Lowest Bit(hdu1196,位运算)

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本文深入探讨了如何将一个正整数转换为二进制形式，并着重解释了如何找到二进制数中从右向左第一个1出现前的那一位所对应的十进制数。通过简单的位操作技巧，实现快速定位并获取该位置的值，适用于初学者理解和实践。

http://acm.hdu.edu.cn/showproblem.php?pid=1196

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=29974#problem/C

Lowest Bit

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

Total Submission(s): 6453 Accepted Submission(s): 4737

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

Author

SHI, Xiaohan

Source

Zhejiang University Local Contest 2005

Recommend

Ignatius.L

解析：

题意：

将一个数化为二进制，求出右边第一个开始到出现第一个1形成的十进制数是多少

思路：

1，暴力。先化为二进制，寻找左边出现的第一个的位置，再求解

2.直接用位运算去掉右起第一个1的左边:x&(x^(x-1));

228 KB 15 ms C++ 198 B

#include<stdio.h>
#include<math.h>
#include <iostream>
using namespace std;
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF&&n!=0)
    {
   printf("%d\n",(n)&(n^(n-1)));
    }
    return 0;
}