## Editorial for DMOPC '18 Contest 3 P3 - Bob and Math Class

Remember to use this editorial only when stuck, and not to copy-paste code from it. Please be respectful to the problem author and editorialist.

Submitting an official solution before solving the problem yourself is a bannable offence.

Authors: r3mark

For the first subtask, check each subarray. You can find the longest block of only Ts or only Fs in a subarray in . There are subarrays, so the final time complexity is .

Time Complexity:

For the second subtask, when considering subarrays, categorize them by their left endpoint. Then you can find the longest block of only Ts or only Fs in each of the subarrays with a certain left endpoint in .

Time Complexity:

The third subtask was meant to reward any badly done solutions or potential or solutions. There is no intended solution.

There are several solutions to the final subtask. The solution discussed below is difficult to come up with, but has a simple implementation.

All indices are -indexed. Let be the length of the block of only Ts or only Fs which the element at index is part of. These values can be computed in .

Now consider any odd-length subarray. Observe that if this subarray is suspicious, then the middle element must be part of the subarray's longest block of only Ts or Fs. We will count the number of odd-length subarrays given its middle element. From the previous observation, it can be seen that the number of odd-length subarrays with middle element at index is .

The reasoning for even-length subarrays is similar. If the and answers are both T or both F, there are even-length subarrays whose two middle elements are at index and . If the answer differs from the answer, then there are even-length subarrays.

Time Complexity: