How to solve this problem using recursion? - c++

Question

My function has to be able to print the sum of the numbers 1+1+1.... to get to N. I can only enter N. For example: sumaRecursiva(6): 1+1+1+1+1+1=6.
This is what I have:
int sumaRecursiva(int y) {
int x=0;
if (x == y){
return y;
}
else {
x += 1;
cout << x << "+" <<endl;
sumaRecursiva(y-1);
}
}

Answer 1

You may want another function to handle the fact that you need fewer + characters in your print than you need 1s
void printSum(int y) {
if (y == 0) return;
cout << 1;
printPlusRecursive(y - 1);
}
void printPlusRecursive(int y) {
if (y == 0) return;
cout << '+' << 1;
printPlusRecursive(y - 1);
}
or shorter, but more complex version:
void printSum(int y, bool firstCall = true) {
if (y == 0) return;
if (firstCall) cout << 1;
else cout << '+' << 1;
printSum(y - 1, false);
}

Answer 2

If you want the output to be exactly 1+1+1+1+1+1=6 then you will need a helper function sumaRecursivaHelper to print the 1+'s and the outer function to print the =n which I believe none of the other answers include.
#include <iostream>
using namespace std;
void sumaRecursivaHelper(int x) {
if (x == 0) {
cout << 1;
}
else {
cout << "1+";
sumaRecursivaHelper(x - 1);
}
}
void sumaRecursiva(int x) {
if (x == 0) {
cout << 0;
}
else {
sumaRecursivaHelper(x);
}
cout << "=" << x;
}
int main()
{
sumaRecursiva(6);
return 0;
}

Answer 3

Your recursive function should return a value.
The value passed to a recursive is usually used to test for the escape condition (not usually used as part of the result).
int sum(int N)
{
if (N == 0) {
return 0; // I like zero as the escape value.
// Though this will give you
// sum(6) = 1 + 1 + 1 + 1 + 1 + 1 + 0 = 6
// If you only want to add ones change
// the escape condition to 1 and return 1
// Though this will prevent you doing sum(0)
}
return 1 + sum(N-1);
}