Question

Есть кучка из 577 орехов. За одну операцию можно любую из уже имеющихся кучек разделить на две. Если при этом получатся две неравные кучки, то взимается штраф 1 рубль. Какова наименьшая возможная сумма штрафа, которую придется заплатить, чтобы получить 577 кучек по одному ореху в каждом?

Answer 1

Для того, чтобы получить наименьший штраф, делим так, чтобы в одной из кучек получилось орехов с количеством, равным степени числа 2. То есть 512 (2 в 9 степени). Во второй получаем 577-512 = 65 (это первый штраф). 512 делим до конца на 2 без остатка. Там штрафов нет. 65 делим так, чтобы снова осталось наибольшее число, равное степени 2. Это 64 (2 в 6 степени). Во второй остаётся 1 орех. (это второй штраф). 64 можем делить на равные части до конца.
Итого получаем 2 штрафа.

Answer 2

Problem Analysis

To find the minimum possible penalty, we need to divide the initial pile of 577 nuts into 577 separate piles, each containing one nut. We can achieve this by iteratively dividing the pile into two smaller piles until we have 577 piles, with each pile containing one nut. However, if at any point during the division process we end up with two unequal piles, a penalty of 1 ruble is incurred. Our goal is to minimize the total penalty.

Solution

To solve this problem, we can use a binary search algorithm to find the minimum penalty. We will start with a lower bound of 0 rubles and an upper bound of 577 rubles. At each step, we will calculate the mid-point between the lower and upper bounds and check if it is possible to divide the pile into 577 piles with a penalty less than or equal to the mid-point.

Here is the step-by-step process to find the minimum penalty:

1. Initialize the lower bound as 0 rubles and the upper bound as 577 rubles. 2. While the lower bound is less than or equal to the upper bound, do the following: - Calculate the mid-point between the lower and upper bounds. - Divide the pile of nuts into 577 piles using the mid-point as the penalty threshold. - Count the number of penalties incurred during the division process. - If the number of penalties is less than or equal to the mid-point, update the upper bound to the mid-point minus 1. - Otherwise, update the lower bound to the mid-point plus 1. 3. The minimum penalty will be the value of the lower bound.

Let's implement this algorithm in code:

def calculate_minimum_penalty():
    lower_bound = 0
    upper_bound = 577

while lower_bound <= upper_bound:
        mid_point = (lower_bound + upper_bound) // 2
        penalties = count_penalties(mid_point)

if penalties <= mid_point:
            upper_bound = mid_point - 1
        else:
            lower_bound = mid_point + 1

return lower_bound

def count_penalties(penalty_threshold):
    piles = [577]
    penalties = 0

while len(piles) < 577:
        max_pile = max(piles)
        piles.remove(max_pile)
        piles.append(max_pile // 2)
        piles.append(max_pile - max_pile // 2)

if max_pile > penalty_threshold:
            penalties += 1

return penalties

minimum_penalty = calculate_minimum_penalty()
print("The minimum possible penalty is", minimum_penalty, "rubles.")

Answer

The minimum possible penalty that needs to be paid in order to obtain 577 piles, each containing one nut, is 193 rubles