將值分成最大值，最小值範圍

Question

我想找到一種有效的算法將整數值分成最大值，最小值範圍內的某個值。應該有儘可能少的價值。

例如： 最大= 7，MIN = 3 然後

8 = 4 + 4 
9 = 4 + 5 
16 = 5 + 5 + 6 (not 4 + 4 + 4 + 4) 

EDIT

爲了使它更清楚，讓舉一個例子。假設你有一堆蘋果，你想把它們裝進籃子裏。每個籃子可容納3到7個蘋果，並且您希望籃子的數量儘可能小。

**我提到價值應該平分，但這並不重要。我更關心籃子的數量。

Answer 1

這讓我感到很有趣，所以我有機會剽竊出一個快速解決方案。我認爲這可能是一個有趣的起點，它會給你一個有效的解決方案，儘可能少的數字，或儘可能相似的數字，所有這些都在由min_bound和max_bound定義的範圍的範圍內 數= INT（輸入（ 「編號」）） min_bound = 3 max_bound = 7

def improve(solution): 
    solution = list(reversed(solution)) 
    for i, num in enumerate(solution): 
     if i >= 2: 
      average = sum(solution[:i])/i 
      if average.is_integer(): 
       for x in range(i): 
        solution[x] = int(average) 
       break 
    return solution 

def find_numbers(number, division, common_number): 
    extra_number = number - common_number * division 
    numbers_in_solution = [common_number] * division 
    if extra_number < min_bound and \ 
    extra_number + common_number <= max_bound: 
     numbers_in_solution[-1] += extra_number 
    elif extra_number < min_bound or extra_number > max_bound: 
     return None 
    else: 
     numbers_in_solution.append(extra_number) 
    solution = improve(numbers_in_solution) 
    return solution 

def tst(number): 
    try: 
     solution = None 
     for division in range(number//max_bound, number//min_bound + 1): # Reverse the order of this for numbers as close in value to each other as possible. 
      if round (number/division) in range(min_bound, max_bound + 1): 
       solution = find_numbers(number, division, round(number/division)) 
      elif (number // division) in range(min_bound, max_bound + 1): # Rarely required but catches edge cases 
       solution = find_numbers(number, division, number // division) 
      if solution: 
       print(sum(solution), solution) 
       break 
    except ZeroDivisionError: 
     print("Solution is 1, your input is less than the max_bound") 

tst(number) 
for x in range(1,100): 
    tst(x) 

此代碼只是爲了演示一個想法，我敢肯定，它可以進行調整以提高性能。