An Easy Problem_位运算

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8214		Accepted: 4918

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

x=n&(-n);//求出n的二进制码右边第一个1；

n+x将原来右边第一个01变成了10；如，n=1001110，x=0000010，则n+x=1010000；同时也带来一个问题，01右边的1全部变成了0；

要找回这些1放在最右边，用亦或y=（n^(n+x)）这样一来01及其右边的1都会变成1，而0不变；

然后z=y/x去掉右边的0；

z>>2因为亦或时由于原来的01变成了10多了两个1，所以向右移动两位；

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
    int n;
    while(~scanf("%d",&n),n)
    {
        int x=n&(-n);//求出n的二进制码右边第一个1；
        printf("%d
",n+x+((n^(n+x))/x>>2));
    }
    return 0;
}