The modulo%operator returns the remainder from the division of the first number by the second. If the remainder is0, the number is divisible by the other number. main.py if9%3==0:# 👇️ this runsprint('number A is divisible by number B')else:print('numebr A is NOT divisible by...
print(f"{number} is divisible by 2 but not by 5") else: print(f"{number} is odd") 在这个示例中,for循环遍历列表中的每个元素,首先判断该元素是否为偶数。如果是偶数,则进一步判断是否能被5整除,并打印相应的信息。通过这种方式,可以实现复杂条件的判断和处理。 3. 实际应用 嵌套if语句与循环结合在需...
下面是一个简单的Python代码示例，展示了if和and、or的混合嵌套使用：python`number = 15 if number >= 10 and :print else:print`在这个例子中，由于数字15满足条件，所以会输出“Number is within the range and divisible by either 3 or 5.”这种混合嵌套的使用使得逻辑判断更加精细和...
}privatestaticbooleanisPrintable(intvalue){returnvalue%2==0&&value%3==0&&value%4==0&&value%5==0&&value%6==0&&value%7==0&&value%8==0&&value%9==0; } 4､通过for循环 publicstaticbooleanisDivisible(intnumber){ for(inti =2; i <=9; i++) { if(num % i !=0) { returnfalse; } ...
if(i %2==0&& i %3==0&& i %4==0&& i %5==0&& i %6==0&& i %7==0&& i %8==0&& i %9==0) { System.out.println(i); } 4、使用for循环实现 如果多个if条件判断的不是数字或数字不连续,可以使用数组来进行遍历。 publicstaticbooleanisDivisible(intnumber){ ...
num=6ifnum%2==0andnum%3==0:print("The number is even and divisible by 3")else:print("The number does not meet both conditions") 1. 2. 3. 4. 5. 6. 在这个示例中,我们首先定义了一个变量num,然后使用and运算符连接了两个条件num % 2 == 0和num % 3 == 0。只有当num既是偶数又能...
注意即使 6 可以被 2 整除,也不会输出number is divisible by 2,更不会输出else块中的number is not divisible by 4, 3, or 2。原因是 Rust 只会执行第一个条件为true的代码块,并且一旦它找到一个以后,甚至都不会检查剩下的条件了。 ?let语句中使用if...
If the number is divisible by three, the function should outputFizz. If the number is divisible by five, the function should outputBuzz. If the number is divisible by both three AND five, the function should outputFizz Buzz. Otherwise, the number entered is simply returned as output. ...
if(a%2 == 0 && a%3 == 0) { printf("The entered number is divisible by both 2 and 3"); } else if(a%2 == 0) { printf("The entered number is divisible by 2"); } else if(a%3 == 0) { printf("The entered number is divisible by 3"); ...
Original number: 200 Check the said number is a Harshad number or not! True Flowchart: Sample Solution-2: Python Code: deftest(n):if(n>0):t=sum(map(int,str(n)))returnnotn%t n=666print("Original number:",n)print("Check the said number is a Harshad number or not!")print(test...