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杭电 1196

时间:2014-08-05 09:40:19      阅读:201      评论:0      收藏:0      [点我收藏+]

标签:杭电   c   acm   二进制   优化   

Lowest Bit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7776    Accepted Submission(s): 5715


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
 题意:
给定一个十进制数字,求转化为二进制后的最后一位
思路:
根据位运算能极大地优化程序,求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
代码如下:
<span style="font-size:14px;">#include<stdio.h>
int main()
{
	int n;
	while(~scanf("%d",&n),n)
	{
		printf("%d\n",n&(-n));
	}
	return 0;
}</span>


杭电 1196,布布扣,bubuko.com

杭电 1196

标签:杭电   c   acm   二进制   优化   

原文地址:http://blog.csdn.net/ice_alone/article/details/38379177

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