zoj2417-------------Lowest Bit

最新推荐文章于 2017-07-23 14:57:16 发布

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本文介绍了一个简单的算法，用于找出给定正整数（1≤A≤100）二进制表示中最低位的非零位，并将其转换回十进制形式输出。通过将输入的整数逐步除以2并取余数的方式，可以得到该整数的二进制表示。随后，从低位开始查找第一个非零位，利用2的幂次方计算出该位对应的十进制数值。

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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

转为二进制从低位找到第一个不为零的位再转化为十进制输出即可

代码：

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
	int A;
	while(cin>>A&&A!=0)
	{
		int a[7];
		for(int i=0;A!=0;i++)
		{
			a[i]=A%2;
			A=A/2;	
		}	
		int i=0;
		while(a[i]==0)
			i++;
		cout<<pow(2,double(i))<<endl;
			
	}
	return 0;	
}