608A - Сайтама громит гостиницу

Author: ed1d1a8d

The minimum amount of time required is the maximum value of t_i + f_i and s, where t_i and f_i are the time and the floor of the passenger respectively.

The initial observation that should be made for this problem is that only the latest passenger on each floor matters. So, we can ignore all passengers that aren't the latest passenger on each floor.

Now, assume there is only a passenger on floor s. Call this passenger a. The time taken for this passenger is clearly t_a + f_a (the time taken to wait for the passenger summed to the time taken for the elevator to reach the bottom).

Now, add in one passenger on a floor lower than s. Call this new passenger b. There are 2 possibilities for this passenger. Either the elevator reaches the passenger's floor after the passenger's time of arrival or the elevator reaches the passenger's floor before the passenger's time of arrival. For the first case, no time is added to the solution, and the solution remains t_a + f_a. For the second case, the passenger on floor s doesn't matter, and the time taken is t_b + f_b for the new passenger.

The only thing left is to determine whether the elevator reaches the new passenger before t_i of the new passenger. It does so if t_a + (f_a - f_b) > t_b. Clearly this is equivalent to whether t_a + f_a > t_b + f_b. Thus, the solution is max of max(t_a + f_a, t_b + f_b).

A similar line of reasoning can be applied to the rest of the passengers. Thus, the solution is the maximum value of t_i + f_i and s.

608B - Сумма расстояний Хэмминга

Author: ed1d1a8d

We are trying to find . Swapping the sums, we see that this is equivalent to .

Summing up the answer in the naive fashion will give an O(n²) solution. However, notice that we can actually find without going through each individual character. Rather, all we need is a frequency count of different characters. To obtain this frequency count, we can simply build prefix count arrays of all characters on b. Let's call this prefix count array F, where F[x][c] gives the number of occurrences of the character c in the prefix [0, x) of b. We can then write . as . This gives us a linear solution, albeit with a slightly large constant factor.

Time Complexity — O(|a| + |b|), Memory Complexity — O(|b|)