集训第六周古典概型期望 D题 Discovering Gold 期望

Description You are in a cave, a long cave! The cave can be represented by a 1 x N grid. Each cell of the cave can contain any amount of gold. Initially you are in position 1. Now each turn you throw a perfect 6 sided dice. If you get X in the dice a…

集训第六周古典概型期望 C题

http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=30728 一个立体方块,每个单位方块都是关闭状态,每次任两个点,以这两点为对角线的一个立方体状态都会进行一次转变,(开变成关,关变成开) 如此进行k次后,问开着的灯的期望值思路:枚举所有的X,Y,Z,此灯被选中的概率为p=((2*(N-x+1)*x-1)*(2*(M-y+1)*y-1)*(2*(Z-z+1)*z-1))/(N*N*M*M*Z*Z) 这一点最后开着的期望值为(…

集训第六周数学概念与方法 J题数论，质因数分解

Description Tomorrow is contest day, Are you all ready? We have been training for 45 days, and all guys must be tired.But , you are so lucky comparing with many excellent boys who have no chance to attend the Province-Final. Now, your task is relaxin…

LightOj-1030 Discovering Gold (期望DP)

You are in a cave, a long cave! The cave can be represented by a 1 x N grid. Each cell of the cave can contain any amount of gold. Initially you are in position 1. Now each turn you throw a perfect 6 sided dice. If you get X in the dice after throwin…

集训第六周数学概念与方法 UVA 11722 几何概型

---恢复内容开始--- http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=31471 题意,两辆火车,分别会在[t1,t2],[s1,s2]的时间段停留在同一个站点w分钟,问两辆火车能够在这个站点相遇的概率. 思路,枚举每一种情况,把两辆火车的相交区间画出来,然后求都在这个区间的概率 #include"iostream" #include"cstdio" #include"cmath&…

CF Gym 100187B A Lot of Joy （古典概型）

题意:给两个一样的只含有26个小写字母的字符串,然后两个分别做一下排列,问如果对应位置的字母相等那么就愉悦值就加一,问愉悦值的期望是多少? 题解:只考虑两个序列相对的位置,那么就相当于固定一个位置,另外一个序列做排列.对于一个字符,假设是a,然后a有Na个,那么a选第一个序列中a对应的位置,发生这个事件的概率为1*(n-1)!/n!,1是选那个位置,(n-1)!是其他的元素排列,n!是总的情况.(注意,不能因为a可以选多个点就乘上x,因为a不可能同时对应两个位置,所以对答案的贡献只有一次) 学习…

集训第六周 E题

E - 期望(经典问题) Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu Submit Status Description Given a dice with n sides, you have to find the expected number of times you have to throw that dice to see all its faces at least once.…

python数学第四天【古典概型】

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集训第六周 O题

Description Gerald got a very curious hexagon for his birthday. The boy found out that all the angles of the hexagon are equal to . Then he measured the length of its sides, and found that each of them is equal to an integer number of centimeters. Th…

集训第六周 M题

Description During the early stages of the Manhattan Project, the dangers of the new radioctive materials were not widely known. Vast new factory cities were built to manufacture uranium and plu- tonium in bulk. Compounds and solutions of these sub…

集训第六周矩阵快速幂 K题

Description In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, … An alternative formula for the Fibonacci sequence is…

集训第六周数学概念与方法计数排列 L题

Description 大家常常感慨,要做好一件事情真的不容易,确实,失败比成功容易多了! 做好“一件”事情尚且不易,若想永远成功而总从不失败,那更是难上加难了,就像花钱总是比挣钱容易的道理一样. 话虽这样说,我还是要告诉大家,要想失败到一定程度也是不容易的.比如,我高中的时候,就有一个神奇的女生,在英语考试的时候,竟然把40个单项选择题全部做错了!大家都学过概率论,应该知道出现这种情况的概率,所以至今我都觉得这是一件神奇的事情.如果套用一句经典的评语,我们可以这样总结:一个人做错一道选择题并不…

集训第六周数学概念与方法数论线性方程 I题

Description The Sky is Sprite. The Birds is Fly in the Sky. The Wind is Wonderful. Blew Throw the Trees Trees are Shaking, Leaves are Falling. Lovers Walk passing, and so are You. ................................Write in English class by yifenfei Gir…

集训第六周数学概念与方法概率 N题

N - 概率 Time Limit:4000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu Submit Status Description As Harry Potter series is over, Harry has no job. Since he wants to make quick money, (he wants everything quick!) so he decided to rob banks.…

集训第六周数学概念与方法数论筛素数 H题

Description 小明对数的研究比较热爱,一谈到数,脑子里就涌现出好多数的问题,今天,小明想考考你对素数的认识. 问题是这样的:一个十进制数,如果是素数,而且它的各位数字和也是素数,则称之为“美素数”,如29,本身是素数,而且2+9 = 11也是素数,所以它是美素数. 给定一个区间,你能计算出这个区间内有多少个美素数吗? Input 第一行输入一个正整数T,表示总共有T组数据(T <= 10000). 接下来共T行,每行输入两个整数L,R(1<= L <= R <=…

