http://acm.hdu.edu.cn/showproblem.php?pid=1196
Lowest Bit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7689 Accepted Submission(s): 5646
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.
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
#include<stdio.h>
int main()
{
int A;
while(scanf("%d",&A)&&A!=0)
{
printf("%d\n",A&(-A));
}
return 0;
}
相关文章(转): http://www.cnblogs.com/zhangshu/archive/2011/08/16/2141396.html
本文详细解释并提供了如何使用位操作技巧来解决HDU 1196 LowestBit问题,即给定正整数A,输出其最低位的值。通过实例演示了输入输出格式,并附上了简洁高效的C语言代码实现,旨在帮助初学者快速掌握位运算在编程中的应用。
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