代码解析 is_armstrong_number函数:该函数接收一个整数参数,并计算其每位数字的位数次方之和。如果这个和等于原数字,则返回True,否则返回False。 find_armstrong_numbers函数:遍历一个给定的范围,使用is_armstrong_number函数检查每个数字,并将符合条件的阿姆斯特朗数收集到列表中。 程序输出:在指定范围内找到的所有阿姆斯特...
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: 实例(Python 3.0+) # Filename : tes...
defis_armstrong_number(num):# 将数转换为字符串num_str=str(num)# 计算数的位数n=len(num_str)# 判断是否为阿姆斯特朗数sum=0fordigitinnum_str:digit_int=int(digit)sum+=digit_int**n# 判断是否超过原始数ifsum>num:returnFalseifsum==num:returnTrueelse:returnFalse 1. 2. 3. 4. 5. 6. 7. ...
Python program to check ifthe number provided by the userisan Armstrong numberornot # take input from the user num = int(input("Enter a number: ")) # initialise sum sum =0 # find the sum of the cube of each digit temp = num whiletemp >0: digit = temp %10 sum += digit **3...
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...
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 ...
Program to add two numbers in Python a = input('Enter first number: ') b = input('Enter second number: ') sum = float(a) + float(b) print('The sum of {0} and {1} is {2}'.format(num1, num2, sum)) Program to Check Armstrong Number in Python An Armstrong number is...
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print(num,"is not an Armstrong number") “Compile”:编译代码,类似于执行了python -m py_compile helloWorld.py,生成.pyc文件。pyc是一种二进制文件,是由py文件经过编译后生成的文件,属于byte code。py文件变成pyc文件后,加载的速度有所提高,而且pyc是一种跨平台的字节码,是由python的虚拟机来执行的,这个是...
num=int(input("Enter a number:")) # initialize sum sum=0# find the sum of the cube of each digit temp=numwhiletemp >0: digit= temp %10sum+= digit **3temp//= 10# display the resultifnum ==sum: print(num,"is an Armstrong number")else: ...