I'm new to java/programming in general and this is a homework assignment. This is what I have so far: When I run it I get the powers of 2 below the n input. example if n = 50, output is 2 + 4 + 8 + 16 + 32 + = -2
I would like the + after 32 to be gone and I don't know how to properly sum it. I would want the sum to = 62 in this case. I tried using string builder to take off the last two characters but that isn't working for me.
import java.util.Scanner;
public class Powers {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n;
System.out.print("Enter the upper limit: ");
n = scan.nextInt();
int sum = 0;
int power = 1;
for (int i = 0; i <= n; i++) {
power = 2 * power;
if (power < n && 0 < power) {
System.out.print(power + " + ");
}
sum = sum + power;
}
System.out.println(" = " + sum);
}
}
There are multiple issues here:
When reaching the upper limit you simply stop doing the output but continue doing the summation.
You use the upper limit as the number of iterations, so in case of 50 in your example, you do a sum of all values between 1 and 2^50, which is the reason why the result is negative, because the sum became larger than the maximum number an int can keep.
Concerning your question how to break a loop, there is break ;-)
Your print is always outputting a + which is why you have the + = in your output. Change the output to something like this:
if (power < n && 0 < power) {
if (i != 0) {
System.out.print(" + ");
}
System.out.print(power);
}
I've added some functionality to your code.
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
System.out.println("Type number:");
Scanner scanner = new Scanner(System.in);
int n = 0;
while (n == 0) { // to ask again if "n" is zero.
n = scanner.nextInt();
}
if (n != 0) {
scanner.close(); // to prevent resource leak
int sum = 0;
int power = 1;
for (int i = 0; i < n; i++) {
power *= 2;
sum += power;
System.out.print(power + " ");
if (sum + power * 2 < 0 | i == n - 1) {
// Should we step to the next iteration?
// If next "sum" will be bigger than the max value for
// integers
// or if this iteration is the last - it will type "sum",
// break "for" cycle and go the next line of code after
// "for" cycle.
// So in this case System.out.print("+ "); won't be
// executed.
System.out.print("= " + sum);
break;
}
System.out.print("+ ");
}
}
}
}
Below is a Archive PROBLEM from SPOJ. Sample testCase is passing, but I am getting W/A on submission. I am missing some testCase(testCases). Need help to figure out what case I am missing and/or what I am doing wrong here.
Ada the Ladybug is playing Game of Divisors against her friend Velvet Mite Vinit. The game has following rules. There is a pile of N stones between them. The player who's on move can pick at least 1 an at most σ(N) stones (where σ(N) stands for number of divisors of N). Obviously, N changes after each move. The one who won't get any stones (N == 0) loses.
As Ada the Ladybug is a lady, so she moves first. Can you decide who will be the winner? Assume that both players play optimally.
Input
The first line of input will contain 1 ≤ T ≤ 10^5, the number of test-cases.
The next T lines will contain 1 ≤ N ≤ 2*10^7, the number of stones which are initially in pile.
Output
Output the name of winner, so either "Ada" or "Vinit".
