Binary to Decimal Java converter - java

I am creating a code that allows you to convert a binary number to a decimal number and vice versa. I have created a code that converts decimal to binary but can not workout how to implement the binary to decimal aspect.
My code for decimal to binary is below:
import java.util.*;
public class decimalToBinaryTest
{
public static void main (String [] args)
{
int n;
Scanner in = new Scanner(System.in);
System.out.println("Enter a positive interger");
n=in.nextInt();
if(n < 0)
{
System.out.println("Not a positive interger");
}
else
{
System.out.print("Convert to binary is: ");
binaryform(n);
}
}
private static Object binaryform(int number)
{
int remainder;
if(number <= 1)
{
System.out.print(number);
return " ";
}
remainder= number % 2;
binaryform(number >> 1);
System.out.print(remainder);
{
return " ";
}
}
}
An explanation to how the binary to decimal code work would help as well.
I have tried the method of the least significant digit*1 then the next least *1*2 then *1*2*2 but can not get it to work.
Thank you #korhner I used your number system with arrays and if statements.
This is my working code:
import java.util.*;
public class binaryToDecimalConvertor
{
public static void main (String [] args)
{
int [] positionNumsArr= {1,2,4,8,16,32,64,128};
int[] numberSplit = new int [8];
Scanner scanNum = new Scanner(System.in);
int count1=0;
int decimalValue=0;
System.out.println("Please enter a positive binary number.(Only 1s and 0s)");
int number = scanNum.nextInt();
while (number > 0)
{
numberSplit[count1]=( number % 10);
if(numberSplit[count1]!=1 && numberSplit[count1] !=0)
{
System.out.println("Was not made of only \"1\" or \"0\" The program will now restart");
main(null);
}
count1++;
number = number / 10;
}
for(int count2 = 0;count2<8;count2++)
{
if(numberSplit[count2]==1)
{
decimalValue=decimalValue+positionNumsArr[count2];
}
}
System.out.print(decimalValue);
}
}


sample:
00000100
0 - 1
0 - 2
1 - 4
0 - 8
0 - 16
0 - 32
0 - 64
0 - 128
Sum values with bit 1 = 4
Good luck!


int decimal = Integer.parseInt("101101101010111", 2);
or if you prefer to doit your self
double output=0;
for(int i=0;i<str.length();i++){
if(str.charAt(i)== '1')
output=output + Math.pow(2,str.length()-1-i);
}


Here is a program which does that.
Make sure the integers you give to int and not too large.
import java.util.Scanner;
public class DecimalBinaryProgram {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while (true){
System.out.println("Enter integer in decimal form (or # to quit):");
String s1 = in.nextLine();
if ("#".equalsIgnoreCase(s1.trim())){
break;
}
System.out.println(decimalToBinary(s1));
System.out.println("Enter integer in binary form (or # to quit):");
String s2 = in.nextLine();
if ("#".equalsIgnoreCase(s2.trim())){
break;
}
System.out.println(binaryToDecimal(s2));
}
}
private static String decimalToBinary(String s){
int n = Integer.parseInt(s, 10);
StringBuilder sb = new StringBuilder();
if (n==0) return "0";
int d = 0;
while (n > 0){
d = n % 2;
n /= 2;
sb.append(d);
}
sb = sb.reverse();
return sb.toString();
}
private static String binaryToDecimal(String s){
int degree = 1;
int n = 0;
for (int k=s.length()-1; k>=0; k--){
n += degree * (s.charAt(k) - '0');
degree *= 2;
}
return n + "";
}
}
Of course for this method binaryToDecimal you can just do:
private static String binaryToDecimal(String s){
int n = Integer.parseInt(s, 2);
return n + "";
}
but I wanted to illustrate how you can do that explicitly.


do you want this?
private double dec(String s, int i) {
if (s.length() == 1) return s.equals("1") ? Math.pow(2, i) : 0;
else return (s.equals("1") ? Math.pow(2, i) : 0) + dec(s.substring(0, s.length() - 1), i - 1);
}
dec("101011101",0);


