620A - Professor GukiZ's Robot

Easy to see that the answer is max(|x₁ - x₂|, |y₁ - y₂|).

С++ solution

Complexity: O(1).

620B - Grandfather Dovlet’s calculator

Let's simply iterate over all the values from a to b and add to the answer the number of segments of the current value x. To count the number of segments we should iterate over all the digits of the number x and add to the answer the number of segments of the current digit d. These values can be calculated by the image from the problem statement and stored in some array in code.

C++ solution

Complexity: O((b - a)logb).

620C - Pearls in a Row

Let's solve the problem greedily. Let's make the first segment by adding elements until the segment will be good. After that let's make the second segment in the same way and so on. If we couldn't make any good segment then the answer is  - 1. Otherwise let's add all uncovered elements at the end to the last segment. Easy to prove that our construction is optimal: consider the first two segments of the optimal answer, obviously we can extend the second segment until the first segment will be equal to the first segment in our construction.

C++ solution

Complexity: O(nlogn).

620D - Professor GukiZ and Two Arrays

We can process the cases of zero or one swap in O(nm) time. Consider the case with two swaps. Note we can assume that two swaps will lead to move two elements from a to b and vice versa (in other case it is similar to the case with one swap). Let's iterate over all the pairs of the values in a and store them in some data structure (in C++ we can user map). Now let's iterate over all the pairs b_i, b_j and find in out data structure the value v closest to the value x = s_a - s_b + 2·(b_i + b_j) and update the answer by the value |x - v|. Required sum we can find using binary search by data structure (*map* in C++ has lower_bound function).

C++ solution

Сложность: O((n² + m²)log(n + m)).