2023-03-18

Title:

# Lowest Bit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8722    Accepted Submission(s): 6428

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

```

Author
SHI, Xiaohan

Source

Recommend
Ignatius.L   |   We have carefully selected several similar problems for you:
1200
1201
1008
1004
1106

Title analysis:

Simple questions. In the process of converting 10 -in -production numbers into 2nd -in numbers, you can constantly determine whether the current bit is non -zero.

corresponding knowledge points:

At this time, you may need to review the process of 10 -in -proof and two -in -making process:

code is as follows:

``````/*
* D.CPP
*
* Created on: March 11, 2015
*/

#include <iOSTREAM>
#include <cstdio>
#include <cmath>

using namespace std;

int Main () {
int n;
While (scanf ("%d", & n)! = EOF, N) {

int T = 0;
While (n%2 == 0) {{
t ++;
n /= 2;
}

Printf ("%d \ n", (int) pow (2+0.0, t));
}

Return 0;
}``````

source