Amicable numbers Java program - java

So I am having this strange output in which only the first number is checked twice while the second number is not even considered.Please help.
Code :-
import java.util.Scanner;
public class Amicable
{
private static int a,b;
private static String m,n;
public static void main()
{
acceptNumbers();
if (firstNumber() == secondNumber())
{
System.out.println(a+" and "+b+" are amicable numbers");
}
else System.out.println(a+" and "+b+" are not amicable numbers");
}
public static void acceptNumbers()
{
Scanner sc = new Scanner(System.in);
int count=0;
System.out.print("Enter two numbers [ separated by a ',' ] : ");
String input = sc.nextLine();
System.out.println();
for (int i = 0; i < input.length(); i++)
{
char c = input.charAt(i);
if (c == ',')
{
count++;
if (count == 1)
{
m = input.substring(0,i);
n = input.substring(0,i);
}
break;
}
}
if (count == 0)
{
System.out.println("Invalid operation : You have entered only 1 number");
}
m = m.trim(); n = n.trim();
a = Integer.valueOf(m);
b = Integer.valueOf(n);
}
public static int firstNumber()
{
int a1,a2=0;
for (int i = 0; i < m.length()-1; i++)
{
a1 = Integer.valueOf(m.charAt(i));
if (a%a1 == 0) a2 = a2+a1;
}
return a2;
}
public static int secondNumber()
{
int b1,b2=0;
for (int i = 0; i < n.length()-1; i++)
{
b1 = Integer.valueOf(n.charAt(i));
if (b%b1 == 0) b2 = b2+b1;
}
return b2;
}
}
And here is the output :-
Enter 2 numbers [ separated by a ',' ] : 248 , 222
248 and 248 are amicable numbers

your m and n are equal, because you have:
m = input.substring(0,i);
n = input.substring(0,i);
change it to:
m = input.substring(0,i);
n = input.substring(i+1);
Btw you are doing a lot of unnecessary stuff, complete solution (I don't care about exceptions):
import java.util.Scanner;
public class Amicable {
public static void main(String args[]) {
try {
Scanner sc = new Scanner(System.in);
System.out.print("Enter two numbers [ separated by a ',' ] : ");
String input = sc.nextLine();
String[] numbers = input.split(",");
int num1 = Integer.parseInt(numbers[0].trim());
int num2 = Integer.parseInt(numbers[1].trim());
int sum1 = 0, sum2 = 0;
for (int i = 1; i <= num1; i++) {
if (num1 % i == 0)
sum1 += i;
}
for (int i = 1; i <= num2; i++) {
if (num2 % i == 0)
sum2 += i;
}
if (sum1 == sum2)
System.out.println(num1 + " and " + num2
+ " are amicable numbers");
else
System.out.println(num1 + " and " + num2
+ " are not amicable numbers");
} catch (Exception e) {
e.printStackTrace();
}
}
}
parts of code from: http://www.daniweb.com/software-development/java/code/304600/amicable-numbers

a and b are derived from m and n, and the latter are initialized to exactly the same value:
m = input.substring(0,i);
n = input.substring(0,i);
Did you mean to set n to
n = input.substring(i+1);
?

m = input.substring(0,i);
n = input.substring(0,i);
m and n are having the same value.
n should be:
n = input.substring(i+1);
And now the second number will be assigned to n.

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.

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.

Binary to Decimal Java converter

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);
}

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