HDU 1196 Lowest Bit(基础题,有个小技巧)

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#HDU1196 #Lowest Bit

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这篇博客介绍了HDU 1196题目，即找出整数的二进制表示中最低位的1，并提供了两种解决方案，包括基础代码实现和优化后的代码展示。

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Lowest Bit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10560 Accepted Submission(s): 7751

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

Sample Output

2
8

Author

SHI, Xiaohan

Source

Zhejiang University Local Contest 2005

原题链接: http://acm.hdu.edu.cn/showproblem.php?pid=1196

题意:一个十进制的数,化成二进制后,最后一个 '1'与其后的 ' 0 '构成二进制数转换成十进制.

思路:一般算法,将其十进制数转换成二进制后,从最后一位往前模拟;

大神做法,看代码二.

AC代码一:

#include <stdio.h>
int main()
{
    int n,i,j,k;
    int a[50];
    while (scanf("%d",&n)!=EOF,n)
    {
        i=0;
        k=1;
        while (n)
        {
            a[i]=n%2;
            n/=2;
            i++;
        }
        for (j=0; j<i; j++)
        {
            if (a[j]==1)
            {
                printf("%d\n",k);
                break;
            }
            k*=2;
        }
    }
    return 0;
}

AC代码二(大神代码)

#include<stdio.h>
int main()
{
    int A,t;
    while(scanf("%d",&A),A)
    {
        t=1;
        while(A)
        {
            if(A%2==1)break;
            A/=2;
            t*=2;
        } 
        printf("%d\n",t);   
    }    
    return 0;
}