标签：c acm 编程 hdoj
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
二进制的原理就是除二余一的时候出现1，所以呀！
直接循环取二的模，如果为0则记录sum加一，否则break；
然后就是循环乘二，也可以在判断里面直接乘二
下面贴下AC代码：#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
int i,j,k;
int t,n,m;
int liu[106];
while(scanf("%d",&n)!=EOF,n)
{
m=0;
while(1)
{
if(n%2==0)
{
m++;
n/=2;
}
else break;
}
t=1;
for(i=1;i<=m;i++)
t*=2;
printf("%d\n",t);
}
return 0;
}
写代码能力有限，如有神牛发现BUG，还请指出，不胜感激！
Lowest Bit------HDOJ杭电1196（想法很重要）,布布扣,bubuko.com
Lowest Bit------HDOJ杭电1196（想法很重要）
标签：c acm 编程 hdoj
原文地址：http://blog.csdn.net/u014231159/article/details/24906411
