• LightOJ 1027 A Dangerous Maze (期望)


    题目链接:LightOJ - 1027

    Description

    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 (i^{th}) door, it can either take you back to the same position where you begun in (x_i) minutes, or can take you out of the maze after (x_i) 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 (i^{th}) integer ((x_i)) is positive, you can assume that the (i^{th}) door will take you out of maze after (x_i) minutes. If it's negative, then the (i^{th}) door will take you back to the beginning position after (abs(x_i)) minutes. You can safely assume that (1 ≤ abs(x_i) ≤ 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

    3
    
    1
    1
    
    2
    -10 -3
    
    3
    3 -6 -9
    

    Sample Output

    Case 1: 1/1
    Case 2: inf
    Case 3: 18/1
    

    Solution

    题意

    你在一个迷宫里,面前有 (n) 扇门,如果第 (i) 扇门的 (X_i) 值为正值,就可以花费 (X_i) 的时间的走出迷宫,否则花费 (abs(X_i)) 的时间又回到原点,且不记得之前走过哪些门。每次等概率选择一扇门,问走出迷宫的时间的期望。不能走出去输出 (inf)

    思路

    设有 (n_1) 扇门能走出迷宫,(n_2) 不能走出迷宫,则 (n_1 + n_2 = n)

    (n_1) 扇门的 (X_i) 值的平均值为 (t_1)(n_2) 扇门的 (X_i) 值的平均值为 (t_2).

    设走出去的期望为 (E)。则 (E = frac{n_1}{n} cdot t_1 + frac{n_2}{n} cdot (t_2 + E))

    化简后为 (E = frac{1}{n_1}sum_{i=1}^n abs(X_i))

    Code

    #include <bits/stdc++.h>
    using namespace std;
    const int maxn = 110;
    
    int x[maxn];
    
    int gcd(int a, int b) {
        return a == 0? b: gcd(b % a, a);
    }
    
    int main() {
        ios::sync_with_stdio(false);
        cin.tie(0);
        int T;
        cin >> T;
        int kase = 0;
        while(T--) {
            int n;
            cin >> n;
            int sum = 0;
            int cnt = 0;
            for(int i = 1; i <= n; ++i) {
                cin >> x[i];
                if(x[i] > 0) {
                    ++cnt;
                    sum += x[i];
                } else {
                    sum += -x[i];
                }
            }
            int d = gcd(sum, cnt);
            if(cnt == 0) printf("Case %d: inf
    ", ++kase);
            else printf("Case %d: %d/%d
    ", ++kase, sum / d, cnt / d);
        }
        return 0;
    }
    
  • 相关阅读:
    Linux查看进程和已知端口是否启动
    plsql安装
    windows中用批处理文件删除n天前的文件
    阿里云服务器25邮件端口问题
    通过rpm安装crontab
    [RHEL7.1]关闭防火墙及SElinux
    有了 itchat, python 调用微信个人号从未如此简单(新增 py3 支持)
    Linux中tty、pty、pts的概念区别
    不用写代码就能实现深度学习?手把手教你用英伟达 DIGITS 解决图像分类问题
    Hadoop 2.7.3 安装配置及测试
  • 原文地址:https://www.cnblogs.com/wulitaotao/p/11668088.html
Copyright © 2020-2023  润新知