The length of a hailstone sequence is the number of terms it contains. For example, the hailstone sequence in example 1 (5, 16, 8, 4, 2, 1) has a length of 6 and the hailstone sequence in example 2 (8, 4, 2, 1) has a length of 4. Write the method hailstoneLength (int n), which returns the length of the hailstone sequence that starts with n. / * * Returns the length of a hailstone sequence that starts with n, * as described in part (a). * Precondition: n > 0 * / public static int hailstoneLength (int n)

Following are the program to this question:

#include / /defining header file

using namespace std;

int hailstoneLength (int n) / /defining method hailstoneLength

{

int t=1; / /defining integer variable assign

while (n!=1) / /define a loop that checks value is not equal to 1

{

if (n%2==0) / / check even number condition

{

n=n/2; / / divide the value by 2 and store its remainder value in n

t++; / /increment value of t by 1.

}

else

{

n=n*3+1; / /calculate and hold value in n

t++; / /increment value of t variable by 1

}

}

return t; / /return value

}

int main () / /defining main method

{

int n; / /defining integer variable

cout<<"Enter any number: "; / /print message

cin>> n; / /input value

cout<
return 0;

}

Output:

Enter any number: 3

8

Explanation:

Program description can be given as follows:

In the given C+ + language program an integer method "hailstoneLength", is declared, that accepts an integer variable "n" in its parameter. Inside the method, an integer variable t is declared, that assign a value that is 1, in the next line, a while loop is declared, that uses if block to check even condition if number is even it divide by 2 and increment t variable value by 1. If the number is odd it will multiply the value by 3 and add 1 and increment t by 1 then it will go to if block to check value again. when value of n is not equal to 1 it will return t variable value. In the main method, an integer variable "n" is used that call the method and print its return value.