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HDU1196_Lowest Bit【位运算】【水题】

栏目:互联网时间:2014-11-23 09:23:53
Lowest Bit


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8043    Accepted Submission(s): 5920

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


题目大意:给你1个数A,求它的2进制表示中最右侧的1表示的数

比如:26的2进制表示为11010,最右侧的1表示的数为00010。

思路:位运算,其实就是求A & (A ^ (A⑴) ),即A & (-A)

比如:26――011010,-A = 111010  A & ()

#include<stdio.h> int main() { int A; while(~scanf("%d",&A) && A) { //int ans = A & ( A ^( A - 1)); int ans = A & (-A); printf("%d ",ans); } return 0; }



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