首页 > 代码库 > HDU 1196 Lowest Bit （数位）

HDU 1196 Lowest Bit （数位）

2024-08-02 04:56:41   217人阅读

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




题意：求一个十进制数的二进制形式的从最后一个不为0的位开始算起的后面的二进制代表的十进制数是多少。


解题思路：利用位运算，用 & 1 判断一个数的二进制形式的最后一位是否为一，每次右移一位，用k纪录右移的次数，重复判断最后一位是否为1，若是1，则要求的结果就是2^k,输出即可。不过注意，在第一次右移前，要先判断原来的数是否为奇数。





AC代码：

#include <iostream>
#include <cstdio>
#include <string>
using namespace std;

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




HDU 1196 Lowest Bit （数位）


联系
我们