Solution of 11933 - Splitting Numbers

Problem Description
source:https://uva.onlinejudge.org/external/119/11933.html

We define the operation of splitting a binary number n into two numbers a(n), b(n) as follows. Let 0 ≤ i1 < i2 < . . . < ik be the indices of the bits (with the least significant bit having index 0) in n that are 1. Then the indices of the bits of a(n) that are 1 are i1, i3, i5, . . . and the indices of the bits of b(n) that are 1 are i2, i4, i6, . . .
For example, if n is 110110101 in binary then, again in binary, we have a = 010010001 and b = 100100100.

Input

Each test case consists of a single integer n between 1 and 231 − 1 written in standard decimal (base 10) format on a single line. Input is terminated by a line containing ‘0’ which should not be processed.

Output

The output for each test case consists of a single line, containing the integers a(n) and b(n) separated by a single space. Both a(n) and b(n) should be written in decimal format.

Sample Input

25 5000 1000000 2000000000 2147483647 0

Sample Output

17 8 4360 640 671808 328192 1378124800 621875200 1431655765 715827882

Solution:

#include <algorithm>
#include <cstdio>
#include <cmath>
#include <iostream>
#include<stdio.h>
#include<stdlib.h>
using namespace std;

int main()
{
    int n,even,temp1,c,sum;
    while(scanf("%d",&n)==1)
    {
        if(n==0)
            break;
        temp1=n;
        c=0;even=0;sum=0;
        while(temp1!=0)
        {
            if(temp1%2==1)
            {
                even++;
                if(even%2==1)
                {
                    sum+=pow(2,c);
                }
            }
            temp1/=2;
            c++;
        }
        printf("%d %d\n",sum,n-sum);
    }
    return 0;
}