我制作了以下程序,它要求用户输入上限并计算并打印每个完美的正方形直到上限。但是,我认为我的 is_perfect_square 函数不是很有效,因为当上限为数千或更多时计算完美平方需要很长时间。我想知道如何让我的程序更有效率,我认为使用数学模块来使用 sqrt 可以工作,但我不是数学家所以请求帮助。 我的程序是:
"""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()
请您参考如下方法:
我会完全颠倒逻辑,首先取上界的平方根,然后打印小于或等于该数的每个正整数的平方:
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(bind) 78 的示例输出:
[1, 4, 9, 16, 25, 36, 49, 64]
这样你就可以避免很多不必要的循环和数学计算。
