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 ...
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...
In this Python article, you learnedHow to Find Armstrong Number Using Recursion in Pythonwith the practical example where we are taking user input to make the program dynamic. You may also like to read: Bijay Kumar I am Bijay Kumar, aMicrosoft MVPin SharePoint. Apart from SharePoint, I st...
实例(Python 3.0+) # Filename : test.py # author by : www.runoob.com # Python 检测用户输入的数字是否为阿姆斯特朗数 # 获取用户输入的数字 num = int(input("请输入一个数字: ")) # 初始化变量 sum sum = 0 # 指数 n = len(str(num)) # 检测 temp = num while temp > 0: digit = ...
Check N digit Armstrong Number #Python program to check n Digits Armstrong Number #Taking input from user number = int(input("Enter a number: ")) # Declarign and intilizing sum variable result = 0 # Declarign and intilizing number of digits variable n = 0 #coping number in another vari...
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。
$ python3 test.py 请输入一个数字: 1634 1634 是阿姆斯特朗数 获取指定期间内的阿姆斯特朗数 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):...
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:"))fornuminrange(lower,upper+1):# 初始化 sumsum=0# 指数n=len(str(num))# 检测temp=numwhiletemp>0:digit=temp%10sum+=digit**ntemp//=10ifnu...