Is there a simpler way to do this? How can I get rid of one of my while loops?

For my CS assignment we were asked to create a program to approximate pi using Viete's Formula. I have done that, however, I don't exactly like my code and was wondering if there was a way I could do it without using two while loops.

(My professor is asking us to use a while loop, so I want to keep at least one!)

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

int main()
{
  double n, x, out, c, t, count, approx;
  printf("enter the number of iterations to approximate pi\n");
  scanf("%lf", &n);

  c = 1;
  out = 1;
  t = 0;
  count = 1;
  x = sqrt(2);

  while (count<=n)
  { 
      t=t+1;

      while (c<t)
      { 
          x=sqrt(2+x);
          c=c+1; 
      }

      out=out*(x/2);
      count=count+1; 
  }

  approx=2/out;
  printf("%lf is the approximation of pi\n", approx);
}

I just feel like my code could somehow be simpler, but I'm not sure how to simplify it.

Solution

Consider how many times the inner loop runs in each iteration of the outer loop

on the first iteration, it does not run at all (c == t == 1)
on each subsequent iteration, it runs exactly once (as t has been incremented once since the last iteration of the outer loop).

So you could replace this inner while with an if:

    if (count > 1) {

once you do that, t and c are completely unnecessary and can be eliminated.

If you change the initial value of x (before the loop), you could have the first iteration calculate it here, thus getting rid of the if too. That leaves a minimal loop:

out = 1;
count = 1;
x = 0;
while (count<=n) {
    x=sqrt(2+x);
    out=out*(x/2);
    count=count+1; 
}