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题目：
Lowest Bit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8722 Accepted Submission(s): 6428
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
Sample Output
Author
SHI, Xiaohan
Source
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题目分析：
简单题。在将10进制数转换成2进制数的过程中不断地判断当前位是否是非零位即可。
相应的知识点：
在这时候可能需要回顾一下10进制转2进制的过程：
代码如下：
/*
* d.cpp
*
* Created on: 2015年3月11日
* Author: Administrator
*/
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int main(){
int n;
while(scanf("%d",&n)!=EOF,n){
int t = 0;
while(n%2 == 0){
t++;
n /= 2;
}
printf("%d\n",(int)pow(2+0.0,t));
}
return 0;
}
(hdu 简单题 128道)Lowest Bit(求一个数的二进制的最后一个非零位对应的十进制数)
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原文地址：http://blog.csdn.net/hjd_love_zzt/article/details/44201135
