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Crazy boy [7]
3 years ago
6

Write a Java program named Problem 3 that prompts the user to enter two integers, a start value and end value ( you may assume t

hat the start value is less than the end value). As output, the program is to display the odd values from the start value to the end value. For example, if the user enters 2 and 14, the output would be 3, 5, 7, 9, 11, 13 and if the user enters 14 and 3, the output would be 3, 5, 7, 9, 11, 13.
Computers and Technology
1 answer:
Mashcka [7]3 years ago
5 0

Answer:

hope this helps

Explanation:

import java.util.Scanner;

public class Problem3 {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       System.out.print("Enter start value: ");

       int start = in.nextInt();

       System.out.print("Enter end value: ");

       int end = in.nextInt();

       if (start > end) {

           int temp = start;

           start = end;

           end = temp;

       }

       for (int i = start; i <= end; i++) {

           if (i % 2 == 1) {

               System.out.print(i);

               if (i == end || i + 1 == end) {

                   System.out.println();

               } else {

                   System.out.print(", ");

               }

           }

       }

   }

}

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schepotkina [342]

Answer:

Explanation:

The program in question would create a new Scanner Object which asks the user for the Username first. It would then save this input into a temporary variable and compare it to the actual username in the database. Since the username is not case sensitive, we would use the toLowerCase() method on both the input and the database username so that they match even if the letters are not the same case structure. If both usernames match then we would move on to ask the user for the Password and compare it with the database password for that user. Since this one is case sensitive we would compare as is. Finally, if both Username and Password match we would print "Hello World" otherwise we would print "Login Failed."

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2 years ago
What are the two categories of problems that we can simply convert<br> to parallel code?
anastassius [24]
Percentage and decimals
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2 years ago
When recording data on a multiple-disk storage system, should we fill a complete disk surface before starting on another surface
Cerrena [4.2K]

The user needs to complete the entire disk surface first before starting another surface.

Explanation:

When you are using a multiple disk storage system to write the data the disk automatically writes the disk based on the algorithm for better efficiency and availability of the disk space.

hence when you are recording a data on a multiple disk storage system, it is recommended to fill the complete disk surface initially before you start the another surface to record the data.

4 0
3 years ago
Show the array that results from the following sequence of key insertions using a hashing system under the given conditions: 5,
sergejj [24]

Answer:

a) Linear probing is one of the hashing technique in which we insert values into the hash table indexes based on hash value.

Hash value of key can be calculated as :

H(key) = key % size ;

Here H(key) is the index where the value of key is stored in hash table.

----------

Given,

Keys to be inserted are : 5 , 205, 406,5205, 8205 ,307

and size of the array : 100.

First key to be inserted is : 5

So, H(5) = 5%100 = 5,

So, key 5 is inserted at 5th index of hash table.

-----

Next key to inserted is : 205

So, H(205) = 205%100 = 5 (here collision happens)

Recompute hash value as follows:

H(key) =(key+i) % size here i= 1,2,3...

So, H(205) =( 205+1)%100 = 206%100 = 6

So, key 205 is inserted in the 6th index of the hash table.

----------

Next Key to be inserted : 406

H(406) = 406%100 = 6(collision occurs)

H(406) =(406+1) %100 = 407%100 = 7

So, the value 406 is inserted in 7the index of the hash table.

-----------------

Next key : 5205

H(5205) = 5205%100 = 5(collision)

So, H(5205) = (5205+1)%100 = 6( again collision)

So, H(5205) = 5205+2)%100 = 7(again collision)

So, H(5205) = (5205+3)%100 = 8 ( no collision)

So, value 5205 is inserted at 8th index of the hash table.

-------------

Similarly 8205 is inserted at 9th index of the hash table because , we have collisions from 5th to 8th indexes.

-------------

Next key value is : 307

H(307) = 307%100 = 7(collision)

So, (307+3)%100 = 310%100 = 10(no collision)  

So, 307 is inserted at 10th index of the hash table.

So, hash table will look like this:

Key       index

5         5

205         6

406         7

5205 8

8205 9

307         10

b) Quadratic probing:

Quadratic probing is also similar to linear probing but the difference is in collision resolution. In linear probing in case of collision we use : H(key) = (key+i)%size but here we use H(key) =( key+i^2)%size.

Applying Quadratic probing on above keys:.

First key to be inserted : 5.

5 will go to index 5 of the hash table.

-------

Next key = 205 .

H(205) = 205%100 = 5(collision)

So. H(key)= (205+1^2)%100 = 6(no collision)

So, 205 is inserted into 6th index of the hash table.

--------

Next key to be inserted 406:

So, 406 %100 = 6 (collision)

(406+1^2)%100 = 7(no collision)

So, 406 is moved to 7th index of the hash table.

----------

Next key is : 5205

So, 5205%100 = 5 (collision)

So, (5205+1^2)%100 = 6 ( again collision)

So, (5205+2^2)%100 = 9 ( no collision)

So, 5205 inserted into 9th index of hash table.

-----------

Next key is 8205:

Here collision happens at 5the , 6the , 9th indexes,

So H(8205) = (8205+4^2)%100 = 8221%100 = 21

So, value 8205 is inserted in 21st index of hash table.

--------

Next value is 307.

Here there is collision at 7the index.

So, H(307) = (307+1^2)%100 = 308%100= 8.

So, 307 is inserted at 8the index of the hash table.

Key           Index

5                  5

205                  6

406                  7

5205                9

8205               21

307                   8

3 0
3 years ago
HOWARD!!!!!! THE PHONE IS RINGING!!!!!!!!!
777dan777 [17]

Answer:

i am confused

Explanation:

7 0
2 years ago
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