Sample Input:
8
1
3
5
6
11
1000001
1000000
29
Sample Output:
Ada
Vinit
Ada
Ada
Vinit
Vinit
Ada
Ada
CODE
import java.io.*;
public class Main
{
public static int max_size = 2 * (int)Math.pow(10,7) + 1;
//public static int max_size = 25;
//public static int max_size = 2 * (int)Math.pow(10,6) + 1;
public static boolean[] dp = new boolean[max_size];
public static int[] lastPrimeDivisor = new int[max_size];
public static int[] numOfDivisors = new int[max_size];
public static void main(String[] args) throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
preprocess();
int t = Integer.parseInt(br.readLine());
while(t > 0)
{
int n = Integer.parseInt(br.readLine());
if(dp[n] == true)
System.out.println("Ada");
else
System.out.println("Vinit");
t--;
}
}
public static void markLastPrimeDivisor()
{
for(int i = 0 ; i < max_size ; i++)
{
lastPrimeDivisor[i] = 1;
}
for(int i = 2 ; i < max_size ; i += 2)
{
lastPrimeDivisor[i] = 2;
}
int o = (int)Math.sqrt(max_size);
for(int i = 3 ; i < max_size; i++)
{
if(lastPrimeDivisor[i] != 1)
{
continue;
}
lastPrimeDivisor[i] = i;
if(i <= o)
{
for(int j = i * i ; j < max_size ; j += 2 * i)
{
lastPrimeDivisor[j] = i;
}
}
}
/*for(int i = 1 ; i < max_size ; i++)
System.out.println("last prime of " + i + " is " + lastPrimeDivisor[i]);*/
}
public static void countDivisors(int num)
{
int original = num;
int result = 1;
int currDivisorCount = 1;
int currDivisor = lastPrimeDivisor[num];
int nextDivisor;
while(currDivisor != 1)
{
num = num / currDivisor;
nextDivisor = lastPrimeDivisor[num];
if(nextDivisor == currDivisor)
{
currDivisorCount++;
}
else
{
result = result * (currDivisorCount + 1);
currDivisorCount = 1;
currDivisor = nextDivisor;
}
}
if(num != 1)
{
result = result * (currDivisorCount + 1);
}
//System.out.println("result for num : " + original + ", " + result);
numOfDivisors[original] = result;
}
public static void countAllDivisors()
{
markLastPrimeDivisor();
for(int i = 2 ; i < max_size ; i++)
{
countDivisors(i);
//System.out.println("num of divisors of " + i + " = " + numOfDivisors[i]);
}
}
public static void preprocess()
{
countAllDivisors();
dp[0] = dp[1] = dp[2] = true;
for(int i = 3 ; i < max_size ; i++)
{
int flag = 0;
int limit = numOfDivisors[i];
//If for any i - j, we get false,for playing optimally
//the current opponent will choose to take j stones out of the
//pile as for i - j stones, the other player is not winning.
for(int j = 1 ; j <= limit; j++)
{
if(dp[i - j] == false)
{
dp[i] = true;
flag = 1;
break;
}
}
if(flag == 0)
dp[i] = false;
}
}
}
There is a subtle bug in your countDivisors() function. It assumes
that lastPrimeDivisor[num] – as the name indicates – returns the
largest prime factor of the given argument.
However, that is not the case. For example, lastPrimeDivisor[num] = 2
for all even numbers, or lastPrimeDivisor[7 * 89] = 7.
The reason is that in
public static void markLastPrimeDivisor()
{
// ...
for(int i = 3 ; i < max_size; i++)
{
// ...
if(i <= o)
{
for(int j = i * i ; j < max_size ; j += 2 * i)
{
lastPrimeDivisor[j] = i;
}
}
}
}
only array elements starting at i * i are updated.
So lastPrimeDivisor[num] is in fact some prime divisor of num, but not
necessarily the largest. As a consequence, numOfDivisors[55447] is computed
as 8 instead of the correct value 6.
Therefore in countDivisors(), the exponent of a prime factor in num
must be determined explicitly by repeated division.
Then you can use that the divisors function is multiplicative. This leads to
the following implementation:
public static void countAllDivisors() {
// Fill the `somePrimeDivisor` array:
computePrimeDivisors();
numOfDivisors[1] = 1;
for (int num = 2 ; num < max_size ; num++) {
int divisor = somePrimeDivisor[num];
if (divisor == num) {
// `num` is a prime
numOfDivisors[num] = 2;
} else {
int n = num / divisor;
int count = 1;
while (n % divisor == 0) {
count++;
n /= divisor;
}
// `divisor^count` contributes to `count + 1` in the number of divisors,
// now use multiplicative property:
numOfDivisors[num] = (count + 1) * numOfDivisors[n];
}
}
}
This program gets 4 input from the user and prints out all the pairs except for the duplicate pairs. I wanted it print only the pairs that sum equals a prime number. I did the entire program there is no compilation error but in the output it prints to all pairs except the duplicate pair(i want it to print only the pair whose sum= a prime number) can anyone tell me what i am doing wrong
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Prime {
public static List<Integer> numbers = new ArrayList<>(); // a list of integers that was accepted, can be acced in the order they were added
public static Scanner input = new Scanner(System.in);
public static void main(String args[])
{
int inputs = 4;
input(inputs);
