题解 | #不想出差的HR#

#include <iostream>
using namespace std;

int main() {
    int a, b, c;
    char c1, c2;
    cin >> a >> c1 >> b >> c2 >> c;
    
    // 计算三个数的异或值
    int k = a ^ b ^ c;
    
    // 如果异或值为0，必须出差
    if (k == 0) {
        cout << 1 << endl;
        return 0;
    }
    
    // 检查每个科室是否可以通过减少人数实现异或为0
    if (a >= (a ^ k)) {
        cout << "A," << (a - (a ^ k)) << endl;
    }
    else if (b >= (b ^ k)) {
        cout << "B," << (b - (b ^ k)) << endl;
    }
    else if (c >= (c ^ k)) {
        cout << "C," << (c - (c ^ k)) << endl;
    }
    return 0;
}

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String[] input = sc.nextLine().split(",");
        int a = Integer.parseInt(input[0]);
        int b = Integer.parseInt(input[1]);
        int c = Integer.parseInt(input[2]);
        
        int k = a ^ b ^ c;
        
        if (k == 0) {
            System.out.println(1);
            return;
        }
        
        if (a >= (a ^ k)) {
            System.out.println("A," + (a - (a ^ k)));
        }
        else if (b >= (b ^ k)) {
            System.out.println("B," + (b - (b ^ k)));
        }
        else if (c >= (c ^ k)) {
            System.out.println("C," + (c - (c ^ k)));
        }
    }
}

a, b, c = map(int, input().split(','))

# 计算异或值
k = a ^ b ^ c

# 如果异或值为0，必须出差
if k == 0:
    print(1)
else:
    # 检查每个科室是否可以通过减少人数实现异或为0
    if a >= (a ^ k):
        print(f"A,{a - (a ^ k)}")
    elif b >= (b ^ k):
        print(f"B,{b - (b ^ k)}")
    elif c >= (c ^ k):
        print(f"C,{c - (c ^ k)}")