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
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...
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...
Here, we have set the lower limit 100 in variable lower and upper limit 2000 in variable upper using Python range(). We have used for loop to iterate from variable lower to upper. In iteration, the value of lower is increased by 1 and checked whether it is an Armstrong number or not...
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 ...
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:...
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: ...