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padilas [110]
3 years ago
6

Write a "for loop" that displays a string in the reverse order (DO NOT use the reverse method) 3.

Computers and Technology
1 answer:
elena55 [62]3 years ago
5 0

Following are the code in java to reverse any string without using reverse function.

import java.util.*; // import package

class Main // class

{

public static void main(String args[])  // main class in java

{

String s, r = "";  // string variable

Scanner out = new Scanner(System.in);  // scanner classes to take input

System.out.println("Enter a string to reverse");

s = out.nextLine();  // input string

int l1 = s.length();  // finding length of string

l1=l1-1;

for ( int i = l1 ; i >= 0 ; i-- )  // for loop to reverse any string

{

r = r + s.charAt(i);

}

System.out.println(" The Reverse String is: "+r);  // print reverse string

  }

}

Explanation:

firstly we input any string ,finding the length of that string then after that iterating over the loop by using for loop and last display that reverse string

output

Enter a string to reverse

san ran

The Reverse String is: nar nas

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A profit of ₹ 1581 is to be divided amongst three partner P,Q and R in ratio 1/3:1/2:1/5.theshareof R will be?​
Burka [1]

Answer:

R's share = 306

Explanation:

Sum of ratio = 1/3 + 1/2 + 1/5

LCM = 30

Sum of ratio = (10 + 15 + 6) / 30 = 31 /30

Profit to be shared = 1581

P = 1/3 ; Q = 1/2 ; R = 1/5

Share of R :

(Total profit / sum of ratio) * R ratio

[1581 / (31/30)] * 1/5

(1581 / 1 * 30/31) * 1/5

51 * 30 * 1/5

51 * 6 = 306

5 0
3 years ago
Implemente a função ao lado, que recebe um preço e um booleano indicando se já está com desconto ou não. Se o preço for maior qu
Andrew [12]

Answer:

function pecoDesconto(preco, estaComDesconto) {

 

 let p = preco;

 let desconto = estaComDesconto;

 if(p > 100 && desconto == false){

   return "Quero pechinchar";

 }else{

   return "Negócio fechado";

 }

}

Explanation:

function pecoDesconto(preco, estaComDesconto) {

 

 // Implemente a função ao lado, que recebe um preço//

 let p = preco;

// variavel que indica desconco//

 let desconto = estaComDesconto;

//Se o preço for maior que 100 e não estiver com desconto, a função deve retornar Quero pechinchar.//

 if(p > 100 && desconto == false){

   return "Quero pechinchar";

   //Caso contrário, deve retornar Negócio fechado

 }else{

   return "Negócio fechado"

 }

}

só te faltou ler com atenção, e um pouco de logica!

7 0
2 years ago
It takes 2 seconds to read or write one block from/to disk and it also takes 1 second of CPU time to merge one block of records.
Alexxx [7]

Answer:

Part a: For optimal 4-way merging, initiate with one dummy run of size 0 and merge this with the 3 smallest runs. Than merge the result to the remaining 3 runs to get a merged run of length 6000 records.

Part b: The optimal 4-way  merging takes about 249 seconds.

Explanation:

The complete question is missing while searching for the question online, a similar question is found which is solved as below:

Part a

<em>For optimal 4-way merging, we need one dummy run with size 0.</em>

  1. Merge 4 runs with size 0, 500, 800, and 1000 to produce a run with a run length of 2300. The new run length is calculated as follows L_{mrg}=L_0+L_1+L_2+L_3=0+500+800+1000=2300
  2. Merge the run as made in step 1 with the remaining 3 runs bearing length 1000, 1200, 1500. The merged run length is 6000 and is calculated as follows

       L_{merged}=L_{mrg}+L_4+L_5+L_6=2300+1000+1200+1500=6000

<em>The resulting run has length 6000 records</em>.

Part b

<u><em>For step 1</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
  • Time_{I/O per block} is 2 sec

So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{2300}{100} \times 2 sec\\T_{I.O}=46 sec

So the input/output time is 46 seconds for step 01.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{2300}{100} \times 1 sec\\T_{CPU}=23 sec

So the CPU  time is 23 seconds for step 01.

Total time in step 01

T_{step-01}=T_{I.O}+T_{CPU}\\T_{step-01}=46+23\\T_{step-01}=69 sec\\

Total time in step 01 is 69 seconds.

<u><em>For step 2</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 6000 for step 02
  • Size_block is 100 as given
  • Time_{I/O per block} is 2 sec

So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{6000}{100} \times 2 sec\\T_{I.O}=120 sec

So the input/output time is 120 seconds for step 02.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 6000 for step 02
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{6000}{100} \times 1 sec\\T_{CPU}=60 sec

So the CPU  time is 60 seconds for step 02.

Total time in step 02

T_{step-02}=T_{I.O}+T_{CPU}\\T_{step-02}=120+60\\T_{step-02}=180 sec\\

Total time in step 02 is 180 seconds

Merging Time (Total)

<em>Now  the total time for merging is given as </em>

T_{merge}=T_{step-01}+T_{step-02}\\T_{merge}=69+180\\T_{merge}=249 sec\\

Total time in merging is 249 seconds seconds

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