. 显然,dp[n][s]=0,因为已经达到目标了.而dp[0][…
A - Collecting Bugs Time Limit:10000MS     Memory Limit:64000KB     64bit IO Format:%I64d & %I64u Submit Status Appoint description:  System Crawler  (2014-05-15) Description Ivan is fond of collecting. Unlike other people who collect post stamps, co…
LOOPS Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others) Total Submission(s): 1864    Accepted Submission(s): 732 Problem Description Akemi Homura is a Mahou Shoujo (Puella Magi/Magical Girl). Homura wants to help h…
dp求期望的题. 题意: 有n个房间,由n-1条隧道连通起来,实际上就形成了一棵树, 从结点1出发,开始走,在每个结点i都有3种可能: 1.被杀死,回到结点1处(概率为ki) 2.找到出口,走出迷宫 (概率为ei) 3.和该点相连有m条边,随机走一条 求:走出迷宫所要走的边数的期望值. 设 E[i]表示在结点i处,要走出迷宫所要走的边数的期望.E[1]即为所求. 叶子结点: E[i] = ki*E[1] + ei*0 + (1-ki-ei)*(E[father[i]] + 1);//因为是到达,…
Collecting Bugs 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…
D - LOOPS Time Limit:5000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Description Akemi Homura is a Mahou Shoujo (Puella Magi/Magical Girl). Homura wants to help her friend Madoka save the world. But because of the plot…
C - Collecting Bugs Time Limit:10000MS     Memory Limit:64000KB     64bit IO Format:%I64d & %I64u Submit Status Description Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material stuff, he collects software b…
题目链接: http://poj.org/problem?id=2096 Collecting Bugs Time Limit: 10000MSMemory Limit: 64000K 问题描述 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…
题目大意: 一个人受雇于某公司要找出某个软件的bugs和subcomponents,这个软件一共有n个bugs和s个subcomponents,每次他都能同时随机发现1个bug和1个subcomponent,问他找到所有的bugs和subcomponents的期望次数. 这道题目要用期望dp来进行统计 假设已经找到i个bug和j个subcomponents,这个状态记为dp[i][j],那么下次查找会出现4种状态:dp[i][j],dp[i+1][j],dp[i][j+1],dp[i+1][j+…
题目链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=25915 题意:求一个数不断地除以他的因子,直到变成1的时候 除的次数的期望. 思路:设一个数的约数有num个,E[n] = (E[a[1]]+1)/num+(E[a[2]]+1)/num+...+(E[a[num]]+1)/num+1  ,而a[num]==n,于是整理得: E[n]=(E[a[1]]+E[a[2]]+...+E[a[num-1]]+num)/(n…