for(int i = 0; i < inputs; i++){
for(int j = 0+i; j < inputs; j++){
if(i != j)
System.out.println(numbers.get(i)+ " + " + numbers.get(j) + " = "+ isPrime(numbers.get(i) + numbers.get(j)));
}
} }
public static int isPrime (int sumPair)
{
boolean primeVal =true ;
if(sumPair < 2)
primeVal= false;
else if(sumPair ==2)
primeVal=true;
else if (sumPair % 2 == 0)
primeVal = false;
else
{
int stop = (int)Math.sqrt(sumPair);
for (int d = 3; d <= stop && primeVal; d = d + 2)
{
if (sumPair % d == 0)
primeVal = false;
}
}
return(sumPair);
}
public static void input(int inputNumbers)
{
while(numbers.size() < inputNumbers){ // while there is more inputs needed
System.out.print("Enter a positive integer: ");
int num = input.nextInt();
if(num > 0) // if the input is valid
numbers.add(num);
else // if the input is not valid
while (num <= 0) { // while the input is not valid
System.out.print("Enter a positive integer: ");
num = input.nextInt(); // get the next int if it wasn't valid
if(num > 0) { // if the input gets valid
numbers.add(num);
}
}
System.out.println("Thank you.");
}
}
public static int sumPair(int num1, int num2)
{
return num1 + num2;
}
}
You go to a lot of trouble to compute a boolean primeVal which tells whether the input be true or false, but you never actually return this value. Try this instead:
public static boolean isPrime (int sumPair) {
boolean primeVal = true;
if (sumPair < 2 || sumPair % 2 == 0) {
primeVal = false;
}
else {
int stop = (int)Math.sqrt(sumPair);
for (int d=3; d <= stop; d += 2) {
if (sumPair % d == 0) {
primeVal = false;
break;
}
}
}
return primeVal;
}
Use this main() method:
public static void main (String args[]) {
int inputs = 4;
input(inputs);
for (int i=0; i < inputs-1; i++) {
for (int j=i+1; j < inputs; j++) {
boolean prime = isPrime(numbers.get(i) + numbers.get(j));
if (prime) {
System.out.println(numbers.get(i) + " + " + numbers.get(j));
}
}
}
}
Your condition to determine if you print the current pair or not is if(i != j), so it will only print if i is different from j (duplicate).
You made a nice isPrime function, you should call it.
You made your isPrime function return an int, and I'm pretty sure it should return a boolean instead. Also, it returns the argument you gave it, so it does virtually nothing except spend time in a potentially expensive operation. Perhaps should it return primeVal instead.
Your code should look like:
for(int i = 0; i < inputs; i++){
for(int j = 0+i; j < inputs; j++){
ni = numbers.get(i);
nj = numbers.get(j);
if(isPrime(ni+nj)){
System.out.println(ni + " + " + nj + " = "+ (ni + nj));
}
}
}
What this does is get the numbers for indexes i and j and store them in ni and nj respectively. Then, it checks it their sum is prime. If it is, then it prints the sum.
My assignment reads:
Write a program that finds all of the prime numbers between one and some number that you allow the user to input.
Find the prime numbers between 0 and the number the user inputs.
Print these prime numbers, one per line.
After each prime number, add a colon and then print out all the non-prime numbers that were eliminated.
Basically, these non-prime numbers are the multiples of the primes.
For example, the output will look like this:
2: 4 6 8 10 12 14 16 18...
3: 9 15 21 27
I did prime numbers. I can not figure out how to calculate and display multiples? Help, please!
package assignment4;
import java.util.Scanner;
public class Assignment4 { /**
* #param args the command line arguments
*/
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
final int START = 1;
System.out.print("Enter the end number number : ");
int end = s.nextInt();
System.out.println("List of prime numbers between " + START + " and "
+ end);
for (int i = START; i <= end; i++) {
if (isPrime(i)) {
System.out.println(i + ":");
}
}
}
public static boolean isPrime(int n) {
if (n <= 1) {
return false;
}
for (int i = 2; i < Math.sqrt(n); i++) {
if (n % i == 0) {
return false;
}
}
return true;}
}
I am assuming that you want to have print the multiples until the end number. You could nest a while loop that multiplies the prime number as follows:
for (int i = START; i <= end; i++) {
if (isPrime(i)) {
System.out.print(i + ": "); //Changed to print so that the multiples are on the same line
int multiple = i * 2;
while (multiple <= end) {
multiple += i;
System.out.print(multiple + ", ");
}
System.out.println("");
}
}
Displaying multiples would be easy if you were sieving using sieve of Erathrosthenes. You create an array of numbers from 2 to n. start with 2, mark non-prime all its multiples: 2 4 6... print them. Choose next prime number (i.e. the number which was not marked unprime) i.e. 3, mark non-prime all its multiples (which haven't been already): 9 15 ... and so on.
