参考:<深度学习500问> 期望 ​在概率论和统计学中,数学期望(或均值,亦简称期望)是试验中每次可能结果的概率乘以其结果的总和.它反映随机变量平均取值的大小. 线性运算: \(E(ax+by+c) = aE(x)+bE(y)+c\) ​推广形式: \(E(\sum_{k=1}^{n}{a_ix_i+c}) = \sum_{k=1}^{n}{a_iE(x_i)+c}\) 函数期望:设\(f(x)\)为\(x\)的函数,则\(f(x)\)的期望为 离散函数: \(E(f(x))=\sum_{k=

题目链接 495. Kids and Prizes Time limit per test: 0.25 second(s)Memory limit: 262144 kilobytes input: standardoutput: standard ICPC (International Cardboard Producing Company) is in the business of producing cardboard boxes. Recently the company organiz

/ 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][

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

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

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

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

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=4870 原题: Rating Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 654    Accepted Submission(s): 415 Special Judge Problem Description    A little

Domination Time Limit: 8 Seconds      Memory Limit: 131072 KB      Special Judge Edward is the headmaster of Marjar University. He is enthusiastic about chess and often plays chess with his friends. What's more, he bought a large decorative chessboar

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

B - Card Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description There are x cards on the desk, they are numbered from 1 to x. The score of the card which is numbered i(1<=i<=x) is i. Every round Bi

Rating Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 872    Accepted Submission(s): 545Special Judge Problem Description A little girl loves programming competition very much. Recently, she 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,

poj 3744 Scout YYF I(递推求期望) 题链 题意:给出n个坑,一个人可能以p的概率一步一步地走,或者以1-p的概率跳过前面一步,问这个人安全通过的概率 解法: 递推式: 对于每个坑,我们可以这么定义一个数组: d[i]代表它安全落在位置i的概率,在这个1到max(a[i])的范围中,只有那些坑是不安全的,答案只需求出所有不掉入坑的概率的连乘即可 矩阵: 由于数字范围巨大,需要对递推式进行矩阵连乘加速 | p 1-p | | d[i] | | d[i+1] | | 1 0 | *

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

题目大意:在nxm的方格中,从(1,1)走到(n,m).每次只能在原地不动.向右走一格.向下走一格,概率分别为p1(i,j),p2(i,j),p3(i,j).求行走次数的期望. 题目分析:状态转移方程很容易得到: E(i,j)=p1(i,j)*E(i,j)+p2(i,j)*E(i,j+1)+p3(i,j)*E(i+1,j). 代码如下: # include<iostream> # include<cstdio> # include<cmath> # include<

题意:一个人若连续进k1个球或连续不进k2个球,游戏结束,给出这个人不进球的概率p(注意:是不进球!!!),求到游戏结束时这个投球个数的期望. 不进球概率为p,进概率 q=1-p.设 f[i] 表示连续 i 次不进结束的期望,t[i]表示连续 i 次进球,距离结束的期望.显然,f[k2]=t[k1]=0;f[i] = q*(f[i+1]+1)+p*(1+t[1]) , t[i] = p*(t[i+1]+1)+q*(1+f[1]). 答案是 p*t[1]+q*f[1]+1.然后就算t[1],f[1

Aeroplane chess Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1667    Accepted Submission(s): 1123 Problem Description Hzz loves aeroplane chess very much. The chess map contains N+1 grids lab

The number of steps Time Limit: 1000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描写叙述 Mary stands in a strange maze, the maze looks like a triangle(the first layer have one room,the second layer have two rooms,the third layer have three rooms -). Now she

题意:有三个骰子,分别有k1,k2,k3个面. 每次掷骰子,如果三个面分别为a,b,c则分数置0,否则加上三个骰子的分数之和. 当分数大于n时结束.求游戏的期望步数.初始分数为0 设dp[i]表示达到i分时到达目标状态(即i = n)的期望,pk为投掷k分的概率, p0为回到0的概率则dp[i] = ∑(pk * dp[i + k]) + dp[0] * p0 + 1 ; 都和dp[0]有关系,而且dp[0]就是我们所求,为常数设 dp[i] = A[i] * dp[0] + B[i]; 即为d

Time travel Problem Description Agent K is one of the greatest agents in a secret organization called Men in Black. Once he needs to finish a mission by traveling through time with the Time machine. The Time machine can take agent K to some point (0