for number in range(1, 10001): if number < 10: #1-9都为阿姆斯特朗数,直接输出即可 print(number) elif number < 100: n1 = number % 10 #取个位数 n2 = int(number/10 % 10) #取十位数 result = n1**2 + n2**2 if number == result: print(number) elif number < 1000: n1 = number...
In this Python tutorial, you will learn how to find Armstrong numbers using recursion in Python. Generally, the Armstrong number is a way to practice looping, conditions, etc., to gain programming skills and improve logical thinking. Although the Armstrong number is not actually used in real-ti...
It's an Armstrong Number. Cubing numbers:3*3*3 + 7*7*7 + 1= 371 (Armstrong Number) Conclusion We will go through lots of tricky logic and a simple one with Visual Studio in Python. We see it's very easy to implement it. Read more articles onPython: Diving Into Python: Chapter 1...
#Python program to check n Digits Armstrong Number#Taking input from usernumber =int(input("Enter a number: "))# Declarign and intilizing sum variableresult =0# Declarign and intilizing number of digits variablen =0#coping number in another variableoriginalNumber = number#In this while loop...
Python3 实例如果一个 n 位正整数等于其各位数字的 n 次方之和,则称该数为阿姆斯特朗数。例如 1^3 + 5^3 + 3^3 = 153。1000 以内的阿姆斯特朗数: 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407。以下代码用于检测用户输入的数字是否为阿姆斯特朗数:...
Python program to check Armstrong number using object oriented approach# Define a class for Checking Armstrong number class Check : # Constructor def __init__(self,number) : self.num = number # define a method for checking number is Armstrong or not def isArmstrong(self) : # copy num ...
ARMSTRONG:Yes. Well, that’s definitely true as a cultural thing. I would agree with that. There is something really important — you want the engineers to outnumber the lawyers. One of the metrics that we track inside the company is the percentage of engineering. ...
Python 阿姆斯特朗数 Python3 实例 如果一个n位正整数等于其各位数字的n次方之和,则称该数为阿姆斯特朗数。例如1^3 + 5^3 + 3^3 = 153。 1000以内的阿姆斯特朗数: 1, 2, 3, 4, 5, 6, 7, 8, 9, 153_来自Python 3教程,w3cschool。
实例(Python 3.0+) # Filename :test.py # author by : www.runoob.com # 获取用户输入数字 lower = int(input("最小值: ")) upper = int(input("最大值: ")) for num in range(lower,upper + 1): # 初始化 sum sum = 0 # 指数 n = len(str(num)) # 检测 temp = num while temp ...
Python3 实例 如果一个n位正整数等于其各位数字的n次方之和,则称该数为阿姆斯特朗数。 例如1^3 + 5^3 + 3^3 = 153。 1000以内的阿姆斯特朗数: 1, 2, 3, 4, 5, 6, 7, 8, 9, 153, 370, 371, 407。 以下代码用于检测用户输入的数字是否为阿姆斯特朗数: ...