Blue Bridge Cup 31-day sprint punch-in problem solution (Day17)

Day17

first question

2021 Demo Tournament

distance and

填空题

public class Main {
    
    	
	public static void main(String[] args) {
    
    
		String s ="LANQIAO";
		int ans = 0;
		char[] c = s.toCharArray();
		for (int i = 0; i < c.length; i++) {
    
    
			for (int j = i + 1; j < c.length; j++)
			ans += Math.abs(c[j] - c[i]);
		}
        
		System.out.println(ans);
	}
}

Question 2

The 11th 2020 Blue Bridge Cup National Championship

diffusion

JavaB组第2题

填空题

Multi-source bfs.

import java.util.LinkedList;
import java.util.Queue;

public class Main {
    
    

     static final int N = 10010;
     static boolean[][] st = new boolean[N][N];

    public static void main(String[] args) {
    
    
        bfs();
    }

    private static void bfs() {
    
    
        int[] dx= {
    
    0, -1, 0, 1}, dy= {
    
    1, 0, -1, 0};
        Queue<PII> q = new LinkedList<>();
        // 因为有可能往负半轴扩展 所以我们在原坐标的基础上往右上移2100位 保证所有的点扩展是都在正半轴
        q.offer(new PII(0 + 2100, 0 + 2100, 0));
        q.offer(new PII(2020 + 2100, 11 + 2100, 0));
        q.offer(new PII(11 + 2100, 14 + 2100, 0));
        q.offer(new PII(2000 + 2100, 2000 + 2100, 0));
        st[0 + 2100][0 + 2100] = true;
        st[2020 + 2100][11 + 2100] = true;
        st[11 + 2100][14 + 2100] = true;
        st[2000 + 2100][2000 + 2100] = true;

        int ans = 0;
        while (!q.isEmpty()) {
    
    
            PII t = q.poll();

            if (t.cnt == 2021)  {
    
    
                System.out.println(ans);
                return;
            }

            ans++;

            for (int i = 0; i < 4; i++) {
    
    
                int x = t.x + dx[i], y = t.y + dy[i];
                if (x < 0 || x >= N || y < 0 || y >= N) continue;
                if (st[x][y]) continue;
                q.offer(new PII(x, y, t.cnt + 1));
                st[x][y] = true;
             }
         }
     }

    static class PII{
    
    
        int x;
        int y;
        int cnt;

        public PII(int x, int y, int cnt) {
    
    
            this.x = x;
            this.y = y;
            this.cnt = cnt;
        }
    }
}

It's timed out, so we can just submit the answer directly.

public class Main {
    
    
    public static void main(String[] args) {
    
    
        System.out.println(20312088);
    }
}

Question 3

The 4th 2013 Blue Bridge Cup Zhenti

wrong ticket

A simulated question, the solution is in the second question of my previous article: Blue Bridge Cup AcWing Study Notes 4-2 Simulation Learning

Question 4

The 9th 2018 Blue Bridge Cup Provincial Championship

multiples problem

贪心+DP

Refer to the solution of the AcWing Xiaodaidai big brother .

import java.util.Arrays;
import java.util.Collections;
import java.util.LinkedList;
import java.util.Scanner;

public class Main {
    
    
    
    static int N = 1010;
    static int INF = 0x3f3f3f3f;
    static int n,m;
    static int[][] f = new int[4][N];
    static LinkedList[] a = new LinkedList[N];
   
    public static void main(String[] args){
    
    
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();
        m = sc.nextInt();
        for (int i = 0; i <= m; i++) a[i] = new LinkedList<Integer>();
        for (int i = 1; i <= n; i++) {
    
    
            int x = sc.nextInt();
            a[x % m].add(x);
        }
        
        for (int i = 0; i <= 3; i++) Arrays.fill(f[i], -INF);
        
        f[0][0] = 0;
        for (int i = 0; i <= m; i++) {
    
    
            Collections.sort(a[i]);
            Collections.reverse(a[i]);
            // 选择前3大
            for(int u = 0; u < Math.min(3, a[i].size()); u++) {
    
    
                int x = (int) a[i].get(u);
                // 从大到小
                for (int j = 3; j >= 1; j--) {
    
    
                    for (int k = 0; k < m; k++)
                        f[j][k] = Math.max(f[j][k], f[j - 1][mod(k - x, m)] + x);
                }
            }
        }
        
        System.out.println(f[3][0]);
    }

    private static int mod(int x,int y) {
    
    
        return (x % y + y) % y;
    }
}