hdu1196 Lowest Bit(二进制)

最新推荐文章于 2021-08-15 23:46:53 发布

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本文介绍了一种算法，用于找出一个正整数在二进制表示中从后往前数第一个为1的数字，通过解析输入的正整数，将其转换为二进制形式并确定最低有效位。

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题目：
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

题意：找出一个数的二进制从后往前数第一个为1 的数字
解题思路：要知道N%2最先除出来的余数是2进制的最低位，所以只要判断什么时候N%2的余数第一次不等于0

代码

//
//  main.cpp
//  hdu1096
//
//  Created by zhan_even on 2018/11/1.
//  Copyright © 2018年 zhan_even. All rights reserved.
//

#include <iostream>
#include <math.h>
using namespace std;

int main(int argc, const char * argv[]) {
    int n,i,num,bin[10000];
    while (cin>>n&&n) {
        int temp = n;
        int m=0;
        while (temp !=0) {
            i = temp%2;
            temp = temp/2;
            bin[m] = i;
            m++;
        }
        
        for(int x=0;x<m;x++){
            
            if (bin[x]==1) {
                num = pow(2,x);
                cout<<num<<endl;
                break;
            }
        }
    }
    return 0;
}