Sample Solution: Python Code: # Prompt the user to enter a number and convert the input to an integernum=int(input("Enter a number: "))# Calculate the remainder when the number is divided by 2mod=num%2# Check if the remainder is greater than 0, indicating an odd numberifmod>0:# P...
# Python program to check if the input number is odd or even. # A number is even if division by 2 gives a remainder of 0. # If the remainder is 1, it is an odd number. num = int(input("Enter a number: ")) if (num % 2) == 0: print("{0} is Even".format(num)) else...
The most common approach to check if a number is even or odd in Python is using themodulo operator (%). The modulo operator returns the remainder after division. An even number is evenly divisible by 2 with no remainder, while an odd number leaves a remainder of 1 when divided by 2. ...
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 ...
"""Function to check if number is even or odd""" return"Even"ifnum%2==0else"Odd" NUM=5 print(f"The number {NUM} is {is_number_even(NUM)}") 通过这段代码,我们添加了一个模块和函数docstring,在结尾处添加了一个新行,并重新命名了上述代码中...
if num & 1 == 0: return “偶数” else: return “奇数” # 测试 print(is_even_odd(2)) # 输出:偶数 print(is_even_odd(7)) # 输出:奇数 “` 以上两种方法都可以用来判断一个数是奇数还是偶数。根据实际情况选择合适的方法即可。 在Python中,判断一个数是奇数还是偶数可以使用以下几种方法: ...
defcheck_even_odd(number):ifnumber%2==0:return"偶数"else:return"奇数" 1. 2. 3. 4. 5. 在这个例子中,check_even_odd函数接收一个参数number。如果number能被2整除,函数将返回“偶数”,否则返回“奇数”。 2.2 多条件的示例 假设我们需要一个函数来判断一个考试成绩等级: ...
We will learn how to check if any given number is even or odd using different techniques, using the subtraction method and using the modulo operator method.
(): if event.type == pygame.QUIT: pygame.quit() sys.exit() elif event.type == pygame.KEYDOWN: if event.key == pygame.K_SPACE or event.key == pygame.K_UP: dino.jump(sounds) elif event.key == pygame.K_DOWN: dino.duck() elif event.type == pygame.KEYUP and event.key == ...
其中if...else语句用来执行需要判断的情形。 代码语言:javascript 代码运行次数:0 运行 AI代码解释 # Assign a numeric value number=70# Check the is more than70or notif(number>=70):print("You have passed")else:print("You have not passed") ...