Parity.java

/**
 * We define the parity of an integer n as the sum of the bits in binary representation computed modulo
 * two. As an example, the number 21 = 101012 has three 1s in its binary representation so it has parity
 * 3(mod2), or 1.
 * In this problem you have to calculate the parity of an integer 1 ≤ I ≤ 2147483647.
 * Input
 * Each line of the input has an integer I and the end of the input is indicated by a line where I = 0 that
 * should not be processed.
 * Output
 * For each integer I in the inputt you should print a line ‘The parity of B is P (mod 2).’, where B
 * is the binary representation of I.
 * Sample Input
 * 1
 * 2
 * 10
 * 21
 * 0
 * Sample Output
 * The parity of 1 is 1 (mod 2).
 * The parity of 10 is 1 (mod 2).
 * The parity of 1010 is 2 (mod 2).
 * The parity of 10101 is 3 (mod 2).
 */

//https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1872

import java.util.Scanner;

public class Parity {

	public static void main(String[] args) {
		while (true) {
			Scanner input = new Scanner(System.in);
			int number = input.nextInt();
			if (number == 0) {
				break;
			}
			String binaryInString = convertToBinary(number);
			int count = 0;
			for (int i = 0; i < binaryInString.length(); i++) {
				if ("1".equals(binaryInString.charAt(i) + "")) {
					count++;
				}
			}
			System.out.println("The parity of " + binaryInString + " is "
					+ count + " (mod 2).");

		}
	}

	private static String convertToBinary(int number) {
		StringBuilder s = new StringBuilder("");
		while (number != 0) {
			s = s.append(number % 2);
			number = number / 2;
		}
		return s.reverse().toString();
	}
}