HDUOJ_1196(二进制-十进制)

HDUOJ_1196(二进制-十进制)

Lowest Bit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9713    Accepted Submission(s): 7134

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

26880

 Sample Output

28

 

题意：把十进制数转换为二进制数时，记录第一个1出现的位置，输出这个1和前面的0组成的二进制数相对应的十进制数。

My  solution:

/*2015.8.26*/

#include<stdio.h>int mi(int j)/*快速幂*/{int ans=1,v=2;while(j>0){if(j%2)ans*=v;v*=v;j/=2;}return ans;}int  main(){int i,j,n,q;while(scanf("%d",&n)==1&&n){q=0,i=0;while(n>0){q++;if(n%2)/* 这里求1最先出现的位置（从左往右）*/{i=q;break;}n/=2;}j=i-1;/*j求的是1后面的0的个数*/ printf("%d\n",mi(j));/*求2的j次方*/ }return 0;}