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Lowest Bit------HDOJ杭电1196(想法很重要)
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.
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
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:写代码能力有限,如有神牛发现BUG,还请指出,不胜感激!#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; }
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