HDU 1196 Lowest Bit （數位）

Lowest Bit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8166 Accepted Submission(s): 5998


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 Zhejiang University Local Contest 2005

題意：求一個十進制數的二進制形式的從最後一個不為0的位開始算起的後面的二進制代表的十進制數是多少。

解題思路：利用位運算，用 & 1 判斷一個數的二進制形式的最後一位是否為一，每次右移一位，用k紀錄右移的次數，重復判斷最後一位是否為1，若是1，則要求的結果就是2^k,輸出即可。不過注意，在第一次右移前，要先判斷原來的數是否為奇數。

AC代碼：

#include 
#include 
#include 
using namespace std;

int main(){
//    freopen("in.txt", "r", stdin);
    int a;
    while(scanf("%d", &a)!=EOF && a){
        int k = 0;        //右移位數
        while(1){
            if(a & 1){
                printf("%d\n", (1<>= 1;
            k ++;
        }    
    }
    return 0;
}