解题思路

• 对于每个宝藏维护个区间，答案一定在这些区间中
• 对于每个区间的端点由小到大排序
• 对于每个点进行判断，若当前位置合法，则该点一定为一个右端点
• 则该点到前一个端点之间均为合法点
• 若前一个点不合法，则一定是某一个区间限制的左端点，所以该点到这个端点之间均未超出范围，使某一宝藏取不到
• 若前一个点合法，则在满足的前提下，还避免了重复

import java.io.*;
import java.math.BigInteger;
import java.util.*;


//implements Runnable
public class Main {
    static long md=(long)998244353;
    static long Linf=Long.MAX_VALUE/2;
    static int inf=Integer.MAX_VALUE/2;
    static int N=200010;
    static int n=0;
    static int m=0;


    static long ans=0;
    static long[] a;
    static long[] b;

    static boolean check(long x){
        PriorityQueue<Long> q=new PriorityQueue<>((o1,o2)->{
            if(o1-o2>0)return 1;
            else if(o1-o2<0)return -1;
            else return 0;
        });
        for(int i=1;i<=n;++i){
            q.add(Math.abs(x-a[i]));
        }
        for(int i=1;i<=n;++i){
            if(q.poll()>b[i])return false;

        }
        return true;
    }
    static void solve() throws Exception{
        AReader input=new AReader();
//        String fileName="C:\\Users\\Lenovo\\Downloads\\055.txt";
//		Scanner input=new Scanner(new FileReader(fileName));

//        BufferedReader input = new BufferedReader(new FileReader(fileName));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        String al="abcdefghijklmnopqrstuvwxyz";
        char[] ac=al.toCharArray();
        n=input.nextInt();
        a=new long[n+1];
        for(int i=1;i<=n;++i)a[i]=input.nextLong();
        b=new long[n+1];
        for(int i=1;i<=n;++i)b[i]=input.nextLong();
        Arrays.sort(b,1,n+1);
        TreeSet<Long> hs=new TreeSet<>();
        for(int i=1;i<=n;++i){
            for(int j=1;j<=n;++j){
                hs.add(a[i]-b[j]-1);//左端点，左开右闭，区分左端点和右端点
                hs.add(a[i]+b[j]);//右端点
            }
        }
        long l=0;
        for(long x:hs){
            if(check(x))ans+=x-l;
            l=x;//左端点,要么是右端点区间去重叠
        }
        out.println(ans);
        out.flush();
        out.close();
    }
    public static void main(String[] args) throws Exception{
        solve();
    }
    //	public static final void main(String[] args) throws Exception {
//		  new Thread(null, new Tx2(), "线程名字", 1 << 27).start();
//	}
//		@Override
//		public void run() {
//			try {
//				//原本main函数的内容
//				solve();
//
//			} catch (Exception e) {
//			}
//		}
    static
    class AReader{
        BufferedReader bf;
        StringTokenizer st;
        BufferedWriter bw;

        public AReader(){
            bf=new BufferedReader(new InputStreamReader(System.in));
            st=new StringTokenizer("");
            bw=new BufferedWriter(new OutputStreamWriter(System.out));
        }
        public String nextLine() throws IOException{
            return bf.readLine();
        }
        public String next() throws IOException{
            while(!st.hasMoreTokens()){
                st=new StringTokenizer(bf.readLine());
            }
            return st.nextToken();
        }
        public char nextChar() throws IOException{
            //确定下一个token只有一个字符的时候再用
            return next().charAt(0);
        }
        public int nextInt() throws IOException{
            return Integer.parseInt(next());
        }
        public long nextLong() throws IOException{
            return Long.parseLong(next());
        }
        public double nextDouble() throws IOException{
            return Double.parseDouble(next());
        }
        public float nextFloat() throws IOException{
            return Float.parseFloat(next());
        }
        public byte nextByte() throws IOException{
            return Byte.parseByte(next());
        }
        public short nextShort() throws IOException{
            return Short.parseShort(next());
        }
        public BigInteger nextBigInteger() throws IOException{
            return new BigInteger(next());
        }
        public void println() throws IOException {
            bw.newLine();
        }
        public void println(int[] arr) throws IOException{
            for (int value : arr) {
                bw.write(value + " ");
            }
            println();
        }
        public void println(int l, int r, int[] arr) throws IOException{
            for (int i = l; i <= r; i ++) {
                bw.write(arr[i] + " ");
            }
            println();
        }
        public void println(int a) throws IOException{
            bw.write(String.valueOf(a));
            bw.newLine();
        }
        public void print(int a) throws IOException{
            bw.write(String.valueOf(a));
        }
        public void println(String a) throws IOException{
            bw.write(a);
            bw.newLine();
        }
        public void print(String a) throws IOException{
            bw.write(a);
        }
        public void println(long a) throws IOException{
            bw.write(String.valueOf(a));
            bw.newLine();
        }
        public void print(long a) throws IOException{
            bw.write(String.valueOf(a));
        }
        public void println(double a) throws IOException{
            bw.write(String.valueOf(a));
            bw.newLine();
        }
        public void print(double a) throws IOException{
            bw.write(String.valueOf(a));
        }
        public void print(char a) throws IOException{
            bw.write(String.valueOf(a));
        }
        public void println(char a) throws IOException{
            bw.write(String.valueOf(a));
            bw.newLine();
        }
    }
}