HDU1196_Lowest Bit【位运算】【水题】
来源：程序员人生 发布时间：2014-11-23 09:23:53 阅读次数：1926次
Lowest Bit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8043 Accepted Submission(s): 5920
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.
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
题目大意：给你1个数A，求它的2进制表示中最右侧的1表示的数
比如：26的2进制表示为11010，最右侧的1表示的数为00010。
思路：位运算，其实就是求A & (A ^ (A⑴) )，即A & (-A)
比如：26――011010，-A = 111010 A & ()
#include<stdio.h>
int main()
{
int A;
while(~scanf("%d",&A) && A)
{
//int ans = A & ( A ^( A - 1));
int ans = A & (-A);
printf("%d
",ans);
}
return 0;
}
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