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).

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

C++

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);
}

What I have tried:

C++

#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);
}

the problem with my code is the test case that n>=6 it sums 6 only once because of the || condition...any ideas how to modify my code to pass that test case?!

Posted 7-Nov-17 2:11am

Member 13476370

Updated 7-Nov-17 2:53am

KarstenK

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CHill60 7-Nov-17 8:39am

What is it about the model answer that you don't understand - you've used all of the constructs in your own code except for the ternary/conditional operator ?: - How to use the Conditional (ternary) ope - C++ Forum[^]

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CPallini · Answer 1 · 2017-11-07T02:53:00

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

C

if (n <= 2)
  return 0;

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

C

if (n == 3)
  return 2;

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

C

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

This means

C

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

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

The intepretation of

C

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

is similar.

C

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

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).

KarstenK · Answer 2 · 2017-11-07T02:40:00

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

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

C++

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

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

C++

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

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

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

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).

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