acm hnu 10038 进制转换

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本文介绍了一个简单的算法问题：给定一个正整数A，输出A的最低位对应的数值。通过将整数转换为二进制形式并找出最低位为1的位置来解决此问题。输入包含一系列正整数，输出则是每个整数的最低位值。

Lowest Bit

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:32768KB

Total submit users: 257, Accepted users: 238

Problem 10038 : No special judgement

Problem description

Given an positive integer A (1 <= A <= 10^9), 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 <= 10⁹). 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
8
0

Sample Output

2
8

Problem Source

HNU 1'st Contest

#include <stdio.h>
#include <math.h>

int getbits(int);
main()
{
    int n;
    scanf("%d",&n);
    while(n!=0){
        printf("%.0f ",pow(2,getbits(n)));
        scanf("%d",&n);
    }

    return 0;
}

int getbits(int n)
{
    if(n==1)return 0;
    int tmp=n,i=0,s[100];
    for (i = 0;i<100;i++)s[i]=-1;
    i = 0;
    while(1){
        if( (s[i++]=tmp%2) == 1)return i-1;
        tmp = tmp/2;
    }
}