Sum of n + last digit on the left

Asked

Viewed 65 times

1

I have a recursion challenge. The goal is to return "n + the sum of the last digit on the left". Example:

Input: 54321

Output: 54326 (54321 + 5)

But the only way I could was like this:

#include <stdio.h>

int somault(int n){
   if(n < 10) return n;
   return somault(n/10);
}

int main() {
   int n;
   scanf("%d", &n);
   printf("%d", n+somault(n));//A soma é realizada aqui, não dentro da função.
   return 0;
}

It is possible to return this value meeting the following criteria?

  • Only one function
  • Recursive function
  • Not using vectors, global variables and/or pointers

I tried other ways, but none returned the correct value. Precisely because I can’t get the value of the last call and only do with the first one. I tried to use static to count or divide increasing the base 10 to catch the last, but did not work.

c recursion
  • 1

    But is it possible to use external variables ? Depending on the entire list of requirements can drastically change the solution and so I advise you to specify them all clearly. With a global variable outside can also get there, but something tells me that also can not do this lol

  • The problem is actually another, but I don’t know if it can change anything. So... Put the whole problem? With the solution you had made? Maybe the one I made based on the other can have solution. But the rules are the same.

  • The problem is another ? But what then ? The restrictions apparently were not all discriminated against. You can always change the question by editing it and making it clearer, as suggested. But if there’s another solution that I had developed with another code, maybe it’s something for another question, hard to say.

  • The solution is to return if n-4 if the leftmost digit is even, or n-3 if it is odd.

  • If you have a solution that solves the problem in the way you expected, then the ideal is to put it as an answer.

2 answers

3

I personally think that these kinds of challenges that force a specific recursive solution are not very good, and end up generating code rather not intuitive and unusual, and that in the end has no real-world applicability.

Still if nothing is indicated in relation to the number of parameters of the function then it can pass receive two parameters, in which it passes the same number in the two. In one of them you walk until you reach the first digit and in the other you keep the original number for the sum:

#include <stdio.h>

int somault(int n, int original){
   if(n < 10) return n + original;
   return somault(n/10, original);
}

int main() {
   int n;
   scanf("%d", &n);
   printf("%d", somault(n, n));
   return 0;
}

Watch it work on Ideone

Reinforcing, this would have no use in real code, but meets the requirements you indicated, and generates the expected response.

0

Solution:

#include <stdio.h>

int somault(int n){
   if(n/10 < 10) return n-n/10;
   return n+(n/10-somault(n/10));
}

int main() {
   int n;
   scanf("%d", &n);
   printf("%d", somault(n));
   return 0;
}

I’ll explain the operation later.

Browser other questions tagged c recursion

You are not signed in. Login or sign up in order to post.