toj 2841 Bitwise Reverse

 

 

2841.   Bitwise Reverse

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Time Limit: 1.0 Seconds    Memory Limit: 65536K
Total Runs: 1520    Accepted Runs: 1025

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Professor Robby invents a powerful encryption method, but he is too lazy to implement it. So he turns to you for help.

In fact, the encryption method is only applied to positive integers. At first, we express the number as binary code, that is, a string only contain '0' and '1', and the first digit can't be '0'. Then we reverse the string. And the last step, we calculate the reversed binary code and express it in decimal again.

For example, we want to encrypt the number 14, we express it as 1110, after reversing it we get 0111, and (0111)₂ = 7. So we get 7.

Input

There is only one line for each test case, containing the positive integer to be encrypted. You can assume the number is not more than 10⁶.

The input is terminated with a zero.

Output

Output one line for each test case, indicating the number after encryption.

Sample Input

5

6

14

0

 

Sample Output

5

3

7

 

Problem Setter: RoBa

Source: Tianjin Metropolitan Collegiate Programming Contest 2007

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Submit   List    Runs   Forum   Statistics

 

                        Show Code  -  Run ID  633955
Submit Time:  2009 - 05 - 09   17 : 10 : 25      Language: GNU C ++      Result: Accepted
    Pid:  2841      Time:  0.00  sec.     Memory:  1208  K.     Code Length:  0.4  K.

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#include  < iostream >
#include  < cmath >
#include  < stack >
using   namespace  std;
stack < int > S;
int  main()
{
     int  n,sum;
     while (scanf( " %d " , & n) != EOF && n != 0 )
    {
        sum = 0 ;
         int  i = 0 ;
         while ( ! S.empty())
            S.pop();
         while (n != 0 )
        {
             int  temp = n % 2 ;
            n = n / 2 ;
            S.push(temp);
        }
         while ( ! S.empty())
        {
             int  temp2 = S.top();
            S.pop();
            sum += ( int )(temp2 * (pow( 2.0 ,i ++ )));
        }
        printf( " %d\n " ,sum);
    }
     return   0 ;
}

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