108A - Palindromic Times

Note that for each hour, there either exists a unique answer or no answer. For instance, at hour 14, the answer is 14: 41 while for hour 17, no answer exists.

Therefore, we can first check for the given hour whether there is a feasible answer or not. If no, then we continue to search for the next hour that has such an answer.

108B - Datatypes

We should first sort the array in an increasing order. Then, for every pair of (a_i, a_i + 1) with a_i < a_i + 1, we check whether there exists such an integer x.

The maximum x that can be represented by a_i bits is 2^a_i - 1. Thus, x² = 2^2a_i - 2^a_i + 1 + 1. One can check that x² can always be represented by a_i + 1 bits as long as a_i + 1 ≥ 2a_i.

108C - Dorm Water Supply

According to the description of the problem, the graph in fact consists of several simple rings and single links. Thus, we start enumerating from node 1 to node n and for each node that has no input pipe, we visit all the nodes along the pipes while recording the minimum diameter. Finally, print out the recorded results.

108D - Basketball Team

We denote the number of his teammates as A and the number of the other students as S. Then, the problem can be solved based on the following three cases.

1) A + S < n - 1: this means that there are not enough students;

2) S < n - 1: this means that the probability is absolutely one;

3) none of the above cases: the answer is just . To compute , I usually transform it into F = e^log(F). In other words, we first calculate , and then obtain F = e^f. I think “log” can guarantee better precision than directly mutiplying all the float numbers together.