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Tems11 [23]
3 years ago
15

4.In order for a driver to graduate from a learner’s license to an operator’s license, how many hours of driving need to be docu

mented? How many nighttime driving hours are required?
Computers and Technology
1 answer:
Illusion [34]3 years ago
8 0

Answer and Explanation

In the United States of America, driver's licenses are issued by each individual state, territory.

In order for a driver to graduate from a learner's license to an operator's license, how many hours of driving you need to be documented is 50 hours, of which 10 hours must be at night.To move on to your intermediate driver's license, you must: Be at least 16 years old .

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Write code to complete PrintFactorial()'s recursive case. Sample output if userVal is 5:5! = 5 * 4 * 3 * 2 * 1 = 120
garik1379 [7]

Answer:

Here is the C++ program:

#include <iostream>  //to use input output functions

using namespace std;   //to access objects like cin cout

void PrintFactorial(int factCounter, int factValue){  // method definition

  int nextCounter = 0;  // initialize value of nextCounter by 0

  int nextValue = 0;   // initialize value of nextValue by 0

  if (factCounter == 0) { //base case 1: 0! = 1          

     cout << "1" << endl;    }  //displays 1 in the output

  else if (factCounter == 1) {  //base case 2: 1! = 1  

     cout << factCounter << " = " << factValue << endl;    } //displays 1 in result    

  else {  //recursive case                          

     cout << factCounter << " * ";

     nextCounter = factCounter - 1;  //decrements factCounter by 1

     nextValue = nextCounter * factValue;  // multiplies values of nextCounter and factValue and assigns the result to nextValue

PrintFactorial( nextCounter,  nextValue);    } }  //recursively calls PrintFactorial method to compute the factorial

int main() {  //start of main function

  int userVal = 0;   //initialize the userVal variable to 0

  userVal = 5;  //assigns 5 to the userVal variable

  cout << userVal << "! = ";  //prints the userVal with ! =

  PrintFactorial(userVal, userVal); } //calls PrintFactorial method to compute the factorial of userVal i.e. 5

Explanation:

I will explain the logic of the program with an example.

The value of variable userVal is 5  

This is passed to the method PrintFactorial()

1st step:

void PrintFactorial(int factCounter, int factValue) becomes:  

PrintFactorial(5, 5)

PrintFactorial method has two base conditions which do not hold because the factCounter is 5 here which is neither 1 nor 0 so the program control moves to the recursive part.  

cout << factCounter << " * ";

In recursive part, the above statement prints the value of factCounter  followed by an asterisk on the output screen. So this prints:

5 *

nextCounter = factCounter - 1;

This statement decrements the value of factCounter by 1 and assigns it to nextCounter as:

nextCounter = 5 - 1;

nextCounter = 4

nextValue = nextCounter * factValue;  multiplies the value of nextCounter by factValue and stores result in nextValue . So

nextValue = 4 * 5  

nextValue = 20

PrintFactorial( nextCounter,  nextValue);  this statement calls the PrintFactorial() recursively to perform the above steps again until the base condition is reached. This statement becomes:  

PrintFactorial(4,20)

2nd step:

PrintFactorial method has two base conditions which do not hold because the factCounter is 4 here which is neither 1 nor 0 so the program control moves to the recursive part.  

cout << factCounter << " * ";

In recursive part, the above statement prints the value of factCounter  followed by an asterisk on the output screen. So this prints:

5* 4 *

nextCounter = factCounter - 1;

This statement decrements the value of factCounter by 1 and assigns it to nextCounter as:

nextCounter = 4 - 1;

nextCounter = 3

nextValue = nextCounter * factValue;  multiplies the value of nextCounter by factValue and stores result in nextValue . So

nextValue = 3 * 20

nextValue = 60

PrintFactorial( nextCounter,  nextValue);   This statement becomes:  

PrintFactorial(3,60)

3rd step:

PrintFactorial method has two base conditions which do not hold because the factCounter is 3 here which is neither 1 nor 0 so the program control moves to the recursive part.  

cout << factCounter << " * ";

In recursive part, the above statement prints the value of factCounter  followed by an asterisk on the output screen. So this prints:

5* 4 * 3 *

nextCounter = factCounter - 1;

This statement decrements the value of factCounter by 1 and assigns it to nextCounter as:

nextCounter = 3 - 1;

nextCounter = 2

nextValue = nextCounter * factValue;  multiplies the value of nextCounter by factValue and stores result in nextValue . So

nextValue = 2 * 60

nextValue = 120

PrintFactorial( nextCounter,  nextValue);  This statement becomes:  

PrintFactorial(2,120)

4th step:

PrintFactorial method has two base conditions which do not hold because the factCounter is 2 here which is neither 1 nor 0 so the program control moves to the recursive part.  

cout << factCounter << " * ";

In recursive part, the above statement prints the value of factCounter  followed by an asterisk on the output screen. So this prints:

5* 4 * 3 * 2 *

nextCounter = factCounter - 1;

This statement decrements the value of factCounter by 1 and assigns it to nextCounter as:

nextCounter = 2 - 1;

nextCounter = 1

nextValue = nextCounter * factValue;  multiplies the value of nextCounter by factValue and stores result in nextValue . So

nextValue = 1 * 120

nextValue = 120

PrintFactorial( nextCounter,  nextValue);  This statement becomes:  

PrintFactorial(1,120)

Now the base case 2 evaluates to true because nextCounter is 1

cout << factCounter << " = " << factValue  statement prints the value of factCounter i.e. 1 followed by an equal sign = followed by value of factValue i..e 120

So the output of the above program with userVal = 5 is:  

5! = 5 * 4 * 3 * 2 * 1 = 120

6 0
2 years ago
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