参看资料：https://www.cnblogs.com/By-ruoyu/p/4713535.html
题目：
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 nspace separated integers. If the ith integer (xi) is positive, you can assume that the ith door 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
3
扫描二维码关注公众号，回复： 5785177 查看本文章
1
1
2
-10 -3
3
3 -6 -9
Sample Output
Case 1: 1/1
Case 2: inf
Case 3: 18/1
题目大意：
有n个门，随机推开第 i 扇门，将得到一个整数 t [ i ]，若该数为正，则经过 t [ i ] 时间后，你将会走出这个门； 若该数为负，则经过 t [ i ] 时间后，你将会回到起点，忘记一切，直到走出某一扇门，求走出这里所用时间的期望。
解题思路：
因为每次都不记得曾经的选择， 所以每次的期望都是一样的。
设，T1为每次走出去所用时间的期望， T2为回到之前所用时间的期望。
则E = p * T1 + (T2 + E) * (1 - p)
化简后得， p * E = p * T1 + T2 * (1 - p)
假设有m个门是走出去的， 则p = m / n
代进上式得：m * E / n = m * T1 / n + (n - m) * T2 / n
m * E = m * T1 + (n - m) * T2
T1 = (sigma x[i]) / m, x[i] > 0, T2 = (sigma |x[i]|) / (n - m), x[i] < 0
E = (sigma x[i] + sigma |x[i]|) / m
实现代码：