HDU 1.2.6 Lowest Bit

英国老鼠_

于 2016-03-10 18:06:06 发布

阅读量245

点赞数

CC 4.0 BY-SA版权

分类专栏： ACM 文章标签： ACM HDU

本文链接：https://blog.youkuaiyun.com/hwj_wayne/article/details/50848897

ACM 专栏收录该内容

41 篇文章

订阅专栏

本文提供了一个算法解决方案，用于找出给定正整数（范围1到100）二进制表示中最低位的值。通过一系列示例说明了如何将十进制数转换为二进制，并从中提取最低位。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Lowest Bit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2297 Accepted Submission(s): 1588

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

#include<stdio.h>  
int main()  
{  
    int A,t;  
    while(scanf("%d",&A),A)  
    {  
        t=1;  
        while(A)  
        {  
            if(A%2==1)break;  
            A/=2;  
            t*=2;  
        }   
        printf("%d\n",t);     
    }      
    return 0;  
}