Luogu problem list: [mathematics 1] basic mathematics problems

（1）P1143 Hexadecimal conversion
java It is very good for binary conversion API, Can achieve unlimited length of any hexadecimal conversion .

import java.math.BigInteger;
import java.util.Scanner;

public class Main {
    
	public static void main(String[] args) {
    
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		sc.nextLine();
		String s=sc.nextLine();
		int m=sc.nextInt();
		System.out.println(new BigInteger(s,n).toString(m).toUpperCase());
	}
}

（2）P1469 Looking for chopsticks
If you know the knowledge of binary , This problem is not difficult . When XOR is performed on each number , Always leave the single number . But this question is disgusting , Only for 4M Of memory , Its original intention is to prevent using the bucket sorting method to do this problem , But it doesn't consider java It's impossible to be in 4M Make it in memory .

import java.util.Scanner;

public class Main {
    
	public static void main(String[] args) {
    
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int res=sc.nextInt();
		for(int i=1;i<n;i++) {
    
			res^=sc.nextInt();
		}
		System.out.println(res);
	}
}

（3）P1100 High low swap
This question is to examine the understanding of binary . Put the high 16 Shift to low position , We can put n Move right 16 position , In this way, the high position goes directly to the low position , And the low order is directly because it is less than 0 And removed . Then we move the lower number to the right 16 position , We can easily get the low position by taking the mold . Finally, we add the two converted numbers to get a new number .

import java.util.Scanner;

public class Main {
    
	public static void main(String[] args) {
    
		Scanner sc=new Scanner(System.in);
		long n=sc.nextInt();
		long x=(n>>16)+((n%65536)<<16);
		System.out.println(x);
	}
}

（4）