Question

The exercise as was written:

inger: 起草一项方案,要求用户输入2个ger、根和粉碎机,如1 <pwr <6和根**pwr等于用户输入的ger。如果不存在这种分类,它就应当为此印制信息。

这是我提出的:

    num_value = int(input("Enter a number value:"))
    result_state = "unknown"
    for root in range (-abs(num_value),abs(num_value)):
        for pwr in range(2,6):
            test_value = root**pwr
            if test_value == num_value:
                print(f {num_value} is equivalent to {root} to the power of {pwr} )
                result_state = "Printed"
    if result_state != "Printed":
        print( I was not able to find a root/power integer pair for this number. )

我所写的守则是行之有效的,但我感到,同样,实现同样结果的方法可能比较有效。我也肯定地感到,我可能认为这只是增加根本范围绝对功能。我试图用开放式的电灯从我自己那里学习,但我希望确保我不会根据我的理解发展盲点。因此,感谢你的帮助!

Answer 1

Here is some improvement of your code:

Change result_state from string to boolean. The comparison of boolean is faster than comprison of string.
Change the range of root from range(-abs(num_value),abs(num_value)) to range(-int(abs(num_value) ** 0.5), int(abs(num_value) ** 0.5) + 1). Since the smallest pwr is 2, if root is greater than the square root of abs(num_value), test_value cannot equal to num_value.
Add break to if-statement, if a pair of root and pwr, when root grows, test_value cannot still equals to num_value.

增订的法典如下:

num_value = int(input("Enter a number value: "))
result_state = False

for root in range(-int(abs(num_value) ** 0.5), int(abs(num_value) ** 0.5) + 1):
    for pwr in range(2, 6):
        test_value = root ** pwr
        if test_value == num_value:
            print(f {num_value} is equivalent to {root} to the power of {pwr} )
            result_state = True
            break  # Exit inner loop if pair found

if result_state != True:
    print( I was not able to find a root/power integer pair for this number. )

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