This is a version of a binary to decimal converter. I have used plenty of comments also. Just taught I would like to share it. Hope it is of some use to somebody.
import java.util.Scanner;
public class BinaryToDecimal
{
public static void main(String args[])
{
Scanner input = new Scanner(System.in);
System.out.println("Enter a binary number: ");
String binary = input.nextLine(); // store input from user
int[] powers = new int[16]; // contains powers of 2
int powersIndex = 0; // keep track of the index
int decimal = 0; // will contain decimals
boolean isCorrect = true; // flag if incorrect input
// populate the powers array with powers of 2
for(int i = 0; i < powers.length; i++)
powers[i] = (int) Math.pow(2, i);
for(int i = binary.length() - 1; i >= 0; i--)
{
// if 1 add to decimal to calculate
if(binary.charAt(i) == '1')
decimal = decimal + powers[powersIndex]; // calc the decimal
else if(binary.charAt(i) != '0' & binary.charAt(i) != '1')
{
isCorrect = false; // flag the wrong input
break; // break from loop due to wrong input
} // else if
// keeps track of which power we are on
powersIndex++; // counts from zero up to combat the loop counting down to zero
} // for
if(isCorrect) // print decimal output
System.out.println(binary + " converted to base 10 is: " + decimal);
else // print incorrect input message
System.out.println("Wrong input! It is binary... 0 and 1's like.....!");
} // main
} // BinaryToDecimal


I've written a converter that accepts both strings and ints.
public class Main {
public static void main(String[] args) {
int binInt = 10110111;
String binString = "10110111";
BinaryConverter convertedInt = new BinaryConverter(binInt);
BinaryConverter convertedString = new BinaryConverter(binString);
System.out.println("Binary as an int, to decimal: " + convertedInt.getDecimal());
System.out.println("Binary as a string, to decimal: " + convertedString.getDecimal());
}
}
public class BinaryConverter {
private final int base = 2;
private int binaryInt;
private String binaryString;
private int convertedBinaryInt;
public BinaryConverter(int b) {
binaryInt = b;
convertedBinaryInt = Integer.parseInt(Integer.toString(binaryInt), base);
}
public BinaryConverter(String s) {
binaryString = s;
convertedBinaryInt = Integer.parseInt(binaryString, base);
}
public int getDecimal() {
return convertedBinaryInt;
}
}


public static void main(String[] args)
{
System.out.print("Enter a binary number: ");
Scanner input = new Scanner(System.in);
long num = input.nextLong();
long reverseNum = 0;
int decimal = 0;
int i = 0;
while (num != 0)
{
reverseNum = reverseNum * 10;
reverseNum = num % 10;
decimal = (int) (reverseNum * Math.pow(2, i)) + decimal;
num = num / 10;
i++;
}
System.out.println(decimal);
}

Related

How to find factorial and show result of counting in console?

public class Car {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
System.out.println(n+"!="+factorial(n));
}
public static int factorial(int num) {
return (num == 0) ? 1 : num * factorial (num - 1);
}
}
how make this code to text in console 3! = 1*2*3 = 6?

Don't use recursion for this. Besides, it isn't really efficient or necessary.
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int fact = 1;
String s = n + "! = 1";
for (int i = 2; i <= n; i++) {
fact *= i;
s += "*" + i;
}
s += " = ";
System.out.println(s + fact);

There can be many ways to do it e.g. you can build the required string or print the trail while calculating the factorial. In the following example, I have done the former.
As an aside, you should check the input whether it is a positive integer.
import java.util.Scanner;
public class Car {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.print("Enter a positive integer: ");
int n = in.nextInt();
if (n >= 0) {
StringBuilder strFact = new StringBuilder();
int fact = factorial(n, strFact);
if (strFact.length() > 0) {
// Delete the last '*'
strFact.deleteCharAt(strFact.length() - 1);
System.out.println(n + "!= " + strFact + " = " + fact);
} else {
System.out.println(n + "!= " + fact);
}
} else {
System.out.println("This is an invalid input.");
}
}
public static int factorial(int num, StringBuilder strFact) {
int fact;
if (num == 0) {
fact = 1;
} else {
fact = num * factorial(num - 1, strFact);
strFact.append(num + "*");
}
return fact;
}
}
A sample run:
Enter an integer: 3
3!= 1*2*3 = 6

