Uva12230Crossing Rivers

问题:

You live in a village but work in another village. You decided to follow the straight path between your house (A) and the working place (B), but there are several rivers you need to cross. Assume B is to the right of A, and all the rivers lie between them. Fortunately, there is one “automatic” boat moving smoothly in each river. When you arrive the left bank of a river, just wait for the boat, then go with it. You’re so slim that carrying you does not change the speed of any boat.

Days and days after, you came up with the following question: assume each boat is independently placed at random at time 0, what is the expected time to reach B from A? Your walking speed is always 1. To be more precise, for a river of length L, the distance of the boat (which could be regarded as a mathematical point) to the left bank at time 0 is uniformly chosen from interval [0, L], and the boat is equally like to be moving left or right, if it’s not precisely at the river bank.

Input

There will be at most 10 test cases. Each case begins with two integers n and D, where n (0 ≤ n ≤ 10) is the number of rivers between A and B, D (1 ≤ D ≤ 1000) is the distance from A to B. Each of the following n lines describes a river with 3 integers: p, L and v (0 ≤ p < D, 0 < L ≤ D, 1 ≤ v ≤ 100). p is the distance from A to the left bank of this river, L is the length of this river, v is the speed of the boat on this river. It is guaranteed that rivers lie between A and B, and they don’t overlap. The last test case is followed by n = D = 0, which should not be processed.

Output

For each test case, print the case number and the expected time, rounded to 3 digits after the decimal point. Print a blank line after the output of each test case.

Sample Input

1 1

0 1 2

0 1

0 0

Sample Output

Case 1: 1.000

Case 2: 1.000

题目大意:

有个人每天要去公司上班,每次会经过N条河,家和公司的距离为D,默认在陆地的速度为1,给出N条河的信息,包括起始坐标p,宽度L,以及船的速度v。船会往返在河的两岸,人到达河岸时,船的位置是随机的(往返中)。问说人达到公司所需要的期望时间。

思路:

过每条河最坏的情况是t=3*L/v; 即去的时候船刚刚走。

过没条河最优的情况是t=L/v;    即去的时候船刚刚来。

由于船是均匀发布的,符合线性性质,所以平均下来,过每条河的时间t=2*L/v。

代码:

 #include"bits/stdc++.h"

 #define db double
#define ll long long
#define vl vector<ll>
#define ci(x) scanf("%d",&x)
#define cd(x) scanf("%lf",&x)
#define cl(x) scanf("%lld",&x)
#define pi(x) printf("%d\n",x)
#define pd(x) printf("%f\n",x)
#define pl(x) printf("%lld\n",x)
#define rep(i, n) for(int i=0;i<n;i++)
using namespace std;
const int NN = 1e6 + ;
const int mod = 1e9 + ;
const int MOD = ;
const db PI = acos(-1.0);
const db eps = 1e-;
const ll INF = 0x3fffffffffffffff;
int n,d;
int main()
{
int t=;
while(scanf("%d%d",&n,&d)==&&n||d){
db ans=;
for(int i=;i<n;i++){
int p,l,v;
ci(p),ci(l),ci(v);
ans+=2.0*l/v;
d-=l;
}
ans+=d;
printf("Case %d: %.3f\n",t++,ans);
puts("");
}
return ;
}

Uva12230Crossing Rivers 数学的更多相关文章

  1. Uva12230Crossing Rivers (数学期望)

    问题: You live in a village but work in another village. You decided to follow the straight path betwe ...

  2. HDU3232 Crossing Rivers 数学期望问题

    Crossing Rivers                                                                                     ...

  3. uva12230Crossing Rivers

    数学期望. 过每条河的时间的可能在[L/v,3*L/v]间均匀分布,数学期望为2*L/v. 然后在加上在陆上走的时间. #include<cstdio> #include<algor ...

  4. Uva - 12230 Crossing Rivers (数学期望)

    你住在村庄A,每天需要过很多条河到另一个村庄B上班,B在A的右边,所有的河都在A,B之间,幸运的是每条船上都有自由移动的自动船, 因此只要到达河左岸然后等船过来,在右岸下船,上船之后船的速度不变.现在 ...

