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vovangra [49]
3 years ago
15

Write code which takes a user input of a String and an integer. The code should print each letter of the String the number of ti

mes the user inputted in reverse order.
Sample run:

Input a String:
code
Input an integer:
3
eeedddoooccc


Note: Write In Java
Thank You...
Computers and Technology
2 answers:
Alexus [3.1K]3 years ago
7 0

import java.util.Scanner;

public class JavaApplication70 {

   public static void main(String[] args) {

       Scanner scan = new Scanner(System.in);

       System.out.println("Input a String:");

       String txt = scan.nextLine();

       System.out.println("Input an integer:");

       int num = scan.nextInt();

       String newTxt = "";

       int w = 0;

       for (int i = txt.length()-1; i >= 0; i--){

           char c = txt.charAt(i);

           while (w < num){

               newTxt += c;

               w++;

           }

           w = 0;

       }

       System.out.println(newTxt);

   }

   

}

I hope this helps!

Nesterboy [21]3 years ago
6 0

Answer:

import java.util.*;  

class Main {  

 public static void main(String args[]) {

   Scanner sc= new Scanner(System.in);

   

   System.out.print("Input a String: ");  

   String str = sc.nextLine();  

   

   System.out.print("Input an integer: ");  

   int num = sc.nextInt();  

   

   for(int i=str.length()-1; i>=0; i--) {

       for(int j=0; j<num; j++) {

           System.out.print(str.charAt(i));

       }

   }

         

 }

}

Explanation:

There you go.

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In python please!! Write the definition of a function named countPos that needs integer values from standard input until there a
jek_recluse [69]

Answer:

Explanation:

The following code is written in Python it doesn't use any loops, instead it uses a recursive function in order to continue asking the user for the inputs and count the number of positive values. If anything other than a number is passed it automatically ends the program.

def countPos(number=input("Enter number: "), counter=0):

   try:

       number = int(number)

       if number > 0:

           counter += 1

           newNumber = input("Enter number: ")

           return countPos(newNumber, counter)

       else:

           newNumber = input("Enter number: ")

           return countPos(newNumber, counter)

   except:

       print(counter)

       print("Program Finished")

countPos()

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A hardware supplier manufactures three kinds of​ clamps, types​ A, B, and C. Production restrictions force it to make 10 more ty
Rufina [12.5K]

Answer:

A = 120

B = 40

C = 70

Solution:

As per the question:

Manufacturer forced to make 10 more type C clamps than the total of A and b:

10 + A + B = C                (1)

Also, 3 times as many type B as type A clamps are:

A = 3B                             (2)

The total no. of clamps produced per day:

A + B + C  = 330              (3)

The no. of each type manufactured per day:

Now, from eqn (1), and (3):

A + B + 10 + A + B = 330

2A + 2B = 320

A + B = 160                       (4)

Now, from eqn (2) and (4):

3B + B = 160

B = 40

Since, A = 3B

A = 3\times 40

A = 120          

Put the values of A and C in eqn (3):

120 + 40 + C = 330

C = 70

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zysi [14]

Solution:

Guest integration services resolves compatibility issues of certain guest operating systems experiencing non-functioning features.

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This is the required answer.

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