方法思路

1. 吸引度为 |点赞数 - 点踩数|，要使其最大化，我们需要让点赞数和点踩数的差值尽可能大
2. 对于每个帖子，我们有两种选择：选择或不选择
3. 如果选择一个帖子，它会对点赞数和点踩数分别产生贡献
4. 我们可以将每个帖子看作是对点赞数和点踩数差值的贡献：(a[i] - b[i])
5. 如果贡献为正，它会增加点赞数与点踩数的差值；如果为负，它会减少差值
6. 因此，我们可以贪心地选择所有贡献为正的帖子来增加点赞数，或选择所有贡献为负的帖子来增加点踩数
7. 最终，取两种情况的最大值

代码实现

Java

import java.util.*;
import java.io.*;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        
        long[] a = new long[n];
        long[] b = new long[n];
        
        String[] likes = br.readLine().split(" ");
        String[] dislikes = br.readLine().split(" ");
        
        for (int i = 0; i < n; i++) {
            a[i] = Long.parseLong(likes[i]);
            b[i] = Long.parseLong(dislikes[i]);
        }
        
        long sumLikes = 0;
        long sumDislikes = 0;
        
        for (int i = 0; i < n; i++) {
            // 如果选择这个帖子会增加点赞数与点踩数的差值
            if (a[i] > b[i]) {
                sumLikes += a[i];
                sumDislikes += b[i];
            }
        }
        
        long maxDiff1 = Math.abs(sumLikes - sumDislikes);
        
        sumLikes = 0;
        sumDislikes = 0;
        
        for (int i = 0; i < n; i++) {
            // 如果选择这个帖子会增加点踩数与点赞数的差值
            if (a[i] < b[i]) {
                sumLikes += a[i];
                sumDislikes += b[i];
            }
        }
        
        long maxDiff2 = Math.abs(sumLikes - sumDislikes);
        
        System.out.println(Math.max(maxDiff1, maxDiff2));
    }
}

Python

def max_attraction():
    n = int(input())
    likes = list(map(int, input().split()))
    dislikes = list(map(int, input().split()))
    
    sum_likes1 = 0
    sum_dislikes1 = 0
    
    # 选择所有 likes > dislikes 的帖子
    for i in range(n):
        if likes[i] > dislikes[i]:
            sum_likes1 += likes[i]
            sum_dislikes1 += dislikes[i]
    
    max_diff1 = abs(sum_likes1 - sum_dislikes1)
    
    sum_likes2 = 0
    sum_dislikes2 = 0
    
    # 选择所有 likes < dislikes 的帖子
    for i in range(n):
        if likes[i] < dislikes[i]:
            sum_likes2 += likes[i]
            sum_dislikes2 += dislikes[i]
    
    max_diff2 = abs(sum_likes2 - sum_dislikes2)
    
    return max(max_diff1, max_diff2)

print(max_attraction())

C++

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

int main() {
    int n;
    cin >> n;
    
    vector<long long> a(n), b(n);
    
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    
    for (int i = 0; i < n; i++) {
        cin >> b[i];
    }
    
    long long sum_likes1 = 0, sum_dislikes1 = 0;
    
    // 选择所有 likes > dislikes 的帖子
    for (int i = 0; i < n; i++) {
        if (a[i] > b[i]) {
            sum_likes1 += a[i];
            sum_dislikes1 += b[i];
        }
    }
    
    long long max_diff1 = abs(sum_likes1 - sum_dislikes1);
    
    long long sum_likes2 = 0, sum_dislikes2 = 0;
    
    // 选择所有 likes < dislikes 的帖子
    for (int i = 0; i < n; i++) {
        if (a[i] < b[i]) {
            sum_likes2 += a[i];
            sum_dislikes2 += b[i];
        }
    }
    
    long long max_diff2 = abs(sum_likes2 - sum_dislikes2);
    
    cout << max(max_diff1, max_diff2) << endl;
    
    return 0;
}