Make a program that reads an integer N and then print the N first odd natural numbers

Question

#include<stdio.h>
#include<locale.h>
int main(){
    setlocale(LC_ALL,"portuguese");
    int n,imp;
    
    printf("Digite um número:\n");
    scanf("%d",&n);
    if(n!=0){//para Zero n ser considerado par
        if(n%2!=0){
            while(imp<n){
                imp=n+2;
                
                printf("%d",imp);
            }
                
                
                
        }
    }
    
    return 0;
}

When I run it displays only the next odd number and stops. But, what should happen is, if the user type 9, the next 9 odd numbers should appear - 11,13,15,17,19,21,23,25,27. If type 3 appears (5,7,9) and so on.

Answer 1

it presents only the next odd number and for

The problem is here:

while(imp<n){
    imp=n+2;

In fact, the problem starts a little earlier because you did not assign any value to imp before using it, then the value of this variable will be undetermined and may not even enter this while. If it "worked" in your case, it’s by coincidence, but you can’t program by coincidence.

Anyway, even if you get into while, right in the first iteration you make the value of imp be equal to n + 2. I mean, right now imp will have a value greater than n and therefore the condition imp < n is no longer satisfied. That’s why the while only runs once.

---

For the program to work, you have to consider its requirements:

• the value read (n) is the amount of numbers, so it should be used as control of the loop
• the n also determines what is the initial value to be printed: is the odd number immediately after n
  • that is, if n is odd, the initial value is n + 2
  • if n is even, the initial value is n + 1

You could do something like:

// obtém o valor inicial a ser impresso
int num;
if (n % 2 == 0) // se n é par, o valor inicial é n + 1
    num = n + 1;
else // se n é ímpar, o valor inicial é n + 2
    num = n + 2;

Or we can use the very result of n % 2 (the rest of the division by 2, i.e., it is zero for even numbers and 1 for odd numbers):

int num = n + 1 + (n % 2);

From there, we know that num is an odd number, so to get the next odd odd, just add 2 to it. I do this n times. It would look like this:

int n;
printf("Digite um numero: \n");
scanf("%d", &n);
// se n é ímpar, soma 2 para encontrar o próximo ímpar, se for par, soma 1
int num = n + 1 + (n % 2); // n % 2 é o resto da divisão por 2, então se for ímpar será 1, se for par será zero

// assim, num é o primeiro número ímpar imediatamente depois de n
for (int i = 0; i < n; i++) { // repito o for n vezes
    printf("%d\n", num);
    num += 2; // como num é ímpar, posso somar 2 para ir para o próximo ímpar
}

You can make it even shorter, taking advantage of the fact that in a for we can initialize and increment multiple variables:

int n;
printf("Digite um numero: \n");
scanf("%d", &n);
for (int i = 0, num = n + 1 + (n % 2); i < n; i++, num += 2) {
    printf("%d\n", num);
}

Answer 2

#include<stdio.h>
#include<locale.h>
int main(){
    setlocale(LC_ALL,"portuguese");
    int n,aux=0,vezes;
    
    printf("Digite um número:\n");
    scanf("%d",&n);
    vezes=n;
    if(n!=0){//para Zero n ser considerado par
        if(n%2!=0){
            while(aux<vezes){
                n+=2;
                printf("%d-",n);
                aux++;
            }
                
        }
    }
    
    return 0;
}

Answer 3

You can use an auxiliary variable and a counter to be able to specify the number of loop repetition times equal to the number typed by the user. And every while loop check that the rest of the split by 2 is different from 0 to see if the number is odd.

I used as an example my code below:

#include "stdio.h"

int main(void) {
    int n, aux, cont = 0;
    printf("Digite um numero: ");
    scanf("%d", &n);
    aux = n;

    while (cont < aux) {
        n++;
        if (n % 2 != 0) {
            printf("%d\t", n);
            cont++;
        }
    }

    printf("\n");
    system("PAUSE");
    return 0;
}