湖大ACM—Lowest Bit

最新推荐文章于 2022-01-17 21:53:49 发布

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最新推荐文章于 2022-01-17 21:53:49 发布

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分类专栏： ACM 文章标签： acm

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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
0

Sample Output
2
8

Problem 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;
}