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frez [133]
3 years ago
15

A coffee shop is considering accepting orders and payments through their phone app and have decided to use public key encryption

to encrypt their customers' credit card information. Is this a secure form of payment?
Yes, public key encryption is built upon computationally hard problems that even powerful computers cannot easily solve.
No, encryption is an issue in that case
Computers and Technology
1 answer:
Charra [1.4K]3 years ago
8 0

Answer: Yes

Explanation:

Public key encryption is the encryption technique that is used for private keys and public keys for securing the system.Public key is used for encryption and private key is for decryption.Public keys can only open content of the system

  • According to the question, public key encryption is secure for coffee shop customer payment process as they are stored on digital certificates in long form for verifying digital signature and encrypting information.Its computation is difficult to crack through power computer access.  
  • Other options is incorrect as encryption is not a problem for payment procedures. Thus, the correct option is yes ,public key encryption is secure method for coffee shop customers .
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/* package whatever; // don't place package name! */

import java.util.*;

import java.lang.*;

import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */

public class Ideone

{

  public static final int STALL_COUNT = 10;

 

public static int find_stall(boolean[] stalls) {

int longest_count = -1;

int longest_length = 0;

int current_count = -1;

int current_length = 0;

boolean inRun = false;

for (int i = 0; i < stalls.length; i++) {

if (inRun && stalls[i]) {

inRun = false;

if (current_length >= longest_length) {

longest_length = current_length;

longest_count = current_count;

}

 

}

else if (!inRun && !stalls[i]) {

inRun = true;

current_count = i;

current_length = 1;

}

else if (inRun && !stalls[i]) {

current_length += 1;

}

}

if (inRun) {

if (current_length >= longest_length) {

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return (longest_length - 1) / 2 + longest_count;

}

 

public static void print_pattern(boolean[] stalls) {

for (int i = 0; i < stalls.length; i++) {

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System.out.print("X ");

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else {

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}

System.out.println();

}

 

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

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for (int i = 0; i < stalls.length; i++) {

stalls[find_stall(stalls)] = true;

print_pattern(stalls);

  }

}

}

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4 years ago
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