蓝桥杯:字母阵列(注意事项)(答案)

  **本题题目过长,题目在代码后面 **

答案

  41

注意事项

  1. 重点:Scanner识别到空格会认为是第二次输入,需要一个for循环来接受输入,如果只接收一次,将会只接收第一个空格前的内容,比如:
    蓝桥杯:字母阵列(注意事项)(答案)_第1张图片

  2. 进行字符判断时,o的大小写

解题思路

  暴力破解8个方向,将其中某一个方向破解后其余七种依照相同的方法依次破解。

代码

import java.util.Scanner;

public class Main { //蓝桥杯要求class命名为Main,且无package
    public static void main(String []args){
        int length = 100; //定义该方阵的长
        char a[][] = new char[length][length];
        Scanner scanner = new Scanner(System.in);
        String str[]=new String[length];
        for(int i=0;i<length;i++){
            str[i] = scanner.next(); //*** 重点:scanner识别到空格会认为是第二次输入!!
        }
        for(int i=0;i<length;i++){ //将字符串方阵转换为char方阵
            for(int j=0;j<length;j++){
                a[i][j] = str[i].charAt(j) ;
            }
        }

        int count =0;
        for(int i=0;i<length;i++){ //开始暴力破解8种方向,想通第一种,后七种就是愉快的'搬砖'
            for(int j=0;j<length;j++){
                if(j<length-6) //1横向
                    if(a[i][j]=='L')
                        if(a[i][j+1]=='A')
                            if(a[i][j+2]=='N')
                                if(a[i][j+3]=='Q')
                                    if(a[i][j+4]=='I')
                                        if(a[i][j+5]=='A')
                                            if(a[i][j+6]=='O')
                                                count++;
                if(i<length-6) //2纵向
                    if(a[i][j]=='L')
                        if(a[i+1][j]=='A')
                            if(a[i+2][j]=='N')
                                if(a[i+3][j]=='Q')
                                    if(a[i+4][j]=='I')
                                        if(a[i+5][j]=='A')
                                            if(a[i+6][j]=='O')
                                                count++;
                if(j>=6) //3反横向
                    if(a[i][j]=='L')
                        if(a[i][j-1]=='A')
                            if(a[i][j-2]=='N')
                                if(a[i][j-3]=='Q')
                                    if(a[i][j-4]=='I')
                                        if(a[i][j-5]=='A')
                                            if(a[i][j-6]=='O')
                                                count++;
                if(i>=6) //4反纵向
                    if(a[i][j]=='L')
                        if(a[i-1][j]=='A')
                            if(a[i-2][j]=='N')
                                if(a[i-3][j]=='Q')
                                    if(a[i-4][j]=='I')
                                        if(a[i-5][j]=='A')
                                            if(a[i-6][j]=='O')
                                                count++;
                if(i<length-6&&j<length-6) //5斜右下
                    if(a[i][j]=='L')
                        if(a[i+1][j+1]=='A')
                            if(a[i+2][j+2]=='N')
                                if(a[i+3][j+3]=='Q')
                                    if(a[i+4][j+4]=='I')
                                        if(a[i+5][j+5]=='A')
                                            if(a[i+6][j+6]=='O')
                                                count++;
                if(i>=6&&j<length-6) //6斜右上
                    if(a[i][j]=='L')
                        if(a[i-1][j+1]=='A')
                            if(a[i-2][j+2]=='N')
                                if(a[i-3][j+3]=='Q')
                                    if(a[i-4][j+4]=='I')
                                        if(a[i-5][j+5]=='A')
                                            if(a[i-6][j+6]=='O')
                                                count++;
                if(i>=6&&j>=6) //7斜左上
                    if(a[i][j]=='L')
                        if(a[i-1][j-1]=='A')
                            if(a[i-2][j-2]=='N')
                                if(a[i-3][j-3]=='Q')
                                    if(a[i-4][j-4]=='I')
                                        if(a[i-5][j-5]=='A')
                                            if(a[i-6][j-6]=='O')
                                                count++;
                if(i<length-6&&j>=6) //8斜左下
                    if(a[i][j]=='L')
                        if(a[i+1][j-1]=='A')
                            if(a[i+2][j-2]=='N')
                                if(a[i+3][j-3]=='Q')
                                    if(a[i+4][j-4]=='I')
                                        if(a[i+5][j-5]=='A')
                                            if(a[i+6][j-6]=='O')
                                                count++;
            }
        }
        System.out.println(count);
        }
    }





~

题目

 标题:字母阵列
  仔细寻找,会发现:在下面的8x8的方阵中,隐藏着字母序列:“LANQIAO”。 SLANQIAO ZOEXCCGB MOAYWKHI BCCIPLJQ SLANQIAO RSFWFNYA XIFZVWAL COAIQNAL 我们约定: 序列可以水平,垂直,或者是斜向; 并且走向不限(实际上就是有一共8种方向)。 上图中一共有4个满足要求的串。 下面有一个更大的(100x100)的字母方阵。 你能算出其中隐藏了多少个“LANQIAO”吗? 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