BaseConversion using Strings and bases 2 - 36

I have a program that prompts the user for a string, an initial base, and a final base. The program works fine for all digits however, when I enter in a string mixed with digits and characters it does not return the correct answer. For example, when I input the string BDRS7OPK48DAC9TDT4, original base: 30, newBase: 36, it should return ILOVEADVANCEDJAVA, but instead I get ILOVEADVANHSC6LTS. I know it's a problem with my algorithm but I cant figure out why it's returning the incorrect conversion.
import java.util.Scanner;
public class BaseConversion {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
String theValue;
String result;
String newNum;
int initialBase;
int finalBase;
String[] parts = args;
if (parts.length > 0) {
theValue = parts[0];
isValidInteger(theValue);
initialBase = Integer.parseInt(parts[1]);
finalBase= Integer.parseInt(parts[2]);
isValidBase(finalBase);
}
else {
System.out.println("Please enter a value: ");
theValue = s.nextLine();
isValidInteger(theValue);
System.out.println("Please enter original base: ");
initialBase = s.nextInt();
System.out.println("Please enter new base: ");
finalBase = s.nextInt();
isValidBase(finalBase);
}
// check it
// isValidInteger(theValue, finalBase);
s.close();
newNum = convertInteger(theValue, initialBase, finalBase);
System.out.println("new number: " + newNum);
}
public static void isValidBase(int finalBase) {
if (finalBase < 2 || finalBase > 36) {
System.out.println("Error: Base must be greater than or equal to 2 & less than or equal to 36");
System.exit(1);
}
}
public static void isValidInteger(String num) {
char chDigit;
num = num.toUpperCase();
for(int d = 0; d < num.length(); d++) {
chDigit = num.charAt(d);
if (!Character.isLetter(chDigit) && !Character.isDigit(chDigit)) {
//System.out.println(chDigit);
System.out.println("Error character is not a letter or number");
System.exit(1);
}
}
}
public static String convertInteger(String theValue, int initialBase, int finalBase) {
double val = 0;
double decDigit = 0;
char chDigit;
// loop through each digit of the original number
int L = theValue.length();
for(int p = 0; p < L; p++) {
// get the digit character (0-9, A-Z)
chDigit = Character.toUpperCase(theValue.charAt(L-1-p));
// get the decimal value of our character
if(Character.isLetter(chDigit)) {
decDigit = chDigit - 'A' + 10;
}
else if (Character.isDigit(chDigit)) {
decDigit = chDigit - '0';
}
else {
System.out.println("Error d");
System.exit(1);
}
// add value to total
val += decDigit * Math.pow(initialBase, p);
}
// determine number of digits in new base
int D = 1;
for( ; Math.pow(finalBase, D) <= val; D++) {}
// use char array to hold new digits
char[] newNum = new char[D];
double pwr;
for(int p = D-1; p >= 0; p--) {
// calculate the digit for this power of newBase
pwr = Math.pow(finalBase, p);
decDigit = Math.floor(val / pwr);
val -= decDigit*pwr;
// store the digit character
if(decDigit <= 9) {
newNum[D - 1 - p] = (char) ('0' + (int)decDigit);
}
else {
newNum[D - 1 - p] = (char) ('A' + (int)(decDigit - 10));
}
}
return new String(newNum);
}
}

The algorithm is correct. Take a closer look instead at the place where you convert the input value to a decimal system and in particular at the limitations of the data type you are using.
Resources that could be helpful:
primitive data types - double point in the list
Floating point arithmetic
Question concerning similar problem
JLS - 4.2.3. Floating-Point Types, Formats, and Values
Hope this points you to the right track.

