编写一个 Python 程序来检查一个数字是不是霓虹数字，或者是否使用 while 循环。如果一个数等于平方数位数之和，它就是霓虹数。例如，9 是一个霓虹数字，因为 92 = 81，8 +1 = 9
在这个 python 例子中，首先，我们找到一个数的平方。接下来，把那个正方形分成几个独立的数字，求出总和。如果总和等于实际数字，它就是一个霓虹数字。
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)Python 程序检查一个数字是不是 neon 数或者不使用递归或者递归函数。
# Python Program to Check Neon Number
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.使用 for 循环和 while 循环打印从 1 到 n 的霓虹数字的 Python 程序。
import math
MinNeon = int(input("Please Enter the Minimum Neon Value = "))
MaxNeon = int(input("Please Enter the Maximum Neon Value = "))
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 Neon Value = 1
Please Enter the Maximum Neon Value = 10000
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