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Svetllana [295]
3 years ago
6

Write a program that prints the question " Do youwant to continue? " and reads a user input. if the userinput is " Y ", " OK", o

r " Why not? ", printout " OK ". if the user input is "No " , then print out " Terminating ", Otherwise print " Bad input". The case of the user input should notmatter. For example, "y" or "yes" are also valid inputs. Write aclass YesNoChecker for this purpose.
Computers and Technology
1 answer:
Nataly_w [17]3 years ago
7 0

Answer:

#include <bits/stdc++.h>

using namespace std;

int main() {

   string in;//string in for taking input...

   cout<<"Do you want to continue ?"<<endl;

   getline(cin,in);//taking input from the user...

   if(in=="y"||in=="Y"||in=="yes"||in=="OK"||in=="Why not?")//conditions..

   cout<<"OK"<<endl;

   else if(in=="No")

   cout<<"terminating"<<endl;

   else

   cout<<"Bad input"<<endl;

return 0;

}

Explanation:

I have taken a string in.

Then taking input from user.

Checking the conditions using if else if ladder.

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Answer:

Written in Java

import java.util.*;

public class Main {

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     else

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  }

  public static void main(String[] args) {

     int num;

     Scanner input = new Scanner(System.in);

     char tryagain = 'y';

     while(tryagain == 'y'){

     System.out.print("Number: ");

     num = input.nextInt();

     System.out.println(num+"! = "+ fact(num));

     System.out.print("Try another input? y/n : ");

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}        

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}

Explanation:

The static method is defines here

  public static int fact(int n) {

This checks if n is 1. If yes, it returns 1

     if (n == 1)

        return n;

If otherwise, it returns the factorial of n, recursively

     else

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  }

The main method starts here. Where the user can continue executing different values of n. The program keep prompting user to try again for another number until user signals for stoppage

  public static void main(String[] args) {

This declares num as integer

     int num;

     Scanner input = new Scanner(System.in);

This initializes tryagain as y

     char tryagain = 'y';

This checks if user wants to check the factorial of a number

     while(tryagain == 'y'){

This prompts user for input

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This gets user input

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This passes user input to the function and also prints the result

     System.out.println(num+"! = "+ fact(num));

This prompts user to try again for another value

     System.out.print("Try another input? y/n : ");

This gets user response

     tryagain = input.next().charAt(0);

}        

Download txt
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