import java.math.BigInteger;
import java.util.Scanner;
public class BaseConversion {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
String theValue;
String result;
String newNum;
int initialBase;
int finalBase;
String[] parts = args;
if (parts.length > 0) {
theValue = parts[0];
isValidInteger(theValue);
initialBase = Integer.parseInt(parts[1]);
finalBase= Integer.parseInt(parts[2]);
isValidBase(finalBase);
}
else {
System.out.println("Please enter a value: ");
theValue = s.nextLine();
isValidInteger(theValue);
System.out.println("Please enter original base: ");
initialBase = s.nextInt();
System.out.println("Please enter new base: ");
finalBase = s.nextInt();
isValidBase(finalBase);
}
// check it
// isValidInteger(theValue, finalBase);
s.close();
newNum = convertInteger(theValue, initialBase, finalBase);
System.out.println("new number: " + newNum);
}
public static void isValidBase(int finalBase) {
if (finalBase < 2 || finalBase > 36) {
System.out.println("Error: Base must be greater than or equal to 2 & less than or equal to 36");
System.exit(1);
}
}
public static void isValidInteger(String num) {
char chDigit;
num = num.toUpperCase();
for(int d = 0; d < num.length(); d++) {
chDigit = num.charAt(d);
if (!Character.isLetter(chDigit) && !Character.isDigit(chDigit)) {
//System.out.println(chDigit);
System.out.println("Error character is not a letter or number");
System.exit(1);
}
}
}
public static String convertInteger(String theValue, int initialBase, int finalBase) {
BigInteger bigInteger = new BigInteger(theValue,initialBase);
String value = bigInteger.toString(finalBase);
value = value.toUpperCase();
return value;
}
}
Here is the correct solution. The problem was with the data type not the algorithm. I hope this helps anyone dealing with the same type of problem.

Not able to generate all the Kaprekar Number in a range

Though the problem seemed simple, here it is :-
A Kaprekar number is a positive whole number n with d digits, such that when we split its square into two pieces - a right hand piece r with d digits and a left hand piece l that contains the remaining d or d−1 digits, the sum of the pieces is equal to the original number (i.e. l + r = n).
The Task
You are given the two positive integers p and q, where p is lower than q. Write a program to determine how many Kaprekar numbers are there in the range between p and q (both inclusive) and display them all.
Input Format
There will be two lines of input: p, lowest value q, highest value
Constraints:
0<p<q<100000
Output Format
Output each Kaprekar number in the given range, space-separated on a single line. If no Kaprekar numbers exist in the given range, print INVALID RANGE.
I could not clear the test cases in the range
22223
99999
In the above range the follwoing numbers should have been generated :-
77778 82656 95121 99999
Here is my code :-
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
Scanner scan = new Scanner(System.in);
int p = scan.nextInt();
int q = scan.nextInt();
boolean exist = false;
if(q <= p){
System.out.println("INVALID RANGE");
}
int m = 0,n = 0;
long sqr = 0;
String numb = "";
String[] digits = new String[2];
for(int i = p; i <= q; i++){
if(i == 1)System.out.print(1 + " ");
else{
sqr = i*i;
numb = String.valueOf(sqr);// Changing it into a string.
if(numb.length() % 2 == 0){
digits[0] = numb.substring(0, numb.length()/2);//Splitting it into two parts
digits[1] = numb.substring(numb.length()/2);
}else{
digits[0] = numb.substring(0, (numb.length() - 1)/2);
digits[1] = numb.substring((numb.length() -1)/2);
}
if(digits[0] == "" )
m = 0;
if(digits[1] == "")
n = 0;
if(!digits[1].equals("") && !digits[0].equals("")){
m = Integer.parseInt(digits[0]);
n = Integer.parseInt(digits[1]);
}
if(i == (m + n) ){ //Testing for equality
System.out.print(i + " ");
exist = true;
}
}
}
if(exist == false){// If exist is never modified print Invalid Range.
System.out.println("INVALID RANGE");
}
}
}

import java.util.*;
public class Kaprekar
{
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
System.out.println("Enter a number");
int num=sc.nextInt();
int sq=num*num; String sq1=Integer.toString(sq);
int mid=(Integer.toString(sq).length())/2;
int rem=sq%((int)Math.pow(10,sq1.length()-mid));
int quo=sq/((int)Math.pow(10,sq1.length()-mid));
int sum=rem+quo;
if(sum==num)
{System.out.println("Kaprekar");}else{System.out.println("Not Kaprecar");}
}
}

