Lowest Bit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7776 Accepted Submission(s): 5715
Problem 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.
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
Author
SHI, Xiaohan
Source
题意：
给定一个十进制数字，求转化为二进制后的最后一位
思路：
依据位运算能极大地优化程序。求二进制的最低位可依据a&(-a)就能直接求出二进制的最后一位
代码例如以下：
<span style="font-size:14px;">#include<stdio.h>
int main()
{
int n;
while(~scanf("%d",&n),n)
{
printf("%d
",n&(-n));
}
return 0;
}</span>