进制转化

Lowest Bit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2965 Accepted Submission(s): 2003

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

Author SHI, Xiaohan

Source Zhejiang University Local Contest 2005

Recommend Ignatius.L   #include <stdio.h> #include <math.h> int main() {     int n, merchant, remainder;     int count;     int ans;     while(scanf("%d", &n)&&n)     {         count=0;         do         {             remainder=n%2;             merchant=n/2;             n=merchant;             ++count;         }while(!remainder);         //printf("%d\n", count);         ans=pow(2,count-1);         printf("%d\n", ans);     }     return 0; }