HDU 1196 Lowest Bit

最新推荐文章于 2017-07-23 14:57:16 发布

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最新推荐文章于 2017-07-23 14:57:16 发布

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分类专栏：寒假训练

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本文介绍了一种算法，用于找出一个正整数（范围1到100）二进制表示中最低位非零位的值，并通过几个示例详细解释了算法的实现方式。

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
#include<iostream>
using namespace std;
int n;
int l(int x)
{
	return x&(-x);
}
int main()
{
	while(scanf("%d",&n)!=EOF&&n)
	{
		printf("%d\n",l(n));
	}
	return 0;
}
就是求第一个不为0的数位，有多少0就是2的多少次方。
#include <iostream>
using namespace std;
int main()
{
    int A,Ans;
    while(cin>>A)
    {
        if(A==0)break;
        Ans = 1;
        while(A%2==0)
        {
            A /= 2;
            Ans *= 2;
        }
        cout<<Ans<<endl;
    }
    return 0;
}
转的一个模拟
#include <stdio.h>  
#include <string.h>  
int set(int n)  
{  
    char str[10000];  
    int k = 0;  
    while(n)  
    {  
        int r = n%2;  
        str[k++] = r+'0';  
        n/=2;  
    }  
    str[k] = '\0';  
    int i;  
    for(i = 0;i<k;i++)  
    {  
        if(str[i]=='1')  
        break;  
    }  
    return i;  
}  
  
int pow(int n)  
{  
    int s = 1;  
    for(int i = 1;i<=n;i++)  
    s*=2;  
    return s;  
}  
int main()  
{  
    int n,k;  
    while(~scanf("%d",&n),n)  
    {  
        k = set(n);  
        k = pow(k);  
        printf("%d\n",k);  
    }  
  
    return 0;  
}