这道题是一道需要用位运算的题目,题目所谓的lowest bit,其实是指1的最低位,我们只需将1和后面的0所构成的一个数以十进制的形式输出即可。
我们在这里需要用到位操作符中的“&”和一个用于试探的数字,这个用于试探的数字要求就是在二进制形式中只出现一个1,即00000001,00000010,00000100等等。让试探数字中的1从最低位开始逐渐向高位移动(这里就要用到另一个操作符“<<”),与输入的数字进行按位与运算,当试探数字中的1和输入数字中最低位的1处于同一位置,则运算结果大于0,此时试探数字便是符合要求的结果,输出试探数字,否则是运算结果是0,试探数字的1向高位移一位。
#include <cstdlib>
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
int a,t;
while(cin>>a&&a!=0)
{
t=1;
while(0==(a&t)) //注意要加上括号,否则不会得出正确结果
{
t<<=1;
}
cout<<t<<endl;
}
system("PAUSE");
return EXIT_SUCCESS;
}
附上原题:
Lowest Bit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)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