Lowest Bit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8166 Accepted Submission(s): 5998

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
Source Zhejiang University Local Contest 2005

题意：求一个十进制数的二进制形式的从最后一个不为0的位开始算起的后面的二进制代表的十进制数是多少。

解题思路：利用位运算，用 & 1 判断一个数的二进制形式的最后一位是否为一，每次右移一位，用k纪录右移的次数，重复判断最后一位是否为1，若是1，则要求的结果就是2^k,输出即可。不过注意，在第一次右移前，要先判断原来的数是否为奇数。

AC代码：

#include 
  
   
#include 
   
     #include 
    
      using namespace std; int main(){ // freopen("in.txt", "r", stdin); int a; while(scanf("%d", &a)!=EOF && a){ int k = 0; //右移位数 while(1){ if(a & 1){ printf("%d\n", (1<
     
      >= 1; k ++; } } return 0; }

HDU 1196 Lowest Bit （数位）

Lowest Bit