HDU1196进制转换

Lowest Bit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12485 Accepted Submission(s): 9114

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

---

思路

先转化为二进制，找到从右往左第一个1处，然后求的这个二进制的十进制数

---

代码

#include <iostream>#include <stdio.h>using namespace std;int main(){    int n;    while(cin>>n)    {        int a[100]={0};        int j=0;        int i;        int w=1;        int number=0;        if(n==0)            break;        for(i = 16; i >= 0; i--)  //十进制转二进制        {            if(n & (1 << i))                a[j]=1;            else                a[j]=0;            ++j;        }        for(i=j-1;i>=0;i--)        {            if(a[i]==1)                break;        }        for(int m=j-1;m>=i;m--)//二进制转十进制        {            number += a[m]*w;            w*=2;        }        cout<<number<<endl;    }    return 0;}