Collecting Bugs Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 5394   Accepted: 2670 Case Time Limit: 2000MS   Special Judge Description Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material s…

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…

Collecting Bugs Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 6237   Accepted: 3065 Case Time Limit: 2000MS   Special Judge Description Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material s…

Collecting Bugs Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 1792   Accepted: 832 Case Time Limit: 2000MS   Special Judge Description Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material st…

偷一波翻译: 工程师可以花费一天去找出一个漏洞——这个漏洞可以是以前出现过的种类,也可能是未曾出现过的种类,同时,这个漏洞出现在每个系统的概率相同.要求得出找到n种漏洞,并且在每个系统中均发现漏洞的期望天数. Translated By 大米饼 求期望天数...那就推式子吧:dp[i][j]表示已经找出了i个漏洞,已经出现在了j个系统中的期望天数 $$dp[i][j]=(dp[i][j]+1)*\frac{i*j}{s*n}+(dp[i][j+1]+1)*\frac{i*(s-j)}{n*s}+…

题目大意:有n种bug,m个程序,小明每天能找到一个bug.每次bug出现的种类和所在程序都是等机会均等的,并且默认为bug的数目无限多.如果要使每种bug都至少找到一个并且每个程序中都至少找到一个bug,小明平均需要找几天? 题目分析:定义状态dp(i,j)表示找到了 i 种程序,并且出现在了 j 个程序中还需要的平均时间.则状态转移方程为: dp(i,j)=i/n*j/m*dp(i,j)+(n-i)/n*j/m*dp(i+1,j)+i/n*(m-j)/m*dp(i,j+1)+(n-i)/n*…

传送门:http://poj.org/problem?id=2096 题面很长,大意就是说,有n种bug,s种系统,每一个bug只能属于n中bug中的一种,也只能属于s种系统中的一种.一天能找一个bug,问找到的bug涵盖所有种类的bug与所有种类的系统期望需要几天. 令f(i, j)为找到了i种bug,j种系统期望的天数,那么今天再找一个bug,有4种情况: ①,bug种类为已找到的i种中的一种,系统种类为已找到的j种中的一种,则概率p1 = (i / n) * (j / s) ②,bug种类…

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现bug需要多少天. 设dp[i][j]为现在发现了i种bug,在j个子系统内,到目标状态需要的期望天数. 则dp[n][s] ＝ 0,因为已经发现了n种bug在s个子系统内,不需要额外的天数了. dp[i][j]可以转移到如下状态: dp[i][j]:新发现的bug在已经发现的i种,已经发现的j个…

Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 5090   Accepted: 2529 Case Time Limit: 2000MS   Special Judge Description Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material stuff, he collect…

期望\(DP\) 方法总结 这个题目太大了,变化也层出不穷,这里只是我的一点心得,不定期更新! 1. 递推式问题 对于无穷进行的操作期望步数问题,一般可用递推式解决. 对于一个问题\(ans[x]\), 我们可以考虑建立逻辑转移: \[ans[now] = Merge(\ \ Function(ans[now])\ ,\ Function(ans[other])\ \ )\] 那么我们进行移项后, \[ans[now]\ Delete\ Function(ans[now])\ \ =\ \ Fu…

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…

[POJ2096]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. E…

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 exac…

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…

Collecting Bugs Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 3523   Accepted: 1740 Case Time Limit: 2000MS   Special Judge Description Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material s…

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…

                                                            Collecting Bugs Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 1899   Accepted: 901 Case Time Limit: 2000MS   Special Judge Description Ivan is fond of collecting. Unlike othe…

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 题意: 有一个程序猿,他每天都会发现一个bug. bug共有n个种类.属于某一个种类的概率为1/n. 有s个子系统,每个bug属于一个系统.属于某一个系统的概率为1/s. 问你发现的bug能够覆盖到n个种类和s个系统的期望天数. 题解: 期望dp转移的套路: 倒着推. 利用性质:期望 = ∑ (P(子期望)*φ(子期望)) 状态表示: dp[i][j] = expectation i:覆盖到i个种类 j:覆盖到j个系统 dp…

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…

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…

B. Collecting Bugs Time Limit: 10000ms Memory Limit: 64000KB 64-bit integer IO format: %lld      Java class name: Main Special Judge   Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material stuff, he collects…

http://poj.org/problem?id=2096 (题目链接) 题意 有一个程序,其中有s个子结构,每个子结构出bug的概率相等.bug总共分成n类,每种bug出现的概率相等.每天找出一个bug,求所有子结构都出现bug并且每种bug都出现过所需要的期望天数. Solution 终于领会了期望dp的一点点... 细节 %.4lf用G++交会Wa.. 代码 // poj2096 #include<algorithm> #include<iostream> #include…

Collecting Bugs Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 2745   Accepted: 1345 Case Time Limit: 2000MS   Special Judge Description Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material s…

Collecting Bugs Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 6400   Accepted: 3128 Case Time Limit: 2000MS   Special Judge Description Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material s…

Collecting Bugs Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 1716   Accepted: 783 Case Time Limit: 2000MS   Special Judge Description Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material st…

Collecting Bugs Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 3064   Accepted: 1505 Case Time Limit: 2000MS   Special Judge Description Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material s…

一直不明白为什么概率是正推,期望是逆推. 现在题目做多了,慢慢好像有点明白了 poj2096 收集bug,  有n个种类的bug,和s个子系统.  每找到一个bug需要一天. 要我我们求找到n个种类的bug,且在每个系统中都找到一个bug的期望天数 设dp[i][j] 为找到i个种类的bug和在j个系统中找到bug后,还需要的期望天数 那么dp[n][s] 肯定是0,而dp[0][0]是我们要求的. 这也就是为什么期望是要逆推. 还有一点就是这一状态的期望会等于   所有(下一状态的的期望*这一…

[总览] [期望dp] 求解达到某一目标的期望花费:因为最终的花费无从知晓(不可能从$\infty$推起),所以期望dp需要倒序求解. 设$f[i][j]$表示在$(i, j)$这个状态实现目标的期望值(相当于是差距是多少). 首先$f[n][m] = 0$,在目标状态期望值为0.然后$f = (\sum f' × p) + w $,$f'$为上一状态(距离目标更近的那个,倒序),$p$为从$f$转移到$f'$的概率(则从$f'$转移回$f$的概率也为$p$),w为转移的花费. 最后输出初始位置…

对于概率dp,我一直都弄得不是特别明白,虽然以前也有为了考试去突击过,但是终究还是掌握得不是很好,所以决定再去学习一遍,把重要的东西记录下来. 1.hdu4405 Description 在一个 \(1*n\) 的格子上掷色子,从 \(0\) 点出发,掷了多少前进几步,同时有些格点直接相连,即若 \(a\) ,\(b\) 相连,当落到 \(a\) 点时直接飞向 \(b\) 点.求走到 \(n\) 或超出 \(n\) 期望掷色子次数 \(n \leq 100000\) Solution 这道题目有…