Lowest Bit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12204 Accepted Submission(s): 8905
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
二进制的最低位可根据a&(-a)就能直接求出二进制的最后一位
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
int k;
while(scanf("%d",&k)!=EOF)
{
if(k==0)
break;
printf("%d
",k&(-k));
}
return 0;
}