如何有效地枚举所有完美正方形？

Question

我做了下面的程序，它要求用户输入上限，并计算并打印到上限的每个完美正方形。然而，我认为我的is_perfect_square函数的效率不高，因为当上界为数千或更多时，计算完美平方需要很长时间。我想知道如何使我的程序更高效，我认为使用数学模块with use sqrt可以工作，但我不是数学家，所以请求帮助。 我的计划是：

"""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

我将完全颠倒逻辑，首先取上界的平方根，然后打印每个小于或等于该数字的正整数的平方：

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的输出示例：

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

这样可以避免很多不必要的循环和数学计算。在

Answer 2

如您所说使用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

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

或者如@PacoH所示：

^{pr2}$