Write a C++ program that implements Algorithm 5.1.1 page 241 using the steps and data structures shown.

ak = k / k + 1 for all integers k 1,

bi = i 1 / i for all integers i 2.

Solution a1 = 1 / 1 + 1 = 1 / 2 b2 = 2 1 / 2 = 1/2 a2 = 2 / 2 + 1= 2 / 3 b3 = 3 1 / 3 = 2 / 3

a3 = 3 / 3 + 1 = 3 / 4 b4 = 4 1 / 4 = 3 / 4 a4 = 4 / 4 + 1 = 4 / 5 b5 = 5 1 / 5 = 4 / 5

a5 = 5 / 5 + 1= 5/6

Use the division algorithm shown in Epp, rather than integer or division operators in C++.

You can use the code from last week for the division algorithm. Write the code to prompt the user for a decimal number and print out the binary result. The program should be set up to repeat the process 4 times. The program should respond gracefully (but not end) if an incorrect input occurs. The program should print out the binary result with space every 4 bits as in our homework.

This is my old code

#include

#include

#include

#include

#include

using namespace std;

// Function to return gcd of a and b using

// Euclidean algorithm

int gcd(int long long a, int b)

{

if (a == 0)

return b;

return gcd(b%a, a);

}

// Driver program to test above function

int main()

{

//long long: -2^63+1 to +2^63-1

long long x;

int y;

// have a check that 1 <= x <= 2^63

// check that x is in the limit of long long

// Keeping it under while makes it ask for the valid value once gain

while (std::cout << "Enter the value of x in range -2^63+1 to +2^63-1 " << " " && !(std::cin >> x))

{

cout << "Bad value!" << " ";

cin.clear();

cin.ignore(numeric_limits::max(), ' ');

};

do {

cout << "Enter the value of Y less than X and greater than 1" << " ";

} while (!(cin >> y && y < x && y >= 1));

cout << "Gcd of x , y " << x << "," << y << " is " << gcd(x, y);

return 0;

}