Changing the type of Loop index i from int to long fixed the problem.
The square computation of i*i was overflowing the upper limit of int so by changing it to long we were able to get the required computation done.
Thanks to Rup for pointing this out.
The code which cleared all the test cases is here :-
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
Scanner scan = new Scanner(System.in);
int p = scan.nextInt();
int q = scan.nextInt();
boolean exist = false;
if(q <= p){
System.out.println("INVALID RANGE");
return;
}
int m = 0,n = 0;
long sqr = 0;
String numb = "";
String[] digits = new String[2];
for(long i = p; i <= q; i++){
if(i == 1)System.out.print(1 + " ");
else{
sqr = i*i;
numb = String.valueOf(sqr);
if(numb.length() % 2 == 0){
digits[0] = numb.substring(0, numb.length()/2);
digits[1] = numb.substring(numb.length()/2);
}else{
digits[0] = numb.substring(0, (numb.length() - 1)/2);
digits[1] = numb.substring((numb.length() -1)/2);
}
if(digits[0] == "" )
m = 0;
if(digits[1] == "")
n = 0;
if(!digits[1].equals("") && !digits[0].equals("")){
m = Integer.parseInt(digits[0]);
n = Integer.parseInt(digits[1]);
}
if(i == (m + n) ){
System.out.print(i + " ");
exist = true;
}
}
}
if(exist == false){
System.out.println("INVALID RANGE");
}
}
}

import java.io.*;
import java.math.*;
import java.util.*;
class Kaprekar {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int p = scanner.nextInt();
int q = scanner.nextInt();
long i,n,s,a,c,k,d,e=0;
for(i=p;i<=q;i++)
{
k=i;d=0;
while(k!=0)
{
d++;
k=k/10;
}
c=0;s=0;n=i*i;
while(n!=0)
{
a=n%10;
n=n/10;
s=s+ (int)Math.pow(10,c)*a;
c++;
if(n+s==i&&(c==d||c==d-1)&&s!=0)
{
System.out.print(i+" ");
e++; break;
}
}
}
if(e==0)
{
System.out.println("INVALID RANGE");
}
scanner.close();
}
}

how to find StickNumbers

So I have to read a sequence of numbers from the console ( 1 to 50 numbers), none of which are equal and print out the numbers for which is true that a|b == c|d (example: 5|32 == 53|2), but I get an NubmferFormatException each time. Why?
import java.util.Scanner;
public class StuckNumbers {
public static void main(String[] args) {
// create Scanner
Scanner input = new Scanner(System.in);
// input count and declare array
System.out.println("input number of numbers");
int count = input.nextInt();
int[] numbers = new int[count];
// check if count is between 1 and 50
if (count < 1 && count > 50) {
System.out.println("Wrong input. Input a number between 1 and 50");
count = input.nextInt();
}
// input n numbers
for (int i : numbers) {
i = input.nextInt();
// check if i = j
for (int j : numbers) {
if (i == j) {
System.out
.println("All numbers must be dist75inct. Try again.");
i = input.nextInt();
}
}
}
for (int i = 0; i < count; i++) {
for (int j = 0; j < count; j++) {
if (stuckNumbers(numbers[i], numbers[j]) == stuckNumbers(
numbers[j], numbers[i])) {
System.out.println(i + "|" + j + " == " + j + "|" + i);
}
}
}
input.close();
}
public static int stuckNumbers(int a, int b) {
String firstNum = "a";
String secondNum = "b";
String res = "ab";
int result = Integer.parseInt(res);
return result;
}
}

Look at these lines:
String res = "ab";
int result = Integer.parseInt(res);
"ab" is not a number, so you're going to get a NumberFormatException when you try to parse it as an integer.

Change the firstNum and SecondNum variables from "a" and "b" to Integer.toString(a); OR String.valueOf(a); and similar for b.
public static int stuckNumbers(int a, int b) {
String firstNum = String.valueOf(a);
String secondNum = String.valueOf(b);
String res = "";
res.concat(firstNum);
res.concat(secondNum);
int result = Integer.parseInt(res);
return result;
}
I hope this will remove any Exception being thrown.

