以上代码定义了一个函数are_numbers_divisible_by_3_and_5,接受一个参数numbers,用于判断其中的数是否能被3和5整除。函数内部定义了一个空列表divisible_numbers,用于存储能被3和5整除的数。然后使用for循环遍历numbers中的每个数,调用is_divisible_by_3_and_5函数进行判断,如果返回True,则将该数添加到divisible_nu...
int() 将其转换为整型# 检查数字是否能被3整除is_divisible_by_3=(number%3==0)# 如果能整除,is_divisible_by_3 为 True# 检查数字是否能被5整除is_divisible_by_5=(number%5==0)# 如果能整除,is_divisible_by_5 为 True# 输出结果ifis_divisible_by_3andis_divisible_by_5:# 如果可以...
answer[i] == "FizzJazz" if i is divisible by 3 and 7. answer[i] == "BuzzJazz" if i is divisible by 5 and 7. answer[i] == "Jazz" if i is divisible by 7. 应用结构写出来的答案如下: classSolution:deffizzBuzz(self,num:int):# 1.initial outputans=[]# 2. Parameters set-up#...
In this update, you moved the condition that checks for numbers divisible by 3 and 5 to the beginning of the function. Now you can run the tests again: Shell $ python test_fizzbuzz.py test_buzz (__main__.TestFizzBuzz.test_buzz) ... ok test_fizz (__main__.TestFizzBuzz.test_fiz...
Here, the list comprehension will first check to see if the numberxis divisible by 3, and then check to see ifxis divisible by 5. Ifxsatisfies both requirements it will print, and the output is: Output [0, 15, 30, 45, 60, 75, 90] ...
For example, 3 and 5 are coprime, while 3 and 6 are not, because they are both divisible by 3.So, the Euclidean formula tells us that if m and n are coprime, and m– n is odd, the triple they generate is primitive. In the following example, we will write a generator expression ...
Print "FizzBuzz" for numbers divisible by both 3 and 5. Print the number itself if it is not divisible by either 3 or 5. 4.What are some common mistakes to avoid when solving the FizzBuzz problem? Common mistakes include: Forgot to check the condition for "FizzBuzz" (multiples of both...
执行上述代码时,结果如下: enter number8 divisible by 2 not divisible by 3 enter number15 divisible by 3 not divisible by 2 enter number12 Divisible by 3 and 2 enter number5 not Divisible by 2 not divisible by 3 Python 3 条件语句 Python 3 IF...ELIF...ELSE 语句 查看...
在基于 Debian 的系统(比如 Ubuntu)中,需要使用libsqlite3-dev才能成功。在基于 Red Hat 的系统(比如 Fedora 或 CentOS)中,需要使用libsqlite3-dev才能成功。 接下来,用./python.exe -c 'import _ctypes'检查_ctypes。如果失败,很可能没有安装libffi割台。
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: ...