Simplest Solution:
boolean[] x = new boolean[N];// x[i] is true when i is not prime
for(int i=2;i<N;i++){
if(!x[i]){
System.out.print(i+" : ");
for(int j=i*i;j<N;j+=i){
if(!x[j])
System.out.print(j+" ");
x[j]=true;
}
System.out.println();
}
}
Please check below answer for your requirement,
package assignment4;
import java.util.*;
import java.lang.*;
import java.io.*;
public class Assignment4
{
public static List<Integer> primeNumbers = new ArrayList<>();
public static String strOtherData[];
public static void main (String[] args) throws java.lang.Exception
{
final int START = 1;
System.out.print("Enter the end number number : ");
int end = s.nextInt();
System.out.println("List of prime numbers between " + START + " and " + end);
for (int i = START; i <= end; i++) {
if (isPrime(i)) {
primeNumbers.add(i);
}
}
strOtherData = new String[primeNumbers.size()];
for (int i = START; i <= end; i++) {
isMultiple(i);
}
int tempCount = 0;
for(Integer currentNumber : primeNumbers)
{
System.out.print(currentNumber + ": \t");
if(strOtherData.length > tempCount)
{
System.out.print(strOtherData[tempCount] + "\n\n");
}
tempCount++;
}
}
public static boolean isPrime(int n) {
if (n <= 1) {
return false;
}
for (int i = 2; i <= n/2; i++)
{
if (n % i == 0)
{
return false;
}
}
return true;
}
public static void isMultiple(int n)
{
if (n <= 1) {
return;
}
if(isPrime(n))
return;
int count = 0;
for(Integer currentInt : primeNumbers)
{
if(strOtherData[count] != null)
{
strOtherData[count] += "," + n;
}
else
{
strOtherData[count] = "" + n;
}
count++;
}
}
}
I tried to check the validation of credit card using Luhn algorithm, which works as the following steps:
Double every second digit from right to left. If doubling of a digit results in a two-digit number, add up the two digits to get a single-digit number.
2 * 2 = 4
2 * 2 = 4
4 * 2 = 8
1 * 2 = 2
6 * 2 = 12 (1 + 2 = 3)
5 * 2 = 10 (1 + 0 = 1)
8 * 2 = 16 (1 + 6 = 7)
4 * 2 = 8
Now add all single-digit numbers from Step 1.
4 + 4 + 8 + 2 + 3 + 1 + 7 + 8 = 37
Add all digits in the odd places from right to left in the card number.
6 + 6 + 0 + 8 + 0 + 7 + 8 + 3 = 38
Sum the results from Step 2 and Step 3.
37 + 38 = 75
If the result from Step 4 is divisible by 10, the card number is valid; otherwise, it is invalid. For example, the number 4388576018402626 is invalid, but the number 4388576018410707 is valid.
Simply, my program always displays valid for everything that I input. Even if it's a valid number and the result of sumOfOddPlace and sumOfDoubleEvenPlace methods are equal to zero. Any help is appreciated.
