How to add all the numbers in the sequence in while?

Question

How to add 1 to 64 in while?

I wanted to add up 1+2+3+4+...+64 print the sum value when arriving at the end.

What I’ve tried so far:

#include <stdlib.h>
#include <stdio.h>

int main(void)
{
    int numero= 1;
    //int resultado;

   while(numero <= 64) {

    printf("%d\n", numero);
    numero++;

    }
return 0;


}

Answer 1

If you want to add it up you should put it in the code. So you need a variable, obviously starting from 0, that takes the sum, and you need to add the new number to each step (the increment in each step you’ve already done), I imagine the impression should only occur at the end so it should stay out of the loop, something like that:

#include <stdio.h>

int main(void) {
    int numero = 1;
    int resultado = 0;
    while (numero < 65) resultado += numero++;
    printf("%d\n", resultado);
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

The ++ is increasing the variable numero at each step of the loop, it adds 1 to the variable itself, the same as numero = numero + 1. And the += is adding the result of the expression to the right in the variable resultado, sum in the variable itself, the same as resultado = resultado + numero. Note that the increment of numero only occurs after because he is a post-increment.

You can do with for, but I think it’s worse:

    int resultado = 0;
    for (int numero = 1; numero < 65; numero++) resultado += numero;
    printf("%d\n", resultado);

Answer 2

How about including at design level the loop, but not at code level?

Well, your question is clearly mathematical. As such, it could be solved mathematically.

You have the numbers 1, 2, 3..., n, for a n any. Be the value of that sum S.

What happens if we take this list and add it to the reverse of it?

1     +  2    +  3    + ... + (n-1) + n
n     + (n-1) + (n-2) + ... + 2     + 1
-------------------------------------------
(n+1) + (n+1) + (n+1) + ... + (n+1) + (n+1)

The first list goes S. Since it is a finite sum, reversing the elements will not alter the final result, so the reverse list is also valid S. This means that the sum of the two lists is 2S. Note that the term (n+1) is repeated n times, therefore:

2S = n*(n+1)

Hence:

S = n*(n+1)/2

Like n is a whole, n*(n+1) necessarily is an even number. Therefore dividing by 2 will result in an integer. To guarantee the value, it costs nothing to force the multiplication precedence before the division:

( n*(n+1) )/2

So, as one wishes to know the value for the sum up to 64, could do (64*65)/2 or leave it in the variable, which would be easier to change to another value in the future:

#include <stdio.h>

int main()
{
    int n = 64;
    printf("%d\n", ( n*(n+1) )/2);

    return 0;
}

Answer 3

You are printing all numbers with this program. You need to create a variable (in my case, "sum") that joins all numbers.

#include <stdio.h>
int main(){
    int numero = 0;
    int soma = 0;

    while(numero < 5) {
        numero++;
        soma = soma + numero;
        printf("%d\n", soma);
    }
}