Lowest Bit
Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:32768KB
Total submit users: 1713, Accepted users: 1586
Problem 10038 : No special judgement
Problem description
Given an positive integer A (1 <= A <= 109), 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
0Sample Output
2
8Problem Source
HNU 1’st Contest
算法目标:将十进制数转化为二进制数,从后往前第一个1开始到最后为该数的最小比特。
解决困难:开始时出现了RE的情况,但是过了两天就一次AC了。一定是OJ的问题…一定是…
代码部分:
#include <stdio.h>
int main()
{
int a,b[50],i=0,j=1,k=1;
scanf("%d",&a);
while (a){
while (a!=0){
b[i]=a%2;
a/=2;
i++;
}
i=1;
while (b[i-1]!=1){
k*=2;
i++;
}
printf("%d\n",k);
scanf("%d",&a);
for(i=0;i<50;i++) b[i]=0;
i=0;k=1;j=1;}
return 0;
}