To check if a number is Armstrong number using java

this is my program:
public class ArmstrongNumber {
public static void main(String args[]) {
int n = 0, temp = 0, r = 0, s = 0;
Scanner in = new Scanner(System.in);
System.out.println("Enter a number ");
if (in.hasNextInt()) {
n = in.nextInt(); // if there is another number
} else {
n = 0;
}
temp = n;
while (n != 0) {
r = n % 10;
s = s + (r * r * r);
n = n / 10;
}
if (temp == s) {
System.out.println(n + " is an Armstrong Number");
} else {
System.out.println(n + " is not an Armstrong Number");
}
}
}
output:
Exception in thread "main" java.lang.NoClassDefFoundError
I tried it using DataInputStream but still getting same error.

// To check the given no is Armstrong number (Java Code)
class CheckArmStrong{
public static void main(String str[]){
int n=153,a, b=0, c=n;
while(n>0){
a=n%10; n=n/10; b=b+(a*a*a);
System.out.println(a+" "+n+" "+b); // to see the logic
}
if(c==b) System.out.println("Armstrong number");
else System.out.println(" Not Armstrong number");
}
}

Find any digit is Armstrong number or not using loop
for(int arm_num = 0 ; arm_num < 100000 ; arm_num++)
{
String[] data = String.valueOf(arm_num).split("(?<=.)");
int lngth = String.valueOf(arm_num).length();
int arm_t_num = 0;
int ary[] = new int[lngth];
for(int i = 0 ; i < lngth ; i++)
{
ary[i] = Integer.parseInt(data[i]);
for(int x = 0 ; x < lngth-1 ; x++)
{
ary[i] = ary[i] * Integer.parseInt(data[i]);
}
arm_t_num+=ary[i];
}
if(arm_num == arm_t_num)
{
System.out.println("Number is ArmStrong : "+arm_num);
}
}

you need to set CLASS_PATH variable and point it to where ever your class file is
then this should work
I have tried it locally, refer my answer to check how to set class path and how to compile and run java code using command prompt

//This is my program to check whether the number is armstrong or not!!
package myprogram2;
public class Myprogram2 {
public static void main(String[] args)
{
String No="407";
int length_no=No.length();
char[] S=new char[length_no];
int[] b = new int[length_no];
int arm=0;
for(int i=0;i<length_no;i++)
{
S[i]=No.charAt(i);
b[i]=Character.getNumericValue(S[i]);
//System.out.print(b[i]);
arm=arm + (b[i]*b[i]*b[i]);
System.out.println(arm);
}
//System.out.println(" is the number \n now Checking for its Armstrong condition");
int orgno = Integer.parseInt(No);
if (orgno==arm)
System.out.println("YESm its an armstrong");
else
System.out.println("\n<<Not an armstrong>>");
//System.out.println(length_no);
System.out.println("Original number is "+orgno);
System.out.println("Sum of cubes "+arm);
}
}

There are a couple of nice String-based solutions and numeric solutions with single-letter variable names.
Consider this to make sense of how it works numerically, which includes a couple of interesting numeric tricks:
import java.io.*;
public class Armstrong
{
public static void main(String args[]) throws IOException
{
InputStreamReader read = new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(read);
int modifiedNumber, originalNumber, modifiedNumberWithUnitsDigitZero,
unitsDigit, runningSum;
System.out.println("Enter your number:");
modifiedNumber = Integer.parseInt(in.readLine());
runningSum = 0;
originalNumber = modifiedNumber;
while(modifiedNumber > 0)
{
modifiedNumberWithUnitsDigitZero = modifiedNumber / 10 * 10;
unitsDigit = modifiedNumber - modifiedNumberWithUnitsDigitZero;
runningSum += unitsDigit * unitsDigit * unitsDigit;
modifiedNumber = modifiedNumber / 10;
}
System.out.println("The number " + originalNumber
+ (originalNumber == runningSum ? " IS" : " is NOT")
+ " an Armstrong number because sum of cubes of digits is " + runningSum);
}
}

import java.util.Scanner;
public class Amst {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
System.out.println("Enter The No. To Find ArmStrong Check");
int i = sc.nextInt();
int sum = 0;
for(int j = i; j>0 ; j = j/10){
sum = sum + ((j%10)*(j%10)*(j%10));
}
if(sum == i)
System.out.println("Armstrong");
else
System.out.println("Not Armstrong");
}
}

