写一个递归函数threeplustwo，它接受参数（n），将所有可被2整除的数字加起来+所有可从范围（1到n-1）整除3的数字。 [英] Write a recursive function threeplustwo that takes argument (n), sums all numbers divisible by 2 + all numbers divisible by 3 from range(1 to n-1).

问题描述

这是一个模型答案代码（我不明白）

this is a model answer code (which I don't understand )

int threePlusTwo (int n)
{
if (n <= 2)
return 0;
if (n == 3)
return 2;
int twoSum = (n - 1) % 2 == 0? (n - 1): 0;
int threeSum = (n - 1) % 3 == 0? (n - 1): 0;
return twoSum + threeSum + threePlusTwo(n - 1);
}

我的尝试：

What I have tried:

#include<stdio.h>
#include<stdlib.h>
int tpt(int n)
{
    if(n==1) return 0;
    if(n%3==0||n%2==0) return n+tpt(n-1);

    return tpt(n-1);
}

我的代码的问题是n> = 6的测试用例因为||它只有6次条件...任何想法如何修改我的代码以通过该测试用例？！

推荐答案

你必须理解文本的含义你的作业并将其翻译成代码：



从范围（1到n-1）转换为：

You must understand the meaning of the text of your homework and tranlate it into code:

"from range (1 to n-1)" translates to:

for( int i = 1; i < n; i++ ) {
  //do magic stuff
}

将所有可分为2的数字相加（一些神奇的东西）

"sum all numbers divisable by 2" (some magic stuff)

if( (i %2) == 0) ) // % is modulo operator
{
  sum = summ + n;
}

提示：对3部分使用else if语句。



祝你好运与你的作业。并编写一些测试来检查正确的结果。

Tip: use a else if statement for the 3 part.

Good luck with your homeworks. And write some tests for checking the correct results.

请记住，你必须考虑（n-1） as上限。让我们看一下如何解释模型程序

Keep in mind you have to consider (n-1) as upper limit. Let's see how the model program could be intepreted

if (n <= 2)
  return 0;

这是微不足道的，因为 1 不能被 2 或<整除code> 3 。

This is trivial, since 1 is not divisible by 2 or 3.

if (n == 3)
  return 2;

这也是微不足道的，因为在 1,2 序列中，只有 2 可以整除。

This is trivial too, since in the 1,2 sequence, there is only 2 which is divisible.

int twoSum = (n - 1) % 2 == 0? (n - 1): 0;

这意味着

This means

int twoSum;
if ( (n-1) % 2 == 0)
  twoSum = (n-1);
else
  twoSum = 0;

即if（n-1）可被 2 整除，然后保留它以便将其添加到总和。



解释

That is "if (n-1) is divisible by 2 then keep it in order to add it to the sum".

The intepretation of

int threeSum = (n - 1) % 3 == 0? (n - 1): 0;

类似。

is similar.

return twoSum + threeSum + threePlusTwo(n - 1);

这样可以添加两倍于 2 和<$ c $的数字c> 3 （请注意问题标题中的这一要求并不明显。）

This makes the trick of adding two times a number which divisible both by 2 and 3 (please note such a requirement is not obvious from the question title).

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