Lowest Bit(数位，二进制)

Lowest Bit
Time Limit: 1000MS Memory Limit: 65536KB
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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.
Example Input

26
88
0
Example Output

2
8

我记得自己之前，当遇到没有思路的题目的时候，找博客看，通常是找自己看得懂的博客~（这不是一个好的习惯，要改！！！）

今天，突然想写一下之前都不会用的二进制了，嘿嘿~

题目是，找出一个十进制数的二进制位是1的最低的一位，输出代表的数（2的*次幂）~

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main()
{
    int n;
    while(~scanf("%d", &n)&&n)
    {
        for(int i=0;i<8;i++)
        {
            if((n>>i)&1)//表示n的二进制数位向右移动i位，然后与1取与，
            //判断最末一位是否是1,如果是，if条件成立,输出
            {
                int b = pow(2, i);//在<math.h>头文件里，
                //表示2的*次幂（但函数的返回值不是int类型，要用一个int类型的数字转换）
                printf("%d\n", b);
                break;
            }
        }
    }
    return 0;
}