For 'N' digit amstrong number
package jjtest;
public class Amstrong {
public static void main(String[] args) {
// TODO Auto-generated method stub
int num=54748;
int c=0;
int temp=num;
int b=1;
int length = (int)(Math.log10(num)+1);
while(num>0){
int r = num%10;
num=num/10;
int a =1;
for(int i=1;i<=length;++i){
b=b*r;
}
c = c + b;
b=1;
}
System.out.println(c);
if(c==temp){
System.out.println("its an amstrong number");
}else{
System.out.println("its not an amstrong number");
}
}
}

This is the simple logic for Armstrong number program :
for (int i = number; i > 0; i = i / 10)
{
remainder = i % 10;
sum = sum + remainder * remainder * remainder;
}
if(sum == number)
{
System.out.println("\n" + number + " is an Armstrong Number\n");
}
Reference :
http://topjavatutorial.com/java/java-programs/java-program-to-check-if-a-number-is-armstrong-number/

import java.util.Scanner;
/* a number is armstrong if the sum of cubes if individual digits of
a number is equal to the number itself.for example, 371 is
an armstrong number. 3^3+7^3+1^3=371.
some others are 153,370,407 etc.*/
public class ArmstrongNumber {
public static void main(String args[]) {
int input, store, output=0, modolus;
Scanner in = new Scanner(System.in);
System.out.println("Please enter a number for ckecking.");
input = in.nextInt();
store = input;
while(input != 0) {
modolus = input % 10;
output = output + (modolus * modolus * modolus);
input = input / 10;
}
System.out.println(output);
if(store == output) {
System.out.println("This is an armstrong number.");
} else {
System.out.println("This is not an armstrong number.");
}
in.close();
}
}

import java.util.*;
public class ArmstrongNumber
{
public static void main( String[] args )
{
int n = 0, temp = 0, r = 0, s = 0;
Scanner in = new Scanner(System.in);
System.out.println("Enter a number ");
if (in.hasNextInt()) {
n = in.nextInt(); // if there is another number
} else {
n = 0;
}
temp = n;
while (n != 0) {
r = n % 10;
s = s + (r * r * r);
n = n / 10;
}
if (temp == s) {
System.out.println(temp + " is an Armstrong Number");
} else {
System.out.println(temp + " is not an Armstrong Number");
}
}
}
You missed to import java.util package
Change n to temp in S.O.P

import java.util.Scanner;
public class AmstrongNumber {
public static void main(String[] args) {
System.out.println("Enter the number");
Scanner scan=new Scanner(System.in);
int x=scan.nextInt();
int temp2=0;
String s1 = Integer.toString(x);
int[] a = new int[s1.length()];
int[] a1 = new int[s1.length()];
for (int i = 0; i < s1.length(); i++){
a[i] = s1.charAt(i)- '0';
int temp1=a[i];
a1[i]=temp1*temp1*temp1;
}
for (int i = 0; i < s1.length(); i++){
temp2=temp2+a1[i];
if(i==s1.length()-1){
if(x==temp2){
System.out.println("Amstrong num");
}else{
System.out.println("Not !");
}
}
}
}
}

private static boolean isArmstrong(int num) {
int totalSum = 0;
int copyNum = num;
while (num != 0) {
int reminder = num % 10;
int cubeOfReminder = reminder * reminder * reminder;
totalSum = totalSum + cubeOfReminder;
num = num / 10;
}
if (copyNum == totalSum)
return true;
return false;
}

public class Testamstrong
{
public static void main(String...strings) {
int num = 153,temp;
temp = num;
if(temp == amstrongNumber(num)) {
System.out.println("Number is amstrong number...");
}
else {
System.out.println("Number is not amstrong number...");
}
}
public static int amstrongNumber(int num) {
int count=0,sum=0;
count = String.valueOf(num).length();
char[] ch = String.valueOf(num).toCharArray();
for(char ch1:ch) {
int num1 = Character.getNumericValue(ch1);
sum += Math.pow(num1, count);
}
return sum;
}
}