集训第六周数学概念与方法概率数论最大公约数 G题

Description There is a hill with n holes around. The holes are signed from 0 to n-1. A rabbit must hide in one of the holes. A wolf searches the rabbit in anticlockwise order. The first hole he get into is the one signed with 0. Then he will get into…

集训第六周数学概念与方法概率 F题

Submit Status Description Sometimes some mathematical results are hard to believe. One of the common problems is the birthday paradox. Suppose you are in a party where there are 23 people including you. What is the probability that at least two peopl…

集训第六周数学概念与方法 UVA 11181 条件概率

http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=18546 题意:有n个人会去超市,其中只有r个人会买东西,每个人独自买东西的概率会给出,问这一群人去买东西,第i个人属于r之中的概率是多少思路:首先得了解什么是条件概率. 条件概率:事件A在事件B成立的基础上再成立的概率,公式为:P(A|B)=P(A*B)/P(B) 可以照着题目案例1进行分析: 输入 0.10 0.20 0.30 输出 0.413043 0.739130…

暑假集训第六周contest1

51Nod - 1413 权势二进制题意:就是讲给出一个数n,让你求最少由多少个像0,1,10,11......这样的二进制数相加构成:样例n=9就是由9个二进制1相加组成,我不懂比赛的时候我为什么没有看懂题目在讲什么:再举个例子吧!n=2333 就有二进制1111,1111,111三个数相加而成: 代码: #include<iostream> #include<cstdio> #include<cmath> using namespace std; int main…

LightOJ 1030 Discovering Gold(期望概率)

正推,到达i的概率为p[i],要注意除了1和n外,到达i的概率并不一定为1 概率表达式为p[i] += p[j] / min(n - j, 6) 从j带过来的期望为exp[i] += exp[j] / min(n - j, 6) 又到达i时有价值val[i],到达i的概率为p[i],故exp[i] += val[i] * p[i] #include<cstdio> #include<iostream> #include<cstdlib> #include<cstr…

2018.10.31 NOIP训练锻造（方程式期望入门题）（期望dp）

传送门根据题目列出方程: fi=pi∗(fi−1+fi−2)+(1−pi)∗(fi+1+fi)f_i=p_i*(f_{i-1}+f_{i-2})+(1-p_i)*(f_{i+1}+f_i)fi=pi∗(fi−1+fi−2)+(1−pi)∗(fi+1+fi) 但这会牵扯到iii之后的状态没法做. 因此考虑如果合成失败会变成一个等级为i−2i-2i−2的武器. 相当于消耗了一个等级为i−1i-1i−1的武器. 因此fi=pi∗(fi−1+fi−2)+(1−pi)∗(fi−1+fi)f…

LightOJ - 1030 Discovering Gold —— 期望

题目链接:https://vjudge.net/problem/LightOJ-1030 1030 - Discovering Gold PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB You are in a cave, a long cave! The cave can be represented by a 1 x N grid. Each cell of the cave can…

JAVA实验报告四及第六周总结

JAVA第六周作业实验报告四第一题 (1)根据下面的要求实现圆类Circle. 1.圆类Circle的成员变量:radius表示圆的半径. 2.圆类Circle的方法成员: Circle():构造方法,将半径置0 Circle(double r):构造方法,创建Circle对象时将半径初始化为r double getRadius():获得圆的半径值 double getPerimeter():获得圆的周长 void disp():将圆的半径和圆的周长,圆的面积输出到屏幕 (2)继承第一题中的…

UVA 11722 几何概型

第六周A题 - 几何概型 Time Limit:1000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Description You are going from Dhaka to Chittagong by train and you came to know one of your old friends is goingfrom city Chittagong to Sylhet. You also know that…

201521123122 《java程序设计》第六周实验总结

201521123122 <java程序设计>第六周实验总结 1. 本周学习总结 1.1 面向对象学习暂告一段落,请使用思维导图,以封装.继承.多态为核心概念画一张思维导图,对面向对象思想进行一个总结. 注1:关键词与内容不求多,但概念之间的联系要清晰,内容覆盖面向对象的核心内容即可. 注2:颜色要少.连线不要复杂,必要的时候要在连线上进行说明. 2. 书面作业 clone方法 1.1 Object对象中的clone方法是被protected修饰,在自定义的类中覆盖clone方法时需要注意什么…

201621123040《Java程序设计》第六周学习总结

1.本周学习总结 1.1 面向对象学习暂告一段落,请使用思维导图,以封装.继承.多态为核心概念画一张思维导图或相关笔记,对面向对象思想进行一个总结 2.书面作业 2.1clone方法 2.1.1在test1包中编写Employee类,在test2包中新建一个Test类,并在其main中新建test1包中的Employee的对象,然后使用clone方法克隆一个新对象,能否成功?为什么? 不成功:clone方法为Object对象protected Object clone();protected不能…