  5. UVA12230 Crossing Rivers (数学期望)

    题目链接 题意翻译 一个人每天需要从家去往公司,然后家与公司的道路是条直线,长度为 \(D\). 同时路上有 \(N\) 条河,给出起点和宽度\(W_i\) , 过河需要乘坐速度为\(V_i\) 的渡 ...

  6. 【整理】简单的数学期望和概率DP

    数学期望 P=Σ每一种状态*对应的概率. 因为不可能枚举完所有的状态,有时也不可能枚举完,比如抛硬币,有可能一直是正面,etc.在没有接触数学期望时看到数学期望的题可能会觉得很阔怕(因为我高中就是这么 ...

  7. hdu 3232 Crossing Rivers(期望 + 数学推导 + 分类讨论,水题不水)

    Problem Description   You live in a village but work in another village. You decided to follow the s ...

  8. UVa 12230 && HDU 3232 Crossing Rivers (数学期望水题)

    题意:你要从A到B去上班,然而这中间有n条河,距离为d.给定这n条河离A的距离p,长度L,和船的移动速度v,求从A到B的时间的数学期望. 并且假设出门前每条船的位置是随机的,如果不是在端点,方向也是不 ...

  9. UVa 12230 - Crossing Rivers(数学期望)

    链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem& ...

随机推荐

  1. Dreams save us. Dreams lift us up and transform us into something better.

    Dreams save us. Dreams lift us up and transform us into something better.梦想能够拯救我们.梦想能够激励我们并让我们成为更好的人 ...

  2. 《Cron表达式详解》

    Cron表达式是一个字符串,字符串以5或6个空格隔开,分为6或7个域,每一个域代表一个含义,Cron有如下两种语法格式: Seconds Minutes Hours DayofMonth Month ...

  3. HCNA配置RIPv1

    1.拓扑图 2.配置 R1 The device is running! ###### <Huawei>sys Enter system view, return user view wi ...

  4. xampp中mysql重置root密码

    1.  停止mysql:用图形化工具或者在cmd命令下输入net stop mysql,在c盘根目录下输入 2.  打开cmd,切换目录到 /xampp/mysql/bin, 运行    mysqld ...

  5. 关于硬盘分区使用exFat格式的优势及劣势(含摘抄)

    优势 可以设置最大32M的簇: 不记录日志. 劣势 无法使用windows的“文件共享”: 通过近期某个文件数量密级任务的测试发现,在大量文件的处理性能上,NTFS比exFAT文件系统的性能高出不少. ...

  6. C盘压缩,电脑无法正常启动的解决方法?

    有时候,我们觉得电脑很卡,因此压缩磁盘来节约资源,前段时间,由于不小心将C盘压缩了,导致电脑无法正常启动,查了一些有关的资料,发现很多人都遇到过类似的问题,如果你不想重装系统的话,那么,现在我说一下我 ...

  7. sql server 拆分字符串,拆分两次(:和;)

    declare @DisciplineID int declare @paramStringVal nvarchar() declare @NPNT nvarchar() declare @Disci ...

  8. Android OpenGL ES 画球体

    近期由于兴趣所向.開始学习OpenGL绘图. 本文以"画球体"为点,小结一下近期所学. > 初识OpenGL ES 接触OpenGL是从Android開始的.众所周知,And ...

  9. C++STL之map映照容器

    map映照容器 map映照容器的元素数据是由一个键值和一个映照数据组成的, 键值与映照数据之间具有一一映照关系. map映照容器的数据结构也是采用红黑树来实现的, 插入元素的键值不允许重复, 比较函数 ...

  10. spfa判负权边

    spfa判负环 如果一个点在spfa中被入队了大于n次 那么,我们就能肯定,有负环出现. 因为一个点入队时,他肯定被更新了一次. 所以........ 如果不存在负权环.这个点最多被更新节点数次 我们 ...