LightOJ 题目1027 - A Dangerous Maze（期望）

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本文介绍了一种计算从危险迷宫中逃脱所需期望时间的算法。面对n个门的选择，每个门背后隐藏着返回起点或直接逃脱的时间值。通过分析输入数据，包括门的数量及其对应的正负值，算法能够计算出成功逃离迷宫所需的平均时间。

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1027 - A Dangerous Maze

PDF (English)

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Forum

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 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^thdoor 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

Output for Sample Input

-10 -3

3 -6 -9

Case 1: 1/1

Case 2: inf

Case 3: 18/1

PROBLEM SETTER: JANE ALAM JAN

题目大意：一个迷宫有n个门，每个对应一个值，正值表示经过这么多秒后直接出迷宫，负值代表这么多秒后回到最开始的地方，问最后出去的期望

思路：一次出去的概率为n1/n期望为n/n1,一次出去的平均时间问sum（t）/n,期望两个相乘即，sum（t）/n1

ac代码

616445	2015-11-18 19:53:53	1027 - A Dangerous Maze	C++	0.000	1688	Accepted

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int gcd(int a,int b)
{
    if(a<b)
    {
        int t=a;
        a=b;
        b=t;
    }
    if(b==0)
        return a;
    return gcd(b,a%b);
}
int main()
{
    int t,c=0;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        int sum=0,n1=0;
        int i;
        for(i=1;i<=n;i++)
        {
            int x;
            scanf("%d",&x);
            if(x<0)
                sum-=x;
            else
            {
                sum+=x;
                n1++;
            }
        }
        if(n1==0)
        {
            printf("Case %d: inf\n",++c);
            continue;
        }
        int tmp=gcd(sum,n1);
        printf("Case %d: %d/%d\n",++c,sum/tmp,n1/tmp);
    }
}