HDU1196_Lowest Bit【ビット演算】【水題】

Lowest Bit
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8043    Accepted Submission(s): 5920
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   Author
SHI, Xiaohan
あなたに1つの数Aをあげて、その2進数の表示の中で最も右の1の表示の数を求めます
例えば、26のバイナリ表現は11010、右端の1表現の数は00010である.
構想:ビット演算、実はA&(A^(A-1))、すなわちA&(-A)を求める
例えば:26——011010,-A=111010 A & ()

#include<stdio.h>

int main()
{
    int A;
    while(~scanf("%d",&A) && A)
    {
        //int ans = A & ( A ^( A - 1));
        int ans = A & (-A);
        printf("%d
",ans);
    }
    return 0;
}