将上面的各个步骤整合在一起,我们就可以得到完整的代码: # 获取用户输入,并将其转换为整数num=int(input("请输入一个整数:"))# 判断是否能被2整除is_divisible_by_2=(num%2==0)# 判断是否能被3整除is_divisible_by_3=(num%3==0)# 输出结果ifis_divisible_by_2andis_divisible_by_3:print(f"{num...
整数序列的生成 我们可以生成一个整数序列,并求出其中能被2整除和能被3整除的数。以下是一个完整的例子。 deffind_divisible_numbers(limit):divisible_by_2=[]divisible_by_3=[]fornumberinrange(1,limit+1):ifnumber%2==0:divisible_by_2.append(number)ifnumber%3==0:divisible_by_3.append(number)ret...
if x%2 == 0: if x%3 == 0: print 'Divisible by 2 and 3' else: print 'Divisible by 2 and not by 3' elif x%3 == 0: print 'Divisible by 3 and not by 2' 上面代码中的elif表示“else if”。在条件语句的测试中使用复合布尔表达式是很方便的,例如: if x < y and x < z: ...
The number 8 is divisible by 2.20212223C.3.2 函数Python支持函数。函数是一种定义一段可在程序中多处被执行的代码的好方法。我们可使用关键词def定义一个函数。[输入]source_code/appendix_c_python/example13_function.pydef rectangle_perimeter(a, b): return 2 * (a + b)print 'Let a rectangle ...
division and returns the remainder of that division. For example, ifx = 3andy = 2, thenx%ywill divide3by2and give us1as a remainder. Thedivisible()function in the following code snippet shows us how we can check whether one number is completely divisible by another number with the%operator...
Use therange(2, 22, 2)to get all even numbers from 2 to 20. (Here astepvalue is 2 to get the even number because even numbers are divisible by 2) Next, useforloop to iterate over each number In each iteration add the current number to the sum variable using thearithmetic operator....
Using the modulo operator with a modulus of 2, you can check any number to see if it’s evenly divisible by 2. If it is evenly divisible, then it’s an even number. Take a look at is_even() which checks to see if the num parameter is even:...
# find even numbers that are divisible by 5 num_list = [y for y in range(100) if y % 2 == 0 if y % 5 == 0] print(num_list) Run Code Output [0, 10, 20, 30, 40, 50, 60, 70, 80, 90] Here, list comprehension checks two conditions: if y is divisible by 2 or not...
# Not recommended if x > 5 and x % 2 == 0: print('x is larger than 5 and divisible by 2!') 在上面的示例中,and 运算符具有最低优先级。因此,可以更清楚地表达if语句如下: # Recommended if x>5 and x%2==0: print('x is larger than 5 and divisible by 2!') 你可以自由选择哪...
Python >>> def is_divisible(a, b): ... return not a % b ... >>> is_divisible(4, 2) True >>> is_divisible(7, 4) False If a is divisible by b, then a % b returns 0, which is falsy in Python. So, to return True, you need to use the not operator....