An Easy Problem

Description

As we known, data stored in the computers is in binary form. The problem we discuss now is about the positive integers and its binary form.

Given a positive integer I, you task is to find out an integer J, which is the minimum integer greater than I, and the number of '1's in whose binary form is the same as that in the binary form of I.

For example, if "78" is given, we can write out its binary form, "1001110". This binary form has 4 '1's. The minimum integer, which is greater than "1001110" and also contains 4 '1's, is "1010011", i.e. "83", so you should output "83".

Input

One integer per line, which is I (1 <= I <= 1000000).

A line containing a number "0" terminates input, and this line need not be processed.

Output

One integer per line, which is J.

Sample Input

1
2
3
4
78
0

Sample Output

2
4
5
8
83

这道题的确很简单：
解决办法为：
找出从最右一个不为0(即最右一个1)开始的连续"1"串(假设k个),将其加 1,再将最右端的k-1个0变为1。

 在进制转换上面，可以偷懒，用bitset，真的很不错噶。

#include  < iostream >
#include  < bitset >
using   namespace  std;
        
int  main()
... {
    unsigned long n;
    while(cin>>n && n!=0)
    ...{    
        int i=0,len=0;
        bitset<32> mybit(n);
        while(!mybit[i]) i++;
        while(mybit[i])
        ...{
            mybit[i]=0;
            mybit[len]=1;
            i++;
            len++;
        }
        mybit[len-1]=0;
        mybit[i]=1;

        cout<<mybit.to_ulong()<<endl;
    }

    return 0;
}