import java.util.Scanner;
public class CreditCardValidation {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int count = 0;
long array[] = new long [16];
do
{
count = 0;
array = new long [16];
System.out.print("Enter your Credit Card Number : ");
long number = in.nextLong();
for (int i = 0; number != 0; i++) {
array[i] = number % 10;
number = number / 10;
count++;
}
}
while(count < 13);
if ((array[count - 1] == 4) || (array[count - 1] == 5) || (array[count - 1] == 3 && array[count - 2] == 7)){
if (isValid(array) == true) {
System.out.println("\n The Credit Card Number is Valid. ");
} else {
System.out.println("\n The Credit Card Number is Invalid. ");
}
} else{
System.out.println("\n The Credit Card Number is Invalid. ");
}
}
public static boolean isValid(long[] array) {
int total = sumOfDoubleEvenPlace(array) + sumOfOddPlace(array);
if ((total % 10 == 0)) {
for (int i=0; i< array.length; i++){
System.out.println(array[i]);}
return true;
} else {
for (int i=0; i< array.length; i++){
System.out.println(array[i]);}
return false;
}
}
public static int getDigit(int number) {
if (number <= 9) {
return number;
} else {
int firstDigit = number % 10;
int secondDigit = (int) (number / 10);
return firstDigit + secondDigit;
}
}
public static int sumOfOddPlace(long[] array) {
int result = 0;
for (int i=0; i< array.length; i++)
{
while (array[i] > 0) {
result += (int) (array[i] % 10);
array[i] = array[i] / 100;
}}
System.out.println("\n The sum of odd place is " + result);
return result;
}
public static int sumOfDoubleEvenPlace(long[] array) {
int result = 0;
long temp = 0;
for (int i=0; i< array.length; i++){
while (array[i] > 0) {
temp = array[i] % 100;
result += getDigit((int) (temp / 10) * 2);
array[i] = array[i] / 100;
}
}
System.out.println("\n The sum of double even place is " + result);
return result;
}
}
You can freely import the following code:
public class Luhn
{
public static boolean Check(String ccNumber)
{
int sum = 0;
boolean alternate = false;
for (int i = ccNumber.length() - 1; i >= 0; i--)
{
int n = Integer.parseInt(ccNumber.substring(i, i + 1));
if (alternate)
{
n *= 2;
if (n > 9)
{
n = (n % 10) + 1;
}
}
sum += n;
alternate = !alternate;
}
return (sum % 10 == 0);
}
}
Link reference: https://github.com/jduke32/gnuc-credit-card-checker/blob/master/CCCheckerPro/src/com/gnuc/java/ccc/Luhn.java
Google and Wikipedia are your friends. Instead of long-array I would use int-array. On Wikipedia following java code is published (together with detailed explanation of Luhn algorithm):
public static boolean check(int[] digits) {
int sum = 0;
int length = digits.length;
for (int i = 0; i < length; i++) {
// get digits in reverse order
int digit = digits[length - i - 1];
// every 2nd number multiply with 2
if (i % 2 == 1) {
digit *= 2;
}
sum += digit > 9 ? digit - 9 : digit;
}
return sum % 10 == 0;
}
You should work on your input processing code. I suggest you to study following solution:
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
boolean repeat;
List<Integer> digits = new ArrayList<Integer>();
do {
repeat = false;
System.out.print("Enter your Credit Card Number : ");
String input = in.next();
for (int i = 0; i < input.length(); i++) {
char c = input.charAt(i);
if (c < '0' || c > '9') {
repeat = true;
digits.clear();
break;
} else {
digits.add(Integer.valueOf(c - '0'));
}
}
} while (repeat);
int[] array = new int[digits.size()];
for (int i = 0; i < array.length; i++) {
array[i] = Integer.valueOf(digits.get(i));
}
boolean valid = check(array);
System.out.println("Valid: " + valid);
}
I took a stab at this with Java 8:
public static boolean luhn(String cc) {
final boolean[] dbl = {false};
return cc
.chars()
.map(c -> Character.digit((char) c, 10))
.map(i -> ((dbl[0] = !dbl[0])) ? (((i*2)>9) ? (i*2)-9 : i*2) : i)
.sum() % 10 == 0;
}
Add the line
.replaceAll("\\s+", "")
Before
.chars()
If you want to handle whitespace.
Seems to produce identical results to
return LuhnCheckDigit.LUHN_CHECK_DIGIT.isValid(cc);
From Apache's commons-validator.
