(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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本文介绍了一种解决整数最低位二进制表示的方法，通过不断除以2并检查余数的方式找到最低位的二进制位，并提供了一个简单的C++实现示例。

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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

2
8

Author

SHI, Xiaohan

Source

Zhejiang University Local Contest 2005

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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;
}