Find Armstrong number using for loops (with example)
import java.util.*;
public class ArmstorngNumber {
public static void main(String args[]) {
int cube, num, quo, n;
int s = 0;
do
{
System.out.println("Enter Your Number");
Scanner sc = new Scanner(System.in);
num = sc.nextInt();//153
n = num;
for (int i = 0; i < 10; i++) {
int rem = num % 10;//3
quo = num / 10; //15
cube = rem * rem * rem;//9
s = s + cube;//0+9
num = quo;//0
}
System.out.println(s);
System.out.println(n);
if (s == n) {
System.out.println("The number is Armstrong");
System.out.println("-------------------------------------");
}
else {
System.out.println("The number is not Armstrong");
System.out.println("-------------------------------------");
}
}
while (n > 0);
}
}

Check the Armstrong number of any number [java] [Armstrong]
import java.util.*;
public class Armstrong {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.print("enter any number?");
int x = sc.nextInt();
int n=0;
int number = x;
int j =x;
int result = 0 ,remainder;
while (x!=0) {
x/=10;
++n;
}
for(;j>0 ;j=j/10) {
remainder=j%10;
result+=Math.pow(remainder, n);
}
if (number==result) {
System.out.print(number +" is Armstrong ");
}
else
System.out.print(number +" is not Armstrong");
}
}

here is my code, please check if this works for you!
import java.util.Scanner;
public class Armstromg {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.print("Please enter the number: ");
int num = sc.nextInt();
int length = 0;
int temp1 = num;
while(temp1 != 0) {
temp1/=10;
length+=1;
}
int result = 1;
int temp2 = num;
for(int i = 1; i <= length; i++) {
temp2 = temp2 % 10;
result*=Math.pow(temp2,length);
}
if(result == num) {
System.out.print("The number is an armstrong number!");
} else {
System.out.print("The number is not an armstrong number");
}
}
}

package Loops;
public class ArmStrongNumber {
public static void main(String[] args) {
int digit1, digit2, digit3;
int number = 153;
int temp = number;
digit1 = number % 10;
number = number / 10;
digit2 = number % 10;
number = number / 10;
digit3 = number % 10;
if ((digit1 * digit1 * digit1) + (digit2 * digit2 * digit2) + (digit3 * digit3 * digit3) == temp) {
System.out.println(+temp + " Number is Armstrong ");
} else {
System.out.println("Number is not Armstrong");
}
}
}

//Not limited to 3 digit integers
public static void main(String[] args)
{
int num = 54748,a,sum=0;
int x = num;
int p =Integer.toString(num).length();
while (num !=0)
{
a = num%10;
num = num/10;
sum = sum + (int) Math.pow(a, p);
}
if (x==sum)
System.out.println("Its an Armstrong number");
else
System.out.println("Not an Armstrong number");
}
}

Scanner input = new Scanner (System.in);
int num , counter = 0 ,temp;
System.out.print("Enter Nmber :");
num = input.nextInt();
int lnum = num;
while ( num != 0 ){
num = num/10 ;
counter++;
}
int store_num_keyboard_input = lnum;
int new_tot = 0;
int c = counter;
while(lnum > 0){
temp = lnum % 10;
lnum = lnum / 10 ;
int m = 0; //m is counter
int tot = 1;
while (m != c){
tot = tot * temp;
m++;
}
new_tot = new_tot + tot;
}
System.out.println("new total "+ new_tot );
if(new_tot == store_num_keyboard_input){
System.out.println(store_num_keyboard_input + " is an Armstrong number" );
}
else{
System.out.println(store_num_keyboard_input + " is not an Armstrong number" );
}

My answer using JAVA 8
tested for..[1, 153, 370, 371, 407]
public class Armstrong {
public static boolean isArmstrong(int num) {
return num == getArmstrongSum(num);
}
public static int getArmstrongSum(int num) {
int pow = String.valueOf(num).length();
return IntStream.iterate(num, i -> i / 10)
.limit(pow)
.map(i -> (int) Math.pow(i % 10, 3))
.sum();
}
public static void main(String[] args) {
System.out.println(isArmstrong(153));
}
}
Thank you.

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