LightOJ - 1027 A Dangerous Maze —— 期望
题目链接:https://vjudge.net/problem/LightOJ-1027
Time Limit: 2 second(s) | Memory Limit: 32 MB |
You are in a maze; seeing n doors in front of you in beginning. You can choose any door you like. The probability for choosing a door is equal for all doors.
If you choose the ith door, it can either take you back to the same position where you begun in xi minutes, or can take you out of the maze after xi minutes. If you come back to the same position, you can't remember anything. So, every time you come to the beginning position, you have no past experience.
Now you want to find the expected time to get out of the maze.
Input
Input starts with an integer T (≤ 100), denoting the number of test cases.
Each case contains a blank line and an integer n (1 ≤ n ≤ 100) denoting the number of doors. The next line contains n space separated integers. If the ith integer (xi) is positive, you can assume that the ithdoor will take you out of maze after xi minutes. If it's negative, then the ith door will take you back to the beginning position after abs(xi) minutes. You can safely assume that 1 ≤ abs(xi) ≤ 10000.
Output
For each case, print the case number and the expected time to get out of the maze. If it's impossible to get out of the maze, print 'inf'. Print the result in p/q format. Where p is the numerator of the result and q is the denominator of the result and they are relatively prime. See the samples for details.
Sample Input |
Output for Sample Input |
3 1 1 2 -10 -3 3 3 -6 -9 |
Case 1: 1/1 Case 2: inf Case 3: 18/1 |
题意:
有n扇门,一扇门要么能在特定时间内把人带出迷宫,要么能在特定时间内把人带会原地,并使人失去记忆(即不知道这扇门是不能带出迷宫的),每扇门被选择的几率是相等的。问走出迷宫的平均时间。
题解:
设期望为EX,有t1扇“正门”,t2扇“负门”,其中t1+t2 = n 。第i扇门所花费的时间为ai
1.当选择的门为正门时,它可以直接走出迷宫,因此:1/n*ai,1/n为选择这扇门的几率,且所有的门被选择的几率也为1/n。
2.当选择的门为负门时,它先花费了ai的时间,然后又重新选择,重新选择然能走出迷宫的时间即为平均时间,因此:1/n*(ai+EX)。
3.综上,EX = ∑1/n*ai + ∑1/n*(ai+EX),其中第一个i的范围为:1<=i<=t1,第二个i:1<=i<=t2。
代码如下:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+;
const int MAXN = 1e6+; int gcd(int a, int b)
{
return b==?a:gcd(b,a%b);
} int main()
{
int T, n, kase = ;
scanf("%d", &T);
while(T--)
{
scanf("%d", &n);
int numP = , sum = ;
for(int i = ; i<=n; i++)
{
int val;
scanf("%d", &val);
sum += abs(val);
if(val>) numP++;
}
if(numP==)
printf("Case %d: inf\n", ++kase);
else
printf("Case %d: %d/%d\n", ++kase, sum/gcd(sum, numP), numP/gcd(sum, numP));
}
}
LightOJ - 1027 A Dangerous Maze —— 期望的更多相关文章
- Lightoj 1027 - A Dangerous Maze 【期望】
1027 - A Dangerous Maze PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB Y ...
- [LightOJ 1027] A Dangerous Maze
A Dangerous Maze You are in a maze; seeing n doors in front of you in beginning. You can choose any ...
- LightOJ 1027 - A Dangerous Maze(求期望)
题目链接:http://lightoj.com/volume_showproblem.php?problem=1027 题意:又一个迷宫,有n个门,每个门又一个值num,如果num>0 说明在n ...
- LightOJ 1027 A Dangerous Maze(期望)题解
题意:n扇门,每扇门后都有一个值x,如果x<0会让你等待-x再重新回到这里选择门,x>0你经过x时间就会被传送走,问你被传送走的期望 思路:假设被传送走的期望为E,那么对于x<0来说 ...
- LightOJ 1027 A Dangerous Maze(期望)
https://cn.vjudge.net/problem/LightOJ-1027 题意:有n扇门,每扇门有个时间ti,选择正数的门可以在ti后带你走出迷宫,负数的门会在ti后带你回到起点,然后重新 ...
- LightOJ 1027 A Dangerous Maze (数学期望)
题意:你面前有 n 个门,每次你可以选择任意一个进去,如果xi是正数,你将在xi后出去,如果xi是负数,那么xi后你将回来并且丢失所有记忆,问你出去的期望. 析:两种情况,第一种是直接出去,期望就是 ...
- LightOj 1027 A Dangerous Maze【概率】
题目链接:http://www.lightoj.com/volume_showproblem.php? problem=1027 题意: 你面前有n个门,每一个相应一个数字,若为正xi.代表xi分钟后 ...
- LightOJ - 1395 A Dangerous Maze (II) —— 期望
题目链接:https://vjudge.net/problem/LightOJ-1395 1395 - A Dangerous Maze (II) PDF (English) Statistic ...
- LightOJ - 1027 Dangerous Maze 期望
你在迷宫中;开始时在你面前看到n扇门.你可以选择你喜欢的任何门.所有门的选择门的概率是相等的. 如果您选择第i个门,它可以让您回到您在xi(xi小于0)分钟内开始的相同位置,也可以在xi(xi大于0) ...
随机推荐
- 用hashmap实现自己的缓存
@SuppressWarnings({"unchecked", "rawtypes"})public class DefaultCache implements ...
- Silverlight 离线安装包
直接下载地址 https://www.microsoft.com/getsilverlight/locale/en-us/html/Microsoft%20Silverlight%20Release% ...
- Hibernate 3 深度解析--苏春波
Hibernate 3 深度解析 Hibernate 作为 Java ORM 模式的优秀开源实现, 当下已经成为一种标准,为饱受 JDBC 折磨的 Java 开发者带来了“福音.快速的版本更新,想 ...
- 关于查看python的trace的方法
lptrace本质上是基于GDB的,进入到进程内存空间,然后执行了一段python指令把当时的trace给print出来 使用工具:https://github.com/khamidou/lptrac ...
- php如何读取ini文件
很多时候,我们使用配置文件来读取配置,那么php如何使用ini文件呢? 代码如下: 例如将:数据库信息存到ini文件中,进行读取. <?php header('content-type:text ...
- overflow滚动条样式设置,ie和webkit内核
<!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...
- MySQL 下优化SQL语句的一些经验
http://java-guru.iteye.com/blog/143377
- generate alphanumeric serial number
generate alphanumeric serial number with the following BADI / Customer Exit: Name of Enhancement : I ...
- python--多种程序分析(2)
1.文件操作有哪些模式?请简述各模式的作用 r模式只读 w模式只写 a模式只添加 r+可读可写 w+可写可读 a+可读可添加 rb 二进制只读 wb 二进制只写 ab 二进制添加 ...
- Android性能优化之中的一个 布局优化
本文为Android性能优化--布局优化,主要介绍使用抽象布局标签(include, viewstub, merge).去除不必要的嵌套和View节点.降低不必要的infalte及其它Layout方面 ...