分号机

import java.io.IOException;
class MC{
    public void run(){
        int cnt = 0;
        for (int i = 9; i >= 0; i--) {
            for (int j = 9; j >= 0; j--) {
                for (int k = 9; k >= 0; k--) {
                    if(i > j && j > k){
                        cnt ++;
                    }
                }
            }
        }
        //120
        System.out.println(cnt);
    }
}
public class Main {
    public static void main(String[] args) throws IOException {
        new MC().run();
    }
}

五星填数

全排列枚举。最后答案要除10，因为一种情况正面可以转动5次，镜像面可以转到5次，但都只是一种情况

import java.io.IOException;
class MC{
    int N = 13, n = 12;
    int[] a = new int[N];
    boolean[] st = new boolean[N];
    int res = 0;

    private boolean check() {
        int s1 = a[1] + a[3] + a[6] + a[9];
        int s2 = a[1] + a[4] + a[7] + a[10];
        int s3 = a[2] + a[3] + a[4] + a[5];
        int s4 = a[2] + a[6] + a[8] + a[10];
        int s5 = a[5] + a[7] + a[8] + a[9];
        return s1 == s2 && s2 == s3 && s3 == s4 && s4 == s5;
    }

    void dfs(int cnt){
        if(cnt >= 11) {
            if(check()){
                res++;
            }
            return;
        }
        for (int i = 1; i <= n; i++) {
            if(st[i] || i == 7 || i == 11) continue;
            st[i] = true;
            a[cnt] = i;
            dfs(cnt + 1);
            st[i] = false;
        }
    }

    public void run(){
        //全排列，10个数
        dfs(1);
        System.out.println(res / 10);
    }
}

public class Main {
    public static void main(String[] args) throws IOException {
        new MC().run();
    }
}

显示二叉树

思路：先看看sv的是什么东西，发现是要填的数值；发现竖线下面是数字，所以是y+1，然后每个数字偏移p2个位置

class BiTree
{
    private int v;
    private BiTree l;
    private BiTree r;

    public BiTree(int v){
        this.v = v;
    }

    public void add(BiTree the){
        if(the.v < v){
            if(l==null) l = the;
            else l.add(the);
        }
        else{
            if(r==null) r = the;
            else r.add(the);
        }
    }

    public int getHeight(){
        int h = 2;
        int hl = l==null? 0 : l.getHeight();
        int hr = r==null? 0 : r.getHeight();
        return h + Math.max(hl,hr);
    }

    public int getWidth(){
        int w = (""+v).length();
        if(l!=null) w += l.getWidth();
        if(r!=null) w += r.getWidth();
        return w;
    }

    public void show(){
        char[][] buf = new char[getHeight()][getWidth()];
        printInBuf(buf, 0, 0);
        showBuf(buf);
    }

    private void showBuf(char[][] x){
        for(int i=0; i<x.length; i++){
            for(int j=0; j<x[i].length; j++)
                System.out.print(x[i][j]==0? ' ':x[i][j]);
            System.out.println();
        }
    }

    private void printInBuf(char[][] buf, int x, int y){
        String sv = "" + v;

        int p1 = l==null? x : l.getRootPos(x);
        int p2 = getRootPos(x);
        int p3 = r==null? p2 : r.getRootPos(p2+sv.length());

        buf[y][p2] = '|';
        for(int i=p1; i<=p3; i++) buf[y+1][i]='-';

        for(int i=0; i<sv.length(); i++) buf[y+1][p2+i]=sv.charAt(i);  //填空位置
        System.out.println();
        if(p1<p2) buf[y+1][p1] = '/';
        if(p3>p2) buf[y+1][p3] = '\\';

        if(l!=null) l.printInBuf(buf,x,y+2);
        if(r!=null) r.printInBuf(buf,p2+sv.length(),y+2);
    }

    private int getRootPos(int x){
        return l==null? x : x + l.getWidth();
    }
}

public class Main
{
    public static void main(String[] args)
    {
        BiTree tree = new BiTree(500);
        tree.add(new BiTree(200));
        tree.add(new BiTree(509));
        tree.add(new BiTree(100));
        tree.add(new BiTree(250));
        tree.add(new BiTree(507));
        tree.add(new BiTree(600));
        tree.add(new BiTree(650));
        tree.add(new BiTree(450));
        tree.add(new BiTree(510));
        tree.add(new BiTree(440));
        tree.add(new BiTree(220));
        tree.show();
    }
}

穿越雷区

穿越雷区

//记录起点和终点，bfs搜索，最先到点B即为最短
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>


using namespace std;
typedef pair<int, int> PII;


const int N = 110;
int pro[4][2] = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};
int dist[N][N];
int g[N][N];
PII A, B;

int main(){

    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i ++ ){
        for (int j = 1; j <= n; j ++ ){
            char c;
            cin >> c;
            if(c == '+') g[i][j] = 1;
            else if(c == 'A') {
                g[i][j] = 3;
                A = {i, j};
            }
            else if(c == 'B') {
                g[i][j] = 6;
                B = {i, j};
            }
            else g[i][j] = 0;
        }
    }
    
    memset(dist, 0x3f, sizeof dist);
    dist[A.first][A.first] = 0;
    queue<PII> q;
    q.push(A);
    while (q.size() > 0){
        PII p = q.front();
        q.pop();
        int x = p.first, y = p.second;
        //遍历邻点
        for (int i = 0; i < 4; i ++ ){
            int tx = x + pro[i][0], ty = y + pro[i][1];
            //判断该点合不合法
            if(tx < 1 || tx > n || ty < 1 || ty > n || dist[tx][ty] != 0x3f3f3f3f || g[x][y] == g[tx][ty]) continue;
            dist[tx][ty] = dist[x][y] + 1;
            q.push({tx, ty});
        }
    }

    if(dist[B.first][B.second] == 0x3f3f3f3f) printf("%d", -1);
    else printf("%d", dist[B.first][B.second]);
}

表格计算