HDU 1196 Lowest Bit

Lowest Bit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5040    Accepted Submission(s): 3673

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

 

Source

Zhejiang University Local Contest 2005

 

Recommend

Ignatius.L

 

分析：只要知道2进制的转换方法就好了，N%2最先除出来的余数是2进制的最低位，所以只要判断什么时候N%2的余数第一次不等于0

代码：

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