Lowest Bit
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.
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
把n化为二进制数，然后找出最低位，例如101000 最低位为1000 为8，
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 using namespace std; 5 int fen(int n) 6 { 7 int i,a; 8 for(i=0;n!=0;i++) 9 { 10 a=n%2; 11 if(a!=0) 12 { 13 return i; 14 } 15 else 16 n=n/2; 17 } 18 } 19 int main() 20 { 21 int n,i; 22 while(cin>>n) 23 { 24 if(n==0) 25 break; 26 else 27 cout<<pow(2,fen(n))<<endl; 28 29 } 30 }