Lowest Bit

Description

Given an positive integer A (1 <= A <= 100), output the lowest bit of A. For example, given A = 26, we can write A in binary form as 11010, so the lowest bit of A is 10, so the output should be 2. Another example goes like this: given A = 88, we can write A in binary form as 1011000, so the lowest bit of A is 1000, so the output should be 8.

Input

Each line of input contains only an integer A (1 <= A <= 100). A line containing "0" indicates the end of input, and this line is not a part of the input data.

Output

For each A in the input, output a line containing only its lowest bit.

Sample Input

26
88
0

Sample Output

2
8


题目意思：我们知道十进制转化成二进制的方法，列如26，二进制则是11010，那么从后面开始算起只由第一位数是1的二进制是10，
其十进制则是2；同理88的
二进制则是1011000，从后面开始算起只由第一位数是1的二进制是1000，其十进制是8。

解题思路：要是直接从题目意思入手可以得到一下的代码：

 1 #include<stdio.h>
 2 #include<math.h>///类似十进制转换成二进制的原理，统计最后面的一个1之后0的个数
 3 int lowest(int n)
 4 {
 5     int x,count=0;
 6     while(n!=0)
 7     {
 8         x=n%2;
 9         if(x==0)
10             count++;
11         else
12             break;///遇到非0就会退出
13         n=n/2;
14 
15     }
16     return count;
17 }
18 int main()
19 {
20     int a,b;
21     while(scanf("%d",&a)!=EOF)
22     {
23         if(a==0)
24             break;
25         b=pow(2,lowest(a));
26         printf("%d\n",b);
27 
28     }
29     return 0;
30 }

其实还可以使用位运算的方法：

 1 #include<stdio.h>///位运算的使用
 2 int main()
 3 {
 4     int n,m;
 5     while(scanf("%d",&n),n)
 6     {
 7          m=n&(n^(n-1));
 8          printf("%d\n",m);
 9     }
10     return 0;
11 }