华为OD机试：分苹果问题高效解法

华为OD机试C卷 - 分苹果问题解析

问题描述
题目要求将一堆苹果分成两堆，使得两堆苹果数量的二进制表示中1的个数相同。需要找出满足条件的分法数量。

解题思路

二进制特性分析
分苹果的核心在于二进制中1的个数。两堆苹果数量的二进制1的个数相同，意味着原始苹果总数的二进制1的个数必须是偶数。若总数为奇数（二进制1的个数为奇数），直接返回0。

数学推导
设总苹果数为 total，其二进制1的个数为 count。若 count 为偶数，分法数量为组合数 $C(count, count/2)$ 或 $2^{count-1}$（推导过程略）。实际分法数为满足条件的子集数。

代码实现

Java 实现

public class DivideApples {
    public static int countWays(int total) {
        int count = Integer.bitCount(total);
        if (count % 2 != 0) return 0;
        return (int) Math.pow(2, count - 1);
    }
}

C++ 实现

#include <bits/stdc++.h>
using namespace std;

int countWays(int total) {
    int count = __builtin_popcount(total);
    if (count % 2 != 0) return 0;
    return pow(2, count - 1);
}

JavaScript 实现

function countWays(total) {
    const count = total.toString(2).split('1').length - 1;
    if (count % 2 !== 0) return 0;
    return Math.pow(2, count - 1);
}

Python 实现

def count_ways(total):
    count = bin(total).count('1')
    if count % 2 != 0:
        return 0
    return 2 ** (count - 1)

复杂度分析

时间复杂度：$O(1)$，仅需计算二进制中1的个数和幂运算。
空间复杂度：$O(1)$，无额外空间使用。

边界条件

输入为0时，直接返回0（无法分割）。
输入为1时，返回0（奇数个1）。
输入为3（二进制11）时，返回2（分法为1+2）。

总结

通过二进制1的奇偶性判断可行性，利用组合数学快速计算分法数。代码简洁高效，适合机试场景。

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