Suppose thenumber is 153. Then, (153 % 10 = 3 ** 3) + is_armstrong(153 // 10 = 15, 3). Here, the is_armstrong() method calls itself,so again, the same code block executes for thedigit 15and so on. This is how we can findArmstrong’s number using recursion in Python. ...
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...
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 = ...
In themain()function, we created four variablesnum,rem,res,tmpthat are initialized with 0. Then we read an integer number from the user and checked the given number is Armstrong or not. After that, we printed the appropriate message on the console screen....
Enter a number: 370 370 IS AN ARMSTRONG NUMBER in the above example a 3 digit number is obtained from the user using the input method. This value is converted into an integer using the int() method and assigned to the num variable. Then the arsum variable is assigned the value of 0...
If the sum and the number are equal, the number is an Armstrong number. Note: In the above program, the cube of a number could be calculated using an exponent operator **. For example, sum += remainder ** 3; Example 2: Check Armstrong Number of n Digits // program to check an ...
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. ...
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: ...
$ 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):...