python 5的倍數(shù)

Sometimes, we need to find the sum of all integers or numbers that are completely divisible by 3 and 5 up to thousands, since thousands are a too large number that’s why it becomes difficult for us. So, here we will do it in Python programming language that solves the problem in just a few seconds. To solve this problem, we will use the range function. So, before going to find the sum we will learn a little bit about range function.

有時(shí)，我們需要找到可以被3和5整除的所有整數(shù)或數(shù)字的總和，直到數(shù)千，因?yàn)閿?shù)千是一個(gè)太大的數(shù)字，這就是為什么我們很難做到這一點(diǎn)。 因此，在這里，我們將使用Python編程語言來完成此任務(wù)，并在幾秒鐘內(nèi)解決問題。 為了解決這個(gè)問題，我們將使用范圍函數(shù)。 因此，在找到總和之前，我們將學(xué)習(xí)一些有關(guān)范圍函數(shù)的知識(shí)。

What is the range function in Python?

Python中的范圍函數(shù)是什么？

The range() is a built-in function available in Python. In simple terms, the range allows them to generate a series of numbers within a given interval. This function only works with the integers i.e. whole numbers.

range()是Python中可用的內(nèi)置函數(shù)。 簡(jiǎn)而言之，該范圍允許它們?cè)诮o定間隔內(nèi)生成一系列數(shù)字。 此函數(shù)僅適用于整數(shù)，即整數(shù)。

Syntax of range() function:

range()函數(shù)的語法：

range(start, stop, step)

It takes three arguments to start, stop, and step and it depends on users choose how they want to generate a sequence of numbers? By default range() function takes steps of 1.

它需要三個(gè)參數(shù)來開始 ， 停止和步進(jìn) ，并且取決于用戶選擇如何生成數(shù)字序列？ 默認(rèn)情況下， range()函數(shù)采用步驟1。

Program:

程序：

# initialize the value of n n=1000 # initialize value of s is zero. s=0 # checking the number is divisible by 3 or 5 # and find their sum for k in range(1,n+1):if k%3==0 or k%5==0: #checking condition s+=k# printing the result print('The sum of the number:',s)

Output

輸出量

The sum of the number: 234168

翻譯自: https://www.includehelp.com/python/find-the-sum-of-all-numbers-below-1000-which-are-multiples-of-3-or-5.aspx

python 5的倍數(shù)

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總結(jié)

以上是生活随笔為你收集整理的python 5的倍数_查找所有低于1000的数字的和，这是Python中3或5的倍数的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問題。

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