LightOJ - 1027 A Dangerous Maze (数学期望）

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本文介绍了一种计算从随机迷宫中逃脱所需期望时间的方法。面对n扇门的选择，每扇门要么将你带回起点，要么直接带你离开迷宫。通过概率分析，给出了一套算法来计算成功逃脱迷宫所需的平均时间。

题目链接：http://lightoj.com/volume_showproblem.php?problem=1027点击打开链接

1027 - A Dangerous Maze

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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^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

Output for Sample Input

-10 -3

3 -6 -9

Case 1: 1/1

Case 2: inf

Case 3: 18/1

p= N1 / N

期望实验次数: E(X)= N / N1

每次实验消耗的时间: T= T1+T2/ N

所需时间的数学期望: E(T)=(N / N1)*( ( T1 + T2) / N) =( T1 + T2) / N1;

#include <bits/stdc++.h>
using namespace std;
int gcd(int x,int y)
{
	if(x<y)
		swap(x,y);
	if(y==0)
		return x;
	return gcd(y,x%y);
}
int main()  
{  
	int n;
	cin >> n;
	for(int cnt=1;cnt<=n;cnt++)
	{
		int t;
		cin >> t;
		int num1=0,num2=0,time1=0,time2=0;
		while(t--)
		{
			int mid;
			cin >> mid;
			if(mid>0)
				num1++,time1+=mid;
			else if(mid<0)
				num2++,time2+=abs(mid);
		}
		if(time1)
			printf("Case %d: %d/%d\n",cnt,(time1+time2)/gcd(time1+time2,num1),num1/gcd(time1+time2,num1));
		else 
			printf("Case %d: inf\n",cnt);
	}
}