Question

我做了以下程序，询问用户输入上限并计算＆amp;将每个完美的正方形打印到上限。但是，我认为我的is_perfect_square函数效率不高，因为当上限为数千或更多时，计算完美正方形需要很长时间。我想知道如何使我的程序更有效率，我认为使用数学模块使用sqrt可以工作，但我不是数学家，所以请求帮助。我的节目是：

"""Print all the perfect squares from zero up to a given maximum."""
import math

def read_bound():
   """Reads the upper bound from the standard input (keyboard).
      If the user enters something that is not a positive integer
      the function issues an error message and retries
      repeatedly"""
   upper_bound = None
   while upper_bound is None:
       line = input("Enter the upper bound: ")
       if line.isnumeric() and int(line) >= 0:
           upper_bound = int(line)
           return upper_bound
       else:
           print("You must enter a positive number.")



def is_perfect_square(num):
   """Return true if and only if num is a perfect square"""
   for candidate in range(1, num):
       if candidate * candidate == num:
           return True



def print_squares(upper_bound, squares):
   """Print a given list of all the squares up to a given upper bound"""


   print("The perfect squares up to {} are: ". format(upper_bound))
   for square in squares:
       print(square, end=' ')



def main():
   """Calling the functions"""
   upper_bound = read_bound()
   squares = []
   for num in range(2, upper_bound + 1):
       if is_perfect_square(num):
           squares.append(num)

   print_squares(upper_bound, squares)


main()

Answer 1

我会完全颠倒逻辑，首先取上边界的平方根，然后打印每个正整数的平方小于或等于该数字：

upper_bound = int(input('Enter the upper bound: '))

upper_square_root = int(upper_bound**(1/2))

print([i**2 for i in range (1, upper_square_root+1)])

绑定78的示例输出：

[1,4,9,16,25,36,49,64]

这样可以避免大量不必要的循环和数学计算。

Answer 2

正如你所说的那样使用math.sqrt：

import math


def is_perfect_square(num):
    """Return true if and only if num is a perfect square"""
    return math.sqrt(num).is_integer()

Answer 3

平方是奇数的部分和：

 1 = 1
 4 = 1 + 3
 9 = 1 + 3 + 5
16 = 1 + 3 + 5 + 7

所以你可以这样做：

   square = 1
   odd = 1
   while square <= upper_bound:
    print(square)
    odd = odd + 2
    square = square + odd

https://ideone.com/dcnEVJ

无需平方根或检查每个号码。它并没有比这更快。

Answer 4

您可以使用sqrt

import math
def is_perfect_square(num):
   """Return true if and only if num is a perfect square"""
    root = math.sqrt(num)
    return int(root) - root == 0

或@PacoH表示：

    return root.is_integer()

如何有效地枚举所有完美的正方形？

4 个答案: