hdu 1196 lowbit裸题

最新推荐文章于 2020-07-24 14:10:52 发布

mazicwong

最新推荐文章于 2020-07-24 14:10:52 发布

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Lowest Bit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11580 Accepted Submission(s): 8459

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

Sample Output

2
8

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int lowbit(int x)
{
	return x&(-x);
}
int main()
{
	int x;
	while (scanf("%d", &x) && x)
		printf("%d\n", lowbit(x));
	return 0;
}

/*
* hdu 1196
* author  : mazciwong
* creat on: 2016-12-16
*/

/*
  lowbit 返回2^t  (t为从右往左第一次出现1的位置)

*/


#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
	int n;
	while (scanf("%d",&n) && n)
	{
		int ans = 1;
		while (n % 2 == 0)
		{
			n /= 2;
			ans *= 2;
		}
		printf("%d\n", ans);
	}
	return 0;
}