Question

x=int(input("Enter an integer:"))
answer=None
cube_root_found=False
for i in range(0,abs(x)+1):
    if i**3 == abs(x):
        answer=i
        cube_root_found=True
if not cube_root_found:
    print(x,"is not a perfect cube")
else:
    if x<0:
        answer=-answer
    print("Cube root of",x,"is",answer)

我无法理解为什么我们在第二行使用了answer = None和cube_root_found = False。另一个问题是我在标题中提到的：我们如何才能使这段代码更有效率？

Answer 1

我的意思是，有一个数学解决方案 - 只是立方根x，而不是测试[0, x+1)中的所有内容。没有理由在这种情况下循环。

x = int(input("Enter an integer: "))
result = x ** (1/3)
if result == int(result):
    print("Cube root of", x, "is", result)
else:
    print(x, "is not a perfect cube")

Answer 2

在break之后添加cube_root_found = True会在找到答案而不是继续时使其短路。

x=int(input("Enter an integer:"))
answer=None
cube_root_found=False
for i in range(0,abs(x)+1):
    if i**3 == abs(x):
        answer=i
        cube_root_found=True
        break
if not cube_root_found:
    print(x,"is not a perfect cube")
else:
    if x<0:
        answer=-answer
    print("Cube root of",x,"is",answer)

Answer 3

不要迭代一堆整数，只需取输入的立方根并检查它是否为整数。

这样的事情：

root = x**(1.0/3.0) return root.is_integer()

Answer 4

answer=None cube_root_found=False

我们这样做是因为，如果x为0，则for循环将执行0次。在这种情况下，如果没有上述代码，我们永远不会为这些变量赋值，

if not cube_root_found:

将失败并显示"NameError: name 'cube_root_found' is not defined消息。

Answer 5

Besides using break, I would suggest limiting range to abs(x)/2 - unless it's either -1, 0 or 1.

Besides, I am not sure whether Python can optimize calling abs on each cycle on the same value - most probably not, so I would pre-store abs(x)

abs_x = abs(x)
upper_limit = 2 if abs_x < 2 else abs_x / 2
for i in range(upper_limit):
    if i**3 == abs_x:
......

Answer 6

checking if int(n**(1./3)) == n**(1./3) does NOT work generally, since 1/3 does not represent nicely in binary. try abs(int(n**(1./3)) == n**(1./3)) < 0.000000000001

Checking within a threshold is a normal way to check for equality in the presence of rounding errors. Note that (n+1) ** (1/3) -> n**(1/3) as n gets large, so this will give incorrect results for very very very large n.

我们如何才能提高此代码的效率？

6 个答案: