# Python program to check if the number provided by the user is an Armstrong number or not # take input from the user num = int(input("Enter a number: ")) # initialize sum sum = 0 # find the sum of the cube of each digit temp = num while temp > 0: digit = temp % 10 sum...
接下来,我们将使用Python编程语言来实现一个算法,来找出所有的水仙花数。 defis_armstrong_number(num):# 将数字转换为字符串以便逐位计算digits=str(num)num_digits=len(digits)# 计算每位数字的 num_digits 次幂的和sum_of_powers=sum(int(digit)**num_digitsfordigitindigits)returnsum_of_powers==numdeffind...
defis_armstrong_number(num):order=len(str(num))sum_of_powers=sum(int(digit)**orderfordigitinstr(num))returnsum_of_powers==numdeffind_armstrong_numbers(start,end):armstrong_numbers=[]fornumberinrange(start,end+1):ifis_armstrong_number(number):armstrong_numbers.append(number)returnarmstrong_numbe...
This is how we can find Armstrong’s number using recursion in Python. Conclusion In this Python article, you learned How to Find Armstrong Number Using Recursion in Python with the practical example where we are taking user input to make the program dynamic. You may also like to read: How...
Enter a number: 0 Zero A number is positive if it is greater than zero. We check this in the expression of if. If it is False, the number will either be zero or negative. This is also tested in subsequent expression.Also Read: Python Program to Check if a Number is Odd or Even ...
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if num == sum: print(num,"is an Armstrong number")else: print(num,"is not an Armstrong number")▍50、用一行Python代码,从给定列表中取出所有的偶数和奇数a = [1,2,3,4,5,6,7,8,9,10]odd, even = [el for el in a if el % 2==1...
The output of Leap Year Program In Python: 2024 is a leap year. Also Read: How to check for Armstrong Number in Python? Different Methods to Find Leap Year In Python Programming There are a lot of different methods to find the leap year in the Python Program. Here are a few: i)...
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num = int(input( “Enter the number:\n”)) order = len(str(num)) sum = 0 temp = num whiletemp > 0: digit = temp % 10 sum += digit ** order temp //= 10 ifnum == sum: print(num, “is an Armstrong number”) else: ...