如何检查一个数字是否可以被另一个数字整除（Python）？

I need to test whether each number from 1 to 1000 is a multiple of 3 or a multiple of 5. The way I thought I'd do this would be to divide the number by 3, and if the result is an integer then it would be a multiple of 3. Same with 5.

我需要测试从1到1000的每个数字是3的倍数还是5的倍数。我认为我这样做的方法是将数字除以3，如果结果是整数那么它会是3的倍数。与5相同。

How do I test whether the number is an integer?

如何测试数字是否为整数？

here is my current code:

这是我目前的代码：

n = 0
s = 0

while (n < 1001):
    x = n/3
    if isinstance(x, (int, long)):
        print 'Multiple of 3!'
        s = s + n
    if False:
        y = n/5
        if isinstance(y, (int, long)):
            s = s + n

    print 'Number: '
    print n
    print 'Sum:'
    print s
    n = n + 1

9 个解决方案

#1

You do this using the modulus operator, %

您可以使用模数运算符执行此操作，％

n % k == 0

evaluates true if and only if n is an exact multiple of k. In elementary maths this is known as the remainder from a division.

当且仅当n是k的精确倍数时才计算true。在小学数学中，这被称为分裂的余数。

In your current approach you perform a division and the result will be either

在您当前的方法中，您执行除法，结果将是

• always an integer if you use integer division, or
• 如果使用整数除法，则始终为整数，或
• always a float if you use floating point division.
• 如果使用浮点除法，它总是一个浮点数。

It's just the wrong way to go about testing divisibility.

这是测试可分性的错误方法。

#2

You can simply use % Modulus operator to check divisibility.
For example: n % 2 == 0 means n is exactly divisible by 2 and n % 2 != 0 means n is not exactly divisible by 2.

您只需使用％Modulus运算符来检查可分性。例如：n％2 == 0表示n可以被2整除，n％2！= 0表示n不能完全被2整除。

#3

You can use % operator to check divisiblity of a given number

您可以使用％运算符来检查给定数字的可分数

The code to check whether given no. is divisible by 3 or 5 when no. less than 1000 is given below:

检查是否给出的代码。当没有时，可以被3或5整除。下面给出的不到1000：

n=0
while n<1000:
    if n%3==0 or n%5==0:
        print n,'is multiple of 3 or 5'
    n=n+1

#4

This code appears to do what you are asking for.

此代码似乎可以满足您的要求。

for value in range(1,1000):
    if value % 3 == 0 or value % 5 == 0:
        print(value)

Or something like

或类似的东西

for value in range(1,1000):
    if value % 3 == 0 or value % 5 == 0:
        some_list.append(value)

Or any number of things.

或任何数量的东西。

#5

I had the same approach. Because I didn't understand how to use the module(%) operator.

我有同样的方法。因为我不明白如何使用模块（％）运算符。

6 % 3 = 0 *This means if you divide 6 by 3 you will not have a remainder, 3 is a factor of 6.

6％3 = 0 *这意味着如果你将6除以3则不会有余数，3则是6。

Now you have to relate it to your given problem.

现在你必须将它与你给定的问题联系起来。

if n % 3 == 0 *This is saying, if my number(n) is divisible by 3 leaving a 0 remainder.

如果n％3 == 0 *这就是说，如果我的数字（n）可被3整除，则留下0余数。

Add your then(print, return) statement and continue your

添加你的then（打印，返回）语句并继续你的

#6

For small numbers n%3 == 0 will be fine. For very large numbers I propose to calculate the cross sum first and then check if the cross sum is a multiple of 3:

对于小数字，n％3 == 0将没问题。对于非常大的数字，我建议先计算交叉和，然后检查交叉和是否是3的倍数：

def is_divisible_by_3(number):
    if sum(map(int, str(number))) % 3 != 0:
        my_bool = False
    return my_bool

#7

Try this ...

尝试这个 ...

public class Solution {

  public static void main(String[] args) {
    long t = 1000;
    long sum = 0;

    for(int i = 1; i<t; i++){
            if(i%3 == 0 || i%5 == 0){
                sum = sum + i;
            }
        }
        System.out.println(sum);    
  }
}

#8

jinja2 template fizzbuz:

jinja2模板fizzbuz：

<form>
  <ol>
    {% for x in range(1,n+1) %}
      {% set fizzbuzz_rpm = x %}
      {% if x % 3 == 0 and x % 5 == 0 %}
        {% set fizzbuzz_rpm="FizzBuzz" %}
      {% elif x % 3 == 0 %}
          {% set fizzbuzz_rpm="Fizz" %}
      {% elif x %5 == 0 %}
          {% set fizzbuzz_rpm="Buzz" %}
      {% endif %}
      <li>{{fizzbuzz_rpm}}</li>
    {% endfor %}
  </ol>
</form>

#9

The simplest way is to test whether a number is an integer is int(x) == x. Otherwise, what David Heffernan said.

最简单的方法是测试数字是否为整数是int（x）== x。否则，大卫赫弗南说。