Python 程序检查一个数是否为霓虹数

用 while 循环编写一个 Python 程序来检查一个数是否为霓虹数。如果一个数等于其平方数的数字之和，则它是一个霓虹数。例如，9 是，因为 9 的平方是 81，8 + 1 = 9

在这个 Python示例中，我们首先找到一个数的平方。接下来，将该平方数分成单独的数字并找到它们的和。如果和等于原始数字，则它是一个霓虹数。

import math

Number = int(input("Enter the Number to Check Neon Number = "))
Sum = 0

squr = math.pow(Number, 2)
print("Square of a Given Digit = %d" %squr)

while squr > 0:
    rem = squr % 10
    Sum = Sum + rem
    squr = squr // 10

print("The Sum of the Digits   = %d" %Sum)

if Sum == Number:
    print("\n%d is a Neon Number." %Number)
else:
    print("%d is Not a Neon Number." %Number)

使用递归或递归函数编写一个 Python 程序来检查一个数是否为霓虹数。

import math
Sum = 0

def neonNumber(squr):
    global Sum
    if squr > 0:
        rem = squr % 10
        Sum = Sum + rem
        neonNumber(squr // 10)
    return Sum

    
Number = int(input("Enter the Number to Check Neon Number = "))

squr = math.pow(Number, 2)
print("Square of a Given Digit = %d" %squr)

Sum = neonNumber(squr)
print("The Sum of the Digits   = %d" %Sum)

if Sum == Number:
    print("\n%d is a Neon Number." %Number)
else:
    print("%d is Not a Neon Number." %Number)

Enter the Number to Check Neon Number = 44
Square of a Given Digit = 1936
The Sum of the Digits   = 19
44 is Not a Neon Number.

Enter the Number to Check Neon Number = 9
Square of a Given Digit = 81
The Sum of the Digits   = 9

9 is a Neon Number.

下面所示的程序将使用 for 循环和 while 循环打印出 1 到 n 之间的霓虹数。

import math

MinNeon = int(input("Please Enter the Minimum = "))
MaxNeon = int(input("Please Enter the Maximum = "))

for i in range(MinNeon, MaxNeon + 1):
    Sum = 0
    squr = math.pow(i, 2)

    while squr > 0:
        rem = squr % 10
        Sum = Sum + rem
        squr = squr // 10

    if Sum == i:
        print(i, end = '  ')

Please Enter the Minimum = 1
Please Enter the Maximum = 10000
1  9