題目：
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

題意：找出一個數的二進位制從後往前數第一個為1 的數字
解題思路：要知道N%2最先除出來的餘數是2進位制的最低位，所以只要判斷什麼時候N%2的餘數第一次不等於0

程式碼

//
//  main.cpp
//  hdu1096
//
//  Created by zhan_even on 2018/11/1.
//  Copyright © 2018年 zhan_even. All rights reserved.
//

#include <iostream>
#include <math.h>
using namespace std;

int main(int argc, const char * argv[]) {
    int n,i,num,bin[10000];
    while (cin>>n&&n) {
        int temp = n;
        int m=0;
        while (temp !=0) {
            i = temp%2;
            temp = temp/2;
            bin[m] = i;
            m++;
        }
        
        for(int x=0;x<m;x++){
            
            if (bin[x]==1) {
                num = pow(2,x);
                cout<<num<<endl;
                break;
            }
        }
    }
    return 0;
}