HDU 1196 Lowest Bit（二进制）

                                        Lowest Bit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10670    Accepted Submission(s): 7839

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

26880

 

Sample Output

28

 

Author

SHI, Xiaohan

 

 题解：从二进制中的最右边的1开始到最后的数转化为十进制。。。

AC代码：

#include<iostream>#include<cstdlib>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<cstdlib>#include<algorithm>typedef long long LL;using namespace std;int main(){int n,a,t;while(cin>>n){if(n==0)break;t=1;while(n){if(n%2==1)break;  //如果是奇数，最后结果肯定等于1 n=n/2;t=t*2;}printf("%d\n",t);}   return 0;}