Algorithm
Problem Name: 2 AD-HOC - beecrowd | 1593
Problem Link: https://www.beecrowd.com.br/judge/en/problems/view/1593
Binary Function
By Bruno Adami, Universidade de São Paulo - São Carlos 
Brazil
Timelimit: 1
We define the parity of an integer as the sum of its bits in the binary form modulo two. Take this example, the number 2110 = 101012 has three 1's in its binary representation and thus it has an odd parity.
In this problem, you need to calculate the number of bits 1 in an integer I given in the input, it is, calculate the number of 1's in the binary representation.
Input
The first line contains an integer T (T <= 100) indicating the number of test cases.
For each case, there will be a single line with the number I (1 ≤ I < 1018* or 1 ≤ I < 101000**).The input number won't have leading zeroes.
*for around 90% of the cases;
**for the other test cases.
Output
Output the number of 1's in the binary representation for each case in a single line.
| Sample Input | Sample Output |
|
3 21 3 123456789123456789123456789 |
3 2 50 |
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
import java.math.BigInteger;
import java.util.Scanner;
public class Main
{
final static BigInteger TWO = new BigInteger("2");
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
BigInteger num;
int t, b, ans;
t = scan.nextInt();
for ( int i = 0; i < t; i++ )
{
num = scan.nextBigInteger();
ans = 0;
while ( num.compareTo(BigInteger.ZERO) != 0 )
{
b = num.mod(TWO).intValue();
num = num.divide(TWO);
if ( b == 1 ) ans++;
}
System.out.println(ans);
}
}
}
Input
21
3
123456789123456789123456789
Output
2
50