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givi [52]
1 year ago
11

The user interface contains two types of user input controls: TextInput, which accepts all characters and Numeric Input, which a

ccepts only digits.
1. Implement the class TextInput that contains:
O Public method def add(c : Char) - concatenates the given character to the current value
O Public method def getValue(): String - returns the current value
Implement the class NumericInput that:
O Inherits from TextInput
O Overrides the add method so that each non-numeric character is ignored
For example, the following code should output "10":
$input = new NumericInput();
$input->add('1');
$input->add('a');
$input->add('0');
echo $input->getValue();
The code skeleton is provided below:
<?php
class TextInput
{
}
class Numericinput
{
{
//$input = new NumericInput();
//$input->add('1');
//$input->add('a');
//$input->add('0');
//echo $input->getValue();
Computers and Technology
1 answer:
Shtirlitz [24]1 year ago
4 0

Answer:

Explanation:

public class Main

{

private static String val; //current val

public static class TextInput

{

public TextInput()

{

val= new String();

}

public void add(char c)

{

if(val.length()==0)

{

val=Character.toString(c);

}

else

{

val=val+c;

}

}

public String getvalue()

{

return val;

}

}

public static class NumericInput extends TextInput

{

Override

public void add(char c)

{

if(Character.isDigit(c))

{

//if character is numeric

if(val.length()==0)

{

val=Character.toString(c);

}

else

{

val=val+c;

}

}

}

}

public static void main(String[] args)

{

TextInput input = new NumericInput();

input.add('1');

input.add('a');

input.add('0');

System.out.println(input.getvalue());

}

}

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Answer:

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Explanation:

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Kaira's company recently switched to a new calendaring system provided by a vendor. Kaira and other users connect to the system,
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2 years ago
We have said that the average number of comparisons need to find a target value in an n-element list using sequential search is
bija089 [108]

Answer:

Part a: If the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b: If the list contains n elements (where n is even) the middle terms are  at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c: The average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

Explanation:

Suppose the list is such that the starting index is 0.

Part a

If list has 15 elements, the middle item would be given at 7th index i.e.

there are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(8,9,10,11,12,13,14) above it. It will have to run 8 comparisons  to find the middle term.

If list has 17 elements, the middle item would be given at 8th index i.e.

there are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(9,10,11,12,13,14,15,16) above it.It will have to run 9 comparisons  to find the middle term.

If list has 21 elements, the middle item would be given at 10th index i.e.

there are 10 indices (0,1,2,3,4,5,6,7,8,9) below it and 10 indices (11,12,13,14,15,16,17,18,19,20) above it.It will have to run 11 comparisons  to find the middle term.

Now this indicates that if the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b

If list has 16 elements, there are two middle terms as  one at would be at 7th index and the one at 8th index .There are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(9,10,11,12,13,14,15) above it. It will have to run 9 comparisons  to find the middle terms.

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If list has 20 elements, there are two middle terms as  one at would be at 9th index and the one at 10th index .There are 9 indices(0,1,2,3,4,5,6,7,8) below it and 9 indices(11,12,13,14,15,16,17,18,19) above it. It will have to run 11 comparisons  to find the middle terms.

Now this indicates that if the list contains n elements (where n is even) the middle terms are  at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c

So the average number of comparisons is given as

((n+1)/2+(n+2)/2)/2=(2n+3)/4

So the average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

6 0
2 years ago
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Answer:

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Explanation:

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Number of computer programmers proficient only in python = 20 - ( 1+1+5) = 13

Number of computer programmers are not proficient in any of these three languages = 100 - ( 37 + 18 + 13 + 1+ 1+ 5+ 6 ) = 100 - 81 = 19

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