needs to be c++ sorry about that. Suppose that the first number of a sequence is

ID: 3905250 • Letter: N

Question

needs to be c++

sorry about that.

Suppose that the first number of a sequence is x, in which x is an integer. Define ao x; an+ 1-an/ 2 if an is even; ant 1 3 an +1 if an is odd. Then, there exists an integer k such that ak-1. Write a program that prompts the user to input the value of x. The program output the integer k such that ak-1 and the numbers a0, a1, a2-… ak. ( For example, if x = 75, then k = 14, and the numbers a0, a1, a2-.., a14, respectively, are 75, 226, 113, 340, 170, 85, 256, 128, 64, 32, 16, 8, 4, 2, 1.) Test your program for the following values of x: 75, 111, 678, 732, 873, 2048, and 65535 1. No documentation (-5) 2. Weak documentation (-3) 3. Should include a clear prompt for how to start by entering an integer, x (-2 if not) 4. Needs a while loop or do while to calculate the new value of x each iteration of the loop (-6 if not, or if for loop used) 5. Must have an if/else if or if/else within the while to determine if x is even or odd (-3 if not) 6. Each iteration of the loop, should output the value of x (-8 if not attempting to output x) 7. After the loop, output k (clearly labeled) (-6 if k is not output at all, -2 if not clearly labeled) 8. Inefficiencies, up to -5 and noted why TEST: Use the integer 13 for x. Output should be: 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 and k is 9 (-6 if output of x values is incorrect) and (-5 if k is incorrect)

Explanation / Answer

Hello Student!

I am happy to help you!

Here goes the code :

*************** Code starts here *********************

#include <iostream>

#include <cmath>

using namespace std;

int main(int argc, const char * argv[]) {

int k, x;

// Input the number

// k is for the size of the array

// x is a0 the first variable

cin >> k >> x;

int arr[k+1];

// initialization of a0 = x and arr = {0}

for (int i = 0 ; i <= k ; i++) {

arr[i] = 0;

}

arr[0] = x;

int size = (k-1)/2;

if (k%2 != 0) {

size = size+1;

}

for (int i = 1 ; i < size ; i++) {

if (arr[i-1]%2 != 0) {

arr[i] = 3*arr[i-1]+1; // for odd values

}else {

arr[i] = arr[i-1]/2; // for even values

}

for (int i = size ; i <= k ; i++) {

// Storing for ak

arr[i] = pow(2,k-i);

}

// The output will be

for (int i = 0 ; i <= k ; i++) {

if (i == k) {

cout << arr[i] << endl;

}else {

cout << arr[i] << ", " ;

}

return 0;

}

********************** code ends here *******************

Test Case : (I/O)

14 75

75, 226, 113, 340, 170, 85, 256, 128, 64, 32, 16, 8, 4, 2, 1