HDU 1196 Lowest Bit

/*
题目大意：将一个数变成一个二进制数，输出最小的一个数
解题思路：小的知识点，了解while循环即可，细心一点应该就可以了
难点详解：熟悉用while语句
关键点：输出后面的最后一个，最小的一个数
解题人：lingnichong
解题时间：2014-08-11 23:14:01
解题感受：水题吧，但也要想清楚
*/

Lowest Bit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7819    Accepted Submission(s): 5755

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

 

#include<stdio.h>int a[1000];int main(){int n;while(scanf("%d",&n),n){int k=0,j,t,s=0,m;while(n){a[k++]=n%2;n/=2;}for(j=0;j<k;j++){if(a[j]==1){t=j;break;}}for(j=0,m=1;j<=t;j++,m*=2){s=s+a[j]*m;}printf("%d\n",s);}return 0;}