Please answer 1 and 2 code is provided (I need a flowchart and analysis) of my code I have provided.

1. Analysis:

Input:

Process:

Output:

2. Design: (flow chart)

Coding: copy your Java source program here

/*

* C201

*/

/**

*

* @author __

*/

import java.util.Scanner;

public class HW06A

{

public static void main(String[] args)

{

Scanner input = new Scanner(System.in); // Create a Scanner

// Prompt the user to enter a credit card number as a long integer

System.out.print("Enter a credit card number as a long integer: ");

long number = input.nextLong();

System.out.println(number + " is " + (isValid(number) ? "valid" : "invalid"));

}

// Return true if the card number is valid

public static boolean isValid(long number)

{

boolean valid =

(getSize(number) >= 13 && getSize(number) <= 16) &&

(prefixMatched(number, 4) || prefixMatched(number, 5) ||

prefixMatched(number, 37) || prefixMatched(number, 6)) &&

((sumOfDoubleEvenPlace(number) + sumOfOddPlace(number)) % 10 == 0);

return valid;

}

// Get the result from Step 2

public static int sumOfDoubleEvenPlace(long number)

{

int sum = 0;

String num = number + "";

for (int i = getSize(number) - 2; i >= 0; i -= 2) {

sum += getDigit(Integer.parseInt(num.charAt(i) + "") * 2);

}

return sum;

}

// Return this number if it is a single digit, otherwise,

//return the sum of the two digits

public static int getDigit(int number)

{

if (number < 9)

return number;

else

return number / 10 + number % 10;

}

// Return sum of odd-place digits in number

public static int sumOfOddPlace(long number)

{

int sum = 0;

String num = number + "";

for (int i = getSize(number) - 1; i >= 0; i -= 2)

{

sum += Integer.parseInt(num.charAt(i) + "");

}

return sum;

}

// Return true if the digit d is a prefix for number

public static boolean prefixMatched(long number, int d)

{

return getPrefix(number, getSize(d)) == d;

}

// Return the number of digits in d

public static int getSize(long d)

{

String num = d + "";

return num.length();

}

// Return the first k number of digits from number. If thenumber of digits in number is less than k, return number.

public static long getPrefix(long number, int k)

{

if (getSize(number) > k)

{

String num = number + "";

return Long.parseLong(num.substring(0, k));

}

return number;

}

}