HDU1196_Lowest Bit【位运算】【水题】

来源：互联网发布：企业网站域名备案流程编辑：程序博客网时间：2024/05/16 08:42

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：给你一个数A，求它的二进制表示中最右边的1表示的数

比如：26的二进制表示为11010，最右边的1表示的数为00010。

思路：位运算，其实就是求A & (A ^ (A-1) )，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\n",ans);    }    return 0;}

0 0