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Nadusha1986 [10]
3 years ago
7

Write a while loop that prints userNum divided by 2 (integer division) until reaching 1. Follow each number by a space.

Computers and Technology
1 answer:
Ksenya-84 [330]3 years ago
3 0

Answer:

   while(userNum>=1){

       System.out.print(userNum/2+" ");

       userNum--;

        }

Explanation:

This is implemented in Java programming language. Below is a complete code which prompts a user for the number, receives and stores this number in the variable userNum.

<em>import java.util.Scanner;</em>

<em>public class TestClock {</em>

<em>    public static void main(String[] args) {</em>

<em>    Scanner in = new Scanner (System.in);</em>

<em>        System.out.println("Enter the number");</em>

<em>    int userNum = in.nextInt();</em>

<em>    while(userNum>=1){</em>

<em>        System.out.print(userNum/2+" ");</em>

<em>        userNum--;</em>

<em>         }</em>

<em>    }</em>

<em>}</em>

The condition for the while statement is userNum>=1 and after each iteration we subtract 1 from the value of   userNum until reaching 1 (Hence userNum>=1)

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You are given a string of n characters s[1 : : : n], which you believe to be a corrupted text document in which all punctuation
alexdok [17]

Answer: provided in the explanation section

Explanation:

Given that:

Assume D(k) =║ true it is [1 : : : k] is valid sequence  words or false otherwise

  • To determine D(n)

now the sub problem s[1 : : : k] is a valid sequence of words IFF s[1 : : : 1] is a valid sequence of words and s[ 1 + 1 : : : k] is valid word.

So, from here we have that D(k) is given by the following recorance relation:

D(k) = ║ false maximum (d[l]∧DICT(s[1 + 1 : : : k]) otherwise

Algorithm:

Valid sentence (s,k)

D [1 : : : k]             ∦ array of boolean variable.

for a ← 1 to m

do ;

d(0) ← false

for b ← 0 to a - j

for b ← 0 to a - j

do;

if D[b] ∧ DICT s([b + 1 : : : a])

d (a) ← True

(b). Algorithm Output

      if D[k] = = True

stack = temp stack           ∦stack is used to print the strings in order

c = k

while C > 0

stack push (s [w(c)] : : : C] // w(p) is the position in s[1 : : : k] of the valid world at // position c

P = W (p) - 1

output stack

= 0 =

cheers i hope this helps !!!

8 0
3 years ago
Which of the following problems is least likely to be solved through grid computing? Financial risk modeling Gene analysis Linea
gregori [183]

Answer:

Linear problems

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Grid computing can be used to solve various issues connected with the use of Computer resources such as Financial risk modelling,Gene analysis etc but least likely to be used to solve Linear problems.

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How many currencies do recognize?
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On a wireless router, what gives out IP addresses?<br> DHCP<br> DNS<br> WPA<br> WPS
attashe74 [19]

Answer:

DHCP

Explanation:

The only option can give IP addresses is the DHCP The Dynamic Host Configuration Protocol, because the DNS (Domain Name System) only translates IP addresses to a hostname, and WPS (Wi-Fi Protected Setup) or WPA (Wi-Fi Protected Access) if for security network, doesn't matter if is a wireless router, in a network always DHCP gives the IP addresses.

7 0
3 years ago
Write a Java program to accept an item's name and price from the user and output them to the console
Anit [1.1K]

Answer:

// program in java.

// package

import java.util.*;

// class definition

class Main

{// main method of the class

public static void main (String[] args) throws java.lang.Exception

{

   try{

    // scanner object to read inputs

Scanner scr=new Scanner(System.in);

 // variables

String name;

   double price;

   System.out.print("Enter item's name: ");

   // read item's name

   name = scr.next();

   System.out.print("Enter item's price: ");

   // read item's price

   price= scr.nextDouble();

   // print name

   System.out.println("name of item is:"+name);

   // print price

   System.out.println("price of item is:"+price);

   }catch(Exception ex){

       return;}

}

}

Explanation:

Read name & price of item from user and assign it to variables "name" & "price" respectively with scanner object.Then print the name and price of item .

Output:

Enter item's name: Apple                                                                                                  

Enter item's price: 100                                                                                                    

name of item is:Apple                                                                                                      

price of item is:100.0

6 0
3 years ago
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