There are two ways to split up your int into List<Integer>
Use %10 as you are using and store it into a List
Convert to a String and then take the numeric values
Here are a couple of quick examples
public static void main(String[] args) throws Exception {
final int num = 12345;
final List<Integer> nums1 = splitInt(num);
final List<Integer> nums2 = splitString(num);
System.out.println(nums1);
System.out.println(nums2);
}
private static List<Integer> splitInt(int num) {
final List<Integer> ints = new ArrayList<>();
while (num > 0) {
ints.add(0, num % 10);
num /= 10;
}
return ints;
}
private static List<Integer> splitString(int num) {
final List<Integer> ints = new ArrayList<>();
for (final char c : Integer.toString(num).toCharArray()) {
ints.add(Character.getNumericValue(c));
}
return ints;
}
I'll use 5 digit card numbers for simplicity. Let's say your card number is 12345; if I read the code correctly, you store in array the individual digits:
array[] = {1, 2, 3, 4, 5}
Since you already have the digits, in sumOfOddPlace you should do something like
public static int sumOfOddPlace(long[] array) {
int result = 0;
for (int i = 1; i < array.length; i += 2) {
result += array[i];
}
return result;
}
And in sumOfDoubleEvenPlace:
public static int sumOfDoubleEvenPlace(long[] array) {
int result = 0;
for (int i = 0; i < array.length; i += 2) {
result += getDigit(2 * array[i]);
}
return result;
}
this is the luhn algorithm implementation which I use for only 16 digit Credit Card Number
if(ccnum.length()==16){
char[] c = ccnum.toCharArray();
int[] cint = new int[16];
for(int i=0;i<16;i++){
if(i%2==1){
cint[i] = Integer.parseInt(String.valueOf(c[i]))*2;
if(cint[i] >9)
cint[i]=1+cint[i]%10;
}
else
cint[i] = Integer.parseInt(String.valueOf(c[i]));
}
int sum=0;
for(int i=0;i<16;i++){
sum+=cint[i];
}
if(sum%10==0)
result.setText("Card is Valid");
else
result.setText("Card is Invalid");
}else
result.setText("Card is Invalid");
If you want to make it use on any number replace all 16 with your input number length.
It will work for Visa number given in the question.(I tested it)
Here's my implementation of the Luhn Formula.
/**
* Runs the Luhn Equation on a user inputed CCN, which in turn
* determines if it is a valid card number.
* #param c A user inputed CCN.
* #param cn The check number for the card.
* #return If the card is valid based on the Luhn Equation.
*/
public boolean luhn (String c, char cn)
{
String card = c;
String checkString = "" + cn;
int check = Integer.valueOf(checkString);
//Drop the last digit.
card = card.substring(0, ( card.length() - 1 ) );
//Reverse the digits.
String cardrev = new StringBuilder(card).reverse().toString();
//Store it in an int array.
char[] cardArray = cardrev.toCharArray();
int[] cardWorking = new int[cardArray.length];
int addedNumbers = 0;
for (int i = 0; i < cardArray.length; i++)
{
cardWorking[i] = Character.getNumericValue( cardArray[i] );
}
//Double odd positioned digits (which are really even in our case, since index starts at 0).
for (int j = 0; j < cardWorking.length; j++)
{
if ( (j % 2) == 0)
{
cardWorking[j] = cardWorking[j] * 2;
}
}
//Subtract 9 from digits larger than 9.
for (int k = 0; k < cardWorking.length; k++)
{
if (cardWorking[k] > 9)
{
cardWorking[k] = cardWorking[k] - 9;
}
}
//Add all the numbers together.
for (int l = 0; l < cardWorking.length; l++)
{
addedNumbers += cardWorking[l];
}
//Finally, check if the number we got from adding all the other numbers
//when divided by ten has a remainder equal to the check number.
if (addedNumbers % 10 == check)
{
return true;
}
else
{
return false;
}
}
I pass in the card as c which I get from a Scanner and store in card, and for cn I pass in checkNumber = card.charAt( (card.length() - 1) );.
