Lowest Bit------HDOJ杭电1196（想法很重要）

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

二进制的原理就是除二余一的时候出现1，所以呀！
直接循环取二的模，如果为0则记录sum加一，否则break；
然后就是循环乘二，也可以在判断里面直接乘二
下面贴下AC代码：
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;

int main()
{
    int i,j,k;
    int t,n,m;
    int liu[106];
    while(scanf("%d",&n)!=EOF,n)
    {
        m=0;
        while(1)
        {
            if(n%2==0)
            {
                m++;
                n/=2;
            }
            else break;
        }
        t=1;
        for(i=1;i<=m;i++)
            t*=2;
        printf("%d\n",t);
    }
    return 0;
}

写代码能力有限，如有神牛发现BUG，还请指出，不胜感激！

Lowest Bit------HDOJ杭电1196（想法很重要）