**C++ Programming Help Please!**

**NOTE: DO the *Challenging* Version!**

**READ THE ENTIRE PROGRAM DESCRIPTION CAREFULLY PLEASE!****Program Info:**The Babylonian algorithm to compute the square root of a number n is as follows: 1. Make a guess at the answer (you can pick n/2 as your initial guess). 2. Compute r = n / guess 3. Set guess = (guess + r) / 2 4. Go back to step 2 for as many iterations as necessary. The more that steps 2 and 3 are repeated, the closer guess will become to the square root of n. Write a program that inputs a double for n and iterates through the Babylonian algorithm 100 times. For a more challenging version, iterate until guess is within 1% of the previous guess, and outputs the answer as a double.

**More Important Info:**

**Babylonian Algorithm**for square root:

**Do the Challenging version!!!**(I would do the simple version first, then add the challenging loop condition) This is a math problem: The challenge is to find the condition that will stop the loop when guess is within 1% of the previous guess. This is not so hard to do on paper. It's a simple inequality:

**| guess - previous_guess | < 1% * previous_guess**You will NOT use absolute value function (if you know what that is) You will have to convert this expression to a simple boolean expression, removing the absolute value.