无法让 python 程序扫描用户输入的数字并检查它是否是回文、阿姆斯特朗数和完全数

如何解决无法让 python 程序扫描用户输入的数字并检查它是否是回文、阿姆斯特朗数和完全数

我想扫描一个数字并在 python 中打印它是回文、阿姆斯特朗数或完全数，我尝试了以下方法但它不起作用，我已经单独检查了它们并且它们按预期工作。 代码：

#palindrome
n=int(input("Enter number:"))
temp=n
rev=0
while(n>0):
    dig=n%10
    rev=rev*10+dig
    n=n//10
if(temp==rev):
    print(n,"The number is a palindrome!")
else:
    print(n,"The number isn't a palindrome!")
#Armstrong
sum = 0
temp = n
while temp > 0:
   digit = temp % 10
   sum += digit ** 3
   temp //= 10
if n == sum:
   print(n,"is an Armstrong number")
else:
   print(n,"is not an Armstrong number")
#Perfect Number
sum1 = 0
for i in range(1,n):
    if(n % i == 0):
        sum1 = sum1 + i
if (sum1 == n):
    print(n,"The number is a Perfect number!")
else:
    print(n,"The number is not a Perfect number!")   

解决方法

回文码修改n，然后你直接重用amstrong和perfect，你要回到最初的数字，为此，请使用您的 temp

#Armstrong
sum = 0
n = temp

同时执行以下操作，出于同样的原因 n 已更改

print(temp,"The number is a palindrome!")

---

但是摆脱这种情况的最好方法以及更好的编码方法是使用方法

def palindrome(n):
    temp = n
    rev = 0
    while n > 0:
        dig = n % 10
        rev = rev * 10 + dig
        n = n // 10
    if temp == rev:
        print(temp,"The number is a palindrome!")
    else:
        print(temp,"The number isn't a palindrome!")

def armstrong(n):
    count = 0
    temp = n
    while temp > 0:
        digit = temp % 10
        count += digit ** 3
        temp //= 10
    if n == count:
        print(n,"is an Armstrong number")
    else:
        print(n,"is not an Armstrong number")

def perfect(n):
    count = 0
    for i in range(1,n):
        if n % i == 0:
            count = count + i
    if count == n:
        print(n,"The number is a Perfect number!")
    else:
        print(n,"The number is not a Perfect number!")

if __name__ == '__main__':
    n = int(input("Enter number:"))
    palindrome(n)
    armstrong(n)
    perfect(n)