堆叠和嵌套的if-else语句不移动到下一个i

Question

我一直在搜索堆叠/嵌套的if-else语句，elif的语句似乎不能满足我的需要。 其目的是打印所有偶数、平方数和立方体数，并突出显示哪些是平方数和立方体数。 这是我的代码：

import math
rn = 0
for n in range(1,100,1):
    if rn is 0:
        if n%2 is 0:
            print(n, " is an even number")           
        else:
            rn = 1         
        if math.sqrt(n) is int:
            print(n, " is a square number")
            if n**(1/3) is int:
                print(n, " is a Cube number too")
                continue
            else:
                print(n, "is only a square number")
                continue

        else:
            rn = 1
    if rn is 1:
        if n**(1/3) is int:
            print(n, " is a Cube only number")
        else:
            rn = 0
            continue

在递增n之前，代码不会移到下一个if语句

输出：

   2  is an even number
   4  is an even number
   6  is an even number

有没有其他方法可以继续或中断，但保持在同一个循环中？ 提前谢谢

Answer 1

import math
rn = 0
d=0.
for n in range(1,100,1):
    if rn is 0:
        if n%2 is 0:
            print(n, " is an even number")
        else:
            rn = 1
        if math.sqrt(d).is_integer():
            print(n, " is a square number")
            if float(n**(1-3)).is_integer():
                print(n, " is a Cube number too")
                continue
            else:
                print(n, "is only a square number")
                continue

        else:
            rn = 1
    if rn is 1:
        if n**(1-3) is int:
            print(n, " is a Cube only number")
        else:
            rn = 0
            continue

这将起作用，使用float.is_integer()，以前的问题是is int为任何输入返回False。还要注意，这将给出从1到100的所有数字>> {} is a square number，因为你在做math.sqrt(d)，而d是0，所以平方根总是整数0