总和数字总线循环 [英] sum numbers busing while loop

问题描述

大家好我想把所有3的倍数加在1到100之间

这就是我试过的东西

hi guys i want to sum all the multiples of 3 between 1 and 100
this is what i tried

#include <iostream>
#include <cmath>
using namespace std;
int main()
{
	int t;
	for (int i=1 ;i<=100; i++)
	{
	cout << i+3<< " ";
	}
	
		
	return 0;
} 

推荐答案

我会做一些更简单的事情 - 但因为这是你的功课，我赢了;不给你代码！

而不是那样，使用一个在1和100/3之间运行的for循环。在循环外声明一个名为sum的变量，并将其设置为零。

然后在循环内，每次将值i * 3添加到sum。

循环后，打印sum的值。

I'd do something simpler - but since this is your homework, I won;t give you the code!
Instead of that, use a for loop that runs with i between 1 and 100/3. Declare a variable called "sum" outside the loop, and set it to zero.
Then inside the loop, add the value "i * 3" to "sum" each time.
After the loop, print the value of "sum".

有很多种方法。以下是一些：



1.将数字3,6,9，...添加到99.

2.添加数字1到33并乘以3.

3. 3 *（33 * 33 + 33）/ 2

4. 1683



我已经添加了这个解决方案作为对Griff解决方案的赞美。如果你选择＃3或＃4，请确保你理解所涉及的数学，并准备解释你对教授的答案。

There are a number of ways. Here are a few:

1. add the numbers 3, 6, 9, ... to 99.
2. add the numbers 1 to 33 and multiply by 3.
3. 3 * (33*33 + 33)/2
4. 1683

I've added this solution as a compliment to Griff's solution. If you pick #3 or #4, make sure you understand the math involved and are prepared to explain your answer to the professor.

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