Okay, this can be solved with a type conversions to string and some Java 8
stuff. Don't forget numbers and the characters representing numbers are not the same. '1' != 1
public static int[] longToIntArray(long cardNumber){
return Long.toString(cardNumber).chars()
.map(x -> x - '0') //converts char to int
.toArray(); //converts to int array
}
You can now use this method to perform the luhn algorithm:
public static int luhnCardValidator(int cardNumbers[]) {
int sum = 0, nxtDigit;
for (int i = 0; i<cardNumbers.length; i++) {
if (i % 2 == 0)
nxtDigit = (nxtDigit > 4) ? (nxtDigit * 2 - 10) + 1 : nxtDigit * 2;
sum += nxtDigit;
}
return (sum % 10);
}
private static int luhnAlgorithm(String number){
int n=0;
for(int i = 0; i<number.length(); i++){
int x = Integer.parseInt(""+number.charAt(i));
n += (x*Math.pow(2, i%2))%10;
if (x>=5 && i%2==1) n++;
}
return n%10;
}
public class Creditcard {
public static void main(String args[]){
Scanner sc=new Scanner(System.in);
String cardno = sc.nextLine();
if(checkType(cardno).equals("U")) //checking for unknown type
System.out.println("UNKNOWN");
else
checkValid(cardno); //validation
}
private static String checkType(String S)
{
int AM=Integer.parseInt(S.substring(0,2));
int D=Integer.parseInt(S.substring(0,4)),d=0;
for(int i=S.length()-1;i>=0;i--)
{
if(S.charAt(i)==' ')
continue;
else
d++;
}
if((AM==34 || AM==37) && d==15)
System.out.println("AMEX");
else if(D==6011 && d==16)
System.out.println("Discover");
else if(AM>=51 && AM<=55 && d==16)
System.out.println("MasterCard");
else if(((S.charAt(0)-'0')==4)&&(d==13 || d==16))
System.out.println("Visa");
else
return "U";
return "";
}
private static void checkValid(String S) // S--> cardno
{
int i,d=0,sum=0,card[]=new int[S.length()];
for(i=S.length()-1;i>=0;i--)
{
if(S.charAt(i)==' ')
continue;
else
card[d++]=S.charAt(i)-'0';
}
for(i=0;i<d;i++)
{
if(i%2!=0)
{
card[i]=card[i]*2;
if(card[i]>9)
sum+=digSum(card[i]);
else
sum+=card[i];
}
else
sum+=card[i];
}
if(sum%10==0)
System.out.println("Valid");
else
System.out.println("Invalid");
}
public static int digSum(int n)
{
int sum=0;
while(n>0)
{
sum+=n%10;
n/=10;
}
return sum;
}
}
Here is the implementation of Luhn algorithm.
public class LuhnAlgorithm {
/**
* Returns true if given card number is valid
*
* #param cardNum Card number
* #return true if card number is valid else false
*/
private static boolean checkLuhn(String cardNum) {
int cardlength = cardNum.length();
int evenSum = 0, oddSum = 0, sum;
for (int i = cardlength - 1; i >= 0; i--) {
System.out.println(cardNum.charAt(i));
int digit = Character.getNumericValue(cardNum.charAt(i));
if (i % 2 == 0) {
int multiplyByTwo = digit * 2;
if (multiplyByTwo > 9) {
/* Add two digits to handle cases that make two digits after doubling */
String mul = String.valueOf(multiplyByTwo);
multiplyByTwo = Character.getNumericValue(mul.charAt(0)) + Character.getNumericValue(mul.charAt(1));
}
evenSum += multiplyByTwo;
} else {
oddSum += digit;
}
}
sum = evenSum + oddSum;
if (sum % 10 == 0) {
System.out.println("valid card");
return true;
} else {
System.out.println("invalid card");
return false;
}
}
public static void main(String[] args) {
String cardNum = "4071690065031703";
System.out.println(checkLuhn(cardNum));
}
}
public class LuhnAlgorithm {
/**
* Returns true if given card number is valid
*
* #param cardNum Card number
* #return true if card number is valid else false
*/
private static boolean checkLuhn(String cardNum) {
int cardlength = cardNum.length();
int evenSum = 0, oddSum = 0, sum;
for (int i = cardlength - 1; i >= 0; i--) {
System.out.println(cardNum.charAt(i));
int digit = Character.getNumericValue(cardNum.charAt(i));
if (i % 2 == 0) {
int multiplyByTwo = digit * 2;
if (multiplyByTwo > 9) {
/* Add two digits to handle cases that make two digits after doubling */
String mul = String.valueOf(multiplyByTwo);
multiplyByTwo = Character.getNumericValue(mul.charAt(0)) + Character.getNumericValue(mul.charAt(1));
}
evenSum += multiplyByTwo;
} else {
oddSum += digit;
}
}
sum = evenSum + oddSum;
if (sum % 10 == 0) {
System.out.println("valid card");
return true;
} else {
System.out.println("invalid card");
return false;
}
}
public static void main(String[] args) {
String cardNum = "8112189875";
System.out.println(checkLuhn(cardNum));
}
}
Hope it may works.
const options = {
method: 'GET',
headers: {Accept: 'application/json', 'X-Api-Key': '[APIkey]'}
};
fetch('https://api.epaytools.com/Tools/luhn?number=[CardNumber]&metaData=true', options)
.then(response => response.json())
.then(response => console.log(response))
.catch(err => console.error(err));